January 2018 Algebra 1 Regents M.C. 1 – 6
January 2018 Algebra 1 Regents M.C. 1 – 6

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JMAP REGENTS BY STATE STANDARD: TOPIC

NY Algebra II Regents Exam Questions from Spring 2015 to January 2018 sorted by State Standard:

Topic

www.jmap.org

TABLE OF CONTENTS

TOPIC CCSS: SUBTOPIC QUESTION NUMBER

GRAPHS AND STATISTICS

S.IC.A.2, S.IC.B.3-6: Analysis of Data ……………………………………. 1-22 S.ID.B.6: Regression …………………………………………………………… 23-25S.ID.A.4: Normal Distributions …………………………………………….. 26-32

PROBABILITY S.CP.A.2, S.CP.B.7: Theoretical Probability …………………………… 33-36 S.CP.A.3-4, S.CP.B.6: Conditional Probability ……………………….. 37-44

RATE F.IF.B.6: Rate of Change ……………………………………………………… 45-50

QUADRATICS A.REI.B.4: Solving Quadratics……………………………………………… 51-55 A.REI.B.4: Complex Conjugate Root Theorem …………………………… 56 G.GPE.A.2: Graphing Quadratic Functions ……………………………. 57-62

SYSTEMS A.REI.C.6: Solving Linear Systems ………………………………………. 63-66 A.REI.C.7, A.REI.D.11: Quadratic-Linear Systems ………………… 67-71 A.REI.D.11: Other Systems ………………………………………………….. 72-81

POWERS

A.SSE.B.3, F.IF.C.8, F.BF.A.1, F.LE.A.2, F.LE.B.5: Modeling Exponential Functions …………………………………………… 82-95 F.IF.B.4: Evaluating Logarithmic Expressions ……………………………. 96 F.IF.C.7: Graphing Exponential and Logarithmic Functions …… 97-100 A.CED.A.1, F.LE.A.4: Exponential Equations ……………………. 101-110

POLYNOMIALS

A.SSE.A.2: Factoring Polynomials ……………………………………. 111-118 A.APR.B.3: Zeros of Polynomials……………………………………… 119-122 F.BF.B.4, F.IF.C.7: Graphing Polynomial Functions ……………. 123-131 A.APR.B.2: Remainder Theorem ………………………………………. 132-140 A.APR.C.4: Polynomial Identities ……………………………………… 141-146

RADICALS N.RN.A.2: Operations with Radicals ………………………………………… 147 A.REI.A.2: Solving Radicals …………………………………………….. 148-154 N.RN.A.1-2: Radicals and Rational Exponents ……………………. 155-164 N.CN.A.2: Operations with Complex Numbers …………………… 165-171

RATIONALS

A.APR.D.6: Undefined Rationals …………………………………………….. 172 A.APR.D.6: Expressions with Negative Expressions ………………….. 173 A.APR.D.6: Rational Expressions ……………………………………… 174-180 A.CED.A.1: Modeling Rationals ……………………………………….. 181-183 A.REI.A.2: Solving Rationals …………………………………………… 184-190

FUNCTIONS F.BF.A.1: Operations with Functions …………………………………. 191-194 F.IF.C.9: Comparing Functions …………………………………………. 195-197 F.BF.A.2: Even and Odd Functions ……………………………………. 198-200 F.BF.B.4: Inverse of Functions ………………………………………….. 201-205

SEQUENCES AND SERIES

F.LE.A.2, F.IF.A.3, F.BF.A.2: Sequences …………………………… 206-218 F.BF.B.6: Sigma Notation ………………………………………………………. 219 A.SSE.B.4: Series ……………………………………………………………. 220-225

TRIGONOMETRY

F.TF.A.1-2: Unit Circle ……………………………………………………. 226-227 F.TF.A.2: Reciprocal Trigonometric Functions …………………………. 228 F.TF.A.2: Reference Angles ……………………………………………………. 229 F.TF.A.2, F.TF.C.8: Determining Trigonometric Functions ….. 230-234 F.TF.C.8: Simplifying Trigonometric Identities …………………………. 235 F.TF.B.5: Modeling Trigonometric Functions …………………………… 236 F.IF.B.4, F.IF.C.7: Graphing Trigonometric Functions …………. 237-252

CONICS G.GPE.A.1: Equations of Conics……………………………………………… 253

Algebra II Regents Exam Questions by State Standard: Topicwww.jmap.org

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Algebra II Regents Exam Questions by Common Core State Standard: Topic

GRAPHS AND STATISTICSS.IC.A.2, S.IC.B.3-6: ANALYSIS OF DATA

1 Anne has a coin. She does not know if it is a fair coin. She flipped the coin 100 times and obtained 73 heads and 27 tails. She ran a computer simulation of 200 samples of 100 fair coin flips. The output of the proportion of heads is shown below.

Given the results of her coin flips and of her computer simulation, which statement is most accurate?1 73 of the computer’s next 100 coin flips will be

heads.2 50 of her next 100 coin flips will be heads.3 Her coin is not fair.4 Her coin is fair.

2 A game spinner is divided into 6 equally sized regions, as shown in the diagram below.

For Miles to win, the spinner must land on the number 6. After spinning the spinner 10 times, and losing all 10 times, Miles complained that the spinner is unfair. At home, his dad ran 100 simulations of spinning the spinner 10 times,

assuming the probability of winning each spin is 16.

The output of the simulation is shown in the diagram below.

Which explanation is appropriate for Miles and his dad to make?1 The spinner was likely unfair, since the number

6 failed to occur in about 20% of the simulations.

2 The spinner was likely unfair, since the spinner should have landed on the number 6 by the sixth spin.

3 The spinner was likely not unfair, since the number 6 failed to occur in about 20% of the simulations.

4 The spinner was likely not unfair, since in the output the player wins once or twice in the majority of the simulations.

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3 The results of simulating tossing a coin 10 times, recording the number of heads, and repeating this 50 times are shown in the graph below.

Based on the results of the simulation, which statement is false?1 Five heads occurred most often, which is

consistent with the theoretical probability of obtaining a heads.

2 Eight heads is unusual, as it falls outside the middle 95% of the data.

3 Obtaining three heads or fewer occurred 28% of the time.

4 Seven heads is not unusual, as it falls within the middle 95% of the data.

4 An orange-juice processing plant receives a truckload of oranges. The quality control team randomly chooses three pails of oranges, each containing 50 oranges, from the truckload. Identify the sample and the population in the given scenario. State one conclusion that the quality control team could make about the population if 5% of the sample was found to be unsatisfactory.

5 Mrs. Jones had hundreds of jelly beans in a bag that contained equal numbers of six different flavors. Her student randomly selected four jelly beans and they were all black licorice. Her student complained and said “What are the odds I got all of that kind?” Mrs. Jones replied, “simulate rolling a die 250 times and tell me if four black licorice jelly beans is unusual.” Explain how this simulation could be used to solve the problem.

6 In a random sample of 250 men in the United States, age 21 or older, 139 are married. The graph below simulated samples of 250 men, 200 times, assuming that 139 of the men are married.

a) Based on the simulation, create an interval in which the middle 95% of the number of married men may fall. Round your answer to the nearest integer. b) A study claims “50 percent of men 21 and older in the United States are married.” Do your results from part a contradict this claim? Explain.

7 Describe how a controlled experiment can be created to examine the effect of ingredient X in a toothpaste.

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8 Which statement(s) about statistical studies is true?I. A survey of all English classes in a high

school would be a good sample to determine the number of hours students throughout the school spend studying.

II. A survey of all ninth graders in a high school would be a good sample to determine the number of student parking spaces needed at that high school.

III. A survey of all students in one lunch period in a high school would be a good sample to determine the number of hours adults spend on social media websites.

IV. A survey of all Calculus students in a high school would be a good sample to determine the number of students throughout the school who don’t like math.

1 I, only2 II, only3 I and III4 III and IV

9 Which statement about statistical analysis is false?1 Experiments can suggest patterns and

relationships in data.2 Experiments can determine cause and effect

relationships.3 Observational studies can determine cause and

effect relationships.4 Observational studies can suggest patterns and

relationships in data.

10 Cheap and Fast gas station is conducting a consumer satisfaction survey. Which method of collecting data would most likely lead to a biased sample?1 interviewing every 5th customer to come into

the station2 interviewing customers chosen at random by a

computer at the checkout3 interviewing customers who call an 800

number posted on the customers’ receipts4 interviewing every customer who comes into

the station on a day of the week chosen at random out of a hat

11 Which scenario is best described as an observational study?1 For a class project, students in Health class ask

every tenth student entering the school if they eat breakfast in the morning.

2 A social researcher wants to learn whether or not there is a link between attendance and grades. She gathers data from 15 school districts.

3 A researcher wants to learn whether or not there is a link between children’s daily amount of physical activity and their overall energy level. During lunch at the local high school, she distributed a short questionnaire to students in the cafeteria.

4 Sixty seniors taking a course in Advanced Algebra Concepts are randomly divided into two classes. One class uses a graphing calculator all the time, and the other class never uses graphing calculators. A guidance counselor wants to determine whether there is a link between graphing calculator use and students’ final exam grades.

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12 The operator of the local mall wants to find out how many of the mall’s employees make purchases in the food court when they are working. She hopes to use these data to increase the rent and attract new food vendors. In total, there are 1023 employees who work at the mall. The best method to obtain a random sample of the employees would be to survey1 all 170 employees at each of the larger stores2 50% of the 90 employees of the food court3 every employee4 every 30th employee entering each mall

entrance for one week

13 A candidate for political office commissioned a poll. His staff received responses from 900 likely voters and 55% of them said they would vote for the candidate. The staff then conducted a simulation of 1000 more polls of 900 voters, assuming that 55% of voters would vote for their candidate. The output of the simulation is shown in the diagram below.

Given this output, and assuming a 95% confidence level, the margin of error for the poll is closest to1 0.012 0.033 0.064 0.12

14 Stephen’s Beverage Company is considering whether to produce a new brand of cola. The company will launch the product if at least 25% of cola drinkers will buy the product. Fifty cola drinkers are randomly selected to take a blind taste-test of products A, B, and the new product. Nine out of fifty participants preferred Stephen’s new cola to products A and B. The company then devised a simulation based on the requirement that 25% of cola drinkers will buy the product. Each dot in the graph shown below represents the proportion of people who preferred Stephen’s new product, each of sample size 50, simulated 100 times.

Assume the set of data is approximately normal and the company wants to be 95% confident of its results. Does the sample proportion obtained from the blind taste-test, nine out of fifty, fall within the margin of error developed from the simulation? Justify your answer. The company decides to continue developing the product even though only nine out of fifty participants preferred its brand of cola in the taste-test. Describe how the simulation data could be used to support this decision.

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15 A study conducted in 2004 in New York City found that 212 out of 1334 participants had hypertension. Kim ran a simulation of 100 studies based on these data. The output of the simulation is shown in the diagram below.

At a 95% confidence level, the proportion of New York City residents with hypertension and the margin of error are closest to1 proportion ≈ .16; margin of error ≈ .012 proportion ≈ .16; margin of error ≈ .023 proportion ≈ .01; margin of error ≈ .164 proportion ≈ .02; margin of error ≈ .16

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16 Seventy-two students are randomly divided into two equally-sized study groups. Each member of the first group (group 1) is to meet with a tutor after school twice each week for one hour. The second group (group 2), is given an online subscription to a tutorial account that they can access for a maximum of two hours each week. Students in both groups are given the same tests during the year. A summary of the two groups’ final grades is shown below:

Group 1 Group 2x 80.16 83.8S x 6.9 5.2

Calculate the mean difference in the final grades (group 1 – group 2) and explain its meaning in the context of the problem. A simulation was conducted in which the students’ final grades were rerandomized 500 times. The results are shown below.

Use the simulation to determine if there is a significant difference in the final grades. Explain your answer.

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17 Fifty-five students attending the prom were randomly selected to participate in a survey about the music choice at the prom. Sixty percent responded that a DJ would be preferred over a band. Members of the prom committee thought that the vote would have 50% for the DJ and 50% for the band. A simulation was run 200 times, each of sample size 55, based on the premise that 60% of the students would prefer a DJ. The approximate normal simulation results are shown below.

Using the results of the simulation, determine a plausible interval containing the middle 95% of the data. Round all values to the nearest hundredth. Members of the prom committee are concerned that a vote of all students attending the prom may produce a 50% − 50% split. Explain what statistical evidence supports this concern.

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18 Ayva designed an experiment to determine the effect of a new energy drink on a group of 20 volunteer students. Ten students were randomly selected to form group 1 while the remaining 10 made up group 2. Each student in group 1 drank one energy drink, and each student in group 2 drank one cola drink. Ten minutes later, their times were recorded for reading the same paragraph of a novel. The results of the experiment are shown below.

Group 1(seconds)

Group 2(seconds)

17.4 23.318.1 18.818.2 22.119.6 12.718.6 16.916.2 24.416.1 21.215.3 21.217.8 16.319.7 14.5

Mean = 17.7 Mean = 19.1

Ayva thinks drinking energy drinks makes students read faster. Using information from the experimental design or the results, explain why Ayva’s hypothesis may be incorrect. Using the given results, Ayva randomly mixes the 20 reading times, splits them into two groups of 10, and simulates the difference of the means 232 times.

Ayva has decided that the difference in mean reading times is not an unusual occurrence. Support her decision using the results of the simulation. Explain your reasoning.

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19 Gabriel performed an experiment to see if planting 13 tomato plants in black plastic mulch leads to larger tomatoes than if 13 plants are planted without mulch. He observed that the average weight of the tomatoes from tomato plants grown in black plastic mulch was 5 ounces greater than those from the plants planted without mulch. To determine if the observed difference is statistically significant, he rerandomized the tomato groups 100 times to study these random differences in the mean weights. The output of his simulation is summarized in the dotplot below.

Given these results, what is an appropriate inference that can be drawn?1 There was no effect observed between

the two groups.3 There is strong evidence to support the

hypothesis that tomatoes from plants planted in black plastic mulch are larger than those planted without mulch.

2 There was an effect observed that could be due to the random assignment of plants to the groups.

4 There is strong evidence to support the hypothesis that tomatoes from plants planted without mulch are larger than those planted in black plastic mulch.

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20 Charlie’s Automotive Dealership is considering implementing a new check-in procedure for customers who are bringing their vehicles for routine maintenance. The dealership will launch the procedure if 50% or more of the customers give the new procedure a favorable rating when compared to the current procedure. The dealership devises a simulation based on the minimal requirement that 50% of the customers prefer the new procedure. Each dot on the graph below represents the proportion of the customers who preferred the new check-in procedure, each of sample size 40, simulated 100 times.

Assume the set of data is approximately normal and the dealership wants to be 95% confident of its results. Determine an interval containing the plausible sample values for which the dealership will launch the new procedure. Round your answer to the nearest hundredth. Forty customers are selected randomly to undergo the new check-in procedure and the proportion of customers who prefer the new procedure is 32.5%. The dealership decides not to implement the new check-in procedure based on the results of the study. Use statistical evidence to explain this decision.

21 Elizabeth waited for 6 minutes at the drive thru at her favorite fast-food restaurant the last time she visited. She was upset about having to wait that long and notified the manager. The manager assured her that her experience was very unusual and that it would not happen again. A study of customers commissioned by this restaurant found an approximately normal distribution of results. The mean wait time was 226 seconds and the standard deviation was 38 seconds. Given these data, and using a 95% level of confidence, was Elizabeth’s wait time unusual? Justify your answer.

22 A public opinion poll was conducted on behalf of Mayor Ortega’s reelection campaign shortly before the election. 264 out of 550 likely voters said they would vote for Mayor Ortega; the rest said they would vote for his opponent. Which statement is least appropriate to make, according to the results of the poll?1 There is a 48% chance that Mayor Ortega will

win the election.2 The point estimate (p) of voters who will vote

for Mayor Ortega is 48%.3 It is most likely that between 44% and 52% of

voters will vote for Mayor Ortega.4 Due to the margin of error, an inference cannot

be made regarding whether Mayor Ortega or his opponent is most likely to win the election.

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S.ID.B.6: REGRESSION

23 Using a microscope, a researcher observed and recorded the number of bacteria spores on a large sample of uniformly sized pieces of meat kept at room temperature. A summary of the data she recorded is shown in the table below.

Hours (x) Average Numberof Spores (y)

0 40.5 101 152 603 2604 11306 16,380

Using these data, write an exponential regression equation, rounding all values to the nearest thousandth. The researcher knows that people are likely to suffer from food-borne illness if the number of spores exceeds 100. Using the exponential regression equation, determine the maximum amount of time, to the nearest quarter hour, that the meat can be kept at room temperature safely.

24 A runner is using a nine-week training app to prepare for a “fun run.” The table below represents the amount of the program completed, A, and the distance covered in a session, D, in miles.

A 49

59

69

89 1

D 2 2 2.25 3 3.25

Based on these data, write an exponential regression equation, rounded to the nearest thousandth, to model the distance the runner is able to complete in a session as she continues through the nine-week program.

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25 The price of a postage stamp in the years since the end of World War I is shown in the scatterplot below.

The equation that best models the price, in cents, of a postage stamp based on these data is1 y = 0.59x − 14.822 y = 1.04(1.43)x

3 y = 1.43(1.04)x

4 y = 24sin(14x) + 25

S.ID.A.4: NORMAL DISTRIBUTIONS

26 Two versions of a standardized test are given, an April version and a May version. The statistics for the April version show a mean score of 480 and a standard deviation of 24. The statistics for the May version show a mean score of 510 and a standard deviation of 20. Assume the scores are normally distributed. Joanne took the April version and scored in the interval 510-540. What is the probability, to the nearest ten thousandth, that a test paper selected at random from the April version scored in the same interval? Maria took the May version. In what interval must Maria score to claim she scored as well as Joanne?

27 The heights of women in the United States are normally distributed with a mean of 64 inches and a standard deviation of 2.75 inches. The percent of women whose heights are between 64 and 69.5 inches, to the nearest whole percent, is1 62 483 684 95

28 The lifespan of a 60-watt lightbulb produced by a company is normally distributed with a mean of 1450 hours and a standard deviation of 8.5 hours. If a 60-watt lightbulb produced by this company is selected at random, what is the probability that its lifespan will be between 1440 and 1465 hours?1 0.38032 0.46123 0.84154 0.9612

29 In 2013, approximately 1.6 million students took the Critical Reading portion of the SAT exam. The mean score, the modal score, and the standard deviation were calculated to be 496, 430, and 115, respectively. Which interval reflects 95% of the Critical Reading scores?1 430± 1152 430± 2303 496± 1154 496± 230

30 The weight of a bag of pears at the local market averages 8 pounds with a standard deviation of 0.5 pound. The weights of all the bags of pears at the market closely follow a normal distribution. Determine what percentage of bags, to the nearest integer, weighed less than 8.25 pounds.

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31 The distribution of the diameters of ball bearings made under a given manufacturing process is normally distributed with a mean of 4 cm and a standard deviation of 0.2 cm. What proportion of the ball bearings will have a diameter less than 3.7 cm?1 0.06682 0.43323 0.86644 0.9500

32 There are 440 students at Thomas Paine High School enrolled in U.S. History. On the April report card, the students’ grades are approximately normally distributed with a mean of 79 and a standard deviation of 7. Students who earn a grade less than or equal to 64.9 must attend summer school. The number of students who must attend summer school for U.S. History is closest to1 32 53 104 22

PROBABILITYS.CP.A.2, S.CP.B.7: THEORETICAL PROBABILITY

33 In contract negotiations between a local government agency and its workers, it is estimated that there is a 50% chance that an agreement will be reached on the salaries of the workers. It is estimated that there is a 70% chance that there will be an agreement on the insurance benefits. There is a 20% chance that no agreement will be reached on either issue. Find the probability that an agreement will be reached on both issues. Based on this answer, determine whether the agreement on salaries and the agreement on insurance are independent events. Justify your answer.

34 Given events A and B, such that P(A) = 0.6, P(B) = 0.5, and P(A∪B) = 0.8, determine whether A and B are independent or dependent.

35 A suburban high school has a population of 1376 students. The number of students who participate in sports is 649. The number of students who participate in music is 433. If the probability that a student participates in either sports or music is 9741376, what is the probability that a student

participates in both sports and music?

36 The probability that Gary and Jane have a child with blue eyes is 0.25, and the probability that they have a child with blond hair is 0.5. The probability that they have a child with both blue eyes and blond hair is 0.125. Given this information, the events blue eyes and blond hair are I: dependent II: independent III: mutually exclusive1 I, only2 II, only3 I and III4 II and III

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S.CP.A.3-4, S.CP.B.6: CONDITIONAL PROBABILITY

37 Sean’s team has a baseball game tomorrow. He pitches 50% of the games. There is a 40% chance of rain during the game tomorrow. If the probability that it rains given that Sean pitches is 40%, it can be concluded that these two events are1 independent2 dependent3 mutually exclusive4 complements

38 A student is chosen at random from the student body at a given high school. The probability that the student selects Math as the favorite subject is 14. The probability that the student chosen is a

junior is 116459. If the probability that the student

selected is a junior or that the student chooses Math

as the favorite subject is 47108, what is the exact

probability that the student selected is a junior whose favorite subject is Math? Are the events “the student is a junior” and “the student’s favorite subject is Math” independent of each other? Explain your answer.

39 The set of data in the table below shows the results of a survey on the number of messages that people of different ages text on their cell phones each month.

If a person from this survey is selected at random, what is the probability that the person texts over 50 messages per month given that the person is between the ages of 23 and 60?

1 157229

2 157312

3 157384

4 157456

40 The results of a poll of 200 students are shown in the table below:

Preferred Music StyleTechno Rap Country

Female 54 25 27Male 36 40 18

For this group of students, do these data suggest that gender and preferred music styles are independent of each other? Justify your answer.

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41 The results of a survey of the student body at Central High School about television viewing preferences are shown below.

Comedy Series Drama Series Reality Series TotalMales 95 65 70 230Females 80 70 110 260Total 175 135 180 490

Are the events “student is a male” and “student prefers reality series” independent of each other? Justify your answer.

42 Data collected about jogging from students with two older siblings are shown in the table below.

Neither SiblingJogs

One SiblingJogs

Both SiblingsJogs

Student DoesNot Jog 1168 1823 1380

Student Jogs 188 416 400

Using these data, determine whether a student with two older siblings is more likely to jog if one sibling jogs or if both siblings jog. Justify your answer.

43 The guidance department has reported that of the senior class, 2.3% are members of key club, K, 8.6% are enrolled in AP Physics, P, and 1.9% are in both. Determine the probability of P given K, to the nearest tenth of a percent. The principal would like a basic interpretation of these results. Write a statement relating your calculated probabilities to student enrollment in the given situation.

44 A study was designed to test the effectiveness of a new drug. Half of the volunteers received the drug. The other half received a sugar pill. The probability of a volunteer receiving the drug and getting well was 40%. What is the probability of a volunteer getting well, given that the volunteer received the drug?

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RATEF.IF.B.6: RATE OF CHANGE

45 Which function shown below has a greater average rate of change on the interval [−2,4]? Justify your answer.

x f(x)−4 0.3125−3 0.625−2 1.25−1 2.50 51 102 203 404 805 1606 320

g(x) = 4×3 − 5×2 + 3

46 Joelle has a credit card that has a 19.2% annual interest rate compounded monthly. She owes a total balance of B dollars after m months. Assuming she makes no payments on her account, the table below illustrates the balance she owes after m months.

m B0 100.0010 1172.0019 1352.0036 1770.8060 2591.9069 2990.0072 3135.8073 3186.00

Over which interval of time is her average rate of change for the balance on her credit card account the greatest?1 month 10 to month 60 3 month 36 to month 722 month 19 to month 69 4 month 60 to month 73

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47 The distance needed to stop a car after applying the brakes varies directly with the square of the car’s speed. The table below shows stopping distances for various speeds.

Speed (mph) 10 20 30 40 50 60 70Distance (ft) 6.25 25 56.25 100 156.25 225 306.25

Determine the average rate of change in braking distance, in ft/mph, between one car traveling at 50 mph and one traveling at 70 mph. Explain what this rate of change means as it relates to braking distance.

48 A cardboard box manufacturing company is building boxes with length represented by x + 1, width by 5 − x, and height by x − 1. The volume of the box is modeled by the function below.

Over which interval is the volume of the box changing at the fastest average rate?1 [1,2]2 [1,3.5]3 [1,5]4 [0,3.5]

49 The function f(x) = 2−0.25x • sin π2 x

represents a

damped sound wave function. What is the average rate of change for this function on the interval [−7,7], to the nearest hundredth?1 -3.662 -0.303 -0.264 3.36

50 The value of a new car depreciates over time. Greg purchased a new car in June 2011. The value, V, of his car after t years can be modeled by the equation

log0.8V

17000

= t . What is the average decreasing

rate of change per year of the value of the car from June 2012 to June 2014, to the nearest ten dollars per year?1 19602 21803 24504 2770

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QUADRATICSA.REI.B.4: SOLVING QUADRATICS

51 The solutions to the equation −12 x 2 = −6x + 20 are

1 −6 ± 2i2 −6 ± 2 193 6 ± 2i4 6 ± 2 19

52 A solution of the equation 2x 2 + 3x + 2 = 0 is

1 −34 + 1

4 i 7

2 −34 + 1

4 i

3 −34 + 1

4 7

4 12

53 The solution to the equation 18×2 − 24x + 87 = 0 is

1 −23 ± 6i 158

2 −23 ± 1

6 i 158

3 23 ± 6i 158

4 23 ± 1

6 i 158

54 The roots of the equation x2 + 2x + 5 = 0 are1 −3 and 12 −1, only3 −1 + 2i and − 1 − 2i4 −1 + 4i and − 1 − 4i

55 The solution to the equation 4×2 + 98 = 0 is1 ±72 ±7i

3 ±7 2

2

4 ±7i 2

2

A.REI.B.4: COMPLEX CONJUGATE ROOT THEOREM

56 Which equation has 1− i as a solution?1 x2 + 2x − 2 = 02 x2 + 2x + 2 = 03 x2 − 2x − 2 = 04 x2 − 2x + 2 = 0

G.GPE.A.2: GRAPHING QUADRATIC FUNCTIONS

57 Which equation represents a parabola with a focus of (0,4) and a directrix of y = 2?1 y = x2 + 32 y = −x2 + 1

3 y = x2

2 + 3

4 y = x2

4 + 3

58 The directrix of the parabola 12(y + 3) = (x − 4)2 has the equation y = −6. Find the coordinates of the focus of the parabola.

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59 Which equation represents the set of points equidistant from line and point R shown on the graph below?

1 y = −18 (x + 2)2 + 1

2 y = −18 (x + 2)2 − 1

3 y = −18 (x − 2)2 + 1

4 y = −18 (x − 2)2 − 1

60 A parabola has its focus at (1,2) and its directrix is y = −2. The equation of this parabola could be1 y = 8(x + 1)2

2 y = 18 (x + 1)2

3 y = 8(x − 1)2

4 y = 18 (x − 1)2

61 Which equation represents a parabola with the focus at (0,−1) and the directrix of y = 1?1 x 2 = −8y2 x 2 = −4y3 x 2 = 8y4 x 2 = 4y

62 What is the equation of the directrix for the parabola −8(y − 3) = (x + 4)2?1 y = 52 y = 13 y = −24 y = −6

SYSTEMSA.REI.C.6: SOLVING LINEAR SYSTEMS

63 Solve the following system of equations algebraically for all values of x, y, and z:

x + 3y + 5z = 45

6x − 3y + 2z = −10

−2x + 3y + 8z = 72

64 Which value is not contained in the solution of the system shown below?

a + 5b − c = −20

4a − 5b + 4c = 19

−a − 5b − 5c = 21 −22 23 34 −3

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65 Solve the following system of equations algebraically for all values of x, y, and z:

x + y + z = 1

2x + 4y + 6z = 2

−x + 3y − 5z = 11

66 For the system shown below, what is the value of z?

y = −2x + 14

3x − 4z = 2

3x − y = 161 52 23 64 4

A.REI.C.7, A.REI.D.11: QUADRATIC-LINEAR SYSTEMS

67 Algebraically determine the values of x that satisfy the system of equations below.

y = −2x + 1

y = −2x 2 + 3x + 1

68 Solve the system of equations shown below algebraically.

(x − 3)2 + (y + 2)2 = 16

2x + 2y = 10

69 What is the solution to the system of equations y = 3x − 2 and y = g(x) where g(x) is defined by the function below?

1 {(0,−2)}2 {(0,−2),(1,6)}3 {(1,6)}4 {(1,1),(6,16)}

70 Consider the system shown below.2x − y = 4

(x + 3)2 + y2 = 8The two solutions of the system can be described as1 both imaginary2 both irrational3 both rational4 one rational and one irrational

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71 Sally’s high school is planning their spring musical. The revenue, R, generated can be determined by the function R(t) = −33t2 + 360t , where t represents the price of a ticket. The production cost, C, of the musical is represented by the function C(t) = 700+ 5t. What is the highest ticket price, to the nearest dollar, they can charge in order to not lose money on the event?1 t = 32 t = 53 t = 84 t = 11

A.REI.D.11: OTHER SYSTEMS

72 Given: h(x) = 29 x3 + 8

9 x2 − 1613 x + 2

k(x) = − 0.7x| | + 5State the solutions to the equation h(x) = k(x), rounded to the nearest hundredth.

73 Which value, to the nearest tenth, is not a solution of p(x) = q(x) if p(x) = x 3 + 3x 2 − 3x − 1 and q(x) = 3x + 8?1 −3.92 −1.13 2.14 4.7

74 Drugs break down in the human body at different rates and therefore must be prescribed by doctors carefully to prevent complications, such as overdosing. The breakdown of a drug is represented by the function N(t) = N0 (e)−rt , where N(t) is the amount left in the body, N0 is the initial dosage, r is the decay rate, and t is time in hours. Patient A, A(t), is given 800 milligrams of a drug with a decay rate of 0.347. Patient B, B(t), is given 400 milligrams of another drug with a decay rate of 0.231. Write two functions, A(t) and B(t), to represent the breakdown of the respective drug given to each patient. Graph each function on the set of axes below.

To the nearest hour, t, when does the amount of the given drug remaining in patient B begin to exceed the amount of the given drug remaining in patient A? The doctor will allow patient A to take another 800 milligram dose of the drug once only 15% of the original dose is left in the body. Determine, to the nearest tenth of an hour, how long patient A will have to wait to take another 800 milligram dose of the drug.

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75 To the nearest tenth, the value of x that satisfies 2x = −2x + 11 is1 2.52 2.63 5.84 5.9

76 When g(x) = 2x + 2 and h(x) = log(x + 1) + 3 are

graphed on the same set of axes, which coordinates best approximate their point of intersection?1 (−0.9,1.8)2 (−0.9,1.9)3 (1.4,3.3)4 (1.4,3.4)

77 Pedro and Bobby each own an ant farm. Pedro starts with 100 ants and says his farm is growing exponentially at a rate of 15% per month. Bobby starts with 350 ants and says his farm is steadily decreasing by 5 ants per month. Assuming both boys are accurate in describing the population of their ant farms, after how many months will they both have approximately the same number of ants?1 72 83 134 36

78 If f(x) = 3 x| | − 1 and g(x) = 0.03×3 − x + 1, an approximate solution for the equation f(x) = g(x) is1 1.962 11.293 (−0.99,1.96)4 (11.29,32.87)

79 The value of a certain small passenger car based on its use in years is modeled by V(t) = 28482.698(0.684) t , where V(t) is the value in dollars and t is the time in years. Zach had to take out a loan to purchase the small passenger car. The function Z(t) = 22151.327(0.778) t , where Z(t) is measured in dollars, and t is the time in years, models the unpaid amount of Zach’s loan over time. Graph V(t) and Z(t) over the interval 0 ≤ t ≤ 5, on the set of axes below.

State when V(t) = Z(t), to the nearest hundredth, and interpret its meaning in the context of the problem. Zach takes out an insurance policy that requires him to pay a $3000 deductible in case of a collision. Zach will cancel the collision policy when the value of his car equals his deductible. To the nearest year, how long will it take Zach to cancel this policy? Justify your answer.

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80 For which values of x, rounded to the nearest hundredth, will x2 − 9|

||| − 3 = log3x?

1 2.29 and 3.632 2.37 and 3.543 2.84 and 3.174 2.92 and 3.06

81 Researchers in a local area found that the population of rabbits with an initial population of 20 grew continuously at the rate of 5% per month. The fox population had an initial value of 30 and grew continuously at the rate of 3% per month. Find, to the nearest tenth of a month, how long it takes for these populations to be equal.

POWERSA.SSE.B.3, F.IF.C.8, F.BF.A.1, F.LE.A.2, F.LE.B.5: MODELING EXPONENTIAL FUNCTIONS

82 A study of the annual population of the red-winged blackbird in Ft. Mill, South Carolina, shows the population, B(t), can be represented by the function B(t) = 750(1.16) t , where the t represents the number of years since the study began. In terms of the monthly rate of growth, the population of red-winged blackbirds can be best approximated by the function1 B(t) = 750(1.012) t

2 B(t) = 750(1.012)12t

3 B(t) = 750(1.16)12t

4 B(t) = 750(1.16)t

12

83 A student studying public policy created a model for the population of Detroit, where the population decreased 25% over a decade. He used the model P = 714(0.75)d , where P is the population, in thousands, d decades after 2010. Another student, Suzanne, wants to use a model that would predict the population after y years. Suzanne’s model is best represented by1 P = 714(0.6500) y

2 P = 714(0.8500) y

3 P = 714(0.9716) y

4 P = 714(0.9750) y

84 lridium-192 is an isotope of iridium and has a half-life of 73.83 days. If a laboratory experiment begins with 100 grams of Iridium-192, the number of grams, A, of Iridium-192 present after t days

would be A = 100 12

t73.83

. Which equation

approximates the amount of Iridium-192 present after t days?

1 A = 100 73.832

t

2 A = 100 1147.66

t

3 A = 100(0.990656) t

4 A = 100(0.116381) t

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85 For a given time, x, in seconds, an electric current, y, can be represented by y = 2.5 1 − 2.7−.10x

.

Which equation is not equivalent?1 y = 2.5− 2.5 2.7−.10x

2 y = 2.5− 2.5 2.72

−.05x

3 y = 2.5− 2.5 12.7.10x

4 y = 2.5− 2.5 2.7−2

2.7.05x

86 Which function represents exponential decay?1 y = 20.3t

2 y = 1.23t

3 y = 12

−t

4 y = 5−t

87 The function M(t) represents the mass of radium over time, t, in years.

M(t) = 100e

ln12

t

1590

Determine if the function M(t) represents growth or decay. Explain your reasoning.

88 Last year, the total revenue for Home Style, a national restaurant chain, increased 5.25% over the previous year. If this trend were to continue, which expression could the company’s chief financial officer use to approximate their monthly percent increase in revenue? [Let m represent months.]1 (1.0525)m

2 (1.0525)12m

3 (1.00427)m

4 (1.00427)m12

89 A payday loan company makes loans between $100 and $1000 available to customers. Every 14 days, customers are charged 30% interest with compounding. In 2013, Remi took out a $300 payday loan. Which expression can be used to calculate the amount she would owe, in dollars, after one year if she did not make payments?

1 300(.30)14365

2 300(1.30)14365

3 300(.30)36514

4 300(1.30)36514

90 According to a pricing website, Indroid phones lose 58% of their cash value over 1.5 years. Which expression can be used to estimate the value of a $300 Indroid phone in 1.5 years?1 300e−0.87

2 300e−0.63

3 300e−0.58

4 300e−0.42

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91 Titanium-44 is a radioactive isotope such that every 63 years, its mass decreases by half. For a sample of titanium-44 with an initial mass of 100 grams, write a function that will give the mass of the sample remaining after any amount of time. Define all variables. Scientists sometimes use the average yearly decrease in mass for estimation purposes. Use the average yearly decrease in mass of the sample between year 0 and year 10 to predict the amount of the sample remaining after 40 years. Round your answer to the nearest tenth. Is the actual mass of the sample or the estimated mass greater after 40 years? Justify your answer.

92 A rabbit population doubles every 4 weeks. There are currently five rabbits in a restricted area. If t represents the time, in weeks, and P(t) is the population of rabbits with respect to time, about how many rabbits will there be in 98 days?1 562 1523 36884 81,920

93 An equation to represent the value of a car after t

months of ownership is v = 32,000(0.81)t

12 . Which statement is not correct?1 The car lost approximately 19% of its value

each month.2 The car maintained approximately 98% of its

value each month.3 The value of the car when it was purchased

was $32,000.4 The value of the car 1 year after it was

purchased was $25,920.

94 The function p(t) = 110e0.03922t models the population of a city, in millions, t years after 2010. As of today, consider the following two statements: I. The current population is 110 million. II. The population increases continuously by approximately 3.9% per year.This model supports1 I, only2 II, only3 both I and II4 neither I nor II

95 A certain pain reliever is taken in 220 mg dosages and has a half-life of 12 hours. The function

A = 220 12

t12

can be used to model this situation,

where A is the amount of pain reliever in milligrams remaining in the body after t hours. According to this function, which statement is true?1 Every hour, the amount of pain reliever

remaining is cut in half.2 In 12 hours, there is no pain reliever remaining

in the body.3 In 24 hours, there is no pain reliever remaining

in the body.4 In 12 hours, 110 mg of pain reliever is

remaining.

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F.IF.B.4: EVALUATING LOGARITHMIC EXPRESSIONS

96 The loudness of sound is measured in units called decibels (dB). These units are measured by first assigning an intensity I0 to a very soft sound that is called the threshold sound. The sound to be measured is assigned an

intensity, I, and the decibel rating, d, of this sound is found using d = 10log II0

. The threshold sound audible to

the average person is 1.0× 10−12 W/m2 (watts per square meter). Consider the following sound level classifications:

Moderate 45-69 dBLoud 70-89 dBVery loud 90-109 dBDeafening >110 dB

How would a sound with intensity 6.3× 10−3 W/m2 be classified?1 moderate 3 very loud2 loud 4 deafening

F.IF.C.7: GRAPHING EXPONENTIAL AND LOGARITHMIC FUNCTIONS

97 Graph y = 400(.85)2x − 6 on the set of axes below.

98 If the function g(x) = ab x represents exponential growth, which statement about g(x) is false?1 a > 0 and b > 12 The y-intercept is (0,a).3 The asymptote is y = 0.4 The x-intercept is (b,0).

99 Which statement about the graph of c(x) = log6x is false?1 The asymptote has equation y = 0.2 The graph has no y-intercept.3 The domain is the set of positive reals.4 The range is the set of all real numbers.

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100 Graph y = log2(x + 3) − 5 on the set of axes below. Use an appropriate scale to include both intercepts.

Describe the behavior of the given function as x approaches -3 and as x approaches positive infinity.

A.CED.A.1, F.LE.A.4: EXPONENTIAL EQUATIONS

101 Seth’s parents gave him $5000 to invest for his 16th birthday. He is considering two investment options. Option A will pay him 4.5% interest compounded annually. Option B will pay him 4.6% compounded quarterly. Write a function of option A and option B that calculates the value of each account after n years. Seth plans to use the money after he graduates from college in 6 years. Determine how much more money option B will earn than option A to the nearest cent. Algebraically determine, to the nearest tenth of a year, how long it would take for option B to double Seth’s initial investment.

102 Monthly mortgage payments can be found using the formula below:

M =

P r12

1 + r12

n

1+ r12

n

− 1

M = monthly paymentP = amount borrowedr = annual interest rate

n = number of monthly payments

The Banks family would like to borrow $120,000 to purchase a home. They qualified for an annual interest rate of 4.8%. Algebraically determine the fewest number of whole years the Banks family would need to include in the mortgage agreement in order to have a monthly payment of no more than $720.

103 What is the solution to 8(2x + 3 ) = 48?

1 x = ln6ln2 − 3

2 x = 0

3 x = ln48ln16 − 3

4 x = ln4− 3

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104 After sitting out of the refrigerator for a while, a turkey at room temperature (68°F) is placed into an oven at 8 a.m., when the oven temperature is 325°F. Newton’s Law of Heating explains that the temperature of the turkey will increase proportionally to the difference between the temperature of the turkey and the temperature of the oven, as given by the formula below:

T = Ta + T0 − Ta

e−kt

Ta = the temperature surrounding the object

T0 = the initial temperature of the object

t = the time in hours

T = the temperature of the object after t hours

k = decay constant

The turkey reaches the temperature of approximately 100° F after 2 hours. Find the value of k, to the nearest thousandth, and write an equation to determine the temperature of the turkey after t hours. Determine the Fahrenheit temperature of the turkey, to the nearest degree, at 3 p.m.

105 A house purchased 5 years ago for $100,000 was just sold for $135,000. Assuming exponential growth, approximate the annual growth rate, to the nearest percent.

106 In New York State, the minimum wage has grown exponentially. In 1966, the minimum wage was $1.25 an hour and in 2015, it was $8.75. Algebraically determine the rate of growth to the nearest percent.

107 Judith puts $5000 into an investment account with interest compounded continuously. Which approximate annual rate is needed for the account to grow to $9110 after 30 years?1 2%2 2.2%3 0.02%4 0.022%

108 If aebt = c, where a, b, and c are positive, then t equals

1 ln cab

2 ln cba

3ln c

a

b

4ln c

a

lnb

109 One of the medical uses of Iodine–131 (I–131), a radioactive isotope of iodine, is to enhance x-ray images. The half-life of I–131 is approximately 8.02 days. A patient is injected with 20 milligrams of I–131. Determine, to the nearest day, the amount of time needed before the amount of I–131 in the patient’s body is approximately 7 milligrams.

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110 A radioactive substance has a mass of 140 g at 3 p.m. and 100 g at 8 p.m. Write an equation in the

form A = A012

th

that models this situation,

where h is the constant representing the number of hours in the half-life, A0 is the initial mass, and A is the mass t hours after 3 p.m. Using this equation, solve for h, to the nearest ten thousandth. Determine when the mass of the radioactive substance will be 40 g. Round your answer to the nearest tenth of an hour.

POLYNOMIALSA.SSE.A.2: FACTORING POLYNOMIALS

111 What is the completely factored form of k 4 − 4k 2 + 8k 3 − 32k + 12k 2 − 48?1 (k − 2)(k − 2)(k + 3)(k + 4)2 (k − 2)(k − 2)(k + 6)(k + 2)3 (k + 2)(k − 2)(k + 3)(k + 4)4 (k + 2)(k − 2)(k + 6)(k + 2)

112 Rewrite the expression

4×2 + 5x

2− 5 4x 2 + 5x

− 6 as a product of four

linear factors.

113 Which factorization is incorrect?1 4k 2 − 49 = (2k + 7)(2k − 7)2 a 3 − 8b 3 = (a − 2b)(a 2 + 2ab + 4b 2 )3 m3 + 3m2 − 4m+ 12 = (m− 2)2 (m+ 3)4 t3 + 5t2 + 6t + t2 + 5t + 6 = (t + 1)(t + 2)(t + 3)

114 The completely factored form of 2d 4 + 6d 3 − 18d 2 − 54d is1 2d(d 2 − 9)(d + 3)2 2d(d 2 + 9)(d + 3)3 2d(d + 3)2(d − 3)4 2d(d − 3)2(d + 3)

115 Factored completely, m5 +m3 − 6m is equivalent to1 (m+ 3)(m− 2)2 (m2 + 3m)(m2 − 2)3 m(m4 +m2 − 6)4 m(m2 + 3)(m2 − 2)

116 Over the set of integers, factor the expression 4×3 − x2 + 16x − 4 completely.

117 Which expression has been rewritten correctly to form a true statement?1 (x + 2)2 + 2(x + 2) − 8 = (x + 6)x2 x4 + 4×2 + 9x2y2 − 36y2 = (x + 3y)2 (x − 2)2

3 x3 + 3×2 − 4xy2 − 12y2 = (x − 2y)(x + 3)2

4 (x2 − 4)2 − 5(x 2 − 4) − 6 = (x2 − 7)(x 2 − 6)

118 Completely factor the following expression: x2 + 3xy + 3×3 + y

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A.APR.B.3: ZEROS OF POLYNOMIALS

119 The zeros for f(x) = x4 − 4×3 − 9×2 + 36x are1 {0,±3,4}2 {0,3,4}3 {0,±3,−4}4 {0,3,−4}

120 The graph of the function p(x) is sketched below.

Which equation could represent p(x)?1 p(x) = (x2 − 9)(x − 2)2 p(x) = x3 − 2×2 + 9x + 183 p(x) = (x2 + 9)(x − 2)4 p(x) = x3 + 2×2 − 9x − 18

121 What are the zeros of P(m) = (m2 − 4)(m2 + 1)?1 2 and − 2, only2 2, − 2, and − 43 −4, i, and − i4 2, − 2, i, and − i

122 The graph of y = f(x) is shown below. The function has a leading coefficient of 1.

Write an equation for f(x). The function g is formed by translating function f left 2 units. Write an equation for g(x).

F.IF.B.4, F.IF.C.7: GRAPHING POLYNOMIAL FUNCTIONS

123 A polynomial equation of degree three, p(x), is used to model the volume of a rectangular box. The graph of p(x) has x intercepts at −2, 10, and 14. Which statements regarding p(x) could be true?A. The equation of p(x) = (x − 2)(x + 10)(x + 14).B. The equation of p(x) = −(x + 2)(x − 10)(x − 14).C. The maximum volume occurs when x = 10.D. The maximum volume of the box is approximately 56.1 A and C2 A and D3 B and C4 B and D

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124 There was a study done on oxygen consumption of snails as a function of pH, and the result was a degree 4 polynomial function whose graph is shown below.

Which statement about this function is incorrect?1 The degree of the polynomial is even.2 There is a positive leading coefficient.3 At two pH values, there is a relative maximum

value.4 There are two intervals where the function is

decreasing.

125 The function below models the average price of gas in a small town since January 1st.G(t) = −0.0049t4 + 0.0923t3 − 0.56t2 + 1.166t + 3.23, where 0 ≤ t ≤ 10.If G(t) is the average price of gas in dollars and t represents the number of months since January 1st, the absolute maximum G(t) reaches over the given domain is about1 $1.602 $3.923 $4.014 $7.73

126 If a, b, and c are all positive real numbers, which graph could represent the sketch of the graph of p(x) = −a(x + b) x 2 − 2cx + c2

?

1

2

3

4

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127 Which graph has the following characteristics?• three real zeros• as x → −∞, f(x) → −∞• as x → ∞, f(x) → ∞

1

2

3

4

128 Find algebraically the zeros for p(x) = x 3 + x2 − 4x − 4. On the set of axes below, graph y = p(x).

129 On the axes below, sketch a possible function p(x) = (x − a)(x − b)(x + c), where a, b, and c are positive, a > b, and p(x) has a positive y-intercept of d. Label all intercepts.

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130 On the grid below, sketch a cubic polynomial whose zeros are 1, 3, and -2.

131 The zeros of a quartic polynomial function h are −1, ±2, and 3. Sketch a graph of y = h(x) on the grid below.

A.APR.B.2: REMAINDER THEOREM

132 The graph of p(x) is shown below.

What is the remainder when p(x) is divided by x + 4?1 x − 42 −43 04 4

133 Use an appropriate procedure to show that x − 4 is a factor of the function f(x) = 2x 3 − 5x 2 − 11x − 4. Explain your answer.

134 Given z x = 6×3 + bx 2 − 52x + 15, z 2 = 35, and z −5 = 0, algebraically determine all the zeros of z x .

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135 Determine if x − 5 is a factor of 2×3 − 4×2 − 7x − 10. Explain your answer.

136 When g(x) is divided by x + 4, the remainder is 0. Given g(x) = x4 + 3×3 − 6×2 − 6x + 8, which conclusion about g(x) is true?1 g(4) = 02 g(−4) = 03 x − 4 is a factor of g(x).4 No conclusion can be made regarding g(x).

137 Which binomial is a factor of x4 − 4×2 − 4x + 8?1 x − 22 x + 23 x − 44 x + 4

138 Given r(x) = x3 − 4×2 + 4x − 6, find the value of r(2). What does your answer tell you about x − 2 as a factor of r(x)? Explain.

139 Which binomial is not a factor of the expression x3 − 11x 2 + 16x + 84?1 x + 22 x + 43 x − 64 x − 7

140 If p(x) = 2×3 − 3x + 5, what is the remainder of p(x) ÷ (x − 5)?1 −2302 03 404 240

A.APR.C.4: POLYNOMIAL IDENTITIES

141 Algebraically prove that the difference of the squares of any two consecutive integers is an odd integer.

142 Algebraically prove that x 3 + 9×3 + 8

= 1+ 1x 3 + 8

,

where x ≠ −2.

143 Mr. Farison gave his class the three mathematical rules shown below to either prove or disprove. Which rules can be proved for all real numbers?

I (m+ p)2 = m2 + 2mp + p 2

II (x + y)3 = x3 + 3xy + y3

III (a 2 + b 2 )2 = (a 2 − b 2 )2 + (2ab)2

1 I, only2 I and II3 II and III4 I and III

144 Algebraically determine the values of h and k to correctly complete the identity stated below.

2×3 − 10×2 + 11x − 7 = (x − 4)(2×2 + hx + 3) + k

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145 Verify the following Pythagorean identity for all values of x and y:

(x2 + y2 )2 = (x2 − y2 )2 + (2xy)2

146 The expression (x + a)(x + b) can not be written as1 a(x + b) + x(x + b)2 x2 + abx + ab3 x2 + (a + b)x + ab4 x(x + a) + b(x + a)

RADICALSN.RN.A.2: OPERATIONS WITH RADICALS

147 Write x3 • x as a single term with a rational exponent.

A.REI.A.2: SOLVING RADICALS

148 Solve algebraically for all values of x: x − 5 + x = 7

149 The solution set for the equation 56− x = x is1 {−8,7}2 {−7,8}3 {7}4 { }

150 Solve the equation 2x − 7 + x = 5 algebraically, and justify the solution set.

151 The speed of a tidal wave, s, in hundreds of miles per hour, can be modeled by the equation s = t − 2t + 6, where t represents the time from its origin in hours. Algebraically determine the time when s = 0. How much faster was the tidal wave traveling after 1 hour than 3 hours, to the nearest mile per hour? Justify your answer.

152 Solve algebraically for all values of x: x − 4 + x = 6

153 The solution set for the equation x + 14 − 2x + 5 = 1 is

1 {−6}2 {2}3 {18}4 {2,22}

154 What is the solution set for x in the equation below?

x + 1 − 1 = x1 {1}2 {0}3 {−1,0}4 {0,1}

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N.RN.A.1-2: RADICALS AND RATIONAL EXPONENTS

155 Explain how 315

2

can be written as the

equivalent radical expression 95 .

156 Explain how (−8)43 can be evaluated using

properties of rational exponents to result in an integer answer.

157 Explain why 8134 equals 27.

158 When b > 0 and d is a positive integer, the

expression (3b)2d is equivalent to

1 1

3bd

2

2 3b

d

3 1

3bd

4 3bd

2

159 Given the equal terms x53 and y56 , determine and

state y, in terms of x.

160 Use the properties of rational exponents to determine the value of y for the equation:

x83

x4

13

= x y, x > 1

161 The expression m2

m13

−12

is equivalent to

1 − m56

2 1

m56

3 −m m5

4 1m m5

162 For x ≠ 0, which expressions are equivalent to one divided by the sixth root of x?

I. x6

x3 II. x16

x13

III. x−16

1 I and II, only2 I and III, only3 II and III, only4 I, II, and III

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163 What does −54×9

y4

23

equal?

19ix6 43

y y23

29ix6 43

y 2 y23

39x 6 43

y y3

49x 6 43

y 2 y23

164 If n = a 5 and m = a, where a > 0, an expression

for nm could be

1 a52

2 a 4

3 a 23

4 a 3

N.CN.A.2: OPERATIONS WITH COMPLEX NUMBERS

165 Write (5 + 2yi)(4− 3i) − (5− 2yi)(4− 3i) in a + bi form, where y is a real number.

166 Simplify xi(i − 7i)2 , where i is the imaginary unit.

167 Elizabeth tried to find the product of (2 + 4i) and (3 − i), and her work is shown below.

(2+ 4i)(3 − i)

= 6 − 2i + 12i − 4i2

= 6 + 10i − 4i2

= 6 + 10i − 4(1)

= 6 + 10i − 4

= 2 + 10iIdentify the error in the process shown and determine the correct product of (2 + 4i) and (3 − i).

168 Given i is the imaginary unit, (2 − yi)2 in simplest form is1 y2 − 4yi + 42 −y2 − 4yi + 43 −y2 + 44 y2 + 4

169 Express 1− i 3 in a + bi form.

170 The expression 6xi3(−4xi + 5) is equivalent to1 2x − 5i2 −24×2 − 30xi3 −24×2 + 30x − i4 26x − 24×2 i − 5i

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171 Which expression is equivalent to (3k − 2i)2 , where i is the imaginary unit?1 9k 2 − 42 9k 2 + 43 9k 2 − 12ki − 44 9k 2 − 12ki + 4

RATIONALSA.APR.D.6: UNDEFINED RATIONALS

172 The function f(x) = x − 3×2 + 2x − 8

is undefined when

x equals1 2 or − 42 4 or − 23 3, only4 2, only

A.APR.D.6: EXPRESSIONS WITH NEGATIVE EXPONENTS

173 The expression −3×2 − 5x + 2x 3 + 2×2 can be rewritten as

1 −3x − 3×2 + 2x

2 −3x − 1×2

3 −3x−1 + 14 −3x−1 + x−2

A.APR.D.6: RATIONAL EXPRESSIONS

174 The expression 6×3 + 17×2 + 10x + 22x + 3 equals

1 3×2 + 4x − 1+ 52x + 3

2 6×2 + 8x − 2+ 52x + 3

3 6×2 − x + 13− 372x + 3

4 3x 2 + 13x + 492 + 151

2x + 3

175 The expression 4x 3 + 5x + 102x + 3 is equivalent to

1 2×2 + 3x − 7+ 312x + 3

2 2×2 − 3x + 7− 112x + 3

3 2×2 + 2.5x + 5+ 152x + 3

4 2×2 − 2.5x − 5− 202x + 3

176 The expression x 3 + 2×2 + x + 6x + 2 is equivalent to

1 x2 + 3

2 x2 + 1 + 4x + 2

3 2×2 + x + 6

4 2×2 + 1+ 4x + 2

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177 Given f(x) = 3×2 + 7x − 20 and g(x) = x − 2, state

the quotient and remainder of f(x)g(x) , in the form

q(x) +r(x)g(x) .

178 Which expression is equivalent to 4x 3 + 9x − 52x − 1 ,

where x ≠ 12 ?

1 2×2 + x + 5

2 2x 2 + 112 + 1

2(2x − 1)3 2×2 − x + 5

4 2×2 − x + 4 + 12x − 1

179 What is the quotient when 10×3 − 3×2 − 7x + 3 is divided by 2x − 1?1 5×2 + x + 32 5×2 − x + 33 5×2 − x − 34 5×2 + x − 3

180 Written in simplest form, c2 − d 2

d 2 + cd − 2c 2 where

c ≠ d , is equivalent to

1 c + dd + 2c

2 c − dd + 2c

3 −c − dd + 2c

4 −c + dd + 2c

A.CED.A.1: MODELING RATIONALS

181 Julie averaged 85 on the first three tests of the semester in her mathematics class. If she scores 93 on each of the remaining tests, her average will be 90. Which equation could be used to determine how many tests, T, are left in the semester?

1 255 + 93T3T = 90

2 255 + 90T3T = 93

3 255 + 93TT + 3 = 90

4 255 + 90TT + 3 = 93

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182 Mallory wants to buy a new window air conditioning unit. The cost for the unit is $329.99. If she plans to run the unit three months out of the year for an annual operating cost of $108.78, which function models the cost per year over the lifetime of the unit, C(n), in terms of the number of years, n, that she owns the air conditioner.1 C(n) = 329.99 + 108.78n2 C(n) = 329.99 + 326.34n

3 C(n) = 329.99+ 108.78nn

4 C(n) = 329.99+ 326.34nn

183 A formula for work problems involving two people is shown below.

1t1+ 1

t2= 1

tb

t1 = the time taken by the first person to complete the job t2 = the time taken by the second person to complete the job tb = the time it takes for them working together to complete the job Fred and Barney are carpenters who build the same model desk. It takes Fred eight hours to build the desk while it only takes Barney six hours. Write an equation that can be used to find the time it would take both carpenters working together to build a desk. Determine, to the nearest tenth of an hour, how long it would take Fred and Barney working together to build a desk.

A.REI.A.2: SOLVING RATIONALS

184 Solve for x: 1x − 1

3 = − 13x

185 What is the solution set of the equation 3x + 25x + 7 − 5 = 3

x ?

1 32 ,7

2 72 ,−3

3 −32 ,7

4 −72 ,−3

186 The focal length, F, of a camera’s lens is related to the distance of the object from the lens, J, and the distance to the image area in the camera, W, by the formula below.

1J + 1

W = 1F

When this equation is solved for J in terms of F and W, J equals1 F −W

2 FWF −W

3 FWW− F

4 1F − 1

W

187 What is the solution, if any, of the equation 2

x + 3 − 34 − x = 2x − 2

x2 − x − 12?

1 -12 -53 all real numbers4 no real solution

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188 To solve 2xx − 2 − 11

x = 8x 2 − 2x

, Ren multiplied

both sides by the least common denominator. Which statement is true?1 2 is an extraneous solution.

2 72 is an extraneous so1ution.

3 0 and 2 are extraneous solutions.4 This equation does not contain any extraneous

solutions.

189 Solve for all values of p: 3p

p − 5 − 2p + 3 =

pp + 3

190 The solutions to x + 3 − 4x − 1 = 5 are

1 32 ±

172

2 32 ±

172 i

3 32 ±

332

4 32 ±

332 i

FUNCTIONSF.BF.A.1: OPERATIONS WITH FUNCTIONS

191 Given: f(x) = 2×2 + x − 3 and g(x) = x − 1Express f(x) •g(x) − [f(x) + g(x)] as a polynomial in standard form.

192 If g(c) = 1 − c 2 and m(c) = c + 1, then which statement is not true?1 g(c) ⋅m(c) = 1+ c − c2 − c3

2 g(c) +m(c) = 2+ c − c 2

3 m(c) − g(c) = c + c2

4m(c)g(c) = −1

1 − c

193 If p(x) = ab x and r(x) = cdx , then p(x) • r(x) equals1 ac(b + d)x

2 ac(b + d)2x

3 ac(bd)x

4 ac(bd)x2

194 A manufacturing company has developed a cost model, C(x) = 0.15×3 + 0.01x 2 + 2x + 120, where x is the number of items sold, in thousands. The sales price can be modeled by S(x) = 30− 0.01x. Therefore, revenue is modeled by R(x) = x •S(x). The company’s profit, P(x) = R(x) −C(x), could be modeled by1 0.15×3 + 0.02x 2 − 28x + 1202 −0.15×3 − 0.02×2 + 28x − 1203 −0.15x 3 + 0.01×2 − 2.01x − 1204 −0.15x 3 + 32x + 120

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F.IF.C.9: COMPARING FUNCTIONS

195 The x-value of which function’s x-intercept is larger, f or h? Justify your answer.

f(x) = log(x − 4)

x h(x)−1 60 41 22 03 −2

196 Which statement regarding the graphs of the functions below is untrue?f(x) = 3sin2x, from -π < x < π

g(x) = (x − 0.5)(x + 4)(x − 2)

h(x) = log2x

j(x) = −|4x − 2|+ 31 f(x) and j(x) have a maximum y-value of

3.3 g(x) and j(x) have the same end

behavior as x → −∞ .2 f(x), h(x), and j(x) have one y-intercept. 4 g(x), h(x), and j(x) have rational zeros.

197 Consider the function h(x) = 2sin(3x) + 1 and the function q represented in the table below.

x q(x)−2 −8−1 00 01 −22 0

Determine which function has the smaller minimum value for the domain [−2,2]. Justify your answer.

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FUNCTIONSF.BF.B.3: EVEN AND ODD FUNCTIONS

198 Functions f, g, and h are given below.

f(x) = sin(2x)

g(x) = f(x) + 1

Which statement is true about functions f, g, and h?1 f(x) and g(x) are odd, h(x) is even.2 f(x) and g(x) are even, h(x) is odd.3 f(x) is odd, g(x) is neither, h(x) is even.4 f(x) is even, g(x) is neither, h(x) is odd.

199 Which equation represents an odd function?1 y = sinx2 y = cos x3 y = (x + 1)3

4 y = e5x

200 Algebraically determine whether the function j(x) = x4 − 3x 2 − 4 is odd, even, or neither.

F.BF.B.4: INVERSE OF FUNCTIONS

201 For the function f x = x − 3 3 + 1, find f −1 x .

202 Given f − 1 (x) = −34 x + 2, which equation

represents f(x)?

1 f(x) = 43 x − 8

3

2 f(x) = −43 x + 8

3

3 f(x) = 34 x − 2

4 f(x) = −34 x + 2

203 What is the inverse of the function y = log3x?1 y = x3

2 y = logx 33 y = 3x

4 x = 3 y

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204 The inverse of the function f(x) = x + 1x − 2 is

1 f − 1 (x) = x + 1x + 2

2 f − 1 (x) = 2x + 1x − 1

3 f − 1 (x) = x + 1x − 2

4 f − 1 (x) = x − 1x + 1

205 What is the inverse of f(x) = −6(x − 2)?

1 f −1 (x) = −2− x6

2 f −1 (x) = 2 − x6

3 f −l (x) = 1−6(x − 2)

4 f −1 (x) = 6(x + 2)

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SEQUENCES AND SERIESF.LE.A.2, F.IF.A.3, F.BF.A.2: SEQUENCES

206 Elaina has decided to run the Buffalo half-marathon in May. She researched training plans on the Internet and is looking at two possible plans: Jillian’s 12-week plan and Josh’s 14-week plan. The number of miles run per week for each plan is plotted below.

Which one of the plans follows an arithmetic pattern? Explain how you arrived at your answer. Write a recursive definition to represent the number of miles run each week for the duration of the plan you chose. Jillian’s plan has an alternative if Elaina wanted to train instead for a full 26-mile marathon. Week one would start at 13 miles and follow the same pattern for the half-marathon, but it would continue for 14 weeks. Write an explicit formula, in simplest form, to represent the number of miles run each week for the full-marathon training plan.

207 The sequence a 1 = 6, a n = 3an − 1 can also be written as1 an = 6 ⋅ 3n

2 an = 6 ⋅ 3n + 1

3 an = 2 ⋅ 3n

4 an = 2 ⋅ 3n + 1

208 Given f(9) = −2, which function can be used to generate the sequence −8,−7.25,−6.5,−5.75,. . .?1 f(n) = −8 + 0.75n2 f(n) = −8 − 0.75(n − 1)3 f(n) = −8.75+ 0.75n4 f(n) = −0.75 + 8(n − 1)

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209 While experimenting with her calculator, Candy creates the sequence 4, 9, 19, 39, 79, …. Write a recursive formula for Candy’s sequence. Determine the eighth term in Candy’s sequence.

210 Simon lost his library card and has an overdue library book. When the book was 5 days late, he owed $2.25 to replace his library card and pay the fine for the overdue book. When the book was 21 days late, he owed $6.25 to replace his library card and pay the fine for the overdue book. Suppose the total amount Simon owes when the book is n days late can be determined by an arithmetic sequence. Determine a formula for a n , the nth term of this sequence. Use the formula to determine the amount of money, in dollars, Simon needs to pay when the book is 60 days late.

211 The eighth and tenth terms of a sequence are 64 and 100. If the sequence is either arithmetic or geometric, the ninth term can not be1 -822 -803 804 82

212 Write an explicit formula for a n , the nth term of the recursively defined sequence below.

a 1 = x + 1

an = x(an − 1 )For what values of x would an = 0 when n > 1?

213 The formula below can be used to model which scenario?

a 1 = 3000

an = 0.80an − 1

1 The first row of a stadium has 3000 seats, and each row thereafter has 80 more seats than the row in front of it.

2 The last row of a stadium has 3000 seats, and each row before it has 80 fewer seats than the row behind it.

3 A bank account starts with a deposit of $3000, and each year it grows by 80%.

4 The initial value of a specialty toy is $3000, and its value each of the following years is 20% less.

214 The population of Jamesburg for the years 2010-2013, respectively, was reported as follows:250,000 250,937 251,878 252,822How can this sequence be recursively modeled?1 j n = 250,000(1.00375)n − 1

2 j n = 250,000+ 937(n − 1)

3 j 1 = 250,000

j n = 1.00375j n − 1

4 j 1 = 250,000

j n = j n − 1 + 937

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215 A recursive formula for the sequence 18,9,4.5,. . . is1 g 1 = 18

gn =12 gn − 1

2 gn = 18 12

n − 1

3 g 1 = 18

gn = 2gn − 1

4 gn = 18 2 n − 1

216 In 2010, the population of New York State was approximately 19,378,000 with an annual growth rate of 1.5%. Assuming the growth rate is maintained for a large number of years, which equation can be used to predict the population of New York State t years after 2010?1 Pt = 19,378,000(1.5) t

2 P0 = 19,378,000

Pt = 19,378,000+ 1.015Pt − 1

3 Pt = 19,378,000(1.015) t − 1

4 P0 = 19,378,000

Pt = 1.015Pt − 1

217 The Rickerts decided to set up an account for their daughter to pay for her college education. The day their daughter was born, they deposited $1000 in an account that pays 1.8% compounded annually. Beginning with her first birthday, they deposit an additional $750 into the account on each of her birthdays. Which expression correctly represents the amount of money in the account n years after their daughter was born?1 an = 1000(1.018)n + 7502 an = 1000(1.018)n + 750n3 a 0 = 1000

an = an − 1 (1.018) + 7504 a 0 = 1000

an = an − 1 (1.018) + 750n

218 At her job, Pat earns $25,000 the first year and receives a raise of $1000 each year. The explicit formula for the nth term of this sequence is an = 25,000 + (n − 1)1000. Which rule best represents the equivalent recursive formula?1 an = 24,000+ 1000n2 an = 25,000+ 1000n3 a 1 = 25,000, an = an − 1 + 10004 a 1 = 25,000, an = an + 1 + 1000

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F.BF.B.6: SIGMA NOTATION

219 Kristin wants to increase her running endurance. According to experts, a gradual mileage increase of 10% per week can reduce the risk of injury. If Kristin runs 8 miles in week one, which expression can help her find the total number of miles she will have run over the course of her 6-week training program?

1 8(1.10)n − 1

n = 1

6

2 8(1.10)n

n = 1

6

38− 8(1.10)6

0.90

48− 8(0.10)n

1.10

A.SSE.B.4: SERIES

220 Using the formula below, determine the monthly payment on a 5-year car loan with a monthly percentage rate of 0.625% for a car with an original cost of $21,000 and a $1000 down payment, to the nearest cent.

Pn = PMT1 − (1+ i)−n

i

Pn = present amount borrowedn = number of monthly pay periodsPMT = monthly paymenti = interest rate per monthThe affordable monthly payment is $300 for the same time period. Determine an appropriate down payment, to the nearest dollar.

221 Alexa earns $33,000 in her first year of teaching and earns a 4% increase in each successive year. Write a geometric series formula, S n , for Alexa’s total earnings over n years. Use this formula to find Alexa’s total earnings for her first 15 years of teaching, to the nearest cent.

222 Jasmine decides to put $100 in a savings account each month. The account pays 3% annual interest, compounded monthly. How much money, S, will Jasmine have after one year?1 S = 100(1.03)12

2 S =100− 100(1.0025)12

1 − 1.00253 S = 100(1.0025)12

4 S =100 − 100(1.03)12

1− 1.03

223 Jim is looking to buy a vacation home for $172,600 near his favorite southern beach. The formula to compute a mortgage payment, M, is

M = P •r(1 + r) N

(1+ r) N − 1 where P is the principal

amount of the loan, r is the monthly interest rate, and N is the number of monthly payments. Jim’s bank offers a monthly interest rate of 0.305% for a 15-year mortgage. With no down payment, determine Jim’s mortgage payment, rounded to the nearest dollar. Algebraically determine and state the down payment, rounded to the nearest dollar, that Jim needs to make in order for his mortgage payment to be $1100.

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Algebra II Regents Exam Questions by Common Core State Standard: Topic

224 A ball is dropped from a height of 32 feet. It bounces and rebounds 80% of the height from which it was falling. What is the total downward distance, in feet, the ball traveled up to the 12th bounce?1 292 583 1204 149

225 Brian deposited 1 cent into an empty non-interest bearing bank account on the first day of the month. He then additionally deposited 3 cents on the second day, 9 cents on the third day, and 27 cents on the fourth day. What would be the total amount of money in the account at the end of the 20th day if the pattern continued?1 $11,622,614.672 $17,433,922.003 $116,226,146.804 $1,743,392,200.00

TRIGONOMETRYF.TF.A.1-2: UNIT CIRCLE

226 Which diagram shows an angle rotation of 1 radian on the unit circle?

1

2

3

4

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227 The terminal side of θ , an angle in standard

position, intersects the unit circle at P −13 ,−

83

.

What is the value of sec θ?1 −3

2 −3 8

8

3 −13

4 −8

3

F.TF.A.2: RECIPROCAL TRIGONOMETRIC FUNCTIONS

228 Using the unit circle below, explain why csc θ = 1y .

F.TF.A.2: REFERENCE ANGLES

229 Which diagram represents an angle, α , measuring 13π20 radians drawn in standard position, and its

reference angle, θ ?

1

2

3

4

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F.TF.A.2, F.TF.C.8: DETERMINING TRIGONOMETRIC FUNCTIONS

230 If the terminal side of angle , in standard position, passes through point (−4,3), what is the numerical value of sinθ ?

1 35

2 45

3 −35

4 −45

231 A circle centered at the origin has a radius of 10 units. The terminal side of an angle, θ , intercepts the circle in Quadrant II at point C. The y-coordinate of point C is 8. What is the value of cosθ ?

1 −35

2 −34

3 35

4 45

232 The hours of daylight, y, in Utica in days, x, from January 1, 2013 can be modeled by the equation y = 3.06sin(0.017x − 1.40) + 12.23. How many hours of daylight, to the nearest tenth, does this model predict for February 14, 2013?1 9.42 10.43 12.14 12.2

233 Using the identity sin2θ + cos2θ = 1, find the value of tanθ , to the nearest hundredth, if cosθ is –0.7 and θ is in Quadrant II.

234 Given that sin2θ + cos2θ = 1 and sinθ = − 25 ,

what is a possible value of cosθ ?

15+ 2

5

2235

33 3

5

4355

F.TF.C.8: SIMPLIFYING TRIGONOMETRIC IDENTITIES

235 If sin2 (32°) + cos2 (M) = 1, then M equals1 32°2 58°3 68°4 72°

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F.TF.B.5: MODELING TRIGONOMETRIC FUNCTIONS

236 The voltage used by most households can be modeled by a sine function. The maximum voltage is 120 volts, and there are 60 cycles every second. Which equation best represents the value of the voltage as it flows through the electric wires, where t is time in seconds?1 V = 120sin(t)2 V = 120sin(60t)3 V = 120sin(60πt)4 V = 120sin(120πt)

F.IF.B.4, F.IF.C.7: GRAPHING TRIGONOMETRIC FUNCTIONS

237 The Ferris wheel at the landmark Navy Pier in Chicago takes 7 minutes to make one full rotation. The height, H, in feet, above the ground of one of the six-person cars can be modeled by

H(t) = 70sin 2π7 t − 1.75

+ 80, where t is time,

in minutes. Using H(t) for one full rotation, this car’s minimum height, in feet, is1 1502 703 104 0

238 A sine function increasing through the origin can be used to model light waves. Violet light has a wavelength of 400 nanometers. Over which interval is the height of the wave decreasing, only?1 (0,200)2 (100,300)3 (200,400)4 (300,400)

239 Based on climate data that have been collected in Bar Harbor, Maine, the average monthly temperature, in degrees F, can be modeled by the equation B(x) = 23.914sin(0.508x − 2.116) + 55.300. The same governmental agency collected average monthly temperature data for Phoenix, Arizona, and found the temperatures could be modeled by the equation P(x) = 20.238sin(0.525x − 2.148) + 86.729. Which statement can not be concluded based on the average monthly temperature models x months after starting data collection?1 The average monthly temperature variation is

more in Bar Harbor than in Phoenix.2 The midline average monthly temperature for

Bar Harbor is lower than the midline temperature for Phoenix.

3 The maximum average monthly temperature for Bar Harbor is 79° F, to the nearest degree.

4 The minimum average monthly temperature for Phoenix is 20° F, to the nearest degree.

240 Relative to the graph of y = 3sinx , what is the shift

of the graph of y = 3sin x + π3

?

1 π3 right

2 π3 left

3 π3 up

4 π3 down

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241 Given the parent function p(x) = cos x, which phrase best describes the transformation used to obtain the graph of g(x) = cos(x + a) − b , if a and b are positive constants?1 right a units, up b units2 right a units, down b units3 left a units, up b units4 left a units, down b units

242 As x increases from 0 to π2 , the graph of the

equation y = 2tanx will1 increase from 0 to 22 decrease from 0 to −23 increase without limit4 decrease without limit

243 Which statement is incorrect for the graph of the

function y = −3cos π3 x − 4

+ 7?

1 The period is 6.2 The amplitude is 3.3 The range is [4,10].4 The midline is y = −4.

244 The volume of air in a person’s lungs, as the person breathes in and out, can be modeled by a sine graph. A scientist is studying the differences in this volume for people at rest compared to people told to take a deep breath. When examining the graphs, should the scientist focus on the amplitude, period, or midline? Explain your choice.

245 On the axes below, graph one cycle of a cosine

function with amplitude 3, period π2 , midline

y = −1, and passing through the point (0,2).

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246 Which graph represents a cosine function with no horizontal shift, an amplitude of 2, and a period of 2π3 ?

1

2

3

4

247 Which equation is represented by the graph shown below?

1 y = 12 cos 2x

2 y = cos x

3 y = 12 cos x

4 y = 2cos 12 x

248 The ocean tides near Carter Beach follow a repeating pattern over time, with the amount of time between each low and high tide remaining relatively constant. On a certain day, low tide occurred at 8:30 a.m. and high tide occurred at 3:00 p.m. At high tide, the water level was 12 inches above the average local sea level; at low tide it was 12 inches below the average local sea level. Assume that high tide and low tide are the maximum and minimum water levels each day, respectively. Write a cosine function of the form f(t) = A cos(Bt), where A and B are real numbers, that models the water level, f(t), in inches above or below the average Carter Beach sea level, as a function of the time measured in t hours since 8:30 a.m. On the grid below, graph one cycle of this function.

People who fish in Carter Beach know that a certain species of fish is most plentiful when the water level is increasing. Explain whether you would recommend fishing for this species at 7:30 p.m. or 10:30 p.m. using evidence from the given context.

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249 The graph below represents the height above the ground, h, in inches, of a point on a triathlete’s bike wheel during a training ride in terms of time, t, in seconds.

Identify the period of the graph and describe what the period represents in this context.

250 Which sinusoid has the greatest amplitude?

12 y = 3sin(θ − 3) + 5

34 y = −5sin(θ − 1) − 3

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251 The resting blood pressure of an adult patient can be modeled by the function P below, where P(t) is the pressure in millimeters of mercury after time t in seconds.

P(t) = 24cos(3πt) + 120On the set of axes below, graph y = P(t) over the domain 0 ≤ t ≤ 2.

Determine the period of P. Explain what this value represents in the given context. Normal resting blood pressure for an adult is 120 over 80. This means that the blood pressure oscillates between a maximum of 120 and a minimum of 80. Adults with high blood pressure (above 140 over 90) and adults with low blood pressure (below 90 over 60) may be at risk for health disorders. Classify the given patient’s blood pressure as low, normal, or high and explain your reasoning.

252 a) On the axes below, sketch at least one cycle of a sine curve with an amplitude of 2, a midline at

y = −32 , and a period of 2π .

b) Explain any differences between a sketch of

y = 2sin x − π3

−32 and the sketch from part a.

CONICSG.GPE.A.1: EQUATIONS OF CIRCLES

253 The equation 4×2 − 24x + 4y 2 + 72y = 76 is equivalent to1 4(x − 3)2 + 4(y + 9)2 = 762 4(x − 3)2 + 4(y + 9)2 = 1213 4(x − 3)2 + 4(y + 9)2 = 1664 4(x − 3)2 + 4(y + 9)2 = 436

ID: A

1

Algebra II Regents Exam Questions by Common Core State Standard: TopicAnswer Section

1 ANS: 3 PTS: 2 REF: 061607aii NAT: S.IC.A.2TOP: Analysis of Data

2 ANS: 3 PTS: 2 REF: 061710aii NAT: S.IC.A.2TOP: Analysis of Data

3 ANS: 2 PTS: 2 REF: 011820aii NAT: S.IC.A.2TOP: Analysis of Data

4 ANS: sample: pails of oranges; population: truckload of oranges. It is likely that about 5% of all the oranges are unsatisfactory.

PTS: 2 REF: 011726aii NAT: S.IC.A.2 TOP: Analysis of Data 5 ANS:

Since there are six flavors, each flavor can be assigned a number, 1-6. Use the simulation to see the number of times the same number is rolled 4 times in a row.

PTS: 2 REF: 081728aii NAT: S.IC.A.2 TOP: Analysis of Data 6 ANS:

138.905± 2 ⋅ 7.95 = 123− 155. No, since 125 (50% of 250) falls within the 95% interval.

PTS: 4 REF: 011835aii NAT: S.IC.A.2 TOP: Analysis of Data 7 ANS:

Randomly assign participants to two groups. One group uses the toothpaste with ingredient X and the other group uses the toothpaste without ingredient X.

PTS: 2 REF: 061626aii NAT: S.IC.B.3 TOP: Analysis of DataKEY: type

8 ANS: 1II. Ninth graders drive to school less often; III.Students know little about adults; IV. Calculus students love math!

PTS: 2 REF: 081602aii NAT: S.IC.B.3 TOP: Analysis of DataKEY: bias

9 ANS: 3 PTS: 2 REF: 011706aii NAT: S.IC.B.3TOP: Analysis of Data KEY: type

10 ANS: 3Self selection causes bias.

PTS: 2 REF: 061703aii NAT: S.IC.B.3 TOP: Analysis of DataKEY: bias

11 ANS: 2 PTS: 2 REF: 081717aii NAT: S.IC.B.3TOP: Analysis of Data KEY: type

12 ANS: 4 PTS: 2 REF: 011801aii NAT: S.IC.B.3TOP: Analysis of Data KEY: bias

ID: A

2

13 ANS: 2

ME = zp(1− p)

n

= 1.96

(0.55)(0.45)900

≈ 0.03

PTS: 2 REF: 081612aii NAT: S.IC.B.4 TOP: Analysis of Data 14 ANS:

Yes. The margin of error from this simulation indicates that 95% of the observations fall within ± 0.12 of the simulated proportion, 0.25. The margin of error can be estimated by multiplying the standard deviation, shown to

be 0.06 in the dotplot, by 2, or applying the estimated standard error formula, p(1 − p)

n

or

(0.25)(0.75)50

and multiplying by 2. The interval 0.25 ± 0.12 includes plausible values for the true proportion of people who prefer Stephen’s new product. The company has evidence that the population proportion could be at least 25%. As seen in the dotplot, it can be expected to obtain a sample proportion of 0.18 (9 out of 50) or less several times, even when the population proportion is 0.25, due to sampling variability. Given this information, the results of the survey do not provide enough evidence to suggest that the true proportion is not at least 0.25, so the development of the product should continue at this time.

PTS: 4 REF: spr1512aii NAT: S.IC.B.4 TOP: Analysis of Data 15 ANS: 2

ME = zp(1− p)

n

= 1.96

(0.16)(0.84)1334

≈ 0.02

PTS: 2 REF: 081716aii NAT: S.IC.B.4 TOP: Analysis of Data 16 ANS:

The mean difference between the students’ final grades in group 1 and group 2 is –3.64. This value indicates that students who met with a tutor had a mean final grade of 3.64 points less than students who used an on-line subscription. One can infer whether this difference is due to the differences in intervention or due to which students were assigned to each group by using a simulation to rerandomize the students’ final grades many (500) times. If the observed difference –3.64 is the result of the assignment of students to groups alone, then a difference of –3.64 or less should be observed fairly regularly in the simulation output. However, a difference of –3 or less occurs in only about 2% of the rerandomizations. Therefore, it is quite unlikely that the assignment to groups alone accounts for the difference; rather, it is likely that the difference between the interventions themselves accounts for the difference between the two groups’ mean final grades.

PTS: 4 REF: fall1514aii NAT: S.IC.B.5 TOP: Analysis of Data 17 ANS:

0.602 ± 2 ⋅ 0.066 = 0.47− 0.73. Since 0.50 falls within the 95% interval, this supports the concern there may be an even split.

PTS: 4 REF: 061635aii NAT: S.IC.B.5 TOP: Analysis of Data

ID: A

3

18 ANS: Some of the students who did not drink energy drinks read faster than those who did drink energy drinks.

17.7− 19.1 = −1.4 Differences of -1.4 and less occur 25232 or about 10% of the time, so the difference is not

unusual.

PTS: 4 REF: 081636aii NAT: S.IC.B.5 TOP: Analysis of Data 19 ANS: 2 PTS: 2 REF: 011709aii NAT: S.IC.B.5

TOP: Analysis of Data 20 ANS:

0.506 ± 2 ⋅ 0.078 = 0.35− 0.66. The 32.5% value falls below the 95% confidence level.

PTS: 4 REF: 061736aii NAT: S.IC.B.5 TOP: Analysis of Data 21 ANS:

Using a 95% level of confidence, x ± 2 standard deviations sets the usual wait time as 150-302 seconds. 360 seconds is unusual.

PTS: 2 REF: 081629aii NAT: S.IC.B.6 TOP: Analysis of Data 22 ANS: 1 PTS: 2 REF: 081722aii NAT: S.IC.B.6

TOP: Analysis of Data 23 ANS:

y = 4.168(3.981)x . 100 = 4.168(3.981)x

log 1004.168 = log(3.981)x

log 1004.168 = x log(3.981)

log 1004.168

log(3.981) = x

x ≈ 2.25

PTS: 4 REF: 081736aii NAT: S.ID.B.6 TOP: RegressionKEY: exponential AII

24 ANS: D = 1.223(2.652) A

PTS: 2 REF: 011826aii NAT: S.ID.B.6 TOP: RegressionKEY: exponential AII

25 ANS: 3The pattern suggests an exponential pattern, not linear or sinusoidal. A 4% growth rate is accurate, while a 43% growth rate is not.

PTS: 2 REF: 011713aii NAT: S.ID.B.6 TOP: RegressionKEY: choose model

ID: A

4

26 ANS:

normcdf(510, 540, 480, 24) = 0.0994 z = 510− 48024 = 1.25

z = 540− 48024 = 2.5

1.25 = x − 51020

x = 535

2.5 = x − 51020

x = 560

535-560

PTS: 4 REF: fall1516aii NAT: S.ID.A.4 TOP: Normal DistributionsKEY: probability

27 ANS: 2

x + 2σ represents approximately 48% of the data.

PTS: 2 REF: 061609aii NAT: S.ID.A.4 TOP: Normal DistributionsKEY: percent

28 ANS: 3

PTS: 2 REF: 081604aii NAT: S.ID.A.4 TOP: Normal DistributionsKEY: probability

29 ANS: 4496± 2(115)

PTS: 2 REF: 011718aii NAT: S.ID.A.4 TOP: Normal DistributionsKEY: interval

30 ANS:

69

PTS: 2 REF: 061726aii NAT: S.ID.A.4 TOP: Normal DistributionsKEY: percent

ID: A

5

31 ANS: 1

PTS: 2 REF: 081711aii NAT: S.ID.A.4 TOP: Normal DistributionsKEY: percent

32 ANS: 3440× 2.3% ≈ 10

PTS: 2 REF: 011807aii NAT: S.ID.A.4 TOP: Normal DistributionsKEY: predict

33 ANS:

This scenario can be modeled with a Venn Diagram: Since P(S ∪ I)c = 0.2, P(S ∪ I) = 0.8. Then, P(S ∩ I) = P(S) +P(I) − P(S ∪ I)

= 0.5 + 0.7 − 0.8

= 0.4

If S and I are independent, then the

Product Rule must be satisfied. However, (0.5)(0.7) ≠ 0.4. Therefore, salary and insurance have not been treated independently.

PTS: 4 REF: spr1513aii NAT: S.CP.A.2 TOP: Theoretical Probability 34 ANS:

P(A∪B) = P(A) + P(B) −P(A∩B)

0.8 = 0.6+ 0.5 − P(A∩B)

P(A∩B) = 0.3

A and B are independent since P(A∩B) = P(A) ⋅P(B)

0.3 = 0.6 ⋅ 0.5

0.3 = 0.3

PTS: 2 REF: 081632aii NAT: S.CP.A.2 TOP: Theoretical Probability 35 ANS:

P(S ∩M) = P(S) + P(M) − P(S ∪M) = 6491376 + 433

1376 − 9741376 = 108

1376

PTS: 2 REF: 061629aii NAT: S.CP.B.7 TOP: Theoretical Probability

ID: A

6

36 ANS: 2The events are independent because P(A and B) = P(A) ⋅P(B)

0.125 = 0.5 ⋅ 0.25

.

If P(A or B) = P(A) +P(B) − P(A and B) = 0.25 + 0.5 −.125 = 0.625, then the events are not mutually exclusive because P(A or B) = P(A) + P(B)

0.625 ≠ 0.5 + 0.25

PTS: 2 REF: 061714aii NAT: S.CP.B.7 TOP: Theoretical Probability 37 ANS: 1

The probability of rain equals the probability of rain, given that Sean pitches.

PTS: 2 REF: 061611aii NAT: S.CP.A.3 TOP: Conditional Probability 38 ANS:

47108 = 1

4 + 116459 − P(M and J)

P(M and J) = 31459

; No, because 31459 ≠ 1

4 ⋅ 116459

PTS: 4 REF: 011834aii NAT: S.CP.A.3 TOP: Conditional Probability 39 ANS: 1

15725 + 47 + 157

PTS: 2 REF: 081607aii NAT: S.CP.A.4 TOP: Conditional Probability 40 ANS:

Based on these data, the two events do not appear to be independent. P(F) = 106200 = 0.53, while

P(F | T) = 5490 = 0.6, P(F | R) = 25

65 = 0.39, and P(F | C) = 2745 = 0.6. The probability of being female are not the

same as the conditional probabilities. This suggests that the events are not independent.

PTS: 2 REF: fall1508aii NAT: S.CP.A.4 TOP: Conditional Probability 41 ANS:

No, because P(M / R) ≠ P(M)

70180 ≠ 230

490

0.38 ≠ 0.47

PTS: 2 REF: 011731aii NAT: S.CP.A.4 TOP: Conditional Probability 42 ANS:

A student is more likely to jog if both siblings jog. 1 jogs: 4162239 ≈ 0.19. both jog: 400

1780 ≈ 0.22

PTS: 2 REF: 061732aii NAT: S.CP.A.4 TOP: Conditional Probability

ID: A

7

43 ANS:

P(P / K) =P(P^K)

P(K) = 1.92.3 ≈ 82.6% A key club member has an 82.6% probability of being enrolled in AP Physics.

PTS: 4 REF: 011735aii NAT: S.CP.B.6 TOP: Conditional Probability 44 ANS:

P(W / D) =P(W^D)

P(D) = .4.5 ≈.8

PTS: 2 REF: 081726aii NAT: S.CP.B.6 TOP: Conditional Probability 45 ANS:

f(4) − f(−2)4− −2 = 80 − 1.25

6 = 13.125

g(4) − g(−2)4− −2 = 179 − −49

6 = 38

g(x) has a greater rate of change

PTS: 4 REF: 061636aii NAT: F.IF.B.6 TOP: Rate of ChangeKEY: AII

46 ANS: 4

(1) B(60) −B(10)

60 − 10 ≈ 28% (2) B(69) −B(19)

69 − 19 ≈ 33% (3) B(72) −B(36)

72 − 36 ≈ 38% (4) B(73) −B(60)

73 − 60 ≈ 46%

PTS: 2 REF: 011721aii NAT: F.IF.B.6 TOP: Rate of ChangeKEY: AII

47 ANS: 156.25− 56.25

70− 50 = 15020 = 7.5 Between 50-70 mph, each additional mph in speed requires 7.5 more feet to stop.

PTS: 2 REF: 081631aii NAT: F.IF.B.6 TOP: Rate of ChangeKEY: AII

48 ANS: 1

(1) 9− 02− 1 = 9 (2) 17− 0

3.5 − 1 = 6.8 (3) 0− 05− 1 = 0 (4) 17 − −5

3.5− 1 ≈ 6.3

PTS: 2 REF: 011724aii NAT: F.IF.B.6 TOP: Rate of ChangeKEY: AII

49 ANS: 3

f(7) − f(−7)7− −7 =

= 2−0.25(7) • sin π2 (7)

− 2−0.25(−7) • sin π2 (−7)

14 ≈ −0.26

PTS: 2 REF: 061721aii NAT: F.IF.B.6 TOP: Rate of ChangeKEY: AII

ID: A

8

50 ANS: 3

log0.8V

17000

= t

0.8t = V17000

V = 17000(0.8) t

17,000(0.8)3 − 17,000(0.8)1

3− 1 ≈ −2450

PTS: 2 REF: 081709aii NAT: F.IF.B.6 TOP: Rate of ChangeKEY: AII

51 ANS: 3

−2 −12 x 2 = −6x + 20

x 2 − 12x = −40

x2 − 12x + 36 = −40 + 36

(x − 6)2 = −4

x − 6 = ±2i

x = 6± 2i

PTS: 2 REF: fall1504aii NAT: A.REI.B.4 TOP: Solving QuadraticsKEY: complex solutions | completing the square

52 ANS: 1

x =−3 ± 32 − 4(2)(2)

2(2) =−3± −7

4 = −34 ±

i 74

PTS: 2 REF: 061612aii NAT: A.REI.B.4 TOP: Solving QuadraticsKEY: complex solutions | quadratic formula

53 ANS: 4

x =8 ± (−8)2 − 4(6)(29)

2(6) =8± −632

12 =8 ± i 4 158

12 = 23 ± 1

6 i 158

PTS: 2 REF: 011711aii NAT: A.REI.B.4 TOP: Solving QuadraticsKEY: complex solutions | quadratic formula

ID: A

9

54 ANS: 3×2 + 2x + 1 = −5+ 1

(x + 1)2 = −4

x + 1 = ±2i

x = −1± 2i

PTS: 2 REF: 081703aii NAT: A.REI.B.4 TOP: Solving QuadraticsKEY: complex solutions | completing the square

55 ANS: 44×2 = −98

x2 = −984

x2 = −492

x = ± −492 = ± 7i

2⋅

22= ±

7i 22

PTS: 2 REF: 061707aii NAT: A.REI.B.4 TOP: Solving QuadraticsKEY: complex solutions | taking square roots

56 ANS: 4If 1− i is one solution, the other is 1+ i. (x − (1− i))(x − (1 + i)) = 0

x2 − x − ix − x + ix + (1− i2 ) = 0

x 2 − 2x + 2 = 0

PTS: 2 REF: 081601aii NAT: A.REI.B.4 TOP: Complex Conjugate Root Theorem

ID: A

10

57 ANS: 4

A parabola with a focus of (0,4) and a directrix of y = 2 is sketched as follows: By inspection, it is determined that the vertex of the parabola is (0,3). It is also evident that the distance, p, between the vertex and the focus is 1. It is possible to use the formula (x − h)2 = 4p(y − k) to derive the equation of the parabola as follows: (x − 0)2 = 4(1)(y − 3)

x 2 = 4y − 12

x2 + 12 = 4y

x 2

4 + 3 = y

or A point (x,y) on the parabola must be the same distance from the focus as it is from the directrix. For any such

point (x,y), the distance to the focus is (x − 0)2 + (y − 4)2 and the distance to the directrix is y − 2. Setting this equal leads to: x2 + y2 − 8y + 16 = y 2 − 4y + 4

x2 + 16 = 4y + 4

x 2

4 + 3 = y

PTS: 2 REF: spr1502aii NAT: G.GPE.A.2 TOP: Graphing Quadratic Functions 58 ANS:

The vertex of the parabola is (4,−3). The x-coordinate of the focus and the vertex is the same. Since the distance from the vertex to the directrix is 3, the distance from the vertex to the focus is 3, so the y-coordinate of the focus is 0. The coordinates of the focus are (4,0).

PTS: 2 REF: 061630aii NAT: G.GPE.A.2 TOP: Graphing Quadratic Functions 59 ANS: 4

The vertex is (2,−1) and p = 2. y = − 14(2) (x − 2)2 − 1

PTS: 2 REF: 081619aii NAT: G.GPE.A.2 TOP: Graphing Quadratic Functions

ID: A

11

60 ANS: 4

The vertex is (1,0) and p = 2. y = 14(2) (x − 1)2 + 0

PTS: 2 REF: 061717aii NAT: G.GPE.A.2 TOP: Graphing Quadratic Functions 61 ANS: 2

The vertex of the parabola is (0,0). The distance, p, between the vertex and the focus or the vertex and the

directrix is 1. y = −14p (x − h)2 + k

y = −14(1) (x − 0)2 + 0

y = −14 x2

PTS: 2 REF: 081706aii NAT: G.GPE.A.2 TOP: Graphing Quadratic Functions 62 ANS: 1

In vertex form, the parabola is y = − 14(2) (x + 4)2 + 3. The vertex is (−4,3) and p = 2. 3 + 2 = 5

PTS: 2 REF: 011816aii NAT: G.GPE.A.2 TOP: Graphing Quadratic Functions 63 ANS:

6x − 3y + 2z = −10

−2x + 3y + 8z = 72

4x + 10z = 62

x + 3y + 5z = 45

6x − 3y + 2z = −10

7x + 7z = 35

4x + 4z = 20

4x + 10z = 62

4x + 4z = 20

6z = 42

z = 7

4x + 4(7) = 20

4x = −8

x = −2

6(−2) − 3y + 2(7) = −10

−3y = −12

y = 4

PTS: 4 REF: spr1510aii NAT: A.REI.C.6 TOP: Solving Linear SystemsKEY: three variables

ID: A

12

64 ANS: 2Combining (1) and (3): −6c = −18

c = 3

Combining (1) and (2): 5a + 3c = −1

5a + 3(3) = −1

5a = −10

a = −2

Using (3): −(−2) − 5b − 5(3) = 2

2 − 5b − 15 = 2

b = −3

PTS: 2 REF: 081623aii NAT: A.REI.C.6 TOP: Solving Linear SystemsKEY: three variables

65 ANS: x + y + z = 1

−x + 3y − 5z = 11

4y − 4z = 12

y − z = 3

2x + 2y + 2z = 2

2x + 4y + 6z = 2

2y + 4z = 0

y + 2z = 0

y = −2z

−2z − z = 3

−3z = 3

z = −1

y − (−1) = 3

y = 2

x + 2 − 1 = 1

x = 0

PTS: 4 REF: 061733aii NAT: A.REI.C.6 TOP: Solving Linear SystemsKEY: three variables

66 ANS: 43x − (−2x + 14) = 16

5x = 30

x = 6

3(6) − 4z = 2

−4z = −16

z = 4

PTS: 2 REF: 011803aii NAT: A.REI.C.6 TOP: Solving Linear SystemsKEY: three variables

67 ANS:

−2x + 1 = −2×2 + 3x + 1

2x 2 − 5x = 0

x(2x − 5) = 0

x = 0, 52

PTS: 2 REF: fall1507aii NAT: A.REI.C.7 TOP: Quadratic-Linear SystemsKEY: AII

ID: A

13

68 ANS:

y = −x + 5

(x − 3)2 + (−x + 5 + 2)2 = 16

x 2 − 6x + 9 + x2 − 14x + 49 = 16

2×2 − 20x + 42 = 0

x2 − 10x + 21 = 0

(x − 7)(x − 3) = 0

x = 7,3

y = −7 + 5 = −2

y = −3 + 5 = 2

PTS: 4 REF: 061633aii NAT: A.REI.C.7 TOP: Quadratic-Linear SystemsKEY: AII

69 ANS: 4y = g(x) = (x − 2)2 (x − 2)2 = 3x − 2

x 2 − 4x + 4 = 3x − 2

x2 − 7x + 6 = 0

(x − 6)(x − 1) = 0

x = 6,1

y = 3(6) − 2 = 16

y = 3(1) − 2 = 1

PTS: 2 REF: 011705aii NAT: A.REI.C.7 TOP: Quadratic-Linear SystemsKEY: AII

70 ANS: 1(x + 3)2 + (2x − 4)2 = 8

x2 + 6x + 9 + 4×2 − 16x + 16 = 8

5x 2 − 10x + 17 = 0

b 2 − 4ac

100− 4(5)(17) < 0

PTS: 2 REF: 081719aii NAT: A.REI.C.7 TOP: Quadratic-Linear SystemsKEY: AII

ID: A

14

71 ANS: 3−33t2 + 360t = 700+ 5t

−33t2 + 355t − 700 = 0

t =−355 ± 3552 − 4(−33)(−700)

2(−33) ≈ 3,8

PTS: 2 REF: 081606aii NAT: A.REI.D.11 TOP: Quadratic-Linear SystemsKEY: AII

72 ANS:

PTS: 2 REF: fall1510aii NAT: A.REI.D.11 TOP: Other SystemsKEY: AII

73 ANS: 4

PTS: 2 REF: 061622aii NAT: A.REI.D.11 TOP: Other SystemsKEY: AII

ID: A

15

74 ANS:

A(t) = 800e−0.347t

B(t) = 400e−0.231t

800e−0.347t = 400e−0.231t

ln2e−0.347t = lne−0.231t

ln2 + lne−0.347t = lne−0.231t

ln2 − 0.347t = −0.231t

ln2 = 0.116t

6 ≈ t

0.15 = e−0.347t

ln0.15 = lne−0.347t

ln0.15 = −0.347t ⋅ lne

5.5 ≈ t

PTS: 6 REF: 061637aii NAT: A.REI.D.11 TOP: Other SystemsKEY: AII

75 ANS: 2

PTS: 2 REF: 081603aii NAT: A.REI.D.11 TOP: Other SystemsKEY: AII

ID: A

16

76 ANS: 2

PTS: 2 REF: 011712aii NAT: A.REI.D.11 TOP: Other SystemsKEY: AII

77 ANS: 2

PTS: 2 REF: 011716aii NAT: A.REI.D.11 TOP: Other SystemsKEY: AII

78 ANS: 2

PTS: 2 REF: 061705aii NAT: A.REI.D.11 TOP: Other SystemsKEY: AII

79 ANS:

At 1.95 years, the value of the car equals the loan balance. Zach can cancel the policy after 6 years.

PTS: 4 REF: 081737aii NAT: A.REI.D.11 TOP: Other SystemsKEY: AII

ID: A

17

80 ANS: 1 PTS: 2 REF: 011814aii NAT: A.REI.D.11TOP: Other Systems KEY: AII

81 ANS: 20e .05t = 30e .03t

23 e .05t

e .05t = e .03t

e .05t

ln 23 = lne−.02t

ln 23 = −.02t lne

ln 23

−.02 = t

20.3 ≈ t

PTS: 2 REF: 011829aii NAT: A.REI.D.11 TOP: Other SystemsKEY: AII

82 ANS: 2

B(t) = 750 1.16112

12t

≈ 750(1.012)12t B(t) = 750 1 + 0.1612

12t

is wrong, because the growth is an annual rate

that is not compounded monthly.

PTS: 2 REF: spr1504aii NAT: A.SSE.B.3 TOP: Modeling Exponential FunctionsKEY: AII

83 ANS: 3

0.75110 ≈.9716

PTS: 2 REF: 061713aii NAT: A.SSE.B.3 TOP: Modeling Exponential FunctionsKEY: AII

84 ANS: 3

12

173.83

≈ 0.990656

PTS: 2 REF: 081710aii NAT: A.SSE.B.3 TOP: Modeling Exponential FunctionsKEY: AII

85 ANS: 4 PTS: 2 REF: 011808aii NAT: A.SSE.B.3TOP: Modeling Exponential Functions KEY: AII

ID: A

18

86 ANS: 4

y = 5−t = 15

t

PTS: 2 REF: 061615aii NAT: F.IF.C.8 TOP: Modeling Exponential Functions 87 ANS:

ln 12

1590 is negative, so M(t) represents decay.

PTS: 2 REF: 011728aii NAT: F.IF.C.8 TOP: Modeling Exponential Functions 88 ANS: 3

1.0525112 ≈ 1.00427

PTS: 2 REF: 061621aii NAT: F.BF.A.1 TOP: Modeling Exponential FunctionsKEY: AII

89 ANS: 4 PTS: 2 REF: 081622aii NAT: F.BF.A.1TOP: Modeling Exponential Functions KEY: AII

90 ANS: 1AP = ert

0.42 = ert

ln0.42 = lne rt

−0.87 ≈ rt

PTS: 2 REF: 011723aii NAT: F.BF.A.1 TOP: Modeling Exponential FunctionsKEY: AII

91 ANS:

A t = 100 0.5 t

63 , where t is time in years, and A t is the amount of titanium-44 left after t years. A 10 − A 0

10− 0 = 89.58132 − 10010 = −1.041868 The estimated mass at t = 40 is 100− 40(−1.041868) ≈ 58.3. The

actual mass is A 40 = 100 0.5 4063 ≈ 64.3976. The estimated mass is less than the actual mass.

PTS: 6 REF: fall1517aii NAT: F.LE.A.2 TOP: Modeling Exponential FunctionsKEY: AII

ID: A

19

92 ANS: 1

P(28) = 5(2)9828 ≈ 56

PTS: 2 REF: 011702aii NAT: F.LE.A.2 TOP: Modeling Exponential FunctionsKEY: AII

93 ANS: 1The car lost approximately 19% of its value each year.

PTS: 2 REF: 081613aii NAT: F.LE.B.5 TOP: Modeling Exponential Functions 94 ANS: 2

The 2010 population is 110 million.

PTS: 2 REF: 061718aii NAT: F.LE.B.5 TOP: Modeling Exponential Functions 95 ANS: 4 PTS: 2 REF: 011805aii NAT: F.LE.B.5

TOP: Modeling Exponential Functions 96 ANS: 3

d = 10log 6.3× 10−3

1.0 × 10−12 ≈ 98

PTS: 2 REF: 011715aii NAT: F.IF.B.4 TOP: Evaluating Logarithmic Expressions 97 ANS:

PTS: 2 REF: 061729aii NAT: F.IF.C.7 TOP: Graphing Exponential FunctionsKEY: AII

98 ANS: 4There is no x-intercept.

PTS: 2 REF: 011823aii NAT: F.IF.C.7 TOP: Graphing Exponential FunctionsKEY: AII

ID: A

20

99 ANS: 1

PTS: 2 REF: 061618aii NAT: F.IF.C.7 TOP: Graphing Logarithmic Functions 100 ANS:

As x → −3, y → −∞ . As x → ∞, y → ∞ .

PTS: 4 REF: 061735aii NAT: F.IF.C.7 TOP: Graphing Logarithmic Functions 101 ANS:

A = 5000(1.045)n

B = 5000 1 + .0464

4n

5000 1 + .0464

4(6)

− 5000(1.045)6 ≈ 6578.87− 6511.30 ≈ 67.57 10000 = 5000 1 + .0464

4n

2 = 1.01154n

log2 = 4n ⋅ log1.0115

n =log2

4log1.0115

n ≈ 15.2

PTS: 6 REF: 081637aii NAT: A.CED.A.1 TOP: Exponential Growth

ID: A

21

102 ANS:

720 =120000 .048

12

1 + .04812

n

1+ .04812

n

− 1

720(1.004)n − 720 = 480(1.004)n

240(1.004)n = 720

1.004n = 3

n log1.004 = log3

n ≈ 275.2 months

275.212 ≈ 23 years

PTS: 4 REF: spr1509aii NAT: A.CED.A.1 TOP: Exponential Growth 103 ANS: 1

8(2x + 3 ) = 48

2x + 3 = 6

(x + 3) ln2 = ln6

x + 3 = ln6ln2

x = ln6ln2 − 3

PTS: 2 REF: 061702aii NAT: F.LE.A.4 TOP: Exponential EquationsKEY: without common base

ID: A

22

104 ANS:

100 = 325+ (68 − 325)e−2k

−225 = −257e−2k

k =

ln −225−257

−2

k ≈ 0.066

T = 325− 257e−0.066t

T = 325− 257e−0.066(7) ≈ 163

PTS: 4 REF: fall1513aii NAT: F.LE.A.4 TOP: Exponential Growth 105 ANS:

A = Pert

135000 = 100000e 5r

1.35 = e5r

ln1.35 = lne5r

ln1.35 = 5r

.06 ≈ r or 6%

PTS: 2 REF: 061632aii NAT: F.LE.A.4 TOP: Exponential Growth 106 ANS:

8.75 = 1.25×49

7 = x49

x = 749 ≈ 1.04

4

PTS: 2 REF: 081730aii NAT: F.LE.A.4 TOP: Exponential Growth 107 ANS: 1

9110 = 5000e30r

ln 911500 = lne30r

ln 91150030 = r

r ≈.02

PTS: 2 REF: 011810aii NAT: F.LE.A.4 TOP: Exponential Growth

ID: A

23

108 ANS: 3

ebt = ca

lnebt = ln ca

bt lne = ln ca

t =ln c

ab

PTS: 2 REF: 011813aii NAT: F.LE.A.4 TOP: Exponential Growth 109 ANS:

7 = 20(0.5)t

8.02

log0.35 = log0.5t

8.02

log0.35 =t log0.5

8.02

8.02log0.35log0.5 = t

t ≈ 12

PTS: 4 REF: 081634aii NAT: F.LE.A.4 TOP: Exponential Decay 110 ANS:

100 = 140 12

5h

log 100140 = log 1

2

5h

log 57 = 5

h log 12

h =5log 1

2

log 57

≈ 10.3002

40 = 140 12

t10.3002

log 27 = log 1

2

t10.3002

log 27 =

t log 12

10.3002

t =10.3002log 2

7

log 12

≈ 18.6

PTS: 6 REF: 061737aii NAT: F.LE.A.4 TOP: Exponential Decay

ID: A

24

111 ANS: 4k 4 − 4k 2 + 8k 3 − 32k + 12k 2 − 48

k 2 (k 2 − 4) + 8k(k 2 − 4) + 12(k 2 − 4)

(k 2 − 4)(k 2 + 8k + 12)

(k + 2)(k − 2)(k + 6)(k + 2)

PTS: 2 REF: fall1505aii NAT: A.SSE.A.2 TOP: Factoring PolynomialsKEY: factoring by grouping

112 ANS: The expression is of the form y 2 − 5y − 6 or (y − 6)(y + 1). Let y = 4x 2 + 5x :

4×2 + 5x − 6

4x 2 + 5x + 1

(4x − 3)(x + 2)(4x + 1)(x + 1)

PTS: 2 REF: fall1512aii NAT: A.SSE.A.2 TOP: Factoring PolynomialsKEY: a>1

113 ANS: 3(m− 2)2(m+ 3) = (m2 − 4m+ 4)(m+ 3) = m3 + 3m2 − 4m2 − 12m+ 4m+ 12 = m3 −m2 − 8m+ 12

PTS: 2 REF: 081605aii NAT: A.SSE.A.2 TOP: Factoring PolynomialsKEY: factoring by grouping

114 ANS: 32d(d 3 + 3d 2 − 9d − 27)

2d(d 2 (d + 3) − 9(d + 3))

2d(d 2 − 9)(d + 3)

2d(d + 3)(d − 3)(d + 3)

2d(d + 3)2(d − 3)

PTS: 2 REF: 081615aii NAT: A.SSE.A.2 TOP: Factoring PolynomialsKEY: factoring by grouping

115 ANS: 4m5 +m3 − 6m = m(m4 +m2 − 6) = m(m2 + 3)(m2 − 2)

PTS: 2 REF: 011703aii NAT: A.SSE.A.2 TOP: Factoring PolynomialsKEY: higher power AII

116 ANS: x2 (4x − 1) + 4(4x − 1) = (x2 + 4)(4x − 1)

PTS: 2 REF: 061727aii NAT: A.SSE.A.2 TOP: Factoring PolynomialsKEY: factoring by grouping

ID: A

25

117 ANS: 11) let y = x + 2, then y2 + 2y − 8

(y + 4)(y − 2)

(x + 2+ 4)(x + 2− 2)

(x + 6)x

PTS: 2 REF: 081715aii NAT: A.SSE.A.2 TOP: Factoring PolynomialsKEY: multivariable

118 ANS: 3×3 + x2 + 3xy + y = x2 (3x + 1) + y(3x + 1) = x 2 + y

(3x + 1)

PTS: 2 REF: 011828aii NAT: A.SSE.A.2 TOP: Factoring PolynomialsKEY: factoring by grouping

119 ANS: 1

x4 − 4x 3 − 9x 2 + 36x = 0

x 3 (x − 4) − 9x(x − 4) = 0

(x3 − 9x)(x − 4) = 0

x(x2 − 9)(x − 4) = 0

x(x + 3)(x − 3)(x − 4) = 0

x = 0,±3,4

PTS: 2 REF: 061606aii NAT: A.APR.B.3 TOP: Zeros of PolynomialsKEY: AII

120 ANS: 1 PTS: 2 REF: 061701aii NAT: A.APR.B.3TOP: Zeros of Polynomials KEY: AII

121 ANS: 4 PTS: 2 REF: 081708aii NAT: A.APR.B.3TOP: Zeros of Polynomials KEY: AII

122 ANS: f(x) = x2 (x + 4)(x − 3); g(x) = (x + 2)2 (x + 6)(x − 1)

PTS: 4 REF: 011836aii NAT: A.APR.B.3 TOP: Zeros of Polynomials 123 ANS: 4

The maximum volume of p(x) = −(x + 2)(x − 10)(x − 14) is about 56, at x = 12.1

PTS: 2 REF: 081712aii NAT: F.IF.B.4 TOP: Graphing Polynomial Functions 124 ANS: 2 PTS: 2 REF: 061620aii NAT: F.IF.B.4

TOP: Graphing Polynomial Functions

ID: A

26

125 ANS: 3

PTS: 2 REF: 011817aii NAT: F.IF.B.4 TOP: Graphing Polynomial Functions 126 ANS: 1

The zeros of the polynomial are at −b, and c. The sketch of a polynomial of degree 3 with a negative leading coefficient should have end behavior showing as x goes to negative infinity, f(x) goes to positive infinity. The multiplicities of the roots are correctly represented in the graph.

PTS: 2 REF: spr1501aii NAT: F.IF.C.7 TOP: Graphing Polynomial FunctionsKEY: AII

127 ANS: 3The graph shows three real zeros, and has end behavior matching the given end behavior.

PTS: 2 REF: 061604aii NAT: F.IF.C.7 TOP: Graphing Polynomial FunctionsKEY: AII

128 ANS:

0 = x2 (x + 1) − 4(x + 1)

0 = (x2 − 4)(x + 1)

0 = (x + 2)(x − 2)(x + 1)

x = −2,−1,2

PTS: 4 REF: 081633aii NAT: F.IF.C.7 TOP: Graphing Polynomial Functions

ID: A

27

129 ANS:

PTS: 2 REF: 081732aii NAT: F.IF.C.7 TOP: Graphing Polynomial FunctionsKEY: AII

130 ANS:

PTS: 2 REF: 011729aii NAT: F.IF.C.7 TOP: Graphing Polynomial Functions 131 ANS:

PTS: 2 REF: 011831aii NAT: F.IF.C.7 TOP: Graphing Polynomial Functions 132 ANS: 3

Since x + 4 is a factor of p(x), there is no remainder.

PTS: 2 REF: 081621aii NAT: A.APR.B.2 TOP: Remainder Theorem

ID: A

28

133 ANS: f(4) = 2(4)3 − 5(4)2 − 11(4) − 4 = 128 − 80 − 44 − 4 = 0 Any method that demonstrates 4 is a zero of f(x) confirms

that x − 4 is a factor, as suggested by the Remainder Theorem.

PTS: 2 REF: spr1507aii NAT: A.APR.B.2 TOP: Remainder Theorem 134 ANS:

0 = 6(−5)3 + b(−5)2 − 52(−5) + 15

0 = −750+ 25b + 260+ 15

475 = 25b

19 = b

z x = 6×3 + 19x 2 − 52x + 15

-5 6 19 -52 15

-30 55 15

6 -11 3 0

6×2 − 11x + 3 = 0

(2x − 3)(3x − 1) = 0

x = 32 , 1

3 ,−5

PTS: 4 REF: fall1515aii NAT: A.APR.B.2 TOP: Remainder Theorem

ID: A

29

135 ANS:

x − 5 2x 3 − 4×2 − 7x − 10

2×3 − 10×2

6×2 − 7x

6×2 − 30x

23x − 10

23x − 115

105

2×2 + 6x + 23 Since there is a remainder, x − 5 is not a factor.

PTS: 2 REF: 061627aii NAT: A.APR.B.2 TOP: Remainder Theorem 136 ANS: 2 PTS: 2 REF: 011720aii NAT: A.APR.B.2

TOP: Remainder Theorem 137 ANS: 1

2 1 0 -4 -4 82 4 0 -8

1 2 0 -4 0Since there is no remainder when the quartic is divided by x − 2, this binomial is a factor.

PTS: 2 REF: 061711aii NAT: A.APR.B.2 TOP: Remainder Theorem 138 ANS:

r(2) = −6. Since there is a remainder when the cubic is divided by x − 2, this binomial is not a factor.2 1 -4 4 6

2 -4 01 -2 0 -6

PTS: 2 REF: 061725aii NAT: A.APR.B.2 TOP: Remainder Theorem 139 ANS: 2

-4 1 -11 16 84-4 60 -304

1 -15 76Since there is a remainder when the cubic is divided by x + 4, this binomial is not a factor.

PTS: 2 REF: 081720aii NAT: A.APR.B.2 TOP: Remainder Theorem 140 ANS: 4

p(5) = 2(5)3 − 3(5) + 5 = 240

PTS: 2 REF: 011819aii NAT: A.APR.B.2 TOP: Remainder Theorem

ID: A

30

141 ANS: Let x equal the first integer and x + 1 equal the next. (x + 1)2 − x 2 = x2 + 2x + 1− x 2 = 2x + 1. 2x + 1 is an odd integer.

PTS: 2 REF: fall1511aii NAT: A.APR.C.4 TOP: Polynomial Identities 142 ANS:

x3 + 9×3 + 8

= x3 + 8×3 + 8

+ 1×3 + 8

x3 + 9×3 + 8

= x3 + 9×3 + 8

PTS: 2 REF: 061631aii NAT: A.APR.C.4 TOP: Polynomial Identities 143 ANS: 4

(x + y)3 = x 3 + 3×2 y + 3xy2 + y3 ≠ x 3 + 3xy + y3

PTS: 2 REF: 081620aii NAT: A.APR.C.4 TOP: Polynomial Identities 144 ANS:

2×3 − 10×2 + 11x − 7 = 2×3 + hx 2 + 3x − 8×2 − 4hx − 12+ k

−2×2 + 8x + 5 = hx 2 − 4hx + k

h = −2

k = 5

PTS: 4 REF: 011733aii NAT: A.APR.C.4 TOP: Polynomial Identities 145 ANS:

(x2 + y2 )2 = (x2 − y2 )2 + (2xy)2

x4 + 2x2y2 + y4 = x4 − 2x2y2 + y4 + 4x2y 2

x4 + 2x2y2 + y4 = x4 + 2x2y2 + y4

PTS: 2 REF: 081727aii NAT: A.APR.C.4 TOP: Polynomial Identities 146 ANS: 2 PTS: 2 REF: 011806aii NAT: A.APR.C.4

TOP: Polynomial Identities 147 ANS:

x3 • x = x13 • x

12 = x

36 • x

36 = x

56

PTS: 2 REF: 061731aii NAT: N.RN.A.2 TOP: Operations with RadicalsKEY: with variables, index > 2

ID: A

31

148 ANS:

x − 5 = −x + 7

x − 5 = x 2 − 14x + 49

0 = x2 − 15x + 54

0 = (x − 6)(x − 9)

x = 6, 9

x − 5 = −9 + 7 = −2 is extraneous.

PTS: 2 REF: spr1508aii NAT: A.REI.A.2 TOP: Solving RadicalsKEY: extraneous solutions

149 ANS: 3

56− x = x

56 − x = x2

0 = x2 + x − 56

0 = (x + 8)(x − 7)

x = 7

−8 is extraneous.

PTS: 2 REF: 061605aii NAT: A.REI.A.2 TOP: Solving RadicalsKEY: extraneous solutions

ID: A

32

150 ANS:

2x − 7

2= (5 − x)2

2x − 7 = 25− 10x + x2

0 = x2 − 12x + 32

0 = (x − 8)(x − 4)

x = 4,8

2(4) − 7 + 4 = 5

1 = 1

2(8) − 7 + 8 = 5

9 ≠ −3

PTS: 4 REF: 081635aii NAT: A.REI.A.2 TOP: Solving RadicalsKEY: extraneous solutions

151 ANS:

0 = t − 2t + 6

2t − 6 = t

4t2 − 24t + 36 = t

4t2 − 25t + 36 = 0

(4t − 9)(t − 4) = 0

t = 94 ,4

2 94

− 6 < 0, so 94 is extraneous.

( 1 − 2(1) + 6) − ( 3 − 2(3) + 6) = 5− 3 ≈ 3.268 327 mph

PTS: 6 REF: 011737aii NAT: A.REI.A.2 TOP: Solving RadicalsKEY: context

152 ANS: x − 4 = −x + 6

x − 4 = x2 − 12x + 36

0 = x2 − 13x + 40

0 = (x − 8)(x − 5)

x = 5, 8

x − 4 = −8 + 6 = −2 is extraneous.

PTS: 2 REF: 061730aii NAT: A.REI.A.2 TOP: Solving RadicalsKEY: extraneous solutions

ID: A

33

153 ANS: 2x + 14 = 2x + 5 + 1

x + 14 = 2x + 5+ 2 2x + 5 + 1

−x + 8 = 2 2x + 5

x2 − 16x + 64 = 8x + 20

x2 − 24x + 44 = 0

(x − 22)(x − 2) = 0

x = 2,22

22+ 14 − 2(22) + 5 = 1

6− 7 ≠ 1

PTS: 2 REF: 081704aii NAT: A.REI.A.2 TOP: Solving RadicalsKEY: advanced

154 ANS: 3x + 1 = x + 1

x + 1 = x 2 + 2x + 1

0 = x2 + x

0 = x(x + 1)

x = −1,0

PTS: 2 REF: 011802aii NAT: A.REI.A.2 TOP: Solving RadicalsKEY: extraneous solutions

155 ANS:

Applying the commutative property, 315

2

can be rewritten as 32

15 or 9

15 . A fractional exponent can be

rewritten as a radical with the denominator as the index, or 915 = 95 .

PTS: 2 REF: 081626aii NAT: N.RN.A.1 TOP: Radicals and Rational Exponents 156 ANS:

Rewrite 43 as 1

3 ⋅ 41, using the power of a power rule.

PTS: 2 REF: 081725aii NAT: N.RN.A.1 TOP: Radicals and Rational Exponents 157 ANS:

The denominator of the rational exponent represents the index of a root, and the 4th root of 81 is 3 and 33 is 27.

PTS: 2 REF: 011832aii NAT: N.RN.A.1 TOP: Radicals and Rational Exponents 158 ANS: 4 PTS: 2 REF: 061601aii NAT: N.RN.A.2

TOP: Radicals and Rational Exponents KEY: variables

ID: A

34

159 ANS:

x53

65

= y56

65

x2 = y

PTS: 2 REF: 011730aii NAT: N.RN.A.2 TOP: Radicals and Rational ExponentsKEY: variables

160 ANS:

x83

x43

= x y

x43 = x y

43 = y

PTS: 2 REF: spr1505aii NAT: N.RN.A.2 TOP: Radicals and Rational ExponentsKEY: numbers

161 ANS: 2

m53

−12

= m−

56 = 1

m56

PTS: 2 REF: 011707aii NAT: N.RN.A.2 TOP: Radicals and Rational ExponentsKEY: variables

162 ANS: 4 PTS: 2 REF: 061716aii NAT: N.RN.A.2TOP: Radicals and Rational Exponents KEY: variables

163 ANS: 4

−54×9

y4

23

=(2 ⋅ −27)

23 x

183

y83

= 223 ⋅ 9×6

y2 ⋅ y23

=9×6 43

y2 y23

PTS: 2 REF: 081723aii NAT: N.RN.A.2 TOP: Radicals and Rational ExponentsKEY: variables

ID: A

35

164 ANS: 4

nm =

a 5

a = a52

a22

= a32 = a 3

PTS: 2 REF: 011811aii NAT: N.RN.A.2 TOP: Radicals and Rational ExponentsKEY: variables

165 ANS: (4 − 3i)(5+ 2yi − 5 + 2yi)

(4− 3i)(4yi)

16yi − 12yi2

12y + 16yi

PTS: 2 REF: spr1506aii NAT: N.CN.A.2 TOP: Operations with Complex Numbers 166 ANS:

xi(−6i)2 = xi(36i2 ) = 36xi3 = −36xi

PTS: 2 REF: 081627aii NAT: N.CN.A.2 TOP: Operations with Complex Numbers 167 ANS:

i2 = −1, and not 1; 10+ 10i

PTS: 2 REF: 011825aii NAT: N.CN.A.2 TOP: Operations with Complex Numbers 168 ANS: 2

(2 − yi)(2 − yi) = 4− 4yi + y2 i2 = −y2 − 4yi + 4

PTS: 2 REF: 061603aii NAT: N.CN.A.2 TOP: Operations with Complex Numbers 169 ANS:

1− i 1− i 1 − i = (1 − 2i + i2 ) 1 − i = −2i 1 − i = −2i + 2i2 = −2 − 2i

PTS: 2 REF: 011725aii NAT: N.CN.A.2 TOP: Operations with Complex Numbers 170 ANS: 2

6xi3 (−4xi + 5) = −24x 2 i4 + 30xi3 = −24x 2 (1) + 30x(−1) = −24x 2 − 30xi

PTS: 2 REF: 061704aii NAT: N.CN.A.2 TOP: Operations with Complex Numbers 171 ANS: 3

(3k − 2i)2 = 9k 2 − 12ki + 4i2 = 9k 2 − 12ki − 4

PTS: 2 REF: 081702aii NAT: N.CN.A.2 TOP: Operations with Complex Numbers

ID: A

36

172 ANS: 1×2 + 2x − 8 = 0

(x + 4)(x − 2) = 0

x = −4,2

PTS: 2 REF: 081701aii NAT: A.APR.D.6 TOP: Undefined Rationals 173 ANS: 4

−3×2 − 5x + 2×3 + 2×2 =

(−3x + 1)(x + 2)x2(x + 2)

= −3xx2 + 1

x2 = −3x−1 + x−2

PTS: 2 REF: 061723aii NAT: A.APR.D.6 TOP: Expressions with Negative ExponentsKEY: variables

174 ANS: 1

PTS: 2 REF: fall1503aii NAT: A.APR.D.6 TOP: Rational Expressions 175 ANS: 2

2x + 3 4×3 + 0x2 + 5x + 102×2 − 3x + 7

4×3 + 6x 2

− 6x 2 + 5x

−6x 2 − 9x

14x + 10

14x + 21

− 11

PTS: 2 REF: 061614aii NAT: A.APR.D.6 TOP: Rational Expressions

ID: A

37

176 ANS: 2

x + 2 x3 + 2×2 + x + 6×2 + 0x + 1

x 3 + 2×2

0x2 + x

0x 2 + 0x

x + 6

x + 2

4

PTS: 2 REF: 081611aii NAT: A.APR.D.6 TOP: Rational Expressions 177 ANS:

x − 2 3x 2 + 7x − 20

3×2 − 6x

13x − 20

13x − 26

6

3x + 13 3x + 13+ 6

x − 2

PTS: 2 REF: 011732aii NAT: A.APR.D.6 TOP: Rational Expressions 178 ANS: 1

2x − 1 4x 3 + 0x 2 + 9x − 52x 2 + x + 5

4x 3 − 2x 2

2x 2 + 9x

2x 2 − x

10x − 5

10x − 5

PTS: 2 REF: 081713aii NAT: A.APR.D.6 TOP: Rational Expressions

ID: A

38

179 ANS: 4

2x − 1 10×3 − 3×2 − 7x + 35×2 + x − 3

10×3 − 5×2

2x 2 − 7x

2x 2 − x

− 6x + 3

−6x + 3

PTS: 2 REF: 011809aii NAT: A.APR.D.6 TOP: Rational Expressions 180 ANS: 3

c2 − d 2

d 2 + cd − 2c2 =(c + d)(c − d)(d + 2c)(d − c) =

−(c + d)d + 2c = −c − d

d + 2c

PTS: 2 REF: 011818aii NAT: A.APR.D.6 TOP: Rational ExpressionsKEY: a > 0

181 ANS: 3 PTS: 2 REF: 061602aii NAT: A.CED.A.1TOP: Modeling Rationals

182 ANS: 3 PTS: 2 REF: 061722aii NAT: A.CED.A.1TOP: Modeling Rationals

183 ANS: 18 + 1

6 = 1tb

; 24tb

8 +24tb

6 =24tb

tb

3tb + 4tb = 24

tb =247 ≈ 3.4

PTS: 2 REF: 011827aii NAT: A.CED.A.1 TOP: Modeling Rationals

ID: A

39

184 ANS:

1x − 1

3 = − 13x

3− x3x = − 1

3x

3 − x = −1

x = 4

PTS: 2 REF: 061625aii NAT: A.REI.A.2 TOP: Solving RationalsKEY: rational solutions

185 ANS: 4

x(x + 7) 3x + 25x + 7 − 5 = 3

x

x(3x + 25) − 5x(x + 7) = 3(x + 7)

3×2 + 25x − 5×2 − 35x = 3x + 21

2×2 + 13x + 21 = 0

(2x + 7)(x + 3) = 0

x = −72 ,−3

PTS: 2 REF: fall1501aii NAT: A.REI.A.2 TOP: Solving RationalsKEY: rational solutions

ID: A

40

186 ANS: 31J = 1

F − 1W

1J = W −F

FW

J = FWW − F

PTS: 2 REF: 081617aii NAT: A.REI.A.2 TOP: Solving RationalsKEY: rational solutions

187 ANS: 12(x − 4)

(x + 3)(x − 4) +3(x + 3)

(x − 4)(x + 3) =2x − 2

x2 − x − 12

2x − 8+ 3x + 9 = 2x − 2

3x = −3

x = −1

PTS: 2 REF: 011717aii NAT: A.REI.A.2 TOP: Solving RationalsKEY: rational solutions

188 ANS: 12x

x − 2xx

−11x

x − 2x − 2

=8

x2 − 2x

2x 2 − 11x + 22 = 8

2x 2 − 11x + 14 = 0

(2x − 7)(x − 2) = 0

x = 72 ,2

PTS: 2 REF: 061719aii NAT: A.REI.A.2 TOP: Solving Rationals 189 ANS:

3pp − 5 =

p + 2p + 3

3p 2 + 9p = p 2 − 3p − 10

2p 2 + 12p + 10 = 0

p 2 + 6p + 5 = 0

(p + 5)(p + 1) = 0

p = −5,−1

PTS: 4 REF: 081733aii NAT: A.REI.A.2 TOP: Solving RationalsKEY: rational solutions

ID: A

41

190 ANS: 1

x − 4x − 1 = 2

x(x − 1) − 4 = 2(x − 1)

x2 − x − 4 = 2x − 2

x2 − 3x − 2 = 0

x =3 ± (−3)2 − 4(1)(−2)

2(1) =3 ± 17

2

PTS: 2 REF: 011812aii NAT: A.REI.A.2 TOP: Solving RationalsKEY: rational solutions

191 ANS:

2x 2 + x − 3

• (x − 1) − 2×2 + x − 3

+ (x − 1)

2x 3 − 2x 2 + x2 − x − 3x + 3

− 2×2 + 2x − 4

2×3 − 3×2 − 6x + 7

PTS: 4 REF: 011833aii NAT: F.BF.A.1 TOP: Operations with Functions 192 ANS: 4

m(c)g(c) = c + 1

1 − c 2 = c + 1(1+ c)(1 − c) =

11− c

PTS: 2 REF: 061608aii NAT: F.BF.A.1 TOP: Operations with Functions 193 ANS: 3 PTS: 2 REF: 011710aii NAT: F.BF.A.1

TOP: Operations with Functions 194 ANS: 2

x(30− 0.01x) − (0.15×3 + 0.01×2 + 2x + 120) = 30x − 0.01×2 − 0.15×3 − 0.01x 2 − 2x − 120

= −0.15x 3 − 0.02x 2 + 28x − 120

PTS: 2 REF: 061709aii NAT: F.BF.A.1 TOP: Operations with Functions 195 ANS:

0 = log10 (x − 4)

100 = x − 4

1 = x − 4

x = 5

The x-intercept of h is (2,0). f has the larger value.

PTS: 2 REF: 081630aii NAT: F.IF.C.9 TOP: Comparing FunctionsKEY: AII

196 ANS: 2h(x) does not have a y-intercept.

PTS: 2 REF: 011719aii NAT: F.IF.C.9 TOP: Comparing Functions

ID: A

42

197 ANS: q has the smaller minimum value for the domain [−2,2]. h’s minimum is −1 2(−1) + 1

and q’s minimum is −8.

PTS: 2 REF: 011830aii NAT: F.IF.C.9 TOP: Comparing FunctionsKEY: AII

198 ANS: 3f(x) = −f(x), so f(x) is odd. g(−x) ≠ g(x), so g(x) is not even. g(−x) ≠ −g(x), so g(x) is not odd. h(−x) = h(x), so h(x) is even.

PTS: 2 REF: fall1502aii NAT: F.BF.B.3 TOP: Even and Odd Functions 199 ANS: 1

The graph of y = sinx is unchanged when rotated 180º about the origin.

PTS: 2 REF: 081614aii NAT: F.BF.B.3 TOP: Even and Odd Functions 200 ANS:

j(−x) = (−x)4 − 3(−x)2 − 4 = x 2 − 3x 2 − 4 Since j(x) = j(−x), the function is even.

PTS: 2 REF: 081731aii NAT: F.BF.B.3 TOP: Even and Odd Functions 201 ANS:

x = y − 3

3+ 1

x − 1 = y − 3

3

x − 13 = y − 3

x − 13 + 3 = y

f −1 (x) = x − 13 + 3

PTS: 2 REF: fall1509aii NAT: F.BF.B.4 TOP: Inverse of FunctionsKEY: equations

202 ANS: 2

x = −34 y + 2

−4x = 3y − 8

−4x + 8 = 3y

−43 x + 8

3 = y

PTS: 2 REF: 061616aii NAT: F.BF.B.4 TOP: Inverse of FunctionsKEY: equations

203 ANS: 3 PTS: 2 REF: 011708aii NAT: F.BF.B.4TOP: Inverse of Functions KEY: equations

ID: A

43

204 ANS: 2

x =y + 1y − 2

xy − 2x = y + 1

xy − y = 2x + 1

y(x − 1) = 2x + 1

y = 2x + 1x − 1

PTS: 2 REF: 081714aii NAT: F.BF.B.4 TOP: Inverse of FunctionsKEY: equations

205 ANS: 2x = −6(y − 2)

−x6 = y − 2

−x6 + 2 = y

PTS: 2 REF: 011821aii NAT: F.BF.B.4 TOP: Inverse of FunctionsKEY: equations

206 ANS: Jillian’s plan, because distance increases by one mile each week. a 1 = 10

an = an − 1 + 1

an = n + 12

PTS: 4 REF: 011734aii NAT: F.LE.A.2 TOP: Sequences 207 ANS: 3 PTS: 2 REF: 081618aii NAT: F.LE.A.2

TOP: Sequences 208 ANS: 3 PTS: 2 REF: 061720aii NAT: F.LE.A.2

TOP: Sequences KEY: AII 209 ANS:

a 1 = 4

an = 2an − 1 + 1

a 8 = 639

PTS: 2 REF: 081729aii NAT: F.LE.A.2 TOP: Sequences 210 ANS:

6.25 − 2.2521− 5 = 4

16 = $.25 fine per day. 2.25− 5(.25) = $1 replacement fee. an = 1.25 + (n − 1)(.25). a 60 = $16

PTS: 4 REF: 081734aii NAT: F.LE.A.2 TOP: Sequences

ID: A

44

211 ANS: 1

d = 18; r = ±54

PTS: 2 REF: 011714aii NAT: F.IF.A.3 TOP: SequencesKEY: term

212 ANS: an = xn − 1 (x + 1) xn − 1 = 0

x = 0

x + 1 = 0

x = −1

PTS: 4 REF: spr1511aii NAT: F.BF.A.2 TOP: Sequences 213 ANS: 4

The scenario represents a decreasing geometric sequence with a common ratio of 0.80.

PTS: 2 REF: 061610aii NAT: F.BF.A.2 TOP: Sequences 214 ANS: 3 PTS: 2 REF: 061623aii NAT: F.BF.A.2

TOP: Sequences 215 ANS: 1

(2) is not recursive

PTS: 2 REF: 081608aii NAT: F.BF.A.2 TOP: Sequences 216 ANS: 4 PTS: 2 REF: 081624aii NAT: F.BF.A.2

TOP: Sequences 217 ANS: 3 PTS: 2 REF: 081724aii NAT: F.BF.A.2

TOP: Sequences 218 ANS: 3 PTS: 2 REF: 011824aii NAT: F.BF.A.2

TOP: Sequences 219 ANS: 1 PTS: 2 REF: 081609aii NAT: F.BF.B.6

TOP: Sigma Notation KEY: represent 220 ANS:

20000 = PMT1 − (1+.00625)−60

0.00625

PMT ≈ 400.76

21000− x = 3001− (1 +.00625)−60

0.00625

x ≈ 6028

PTS: 4 REF: 011736aii NAT: A.SSE.B.4 TOP: Series 221 ANS:

Sn =33000 − 33000(1.04)n

1 − 1.04 S 15 =33000− 33000(1.04)15

1 − 1.04 ≈ 660778.39

PTS: 4 REF: 061634aii NAT: A.SSE.B.4 TOP: Series 222 ANS: 2 PTS: 2 REF: 061724aii NAT: A.SSE.B.4

TOP: Series

ID: A

45

223 ANS:

M = 172600 •0.00305(1+ 0.00305)12 ⋅ 15

(1 + 0.00305)12 ⋅ 15 − 1≈ 1247 1100 = (172600− x) •

0.00305(1 + 0.00305)12 ⋅ 15

(1+ 0.00305)12 ⋅ 15 − 1

1100 ≈ (172600− x) • (0.007228)

152193 ≈ 172600 − x

20407 ≈ x

PTS: 4 REF: 061734aii NAT: A.SSE.B.4 TOP: Series

ID: A

1

Algebra II Regents Exam Questions by Common Core State Standard: TopicAnswer Section

224 ANS: 4

d = 32(.8)b − 1 Sn =32 − 32(.8)12

1 −.8 ≈ 149

PTS: 2 REF: 081721aii NAT: A.SSE.B.4 TOP: Series 225 ANS: 2

S 20 =.01 −.01(3)20

1− 3 = 17,433,922

PTS: 2 REF: 011822aii NAT: A.SSE.B.4 TOP: Series 226 ANS: 1 PTS: 2 REF: 081616aii NAT: F.TF.A.1

TOP: Unit Circle 227 ANS: 1 PTS: 2 REF: 011815aii NAT: F.TF.A.2

TOP: Unit Circle 228 ANS:

csc θ = 1sinθ , and sinθ on a unit circle represents the y value of a point on the unit circle. Since y = sinθ ,

csc θ = 1y .

PTS: 2 REF: 011727aii NAT: F.TF.A.2 TOP: Reciprocal Trigonometric Relationships 229 ANS: 4 PTS: 2 REF: 081707aii NAT: F.TF.A.2

TOP: Reference Angles 230 ANS: 1

A reference triangle can be sketched using the coordinates (−4,3) in the second quadrant to find the value of sinθ .

PTS: 2 REF: spr1503aii NAT: F.TF.A.2 TOP: Determining Trigonometric FunctionsKEY: extension to reals

ID: A

2

231 ANS: 1

PTS: 2 REF: 061617aii NAT: F.TF.A.2 TOP: Determining Trigonometric FunctionsKEY: extension to reals

232 ANS: 2 PTS: 2 REF: 011804aii NAT: F.TF.A.2TOP: Determining Trigonometric Functions KEY: radians

233 ANS:

sin2θ + (−0.7)2 = 1

sin2θ =.51

sinθ = ± .51

Since θ is in Quadrant II, sinθ = .51 and tanθ = sinθcosθ =

.51−0.7 ≈ −1.02

PTS: 2 REF: 081628aii NAT: F.TF.C.8 TOP: Determining Trigonometric Functions 234 ANS: 2

cosθ = ± 1 −− 2

5

2

= ± 2525 − 2

25 = ±235

PTS: 2 REF: 061712aii NAT: F.TF.C.8 TOP: Determining Trigonometric Functions 235 ANS: 1 PTS: 2 REF: 011704aii NAT: F.TF.C.8

TOP: Simplifying Trigonometric Expressions 236 ANS: 4

period = 2πB

160 = 2π

B

B = 120π

PTS: 2 REF: 061624aii NAT: F.TF.B.5 TOP: Modeling Trigonometric Functions

ID: A

3

237 ANS: 3

H(t) is at a minimum at 70(−1) + 80 = 10

PTS: 2 REF: 061613aii NAT: F.IF.B.4 TOP: Graphing Trigonometric FunctionsKEY: maximum/minimum

238 ANS: 2 PTS: 2 REF: 081610aii NAT: F.IF.B.4TOP: Graphing Trigonometric Functions KEY: increasing/decreasing

239 ANS: 4Bar Harbor Phoenix

Minimum 31.386 66.491Midline 55.3 86.729Maximum 79.214 106.967Range 47.828 40.476

PTS: 2 REF: 061715aii NAT: F.IF.B.4 TOP: Graphing Trigonometric FunctionsKEY: maximum/minimum

240 ANS: 2 PTS: 2 REF: 011701aii NAT: F.IF.B.4TOP: Graphing Trigonometric Functions

241 ANS: 4 PTS: 2 REF: 061706aii NAT: F.IF.B.4TOP: Graphing Trigonometric Functions

242 ANS: 3 PTS: 2 REF: 081705aii NAT: F.IF.B.4TOP: Graphing Trigonometric Functions KEY: increasing/decreasing

243 ANS: 4

As the range is [4,10], the midline is y = 4+ 102 = 7.

PTS: 2 REF: fall1506aii NAT: F.IF.C.7 TOP: Graphing Trigonometric FunctionsKEY: mixed

244 ANS: Amplitude, because the height of the graph shows the volume of the air.

PTS: 2 REF: 081625aii NAT: F.IF.C.7 TOP: Graphing Trigonometric FunctionsKEY: mixed

ID: A

4

245 ANS:

PTS: 2 REF: 061628aii NAT: F.IF.C.7 TOP: Graphing Trigonometric FunctionsKEY: graph

246 ANS: 3(3) repeats 3 times over 2π .

PTS: 2 REF: 011722aii NAT: F.IF.C.7 TOP: Graphing Trigonometric FunctionsKEY: recognize

247 ANS: 1 PTS: 2 REF: 061708aii NAT: F.IF.C.7TOP: Graphing Trigonometric Functions KEY: identify

248 ANS:

The amplitude, 12, can be interpreted from the situation, since the water level has a minimum of −12 and a maximum of 12. The value of A is −12 since at 8:30 it is low tide. The period of the function is 13 hours, and is expressed in the function through the parameter B. By experimentation with

technology or using the relation P = 2πB (where P is the period), it is determined that B = 2π

13 .

f(t) = −12cos 2π13 t

In order to answer the question about when to fish, the student must interpret the function and determine which choice, 7:30 pm or 10:30 pm, is on an increasing interval. Since the function is increasing from t = 13 to t = 19.5 (which corresponds to 9:30 pm to 4:00 am), 10:30 is the appropriate choice.

PTS: 6 REF: spr1514aii NAT: F.IF.C.7 TOP: Graphing Trigonometric FunctionsKEY: graph

ID: A

5

249 ANS:

period is 23. The wheel rotates once every 2

3 second.

PTS: 2 REF: 061728aii NAT: F.IF.C.7 TOP: Graphing Trigonometric FunctionsKEY: period

250 ANS: 4 PTS: 2 REF: 081718aii NAT: F.IF.C.7TOP: Graphing Trigonometric Functions KEY: amplitude

251 ANS:

The period of P is 23, which means the patient’s blood pressure reaches a high every 2

3

second and a low every 23 second. The patient’s blood pressure is high because 144 over 96 is greater than 120

over 80.

PTS: 6 REF: 011837aii NAT: F.IF.C.7 TOP: Graphing Trigonometric FunctionsKEY: graph

252 ANS:

Part a sketch is shifted π3 units right.

PTS: 4 REF: 081735aii NAT: F.IF.C.7 TOP: Graphing Trigonometric FunctionsKEY: graph

ID: A

6

253 ANS: 44(x2 − 6x + 9) + 4(y2 + 18y + 81) = 76+ 36+ 324

4(x − 3)2 + 4(y + 9)2 = 436

PTS: 2 REF: 061619aii NAT: G.GPE.A.1 TOP: Equations of CirclesKEY: completing the square

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