For a science fair project, Marisela is tracking the growth of two bean plants. Plant A is 8 cm tall and growing at a rate of 2 cm per day. Plant B is 1 cm tall and growing at a rate of 3 cm per day. Marisela hypothesized that the two plants will be the same height after 7 days. Was her hypothesis reasonable?

A. Yes, because if H(x) represents the height of each plant as a function of x days, then H(x)= 8 + 2x and H(x)= 1 + 3x. And, for x = 7 days, then plant A will be 22 cm tall and plant B will be 22 cm tall.
Correct! Both plants will be 22 cm tall after 7 days.

B. Yes, because if H(x) represents the height of each plant as a function of x days, then H(x) = 8 + 2x and H(x) = 1 + 3x. And, for x = 2 days, then plant A will be 12 cm tall and plant B will be 7 cm tall.
Incorrect. After 2 days, plant A will be 12 cm tall and plant B will be 7 cm tall.

C. No, because if H(x) represents the height of each plant as a function of x days, then H(x) = 2 + 8x and H(x) = 3 + 1x. And, for x = 7 days, then plant A will be 58 cm tall and plant B will be 10 cm tall.
Incorrect. The growth of plant A can be represented with the function H(x) = 8 + 2x and plant B can be represented with H(x)=1 + 3x.

D. No, because if H(x) represents the height of each plant as a function of x days, then H(x) = 8 + 2x and H(x) = 1 + 3x. And, for x = 7 days, then plant A will be 14 cm tall and plant B will be 21 cm tall.
Incorrect. After 7 days, plant A will be 8 + 2(7), or 22 cm tall. After 7 days, plant B will be 1 + 3(7), or 22 cm tall.

Xavier is attending a basketball game and plans to park in one of the parking garages adjacent to the arena. Easy Parking charges a flat rate of $14 plus $3.50 per hour. Quick Parking charges no flat rate, but only an hourly charge of $7. Xavier wants to find the duration of the time he would have to park in each garage to pay the same rate. He estimates that if he parks his car for 2 hours, it would cost the same amount in both garages. Is his estimate reasonable?

A. Yes, because if C(x) represents the charge to park in each garage as a function of x hours, then C(x)= 14 + 3.50x and H(x) = 7x. And, for x = 2 hours, then the charge to park at East Parking would be $14 and the charge to park at Quick Parking would also be $14.
Incorrect. To calculate the charge for Easy Parking use C(x) = 14 + 3.50x, or 14 + 3.50(2). To calculate the charge for Quick Parking use C(x) = 7x, or 7(2).

B. Yes, because if C(x) represents the charge to park in each garage as a function of x hours, then C(x)= 14 + 3.50x and H(x) = 7x. And, for x = 2 hours, then the charge to park at East Parking would be $21 and the charge to park at Quick Parking would also be $21.
Incorrect. To calculate the charge for Easy Parking use C(x) = 14 + 3.50x, or 14 + 3.50(2). To calculate the charge for Quick Parking use C(x) = 7x, or 7(2).

C. No, because if C(x) represents the charge to park in each garage as a function of x hours, then C(x)= 14 + 3.50x and H(x) = 7x. And, for x = 2 hours, then the charge to park at East Parking would be $21 and the charge to park at Quick Parking would be $14.
Correct! After 2 hours, the charge to park at Easy Parking would be $21 and the charge to park at Quick Parking would also be $14. In order to pay the same amount in both garages, Xavier would have to park for 4 hours.

D. No, because if C(x) represents the charge to park in each garage as a function of x hours, then C(x)= 3.50 + 14x and H(x) = 7x. And, for x = 2 hours, then the charge to park at East Parking would be $31.50 and the charge to park at Quick Parking would be $14.
Incorrect. To calculate the charge for Easy Parking use C(x) = 14 + 3.50x, or 14 + 3.50(2).

The Greene family and the Carson family plan to take a vacation together next summer. The Greene family has $75 saved so far for the trip and plans to save an additional $10 per week. The Carson family has $25 saved and plans to save $15 more each week. Mr. Greene estimated that the two families would have the same amount saved after 10 weeks. Is Mr. Greene's estimate reasonable?

A. Yes, because if S(x) represents the total amount saved by each family as a function of x weeks, then S(x)= 75 + 10x and S(x) = 25 + 15x. And, for x = 10 weeks, then the Greene Family would have $100 saved and the Carson Family would also have $100 saved.
Incorrect. To calculate the amount of money the Greene family would have saved use S(x) = 75 + 10x, or 75 + 10(10). To calculate the amount of money the Carson family would have saved use S(x) = 25 + 15x, or 25 + 15(10).

B. Yes, because if S(x) represents the total amount saved by each family as a function of x weeks, then S(x)= 75 + 10x and S(x) = 25 + 15x. And, for x = 10 weeks, then the Greene Family would have $175 saved and the Carson Family would also have $175 saved.
Correct! After 10 weeks, the Greene family would have $175 saved and the Carson family would also have $175 saved.

C. No, because if S(x) represents the total amount saved by each family as a function of x weeks, then S(x)= 10 + 75x and S(x) = 15 + 25x. And, for x = 10 weeks, then the Greene Family would have $760 saved and the Carson Family would have $265 saved.
Incorrect. To calculate the amount of money the Greene family would have saved use S(x) = 75 + 10 and to calculate the amount of money the Carson family would have saved use S(x) = 25 + 15x.

D. No, because if S(x) represents the total amount saved by each family as a function of x weeks, then S(x)= 10 + 75x and S(x) = 15 + 25x. And, for x = 5 weeks, then the Greene Family would have $375 saved and the Carson Family would also have $375 saved.
Incorrect. Mr. Greene estimated that the two families would have saved the same amount of money after 10 weeks.

Stephen and Beth are paying off the loans for their cars. Stephen owes $20,000 and pays $600 per month. Beth owes $14,000 and pays $400 per month. Stephen estimated that the two would owe the same amount after 30 months. Beth estimated that the two would owe the same amount after 24 months. Whose estimate is accurate?

A. Stephen, because if L(x) represents the amount still owed as a function of x months, then L(x)= 20,000 – 600x and L(x) = 14,000 − 400x. And, for x = 30 months, then Stephen would owe $2,000 and Beth would also owe $2,000.
Correct! After 30 months, both Stephen and Beth would both owe $2,000.

B. Stephen, because if L(x) represents the amount still owed as a function of x months, then L(x)= 20,000 + 600x and L(x) = 14,000 + 400x. And, for x = 30 months, then Stephen would owe $38,000 and Beth would also owe $38,000.
Incorrect. To calculate the amount owed by Stephen use L(x) = 20,000 − 600x. To calculate the amount owed by Beth use L(x) = 14,000 − 400x.

C. Beth, because if L(x) represents the amount still owed as a function of x months, then L(x)= 20,000 – 600x and L(x) = 14,000 – 400x. And, for x = 24 months, then Stephen would owe $5,600 and Beth would also owe $5,600.
Incorrect. To calculate the amount of money the Greene family would have saved use S(x) = 75 + 10 and to calculate the amount of money the Carson family would have saved use S(x) = 25 + 15x.

D. Beth, because if L(x) represents the amount still owed as a function of x months, then L(x)= 20,000 – 600x and L(x) = 14,000 – 400x. And, for x = 24 months, then Stephen would owe $4,400 and Beth would also owe $4,400.
Incorrect. Mr. Greene estimated that the two families would have saved the same amount of money after 10 weeks.

Two athletes are running in a long distance race. Runner A is 15 miles from the finish line, traveling at a rate of 3 miles per hour. Runner B is 17 miles from the finish line, traveling at a rate of 4 miles per hour. The announcer estimated that Runner B would catch runner A in 3 hours, when the two are 9 miles from the finish line. Was the announcer's estimate reasonable?

A. Yes, because if D(x) represents the distance of each runner from the finish line as a function of x hours, then D(x)=15 – 3x and D(x)= 17 – 4x. And, for x = 3 hours, then runner A would be 9 miles from the finish line and runner B would also be 9 miles from the finish line.
Incorrect. To calculate the distance of runner A from the finish line at time x = 3 hours, use D(x) =15 − 3x, or 15 − 3(3). To calculate the distance of runner B from the finish line at time x = 3 hours, use D(x) = 17 − 4x, or 17 − 4(3).

B. Yes, because if D(x) represents the distance of each runner from the finish line as a function of x hours, then D(x) = 3 – 15x and D(x) = 4 – 17x. And, for x = 3 hours, then runner A would be 9 miles from the finish line and runner B would also be 9 miles from the finish line.
Incorrect. To calculate the distance of runner A from the finish line, use D(x)=15 − 3x. To calculate the distance of runner B from the finish line, use D(x)=17 − 4x.

C. No, because if D(x) represents the distance of each runner from the finish line as a function of x hours, then D(x) = 15 – 3x and D(x)= 17 – 4x. And, for x = 3 hours, then runner A would be 9 miles from the finish line and runner B would be 5 miles from the finish line.
Incorrect. To calculate the distance of runner A from the finish line at time x = 3 hours, use D(x ) = 15 − 3x, or 15 − 3(3).

D. No, because if D(x) represents the distance of each runner from the finish line as a function of x hours, then D(x) = 15 – 3x and D(x) = 17 – 4x. And, for x = 3 hours, then runner A would be 6 miles from the finish line and runner B would be 5 miles from the finish line.
Correct! After 3 hours, runner A would be 6 miles from the finish line and runner B would be 5 miles from the finish line. Therefore, the announcer’s estimate was not reasonable.