Penny's parents gave her $50 to spend on new video games. Used games are $7 and new games are $12. Part 1: What is the system of inequalities that represent this situation? Part 2: What is the maximum amount of used games that she could buy? Part 3: What are the minimum amount of new games that she could buy? Part 4: What are two possible combinations of used and new games she can purchase?

The correct answers are Part 1: 7x + 12y 50, x,y 0; Part 2: 7; Part 3: 0: Part 4: 2 old and 3 new video games.

Step-by-step explanation:

Penny's parents gave her $50 to buy new video games.

Price of used games are $7 and new games are $12.

Let Penny buy x number of old video games and y number of new video games.

Part 1:

Total price she spent on buying the video games are 7x + 12y.

This amount should be less than or equal to the amount of money she possess. therefore 7x + 12y 50, x, y 0.

Part 2:

Maximum number of used game she can buy can be given when she spends all her money just on used games. Therefore y = 0. This implies x .

Thus the maximum number of used game she can buy is 7 where she does not buy any new game and has $1 left with her after the purchase.

Part 3:

Minimum number of new games that Penny can buy is zero. She can not buy any new games and spent all her money purchasing old games.

Part 4:

The possible combination in which she can purchase both the video games is 2 old games and 3 new games.

step-by-step explanation:

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part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).