The price of gas started out at 100¢/gallon on the 1st of the month. Every day since then, it has gone up 2¢/gallon. My car takes 10 gallons of gas. (As you might have guessed, these numbers are all fictional.) Let d equal the date (so the 1st of the month is 1, and so on). Let g equal the price of a gallon of gas, in cents. Let c equal the total price required to ll up my car, in cents. a. Write a function g (d) that gives the price of gas on any given day of the month.
b. Write a function c (g) that tells how much money it takes to ll up my car, as a function of the price of a gallon of gas.
c. Write a composite function c (g (d)) that gives the cost of filling up my car on any given day of the month.
d. How much money does it take to ll up my car on the 11th of the month? First, translate this question into function notation then solve it for a number.
e. On what day does it cost 1,040¢ (otherwise known as $10.40) to ll up my car? First, translate this question into function notation then solve it for a number.

Explained below.

Step-by-step explanation:

At the start of every month the price of gas started out at 100¢/gallon.

Then every day since, it has gone up 2¢/gallon.

The variables are denoted as follows:

d = the date

g = price of a gallon of gas

c = total price required to fill up my car, in cents.

(a)

The function that gives the price of gas on any given day of the month is:

g (d) = 100 + 2(d - 1)

The (d - 1) represent the number of times the price has gone up by 2¢/gallon.

(b)

The function that tells how much money it takes to fill up my car, as a function of the price of a gallon of gas is:

c (g) = 10 × g

Since the car takes 10 gallons of gas.

(c)

The composite function that gives the cost of filling up my car on any given day of the month is:

c (g (d)) = 10 × g (d)

            = 10 [100 + 2(d - 1)]

c (g (d)) = 1000 + 20 (d - 1)

(d)

On the 11th day the price of gas has increased by 2¢/gallon for the past 10 days.

Compute the total price it takes to fill up my car on the 11th of the month as follows:

c (g (d)) = 1000 + 20 (d - 1)

c (g (11)) = 1000 + 20 (11 - 1)

             = 1000 + (20 × 20)

             = 1000 + 400

             = 1400¢

Thus, the it takes to fill up my car on the 11th of the month is 1400¢.

(e)

The total price to fill the car on the nth day is 1040¢.

Compute the value of n as follows:

      c (g (d)) = 1000 + 20 (d - 1)

          1040 = 1000 + 20 (n - 1)

1040 - 1000 = 20 (n - 1)

              40 = 20 (n - 1)

          (n - 1) = 2

                 n = 3

Thus, the day on which it cost 1,040¢ to fill up the car is the 3rd day.

should have been evaluated with one odd value and one even value. the expressions have equivalent values when . when , the first expression has a value of 3 and the second

step-by-step explanation: