Question

1. a car has a 16 - gallon fuel tank. when driven on a highway, it has a gas mileage of 30 miles per gallon. the gas mileage (also called \fuel efficiency\) tells us the number of miles the car can travel for a particular amount of fuel (one gallon of gasoline, in this case). after filling the gas tank, the driver got on a highway and drove for a while.

a. write an equation that represents the relationship between the distance the car has traveled in miles, d, and the amount of gas left in the tank in gallons, x.

b. how many miles has the car traveled if it has the following amounts of gas left in the tank? use a calculator if you need!

i. 15 gallons

ii. 10 gallons

iii. 2.5 gallons

c. write an equation that makes it easier to find the amount of gas left in the tank, x, if we know the car has travelled d miles.

d. how many gallons are left in the tank when the car has traveled the following distances on the highway?

i. 190 miles

ii. 246 miles

1. the area a of a rectangle is represented by the formula a = l × w where l is the length and w is the width.

write an equation that makes it easy to find the width of the rectangle if we know the area and the length.

Explanation:

Step1: Find the relationship for part a

The car has a 16 - gallon tank and 30 miles per gallon. The amount of gas used is $16 - x$. The distance $d$ is the number of gallons used times the miles per gallon. So $d=30(16 - x)$.

Step2: Solve for part b - i

Substitute $x = 15$ into $d=30(16 - x)$. Then $d=30\times(16 - 15)=30\times1 = 30$ miles.

Step3: Solve for part b - ii

Substitute $x = 10$ into $d=30(16 - x)$. Then $d=30\times(16 - 10)=30\times6 = 180$ miles.

Step4: Solve for part b - iii

Substitute $x = 2.5$ into $d=30(16 - x)$. Then $d=30\times(16 - 2.5)=30\times13.5 = 405$ miles.

Step5: Rearrange the equation for part c

Starting from $d = 30(16 - x)$, first expand: $d=480-30x$. Then solve for $x$: $30x=480 - d$, so $x = 16-\frac{d}{30}$.

Step6: Solve for part d - i

Substitute $d = 190$ into $x = 16-\frac{d}{30}$. Then $x=16-\frac{190}{30}=16-\frac{19}{3}=\frac{48 - 19}{3}=\frac{29}{3}\approx9.67$ gallons.

Step7: Solve for part d - ii

Substitute $d = 246$ into $x = 16-\frac{d}{30}$. Then $x=16-\frac{246}{30}=16 - 8.2 = 7.8$ gallons.

Step8: Solve for problem 7

Given $A=l\times w$, solve for $w$. Divide both sides by $l$ (assuming $l
eq0$), so $w=\frac{A}{l}$.

Answer:

a. $d = 30(16 - x)$
b.
i. 30 miles
ii. 180 miles
iii. 405 miles
c. $x = 16-\frac{d}{30}$
d.
i. $\frac{29}{3}\approx9.67$ gallons
ii. 7.8 gallons

1. $w=\frac{A}{l}$