Question

when rachel puts gas in her car, she records the distance that she has driven since her last fill-up and the number of gallons of gas that her car has used. she uses this information to predict the amount of gas that her car will use to travel different distances.

distance, in miles (d)	gas, in gallons (g)
70	2.8
109	4.36
440	17.6

gas used by rachel’s car

complete the equation so that it models the relationship between the number of gallons of gas that the car uses, ( g ), and the number of miles the car has driven, ( d ). click the arrows to choose an answer from each menu.

( g = ) choose... ( + ) choose... ( cdot d )

Explanation:

Step1: Find the rate (slope)

To find the relationship between \( g \) (gallons) and \( d \) (miles), we calculate the rate (slope) by dividing gallons by miles for each data point. For example, using the first point: \(\frac{0.56}{14} = 0.04\). Checking another point: \(\frac{2.8}{70} = 0.04\), \(\frac{4.36}{109} = 0.04\), \(\frac{17.6}{440} = 0.04\). So the rate is 0.04.

Step2: Form the equation

The relationship is linear, so the equation is \( g = 0.04d \) (since it's a direct variation, the form is \( g = kd \) where \( k \) is the rate we found).

Answer:

\( g = 0.04d \) (So the first "Choose" is 0.04 and the second "Choose" is a multiplication sign, but in the equation form, it's \( g = 0.04d \))