QUESTION IMAGE
Question
let g be the number of gallons of gasoline that remain in a cars gasoline tank after the car has been driven d miles since the tank was filled. some pairs of values of d and g are shown in the following table.
| d (miles) | g (gallons) |
|---|---|
| 40 | 9 |
| 60 | 7 |
| 80 | 5 |
| 100 | 3 |
| 120 | 1 |
complete parts a. to e.
charts c and d are shown with g on the y - axis and d on the x - axis, with points plotted
b. estimate how much gasoline is in the tank after the driver has gone 110 miles since last filling up.
after the driver has gone 110 miles since last filling up, there will be 2 gallons of gasoline in the tank.
c. estimate the number of miles driven since the tank was last filled if 4 gallons of gasoline remain in the tank.
the driver drove \\(\square\\) miles since the tank was last filled if 4 gallons of gasoline are in the tank.
Step1: Find the slope of the linear relationship
From the table, when \( d = 20 \), \( g = 11 \); when \( d = 40 \), \( g = 9 \). The slope \( m=\frac{\Delta g}{\Delta d}=\frac{9 - 11}{40 - 20}=\frac{-2}{20}=-0.1 \). So the equation is \( g - g_1=m(d - d_1) \). Using \( (20,11) \), we get \( g - 11=-0.1(d - 20) \), which simplifies to \( g=-0.1d + 13 \).
Step2: Solve for \( d \) when \( g = 4 \)
Substitute \( g = 4 \) into \( g=-0.1d + 13 \):
\( 4=-0.1d + 13 \)
Subtract 13 from both sides: \( 4 - 13=-0.1d \)
\( -9=-0.1d \)
Divide both sides by -0.1: \( d=\frac{-9}{-0.1}=90 \).
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