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.
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 \\(\square\\) gallons of gasoline in the tank.

Explanation:

Step1: Find the slope (rate of change)

The slope \( m \) between two points \((d_1, g_1)\) and \((d_2, g_2)\) is \( m=\frac{g_2 - g_1}{d_2 - d_1} \). Using \((20, 11)\) and \((40, 9)\), \( m=\frac{9 - 11}{40 - 20}=\frac{-2}{20}=-0.1 \). So the equation is \( g - g_1 = m(d - d_1) \), using \((20, 11)\): \( g - 11=-0.1(d - 20) \), which simplifies to \( g=-0.1d + 13 \).

Step2: Substitute \( d = 110 \)

Substitute \( d = 110 \) into \( g=-0.1d + 13 \): \( g=-0.1(110)+13=-11 + 13 = 2 \). Alternatively, observe the pattern: every 20 miles, \( g \) decreases by 2. At \( d = 100 \), \( g = 3 \). At \( d = 110 \) (10 miles more, so \( g \) decreases by \( 0.1\times10 = 1 \)? Wait, no, the rate is 0.1 gallons per mile (since slope is -0.1, meaning 0.1 gallons per mile used). Wait, from \( d = 100 \) (g=3) to \( d = 110 \), that's 10 miles, so gasoline used is \( 0.1\times10 = 1 \)? No, wait the slope is -0.1, so \( \Delta g=-0.1\Delta d \). So from \( d = 100 \) to \( d = 110 \), \( \Delta d = 10 \), so \( \Delta g=-0.1\times10=-1 \), so \( g = 3-1 = 2 \).

Answer:

2