Math 10 question? Please anyone know?

At 0km she had 58𝑙 (0,58) and at 464km she had 0𝑙 (464,0)

V/58 + d/464 = 1 so V(d) = -1/8 d + 58

d(V) = 464 - 8V so when 12𝑙 remain Jana has traveled 464 - 8*12 = 368km

(and has 8*12 = 96km to empty the tank)

If Jana is driving the vehicle at a constant speed, it volume of gas that it burns each mile will be constant. To determine this number, use the following equation.

V = k * d

58 = k * 464

k = 58 ÷ 464 = 0.125 L/km or ⅛ L/km

Determine the distance traveled when 12 L of gas is left

V = 58 – 12 = 46

46 = 0.125 * d

d = 368 km

I hope this is helpful for you.

Hint:

(V=58 , d= 0 )

(V=0 , d= 464 )

write an equation with given points

i) Consumption = 58L/464km = 1/8 L/km

V = 58 - d/8

ii) 12 = 58 - d/8

d = 368 km

With d = 0, the tank is 58L, so we have one data point: (0, 58)

Then after 464 km, the tank is empty, so this data point is: (464, 0)

To get the equation between the points we need to find the slope first which is the change in y over the change in x:

m = (58 - 0) / (0 - 464)

m = 58 / (-464)

m = -1/8

Then we have the intercept already from the first data point:

b = 58

So the equation is:

V = (-1/8)d + 58

For the volume of gas remaining after "d" km has been driven.

What's the distance traveled with 12 L of gas remains? Solve for d when V = 12:

12 = (-1/8)d + 58

-46 = (-1/8)d

46(8) = d

d = 368 km