Express the monthly cost C as a function of the distance driven d, assuming that a linear relationship gives a

The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May it cost her $380 to drive 480 mi and in June it cost her $460 to drive 800 mi.

Express the monthly cost C as a function of the distance driven d, assuming that a linear relationship gives a
Note: "Linear" means straight-line. So we can plot a straight line graph to find the relationship.


Let d = # of miles driven


Let C = cost





So we have:


480 miles = $380


800 miles = $460





Then the points on a graph are: (480,380) and (800,460).





Find the equation of a line passing through these points and you're done.


Gradient = (C2-C1) / (d2-d1)


= (460-380) / (800-480)


= 80/320


= 0.25





So the equation of a line is:


C-C1 = m(d-d1), where m is the gradient and (d1,C1) is a point.


Let's use: (480,380) as the point (d1,C1)


Then the equation of the line is:


C - 380 = 0.25(d-480)


C - 380 = 0.25d - 120


Add 380 to both sides:





C = 0.25d + 260





The equation of the line is the same as the linear relationship. Therefore our final answer is:





C = 0.25d + 260
Reply:C = ($460 - $380)(d - 480)/(800 - 480) + $380
Reply:If it is a linear function then


Cost, c = kd + p, where d = distance, k ,p are constants





Eqn 1: 380 = 480k + p, for May


Eqn 2: 460 = 800k + p, for June





Eqn 2 - Eqn1:


80 = 320k


k = 1/4





Now put k=1/4 into 1


380 = 480/4 + p


p = 380 - 120 = 260





hence, c = d/4 + 260
Reply:For 480 miles we have 380$


For 800 miles we have 460$





We can subtract the two giving:


For each additional:


320 miles - 80$


160 miles - 40$


480 miles - 120$





Subtract this from the first item:


0 miles - 260$





C = md + 260 (m is gradient)


m = 40 / 160


m = 1 / 4


m = 0.25





C = 0.25d + 260