The monthly cost of driving a car depends on the number of miles driven. lynn found that in may it cost her $494 to drive 380 mi and in june it cost her $638 to drive 860 mi. (a) express the monthly cost c as a function of the distance driven, d, assuming that a linear relationship gives a suitable model.

C = 0,3x + 380

Step-by-step explanation: Assuming that a linear relationship is suitable model, C is a function of d, we can say that:

C = d.x + k. where x is what we call a variable cost and k is a fixed cost. We have to assume a fixed cost. If it does not exist, the value will be zero.

This way: for may we have 494 = 380x+k and for June we have 638 = 860x+k.

Let's subtract them

638 = 860x+k

494 = 380x+k (-)

638 = 860x+k

-494 = -380x-k

144 = 480x

x=144/480

x=0,3

For k, we substitute in one of the equations

638 = 860.0,3 + k

638 = 258 + k

k = 380

∴ C = 0,3x+380

in problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. to find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

hope it

3.how is solving radical equations different than solving linear equations?

step-by-step explanation: