SOLUTION: A rental truck costs $50 plus $.50 per mile A) write an equation that gives the cost of driving the truck x miles this is what I came up with $50= .50 + x B) Use the Inte

Algebra ->  Graphs -> SOLUTION: A rental truck costs $50 plus $.50 per mile A) write an equation that gives the cost of driving the truck x miles this is what I came up with $50= .50 + x B) Use the Inte      Log On


   

Question 455188: A rental truck costs $50 plus $.50 per mile
A) write an equation that gives the cost of driving the truck x miles
this is what I came up with $50= .50 + x
B) Use the Intersection of graphs method to determine the number of miles that the truck is driven if the rental cost is $80.
C) Solve part (b) numerically with a table of values.
Found 2 solutions by stanbon, richwmiller:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A rental truck costs $50 plus $.50 per mile
A) write an equation that gives the cost of driving the truck x miles
this is what I came up with $50= .50 + x
----
Ans: C(x) = 50 + 0.50x where x is the number of miles.
-------------------------------------------------------------
B) Use the Intersection of graphs method to determine the number of miles that the truck is driven if the rental cost is $80.
----
graph%28400%2C300%2C-10%2C100%2C-10%2C100%2C50%2B0.5x%2C80%29
------------------------------------
C) Solve part (b) numerically with a table of values.
C(x) = 50 + 0.50x
If x = 55, C(55) = $77.50
If x = 60, C(60) = $80
If x = 65, C(65) = $82.50
=============================
Cheers,
Stan H.
=================

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Where is the final cost y?
let's think about this problem
first you have to fork over 50 for the truck for the day.
+50
then for every miles you drive it you have to pay 50 cents
.50*x
y is the final cost
y=.50x+50
then for the 80 dollars
80=.50x+50
30=.5x
remember .5=1/2
30=x/2
60=x