The cost of maintaining car A is $300 per year plus $.014 per mile driven. The cost of maintaining car B is $200 per year plus $
.025 per mile driven. How many miles are driven in the first year in order for the total costs between car A and B to be equal?Car A = 300+.014xCar B =200+.025xSolve the following equation to determine when costs are equal300+.014x=200+.025x.011x=100
1 answer:
You might be interested in
Do recall that squaring and the *radical sign* cancel each other out... like so:(
)
= a
When you put it that way, it isn't enough :P
()= a
(
)=?
so you start with
()= 
8x+1=25 <-- subtract 1 to both sides
8x=24 <- divide 8 to both sides
x= 3
To find out if it's an extraneous solution ask yourself: It mustn't result in a radical that I like to call... 'illegal'. Plug it into the radicand 8x+1 and make sure you get something that is not a negative number.... so, DO you get a negative number when you plug in x = 3 into the radicand?
(extraneous solution is a invalid solution)
x=3 not extraneous
When you put it that way, it isn't enough :P
(
(
so you start with
(
8x+1=25 <-- subtract 1 to both sides
8x=24 <- divide 8 to both sides
x= 3
To find out if it's an extraneous solution ask yourself: It mustn't result in a radical that I like to call... 'illegal'. Plug it into the radicand 8x+1 and make sure you get something that is not a negative number.... so, DO you get a negative number when you plug in x = 3 into the radicand?
(extraneous solution is a invalid solution)
x=3 not extraneous