Truck A costs $600 more that Truck B. However, it burns 0.06 gallon/mile compared to 0.07 gallon/mile for truck B. Both trucks have a useful life of 10 years. Salvage value of Truck B is $200 less than Truck A’s. The cost of fuel is $2.25/gallon. If the value of $ remains the same in 10 years, what is the number of miles driven per year (X) beyond which Truck A is a better option than Truck B?

miles driven per year (or more) makes Truck A a better option.

Let us represent and the cost and salvage value of Truck B.

Let represent the number of miles per year.

Now,

Truck A's cost for years is  

Truck B's cost for years is  

Here we need to find the value of such that Truck A's cost is lower.

Therefore,  

So, miles driven per year (or more) makes Truck A a better option.

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  1,778

Step-by-step explanation:

Let C and S represent the cost and salvage value of Truck B, respectively. Let X represent the number of miles per year that answers the question.

Truck A's cost for 10 years is ...

  A = C+600 +(0.06)(2.25)(10X) -(S+200)

Truck B's cost for 10 years is ...

  B = C +(0.07)(2.25)(10X) -S

We want to find X such that Truck A's cost is lower, so ...

  C -S +1.35X +400 < C -S +1.575X

  400 < 0.225X

  1777 7/9 < X

1778 miles driven per year (or more) makes Truck A a better option.

correct answers below:

the 1st digit is in the 100s n its 1

step-by-step explanation:

3,589÷18 = 199.389

so the first digit of the quotient is in the 100s n its 1