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kobusy [5.1K]
3 years ago
10

Meredith drives at a speed of 32 miles per hour, and she is currently out of gas. A gallon of gas costs $\$2.15$, and her car go

es 27 miles per gallon of gas. If she has $\$17.20$ to pay for gas, for how many hours can she drive?
Mathematics
2 answers:
Fudgin [204]3 years ago
4 0

Answer:

Step-by-step explanation:

2.15 per gallon

27 miles per gallon

has 17.20

drives at speed of 32 per hour

how many hours can she drive

First we need to find how much gas she can buy. To do that, we need to solve:

\frac{17.20}{2.15} , which equals  8

So now we know that she can buy 8 gallons of gas.

Now we need to find out how much gas she will use going 27 miles

27 x 8 = 216

she can go 216 miles on 8 gallons of gas.

Now for the fun part

\frac{216}{32} = 6.75

So, I believe she can drive for 6 hours and 45 minutes.

Let me know if incorrect.

Hope this helps :)

a_sh-v [17]3 years ago
3 0

Answer:

6 hours and 45 minutes

Step-by-step explanation:

First, we need to find out how many gallons she can buy. Since we know she has $17.20, and 1 gallon = $2.15, we can divide 17.20/2.15 to get 8 gallons. Now, since we <em>also </em>know that her car goes 27 miles per gallon, we have to multiply 27 by 8 to get 216, which means she can drive for 216 hours in total. Since Meredith drives 32 miles per hour, we can divide 216 by 32 to get 6.75 hours, which is 6 hours and 45 minutes.

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Answer:

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Step-by-step explanation:

The process of simplifying an expression generally involves eliminating parentheses, collecting terms, simplifying fractions, rationalizing denominators, removing appropriate factors from under radicals, and generally putting expressions into general form.

So, to accomplish what the problem is asking, you can create an expression that requires you accomplish as many or as few of these steps as you may like.

In general, you will want to both "do" and "undo" whatever operation(s) you may choose. For example, consider a couple of expressions "a" and "b". For this example, we want our final result to simplify to "a".

If we choose to add "b", we might have ...

  (a+b) -b . . . . . we have both added and subtracted b, so the simplified result is "a"

If we choose to multiply by "b", we might have ...

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We can solve both these problems at once by using ...

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