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# Tutor profile: Robert N.

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Robert N.
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## Questions

### Subject:Pre-Algebra

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Question:

Consider the equation $$5x - x = -12$$. Solve the equation for x.

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Robert N.

The original equation is $$5x - x = -12$$. We can simply this to $$4x = -12$$. Next, we can divide 4 by both sides $$4/4x = -12/4$$ and get $$x = -3$$. This is our answer, but we can double check it by plugging $$x = -3$$ back into the original equation. $$5(-3) - (-3) = -12$$ simplify $$-15 + 3 = -12$$ simplify again $$-12 = -12$$. This is true, so we know that $$x = -3$$ is correct!

### Subject:Basic Math

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Question:

John's new car gets him 25 miles per gallon. If John's car has a full tank of 14 gallons of gas, How far can John drive?

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Robert N.

We want to determine number of miles. For every gallon of gas, John can drive 25 miles. Therefore, we can multiply the gallons of gas he has by 25 to determine how far he can go. $$distance = 25 miles/gallon * 14 gallons = 350 miles$$. So we know John can drive 350 miles before he runs out of gas.

### Subject:Calculus

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Question:

The curve $$x = 2y - y^{2}$$ is above the curve $$x = y^{2} - 4y$$. Determine the area between these two curves using integration.

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Robert N.

First we must find where these curves intersects by setting them equal to each other. Therefore, $$2y - y^{2} = y^{2} - 4y$$. Next, we move all terms to one side, $$2y^{2} - 6y = 0$$. Finally, we can simply and solve for y $$2y(y - 3) = 0$$. Therefore, $$2y = 0$$ and $$y - 3 = 0$$. Now we know $$y = 0$$ and $$y = 3$$ are our intersections and hence we should take the integral in the interval $$0 <= y <= 3$$. To find the integral we must subtract the top curve by the bottom leaving $$\int_0^3\mathrm{(2y - y^{2}) - (y^{2} - 4y)},\mathrm{d}y$$. Therfore, the area A = $$\int_0^3\mathrm{(6y - 2y^{2})},\mathrm{d}y$$ = $$[3y^{2} - 2/3y^{3}]_0^3$$ = $$[3(3^{2}) - 2/3(3^{3})] - [3(0^{2}) - 2/3(0^{3})]$$ = [27 - 18] - [0 - 0] = 9. Therefore, the area between these two curves is 9.

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