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# Tutor profile: Kelly V.

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Kelly V.
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## Questions

### Subject:Geometry

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Question:

Find the area of the equilateral triangle inscribed in a circle with a radius of 4 units.

Inactive
Kelly V.

Side note: This problem will be easier to solve if you draw a picture of the information as we go through the problem. Prerequisite knowledge: For any equilateral triangle: 1. The radius of the incircle$$=\frac{side}{2\sqrt{3}}$$ 2. The radius of the circumcircle$$=\frac{side}{\sqrt{3}}$$ Since we are given the radius of the circumcircle (the triangle is inscribed in the circle), we will use #2 to find the area of the equilateral triangle. $$4=\frac{s}{\sqrt{3}}$$ $$4\sqrt{3}=s$$ $$area=\frac{1}{2}bh$$ We now know the base, but we still have to solve for the height. The height of the triangle will divide the base in half and will, by definition, be perpendicular to the base. Therefore, we can solve for the height using the Pythagorean Theorem. $$a^{2}+b^{2}=c^{2}$$ We know the measures of the $$a^{2}$$ and $$c^{2}$$, so we can solve for the height, $$b^{2}$$. $$(2\sqrt{3})^{2}+b^{2}=(4\sqrt{3})^{2}$$ $$(4*3)+b^{2}=16*3$$ $$12+b^{2}=48$$ $$b^{2}=36$$ $$b=6$$ Now find the area. $$A=\frac{1}{2}bh$$ $$A=\frac{1}{2}(4\sqrt{3})(6)$$ $$A=12\sqrt{3}$$

### Subject:Calculus

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Question:

Find the derivative: $$f(x)=x^{4}+3x^{3}-16x^{2}+225x-15$$

Inactive
Kelly V.

$$f'(x)=4x^{3}-9x^{2}-32x+225$$

### Subject:Algebra

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Question:

Kelly is older than Parker. Next year Parker will be exactly half Kelly's age. Six years ago Kelly was three times as old as Parker. How old are Parker and Kelly?

Inactive
Kelly V.

Step one: Write the equations given in the problem. $$P+1=\frac{1}{2}(K+1)$$ $$K-6=3(P-6)$$ Step two: Solve one of the equations for one variable in order to substitute. $$K-6=3(P-6)$$ $$K-6=3P-18$$ $$K=3P-12$$ Step three: Substitute the final equation from step two into the original equation not solved in step two. Then solve. $$P+1=\frac{1}{2}(K+1)$$ $$P+1=\frac{1}{2}((3P-12)+1)$$ $$P+1=\frac{1}{2}(3P-11)$$ $$P+1=\frac{3}{2}P-\frac{11}{2}$$ $$\frac{13}{2}=\frac{1}{2}P$$ $$13=P$$ Step four: Substitute the answer from step three into the final equation from step two to solve for the final variable. $$K=3P-12$$ $$K=3(13)-12$$ $$K=39-12$$ $$K=27$$ Final Answer: Parker is 13 and Kelly is 27.

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