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Tutor profile: Adam K.

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Adam K.
Tutor for 4+ years
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Questions

Subject: Trigonometry

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Question:

1. Convert the following to radians: 45 degrees 2. Find all six trigonometric functions of the angle measure.

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Adam K.
Answer:

To convert, multiple 45 and pi then divide it by 180. The answer is $$ \frac{45\pi} {180} $$. Finally simplify to $$ \frac{\pi} {4} $$. Next step is to find the following: $$sin(\frac{\pi} {4}) = \frac{\sqrt2} {2} $$ $$cos(\frac{\pi} {4}) = \frac{\sqrt2} {2} $$ $$tan(\frac{\pi} {4}) = 1 $$ $$csc(\frac{\pi} {4}) = \sqrt2$$ $$sec(\frac{\pi} {4}) = \sqrt2$$ $$cot(\frac{\pi} {4}) = 1$$

Subject: Calculus

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Question:

Find the derivative to the following equation: $$ f(x) = \sqrt x^3e^x $$

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Adam K.
Answer:

The first step to do when finding the derivative is to evaluate which rules need to be used. The rule for this problem will be the product rule. Remember that the product rules states: If $$ f(x) = g(x) * h(x), then f'(x) = g'(x)*h(x) + h'(x) * g(x) $$ $$ g(x) = \sqrt {x^3} $$ and $$ h(x) = e^x $$ $$ g'(x) =\frac{3\sqrt x }{2} $$ and $$ h'(x) = e^x $$ $$ f'(x) = \frac{3\sqrt x }{2}*e^x + \sqrt {x^3}*e^x$$ After simplification, $$ f'(x) = e^x(\frac{3\sqrt x }{2} + \sqrt {x^3}) $$

Subject: Algebra

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Question:

A boy, who is currently 5 foot 1 inch, is growing at a rate of .75 inches per year. His growth spurt only lasted for 4 years. At the end of his growth spurt, how tall was the boy, in inches?

Inactive
Adam K.
Answer:

Since the final answer will be in inches, first convert 5 foot 1 inch to 61 inches. This is the boy's initial height prior to the growth spurt. To find out how much the boy grew, multiply .75 inches per year and 4 years to get 3 inches. Finally, add 3 inches to the 61 inches to show that the boy is now 64 inches tall.

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