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# Tutor profile: Kevin F.

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Kevin F.
American Math Tutor with B.Sc in Pure & Applied Mathematics from UCF
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## Questions

### Subject:Trigonometry

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Question:

Convert radians into degrees for the below number: $$\frac{5\pi}{6}$$ radians

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Kevin F.

Remember that radians and degrees are just different units of measurements for angles, and that 2$$\pi$$ radians is actually equal to 360 degrees, a full circle. There are many ways to solve this, but one easy trick we will use is that we know: $$\frac{2\pi}{360} = 1$$ and $$\frac{360}{2\pi} = 1$$ since these are the angles of a full circle. Using this trick will allow us to switch our units of measurement, and in fact, physicists do this type of trick all the time with inches, miles, kilograms, pounds, etc. If we decide to multiply $$\frac{5\pi}{6}$$ and $$\frac{360}{2\pi}$$ together, the radians will disappear and we will be left with our answer: $$(\frac{5\pi}{6} )( \frac{360}{2\pi} ) = \frac{1800}{12} = 150$$ degrees

### Subject:Calculus

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Question:

Compute the below integral: $$\int_{-11}^{0} 4k+sin(k) \,dk$$

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Kevin F.

$$\int_{-11}^{0} 4k+sin(k) \,dk$$ $$= \frac{4k^2}{2} - cos(k)$$ (remember the integral of sine) $$= 2k^2 - cos(k)$$ $$= - 1 - (242-cos(-11))$$ $$= - 243-cos(11)$$ (recall the even-odd identity of cosine)

### Subject:Algebra

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Question:

Solve this equation for h: $$15h^{2} = -3$$

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Kevin F.

First, we want to get $$h$$ alone, so we will divide both sides by 15 which will simply the left-hand side: $$75h^{2} = -3 \implies h^{2} = \frac{-3}{75} \implies h^{2} = \frac{-1}{25}$$ We still need to take the square-root of $$h^{2}$$ to get $$h$$ alone, so we will square-root both sides: $$h = \sqrt{\frac{-1}{25}}$$ We also recall that $$\sqrt{\frac{m}{n}} = \frac{\sqrt{m}}{\sqrt{n}}$$, so we can actually write: $$h = \sqrt{\frac{-1}{25}} \implies h = \frac{\sqrt{-1}}{\sqrt{25}}$$ The square-root is 25 is definitely 5, but what is the square-root of -1? You may remember that this is equal to the imaginary number $$i$$. There is no real number that you can square to get -1, but mathematicians still wanted to assign something to $$\sqrt{-1}$$, so they invented the imaginary number. And this is used all the time in electricity and signal processing in very cool ways. Even fractals have been made by the peculiarities of $$i$$. So now we have: $$h = \frac{\sqrt{-1}}{\sqrt{25}} \implies h = \frac{\pm i}{5}$$ , and $$i$$ is plus-or-minus because negative $$i$$ times negative $$i$$ equals -1, the same reason that a negative times a negative equals a positive. So, $$h$$ is solved! :) I can explain this a lot more playfully on video. :)

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