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# Tutor profile: Tanner L.

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Tanner L.
Mechanical Engineering Student at NC State
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## Questions

### Subject:Trigonometry

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Question:

A right triangle has an angle angle of 23deg and a side length of 5.2m adjacent to it. Find the hypotenuse of the triangle.

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Tanner L.

This problem requires the use of the inverse cosine trigonometric function. We know that the cosine function can be written as such: $$\cos{\theta} = \frac{adjacent}{hypotenuse}$$ Plugging in what we know, we get the following. For the purposes of this problem, we'll refer to the hypotenuse length as $$H$$: $$\cos(23\deg) = \frac{5m}{H}$$ Rearranging the above equation to isolate $$H$$, we get the following: $$H = \frac{5m}{\cos(23\deg)}$$ Finally solving the above equations gives: $$H = 5.43m$$

### Subject:Mechanical Engineering

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Question:

Using a 2D x and y plane as a frame of reference, there are 3 forces that all act at the origin (0,0). One force of 50N acts at $$60^o$$ from the positive x-axis and another force of 30N acts at $$120^o$$ from the positive x-axis. Find the magnitude and direction of the third in order to maintain equilibrium.

Inactive
Tanner L.

The best way to approach this problem is to break the given forces into x and y vectors, then summing the forces in the x and y direction to find the required vectors to maintain equilibrium. First, lets break the given forces into vectors: Force 1 x-direction: $$(50N)\cos{60} = 25N$$ y-direction: $$(50N)\sin{60} = 43.3N$$ Force 2 x-direction: $$(30N)\cos{120} = -15N$$ y-direction: $$(30N)\sin{120} = 25.9N$$ Now summing the forces in the x and y direction gives: $$\sum{Fx +25N -15N}$$ $$Fx = 15N -25N$$ $$Fx = -10N$$ $$\sum{Fy +43.3N + 25.9N}$$ $$Fy = -43.3N - 25.9N$$ $$-69.2N$$ Now we just need to convert these x and y components into a magnitude with a direction: $$\vert F \vert = \sqrt{(-10N^2) + (-69.2N^2)}$$ $$\vert F \vert$$ = 69.92N $$\arctan({\frac{-69.2N}{-10N})} = 81.77\deg + 180\deg = 261.78\deg$$ *The reason for adding 180deg to our final angle is because our x and y components of our force vector show the vector acting in the third quadrant of the plane, therefore 81.77deg would not make sense.

### Subject:Calculus

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Question:

Find the following indefinite integral: $$\int\mathrm{6x }^{5} - 18x^2 + 7\mathrm{d}x$$

Inactive
Tanner L.

To make solving this integral easier, we can look at this as three separate integrals by breaking up each of the individual terms like so: $$\int\mathrm{6x }^{5}\mathrm{d}x - \int_0^\infty \mathrm18x^2\mathrm{d}x + \int_0^\infty7\mathrm{d}x$$ Analyzing the first integral, we simply raise the power by one and divide by the new value like so: $$\int\mathrm{6x }^{5}\mathrm{d}x = \frac{6x^6}{6} = x^6$$ Analyzing the second integral using the same method: $$-\int\mathrm{18x }^{2}\mathrm{d}x = \frac{-18x^3}{3} = -6x^3$$ The third term is an integer value with no variable, so we can simply add an $$x$$ to this term like so: $$\int\mathrm{7} \mathrm{d}x = 7x$$ Now, because this is an indefinite integral, we must also add a constant variable to the end of our solution to represent the possibility of a constant in our general solution. With this in mind, the final solution becomes: $$x^6 - 6x^3 + 7x + c$$

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