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# Tutor profile: Hannah M.

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Hannah M.
Algebra, Trigonometry, and Pre-Calculus Tutor
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## Questions

### Subject:Pre-Calculus

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Question:

A 30 meter wide river runs from west to east, and its current has a speed of 4 m/s. A man begins paddling a boat directly north at 3 m/s, starting from the south side of the river. a) How long does it take the man to cross the river? b) When he reaches the other side, how far down the shore will he be from his original position? c) What will the direction and magnitude of his velocity be as he travels across the river?

Inactive
Hannah M.

a) 10 seconds; while the current of the river will push the man sideways, he will still be traveling across the river at 3 m/s, so he will reach the other side in 10 seconds. b) 40 meters; because it took him 10 seconds to cross the river, and the current is acting on him with a speed of 4 m/s, he will end up 40 meters down the river on the other side. c) His motion can be modeled by the vector $$4i + 3j$$, since he is being pushed in the positive x direction (east) at a speed of 4 m/s, and he is paddling in the positive y direction (north) at a speed of 3 m/s. The magnitude of his velocity is the magnitude of this vector, which can be found using the pythagorean theorem. $$\sqrt{4^{2}+3^{2}}$$ = 5. The angle at which he is traveling is found by using the tangent function. The base of the triangle modeling his motion has length 4, and its height is 3. Therefore, the angle can be found using $$\tan^{-1}(\frac{3}{4})$$, which is equal to 36.870 degrees.

### Subject:Calculus

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Question:

Integrate $$\int_0^\pi x\sin(x)\mathrm{d}x$$ using integration by parts.

Inactive
Hannah M.

Integration by parts is useful when the equation being integrated is in the form $$\int udv$$, and both parts of the equation (in this case, $$x$$ and $$\sin(x)$$), can be easily integrated or have their derivative taken. The equation for integration by parts is $$\int udv = uv - \int vdu$$ . In this case, $$u = x$$, so $$du$$, or the derivative of u, is equal to $$1$$. $$dv$$, or the derivative of v, equals $$\sin(x)$$, so $$v = -\cos(x)$$. Now using the integration by parts formula, the new equation looks like this $$(x)(-\cos(x)) - \int (-\cos(x))(1)$$ or $$-x\cos(x) + \int \cos(x)$$ . The integral of $$\cos(x)$$ is $$\sin(x)$$, so now we are evaluating $$-x\cos(x) + \sin(x)$$ from 0 to $$\pi$$. $$-\pi\cos(\pi) + \sin(\pi) + 0\cos(0) - \sin(0)$$ $$= -\pi(-1) + 0 + 0 + 0$$ $$= -\pi$$

### Subject:Algebra

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Question:

See the equation $$\frac{x^{2}-5x+4}{x^{2}+2x-3}$$ . a) At which x-value is there a hole in the graph of this function? b) At which x-value is there a vertical asymptote in the graph of this function?

Inactive
Hannah M.

The first step to solving this problem is factoring the polynomials in the numerator and the denominator. After factoring the polynomials, the function looks like this: $$\frac{(x-1)(x-4)}{(x-1)(x+3)}$$ . a) A hole occurs in a function when both the numerator and the denominator are equal to 0. Because after factoring, the term $$x-1$$ is in both the numerator and denominator, we set $$x-1$$ equal to 0. When solving this, we find that $$x = 1$$, so a hole occurs at $$x = 1$$. b) A vertical asymptote occurs in a function when just the denominator is equal to 0. First, we set the denominator equal to zero and solve using the factored function and the zero product property. $$(x - 1)(x + 3) = 0$$ After solving, we find that the denominator is equal to 0 at $$x = 1$$ and $$x = -3$$. From part A, we already know that there is a hole at $$x = 1$$, so the vertical asymptote occurs at $$x = -3$$.

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