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Tutor profile: Jenel F.

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Jenel F.
STEM Tutor and Undergraduate Student Researcher
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Questions

Subject: Pre-Calculus

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Question:

Evaluate the limit: $$ \lim_{t\rightarrow 4}\frac{t^2-16}{t^2-3t-4} $$

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Jenel F.
Answer:

In this problem we can't directly plug in $$ t=4 $$ into the fraction because it will result in a zero denominator. We can simplify the fraction by factoring the polynomials. $$ \begin{align*} \frac{t^2-16}{t^2-3t-4}&=\frac{(t+4)(t-4)}{(t+1)(t-4)}\\ &=\frac{t+4}{t+1} \end{align*} $$ Now we can plug in $$ t=4 $$: $$ \begin{align*} \lim_{t\rightarrow 4}\frac{t^2-16}{t^2-3t-4}&=\lim_{t\rightarrow 4}\frac{t+4}{t+1}\\ &=\frac{4+4}{4+1}\\ &=\frac{8}{5} \end{align*} $$

Subject: Calculus

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Question:

Compute the integral: $$ \int x^5\ln(x)dx $$

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Jenel F.
Answer:

For this problem we need to use the method of integration by parts following this formula $$ \int udv=uv-\int vdu $$ Let $$ u=ln(x) $$, and $$ dv=x^5dx $$ Then $$ du=\frac{1}{x}dx $$, and $$ v=\frac{1}{6}x^6 $$ We substitute $$ u $$ and $$ v $$ into the formula above and integrate to get: $$ \begin{align*} \int x^5\ln(x)dx&=\frac{1}{6}x^6ln(x)-\int \frac{1}{6}x^6\frac{1}{x}dx\\ &=\frac{1}{6}x^6ln(x)-\frac{1}{6}\int x^5dx\\ &=\frac{1}{6}x^6ln(x)-\frac{1}{6}\left(\frac{1}{6}x^6\right)+C\\ &=\frac{1}{6}x^6ln(x)-\frac{1}{12}x^6+C\\ \end{align*} $$

Subject: Physics

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Question:

A ball is thrown horizontally from a height of 44.1 m with an initial velocity of 2 m/s. How long is the ball in the air?

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Jenel F.
Answer:

In this problem we are given the initial height $$ y_{i} = 44.1 m $$, and the initial velocity $$ v_{i} = 2 m/s $$, however, we are told in the problem that the ball is thrown horizontally, therefore $$ v_{iy} = 0 m/s $$ We know that gravity is constant $$ g = 9.8 m/s^2 $$ We also know that the ball will eventually land, and that means $$ y_{f} = 0 m $$ In order to find the final time we need to use one of the four kinematic equations that describe an object's projectile motion. The equation that best fits our knowns and unknowns is: $$ y_{f}-y_{i}=v_{iy}\Delta t-\frac{1}{2}g\Delta t^2 $$ We can plug in the numbers to get: $$ 0m-44.1m=(0m/s)\Delta t-\frac{1}{2}(9.8m/s^2)\Delta t^2 $$ We can simplify to get: $$ 44.1m=(4.9m/s^2)\Delta t^2 $$ Solving for $$ \Delta t $$ we get: $$ \Delta t=\sqrt{\frac{44.1m}{4.9m/s^2}}=3s $$ The ball is in the air for 3 seconds.

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