Enable contrast version

# Tutor profile: Jenel F.

Inactive
Jenel F.
STEM Tutor and Undergraduate Student Researcher
Tutor Satisfaction Guarantee

## Questions

### Subject:Pre-Calculus

TutorMe
Question:

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

Inactive
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

TutorMe
Question:

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

Inactive
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

TutorMe
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?

Inactive
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.

## Contact tutor

Send a message explaining your
needs and Jenel will reply soon.
Contact Jenel

## Request lesson

Ready now? Request a lesson.
Start Lesson

## FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage
Made in California by Zovio
© 2013 - 2021 TutorMe, LLC
High Contrast Mode
On
Off