Enable contrast version

Tutor profile: Jiwoo C.

Inactive
Jiwoo C.
Computer Science major at Cornell University
Tutor Satisfaction Guarantee

Questions

Subject: Pre-Calculus

TutorMe
Question:

Let $$ f$$ be a function on the real numbers such that $$f(x-4) = 9x^3 + 1.$$ What is $$f(6)$$?

Inactive
Jiwoo C.
Answer:

We let $$x-4 = 6$$ and find $$ x = 10$$. Then, $$ f(6) = f(10 - 4) = 9(10)^3 + 1 = 9001$$.

Subject: Discrete Math

TutorMe
Question:

Prove that $$ x^2 \equiv y^2 \hspace{.02in} (\text{mod } p)$$ if and only if $$ x \equiv y \hspace{.02in} (\text{mod }p)$$ or $$ x \equiv -y \hspace{.02in} (\text{mod }p)$$ where $$p$$ is prime.

Inactive
Jiwoo C.
Answer:

We recall that $$ a \equiv b \hspace{.02in} (\text{mod } n)$$ if and only if $$n \mid (a-b)$$. Then, $$ x^2 \equiv y^2 \hspace{.02in} (\text{mod } p)$$ if and only if $$ p \mid (x^2-y^2)$$. By factoring the difference of squares, we get $$ x^2 \equiv y^2 \hspace{.02in} (\text{mod } p)$$ if and only if $$ p \mid (x+y)(x-y)$$. Since $$p$$ is prime, either $$ p \mid (x+y)$$ or $$ p \mid (x-y) $$. Using the same identity that we started this proof with, $$ p \mid (x+y)$$ if and only if $$ x \equiv y \hspace{.02in} (\text{mod } p)$$, and $$ p \mid (x-y)$$ if and only if $$ x \equiv -y \hspace{.02in} (\text{mod } p)$$. Therefore, we conclude that for a prime $$p$$, $$ x^2 \equiv y^2 \hspace{.02in} (\text{mod } p)$$ if and only if $$ x \equiv y \hspace{.02in} (\text{mod }p)$$ or $$ x \equiv -y \hspace{.02in} (\text{mod }p)$$ .

Subject: Algebra

TutorMe
Question:

Identify the degree of the polynomial $$ 21x^3 - 9x^2 + 63x - 27$$ and find its real zeroes.

Inactive
Jiwoo C.
Answer:

The degree is the value of the highest powered variable in the polynomial, so the degree of the given polynomial is 3. To find the zeroes, we can graph the function or factor the polynomial. We demonstrate factoring the polynomial: Next, we notice that both the first and the third terms are divisible by 7 and that both the second and third terms are divisible by 3, so we consider factoring by grouping. 21 and 9 are both divisible by 3, and 63 and 27 are both divisible by 9, and factoring those out of the coefficients give us $( 3(7x^3 - 3x^2) + 9(7x - 3) $) We also factor out $$ x^2 $$ from the first two terms $( 3x^2(7x - 3) + 9(7x - 3) $) And by grouping we get $( (3x^2 + 9)(7x - 3)$) Now by letting each of those terms equal to zero $( 3x^2 + 9 = 0, \quad 7x-3 = 0 $) we find $$ x^2 = -3$$ and $$x = \frac{3}{7}$$ so the only real solution to the given polynomial is $$x = \frac{3}{7}$$.

Contact tutor

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

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