Jeremy D.

Princeton Student, Tutor for Six Years

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SAT

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Question:

If $$f(x)$$ is a second degree polynomial with integer coefficients, where $$f(0)=6, f(1)=4$$, and $$f(2)=0$$, what is the value of $$f(3)$$?

Jeremy D.

Answer:

Since we are given that $$f(x)$$ is a quadratic, we can write it as: $$f(x)=ax^2+bx+c$$, where $$a,b,c$$ are integers. From the initial values provided, we have the following equations: $$c=6$$ $$a+b+c=4$$ $$4a+2b+c=0$$ Solving for $$a$$ and $$b$$ gives us that $$a=-1,b=-1,c=6$$. Thus, $$f(3)=(-1)(3^2)-(3)+6=\boxed{-6}$$.

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Question:

Colleges often want to know: why do you want to attend our university?

Jeremy D.

Answer:

For me, I wanted to attend Princeton because of their focus on undergraduates' education, the research opportunities provided in their junior and senior theses, and their quality faculty, specifically in the math department. I was also intrigued by the Operations Research and Financial Engineering Department, which was a major unique to Princeton, and the classes in that department such as Probability and Stochastic Systems.

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Question:

In the real numbers, what is the the solution to the equation: $$2^{4x+3}=8^{5-2x}$$?

Jeremy D.

Answer:

We can rewrite the right side of the equation as: $$(2^3)^{5-2x}=2^{15-6x}$$, giving us: $$2^{4x+3}=2^{15-6x}$$. Taking the log base 2 of both sides, we get: $$4x+3=15-6x$$. Solving for $$x$$ gives that $$x=\frac{9}{5}$$.

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