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Tutor profile: Raj T.

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Raj T.
3rd Year at UC Berkeley, 3 Years of Tutoring Experience
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Questions

Subject: SAT

TutorMe
Question:

$(x^2 + y^2 = 153$)Given the above equation and that $$y = -4x$$, what is the value of $$x^2$$?

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Raj T.
Answer:

Substituting $$y = -4x$$ into the first equation, we have:$(x^2 + (-4x)^2 = x^2+ 16x^2 = 153$)Now, simplifying, we get:$(17x^2 = 153$)Dividing by $$17$$ on both sides, we get: $(x^2 = 9$)At this point, you might be tempted to go ahead and fully solve for $$x$$. However, on the SAT, it's important to keep in mind what the question is actually asking. The question asks for the value of $$x^2$$, which we already know to be $$9$$, so we are done. The answer is $$x^2 = 17$$.

Subject: Calculus

TutorMe
Question:

The position of an object at any time $$t$$ is given by $(s(t) = 3t^4 - 40t^3 + 126t^2 - 9$)a) Determine the velocity of the object at any time $$t$$. b) At what times, if any, will the object not be moving? c) Determine the acceleration of the object at any time $$t$$.

Inactive
Raj T.
Answer:

a) A function for the velocity of the object, $$v(t)$$, is the same as the derivative of the position function, or $$s(t)$$. $(v(t) = s'(t) = \frac{d}{dt} (3t^4 - 40t^3 + 126t^2 - 9)$)Now, applying the power rule to each individual term of $$s(t)$$, we get: $(v(t) = 12t^3 - 120t^2 + 252t$)b) In order for the object to not be moving, we need the velocity to be zero. Thus, by setting $$v(t)$$ equal to $$0$$ and solving, we can find the times when the object is stationary. $(v(t) = 12t^3 - 120t^2 + 252t = 0$)Factoring, we get: $(0 = 12t(t-3)(t-7)$)Now, applying the zero product rule, we have either $$12t = 0$$, $$t-3 = 0$$, or $$t-7 = 0$$. Solving, we get that the object is not moving at $$t = 0, 3, 7$$. c) Given the velocity function, the acceleration of the given object is just the derivative of this function. $(a(t) = v'(t) = \frac{d}{dt} (12t^3 - 120t^2 + 252t)$) Applying the power rule once again, we get: $(a(t) = 36t^2 - 240t + 252$)

Subject: Algebra

TutorMe
Question:

The sum of $$3$$ times a number and $$2$$ less than $$4$$ times the same number is equal to $$61$$. Write an equation and solve to determine the value of this unknown number.

Inactive
Raj T.
Answer:

In this problem, we are trying to find the value of an unknown number, which we can do by setting up an equation. Calling the unknown number $$x$$, we can use the problem statement to write an equation. We are given that the sum of $$3$$ times a number, or $$3x$$, and $$2$$ less than $$4$$ times the same number, or $$4x-2$$, is $$61$$. Then, we can write the following equation: $(3x + (4x - 2) = 61$)Combining like terms, we have: $(7x - 2 = 61$)Now, in order to solve for our unknown, we need to isolate $$x$$. We can do this by first adding $$2$$ to both sides, giving us: $(7x = 63$) Dividing by $$7$$ on both sides next, we get: $(x = 9$)Thus, the value of the unknown number we are looking for is $$9$$.

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