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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

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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.

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

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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.

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

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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.

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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