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Christian F.

First-year student at Princeton University

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Spanish

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Question:

Consider the sentence, “El mono come bananas todo los días”. Identify the verb and its tense and rewrite the sentence in the past imperfect tense.

Christian F.

Answer:

The verb in the sentence is "come," from the infinitive (infinitivo) "comer". This is in the present tense. To rewrite the sentence in the past imperfect tense (el pretérito imperfecto), we must change only the verb to the past imperfect tense. The past imperfect tense of "comer" is comía, so the sentence would read: "El mono comía bananas todo los días".

Calculus

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Question:

Find the equation of the tangent line to the curve $$2xy+y^{2}=5x-6$$ at $$x=1$$.

Christian F.

Answer:

To find the equation of the tangent line, we must first find the slope of the tangent line. The slope of the tangent line for a curve at a given point is equal to the first derivative of the curve at that same point. Thus, we must find the first derivative $$\frac{dy}{dx}$$ and substitute into it the given point. Next, we can use the point-slope form and the given point to find the equation of the tangent line. 1) Differentiating, using the power and product rules, we get: $$2x*\frac{dy}{dx}+y*2+2y*\frac{dy}{dx}=5$$. 2) We now want to solve for $$\frac{dy}{dx}$$, our first derivative. Thus, we keep all terms with $$\frac{dy}{dx}$$ to one side and all others to the other side: $$2x(\frac{dy}{dx})+2y(\frac{dy}{dx})=5-2y$$. Factoring out $$\frac{dy}{dx}$$ we get: $$\frac{dy}{dx}(2x+2y)=5-2y$$. Finally, dividing both sides by $$2x+2y$$ we get $$\frac{dy}{dx}=\frac{5-2y}{2x+2y}$$. 3) Now that we found the first derivative, we can substitute into it the given point at which we want the slope of the tangent line for the curve. But since we only have an $$x$$ value, we must first find the $$y$$ value by substituting $$x=1$$ into the original equation: $$2(1)y+y^{2}=5(1)-6$$, or $$2y+y^{2}=-1$$. Moving all terms to solve as a quadratic, we get $$y^{2}+2y+1=0$$. Factoring, we get: $$(y+1)^{2}=0$$. Using the zero product property, $$y+1=0$$, so $$y=-1$$. 4) Thus, the point we must substitute into the derivative is (1, -1). Doing this, we get: $$\frac{dy}{dx}=\frac{5-2(-1)}{2(1)+2(-1)}$$. Since we get a number over zero, the first derivative and the slope of the tangent line is undefined at this point. The equation of the tangent at this point will then be a vertical line of the form $$x=b$$, where $$b$$ is the $$x$$-coordinate of the given point. Thus, the equation of the tangent line would be $$x=1$$.

Algebra

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Question:

Find the value of: $$\sqrt{6+\sqrt{6+\sqrt{6+...}}}$$

Christian F.

Answer:

1) First, set the expression equal to a variable $$x$$: $$x$$= $$\sqrt{6+\sqrt{6+\sqrt{6+...}}}$$. 2) Next, since the square root repeats infinitely, and we've now set $$x$$ to be equal to it, the equation can be rewritten as $$x= \sqrt{6+x}$$. 3) Now we can solve this equation. Squaring both sides, we get: $$x^{2}= 6+x$$. 4) Next, subtract $$6+x$$ from both sides of the equation to be able to set it to zero and solve as a quadratic equation: $$x^{2}-x-6=0$$. 5) Factoring, we get: $$(x-3)(x+2)=0$$. By the zero product property, $$x-3=0$$ and $$x+2=0$$, which yields solutions $$x=3$$ and $$x=-2$$. 6) But since we squared a square root in step 3, we must check for extraneous solutions by substituting our solutions before we squared both sides of the equation. Substituting $$x=3$$ into $$x= \sqrt{6+x}$$, we see that this solution works: $$3= \sqrt{6+3}=\sqrt{9}=3$$. Substituting $$x=-2$$, we see that this solution does not work: $$-2\neq{\sqrt{6+(-2)}=\sqrt{4}=2}$$. Therefore, our only solution is $$x=3$$, and the value of the expression is 3.

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