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# Tutor profile: Natalie O.

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Natalie O.
UC Berkeley graduate and data engineer
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## Questions

### Subject:GRE

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Question:

For some positive integers $$a$$, $$b$$, and $$c$$: if $$a$$ is a factor of $$b$$, and $$b$$ is a factor of $$c$$, which of the following statements must be true (choose all that apply)? A. $$a^2$$ is a factor of $$c^2$$ B. $$b-2$$ is a factor of $$c-2$$ C. $$a$$ is a factor of $$bc$$

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Natalie O.

If $$a$$ is a factor of $$b$$, then we can come up with some other number d such that $$ad = b$$. Likewise, $$b$$ is a factor of $$c$$, we can come up with some $$e$$ such that $$c = be$$. Since in A. we are trying to find a relationship between $$a$$ and $$c$$, we can put $$c$$ in terms of $$a$$ by substituting $$ad$$ for $$b$$: $$c = ade$$. Squaring both sides, we get $$c^2 = (ade)^2 = a^2d^2e^2$$. Since $$a^2$$ is clearly a factor of $$c^2$$. We can select A. as a correct answer. Unfortunately, B. can be trickier for a student less familiar with ideas about factors. Simply put, since we are neither multiplying nor dividing, we cannot be sure that subtracting two, even if from both $$c$$ and its factor, will allow $$b-2$$ to remain a factor of $$c-2$$. An easy way to test this on an exam is to plug in some "easy" numbers. If $$b=2$$ and $$c=6$$, $$c-2=4$$ and $$b-2=0$$, so B is false. If you're still unconvinced because 0 is a weird number, try $$b=11$$ and $$c=55$$. $$c-2=53$$ and $$b-2=9$$, 53 is not divisible by 9, reaffirming that this rule does not hold for all positive integers. For C., we know that $$a$$ is a factor of $$b$$, which is sufficient to prove that $$a$$ is a factor of $$b$$ times any positive integer. Therefore, C. is correct as well as A.

### Subject:Linguistics

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Question:

Gerunds occur when a verb gets a special derivational -ing suffix and is then allowed to appear in different parts of sentences than verbs normally can. Make an argument for what part of speech gerunds might belong to. Examples: $$(1)$$ Crucifying me is completely uncalled for. $$(2)$$ Eating crayons is/*are safe. $$(3)$$ There was no singing after that day. $$(4)$$ Not singing songs is boring.

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Natalie O.

I will argue that these words belong to a subgroup of nouns. Sentences $$(1)$$ and $$(2)$$ create the following pieces of evidence for and against: Evidence A: Can act as sentence subject. Evidence B: Must act as a singular subject regardless of the amount of items described. This indicates that the subject of a sentence like 2. is the action, not the crayons. Evidence C: Acts on nouns semantically like verbs but semantically unlike adjectives. Sentences $$(3)$$ and $$(4)$$ support the following pieces of evidence: Evidence D: Cannot be negated with no(t) in some contexts. Evidence E: Can take not to negate. Evidence for gerunds belonging to the noun class includes observations A and B. Traits that distinguish gerunds from our strict noun definition include C, D and E. If gerunds are nouns, they are being treated as mass nouns and not counting nouns, as we see expressed in the to be verb: $$(5)$$ *She very studies. $$(6)$$ She is very enticing. $$(7)$$ *She is very studying. $$(8)$$ She is studying a lot. A distinction needs to be made, however, between two misleadingly homophonic constructions one can run into here. The adjectival derivational suffix -ing can blur the line between gerund and adjective, but this example will also demonstrate why gerunds are not adjectives, at least semantically. The difference between these cases is both semantic and distributional, in that gerunds also do not typically occur where these adjectives do, as we see with the very case in $$(5)$$-$$(6)$$. I think the most direct method for solving this is a semantic argument. $$(6)$$ means literally that the nature of the object is that which lends to enticement. However, were one to reconstrue the meaning to resemble a different meaning, one that would occur from a gerund, along the lines of "she makes a great effort to entice people around her," this would be ungrammatical, and would resemble the construction in $$(5)$$. Clearly, these gerunds are not verbs, which is a bit superfluous as gerunds are necessarily derived from verbs, thereby distinguishing themselves from the category. Summarily, we can see the difference between a gerund’s verb origins and its new position, both distributionally and semantically, in the following sentence: $$(9)$$ I don’t like wisecracking, however I love to wisecrack. Lastly, gerunds are quite unlikely to be adverbs. Referencing the semantic difference between the clauses in example sentence $$(9)$$, and noting that it violates the rule that adverbs cannot immediately follow verbs, we can rule out adverbial distribution. By process of elimination, the only remaining candidate for gerund classification is noun. However, we must note that like the subclass of mass nouns, there are notable exceptions such as observations A and B. Therefore I postulate that gerunds are a subclass of nouns with an extremely similar but slightly different distribution.

### Subject:Calculus

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Question:

What is the solution to the indefinite integral $$\int\frac{x^{2}}{x^3+4}dx$$?

Inactive
Natalie O.

This question may look daunting at first, but in calculus, everything is about trying to simplify the work you're going to need to do. Luckily for us, many problems like this one allow for an easy technique called u-substitution. Let's split up the problem into $$\int\frac{1}{x^3+4}x^2dx$$. Now let's tackle that denominator by declaring that $$u=x^3+4$$. Since we are going to need it later, we can take the derivative of u with respect to x, that is, $$\frac{du}{dx} = 3x^2$$. While $$\frac{du}{dx}$$ is not technically a fraction, as a shorthand, we can get $$du$$ on its own by "multiplying" both sides of the equation by $$dx$$: $$du = 3x^2 dx$$. As it stands right now with our u-substitution, since $$u=x^3+4$$, we have $$\int\frac{1}{u}x^2dx$$. What a mess. In particular, we are integrating with respect to x over our u variable, which isn't doing anything for us! Clearly, it would help us if we could replace that $$dx$$ with a $$du$$ equivalent term. As you probably guessed, that is precisely what we were setting up for by finding $$du$$. But notice that our "leftover" term still in the integral isn't $$dx$$, but $$x^2 dx$$. How can we change our $$du$$ to look like that? Easy - divide it by 3. $$\frac{du}{3} = x^2 dx$$. A perfect fit. Now we can substitute out all of our $$x$$ terms in our integral and get $$\int\frac{1}{u}\frac{1}{3} du$$. That's easy enough to find an antiderivative of! Simplifying, we get $$\frac{1}{3}\int\frac{1}{u}du = \frac{1}{3}\ln |u| + C$$. What was $$u$$ again? Right, $$x^3+4$$. So our final answer to this indefinite integral question is $$\int\frac{x^{2}}{x^3+4}dx$$ = $$\frac{1}{3}\ln |x^3+4| + C$$

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