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Tutor profile: Rachael A.

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Rachael A.
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Questions

Subject: Set Theory

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Question:

Let $$R = \{a,b,c \}, S = \{ a,b,d \}, T = \{ e,f \}$$. List the elements in the set $$ (R \cup S) \cap (R \cup T)$$.

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Rachael A.
Answer:

SETUP: If we are asked to find the set $$ (R \cap S) \cup (R \cap T)$$, this means we are asked to find the union of the set $$(R \cap S)$$ with the set $$(R \cap T)$$. In order to do this, I first need to find the elements of the sets $$(R \cap S)$$ and $$(R \cap T)$$. That means that step one is to find the set $$(R \cap S)$$, step two is to find the set $$(R \cap T)$$, and step three is to find the union. STEP ONE: Find $$ (R \cap S) $$. By definition of intersection, the elements of $$ (R \cap S)$$ are all the elements that appearing in both $$R$$ and $$S$$. I start by looking at the elements in $$R$$ and check to see if they are also in $$S$$. The element $$a$$ appears in $$R$$ and also in $$S$$, the element $$b$$ appears in $$R$$ and also in $$S$$, but the element $$c$$ appearing in $$R$$ does not appear in $$S$$. Therefore the only elements appearing in both $$R$$ and $$S$$ are $$a$$ and $$b$$. This means $$ (R \cap S) = \{ a, b\}$$. STEP TWO: Find $$ (R \cap T) $$. By definition of intersection, the elements of $$ (R \cap T) $$ are those appearing in both $$R$$ and $$T$$. There are no elements in $$R$$ that appear in $$T$$ so this set is empty. In other words, $$ (R \cap T) = \emptyset$$. STEP THREE: Find $$ (R \cap S) \cup (R \cap T)$$. It is easiest to start by listing the elements of $$ (R \cap S)$$ and $$(R \cap T)$$ next to each other, so we can use this to compare. We can do this because of the work we did in steps one and two. $$ (R \cap S) = \{ a, b\}$$. $$ (R \cap T) = \emptyset$$. We are asked to find the union of these two sets. By definition of union, the elements in this set are all the elements that appear in either $$ (R \cap S)$$ or $$ (R \cap T)$$. Since $$a$$ and $$b$$ are in $$ (R \cap S)$$, they will be elements in our set. We next add in all the elements of $$ (R \cap T)$$. Since $$ (R \cap T)$$ is empty, there are no more elements that will be in our set. Therefore $$ (R \cap S) \cup (R \cap T) = \{ a, b\}$$.

Subject: Discrete Math

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Question:

An urn contains balls numbered 1 through 9. Suppose that you draw a ball from the urn, observe its number, do notreplace the ball, draw again. You repeat this until you have drawn a ball four times. What's the number of possible ways you can have performed this task?

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Rachael A.
Answer:

There are four actions: to draw the first ball, to draw the second ball, to draw the third ball, to draw the fourth ball. On the first draw, there are still 9 options. However, on the second draw, because you kept your first ball, you only have 8 balls left in the urn to choose from, so there are only 8 options. The same thing holds for the next two draws, in that there is one less ball to choose from. Therefore the total number of possible outcomes is $$9 \cdot 8 \cdot 7 \cdot 6$$.

Subject: Calculus

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Question:

Use the comparison test to determine whether the following integral diverges or converges: $$\int_{0}^1 ln(x^{1/x})dx $$

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Rachael A.
Answer:

The integral diverges. Note first that $$\int_{0}^1 ln(x^{1/x})dx = \int_{0}^1 \frac{ln(x)}{x} dx \leq \int_0^1 ln(x)dx$$ since a negative value times a number larger than 1 is a larger negative value. But \begin{align*} \int_0^1 ln(x)dx &= \lim_{t \rightarrow 0} x ln(x) - x |_{t}^1 \\ &= 0 ln(1) - 1 - \lim_{t \rightarrow 0} t ln(t) - t \\ &= -1 - \lim_{t \rightarrow 0} \frac{ln(t)}{t^{-1}} - 0 \\ &= -1 - \lim_{t \rightarrow 0} \frac{1/t}{-t^{-2}} = - \infty \end{align*} by L'Hospital's rule.

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