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# Tutor profile: Travis H.

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Travis H.
Tutor for 7+ years, professor at Sacramento State University
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## Questions

### Subject:Basic Math

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Question:

Suppose we had two medium pizzas, one pepperoni and the other Hawaiian. The pepperoni pizza was divided evenly into 6 slices, while the Hawaiian pizza into 4 slices. John ate two slices of pepperoni and one slice of Hawaiian. How much pizza did John end up eating?

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Travis H.

The first step to solving a problem is to give numerical values to certain parts of the problem. We know John ate two slices of a six-slice pizza, or $$\frac{2}{6}$$ the pizza. Similarly, we know John ate one slice of a four-slice pizza, or $$\frac{1}{4}$$ the pizza. How much pizza John ate is synonymous to asking what is the sum of $$\frac{2}{6}$$ and $$\frac{1}{4}$$. To add fractions, we need to have common denominators. Upon exploration, we find that our lowest common denominator is $$12$$. So we have \begin{align*} \frac{2}{6} + \frac{1}{4} &= \left (\frac{2}{6} \cdot 1 \right ) + \left (\frac{1}{4} \cdot 1 \right )\\ &= \left (\frac{2}{6}\cdot \frac{2}{2} \right )+ \left ( \frac{1}{4} \cdot \frac{3}{3} \right )\\ &= \frac{4}{12} + \frac{3}{12}\\ &= \frac{4 + 3}{12}\\ &= \frac{7}{12} \end{align*} We can safely and confidently conclude that John eating $$\frac{2}{6}$$ of one pizza and $$\frac{1}{4}$$ of another that he ate a total of $$\frac{7}{12}$$ of a pizza.

### Subject:Calculus

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Question:

Change-of-variables are often used in mathematics but certainly can feel daunting and confusing. Let's run through a classic example and see how it should feel. Suppose we are asked to evaluate the integral $$\int \sqrt{3x + 1} \, dx$$. How might we use a change of variable to solve?

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Travis H.

In this classic problem, a strong candidate for our $$u$$-substitution appears to be the $$3x + 1$$ underneath the radical. Let's see what happens when we use that for our substitution. First, we let $$u = 3x + 1$$, which leads to its derivative being $$du = 3 \, dx$$, which leads to $$\frac{1}{3} du = dx$$ Following through, our integral now is \begin{align*} \int \sqrt{3x + 1} \, dx &= \int \sqrt{u} \cdot \frac{1}{3} \, du \\ &= \frac{1}{3} \int u^{\frac{1}{2}} \, du \\ &= \frac{1}{3} \cdot \left (\frac{2}{3} u^\frac{3}{2} \right )\\ &= \frac{2}{9} u^{\frac{3}{2}}\\ &= \frac{2}{9} (3x +1)^{\frac{3}{2}} + C\\ \end{align*} It is always worth checking your answer by taking the derivative of your just-found solution. We hope to get back to where we started! If the derivative of our solution is the original integrand, then we've done everything correctly. To begin, we have \begin{align*} \frac{d}{dx} \bigg [\frac{2}{9} (3x +1)^{\frac{3}{2}} + C \bigg ] &= \frac{d}{dx} \bigg [\frac{2}{9} (3x +1)^{\frac{3}{2}} \bigg ] + \frac{d}{dx} \bigg [C \bigg ]\\ &= \frac{d}{dx} \bigg [\frac{2}{9} (3x +1)^{\frac{3}{2}} \bigg ] + 0\\ &= \frac{d}{dx} \bigg [\frac{2}{9} (3x +1)^{\frac{3}{2}} \bigg ]\\ &= \bigg [\frac{3}{2} \cdot \frac{2}{9} (3x +1)^{\frac{1}{2}} \cdot 3 \bigg ]\\ &= \bigg [\frac{18}{18} \cdot (3x +1)^{\frac{1}{2}} \bigg ]\\ & = (3x +1)^{\frac{1}{2}} \end{align*} Just as we wanted!

### Subject:Algebra

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Question:

Let's explore the common algebraic mistake: $$(x + y)^2 = x^2 + y^2$$. This is not true, but let's confirm why.

Inactive
Travis H.

Let's recall our properties of exponents. While it is true that $$(xy)^2 = x^2y^2$$ via the distributive property for exponents, it does not apply when there is a sum inside the parenthetical. We must recall that something raised to a power is a shorthand for repeated multiplication. That is, \begin{align*} (x+y)^2 &= (x+y)\cdot(x+y)\\ &= (x\cdot x) + (x\cdot y) + (y\cdot x) + (y\cdot y) \, \, \, \text{ (FOIL) }\\ &= x^2 + 2xy + y^2 \end{align*} So remember! While it may seem natural to bring those exponents down to each term, we must be delicate with our exponents and the properties that come with them.

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