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# Tutor profile: Justin W.

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Justin W.
Mechanical Engineering Student at Brigham Young University
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## Questions

### Subject:Spanish

TutorMe
Question:

Translate the following: Lisa went out to purchase tomatoes. On the way she met Juan. He was collecting coins on the ground. Lisa asked him, "What are you doing here?" Juan replied, "I'm looking for my lucky peso; I lost it and can't find it anywhere."

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Justin W.

Lisa se salió para comprar tomates. En camino se encontró con Juan. El estaba recogiendo monedas de la tierra. Lisa le preguntó, "¿Que estás haciendo aquí?" Respondió Juan, "Estoy buscando por mi peso de suerte. Se me lo perdió y no lo puedo encontrar en ningún lado."

### Subject:Calculus

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Question:

What is the area of the region bounded by the x axis, x = pi/4, and y = -sin(x)?

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Justin W.

This is an area problem, which means integration!!! Yay!! So we start off by sketching our region to get an idea of what's going on. We notice that y = 0 (the x axis) is "above" y = -sin(x), so we know that our equation we need to integrate is 0 - (-sin(x)) = sin(x). Sweet! Now we just need our bounds of integration! To do that we notice that -sin(x) and the x axis intersect at x = 0 (which we can obtain by setting y = 0 equal to y = -sin(x) and solving for x when it's not so obvious). And we note that the line x = pi/4 provides our bound on the right side. We then get to integrate! The integral of sin(x) is -cos(x) + C. We then use the First Fundamental Theorem of Calculus and solve for the area. A = -cos(pi/4) - (-cos(0)) = -√(2)/2+ 1 ≈ 0.293

### Subject:Algebra

TutorMe
Question:

Find the roots of the equation y = x^2 - 4x + 1.

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Justin W.

First we notice that we cannot factor the right side of the equation. Thus we have two options, complete the square or use the quadratic equation. I'll demonstrate both. To use the method of completing the square we find the number that when added with x^2 - 4x will produce a term of the form (x + a)^2 when factored. In this case that number is 4, so we add 4 to both sides giving us: y + 4 = (x^2 - 4x + 4) +1. (Remember we don't want to add the 1 and 4 together. That way we can factor our equation.) By factoring we get y + 4 = (x - 2)^2 +1. We then subtract 1 from both sides and our equation becomes: y + 3 = (x - 2)^2. To find the roots we now need to set y = 0. Our equation now becomes 3 = (x - 2)^2. Now we solve for x by taking the square root of both sides. Thus ±√(3) = x - 2. (Remember that when we square root both sides we must include the plus or minus!!) By adding 2 to both sides we get x = 2 + √(3) and x = 2 - √(3) as the roots. With the method of the quadratic equation we must identify a, b, and c. The coefficient of the x^2 term is 1, therefore a = 1; the coefficient of the x term is -4, therefore b = -4; what is left is the c term, therefore c = 1. We then recall that the quadratic equation is x = [-b ±√{b^2 - 4ac}]/{2a} and we substitute in our values of a, b, and c to get x = [-(-4) ±√{(-4)^2 - 4(1)(1)}]/{2(1)}. When we simplify we get x = [4 ±√{16 - 4}]/{2} = 2 ±[√{12}]/2 = 2 ± √(3) which is the same as above!! Yay!

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