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Tutor profile: Matthew S.

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Matthew S.
Math Tutor for two years
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Questions

Subject: Trigonometry

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Question:

Describe the transformations from the base function cos(x) of the function below 4cos(2x + pi) - 10

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Matthew S.
Answer:

First, look at the coefficient in front of the cosine. That is the amplitude 4 Second, look to the constant outside the cosine function, that is the vertical shift -10 Third, look inside the cosine, at the coefficient in front of the x. Since this is a cosine function, our normal period is 2pi, so we will have to divide that by the coefficient to find our new period 2pi/2 = pi Lastly, we look at the number inside the cosine being added to our x-term. That will tell us our phase shift. However, we must also divide this number by our x-coefficient. pi/2 = pi/2 This is a shift to the left, because our phase shift is positive. A negative would shift to the right. So Amplitude: 4 Vertical Shift: -10 Period: pi Phase Shift: pi/2 to the left

Subject: Calculus

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Question:

Take the indefinite integral of int[ 1 / (x^2+4x+4) ]dx

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Matthew S.
Answer:

First, factor the denominator int[ 1/(x+2)(x+2) ]dx then, simplify int[ 1/(x+2)^2 ]dx then, u substitution u = x+2 du = dx int[ 1/u^2 ] du simplify int[ u^-2 ]du integrate (u^-1)/-1 +C simplify -1/u +C return to x -1/x+2 +C

Subject: Algebra

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Question:

Solve the following quadratic equation by factoring: 5x^2 + 12x + 4

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Matthew S.
Answer:

First, multiply the coefficient in front of the x^2 (5), and the constant (4) 5 * 4 = 20 then, determine two sets of numbers that multiply to get the product (20), but add to get the coefficient in front of the x (12). 5 * 4 = 20, 5 + 4 = 9 [X] 10 * 2 = 20, 10 + 2 = 12 [O] divide the coefficient in front of the x (12) into both 10 and 2 5x^2 + 10x + 2x + 4 Now, try to factor both sets (5x^2 and 10x | 2x and 4) such that you get the same thing in the parenthesis 5x(x + 2) + 2(x + 2) Now copy the (x+2), and take the 5x and +2 and put it in its own set of parenthesis (x+2)(5x+2) solutions: x = -2, and -2/5 CHECK! notice, how if you distribute your factored form, you return to your original equation: FOIL: (x+2)(5x+2) = 5x^2 +2x +10x + 4 SIMPLIFY: 5x^2 +12x +4 also, you can plug the solutions back into the original equation, and your equation should equal zero: 5(-2)^2 + 12(-2) + 4 = 5*4 - 24 +4 = 20 -24 + 4 = 0 5(-2/5)^2 + 12(-2/5) + 4 = 5*4/25 - 24/5 + 4 = 4/5 - 24/5 + 20/5 = 0

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