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Ethan M.
BS in Mathematics with Aspirations of PHD
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Geometry
TutorMe
Question:

If you are given a right triangle in which the hypotenuse has a length of 5 units and another side is said to have the length of 3 units, what is the length (in units) of the remaining side of the triangle?

Ethan M.

This is simple enough to solve the Pythagoras' equation a^2+b^2=c^2 in which c is the length of the hypotenuse and a and b are the lengths of the other two sides. For the above mentioned triangle we have two sides of the triangle: the hypotenuse and one other. Hence b = 3 and c = 5. By plugging these into the equation we can solve for the remaining side. a^2+3^2 = 5^2 Therefore a = sqrt(5^2-3^2) by some simple algebraic manipulation. Hence a = sqrt(16) = 4 and the length of the final side is 4 units.

Calculus
TutorMe
Question:

Use the definition of derivative to find the derivative of the equation f(x) = 3x-9.

Ethan M.

First we need to know the definition of derivative. The derivative of f(x) with respect x is the function f'(x) and is defined as f^' (x)=lim┬(h→0)⁡〖(f(x+h)-f(x))/h〗 This means that the derivative of f(x) = 3x-9 will be as follows. f^' (x)=lim┬(h→0)⁡〖([3(x+h)-9]-(3x-9))/h〗 In this case f(x+h) simply means to take the value of f(x) evaluated at x+h. In the equation above that turned out to be 3(x+h)-9. After simplification of the above equation we are left with f^' (x)=lim┬(h→0)⁡3 And since f(x)=3 is not dependent upon h we result in f'(x) = 3.

Algebra
TutorMe
Question:

Solve for the roots of the given quadratic equations: (a) x^2-x-6 (b)3x^2-9x+2

Ethan M.

(a) The first step in solving this equation is to check if it can be factored. In this case I can since I can find two number that multiply to give me -6 and add to get -1. these particular numbers are -3 and 2. Therefore x^2-x-6 can be factored into (x-3)(x+2). Now we are looking for all be points in which the given equation returns zero. So, we want to set each of the factors x-3 and x+2 equal to zero and solve for x. x-3 = 0 and x+2=0. Now add three two both sides and subtract 2 from both sides of the equations respectively. Then we achieve the result x=3 and x=-2. Hence these are the roots of the equation. (b) For this particular equation finding a factorization is not quite as easy. In this case we can use a convenient equation that has been proven true by other mathematicians. This is called the quadratic equation. For any equation in the form Ax^2+Bx+C we can use the following equation to find the roots: [-B +/- sqrt(B^2-4AC)]/2A By plugging in numbers we get, [-(-9) +/- sqrt((-9)^2-4(3)(2))]/2(3) = [9 +/- sqrt(81-24)]/6 =[9 +/- sqrt(57)]/6 The symbol +/- simply means that you will perform both addition and subtraction and result in two possible answers. In this case [9+sqrt(57)]/6 =2.76 and [9-sqrt(57)]/6 =0.24 These answers are rounded to two decimal places.

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