Cj A.

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Pre-Calculus

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Question:

How do you find the limit of |x|/x as x approaches 0?

Cj A.

Answer:

For x<0, |x|/x=−x/x=−1 For x>0, |x|/x=x/x=1 Thus limx→0−|x|/x=−1 limx→0+|x|/x=1 So the limit does not exist.

Calculus

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Question:

How do you find the second derivative of x^2*y^2=1?

Cj A.

Answer:

First we need to find the first derivative of the function. We will do this using implicit differentiation. With this particular function we will use the product rule. 2xy^2+x^2(2y)(dydx)=0 Now subtract 2xy^2 from both sides x^2(2y)dydx=−2xy^2 Divide both sides by x^2(2y) dydx=−2xy2x^2(2y) Which simplifies to dydx=−yx Now for the second derivative we will use the quotient rule d^2ydx^2=(x(−1)dydx)−(1(−y))x^2 d^2ydx^2=−xdydx+yx^2=y−xdydxx^2 plug dydx into the right hand side d^2ydx^2=y−x(−yx)x^2=y+yx2=2yx^2 d^2ydx^2=2yx^2

Algebra

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Question:

Solve this system of equations 4x + y = 16 (1) 2x + 3y = 18 (2)

Cj A.

Answer:

First thing you want to do is create like terms in both equations. So go ahead and multiple equation (2) by 2 That's going to give you 4x + 6y = 36. Now we have the like term (4x) in both equations and we can subtract to temporarily cancel out x. Subtract (1) from (2) (4x-4x) + (6y-y) = (36-16) 5y = 20 Solve for y => y = 20/5 = 4. Now plug y=4 in Equation (1) 4x + 4 = 16 x = 12/4 = 3.

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