Monica G.

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Pre-Calculus

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Question:

Prove that [cot(x)] / [csc(x)] = cos(x)

Monica G.

Answer:

First lets try to write everything in sines and cosines cot(x) = cos(x) / sin(x) csc(x) = 1 /sin(x) So now we have: [ cos(x) / sin(x) ] / [ 1 /sin(x) ] Since we have a fraction divided by a fraction we can multiply by the reciprocal [ cos(x) / sin(x) ] * [ sin(x) / 1 ] Our sin(x) would cancel out leaving us with [ cos(x) / 1 ] * [ 1 / 1 ] Which would be: cos(x) * 1 which is just cos(x) Which is what we wanted to prove.

Calculus

TutorMe

Question:

Find the lim [3x^2 + 4x] / [x] x-> 0

Monica G.

Answer:

We cannot plug in 1 into the problem because we would get a zero in he bottom giving us an indeterminate form. So lets see if we can split it up: If we do we have: lim [3x^2 /x ] + [4x / x] x->0 From here we can simplify So we have lim [3x + 4] x- >0 Now if we plug in zero we get..... 3(0) + 4 = 4 (Note we don't write the limit after plugging in the number) So lim [3x^2 + 4x] / [x] = 4 x-> 0

Algebra

TutorMe

Question:

Solve 3x + 4 = 19

Monica G.

Answer:

When we are asked to solve an equation we need to leave that variable alone. So first we need to move the 4. We will do this by subtracting it from both sides. 3x + 4 = 19 -4 -4 This will leave us with: 3x = 15 Now to get rid of the 3 we need to divide by it on both sides leaving us with..... x = 5

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