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Tutor profile: Nigel C.

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Nigel C.
Adjunct Mathematics Instructor with 4 years of tutoring experience
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Questions

Subject: Pre-Calculus

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Question:

Solve for x: $$3 = log_4(x)$$

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Nigel C.
Answer:

Well we have a property that says: $$y = log_b(x)$$ <==> $$x = b^y$$ so $$3 = log_4(x)$$ <==> $$x = 4^3$$ ==> $$x = 64$$

Subject: Calculus

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Question:

Find the derivative of 5x^2

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Nigel C.
Answer:

to find the derivative of 5x^2: move the 2 to the front of the "x" and multiply with the 5 to get 10 then subtract 1 from 2 to get the new exponent of "x" which is 1 thus we get 10x^1 = 10x

Subject: Algebra

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Question:

Word problem: John has 100ft of copper wire that can be used to create a rectangular fence to keep his cow in. If the length of the fence is 20ft longer than the width. How long should the length and width be? What is the area of the fence?

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Nigel C.
Answer:

First we need to define our length and width. Let "x" be the width and "y" be the length Now the length is said to be 20ft more than the width so we can write the length in terms of the width I.e.: y = x + 20 Now we need to figure of the dimension of the fence. (hint: the perimeter is length + length + width + width) now our perimeter which is the amount of the copper wire we have is 100ft so: 100 = length + length + width + width 100 = (x + 20) + (x + 20) + x + x now we combine our like terms 100 = 4x + 40 now we solve for "x" first by subtracting 40 from both sides of our equation 60 = 4x then divide both sides by 4 15 = x thus out width is 15ft and it follows that our length is "length" = x + 20 = 15 + 20 = 35 so length is 35 now to find our area of our fence (Hint: Area = Length * Width) so Area = 35 * 15 = 525ft^2

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