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Tutor profile: Nancy W.

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Nancy W.
Math and Test Prep Tutor
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Questions

Subject:SAT

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Question:

The length of a rectangle is decreased by 25% and the width is doubled. How does the area of the rectangle change?

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Nancy W.

The area of a rectangle is $$A = lw$$. Let's say the old rectangle has area $$A _{old} = l_{old} w_{old}$$ and the new triangle has area $$A_{new} = l_{new} w_{new}$$. The new length is 25% less than the old length: $$l_{new} = (1-0.25) l_{old} = 0.75 l_{old}$$ And the new width is doubled: $$w_{new} = 2 w_{old}$$ Substituting, we find the area of the new triangle is $( A_{new} = l_{new} w_{new} = 0.75\cdot 2\cdot l_{old}w_{old}$)$( A_{new} = 1.5 A_{old}$) Thus we find that the new rectangle has an area 50% larger than that of the old rectangle.

Subject:Calculus

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Question:

Find the derivative of the function $( \frac{\partial}{\partial x} (x^2 + 8x + 16)$).

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Nancy W.

You may tackle each term separately, since the derivative operator acts linearly: $( \frac{\partial}{\partial x} x^2 + \frac{\partial}{\partial x} 8 x + \frac{\partial}{\partial x} 16$)$( 2x + 8$)You may also simplify into:$(2(x + 4)$)

Subject:Algebra

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Question:

Solve for x: $$x^2 + 8x + 16 = 0$$

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Nancy W.

The simple solution is to recall the formula $(a^2 + 2ab + b^2 = (a+b)^2$) where $$a = x, b = 4$$. Thus, $(x^2 + 2\cdot x \cdot 4 + 4^2 = (x + 4)^2 = 0$) $( x + 4 = 0$) So $$x = -4$$. A more lengthy solution would be to use the quadratic formula: $(x = \frac{-b \pm \sqrt{b^2 - 4 ac}}{2a}$) where $$a = 1, b = 8, c =16$$. This yields: $(x = \frac{-8 \pm \sqrt{64 - 4 *16}}{2}$) $(x = \frac{-8 \pm \sqrt{0}}{2}$) $(x = -4$)

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