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Tutor profile: Emma D.

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Emma D.
Graduate Student with Undergraduate Degree in Mathematics
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Questions

Subject: Microsoft Excel

TutorMe
Question:

What are two ways to find the average of 2 and 4 using excel formulas?

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Emma D.
Answer:

=sum(2,4)/2 =mean(2,4)

Subject: Calculus

TutorMe
Question:

Find the derivative of $x^2$. $$ f(x) = x^2 $$

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Emma D.
Answer:

We can solve for $$f(x + \Delta x)$$, or x + the change in x. We will start with the slope formula, $$\frac{\Delta y}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta x}$$ Since we are solving for the derivative of $$x^2$$, we will take $$f(x+\Delta x) = (x + \Delta x)^2$$. We can multiply this out. $$(x+\Delta x)^2 = (x+\Delta x) \cdot (x + \Delta x)$$ $$ = x^2 + 2x \Delta x + (\Delta x)^2$$ We can now plug in what we have for $$f(x)$$ and $$f(x + \Delta x)$$ into the slope formula and solve. $$\frac{\Delta y}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta x}$$ $$\frac{x^2 + 2x \Delta x + (\Delta x)^2 - x^2}{\Delta x}$$ $$\frac{2x \Delta x + (\Delta x)^2}{\Delta x}$$ $$2x + \Delta x$$ As $$\Delta x$$ approaches 0, we will be left with just $$2x$$.

Subject: Algebra

TutorMe
Question:

Solve for x and y. $$ x + 3y = 110 $$ $$ 2x + y = 120 $$

Inactive
Emma D.
Answer:

Subtract 3y from the left side of the first equation to isolate x, and subtract y from the second equation to isolate 2x. $$ x = 110 - 3y$$ $$ 2x = 120 - y$$ Divide the second equation by 2 on the left and right side to isolate x. $$x = 110 - 3y$$ $$ \frac{1}{2}\cdot x = \frac{1}{2} ( 120 - y ) $$ Since x = x, we can say that the two equations on their right sides are now equal. Set the two equations equal to eachother. $$x = 110 - 3y = x = 60 - \frac{1}{2}y$$ $$110 - 3y = 60 - \frac{1}{2} $$ Subtract 60 from both sides, and add 3y to both sides. $$110 - 3y - 60 + 3y = 60 - \frac{1}{2} - 60 + 3y$$ $$50 = \frac{5}{2}y$$ Divide both sides by the coefficient in front of y. Since we have \frac{5}{2}, we will multiply both sides by \frac{2}{5}. $$\frac{2}{5}\cdot 50 = \frac{2}{5} \cdot \frac{5}{2} y$$ $$\frac{100}{5} = \frac{10}{10}y$$ $$20 = y$$ Now that we have y = 20, let's plug that in to either of our original equations and solve for x. $$x + 3(20) = 110$$ $$x = 110 - 3(20)$$ $$x = 110 - 60$$ $$x = 50$$. x = 50. We can plug into the second equation to double check. $$ 2x + 20 = 120$$ $$2x = 120 - 20$$ $$2x = 100$$ $$ \frac{1}{2} \cdot 2x = \frac{1}{2} \cdot 100$$ $$ x = 50$$

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