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Tutor profile: Michael H.

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Michael H.
Jack of all Trades with an interest in spreading knowledge and education
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Questions

Subject: Microsoft Excel

TutorMe
Question:

After creating a chart, how can you change the data displayed?

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Michael H.
Answer:

Right click the chart and click "Select Data" -or- With the chart selected, click on the Design tab on the ribbon and click "Select Data" Either use the chart data range to select the data to display and allow excel to best guess how to display it, or manually add series one at a time and a list of categories for the horizontal axis.

Subject: Linear Algebra

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Question:

$$\textbf{A} = \bigg [\begin{matrix}-1 & 7 \\ 8 & -1 \end{matrix}\bigg]$$ $$\textbf{u} = \bigg [\begin{matrix}1 & 0 \\ 2 & 1 \end{matrix}\bigg] \,\, \textbf{v} = \bigg [\begin{matrix}2 & -3 \\ 0 & 2 \end{matrix}\bigg] \,\, \textbf{w} = \bigg [\begin{matrix}0 & 1 \\ 2 & 0 \end{matrix}\bigg]$$ If $$\textbf{A}$$ is a linear combination of $$\textbf{u},\textbf{v},$$ and $$\textbf{w}$$, then there exist $$\textbf{c}_1,\textbf{c}_2,$$ and $$\textbf{c}_3$$ such that $$\textbf{c}_1\textbf{u} + \textbf{c}_2\textbf{v} + \textbf{c}_3\textbf{w} = \textbf{A}$$ Find $$\textbf{c}_1,\textbf{c}_2,$$ and $$\textbf{c}_3$$ or prove that $$\textbf{A}$$ is not a linear combination of $$\textbf{u},\textbf{v},$$ and $$\textbf{w}$$.

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Michael H.
Answer:

Using scalar multiplication and matrix addition, we end up with the matrix $$\bigg [\begin{matrix}\textbf{c}_1 + 2\textbf{c}_2 & -3\textbf{c}_2 + \textbf{c}_3 \\ 2\textbf{c}_1 + 2\textbf{c}_2 & \textbf{c}_1 + 2\textbf{c}_2 \end{matrix}\bigg]$$ By setting this equal to $$\textbf{A}$$, we get the system of linear equations: $$\textbf{c}_1 + 2\textbf{c}_2 = -1 \\ -3\textbf{c}_2 + \textbf{c}_3 =7\\ 2\textbf{c}_1 + 2\textbf{c}_2 =8\\ \textbf{c}_1 + 2\textbf{c}_2=-1$$ By solving these equations simultaneously, the solution is $$\textbf{c}_1 = 3,\textbf{c}_2 = -2,$$ and $$\textbf{c}_3 = 1$$

Subject: Calculus

TutorMe
Question:

Walk through the steps to finding the volume of the shape created by rotating the area enclosed by the following functions around the y-axis: $$y(x) = x^2$$ $$g(x) = x^{\frac{1}{2}}$$ Use the shell method.

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Michael H.
Answer:

The two functions intersect at (0,0) and (1,1) giving the radius of the shells from 0 to 1. The height of the shells is g(x) - f(x). The integral is $$2\pi\int^1_0 x^{\frac{1}{2}} - x^2 dx$$ Calculating the indefinite integral gives $$2\pi [(\frac{2x^\frac{3}{2}}{3} - \frac{x^3}{3})|^1_0$$] Plugging in the values 0 and 1 leaves $$2\pi[(\frac{2}{3} - \frac{1}{3}) - (0 - 0)]$$ Simplifies to $$\frac{2\pi}{3} \approx 2.094$$

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