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# Tutor profile: James E.

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James E.
Library Science Masters Student with extensive Math Experience
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## Questions

### Subject:Pre-Calculus

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Question:

What size matrix do you get when you multiply a 2x3 Matrix with a 3x2 Matrix? How about a 3x2 and a 2x3? How about two 2x3 matrices? What is the result when you multiply the matrix $$\begin{pmatrix} 1 & 3 & 2 \\ 2 & 3 & 2 \end{pmatrix}$$ by the matrix $$\begin{pmatrix} 1 & 0 \\ 0 & 2 \\ 1 & 0 \end{pmatrix}$$

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James E.

There is an easy rule of thumb for the size of the resulting matrix and it is that you have the same number of rows as matrix A and columns as Matrix B. Thus you would get a 2x2 and 3x3 matrix respectively for the first two questions. As for the third, you cannot multiply two 2x3 Matrices as Matrix A has more columns than Matrix B has rows. Finally the general formula for multiplying a 2x3 and a 3x2 matrix is as follows: $$\begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix} * \begin{pmatrix} g & h \\ i & j \\ k & l \end{pmatrix} = \begin{pmatrix} ag+bi+ck & ah+bj+cl \\ dg+ei+fk & dh+ej+fl \end{pmatrix}$$ With our specific situation this translates to the following results: $$\begin{pmatrix} 1 & 3 & 2 \\ 2 & 3 & 2 \end{pmatrix} *\begin{pmatrix} 1 & 0 \\ 0 & 2 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1*1+3*0+2*1 & 1*0+3*2+2*0 \\ 2*1+3*0+2*1 & 2*0+3*2+2*0 \end{pmatrix} = \begin{pmatrix} 3 & 6 \\ 4 & 6 \end{pmatrix}$$

### Subject:Calculus

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Question:

What is the product rule for derivatives? What is the Derivative of $$X^3Cos(x)$$?

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James E.

The product rule is that $$\frac{d}{dx}(u(x)v(x))=u'(x)v(x)+u(x)v'(x)$$. This is the rule needed to answer the second part which is fairly simple. $$\frac{d}{dx}Cos(x)=-Sin(x)$$ and $$\frac{d}{dx}x^3=3x^2$$. Thus via simple substitution $$\frac{d}{dx}X^3Cos(x) = 3x^2Cos(x)-x^3Sin(x)$$ Which can also be written as $$x^2(3Cos(x)-xSin(x))$$

### Subject:Differential Equations

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Question:

What is the general solution to $$\frac{4}{x}\frac{dy}{dx}$$ = $$4x+12+\frac{4}{x}$$? What is the specific solution given $$f(1)=4$$?

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James E.

So first you need to get the $$\frac{dy}{dx}$$ alone on its own side of the equation. That is relatively easy you can simply divide both sides of the equation by $$\frac{4}{x}$$. This gives us $$\frac{dy}{dx}$$ = $$X^2 + 3x + 1$$. This is simple to solve, you can simply take the shortcut of going through each individual value and taking the reverse derivative giving us $$y$$ = $$\frac{x^{3}}{3} + \frac{3x^2}{2}+x+C$$. This is the general solution. To get the particular solution we need to plug in 1 and solve for C. By doing that we get $$\frac{1}{3}+\frac{3}{2}+1+C=4$$ which is the same as $$\frac{1}{3}+C=\frac{3}{2}$$ this is the same as $$\frac{2}{6}+C=\frac{9}{6}$$ which means $$\frac{7}{6}=C$$ thus our particular solution is $$y=\frac{x^3}{3}+\frac{3x^2}{2}+x+\frac{7}{6}$$.

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