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3 years ago

7x+3y=m (solve for y)

Mathematics

2 answers:

tia_tia [17]3 years ago

7 0

7x+3y=m

3y=m-7x

y= (m-7x)/3

y = m/3 - 7x/3

allochka39001 [22]3 years ago

6 0

Lets solve for y

7x+3y=m

Step 1: Add -7x to both sides.

7x+3x+-7x=m+-7x

Step 2: Divide both sides by 3.

$\frac{3y}{3}$ = $\frac{m-7x}{3}$

y= $\frac{1}{3}$ m + $\frac{-7}{3}$ x

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A= [1, 3; 2, 1], B=[3, 6; -1, 1]. Find AB & BA if possible

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Answer:

$AB\Rightarrow \quad \begin{bmatrix}0 & 9\\ 5 & 13\end{bmatrix}\\BA\Rightarrow \quad \begin{bmatrix}15 & 15\\ 1 & -2\end{bmatrix}$

Step-by-step explanation:

For two matrix P and Q, the product, say PQ is defined when:

The number of columns of P = The number of rows of Q

Since A is a 2×2 matrix and B is also a 2×2 matrix

Thus both AB and BA are possible.

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$AB\Rightarrow\begin{bmatrix}1 & 3\\ 2 & 1\end{bmatrix}\begin{bmatrix}3 & 6\\ -1 & 1\end{bmatrix}\\AB\Rightarrow\quad \begin{bmatrix}3\times 1+3\times (-1) & 6\times 1+3\times 1\\3\times 2+1\times (-1) & 6\times 2+1\times 1\end{bmatrix}\\AB\Rightarrow \quad \begin{bmatrix}0 & 9\\ 5 & 13\end{bmatrix}$

BA is:

$BA\Rightarrow\begin{bmatrix}3 & 6\\ -1 & 1\end{bmatrix}\begin{bmatrix}1 & 3\\ 2 & 1\end{bmatrix}\\BA\Rightarrow\quad \begin{bmatrix}3\times 1+6\times 2 & 3\times 3+6\times 1\\(-1)\times 1+1\times 2 & (-1)\times 3+1\times 1\end{bmatrix}\\BA\Rightarrow \quad \begin{bmatrix}15 & 15\\ 1 & -2\end{bmatrix}$

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$\frac{3\pi}{4}$

(Assuming you want your answer in radians)

If you want the answer in degrees just multiply your answer in radians by $\frac{180^\circ}{\pi}$ giving you:

$\frac{3\pi}{4} \cdot \frac{180^\circ}{\pi}=\frac{3(180)}{4}=135^{\circ}$ .

We can do this since $\pi \text{ rad }=180^\circ$ (half the circumference of the unit circle is equivalent to 180 degree rotation).

Step-by-step explanation:

$\cos^{-1}(x)$ is going to output an angle measurement in $[0,\pi]$ .

So we are looking to solve the following equation in that interval:

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The solution to the equation is $\frac{3\pi}{4}$ .

So we are saying that $\cos(\frac{3\pi}{4})=\frac{-\sqrt{2}}{2}$ implies $\cos^{-1}(\frac{-\sqrt{2}}{2})=\frac{3\pi}{4}$ since $\frac{3\pi}{4} \in [0,\pi]$ .

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