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3 years ago

In which quadrant does the point (25 , 21) lie? A. Quadrant I B. Quadrant IV C. Quadrant III D. Quadrant II

Mathematics

2 answers:

adell [148]3 years ago

8 0

A) Quadrant 1

Not 2,3,4

bija089 [108]3 years ago

3 0

Answer:

Step-by-step explanation:

Both of the coordinates are positive. The quadrant that has points with positive x and y coordinates is Quadrant I.

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Just use the formula:
$slope(m) = \frac{y_2-y_1}{x_2-x_1}$

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3 years ago

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<img src="https://tex.z-dn.net/?f=%5Cleft.%20%5Cbegin%7Bcases%7D%20%7B%208%20x%20%2B%202%20y%20%3D%2046%20%7D%20%5C%5C%20%7B%207

Brut [27]

$\huge \boxed{\mathfrak{Question} \downarrow}$

$\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right.$

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\left. \begin{cases} { 8 x + 2 y = 46 } \\ { 7 x + 3 y = 47 } \end{cases} \right.$

First, write both the equations in its standard form.

$8x+2y=46\\ 7x+3y=47$

Now, write the equations in form of matrix.

$\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}46\\47\end{matrix}\right)$

Then, multiply the equation towards the left by using the inverse of matrix $\left(\begin{matrix}8&2\\7&3\end{matrix}\right)$

$\sf \: inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}8&2\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)$

The product of the matrix & its inverse will be the identity matrix.

$\sf\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)$

Now, multiply the matrices that lie on the left-hand side of the equal sign.

$\sf\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&2\\7&3\end{matrix}\right))\left(\begin{matrix}46\\47\end{matrix}\right)$

For the 2 × 2 matrix $\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$ , the inverse matrix is ⇨ $\left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right)$ .

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{8\times 3-2\times 7}&-\frac{2}{8\times 3-2\times 7}\\-\frac{7}{8\times 3-2\times 7}&\frac{8}{8\times 3-2\times 7}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)$

Do the calculations.

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}&-\frac{1}{5}\\-\frac{7}{10}&\frac{4}{5}\end{matrix}\right)\left(\begin{matrix}46\\47\end{matrix}\right)$

Multiply the matrices.

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}\times 46-\frac{1}{5}\times 47\\-\frac{7}{10}\times 46+\frac{4}{5}\times 47\end{matrix}\right)$

Do the arithmetics again.

$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{22}{5} \\\frac{27}{5}\end{matrix}\right)$

Finally, extract the matrix elements x & y & write them separately.

$\large \boxed{ \boxed{ \bf \: x=\frac{22}{5},y=\frac{27}{5} }}$

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2 years ago

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