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3 years ago

12

4x^2+4y^2+20x-16y+37=0 standard form

1 answer:

zhenek [66]3 years ago

5 0

The answer in standard form is (x+5/2)^2+(y-2)^2=1

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Find the midpoint of line AB.<br> A(-4,2) B(0,-4)

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3 years ago

Find the matrix product, if possible. Show your work please

steposvetlana [31]

$\left[\begin{array}{ccc}-1&3\\5&4\end{array}\right] \cdot \left[\begin{array}{ccc}0&-2&6\\1&-3&2\end{array}\right]\\\\= \left[\begin{array}{ccc}(-1)(0)+(3)(1)&(-1)(-2)+(3)(-3)&(-1)(6)+(3)(2)\\(5)(0)+(4)(1)&(5)(-2)+(4)(-3)&(5)(6)+(4)(2)\end{array}\right]\\\\= \left[\begin{array}{ccc}0+3&2-9&-6+6\\0+4&-10-12&30+8\end{array}\right]\\\\ =\left[\begin{array}{ccc}3&-7&0\\4&-22&38\end{array}\right]$

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3 years ago

The answer is p = -2/3 and q= -2

igomit [66]

$y=\dfrac{p}{q+x}\\\\\dfrac{1}{y}=\dfrac{1}{\frac{p}{q+x}}=\dfrac{q+x}{p}$

We have x and y intercept:

(2, 0) and (0, 3).

Substitute the coordinates of points to the equation:

$(2,\ 0)\\\\0=\dfrac{q+2}{p}\to q+2=0\to q=-2\\\\(0,\ 3)\\3=\dfrac{q+0}{p}\to3p=q\to3p=-2\to p=-\dfrac{2}{3}$

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3 years ago