Question #85711

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Question #85711

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Answer 1 · 2017-01-18T01:41:03

$r = \frac{23 \cdot 5}{x + 7} = \frac{115}{x + 7} \Rightarrow r^{2} = \frac{115^{2}}{{(x + 7)}^{2}} = \frac{13225}{x^{2} + 14 x + 49}$ $r=(23*5)/(x+7)=115/(x+7) => r^2=115^2/(x+7)^2=13225/(x^2+14x+49)$
$= \frac{23^{2} \cdot 5^{2}}{{(x + 7)}^{2}}$ $=(23^2*5^2)/(x+7)^2$

Explanation:

$r = \frac{23 \cdot 5}{x + 7} \Rightarrow r^{2} = {(\frac{23 \cdot 5}{x + 7})}^{2} = \frac{{(23 \cdot 5)}^{2}}{{(x + 7)}^{2}} = \frac{{(23)}^{2} {(5)}^{2}}{(x + 7) (x + 7)}$ $r=(23*5)/(x+7) => r^2=((23*5)/(x+7))^2=(23*5)^2/(x+7)^2=((23)^2(5)^2)/((x+7)(x+7))$

$= \frac{{(20 + 3)}^{2} (25)}{x \cdot x + 7 x + 7 x + 7 (7)}$ $=((20+3)^2(25))/(x*x+7x+7x+7(7))$

$= \frac{((20) (20) + (3) (20) + (3) (20) + (3) (3)) (25)}{x^{2} + 14 x + 49}$ $=(((20)(20)+(3)(20)+(3)(20)+(3)(3))(25))/(x^2+14x+49)$

$= \frac{(400 + 60 + 60 + 9) (25)}{x^{2} + 14 x + 49}$ $=((400+60+60+9)(25))/(x^2+14x+49)$

$= \frac{(400 + 120 + 9) (25)}{x^{2} + 14 x + 49} = \frac{(529) (25)}{x^{2} + 14 x + 49}$ $=((400+120+9)(25))/(x^2+14x+49)=((529)(25))/(x^2+14x+49)$

$= \frac{(529) (20 + 5)}{x^{2} + 14 x + 49} = \frac{529 (20) + 529 (5)}{x^{2} + 14 x + 49}$ $=((529)(20+5))/(x^2+14x+49)=(529(20)+529(5))/(x^2+14x+49)$

$= \frac{2 (5290) + 5290 (\frac{1}{2})}{x^{2} + 14 x + 49} = \frac{10580 + 2645}{x^{2} + 14 x + 49}$ $=(2(5290)+5290(1/2))/(x^2+14x+49)=(10580+2645)/(x^2+14x+49)$

$\frac{13225}{x^{2} + 14 x + 49}$ $13225/(x^2+14x+49)$

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Question #85711

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