If $a = 12, b = 9$ $a=12, b=9$ , and $c = 4$ $c=4$ , what is $\frac{b^{2} - 2 c^{2}}{a + c - b}$ $(b^2-2c^2)/(a+c-b)$ ?

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If $a = 12, b = 9$ $a=12, b=9$ , and $c = 4$ $c=4$ , what is $\frac{b^{2} - 2 c^{2}}{a + c - b}$ $(b^2-2c^2)/(a+c-b)$ ?

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Answer 1 · 2018-03-30T12:29:43

See a solution process below:

Explanation:

Substitute:

$12$ $color(red)(12)$ for each occurrence of $a$ $color(red)(a)$
$9$ $color(blue)(9)$ for each occurrence of $b$ $color(blue)(b)$
$4$ $color(green)(4)$ for each occurrence of $c$ $color(green)(c)$

and then calculate the result:

$\frac{b^{2} - 2 c^{2}}{a + c - b}$ $(color(blue)(b)^2 - 2color(green)(c)^2)/(color(red)(a) + color(green)(c) - color(blue)(b))$ becomes:

$\frac{9^{2} - (2 \cdot 4^{2})}{12 + 4 - 9} \Rightarrow$ $(color(blue)(9)^2 - (2 * color(green)(4)^2))/(color(red)(12) + color(green)(4) - color(blue)(9)) =>$

$\frac{81 - (2 \cdot 16)}{16 - 9} \Rightarrow$ $(81 - (2 * 16))/(16 - color(blue)(9)) =>$

$\frac{81 - 32}{7} \Rightarrow$ $(81 - 32)/7 =>$

$\frac{49}{7} \Rightarrow$ $49/7 =>$

$7$ $7$

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If $a = 12, b = 9$ $a=12, b=9$ , and $c = 4$ $c=4$ , what is $\frac{b^{2} - 2 c^{2}}{a + c - b}$ $(b^2-2c^2)/(a+c-b)$ ?

1 Answer

Explanation:

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If a=12,b=9a=12, b=9, and c=4c=4, what is b2−2c2a+c−b(b^2-2c^2)/(a+c-b)?

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If $a = 12, b = 9$ $a=12, b=9$ , and $c = 4$ $c=4$ , what is $\frac{b^{2} - 2 c^{2}}{a + c - b}$ $(b^2-2c^2)/(a+c-b)$ ?