Let $F (x) = x^{2} - 20$ $F(x)=x^2-20$ and $G (x) = 14 - x$ $G(x)=14-x$ , how do you find $(\frac{F}{G}) (7)$ $(F/G)(7)$ ?

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Let $F (x) = x^{2} - 20$ $F(x)=x^2-20$ and $G (x) = 14 - x$ $G(x)=14-x$ , how do you find $(\frac{F}{G}) (7)$ $(F/G)(7)$ ?

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Answer 1 · 2017-04-09T19:08:05

https://www.mathsisfun.com/sets/functions-operations.html

See the entire solution process below:

Explanation:

First,

$(\frac{F}{G}) (x) = \frac{x^{2} - 20}{14 - x}$ $(F/G)(x) = (x^2 - 20)/(14 - x)$

To find $(\frac{F}{G}) (7)$ $(F/G)(7)$ we need to substitute $7$ $color(red)(7)$ for each occurrence of $x$ $color(red)(x)$ in $(\frac{F}{G}) (x)$ $(F/G)(x)$ :

$(\frac{F}{G}) (x) = \frac{x^{2} - 20}{14 - x}$ $(F/G)(color(red)(x)) = (color(red)(x)^2 - 20)/(14 - color(red)(x))$ becomes:

$(\frac{F}{G}) (7) = \frac{7^{2} - 20}{14 - 7}$ $(F/G)(color(red)(7)) = (color(red)(7)^2 - 20)/(14 - color(red)(7))$

$(\frac{F}{G}) (7) = \frac{49 - 20}{7}$ $(F/G)(color(red)(7)) = (49 - 20)/7$

$(\frac{F}{G}) (7) = \frac{29}{7}$ $(F/G)(color(red)(7)) = 29/7$

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Let $F (x) = x^{2} - 20$ $F(x)=x^2-20$ and $G (x) = 14 - x$ $G(x)=14-x$ , how do you find $(\frac{F}{G}) (7)$ $(F/G)(7)$ ?

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Explanation:

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Let F(x)=x2−20F(x)=x^2-20 and G(x)=14−xG(x)=14-x, how do you find(FG)(7) (F/G)(7)?

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Explanation:

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Let $F (x) = x^{2} - 20$ $F(x)=x^2-20$ and $G (x) = 14 - x$ $G(x)=14-x$ , how do you find $(\frac{F}{G}) (7)$ $(F/G)(7)$ ?