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VMariaS [17]
3 years ago
6

Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. If f(X)=25-x^2 and

g(X)=x+5, what is (f/g)(x)? Write your answer in simplest form. When f(X)=25-x^2 and g(X)=x+5, (f/g)(x) .
Mathematics
2 answers:
Licemer1 [7]3 years ago
8 0

f(x)=25-x^2\\\\g(x)=x+5\\\\\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}=\dfrac{25-x^2}{x+5}=\dfrac{5^2-x^2}{x+5}=\dfrac{(5-x)(5+x)}{5+x}=5-x\\\\for\ x\neq-5

Arada [10]3 years ago
6 0

Answer: -x+5

Step-by-step explanation:

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padilas [110]

First picture:

You got the first two right (good job!), so I'll just tackle the third one. If you have to compute f(x-4), it means that you have to substitute x-4 in place of x. Differently from the first two points, this will generate a new function, rather than a specific value:

f(x) = \sqrt{x+4}+2 \implies f(x-4) = \sqrt{(x-4)+4}+2 = \sqrt{x}+2

Second picture:

Given the point (x,y) highlighted in the picture, you can deduce that the base is 2x units long (since it spans from (-x,0) to (x,0)) and the height is y units long (because it spans from (x,0) to (x,y)). So, the area of the rectangle is the multiplication between base and height:

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The domain of this function is given by the domain of the square root: we want its argument to be non, negative, so we have

36-x^2 \geq 0 \iff x^2 \leq 36 \iff -6 \leq x \leq 6

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You can only see the answer 0 < x < 6 because, if you choose x=0 or x=6, the rectangle degenerates to a segment, and your exercise doesn't like this scenario, apparently

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