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Marta_Voda [28]
3 years ago
9

If f(x) = -x^2 – 1, and g(x) = x + 5, then g(f (x)) = -x^2+[ ?]x + [ ?]

Mathematics
1 answer:
trasher [3.6K]3 years ago
5 0

Answer:

answer is -x^2-1+5

Step-by-step explanation:

so you can do it .

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Answer:

f(x) =  {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x  \rightarrow \: x = sec \: y\\  \frac{dx}{dx}  =  \frac{d(sec \: y)}{dx}  \\ 1 = \frac{d(sec \: y)}{dx} \times  \frac{dy}{dy}  \\ 1 = \frac{d(sec \: y)}{dy} \times  \frac{dy}{dx}  \\1 = tan \: y.sec \: y. \frac{dy}{dx}  \\ \frac{dy}{dx} =  \frac{1}{tan \: y.sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ \sqrt{( {sec}^{2}   \: y - 1}) .sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\   \therefore  \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx}  =  \frac{1}{ |5x |  \sqrt{25 {x }^{2}  - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} }

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If f(x) = StartFraction x Over 2 EndFraction + 8, what is f(x) when x = 10?
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Answer:

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Step-by-step explanation:

test edge2020

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2 years ago
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Can someone help me with this please?
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<em>Answer:</em>

<em>D I believe </em>

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Some of the basic rule In statistics to represent data using charts are as follows:

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A pie chart is a circular chart divided into wedge-like portions and is basically used to display percentage or proportion of categories in the data The percentage represented by each category is shown by the corresponding slice/portion of the whole pie.

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