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Kamila [148]
3 years ago
10

Verify inverse functions

Mathematics
1 answer:
ziro4ka [17]3 years ago
4 0

Answer:

Step-by-step explanation:

Hello

(fog)(x)=f(g(x))=10(x-6)/10+6=x-6+6=x

(gof)(x)=(10x-6+6)/10=10x/10=x

So YES the functions are inverse

hope this helps

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One more time!
CaHeK987 [17]
Since q(x) is inside p(x), find the x-value that results in q(x) = 1/4

\frac{1}{4} = 5 - x^2\ \Rightarrow\ x^2 = 5 - \frac{1}{4}\ \Rightarrow\ x^2 = \frac{19}{4}\ \Rightarrow \\&#10;x = \frac{\sqrt{19} }{2}

so we conclude that
q(\frac{\sqrt{19} }{2} ) = 1/4

therefore

p(1/4) = p\left( q\left(\frac{ \sqrt{19} }{2} \right)  \right)

plug x=\sqrt{19}/2 into p( q(x) ) to get answer

p(1/4) = p\left( q\left( \frac{ \sqrt{19} }{2} \right) \right)\ \Rightarrow\ \dfrac{4 - \left(  \frac{\sqrt{19} }{2}\right)^2 }{ \left(  \frac{\sqrt{19} }{2}\right)^3 } \Rightarrow \\ \\ \dfrac{4 - \frac{19}{4} }{ \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{8\left(4 - \frac{19}{4}\right) }{ 8 \cdot \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{32 - 38}{19\sqrt{19}} \Rightarrow \dfrac{-6}{19\sqrt{19}} \cdot \frac{\sqrt{19}}{\sqrt{19}}\Rightarrow

\dfrac{-6\sqrt{19} }{19 \cdot 19} \\ \\ \Rightarrow  -\dfrac{6\sqrt{19} }{361}

p(1/4) = -\dfrac{6\sqrt{19} }{361}
3 0
3 years ago
How long will it take for a falling object to reach 108 m/s if its initial velocity is 10 m/s?
defon

Answer:

t=10 s

Step-by-step explanation:

v=u+gt

108=10+9.8t

108-10=9.8t

9.8t=98

t=\frac{98}{9.8} =\frac{980}{98} =10

4 0
1 year ago
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