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Bond [772]
3 years ago
15

If f(g(x)) = (x-3)^4 what is f(x) and g(x)

Mathematics
1 answer:
Lelechka [254]3 years ago
6 0

One possible solution is

f(x) = x^4

g(x) = x-3

Since

f(x) = x^4

f(g(x)) = ( g(x) )^4

f(g(x)) = ( x-3 )^4

=================================================

Another possible solution could be

f(x) = x^2

g(x) = (x-3)^2

Because

f(x) = x^2

f(g(x)) = ( g(x) )^2

f(g(x)) = ( (x-3)^2 )^2

f(g(x)) = (x-3)^(2*2)

f(g(x)) = (x-3)^4

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Easy but i'm confused yikes. (picture attached)
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469 miles

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in 2003 the state pension was increased by 2% to £78.03. What was the state pension before this increase?
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mixer [17]

Answer:

\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.

Step-by-step explanation:

Given:

\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1

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Multiply this equation by \frac{\sqrt{2}}{2}:

-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}

The general solution is

x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.

4 0
3 years ago
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telo118 [61]

Answer:

48 months

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