Someone pls answer this

Mathematics

2 answers:

soldier1979 [14.2K]3 years ago

7 0

Answer:

A=B*H

A=8*3

=24 UNIT^2..

Gala2k [10]3 years ago

7 0

Someone already answered

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1 х Given g(x)=

yarga [219]

(a) Since $g(x)=\sqrt[3]{x}$ and $h(x) = \frac1{x^3}$ , we have

$(g\circ h)(x) = g(h(x)) = g\left(\dfrac1{x^3}\right) = \sqrt{3}{\dfrac1{x^3}} = \dfrac1x$

We're given that

$(f \circ g \circ h)(x) = f(g(h(x))) = f\left(\dfrac1x\right) = \dfrac x{x+1}$

but we can rewrite this as

$\dfrac x{x+1} = \dfrac{\frac xx}{\frac xx + \frac1x} = \dfrac1{1+\frac1x}$

(bear in mind that we can only do this so long as x ≠ 0) so it follows that

$f\left(\dfrac1x\right) = \dfrac1{1+\frac1x} \implies \boxed{f(x) = \dfrac1{1+x}}$

(b) On its own, we may be tempted to conclude that the domain of $(f\circ g\circ h)(x) = \frac1{1+x}$ is simply x ≠ -1. But we should be more careful. The domain of a composite depends on each of the component functions involved.