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Svetach [21]
3 years ago
10

Could I get the answer, all work and explanation for this problem?

Mathematics
1 answer:
ratelena [41]3 years ago
8 0
Okay so this is how i did it but idk if it’s right.
so i multiples the top and bottom by square root of 3 to get rid of the square root of 3
then i was left with 8-1/2(1) and got 7/2

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How to show these 2 problems are inverses
nataly862011 [7]
\bf \begin{cases}
f(x)=\sqrt[3]{7x-2}\\\\
g(x)=\cfrac{x^3+2}{7}
\end{cases}\\\\
-----------------------------\\\\
now
\\\\
f[\ g(x)\ ]\implies f\left[ \frac{x^3+2}{7} \right]\implies \sqrt[3]{7\left[ \frac{x^3+2}{7} \right]-2}\implies \sqrt[3]{x^3+2-2}
\\\\\\
\sqrt[3]{x^3}\implies x\\\\
-----------------------------\\\\
or
\\\\
g[\ f(x)\ ]\implies g\left[\sqrt[3]{7x-2}\right]\implies \cfrac{\left[\sqrt[3]{7x-2}\right]^3+2}{7}
\\\\\\
\cfrac{7x-2+2}{7}\implies \cfrac{7x}{7}\implies x

thus f[ g(x) ] = x indeed, or g[ f(x) ] =x, thus they're indeed inverse of each other
8 0
3 years ago
Plz help
patriot [66]

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

I did it on edg

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