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grandymaker [24]
3 years ago
8

Please answer this

Mathematics
2 answers:
omeli [17]3 years ago
5 0

Answer:

C  is the corrsponding choose

Step-by-step explanation:

Sorry i got the wrong queston :(

KatRina [158]3 years ago
3 0

Check the picture below.

so we can simply use the pythagorean theorem for each triangle and get "w" and "z".

\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \sqrt{15^2-13^2}=w\implies \sqrt{225-169}=w\implies \sqrt{56}=w\implies 7.48\approx w \\\\\\ \sqrt{10^2-4^2}=z\implies \sqrt{100-16}=z\implies \sqrt{84}=z\implies 9.17\approx z \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{w+z}{16.65}~\hfill

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Answer:

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Step-by-step explanation:

Given problem is \sqrt[3]{x}\cdot\sqrt[3]{x^2}.

Now we need to simplify this problem.

\sqrt[3]{x}\cdot\sqrt[3]{x^2}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}

Apply formula

\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}

so we get:

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}

\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x

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