Question

PLEASE HELP Need help The question is in the photo will give points

Answer 1

Answer:

6

Step-by-step explanation:

Given expression:

$\dfrac{2}{\sqrt[6]{8}} \cdot \sqrt{2}-\left(-\dfrac{18}{\sqrt{81}}-2\right)$

Rewrite  81 as 9² :

$\implies \dfrac{2}{\sqrt[6]{8}} \cdot \sqrt{2}-\left(-\dfrac{18}{\sqrt{9^2}}-2\right)$

$\textsf{Apply radical rule} \quad \sqrt{a^2}=a, \quad a \geq 0:$

$\implies \dfrac{2}{\sqrt[6]{8}} \cdot \sqrt{2}-\left(-\dfrac{18}{9}-2\right)$

$\implies \dfrac{2}{\sqrt[6]{8}} \cdot \sqrt{2}-\left(-2-2\right)$

$\implies \dfrac{2}{\sqrt[6]{8}} \cdot \sqrt{2}-\left(-4\right)$

$\implies \dfrac{2}{\sqrt[6]{8}} \cdot \sqrt{2}+4$

$\implies \dfrac{2\sqrt{2}}{\sqrt[6]{8}}+4$

Rewrite 8 as 2³ :

$\implies \dfrac{2\sqrt{2}}{\sqrt[6]{2^3}}+4$

$\textsf{Apply radical rule} \quad \sqrt[n]{a}=a^{\frac{1}{n}:$

$\implies \dfrac{2\sqrt{2}}{\left(2^3\right)^{\frac{1}{6}}}+4$

$\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\implies \dfrac{2\sqrt{2}}{2^{\frac{3}{6}}}+4$

$\implies \dfrac{2\sqrt{2}}{2^{\frac{1}{2}}}+4$

$\textsf{Apply radical rule} \quad a^{\frac{1}{2}}=\sqrt{a}, \quad a \geq 0:$

$\implies \dfrac{2\sqrt{2}}{\sqrt{2}}+4$

Cancel the common factor √2 :

$\implies 2+4$

$\implies 6$

Learn more about radicals here:

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