\frac{\sqrt{10^2-2^6}}{\sqrt{\frac{3^3-2^3}{2^2}}}

asked by guest
on Nov 25, 2024 at 6:40 pm



You asked:

Evaluate the expression: \(\frac{\sqrt{{10}^{2} - {2}^{6}}}{\sqrt{\frac{{3}^{3} - {2}^{3}}{{2}^{2}}}}\)

MathBot Answer:

\[\frac{\sqrt{{10}^{2} - {2}^{6}}}{\sqrt{\frac{{3}^{3} - {2}^{3}}{{2}^{2}}}} = \frac{12 \sqrt{19}}{19} \approx 2.75298880644674119088651493717449\]

\[\begin{aligned}\frac{\sqrt{{10}^{2} - {2}^{6}}}{\sqrt{\frac{{3}^{3} - {2}^{3}}{{2}^{2}}}}& = \sqrt{10^{2} - 2^{6}} \cdot \frac{1}{\sqrt{\left(3^{3} - 2^{3}\right) \cdot \frac{1}{2^{2}}}}\\& = \sqrt{100 - 2^{6}} \cdot \frac{1}{\sqrt{\left(3^{3} - 2^{3}\right) \cdot \frac{1}{2^{2}}}}\\& = \sqrt{100 - 64} \cdot \frac{1}{\sqrt{\left(3^{3} - 2^{3}\right) \cdot \frac{1}{2^{2}}}}\\& = \sqrt{36} \cdot \frac{1}{\sqrt{\left(3^{3} - 2^{3}\right) \cdot \frac{1}{2^{2}}}}\\& = 6 \cdot \frac{1}{\sqrt{\left(3^{3} - 2^{3}\right) \cdot \frac{1}{2^{2}}}}\\& = 6 \cdot \frac{1}{\sqrt{\left(27 - 2^{3}\right) \cdot \frac{1}{2^{2}}}}\\& = 6 \cdot \frac{1}{\sqrt{\left(27 - 8\right) \cdot \frac{1}{2^{2}}}}\\& = 6 \cdot \frac{1}{\sqrt{19 \cdot \frac{1}{2^{2}}}}\\& = 6 \cdot \frac{1}{\sqrt{\frac{19}{4}}}\\& = 6 \cdot \frac{1}{\frac{1}{2} \cdot \sqrt{19}}\\& = \frac{12}{19} \cdot \sqrt{19}\end{aligned}\]

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