Evaluate
\frac{3\sqrt{3}+163}{40}\approx 4.204903811
Quiz
Arithmetic
\frac { 9 \sqrt { 18 } + 9 \sqrt { 6 } + 32 \cdot 5 \sqrt { 54 } } { 120 \sqrt { 6 } }
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\frac{9\times 3\sqrt{2}+9\sqrt{6}+5\times 32\sqrt{54}}{120\sqrt{6}}
Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.
\frac{27\sqrt{2}+9\sqrt{6}+5\times 32\sqrt{54}}{120\sqrt{6}}
Multiply 9 and 3 to get 27.
\frac{27\sqrt{2}+9\sqrt{6}+160\sqrt{54}}{120\sqrt{6}}
Multiply 5 and 32 to get 160.
\frac{27\sqrt{2}+9\sqrt{6}+160\times 3\sqrt{6}}{120\sqrt{6}}
Factor 54=3^{2}\times 6. Rewrite the square root of the product \sqrt{3^{2}\times 6} as the product of square roots \sqrt{3^{2}}\sqrt{6}. Take the square root of 3^{2}.
\frac{27\sqrt{2}+9\sqrt{6}+480\sqrt{6}}{120\sqrt{6}}
Multiply 160 and 3 to get 480.
\frac{27\sqrt{2}+489\sqrt{6}}{120\sqrt{6}}
Combine 9\sqrt{6} and 480\sqrt{6} to get 489\sqrt{6}.
\frac{\left(27\sqrt{2}+489\sqrt{6}\right)\sqrt{6}}{120\left(\sqrt{6}\right)^{2}}
Rationalize the denominator of \frac{27\sqrt{2}+489\sqrt{6}}{120\sqrt{6}} by multiplying numerator and denominator by \sqrt{6}.
\frac{\left(27\sqrt{2}+489\sqrt{6}\right)\sqrt{6}}{120\times 6}
The square of \sqrt{6} is 6.
\frac{\left(27\sqrt{2}+489\sqrt{6}\right)\sqrt{6}}{720}
Multiply 120 and 6 to get 720.
\frac{27\sqrt{2}\sqrt{6}+489\left(\sqrt{6}\right)^{2}}{720}
Use the distributive property to multiply 27\sqrt{2}+489\sqrt{6} by \sqrt{6}.
\frac{27\sqrt{2}\sqrt{2}\sqrt{3}+489\left(\sqrt{6}\right)^{2}}{720}
Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.
\frac{27\times 2\sqrt{3}+489\left(\sqrt{6}\right)^{2}}{720}
Multiply \sqrt{2} and \sqrt{2} to get 2.
\frac{54\sqrt{3}+489\left(\sqrt{6}\right)^{2}}{720}
Multiply 27 and 2 to get 54.
\frac{54\sqrt{3}+489\times 6}{720}
The square of \sqrt{6} is 6.
\frac{54\sqrt{3}+2934}{720}
Multiply 489 and 6 to get 2934.
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