What is 7 over the square root of 27?

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Answer 1 · 2015-07-24T16:18:21

$\frac{7 \sqrt{3}}{9}$ $(7 sqrt(3))/9$

Start by writing your expression, which features $7$ $7$ in the numerator and $\sqrt{27}$ $sqrt(27)$ in the denominator.

$\frac{7}{\sqrt{27}}$ $7/(sqrt(27)$

Now, the important thing to realize here is that you can write $27$ $27$ as

$27 = 9 \cdot 3 = 3 \cdot 3 \cdot 3 = 3^{2} \cdot 3$ $27 = 9 * 3 = 3 * 3 * 3 = 3""^2 * 3$

This means that the denominator becomes

$\sqrt{27} = \sqrt{3^{2} \cdot 3} = \sqrt{3^{2}} \cdot \sqrt{3} = 3 \sqrt{3}$ $sqrt(27) = sqrt(3""^2 * 3) = sqrt(3""^2) * sqrt(3) = 3sqrt(3)$

The expression is now

$\frac{7}{3 \cdot \sqrt{3}}$ $7/(3 * sqrt(3))$

Next, you have to rationalize the denominator, which you can do by multiplying the numerator and the denominator by $\sqrt{3}$ $sqrt(3)$ , to get

$(7 * sqrt(3))/(3 * underbrace(sqrt(3) * sqrt(3))_(color(blue)("=3))) = (7 * sqrt(3))/(3 * color(blue)(3)) = color(green)((7 sqrt(3))/9)$