Question

tiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtiredtired

Answer 1

remember, we can't have square roots in the denominator, so you have to multiply both numerator and denominator by them

Answer 2

[ 3 * sqrt(2) ]/ [ sqrt(2) * sqrt(2) ]

Answer 3

because \[\large 2 = \sqrt{2} * \sqrt{2} \] and vise versa. Let's say you had \[\large \frac{\sqrt{2}}{2}\] Using 2 = sqrt(2) * sqrt(2) how would you evauate that?

Answer 4

evaluate*

Answer 5

No no. remember how we just stated \(\large 2= \sqrt{2}*\sqrt{2}\) ?\[\large \frac{\sqrt{2}}{2} = \large \frac{\sqrt{2}}{\sqrt{2}* \sqrt{2}}\]\[\large \frac{\cancel{\sqrt{2}}}{\sqrt{2}* \cancel{\sqrt{2}}}\]\[\large \frac{1}{\sqrt{2}}= \frac{1}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}\]... I meant the other way around xD 2/sqrt(2) haha But do you see how the process works?... lol.

Answer 6

If you take any number, the number will be a multiple of 2 square roots.

Answer 7

any whole integer.

Answer 8

that's right

Answer 9

i've got to head off now, hope this helped! :)

Answer 10

No problem! :)