Question

Radicals

Answer 1

I can't post it ghh

Answer 2

screenshot it?

Answer 3

It;s from a book

Answer 4

oh welp

Answer 5

is it too complex to just type out?

Answer 6

it is, let me draw it

Answer 7

\[ \sqrt{50^y4 z^5 } \] over \( \color{eneter color name}{√ −10yz } \)

Answer 8

jeez, now it's doesn't loads for me /.-

Answer 9

So a square root divided by a square root can be combined to a be a single square root. So basically its normal simplification just under a square root. Also is that actually supposed to be 50^y?

Answer 10

\[\sqrt((50^y4z^5)/(-10yz))\]

Answer 11

No basically it’s \( \color{eneter color name}{√ 50y^4 z^5 } \)

My answer was \( \color{eneter color name}{−yz^/22 √2y } \)

Answer 12

oh so you didnt mean to put the y as an exponent

Answer 13

no y is not an exponent

Answer 14

rip my drawing abilities

Answer 15

\[\sqrt{\frac{50y^4z^5}{-10yz}}\]
\[\frac{50y^4z^5}{-10yz}=-5y^3z^4\]
\[=\sqrt{-5y^3z^4}\]
\[=\sqrt{-5y^3z^4}\]
\[\mathrm{Apply\:exponent\:rule}:\quad \sqrt{-a}=\sqrt{-1}\sqrt{a}\]
\[\sqrt{-5y^3z^4}=\sqrt{-1}\sqrt{5y^3z^4}\]
\[\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\]
\[=i\sqrt{5y^3z^4}\]

Answer 16

Then...

\[\mathrm{Apply\:exponent\:rule}:\quad \sqrt{a\cdot \:b}=\sqrt{a}\sqrt{b}\]
\[\sqrt{5y^3z^4}=\sqrt{5}\sqrt{z^4}\sqrt{y^3}\]
\[=\sqrt{5}i\sqrt{z^4}\sqrt{y^3}\]
\[\sqrt{z^4}:\quad z^2\]
\[\sqrt{y^3}:\quad y^{\frac{3}{2}}\]
\[=\sqrt{5}iy^{\frac{3}{2}}z^2\]

Answer 17

@AloneS are you gotit

Answer 18

Go it?
Yeah

Answer 19

smarduha!

Answer 20

Or radical bb

Answer 21

Ricking auto correct