For the first one is has greater then 3 terms
The middle one is has exactly one term
And the last one is has two terms
I believe I hope this helps
![\bf \cfrac{\sqrt[4]{63}}{4\sqrt[4]{6}}\qquad \begin{cases} 63=3\cdot 3\cdot 7\\ 6=2\cdot 3 \end{cases}\implies \cfrac{\sqrt[4]{3\cdot 3\cdot 7}}{4\sqrt[4]{2\cdot 3}}\implies \cfrac{\underline{\sqrt[4]{3}}\cdot \sqrt[4]{3}\cdot \sqrt[4]{7}}{4\sqrt[4]{2}\cdot \underline{\sqrt[4]{3}}} \\\\\\ \cfrac{\sqrt[4]{3}\cdot \sqrt[4]{7}}{4\sqrt[4]{2}}\implies \cfrac{\sqrt[4]{3\cdot 7}}{4\sqrt[4]{2}}\implies \cfrac{\sqrt[4]{21}}{4\sqrt[4]{2}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B63%7D%7D%7B4%5Csqrt%5B4%5D%7B6%7D%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0A63%3D3%5Ccdot%203%5Ccdot%207%5C%5C%0A6%3D2%5Ccdot%203%0A%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B3%5Ccdot%203%5Ccdot%207%7D%7D%7B4%5Csqrt%5B4%5D%7B2%5Ccdot%203%7D%7D%5Cimplies%20%5Ccfrac%7B%5Cunderline%7B%5Csqrt%5B4%5D%7B3%7D%7D%5Ccdot%20%5Csqrt%5B4%5D%7B3%7D%5Ccdot%20%5Csqrt%5B4%5D%7B7%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%5Ccdot%20%5Cunderline%7B%5Csqrt%5B4%5D%7B3%7D%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%5Csqrt%5B4%5D%7B3%7D%5Ccdot%20%5Csqrt%5B4%5D%7B7%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B3%5Ccdot%207%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D)
![\bf \textit{now, rationalizing the denominator}\\\\ \cfrac{\sqrt[4]{21}}{4\sqrt[4]{2}}\cdot \cfrac{\sqrt[4]{2^3}}{\sqrt[4]{2^3}}\implies \cfrac{\sqrt[4]{21}\cdot \sqrt[4]{8}}{4\sqrt[4]{2}\cdot \sqrt[4]{2^3}}\implies \cfrac{\sqrt[4]{21\cdot 8}}{4\sqrt[4]{2\cdot 2^3}}\implies \cfrac{\sqrt[4]{168}}{4\sqrt[4]{2^4}} \\\\\\ \cfrac{\sqrt[4]{168}}{4\cdot 2}\implies \cfrac{\sqrt[4]{168}}{8}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bnow%2C%20rationalizing%20the%20denominator%7D%5C%5C%5C%5C%0A%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B2%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B2%5E3%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%7D%5Ccdot%20%5Csqrt%5B4%5D%7B8%7D%7D%7B4%5Csqrt%5B4%5D%7B2%7D%5Ccdot%20%5Csqrt%5B4%5D%7B2%5E3%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B21%5Ccdot%208%7D%7D%7B4%5Csqrt%5B4%5D%7B2%5Ccdot%202%5E3%7D%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B168%7D%7D%7B4%5Csqrt%5B4%5D%7B2%5E4%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%5Csqrt%5B4%5D%7B168%7D%7D%7B4%5Ccdot%202%7D%5Cimplies%20%5Ccfrac%7B%5Csqrt%5B4%5D%7B168%7D%7D%7B8%7D)
and is all you can simplify from it.
so... all we did, was rationaliize it, namely, "getting rid of the pesky radical at the bottom", we do so by simply multiplying it by something that will raise the radicand, to the same degree as the root, thus the radicand comes out.
Step-by-step explanation:
From the figure it is clear that:
x+110°=180
x=180-110
x=70
For y:
the left exterior angle of line q is also70.
By the definition of alternate exterior angle:
y=70
<u>Note</u><u>:</u><u>i</u><u>f</u><u> </u><u>you</u><u> </u><u>n</u><u>e</u><u>e</u><u>d</u><u> </u><u>t</u><u>o</u><u> </u><u>a</u><u>s</u><u>k</u><u> </u><u>a</u><u>n</u><u>y</u><u> </u><u>question</u><u> </u><u>please</u><u> </u><u>let</u><u> </u><u>me</u><u> </u><u>know</u><u>.</u>