When finding the slope between two points, we simply take (y2-y1) / (x2-x1)
For this, we can substitute the following to get this :
y2-y1 and x2-x1
1 - 4 and 7 - 1
-3 and 6
-3 / 6 gets you -1/2
Therefore the answer is :
The slope of the line is -1/2
Answer:
2 and 6 for the circles.
-18 for the square.
Step-by-step explanation:
-3 times a number is -6.

The number is 2.
2 times a number is 12.

The number is 6.
6 times -3.

The number is -18.
Answer:♀️
Step-by-step explanation:
![\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.