So we have a 6 at the bottom, and the root is 3, so hmm how to take it out, simple enough, just let's get something to make the 6 a 6³, so it comes out of the root
so
![\bf \cfrac{2+\sqrt[3]{3}}{\sqrt[3]{6}}\cdot \cfrac{\sqrt[3]{6^2}}{\sqrt[3]{6^2}}\implies \cfrac{(2+\sqrt[3]{3})(\sqrt[3]{6^2})}{(\sqrt[3]{6})(\sqrt[3]{6^2})}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3}\cdot \sqrt[3]{36}}{\sqrt[3]{6^3}} \\\\\\ \cfrac{2\sqrt[3]{36}+\sqrt[3]{3\cdot 36}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{108}}{6}\implies \cfrac{2\sqrt[3]{36}+\sqrt[3]{3^3\cdot 4}}{6} \\\\\\ \cfrac{2\sqrt[3]{36}+3\sqrt[3]{ 4}}{6}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B2%2B%5Csqrt%5B3%5D%7B3%7D%7D%7B%5Csqrt%5B3%5D%7B6%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%5B3%5D%7B6%5E2%7D%7D%7B%5Csqrt%5B3%5D%7B6%5E2%7D%7D%5Cimplies%20%5Ccfrac%7B%282%2B%5Csqrt%5B3%5D%7B3%7D%29%28%5Csqrt%5B3%5D%7B6%5E2%7D%29%7D%7B%28%5Csqrt%5B3%5D%7B6%7D%29%28%5Csqrt%5B3%5D%7B6%5E2%7D%29%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B%5Csqrt%5B3%5D%7B3%7D%5Ccdot%20%5Csqrt%5B3%5D%7B36%7D%7D%7B%5Csqrt%5B3%5D%7B6%5E3%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B%5Csqrt%5B3%5D%7B3%5Ccdot%2036%7D%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B%5Csqrt%5B3%5D%7B108%7D%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B%5Csqrt%5B3%5D%7B3%5E3%5Ccdot%204%7D%7D%7B6%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B2%5Csqrt%5B3%5D%7B36%7D%2B3%5Csqrt%5B3%5D%7B%204%7D%7D%7B6%7D)
Answer:
you can eat this noddles then you will got answer
Answer:
We are given an area and three different widths and we need to determine the corresponding length and perimeter.
The first width that is provided is 4 yards and to get an area of 100 we need to multiply it by 25 yards. This would mean that our length is 25 yards and our perimeter would be 2(l + w) which is 2(25 + 4) = 58 yards.
The second width that is given is 5 yards and in order to get an area of 100 yards we need to multiply by 20 yards. This would mean that our length is 20 yards and our perimeter would be 2(l + w) which is 2(20 + 5) = 50 yards.
The final width that is given is 10 yards and in order to get an area of 100 yards we need to multiply by 10. This would mean that our length is 10 yards and our perimeter would be 2(l + w) which is 2(10 + 10) = 40 yards.
Therefore the field that would require the least amount of fencing (the smallest perimeter) is option C, field #3.
<u><em>Hope this helps!</em></u>
The answer for number 3 is 3.c
So slope intercept form y=mx+b slope is m
m=6 so
y=6x+b
this is not the legit way but it works so
put in 3 for x and 4 for y and solve for b
4=6(3)+b
4=18+b
-14=b
b=-14
the slope intercep form is y=6x-14
