Here is the answer I believe. It goes in a x all it did was added to to each x and y (-3,0)
Answer:
![\frac{\sqrt[3]{16y^4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B16y%5E4%7D%7D%7Bx%5E2%7D)
Step-by-step explanation:
The options are missing; However, I'll simplify the given expression.
Given
![\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B32x%5E3y%5E6%7D%7D%7B%5Csqrt%5B3%5D%7B2x%5E9y%5E2%7D%20%7D)
Required
Write Equivalent Expression
To solve this expression, we'll make use of laws of indices throughout.
From laws of indices ![\sqrt[n]{a} = a^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%7D%20%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
So,
gives
![\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}](https://tex.z-dn.net/?f=%5Cfrac%7B%2832x%5E3y%5E6%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D%7B%282x%5E9y%5E2%29%5E%5Cfrac%7B1%7D%7B3%7D%7D)
Also from laws of indices
![(ab)^n = a^nb^n](https://tex.z-dn.net/?f=%28ab%29%5En%20%3D%20a%5Enb%5En)
So, the above expression can be further simplified to
![\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}](https://tex.z-dn.net/?f=%5Cfrac%7B%2832%5E%5Cfrac%7B1%7D%7B3%7Dx%5E%7B3%2A%5Cfrac%7B1%7D%7B3%7D%7Dy%5E%7B6%2A%5Cfrac%7B1%7D%7B3%7D%7D%29%7D%7B%282%5E%5Cfrac%7B1%7D%7B3%7Dx%5E%7B9%2A%5Cfrac%7B1%7D%7B3%7D%7Dy%5E%7B2%2A%5Cfrac%7B1%7D%7B3%7D%7D%29%7D)
Multiply the exponents gives
![\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}](https://tex.z-dn.net/?f=%5Cfrac%7B%2832%5E%5Cfrac%7B1%7D%7B3%7Dx%2Ay%5E%7B2%7D%29%7D%7B%282%5E%5Cfrac%7B1%7D%7B3%7Dx%5E%7B3%7D%2Ay%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%29%7D)
Substitute
for 32
![\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}](https://tex.z-dn.net/?f=%5Cfrac%7B%282%5E%7B5%2A%5Cfrac%7B1%7D%7B3%7D%7Dx%2Ay%5E%7B2%7D%29%7D%7B%282%5E%5Cfrac%7B1%7D%7B3%7Dx%5E%7B3%7D%2Ay%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%29%7D)
![\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}](https://tex.z-dn.net/?f=%5Cfrac%7B%282%5E%7B%5Cfrac%7B5%7D%7B3%7D%7Dx%2Ay%5E%7B2%7D%29%7D%7B%282%5E%5Cfrac%7B1%7D%7B3%7Dx%5E%7B3%7D%2Ay%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%29%7D)
From laws of indices
![\frac{a^m}{a^n} = a^{m-n}](https://tex.z-dn.net/?f=%5Cfrac%7Ba%5Em%7D%7Ba%5En%7D%20%3D%20a%5E%7Bm-n%7D)
This law can be applied to the expression above;
becomes
![2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}](https://tex.z-dn.net/?f=2%5E%7B%5Cfrac%7B5%7D%7B3%7D-%5Cfrac%7B1%7D%7B3%7D%7Dx%5E%7B1-3%7D%2Ay%5E%7B2-%5Cfrac%7B2%7D%7B3%7D%7D)
Solve exponents
![2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}](https://tex.z-dn.net/?f=2%5E%7B%5Cfrac%7B5-1%7D%7B3%7D%7D%2Ax%5E%7B-2%7D%2Ay%5E%7B%5Cfrac%7B6-2%7D%7B3%7D%7D)
![2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}](https://tex.z-dn.net/?f=2%5E%7B%5Cfrac%7B4%7D%7B3%7D%7D%2Ax%5E%7B-2%7D%2Ay%5E%7B%5Cfrac%7B4%7D%7B3%7D%7D)
From laws of indices,
; So,
gives
![\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B2%5E%7B%5Cfrac%7B4%7D%7B3%7D%7D%2Ay%5E%7B%5Cfrac%7B4%7D%7B3%7D%7D%7D%7Bx%5E2%7D)
The expression at the numerator can be combined to give
![\frac{(2y)^{\frac{4}{3}}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%282y%29%5E%7B%5Cfrac%7B4%7D%7B3%7D%7D%7D%7Bx%5E2%7D)
Lastly, From laws of indices,
; So,
becomes
![\frac{\sqrt[3]{(2y)}^{4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B%282y%29%7D%5E%7B4%7D%7D%7Bx%5E2%7D)
![\frac{\sqrt[3]{16y^4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B16y%5E4%7D%7D%7Bx%5E2%7D)
Hence,
is equivalent to ![\frac{\sqrt[3]{16y^4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B16y%5E4%7D%7D%7Bx%5E2%7D)
Check the picture below.
so... you can pretty much see how long RS and QT are, you can just count the units off the grid.
now, let's find QR's length
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &Q&(~ 8 &,& 8~) % (c,d) &R&(~ 14 &,& 16~) \end{array}~~ % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ QR=\sqrt{(14-8)^2+(16-8)^2}\implies QR=\sqrt{6^2+8^2} \\\\\\ QR=\sqrt{36+64}\implies QR=\sqrt{100}\implies QR=10](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26Q%26%28~%208%20%26%2C%26%208~%29%20%0A%25%20%20%28c%2Cd%29%0A%26R%26%28~%2014%20%26%2C%26%2016~%29%0A%5Cend%7Barray%7D~~%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AQR%3D%5Csqrt%7B%2814-8%29%5E2%2B%2816-8%29%5E2%7D%5Cimplies%20QR%3D%5Csqrt%7B6%5E2%2B8%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AQR%3D%5Csqrt%7B36%2B64%7D%5Cimplies%20QR%3D%5Csqrt%7B100%7D%5Cimplies%20QR%3D10)
and let's also find the length for ST
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &S&(~ 20 &,& 16~) % (c,d) &T&(~ 22 &,& 8~) \end{array}~ d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ ST=\sqrt{(22-20)^2+(8-16)^2}\implies ST=\sqrt{2^2+(-8)^2} \\\\\\ ST=\sqrt{4+64}\implies ST=\sqrt{68}\implies ST=\sqrt{4\cdot 17} \\\\\\ ST=\sqrt{2^2\cdot 17}\implies ST=2\sqrt{17}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26S%26%28~%2020%20%26%2C%26%2016~%29%20%0A%25%20%20%28c%2Cd%29%0A%26T%26%28~%2022%20%26%2C%26%208~%29%0A%5Cend%7Barray%7D~%20%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AST%3D%5Csqrt%7B%2822-20%29%5E2%2B%288-16%29%5E2%7D%5Cimplies%20ST%3D%5Csqrt%7B2%5E2%2B%28-8%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AST%3D%5Csqrt%7B4%2B64%7D%5Cimplies%20ST%3D%5Csqrt%7B68%7D%5Cimplies%20ST%3D%5Csqrt%7B4%5Ccdot%2017%7D%0A%5C%5C%5C%5C%5C%5C%0AST%3D%5Csqrt%7B2%5E2%5Ccdot%2017%7D%5Cimplies%20ST%3D2%5Csqrt%7B17%7D)
so, add the lengths of all sides, and that's the perimeter of the trapezoid.