To simplify
![\sqrt[4]{\dfrac{24x^6y}{128x^4y^5}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B24x%5E6y%7D%7B128x%5E4y%5E5%7D%7D)
we need to use the fact that
![\sqrt[4]{x^4}=|x|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E4%7D%3D%7Cx%7C)
Why the absolute value? It's because
.
We start by rewriting as
![\sqrt[4]{\dfrac{2^23x^6y}{2^6x^4y^5}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B2%5E23x%5E6y%7D%7B2%5E6x%5E4y%5E5%7D%7D)
![\sqrt[4]{\dfrac{2^23x^4x^2y}{2^42^2x^4y^4y}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B2%5E23x%5E4x%5E2y%7D%7B2%5E42%5E2x%5E4y%5E4y%7D%7D)
Since
, we have
, and the above reduces to
![\sqrt[4]{\dfrac{3x^2y}{2^4y^4y}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B3x%5E2y%7D%7B2%5E4y%5E4y%7D%7D)
Then we pull out any 4th powers under the radical, and simplify everything we can:
![\dfrac1{\sqrt[4]{2^4y^4}}\sqrt[4]{\dfrac{3x^2y}{y}}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%5Csqrt%5B4%5D%7B2%5E4y%5E4%7D%7D%5Csqrt%5B4%5D%7B%5Cdfrac%7B3x%5E2y%7D%7By%7D%7D)
![\dfrac1{|2y|}\sqrt[4]{3x^2}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%7C2y%7C%7D%5Csqrt%5B4%5D%7B3x%5E2%7D)
where
allows us to write
, and this also means that
. So we end up with
![\dfrac{\sqrt[4]{3x^2}}{2y}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B4%5D%7B3x%5E2%7D%7D%7B2y%7D)
making the last option the correct answer.
Answer:
B) 116°
Step-by-step explanation:
y=180-54-64=62
x=180-62-64=54
62+54=116
Answer : 96
x – y = 16
--------> equation 1

x is the higher grade and y is the lower grade
We solve the first equation for y
x - y = 16
-y = 16 -x ( divide each term by -1)
y = -16 + x
Now substitute y in second equation



Take common denominator to combine fractions


Add 8 on both sides

Multiply both sides by 
x = 96
We know x is the higher grade
96 is the higher grade of Jose’s two tests.
Chegg porter averages 21 points per basketball
game with a standard deviation of 4.5 points.
suppose porter's points per basketball game
are normally distributed. let x