Sqrt(x+4-3)=1
sqrt (x+1) = 1
x+1 = 1^2
x=1-1
x=0
This is how you would get x=0
Hope I helped :)
Answer:
D. ![xy\sqrt[3]{9y}](https://tex.z-dn.net/?f=xy%5Csqrt%5B3%5D%7B9y%7D)
Step-by-step explanation:
![\sqrt[3]{9x^3y^4}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B9x%5E3y%5E4%7D)
![\sqrt[3]{9}\sqrt[3]{x^3}\sqrt[3]{y^4}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B9%7D%5Csqrt%5B3%5D%7Bx%5E3%7D%5Csqrt%5B3%5D%7By%5E4%7D)
The
cancels out to become x:
![\sqrt[3]{9}x\sqrt[3]{y^4}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B9%7Dx%5Csqrt%5B3%5D%7By%5E4%7D)
Split the 
![\sqrt[3]{9}x\sqrt[3]{y^3}\sqrt[3]{y^1}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B9%7Dx%5Csqrt%5B3%5D%7By%5E3%7D%5Csqrt%5B3%5D%7By%5E1%7D)
![\sqrt[3]{y^3} =y](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%5E3%7D%20%3Dy)
![xy\sqrt[3]{9} \sqrt[3]{y}](https://tex.z-dn.net/?f=xy%5Csqrt%5B3%5D%7B9%7D%20%5Csqrt%5B3%5D%7By%7D)
Put the cube root of y and cube root of 9 together:
![xy\sqrt[3]{9y}](https://tex.z-dn.net/?f=xy%5Csqrt%5B3%5D%7B9y%7D)
Okay you have 2 sum unknow lets call them X+Y
so X+Y=33, 4X=108
Get X by it self then plug it in 4X=108
X=-Y+33
4(-y+33)= 108
-4y+132=108 Then subtract the 132
-4y=-24 Y=6 then to find X subtract 33-6=27
Y=6 X=27 if you check it works out