((2y)^3(y^4))/(8y)^2

Simple and best practice solution for ((2y)^3(y^4))/(8y)^2 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework.

If it's not what You are looking for type in the equation solver your own equation and let us solve it.

Solution for ((2y)^3(y^4))/(8y)^2 equation: 


D( y )

(8*y)^2 = 0

(8*y)^2 = 0

(8*y)^2 = 0

64*y^2 = 0 // : 64

y^2 = 0

y = 0

y in (-oo:0) U (0:+oo)

((2*y)^3*y^4)/((8*y)^2) = 0

1/8*y^5 = 0 // : 1/8

y^5 = 0

y = 0

y in { 0} 

y belongs to the empty set

See similar equations:

| p(x)=[x-(1+i)][x-(1-i)](x^2-2)(x+3) | | 3(2x+6)=6x-10 | | 15=-3n+1+10n | | s=3.5+-2.5t | | 5e^4x+3=25 | | Lnx-3=2 | | x+(x-5)+(x+25)=176 | | -2t+8t+5=17 | | 63-9n=12-2n | | 2s+5t=7 | | 106x-1=181 | | 8+1.50s=20 | | 3x+y-375=0 | | -6x-8=4x-98 | | 4(2c-3)-2=-6-8(1-e) | | 4(5x+3)=252 | | -6(h-2)=30 | | 60x=26+8x | | -5(-4x+4)=16 | | 33=x/8 | | k^2+4=6 | | p(x)=[x-(1+i)][x-(1-i)] | | f(19)=-64+36x-12x^2+x^3 | | 15p+9p=72 | | 0.92*5.6= | | 5x+8y=131 | | f(-4)=x^(4)/6-10x | | 9+4z=4z-3 | | -2[9+3x]=-36 | | x+17=5x-23 | | -6k-7=(6k-7) | | 16x^3-9/x=0 |

Equations solver categories