Product Rules of Powers
082
There are no steps to this problem it just equals to 0 because every time you multiply 0 to something it always equals 0. There is no value so in this case 082 equals 0.
x5/x3
x5-3
=x2
In the quotient rules of powers when there is a sign of division you subtract the exponents that are combine alike terms. For example in this case its x over x so whatever exponents they have you subtract it.
4-2/3-3
4-2/3-3
=33/42
2y2 * 3x
2y2 * 3x
2*3 * y2 * x
=6y2x
x4y3 * (2y2)0
x4y3 * (2y2)0
x4y3 * 20y0/1
=x4y3
4x3 * 2x3
4x3 * 2x3
4x * 2x= 8x
8x3+3
=8x6
x3y2/x2y5
x3y2/x2y5
x3-2 =x1
y2-5 =y-3
=x1y-3
(4xy)-1
(4xy)-1
4xy/1 on the 4, x, and y there is an invisible negative 1 located
=1/4xy
(4r0)4
(4r0)4
44 r8
=256
(2x0y2)-3 * 2yx3
(2x0y2)-3 * 2yx3
2-3x0y-6 * 2yx3 0+3=3 & -6+1=-5
2x3/8y5 =1x3/4y5
=x3/4y5
(x2y) * (x3y4)
(x2y) * (x3y4)
=x2+3 =x5
=y1+5 =y5
=x5y5
x-1/4x4
x-1/4x4
x-1-4/4 = x-5/4
= 1/4x5
4r-3 * 2r2
4r-3 * 2r2
8r-3+2
8r-1/1 =8/r1 r-2/d-1 =d1/r2
=8/r
(2x4y-3)-1
(2x4y-3)-1
2-1x-4y3/1 =y3/22x4
=y3/4x4
2h3j-3k4/3jk
2h3j-3k4/3jk
2h3j-3-1k4-1/3
2h3j-4k3/3
=2h3k3/3j4
(2x2)-4
(2x2)-4
2*2*2*2=16
1/16x+8
4x0y-2z3/4x
4x0y-2z3/4x
0 crosses out the 4x on the numerator and denominator
=y-2z3/x
=z3/xy2
x2y-4 * x3y2
x2y-4 * x3y2
=x2+3y-4+2
=x5y-2
=x5/y2
2x4y-4z-3/3x2y-3z4
2x4y-4z-3/3x2y-3z4
=2x4-2y-4-(-3)z-3-4/3
=2x2y-1z-7/3
=2x2/3y1z7
(2hj2k-2 * h4j-1k4)0/2h-3j-4k-2
(2hj2k-2 * h4j-1k4)0/2h-3j-4k-2
(2h5jk2)0/2h-3j-4k-2
=h3j4k2/2
(2x-2y3z-4)2
(2x-2y3z-4)2
2x2x-4y6z-8
4x-4y6z-8
=4y6/x4z8
4m4n3p3/3m2n2p4
4m4n3p3/3m2n2p4
=4m4-2n3-2p3-4/3 =4m2np-1/3
=4m2n/3
4a3b2 * 3a-4b-3
4a3b2 * 3a-4b-3
12a3+(-4)b2+(-3)
12a-1, b-1
=12/ab
3x3y-1z-1/x-4y0z0
3x3y-1z-1/x-4y0z0
3x3-(-4)y-1-0z-1-0
3x7y-1z-1/1
=3x7/yz
(2pm-1q0)-4 * 2m-1p3/2pq2
(2pm-1q0)-4 * 2m-1p3/2pq2
2m3/32p2q2
=m3/16p2q2