One because if x3 = 1 then x must equal 1
The commutative law of multiplication.
1
X3-5=3 X3=3+5 X3=8 cube root of X3=cube root of 8 X=1.68 I hope this could help. There's not really any good ways to write algebra in a normal text box.
x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
One because if x3 = 1 then x must equal 1
The commutative law of multiplication.
x3 + 8 = 0, then x3 = -8, therefore x = -2 -2 x -2 = 4 4 x -2 = -8
1
X3-5=3 X3=3+5 X3=8 cube root of X3=cube root of 8 X=1.68 I hope this could help. There's not really any good ways to write algebra in a normal text box.
x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0
y5(x3 - 8)
2x - y = 8 x + y = 1 These are your two equations. They will have two solutions since you have two variables. The solutions are x=3 and y=-2
107 divided by 12 equals 8 with a remainder of 11. There are many other possible solutions as well.
They are: (3, 1) and (-11/5, -8/5)
using the t-table determine 3 solutions to this equation: y equals 2x
One.