A)Circle
B)Hexagon
C)rectangle
D)Square
E)Trapezoid
F)Triangle
Jessica's result means that any value of x (or any variable she used) would not be a solution to the equation. When something like 0 = 2 comes up, they answer is no solution.
For Your Information:
If you get 2=2 or 0=0 or anything like that, the solutions are all real numbers. Any value of x (or any variable used) will be a solution to the equation.
Please comment to let me know you understand. I hope this was helpful!
A. 2(z2 + 1)(z2 – 1)
B. 2(z2 + 1)(z – 1)(z + 1)
C. Prime
D. 2(z2 + 1)2
Answer:
The correct answer is B,
Step-by-step explanation:
Since the higher value of z's power, is 4, an even number, and the second number is negative, we can think that this binomial is made of a difference of squares, so that is what we are going to factorize.
First, extract the common number (if any), the number 2, we have now>
This is convenient since "1" is a wonderful number that has this feature> No matters what n's value is,
so the first equation can be written as
The later termn, can also be factorized, using the same as befre.
Remember that