Wx+yz=bc
-yz. -yz
Wx=bc-yz
Wx/W=bc-yz/W
x=bc-yz/W
A: (x + 5i)^2
= (x + 5i)(x + 5i)
= (x)(x) + (x)(5i) + (5i)(x) + (5i)(5i)
= x^2 + 5ix + 5ix + 25i^2
= 25i^2 + 10ix + x^2
B: (x - 5i)^2
= (x + - 5i)(x + - 5i)
= (x)(x) + (x)(- 5i) + (- 5i)(x) + (- 5i)(- 5i)
= x^2 - 5ix - 5ix + 25i^2
= 25i^2 - 10ix + x^2
C: (x - 5i)(x + 5i)
= (x + - 5i)(x + 5i)
= (x)(x) + (x)(5i) + (- 5i)(x) + (- 5i)(5i)
= x^2 + 5ix - 5ix - 25i^2
= 25i^2 + x^2
D: (x + 10i)(x - 15i)
= (x + 10i)(x + - 15i)
= (x)(x) + (x)(- 15i) + (10i)(x) + (10i)(- 15i)
= x^2 - 15ix + 10ix - 150i^2
= - 150i^2 + 5ix + x^2
Hope that helps!!!
Answer:
The answer is 12.4
Step-by-step explanation:
Because it is this way.
To simplify that expression you'll join like terms.
8x + 4 - 5
8x - 1
