Answer:
-5(1-I) is the answer
Step-by-step explanation:
(1+2i)(1+3i)
1(1+3i)+2i(1+3i)
1+3i+2i+6i^2
As i^2=-1
1+5i+6(-1)
1-6+5i
-5+5i
Taking -5 as common
-5(1-i)
I hope this will help you :)
Answer: 23600000
Step-by-step explanation:
It is not true because there C and E are on different sides, if that makes sense (which means your answer is correct)
#1. B
<span>(z * z^2 + z * 2z + z * 4) – (-2 *z^2 – (-2) 2z – (-2) 4)
Z^3 + 2z^2 + 4z – 2z^2 -4z – 8
Z^3 + 2z^2 – 2z^2 + 4z – 4z – 8
Z^3 - 8
</span>
#2 and #3. D
<span>(x + y)(x + 2)
x^2 + 2x + yx + 2y
</span>
#4. D.
<span>(x - 7)(x + 7)(x- 2)
x^2 + 7x – 7x -49
x^2 + x – 49
x^2 -49
(x^2 – 49 ) (x – 2)
x^3 – 2x^2 – 49x + 98
</span>
#5. C
(y - 4) = 0
y = 4
(x + 3)= 0
x = -3
#6. A and B
Answer:
x=2.5
Step-by-step explanation:
Solve for x in this systems of equations, we already have two values which equal y so half the work is finished for us.
Starting with the equation:
Add x to both sides to remove the negative x on the right side of the equation:
Subtract 12 from both sides to remove the 12 on the left side of the equation:
Divide both sides by 2 to cancel out the x coefficient
