Answer:
AyyDaaaaaaaaaaaaaaaaaaaaaaaaaa
I think it’s A because the s is between P&Q and S&C if it’s wrong I’m sorry!
Answer:
Step-by-step explanation:
Left
When a square = a linear, always expand the squared expression.
x^2 - 2x + 1 = 3x - 5 Subtract 3x from both sides
x^2 - 2x - 3x + 1 = -5
x^2 - 5x +1 = - 5 Add 5 to both sides
x^2 - 5x + 1 + 5 = -5 + 5
x^2 - 5x + 6 = 0
This factors
(x - 2)(x - 3)
So one solution is x = 2 and the other is x = 3
Second from the Left
i = sqrt(-1)
i^2 = - 1
i^4 = (i^2)(i^2)
i^4 = - 1 * -1
i^4 = 1
16(i^4) - 8(i^2) + 4
16(1) - 8(-1) + 4
16 + 8 + 4
28
Second from the Right
This one is rather long. I'll get you the equations, you can solve for a and b. Maybe not as long as I think.
12 = 8a + b
<u>17 = 12a + b Subtract</u>
-5 = - 4a
a = - 5/-4 = 1.25
12 = 8*1.25 + b
12 = 10 + b
b = 12 - 10
b = 2
Now they want a + b
a + b = 1.25 + 2 = 3.25
Right
One of the ways to do this is to take out the common factor. You could also expand the square and remove the brackets of (2x - 2). Both will give you the same answer. I think expansion might be easier for you to understand, but the common factor method is shorter.
(2x - 2)^2 = 4x^2 - 8x + 4
4x^2 - 8x + 4 - 2x + 2
4x^2 - 10x + 6 The problem is factoring since neither of the first two equations work.
(2x - 2)(2x - 3) This is correct.
So the answer is D
