1.)
=(x-8i)(x+8i)
x^2+8ix-8ix-64i^2
x^2-64i^2
x^2-64(-1)
x^2+64
2.)
=(4x-7i)(4x+7i)
16x^2+28ix-28ix-49i^2
16x^2-49i^2
16x^2-49(-1)
16x^2+49
3.)
=(x+9i)(x+9i)
x^2+9ix+9ix+81i^2
x^2+18ix+81(-1)
x^2+18ix-81
4.)
=(x-2i)(x-2i)
x^2-2ix-2ix+4i^2
x^2-4ix+4(-1)
x^2-4ix-4
5.)
=[x+(3+5i)]^2
(x+5i+3)^2
(x+5i+3)(x+5i+3)
x^2+5ix+3x+5ix+25i^2+15i+3x+15i+9
x^2+6x+10ix+30i+25i^2+9
x^2+6x+10ix+30i+25(-1)+9
x^2+6x+10ix+30i-25+9
x^2+6x+10ix+30i-16
Hope this helps :)
X+y = 4, therefore y = 4-x when you rearrange the equation.
You can subsitute this in: x-(4-x) = 6
-4+x = 6-x
x=10-x
2x = 10
x = 5
Substitute x into the previous equation:
5+y = 4
y = 4-5
y = -1
15:82 I have to type this so I can send in the answer
<h3>
Answer: x^2 + y^2 = 40</h3>
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Explanation:
The center is the origin, so (h,k) = (0,0).
The point (x,y) = (2,6) is on the circle's edge. We'll use these four values to find the value of r^2
(x-h)^2 + (y-k)^2 = r^2
(2-0)^2 + (6-0)^2 = r^2
2^2 + 6^2 = r^2
4 + 36 = r^2
40 = r^2
r^2 = 40
We don't need to solve for r itself. If you wanted to, you'd get the radius to be r = sqrt(40) = 2*sqrt(10) = 6.324555 approximately.
------------------
Since h = 0, k = 0, and r^2 = 40, we can say:
(x-h)^2 + (y-k)^2 = r^2
(x-0)^2 + (y-0)^2 = 40
x^2 + y^2 = 40
Answer:
The answer for this problem is
C) y = -x + 6
Step-by-step explanation:
I know this because I just did the test
