First, we are going to use the Pythagorean identity:

to rewrite our expression. Solving for

in our Pythagorean identity we get that

, so we can rewrite our expression as follows:

Next, we are going to use the trig identity:

to completely simplify our expression:

We can conclude that
it is C. i think because this i looked it up trust me
3- 2b +4 = 2-7b
3+4 = 2 - 5b
7 = 2 -5b
5 = -5b
-5b /-5 = -1
b = -1
<span>x2 +8x +4y +4 = 0
</span>4y=<span> -x2 -8x -4</span>y = -.25*x^2 -2x -1
a = -.25b = -2c = -1
x position of vertex:
h = -b / 2a
h = 2 / 2*-.25h = 2 / -.5h = -4
y position of vertex:
k = ah^2 + bh + ck = -.25*-4^2 + -2*-4 + -1k = -4 +8 -1k = 3
VERTEX = (-4, 3)**************************************************************************
x value of focus =x value of vertex = -4
y value of focus =(1 (-b^2 -4ac)) / 4a
a = -.25 b = -2 c =-1
y value = (1 (-4 -4*-.25*-1)) / 4*-.25
y value = (1 (-4 -4*-.25*-1)) / -1
y value = (1 -4 +1) / -1y value = (-2 / -1)y value = 2
focus value = (-4, 2)
Answer is the last one.
Answer: the only thing there is 1x29
Step-by-step explanation: hope it helps