So, f[x] = 1/4x^2 - 1/2Ln(x)
<span>thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x </span>
<span>thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) </span>
<span>thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 </span>
<span>thus Sqrt[...] = (x/2 + 1/2x) </span>
Answer:
The measure of the Reference angle is 30vdegrees and the second answer should be Tan(0)=1
Step-by-step explanation:
Answer:
R(16,-8)
Step-by-step explanation:
if (x1,y1) and (x2,y2) are coordinates of two points and (x,y) the coordinate of midpoint.
then x=(x1+x2)/2
and y=(y1+y2)/2
let coordinates of R are (x,y)
4=(-8+x)/2
8=-8+x
x=8+8=16
-3=(2+y)/2
-6=2+y
y=-6-2=-8
so R is (16,-8)
