<span>y = tan^−1(x2/4)</span>
tan(y) = x2/4
sec2(y) = x/2
y′ = xcos^2(y)/2
<span>cos^2(y) = <span>16x2+16</span></span>
<span>y′ = <span>8x/(<span>x2+16)
let u be x2+16
du is 2x dx
dy = 4 du / u
y = 4 ln (</span></span></span>x2 <span>+ 16)
y at x =0 = </span> 4 ln (<span>16) = 11.09</span>
Pythagoras theorem I guess
Answer:
-1
Step-by-step explanation:
The midline is the average of the maximum and minum.
0.5 = (y + 2) / 2
1 = y + 2
y = -1
The minimum is at y=-1.
9 2/3 - 2 8/10 = 16 13/15
