<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>
16r^2+ 2r -4r
16r^2 - 2r
2r ( 8r -1)
Answer:
Step-by-step explanation:
<u>Solving in steps:</u>
- - 8 3/4 ÷ 2 1/6 =
- - 35/4 ÷ 13/6 =
- - 35/4 × 6/13 =
- - 35/2 × 3/13 =
- - 105/26 =
- - 4 1/26
I'm sorry but do you ha e a picture of a flowchart
Answer: p = −15
Step-by-step explanation:
Add 11p to both sides.
−10p+11p=−11p−15+11p
p=−15
