<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>
The answer is B! b^4. If we examine the expression given, we have a^3 * b^3 * a^-3 * b= a^3/a^3 * b^3 * b = 1 * b^4
Hope this helps! Any questions please feel free to ask!!
Thank you!!
Answer:
2317.2
Step-by-step explanation:
2327.2-10
=2317.2
Answer:
IT CAN BE MANY THINGS
Step-by-step explanation:
