<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>
Answer:
For the second one: NO congruent to MO.
For the third one: angle N congruent to angle O.
:)
Answer:67738373783
Step-by-step explanation:
Answer:
The correct value is t* = 2.797.
Step-by-step explanation:
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 25 - 1 = 24
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a two-tailed confidence level of 
. So we have T = 2.797.
Less then or equal to, and your sign is pointing the wrong way I believe.
