. Correct option C)5.38
<u>Step-by-step explanation:</u>
Here we have , Find the value of x. O is the center of the circle. Round your answer to the nearest hundredth. picture is attached . Let's find out:
In the given figure , Let's draw a line from center to the point where cord of length 8 unit is touching the circle or intersecting or , where this chord finishes . This line is radius of circle and is denoted by x , So now we have a right angle triangle with dimensions as :

By Pythagoras Theorem ,

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Therefore ,
. Correct option C)5.38
Answer:
Its 48 °
Since its a triangle, the total angle will be 180°
So,
x° + 42°+90°=180°(90° as one angle in triangle is 90°)
x°=180°-90°-42°
x°=48°
Answer:
We want to find:
![\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7Bn%21%7D%20%7D%7Bn%7D)
Here we can use Stirling's approximation, which says that for large values of n, we get:

Because here we are taking the limit when n tends to infinity, we can use this approximation.
Then we get.
![\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} = \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7Bn%21%7D%20%7D%7Bn%7D%20%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7B%5Csqrt%7B2%2A%5Cpi%2An%7D%20%2A%28%5Cfrac%7Bn%7D%7Be%7D%20%29%5En%7D%20%7D%7Bn%7D%20%3D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7Bn%7D%7Be%2An%7D%20%2A%5Csqrt%5B2%2An%5D%7B2%2A%5Cpi%2An%7D)
Now we can just simplify this, so we get:
![\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B1%7D%7Be%7D%20%2A%5Csqrt%5B2%2An%5D%7B2%2A%5Cpi%2An%7D%20%5C%5C)
And we can rewrite it as:

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.
Thus:

Answer:
first option: Harulo is correct because the angle is coterminal with 3π / 4 and the reference angle is π / 4.
Explanation:
1) 19 π/4 = 4π + 3 π/4, which is 4 complete turns and 3/4 of turn.
2) 3 π/4 is in the third quadrant, so the reference angle is π - 3 π/4 = π/4
3) sin (π/4) = sin (3π/4) = (√2) / 2
4) csc (π/4) = 1 / [ sin (3π/4) ] = 1 / [ (√2) / 2 ] = 2 / (√2) = √2, which shows the validity of the statement and csc(19π/4) = √2.