<em>The </em><em>value</em><em> </em><em>of</em><em> </em><em>u</em><em> </em><em>is</em><em> </em><em>-</em><em>3</em><em>4</em><em>3</em>
<em>pl</em><em>ease</em><em> see</em><em> the</em><em> attached</em><em> picture</em><em> for</em><em> full</em><em> solution</em>
<em>Hope</em><em> </em><em>it</em><em> helps</em>
<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>
Answer:
5,333 square yards.
Step-by-step explanation:
Football fields have standard 100 yards length, if you walk the perimeter of a football field you will walk the twice the length and twice the width. Therefore, the width of the field can be found as follows:
The area of the playing field is the product of the length by the width
The area of the playing field is 5,333 square yards.
Answer:
Step-by-step explanation:
Given that the number of orders received daily by an online vendor of used CDs (say X) is normally distributed with mean 270 and standard deviation 16.
X is N(270, 16)
To find percentage of days will the company have to hire extra help or pay overtime, we can find probability using std normal table
X>302 means

P(Z>2) = 0.25
Thus 25% of days will the company have to hire extra help or pay overtime
The total length of the fencing is 248 ft; we call this the "perimeter" of the garden. Since P = 2W + 2L in general, here:
P = 248 ft = 2W + 2(W+6), or 248 ft = 4W + 12.
Subtracting 12 from both sides of this equation: 236 ft = 4W. Then W = 59 ft, which means that L = 65 ft.
If you're using the app, try seeing this answer through your browser: brainly.com/question/2799412_______________
Let

(that is the range of the inverse sine function).
So,
![\mathsf{sin\,\theta=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\theta=x\qquad\quad(i)}](https://tex.z-dn.net/?f=%5Cmathsf%7Bsin%5C%2C%5Ctheta%3Dsin%5C%21%5Cleft%5Bsin%5E%7B-1%7D%28x%29%5Cright%5D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bsin%5C%2C%5Ctheta%3Dx%5Cqquad%5Cquad%28i%29%7D)
Square both sides:

Since

then

is positive. So take the positive square root and you get

Then,
![\mathsf{tan\,\theta=\dfrac{sin\,\theta}{cos\,\theta}}\\\\\\ \mathsf{tan\,\theta=\dfrac{x}{\sqrt{1-x^2}}}\\\\\\\\ \therefore~~\mathsf{tan\!\left[sin^{-1}(x)\right]=\dfrac{x}{\sqrt{1-x^2}}\qquad\qquad -1\ \textless \ x\ \textless \ 1.}](https://tex.z-dn.net/?f=%5Cmathsf%7Btan%5C%2C%5Ctheta%3D%5Cdfrac%7Bsin%5C%2C%5Ctheta%7D%7Bcos%5C%2C%5Ctheta%7D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7Btan%5C%2C%5Ctheta%3D%5Cdfrac%7Bx%7D%7B%5Csqrt%7B1-x%5E2%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%20%5Ctherefore~~%5Cmathsf%7Btan%5C%21%5Cleft%5Bsin%5E%7B-1%7D%28x%29%5Cright%5D%3D%5Cdfrac%7Bx%7D%7B%5Csqrt%7B1-x%5E2%7D%7D%5Cqquad%5Cqquad%20-1%5C%20%5Ctextless%20%5C%20x%5C%20%5Ctextless%20%5C%201.%7D)
I hope this helps. =)
Tags: <em>inverse trigonometric function sin tan arcsin trigonometry</em>