Solve each equation for y 2x+3y=18 plus show work
2 answers:
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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>
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Let
(that is the range of the inverse sine function).
So,
Square both sides:
Since
Then,
I hope this helps. =)
Tags: <em>inverse trigonometric function sin tan arcsin trigonometry</em>
<h3>Answer: B. Graph is nearly symmetrical</h3>
Explanation:
Given information:
- A number line going from 2 to 11.
- 0 dots are above 2.
- 0 dots are above 3.
- 1 dot is above 4.
- 2 dots are above 5.
- 4 dots are above 6.
- 4 dots are above 7.
- 3 dots are above 8.
- 2 dots are above 9.
- 2 dots are above 10.
- 0 dots are above 11.
From that we can see the data set is {4,5,5,6,6,6,6,7,7,7,7,8,8,8,9,9,10,10} which produces the dot plot you see in the image attachment below.
It's a bit tricky to see, but the graph is nearly symmetrical. If we were to remove the blue points in the dot plot I provided, then we'll get a perfectly symmetrical distribution. Symmetrical means one half is a mirror copy of the the other half. The center line of a symmetrical distribution is both the mean and median.