Step-by-step explanation:
Hi,
I agree with what you did. You could also add y on both sides and get 3x = y + 5 and then divided by three to get...
x = 5 + y /3
(the 5 and y are being divided by 3)
But that's just complicated, it works, but it's complicated. What you did is correct and what I gave you is the value of x, but I'm 100 percent sure yours is correct.
Hope this helps
Answer:
9) v= -12
10) n= -7
Step-by-step explanation:
9) v - 15 = -27
Add 15 from both sides
i.e. v - 15 + 15 = -27 + 15
v = -12
10) n + 16 = 9
Subtract 16 from both sides
i.e. n + 16 - 16 = 9 - 16
n = -7
Hope this helps!!!
Answer:
i think it is. r = -s + 3z could be wrong sorry if it is
1. 2x+3<-3 = x<-3
You get this by subtracting 3 from both sides, which gives you -6. Then divide 2 from both sides.
2. -3x<-6 = x>2
You divide -3 from both sides. You must switch the sign because any time you multiply or divide a negative integer from both sides of an inequality the sign is switched.
3. -5x<5 = x>1
See number 2
4. 5-2x<11 = x>-3
Subtract 5 from both sides to get -2x<6. Then, divide both sides by -2 and switch the sign.
5. x+2<5 = x<3
Subtract 2 from both sides
6. 2-3x<5 = x>-1
Subtract 2 from both sides to get -3x<3. Then, divide -3 from both sides and switch the sign.
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>
