In your calculator, input arcsin(7/12). Make sure that your calculator is in degree mode. The answer is 35.69 degrees.
Here the question is simple.
All because, we only need to find the value of x.
We are given two equations.
5x + 6 = 10 and 10x + 3 =?
So, we will find the the value of x in the first equation, so that we can substitute the value of x in the second one and there we are with the answer.
5x + 6 = 10
For finding the value of x, all we have to do is,
Transpose the number 6 to 10
Therefore. 5x = 10 - 6 ( Take the equal sign as The Magic Bridge on which if anyone crosses it , will change its sign.)
So we have,
5x = 4
So x = 4/5 ( Multiplication will change to division after crossing the equal sign)
( Doubtful? Substitute the value of x and try!)
Now that we got the value of x,
We can just simply substitute the value of x in the second equation.
10x + 3 = ?
x = 4/5
10*4/5 +3 => 5 and 10 get canceled to 2 at the numerator.
By normal multiplication and then addition, we will get,
8 + 3 = 11
Hope this helps!!!! :)
Make a substitution:
Then the system becomes
![\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bu%7D%7D%7Bu-v%7D%2B%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bu%7D%7D%7Bu%2Bv%7D%3D%5Cdfrac%7B81%7D%7B182%7D%5C%5C%5C%5C%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bv%7D%7D%7Bu-v%7D-%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bv%7D%7D%7Bu%2Bv%7D%3D%5Cdfrac1%7B182%7D%5Cend%7Bcases%7D)
Simplifying the equations gives
![\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bu%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac%7B81%7D%7B182%7D%5C%5C%5C%5C%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac1%7B182%7D%5Cend%7Bcases%7D)
which is to say,
![\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bu%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac%7B81%5Ctimes4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7Bu%5E2-v%5E2%7D)
![\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81](https://tex.z-dn.net/?f=%5Cimplies%5Csqrt%5B3%5D%7B%5Cleft%28%5Cdfrac%20uv%5Cright%29%5E4%7D%3D81)
Substituting this into the new system gives
![\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7B%28%5Cpm27v%29%5E2-v%5E2%7D%3D%5Cdfrac1%7B182%7D%5Cimplies%5Cdfrac1%7Bv%5E2%7D%3D1%5Cimplies%20v%3D%5Cpm1)
Then
(meaning two solutions are (7, 13) and (-7, -13))
Answer:
a=4220-2t
This would be if we are finding the altitude of the submarine from the surface of the water(how many meters away from the surface it is). I don't know what it would be if you want to know from the bottom of the ocean.
Answer:
64 textbooks
Step-by-step explanation:
the first two teachers had 21 students each that's 21+21=42
the other teacher had 22 students so apparently all the textbooks handed to the students are,21+21+22=64
