How to solve:
( x , y )
Plug the values in to the equation and solve
-3x+5y=2x+3y
(2,4) plugged in would be
-3(2)+5(4)=2(2)+3(4)
All you have to do is solve this and if the two sides are not equal(like 4=20), then it is not a solution. Just plug in (3,3) next.
Hope this helped!!!
Answer: y>-3 and x is all real numbers
345235235232352352353523523535
Step-by-step explanation:
You have to do first and second derivatives :
y = a sin 3x + b cos 3x
dy/dx = 3a cos 3x - 3b sin 3x
d²y/dx² = - 9a sin 3x - 9b cos 3x
Next, you have to compare it :
let d²y/dx² = ky,
-9a sin 3x - 9b cos 3x = k(a sin 3x + b cos 3x)
-9(a sin 3x + b cos 3x) = k(a sin 3x + b cos 3x)
Therefore, k is equals to 9.
Given:
Consider the given expression is
To find:
The radical form of given expression.
Solution:
We have,
![(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{2^3}\sqrt[10]{x^9}\sqrt[5]{y^3}](https://tex.z-dn.net/?f=%284x%5E3y%5E2%29%5E%7B%5Cfrac%7B3%7D%7B10%7D%7D%3D%5Csqrt%5B5%5D%7B2%5E3%7D%5Csqrt%5B10%5D%7Bx%5E9%7D%5Csqrt%5B5%5D%7By%5E3%7D)
![[\because x^{\frac{1}{n}}=\sqrt[n]{x}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20x%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%3D%5Csqrt%5Bn%5D%7Bx%7D%5D)
![(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{8y^3}\sqrt[10]{x^9}](https://tex.z-dn.net/?f=%284x%5E3y%5E2%29%5E%7B%5Cfrac%7B3%7D%7B10%7D%7D%3D%5Csqrt%5B5%5D%7B8y%5E3%7D%5Csqrt%5B10%5D%7Bx%5E9%7D)
Therefore, the required radical form is ![\sqrt[5]{8y^3}\sqrt[10]{x^9}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B8y%5E3%7D%5Csqrt%5B10%5D%7Bx%5E9%7D)
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