<h3><u>Answer</u> :</h3>
![\bigstar\:\boxed{\bf{\purple{x^{\frac{m}{n}}}=\orange{(\sqrt[n]{x})^m}}}](https://tex.z-dn.net/?f=%5Cbigstar%5C%3A%5Cboxed%7B%5Cbf%7B%5Cpurple%7Bx%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%7D%3D%5Corange%7B%28%5Csqrt%5Bn%5D%7Bx%7D%29%5Em%7D%7D%7D)
Let's solve !
![:\implies\sf\:(\sqrt[2]{25})^3](https://tex.z-dn.net/?f=%3A%5Cimplies%5Csf%5C%3A%28%5Csqrt%5B2%5D%7B25%7D%29%5E3)
<u>Hence, Oprion-D is correct</u> !
Where's your options? give the full info please!
Answer:
The volume(V) of gas will increase when the pressure of the gas decreases assuming all the other variables are held constant.
Step-by-step explanation:
P = Pressure of the ideal gas
V = Volume of the ideal gas
n = Moles of the ideal gas
R = Universal gas constant
T = Temperature of the ideal gas
Answer:
The correct option is;
Increasing one fifth unit/sec
Step-by-step explanation:
The equation that gives the curve of the particle of the particle is y = 5·x² - 1
The rate of decrease of the y value dy/dt = 2 units per second
We have;
dy/dx = dy/dt × dt/dx
dy/dx = 10·x
dy/dt = 2 units/sec
dt/dx = (dy/dx)/(dy/dt)
dx/dt = dy/dt/(dy/dx) = 2 unit/sec/(10·x)
When x = 1
dx/dt = 2/(10·x) = 2 unit/sec/(10 × 1) = 1/5 unit/sec
dx/dt = 1/5 unit/sec
Therefore, x is increasing one fifth unit/sec.
