Answer:
The equivalent will be:
![\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B7%5D%7Bx%5E2%7D%7D%7B%5Csqrt%5B5%5D%7By%5E3%7D%7D%3D%5Cleft%28%5C%3Ax%5E%7B%5Cfrac%7B2%7D%7B7%7D%7D%5Cright%29%5Cleft%28y%5E%7B-%5Cfrac%7B3%7D%7B5%7D%7D%5Cright%29)
Therefore, option 'a' is true.
Step-by-step explanation:
Given the expression
![\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B7%5D%7Bx%5E2%7D%7D%7B%5Csqrt%5B5%5D%7By%5E3%7D%7D)
Let us solve the expression step by step to get the equivalent
as
![\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}](https://tex.z-dn.net/?f=%5Csqrt%5B7%5D%7Bx%5E2%7D%3D%5Cleft%28x%5E2%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B7%7D%7D)
∵ ![\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%5Cquad%20%5Csqrt%5Bn%5D%7Ba%7D%3Da%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
also
![\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7By%5E3%7D%3D%5Cleft%28y%5E3%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D)
∵
so the expression becomes
∵ 
Thus, the equivalent will be:
Therefore, option 'a' is true.
Answer:
B) 4
Step-by-step explanation:
(f-g)(x) = x² - 4x - 1
5² - 4(5) - 1 = 4
5% interest compounded annually for 3 years multiplies the initial balance by
... 1.05^3 = 1.157625
5% simple interest for 3 years multiplies the initial balance by 1 + 3*0.05 = 1.15.
The difference of these multipliers is
... 1.157625 - 1.15 = 0.007625
If the difference in account balance is Rs 61.00, then the invested principal amount P is
... P*0.007625 = 61.00
,.. P = 61.00/0.007625 = 8000.00
The sum of money is Rs 8000.
Answer:
Expert? Sorry, not me, but what sort of questions do you have?
Step-by-step explanation:
I'd start by writing an equation for each of the right triangles. (Pythagorean theorem)
y² + 9² = z²
x² + z² = (4+9)²
4² + y² = x²
we want to find z so combine the equations by substituting the other variables x,y out.
substitute y² for (x² - 4²) in 1st equation.
(x² - 4²) + 9² = z²
now by rearranging the 2nd equation we can substitute x² for (13² - z²)
(13² - z²) - 4² + 9² = z²
169 - z² - 16 + 81 = z²
234 - z² = z²
234 = 2z²
234/2 = z²
117 = z²
√(117) = z
√(9*13) = z
3√(13) = z
13 goes in the box
