The coeeficients are 4 and 7
Answer:10
Step-by-step explanation:
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
![\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)
as
∵ ![\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)
![\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aexponent%5C%3Arule%3A%5C%3A%7D%5Cleft%28a%5Eb%5Cright%29%5Ec%3Da%5E%7Bbc%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200)
![=x^{2\cdot \frac{1}{7}}](https://tex.z-dn.net/?f=%3Dx%5E%7B2%5Ccdot%20%5Cfrac%7B1%7D%7B7%7D%7D)
![=x^{\frac{2}{7}}](https://tex.z-dn.net/?f=%3Dx%5E%7B%5Cfrac%7B2%7D%7B7%7D%7D)
also
∵ ![\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)
![\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aexponent%5C%3Arule%3A%5C%3A%7D%5Cleft%28a%5Eb%5Cright%29%5Ec%3Da%5E%7Bbc%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200)
![=y^{3\cdot \frac{1}{5}}](https://tex.z-dn.net/?f=%3Dy%5E%7B3%5Ccdot%20%5Cfrac%7B1%7D%7B5%7D%7D)
![=y^{\frac{3}{5}}](https://tex.z-dn.net/?f=%3Dy%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D)
so the expression becomes
![\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B%5Cfrac%7B2%7D%7B7%7D%7D%7D%7By%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D%7D)
![\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aexponent%5C%3Arule%7D%3A%5Cquad%20%5C%3Aa%5E%7B-b%7D%3D%5Cfrac%7B1%7D%7Ba%5Eb%7D)
∵ ![\:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}](https://tex.z-dn.net/?f=%5C%3A%5Cfrac%7B1%7D%7By%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D%7D%3Dy%5E%7B-%5Cfrac%7B3%7D%7B5%7D%7D)
Thus, 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.
Answer:
21
Step-by-step explanation:
From the given statements, we know that ![z=6.](https://tex.z-dn.net/?f=z%3D6.)
Therefore, substituting this in, we have
![z+15=6+15=\boxed{21}.](https://tex.z-dn.net/?f=z%2B15%3D6%2B15%3D%5Cboxed%7B21%7D.)