The quotient of the start fraction 7 superscript 7 negative 6 over 7 squared end fractions is,
![\dfrac{1}{7^8}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B7%5E8%7D)
<h3>What is the quotient?</h3>
Quotient is the resultant number which is obtained by dividing a number with another. Let a number<em> a</em> is divided by number <em>b.</em> Then the quotient of these two number will be,
![q=\dfrac{a}{b}](https://tex.z-dn.net/?f=q%3D%5Cdfrac%7Ba%7D%7Bb%7D)
Here, (<em>a, b</em>) are the real numbers.
The number given in the problem is,
![\dfrac{7^{-6}}{7^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B7%5E%7B-6%7D%7D%7B7%5E2%7D)
Let the quotient of these number is n. Therefore,
![n=\dfrac{7^{-6}}{7^2}\\](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7B7%5E%7B-6%7D%7D%7B7%5E2%7D%5C%5C)
Now if the power of numerator is negative, then it can be written in denominator with positive power of the same number. Therefore, the above equation written as,
![n=\dfrac{1}{7^2\times7^{6}}\\n=\dfrac{1}{7^{2+6}}\\n=\dfrac{1}{7^8}](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7B1%7D%7B7%5E2%5Ctimes7%5E%7B6%7D%7D%5C%5Cn%3D%5Cdfrac%7B1%7D%7B7%5E%7B2%2B6%7D%7D%5C%5Cn%3D%5Cdfrac%7B1%7D%7B7%5E8%7D)
Hence, the quotient of the start fraction 7 superscript 7 negative 6 over 7 squared end fractions is,
![\dfrac{1}{7^8}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B7%5E8%7D)
Learn more about the quotient here;
brainly.com/question/673545