The quotient of the start fraction 7 superscript 7 negative 6 over 7 squared end fractions is,
[tex]\dfrac{1}{7^8}[/tex]
Quotient is the resultant number which is obtained by dividing a number with another. Let a number a is divided by number b. Then the quotient of these two number will be,
[tex]q=\dfrac{a}{b}[/tex]
Here, (a, b) are the real numbers.
The number given in the problem is,
[tex]\dfrac{7^{-6}}{7^2}[/tex]
Let the quotient of these number is n. Therefore,
[tex]n=\dfrac{7^{-6}}{7^2}\\[/tex]
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,
[tex]n=\dfrac{1}{7^2\times7^{6}}\\n=\dfrac{1}{7^{2+6}}\\n=\dfrac{1}{7^8}[/tex]
Hence, the quotient of the start fraction 7 superscript 7 negative 6 over 7 squared end fractions is,
[tex]\dfrac{1}{7^8}[/tex]
Learn more about the quotient here;
https://brainly.com/question/673545
Answer:
It would be A. [tex]\frac{1}{7^{8} }[/tex].
Step-by-step explanation:
Hope this helps!
