Answer:
[tex]\large\boxed{\dfrac{4^{-\frac{11}{3}}}{4^{-\frac{2}{3}}}=\dfrac{1}{64}}[/tex]
Step-by-step explanation:
[tex]\dfrac{4^{-\frac{11}{3}}}{4^{-\frac{2}{3}}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=4^{-\frac{11}{3}-\left(-\frac{2}{3}\right)}=4^{-\frac{11}{3}+\frac{2}{3}}=4^{-\frac{9}{3}}=4^{-3}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=\dfrac{1}{4^3}=\dfrac{1}{64}[/tex]
The simplification of the expression is [tex]\dfrac{1}{64}[/tex].
Exponentiation(the process of raising some number to some power) have some basic rules as:
[tex]a^{-b} = \dfrac{1}{a^b}\\\\a^0 = 1 (a \neq 0)\\\\a^1 = a\\\\(a^b)^c = a^{b \times c}\\\\ a^b \times a^c = a^{b+c} \\\\[/tex]
Given ;
[tex]\dfrac{(4^{-11/3})}{(4^{-2/3})}[/tex]
We know that
[tex]\dfrac{a^m}{a^n} = a^{m-n}[/tex]
[tex]\dfrac{(4^{-11/3})}{(4^{-2/3})} = 4^({-11/3 + 2/3})\\\\\\= 4 ^{-9/3}\\\\= 4^{-3}\\\\[/tex]
Hence, [tex]\dfrac{1}{4^3} = \dfrac{1}{64}[/tex]
Learn more about exponentiation here:
https://brainly.com/question/26938318
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