Question

[(64)^2/3 * 2^-2 ÷ 7^0]^-1/2​

Answer 1

Answer:

Thank you guys.

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Answer 2

Given:

[(64)^2/3 * 2^-2 ÷ 7^0]^-1/2​

To find:

The value of [(64)^2/3 * 2^-2 ÷ 7^0]^-1/2​

Solution:

The value of [(64)^2/3 * 2^-2 ÷ 7^0]^-1/2​ is 1/2.

We can find the value by following the given process-

We know that the given expression can be solved by using the rules of BODMAS.

We will first solve the terms within the bracket and then will perform division, multiplication, addition, and subtraction, in that order.

So, we are given [(64)^2/3 * 2^-2 ÷ 7^0]^-1/2​.

=[$64^{2/3}$ * $2^{-2}$ ÷ $7^{0}$$]^{-1/2}$

=[$(4^{3}) ^{2/3}$ × $1/2^{2}$ ÷ 1$]^{-1/2}$

We will divide $1/2^{2}$ in the bracket by 1 and then multiply it with $4^{2}$.

Using $(m^{a}) ^{b}=m^{ab}$,

=[$4^{2}$×1/4$]^{-1/2}$

After solving the bracket, we will use the property $m^{-n}$ = $1/m^{n}$.

=$4^{-1/2}$

=1/$4^{1/2}$

=1/$\sqrt4}$

=1/2

Therefore, the value of [(64)^2/3 * 2^-2 ÷ 7^0]^-1/2​ is 1/2.