Question

i’m lost again can somebody help asap ??

Answer 1

Fully detailed and equipped information process right below. The answer for this question is "25" (simplified form).

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(5^{- 5} \times 2^8 \times 1 \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(2^8 \times 1 \times \dfrac{1}{5^5} \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(1 \times \dfrac{2^8 \times 1}{5^5} \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(1 \times \dfrac{256}{3125} \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(\dfrac{3125}{256} \Bigg)^2}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{1}{2^3 \times 5^2} \Bigg)^4 \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{1^4}{(2^3 \times 5^2)^4} \Bigg) \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{1^4}{(5^2)^4 \times (2^3)^4} \Bigg) \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \dfrac{1}{2^{12} \times 5^8} \times \dfrac{3125^2}{256^2}}$

$\mathbf{\dfrac{3125^2 \times 1 \times 2^{28}}{256^2 \times 5^8 \times 2^{12}}}$

$\mathbf{\dfrac{2^{(28 - 12)} \times 3125^2}{256^2 \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times 3125^2}{256^2 \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times (5^5)^2}{(2^8)^2 \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times 5^{10}}{2^{16} \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times 5^{10 - 8}}{2^{16}}}$

$\mathbf{\dfrac{5^{10 - 8}}{1}}$

$\mathbf{\therefore \quad 5^2}$

$\mathbf{\therefore \quad 25}$

$\boxed{\mathbf{\underline{\therefore \quad Final \: \: Answer \: \: is: \: 25}}}$

Hope it helps.