Answer:
25
Step-by-step explanation:
The applicable rules of exponents are ...
(a^b)^c = a^(bc)
(a^b)(a^c) = a^(b+c)
(a^b)/(a^c) = a^(b-c)
a^0 = 1 . . . for a≠0
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[tex](2^8\cdot5^{-5}\cdot19^0)^{-2}\cdot\left(\dfrac{5^{-2}}{2^3}\right)^4\cdot2^{28}\\\\=2^{8(-2)-3(4)+28}\cdot5^{(-5)(-2)+(-2)(4)}=2^{-16-12+28}\cdot5^{10-8}\\\\=2^0\cdot5^2=\boxed{25}[/tex]
