$(2^8\cdot3^{-5}\cdot6^0)^{-2}\cdot\left(\dfrac{3^{-2}}{2^3}\right)^4\cdot2^{28}\\\\=(2^8)^{-2}\cdot(3^{-5})^{-2}\cdot1^{-2}\cdot\dfrac{(3^{-2})^4}{(2^3)^4}\cdot2^{28}\\\\=2^{-16}\cdot3^{10}\cdot\dfrac{3^{-8}}{2^{12}}\cdot2^{28}\\\\=2^{-16}\cdot2^{28}\cdot2^{-12}\cdot3^{10}\cdot3^{-8}\\\\=2^{-16+28+(-12)}\cdot3^{10+(-8)}\\\\=2^0\cdot3^2=1\cdot9=\boxed{9}\\\\Used:\\\\(a\cdot b)^n=a^n\cdot b^n\\\\(a^n)^m=a^{n\cdot m}\\\\a^n\cdot a^m=a^{n+m}\\\\a^{-n}=\dfrac{1}{a^n}$

4 0

3 years ago

Read 2 more answers

A bag contains 5 red marbles, 6 blue marbles and 9 yellow marbles.

lilavasa [31]

Answer:

Step-by-step explanation:

The total number of marbles is 20

Question A

P(Red) = 4/20 = 1/5

P(yellow) = 9/20

P(Red then Yellow) = 1/5 * 9/20 = 9 / 100

Question B

P(B) = 6/20

P(B again) = 5/19 because you have 1 less total plus 1 less blue.

P(B and B) = 6/20 * 5/19

P(B and B) = 30/ 380 = 3/38

7 0

3 years ago

Is 85/95 greater than 64/76

elena-14-01-66 [18.8K]

The answer is yes I believe

5 0

3 years ago