\(\dfrac{(6.804)^6\cdot(1.701)^{-13}}{2^{19}\cdot(3.402)^{-7}}=\)

M7MBA wrote: ↑

Thu Jul 30, 2020 11:37 am

\(\dfrac{(6.804)^6\cdot(1.701)^{-13}}{2^{19}\cdot(3.402)^{-7}}=\)

A. \(0.8502\)
B. \(1.000\)
C. \(1.7010\)
D \(3.402\)
E. \(6.804\)

[spoiler]OA=B[/spoiler]

Solution:

Notice that 1.701 is exactly half of 3.402, let’s simplify (1.701)^(-13)/(3.402)^(-7) first:

(1.701)^(-13)/(2 x 1.701)^(-7) = (1.701)^(-13)/(2^(-7) x 1.701^(-7)) = (1.701)^(-6)/2^(-7)

So, we have:

[(6.804)^6 x (1.701)^(-6)]/[2^19 x 2^(-7)]

Now, also notice that 6.804 is exactly 4 times of 1.701, we have:

[(4 x 1.701)^6 x (1.701)^(-6)]/[2^19 x 2^(-7)]

[4^6 x (1.701)^6 x (1.701)^(-6)]/[2^12]

[4^6]/[2^12] = (2^2)^6/2^12 = 2^12/2^12 = 1

Answer: B