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

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

by M7MBA » Thu Jul 30, 2020 11:37 am

00:00

A

B

C

D

E

## Global Stats

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

A. $$0.8502$$
B. $$1.000$$
C. $$1.7010$$
D $$3.402$$
E. $$6.804$$

[spoiler]OA=B[/spoiler]

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

by terminator12 » Thu Jul 30, 2020 5:53 pm
Let 1.701 be x

Hence, the equation becomes:
$$\frac{\left(4x\right)^6\cdot\left(2x\right)^7}{2^{19}\cdot x^{13}}$$

We see the powers of 2 and x cancel off in numerator and denominator and we remain with 1

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

by [email protected] » Sun Aug 02, 2020 8:49 am
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