\(\dfrac{(6.804)^6\cdot(1.701)^{-13}}{2^{19}\cdot(3.402)^{-7}}=\)
A. \(0.8502\)
B. \(1.000\)
C. \(1.7010\)
D \(3.402\)
E. \(6.804\)
[spoiler]OA=B[/spoiler]
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\(\dfrac{(6.804)^6\cdot(1.701)^{-13}}{2^{19}\cdot(3.402)^{-7}}=\)
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terminator12
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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
Hence, answer is B.
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
Hence, answer is B.
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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
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