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z*10^k=6*10^m
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06 Nov 2019, 09:29
06 Nov 2019, 09:29
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\(z \times 10^k=6 \times 10^m\)
\(m=k+2\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
\(m=k+2\)
Quantity A |
Quantity B |
\(z\) |
\(60\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Re: z*10^k=6*10^m
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06 Nov 2019, 09:44
06 Nov 2019, 09:44
1
We are given z*10^k=6*10^m
substituting k+1 for m
we have z*10^k=6*10^(k+2)
dividing both sides by 10
z=(6*10^(k+2))/10^k
reminder x^y/x^z equals x^(y-z)
so, we have z=6*10^(k+2-K)
z=6*100 which is higher than 60
substituting k+1 for m
we have z*10^k=6*10^(k+2)
dividing both sides by 10
z=(6*10^(k+2))/10^k
reminder x^y/x^z equals x^(y-z)
so, we have z=6*10^(k+2-K)
z=6*100 which is higher than 60
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z*10^k=6*10^m
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06 Nov 2019, 12:40
06 Nov 2019, 12:40
2
Carcass wrote:
\(z \times 10^k=6 \times 10^m\)
\(m=k+2\)
\(m=k+2\)
Quantity A |
Quantity B |
\(z\) |
\(60\) |
GIVEN: If \(m=k+2\), then we know that \(m-k=2\)
We'll use this information shortly!
GIVEN: \(z \times 10^k=6 \times 10^m\)
Divide both sides by \(10^k\) to get: \(z=\frac{6 \times 10^m}{10^k}\)
Apply Quotient Law to get: \(z=6 \times 10^{m-k}\)
Replace \(m-k\) with \(2\) to get: \(z=6 \times 10^2\)
Simplify: \(z = 600\)
We get:
QUANTITY A: 600
QUANTITY A: 60
Answer: A
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