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Which integer values of b would give the number 2002 ÷ 10−b a value be
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20 Feb 2021, 08:13
20 Feb 2021, 08:13
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Question Stats:
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Which integer values of \(b\) would give the number \(2002 ÷ 10^{−b}\) a value between 1 and 100?
A. {−2, −3}
B. {2, −3}
C. {−2, 3}
D. {2, 3}
E. {−2, 2}
A. {−2, −3}
B. {2, −3}
C. {−2, 3}
D. {2, 3}
E. {−2, 2}
Which integer values of b would give the number 2002 ÷ 10−b a value be
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20 Feb 2021, 09:00
20 Feb 2021, 09:00
1
\(2002 ÷ 10^{−b}\)
We need the value between 1 - 100
Let's simplify it.
\(\frac{2002}{10^{-b}}\)
\(2002 * 10^b\)
Put b = -2, \(2002 * 10^{−2}\) = 20.02
Put b = -3, \(2002 * 10^{−3}\) = 2.002
Answer A
We need the value between 1 - 100
Let's simplify it.
\(\frac{2002}{10^{-b}}\)
\(2002 * 10^b\)
Put b = -2, \(2002 * 10^{−2}\) = 20.02
Put b = -3, \(2002 * 10^{−3}\) = 2.002
Answer A
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Which integer values of b would give the number 2002 ÷ 10−b a value be
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20 Feb 2021, 12:27
20 Feb 2021, 12:27
GeminiHeat wrote:
Which integer values of \(b\) would give the number \(2002 ÷ 10^{−b}\) a value between 1 and 100?
A. {−2, −3}
B. {2, −3}
C. {−2, 3}
D. {2, 3}
E. {−2, 2}
A. {−2, −3}
B. {2, −3}
C. {−2, 3}
D. {2, 3}
E. {−2, 2}
\(1 < \frac{2002}{10^{-b}} < 100\)
\(1 < (2002)(10^b) < 100\)
The only values of \(b\) satisfying this inequality would be: \(-2\) and \(-3\)
For \(b = -2\)
\(1 < 20.20 < 100\)
For \( b = -3;\)
\(1 < 2.002 < 100\)
Hence, option A
