If t = 1 / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point?
a) 3
b) 4
c) 5
d) 6
e) 9
how many zeros
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- ssmiles08
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1 / (2^9 * 5^3) can also be re-written as 1 / (2^6 *2^3 * 5^3)
1 / [2^6*(2^3 * 5^3)] = 1/2^6 * 1/10^3
1/1000 = .001 = 2 zeros.
1/2^6 = 1/64 (quick division tells you that the quotient is .01xyz)
so it 1/64 = .01 = 1/100
1/100 * 1/1000 = 1/100000 = .00001 = 4 zeros
IMO B.
1 / [2^6*(2^3 * 5^3)] = 1/2^6 * 1/10^3
1/1000 = .001 = 2 zeros.
1/2^6 = 1/64 (quick division tells you that the quotient is .01xyz)
so it 1/64 = .01 = 1/100
1/100 * 1/1000 = 1/100000 = .00001 = 4 zeros
IMO B.
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If you got it till 1/2^6 x 10^-3, then all you need to do is multiplication and division.uptowngirl92 wrote:I did'nt understand the last part//
Got it till here...1/2^6 x 10^-3
2^6 = 64
1/64 ~ .015 round it up to .02 if you want...the numerator's exact number does not matter.
.02 = 2/100
10^-3 = 1/1000
2/100 * 1/1000 = 2/100000 (when you multiply you add the zeros in the denominators.)
2/100000 = .00002 (there are 4 zeros between the decimal point and 2)
You got a dream... You gotta protect it. People can't do somethin' themselves, they wanna tell you you can't do it. If you want somethin', go get it. Period.