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Lana:
I'm not sure where you are having a problem but my guess is on the last part of A6. Looking back at Problem A3 (same video 7--note the problem comes down to seeing that since the decimal expansion of 1/2 = 0.5 that 1/2^n = 0.5^n) the last part of the answer in the table is misleading (in my opinion) and the piece below the table is a "better" way of stating the answer to finding the decimal expansion of 1/2^n:
Notice that the decimal will include the power of 5, with some leading zeros. For example, 55 is 3,125, so "3125" shows up in the decimal, with enough leading zeros for it to comprise five digits: .03125. Similarly, 56 is 15,625, so the decimal is .015625 (six digits).
A similar answer is given to problem A5 (again the last "table entry" is misleading in my opinion) where you are looking for the decimal expansion of 1/5^n:
Notice that the decimal will include the power of 2, with some leading zeros. For example, 25 is 32, so the decimal is .00032 (five digits).
Now in the case of 1/(2^n5^m)--your problem A7. Consider this as (0.5)^n * (0.2)^m. Now if m > n you can rewrite this as:
(0.5)^n * (0.2)^m = (0.5)^(n-m) * (0.5)^m * (0.2)^m = (0.5)^(n-m) (0.1)^m (since (0.5)(0.2) = (0.1). Now this can be rewritten as 1/2^(n-m) * 1/10^m <--the answer they give (with fractions).
If I try to write this rule like the statements included above it would look like:
Notice that if there are more 2's in the denominator than 5's the decimal will include powers of 5, with some leading zeros. For example, 1/5326 will have 6 decimal places (maximum of the exponents), and will include 56-3 = 53 = 125 so the decimal is .000125 (six digits).
A similar technique gives the result for n > m.
I hope this helps.
Andy Talmadge
Lana Forecast <lanaforecast@comcast.net> wrote:
HI,
On Page 137 ,I am having trouble filling out a formula for decimal
represantion of Problem A6. This is for Video 7.
thanks,
Lana Davis
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