Question
math - Andy, Thursday, October 25, 2007 at 7:37pm
can you please help me with this math problem? If you can here is the question:
Seventy-five years ago, pharmacists weighed medicine on balances like the ones you have been using. The mass pieces were very expensive, so a pharmacist would buy as few mass pieces as possible. If a pharmacist had 1-g, 3-g, and 9-g mass pieces, he or she could weigh out any number of grams from 1g to 13g. Show how you could measure all of the masses from ig to 13g using only the three mass pieces given
can you please help me with this math problem? If you can here is the question:
Seventy-five years ago, pharmacists weighed medicine on balances like the ones you have been using. The mass pieces were very expensive, so a pharmacist would buy as few mass pieces as possible. If a pharmacist had 1-g, 3-g, and 9-g mass pieces, he or she could weigh out any number of grams from 1g to 13g. Show how you could measure all of the masses from ig to 13g using only the three mass pieces given
Answers
for 1 g, use the 1g piece on one side top balance the material bing measured.
For 2 g, put 3g on one side and 1 on the other. Add material to the side with the 1 g mass.
For 3 g, use the 3g piece to balance.
For 4 g, put the 1 g and 3 g together on one side
For 5 g, Put 9 g on one side and 3 + 1 on the other
Similarly, 6 = 9 - 3
7 = 9 + 1 - 3
8 = 9 - 1
9 = 9
10 = 9 + 1
11 = 9 + 3 - 1
12= 9 + 3
13 = 9+3+1 (all on the same side)
For 2 g, put 3g on one side and 1 on the other. Add material to the side with the 1 g mass.
For 3 g, use the 3g piece to balance.
For 4 g, put the 1 g and 3 g together on one side
For 5 g, Put 9 g on one side and 3 + 1 on the other
Similarly, 6 = 9 - 3
7 = 9 + 1 - 3
8 = 9 - 1
9 = 9
10 = 9 + 1
11 = 9 + 3 - 1
12= 9 + 3
13 = 9+3+1 (all on the same side)
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