one circumference has 100 one digit numbers, the other circumference has 1000 1 digit numbers( the circumferences are of different radii) how many two digit numbers do you arrive at if you rotate the circumference with 1000 1 digit # (stopping at each #) also if you rotate the crcumference w/100 one digit #'s
2007-07-07 19:06:12 · 5 answers · asked by Book of Changes 3 in Science & Mathematics ➔ Mathematics
I'm not sure I have your problem correctly, but it looks like you have two "wheels of fortune", one with 100 one-digit numbers (from 0 to 9) and the other with 1000 one-digit numbers. If you consider one wheel the "tens" digit determiner, and the other wheel the "units" digit determiner, you should wind up with 100 numbers. If you exclude 0, you drop to 81 numbers.
2007-07-07 19:37:11 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
If you are working with base 10 there are only ten one digit numbers. There would be repetition on each dial. Assuming each dial has all ten numbers the number of unique combinations would be:
10*10 = 100
If you don't want a leading zero in the tens place the number drops to
9*10 = 90
2007-07-08 03:53:10 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
If i understand correctly you can only get 100 numbers (00-99) since lots of the numbers in both wheels would repeat themselves... But i dont think i understand you right...
2007-07-08 02:39:34 · answer #3 · answered by jsos88 2 · 0⤊ 0⤋
0.0
If the circles have the same axis, then none will line up.
2007-07-08 02:16:39 · answer #4 · answered by Neil S 4 · 0⤊ 0⤋
whats the question???
2007-07-08 02:10:45 · answer #5 · answered by Anonymous · 0⤊ 0⤋