Answer:
10,000
Step-by-step explanation:
There are 10,000 possible combinations that the digits 0-9 can be arranged into to form a four-digit code.
Answer:
m = 3/5
Step-by-step explanation:
use the formula so m = -5 - -2 / 5 - 10 , so m = -5+2 / -5
m = -3/-5 so m = 3/5
Let's begin by listing the first few multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 38, 40, 44. So, between 1 and 37 there are 9 such multiples: {4, 8, 12, 16, 20, 24, 28, 32, 36}. Note that 4 divided into 36 is 9.
Let's experiment by modifying the given problem a bit, for the purpose of discovering any pattern that may exist:
<span>How many multiples of 4 are there in {n; 37< n <101}? We could list and then count them: {40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100}; there are 16 such multiples in that particular interval. Try subtracting 40 from 100; we get 60. Dividing 60 by 4, we get 15, which is 1 less than 16. So it seems that if we subtract 40 from 1000 and divide the result by 4, and then add 1, we get the number of multiples of 4 between 37 and 1001:
1000
-40
-------
960
Dividing this by 4, we get 240. Adding 1, we get 241.
Finally, subtract 9 from 241: We get 232.
There are 232 multiples of 4 between 37 and 1001.
Can you think of a more straightforward method of determining this number? </span>
Answer:
The actual answer is 2.26, but since that is not an option, I am going to say that you should choose the very last answer at the bottom, π r².
Step-by-step explanation:
