Answer:
896
Step-by-step explanation:
Let's talk first about how many 3 digit numbers there are. The first 3 digit number is 100 and the last is 999. So there are 999-100+1 numbers that are 3 digits long. That simplifies to 900.
Now let's find how many of those have a sum for the digits being 1, then 2 ? Then take that sum away from the 900 to see how many 3 digit numbers have the sum of their digits being more than 2.
3 digit numbers with sum of 1:
The first and only number is 100 since 1+0+0=1.
We can't include 010 or 001 because these aren't really three digits long.
3 digit numbers with sum of 2:
The first number is 101 since 1+0+1=2.
The second number is 110 since 1+1+0=2.
The third number is 200 since 2+0+0=2.
That's the last of those. We could only use 0,1, and 2 here.... Anything with a 3 in it would give us something larger than or equal to 3.
So there are 900-1-3 numbers who are 3 digits long and whose sum of digits is greater than 2.
This answer simplifies to 896.
52 ×.65= X
The answer would be $80. Hope i could be helpful :)
Answer:just start at 7 and go out 2 up 1
Step-by-step explanation:
Edit:sorry ment -7
Answer:
Greater than 21
Less than 45
Step-by-step explanation:
You add and subtract the two numbers you are given.
33 - 12 = 21
The third side has to be bigger than 21.
33 + 12 = 45
The third side has to be less than 45.
It can be written in one math sentence:
21 < x < 45
