Answer:
2PA
Step-by-step explanation:
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.
Answer: 2 because of the constant proportionality
Step-by-step explanation: hope this helps!
We can check this by seeing if Pythagorean's theorem applies. The longest side is the hypotenuse, so let's see if this is true:
This is, indeed correct, thus, this
<em>is </em><em>
a right triangle.</em>
Oh Yes It Does Yes It dose
