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:
1
Step-by-step explanation:
Answer:
n = 6
Step-by-step explanation:
Two intersecting chords. The product of the parts of one chord is equal to the product of the parts of the other chord, that is
7(n + 4) = 5(n + 8) ← distribute parenthesis on both sides
7n + 28 = 5n + 40 ( subtract 5n from both sides )
2n + 28 = 40 ( subtract 28 from both sides )
2n = 12 ( divide both sides by 2 )
n = 6
Solve for h:V = π h r^2
V = h π r^2 is equivalent to h π r^2 = V:π h r^2 = V
Divide both sides by π r^2:
Answer: h = V/(π r^2)
