<h3>Given</h3>
4 hundreds flats; 5 tens rods; 2 ones cubes
<h3>Find</h3>
The number of hundreds flats in each of 2 equal piles
<h3>Solution</h3>
When 4 flats are divided into two equal groups, each group will have ...
... 2 flats
_____
You can imagine doing this the way a card dealer might: first put 1 flat in each of 2 piles, then do the same for the remaining 2 flats. Each pile will end up with 2 flats.
— — — — —
You will have a problem if you continue with the tens rods. There is an odd number of those, so one of them will have to be exchanged for 10 ones cubes.
Hello from MrBillDoesMath!
Answer:
m^2 + 11m - 11, which is the first choice
Discussion:
6m + (m-2)(m+7) +3 =
6m + (m^2 + 7m - 2m -14) + 3 =
(6m + 7m - 2m) + m^2 + (-14 + 3) = combine similar terms
11m + m^2 -11 =
m^2 + 11m - 11
which is the first choice
Thank you,
MrB
Answer:
46.6
Step-by-step explanation:
Answer:
128
Step-by-step explanation:
You would multiply 79 by 1.62 to get the answer.
This would give you 127.98 which you would round up to 128.
We are to select 2 people to sit in first row from 6.
This is a problem of Combinations. We are find permutation of 6 people taken 2 at a time. This can be represented as 6C2.
So, the answer is option B