Answer: There will be 4 teams such that each team has same number of graders of both.
Step-by-step explanation:
Since we have given that
Number of 6th graders = 36
Number of 7th graders = 40
We need to find the greatest number of teams that the students conform if each team has the same number of 6th grader and the same number of 7th graders.
So, we will find H.C.F. for the greatest number of teams.
H.C.F. of 36 and 40 =4
Hence, there will be 4 teams such that each team has same number of graders of both.
Answer:
-80, -110
Step-by-step explanation:
Hope the answer I gave you helps!!!
:)
<span>n + (n + 1) + (n + 2) = 36
Let's first create the equation ourselves and then see if it fits any of the available options. We want 3 consecutive integers that add up to 36. Let's use the value "n" to represent the first number in the sequence.
n
Now the next number will be n+1 and the third will be n+2. And since we want the sum, we get
n + n + 1 + n + 2 = ?
And finally, we want the sum to be 36, so our final equation is
n + n + 1 + n + 2 = 36
With this equation in mind, let's look at the available choices.
n + (n + 2) + (n + 4) = 36
Well, it has the 36 OK, but it has +2 and +4 where we have +1 and +2. So it won't fit and it's wrong.
n + (n + 1) + (n + 3) = 36
Closer. Got the 36 and the +1. But it has +3 instead of +2, so not a fit either.
n + (n + 1) + (n + 2) = 36
This looks good. Has a couple of extra parenthesis, but they don't affect the final answer and it has the +1, +2 and the 36. This one is correct.
n + (n â’ 1) + (n â’ 3) = 36
Hmm. The n - 1 has a possibility. Perhaps they want n to be the middle of the 3 numbers to add. If that's the case, then the other number should be n+1 which it isn't. Maybe they want the n to be the last number. If that's the case, then the third number should be n-2. But that doesn't work either. So this equation won't work.
So of the 4 choices, the only answer that works is "n + (n + 1) + (n + 2) = 36"</span>
It is the last choice or #3 or a C.
