An <u>example of a problem</u> where I <em>would not</em> group the addends differently is:
3+2+4.
An <u>example of a problem</u> where I <em>would</em> group the addends differently is:
2+5+8.
Explanation:
In the <u>first problem</u>, I would not group the addends differently before adding. This is because I cannot make 5 or 10 out of any of the numbers. We group addends together to make "easier" numbers for us to add, such as 5 and 10. If we cannot do that, there is no reason to regroup the addends.
In the <u>second problem</u>, I would regroup like this:
2+8+5
This is because 2+8=10, which makes the problem "easier" for us to add. Since we can get a number like this that shortens the process, we can regroup the addends.
Answer:
73
Step-by-step explanation:
ikp me le rehat se me cave koken o budalla dreqi se sna le rehat fare jevg dreqi i shkallum i ri
To solve for C, you have to get both of the numbers with a C on the same side. So, subtract the 5c from the right side onto the left side.
Then, divide by 3 on each side.
So, c = 7.
Answer:
28/9
Step-by-step explanation:
22/9 = 2.444444...
24/9 = 2.66666...
26/9 = 2.88888...
28/9 = 3.1111111...
Answer:
(3)
Step-by-step explanation:
the sum of n and 5 is (n + 5), then multiply by 3 gives
3(n + 5)
The result is at least 17 which means it must be greater than or equal to 17
3(n + 5) ≥ 17
