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.
SERIOUSLY?!?!?! the anser is the second one there are noe letters so b
0.65, or in fraction form 65/100
Answer:
h = -9
Step-by-step explanation:
Try to make the left hand side and the right hand side equal. By dividing like
h = (-27x+72)/(3x-8)
= [-9(3x-8)]/(3x-8)
= -9
When both sides are equal, you can't find the value of x as there are infinitely many solutions of x.
Answer:
No, because it has repeating domains (x-values) which is -2
Step-by-step explanation:
A function should not have repeating x-values.
