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.
Given:
A line passes through the points (2,-1) and (1,-5).
To find:
The equation of the line.
Solution:
If a line passes through the two points, then the equation of the line is

The line passes through the points (2,-1) and (1,-5). So, the equation of the line is




Using distributive property, we get


Subtract 1 from both sides.


Therefore, the correct option is B.
Step-by-step explanation:
I think that is the right answer I hope ut helps
Answer:
hgjjjjjjjjjjjjjjjjggggg
Step-by-step explanation:
hggggggggggggggggggjgjjj
Answer:
The numbers needed to fill each box are in the image attached
Step-by-step explanation:
The probability of the coin landing on heads is 2/7, so the probability of it landing on tails is 1 - 2/7 = 5/7
The probability of landing heads 2 times is:
P = (2/7) * (2/7) = 4/49
The probability of landing heads and then tails is:
P = (2/7) * (5/7) = 10/49
The probability of landing tails and then heads is:
P = (5/7) * (2/7) = 10/49
The probability of landing tails 2 times is:
P = (5/7) * (5/7) = 25/49
The numbers needed to fill each box are in the image attached.