Explain how (13 + 10) + 5 is solved differently from 13 + (10 + 5). a) for (13 + 10) + 5, you add 13 + 5 first, then add 10. for 13 + (10 + 5), you add the 10 + 13 first, then add 5. eliminate b) for (13 + 10) + 5, you add 13 + 10 first, then add 5. for 13 + (10 + 5), you add the 10 + 5 first, then add 13. c) for (13 + 10) + 5, you add 13 - 10 first, then add 5. for 13 + (10 + 5), you add the 10 - 5 first, then add 13. d) for (13 + 10) + 5, you add 13 + 10 first, then subtract 5. for 13 + (10 + 5), you add the 10 + 5 first, then subtract 13.

B)For (13 + 10) + 5, you add 13 + 10 first, then add 5. For 13 + (10 + 5), you add the 10 + 5 first, then add 13.

Step-by-step explanation:

Apply the associative property of addition.

Let , then

In this case a=13,b=10,c=5.

When we substitute into the above relation, we get:

For the LHS, you add 13+5 first, then add 5

For the RHS, you add 10+5 first, then add 13

The correct answer is B

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>

first one is a: renaming a fraction. second one is c: de-c

step-by-step explanation:

my explanation is that these were the answers on edg : )