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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.

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kudzordzifrancis

Answer:

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 [tex]a,b,c\in \mathbb R[/tex], then [tex](a+b)+c=a+(b+c)[/tex]

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

When we substitute into the above relation, we get:

[tex](13+10)+5=13+(10+5)[/tex]

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

Answer:

The answer is B

Step-by-step explanation:

Simple use P.E.M.D.A.S

Parentheses are always first, you add 13+10 then evaluate everything else.

Have a great day.

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