Bridges in mathematics second edition home connection

Mathematics

1 answer:

guajiro [1.7K]3 years ago

3 0

What be more specificx or take a picture

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I got c which is 100 but I don’t know how jerry got A

Anna007 [38]

Choice C is the correct answer because

$\frac{(6+2)^3-12}{5}\\\\\frac{(8)^3-12}{5}\\\\\frac{512-12}{5}\\\\\frac{500}{5}\\\\100\\\\$

So in short, $\frac{(6+2)^3-12}{5}=100\\\\$

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The mistake Jerry likely made was that he only cubed the 2 and didn't realize the 6 was part of that cubing process. It seems he didn't add first and decided to cube before adding.

This is probably what steps Jerry did

$\frac{6+2^3-12}{5}\\\\\frac{6+8-12}{5}\\\\\frac{14-12}{5}\\\\\frac{2}{5}\\\\$

But as mentioned, those steps are incorrect because the 6 is part of the cubing operation. In other words, Jerry should have added the 6+2 first before cubing afterward (due to PEMDAS determining the order of operations).

Or you could think of it like this

$\frac{(6+2)^3-12}{5}\\\\\frac{(6+2)(6+2)(6+2)-12}{5}\\\\\frac{(8)(8)(8)-12}{5}\\\\\frac{512-12}{5}\\\\\frac{500}{5}\\\\100$