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

I Got C Which Is 100 But I Dont Know How Jerry Got A class=

Sagot :

Choice C is the correct answer because

[tex]\frac{(6+2)^3-12}{5}\\\\\frac{(8)^3-12}{5}\\\\\frac{512-12}{5}\\\\\frac{500}{5}\\\\100\\\\[/tex]

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

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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

[tex]\frac{6+2^3-12}{5}\\\\\frac{6+8-12}{5}\\\\\frac{14-12}{5}\\\\\frac{2}{5}\\\\[/tex]

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

[tex]\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[/tex]