The answer for this question is 51 but I can't seem to get it.

4^3-5[(28-4)/2^3]+√100/25

Any help would be greatly appreciated!

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- Sep 18th 2012, 06:58 PMmac217BEDMAS Question.
The answer for this question is 51 but I can't seem to get it.

4^3-5[(28-4)/2^3]+√100/25

Any help would be greatly appreciated! - Sep 18th 2012, 07:05 PMProve ItRe: BEDMAS Question.
- Sep 18th 2012, 07:05 PMMarkFLRe: BEDMAS Question.
We are given to evaluate:

$\displaystyle 4^3-5\left(\frac{28-4}{2^3} \right)+\sqrt{\frac{100}{25}}$

Where do you think we should begin first? - Sep 18th 2012, 07:07 PMmac217Re: BEDMAS Question.
Here are my steps,

=4^3-5[24/2^3]+√100/25

=4^3-5x3+√100/25

=64-5x3+√100/25

=64-5x3+4

=64-15+4

=64-19

=45 - Sep 18th 2012, 07:11 PMmac217Re: BEDMAS Question.
The brackets.

28-4=24

2^3=8 - Sep 18th 2012, 07:11 PMDIOGYKRe: BEDMAS Question.
Hello mac217.

$\displaystyle 4^3-5(\frac{28-4}{2^3})+\sqrt{\frac{100}{25}}=64-5(\frac{24}{8})+\sqrt{4}=64-5*3+2=64-15+2=51$. I hope this helps. - Sep 18th 2012, 07:12 PMmac217Re: BEDMAS Question.
Thank you so much Diogyk!

- Sep 18th 2012, 07:18 PMMarkFLRe: BEDMAS Question.
Why do you come along after others are already trying to engage a poster and give a full solution?

When I made my first post here, I did not see another had already posted because we posted at the same time, but because the other poster was first, I would then let them continue to engage the OP. - Sep 18th 2012, 07:35 PMDIOGYKRe: BEDMAS Question.
- Sep 18th 2012, 07:40 PMMarkFLRe: BEDMAS Question.
Thank you for your consideration and maturity. (Cool)