1. BEDMAS 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!

2. Re: BEDMAS Question.

Originally Posted by mac217
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!
Would you show us what you did please?

3. Re: BEDMAS Question.

We are given to evaluate:

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

Where do you think we should begin first?

4. Re: 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

5. Re: BEDMAS Question.

The brackets.
28-4=24
2^3=8

6. Re: BEDMAS Question.

Hello mac217.
$4^3-5(\frac{28-4}{2^3})+\sqrt{\frac{100}{25}}=64-5(\frac{24}{8})+\sqrt{4}=64-5*3+2=64-15+2=51$. I hope this helps.

7. Re: BEDMAS Question.

Thank you so much Diogyk!

8. Re: BEDMAS Question.

Originally Posted by DIOGYK
Hello mac217.
$4^3-5(\frac{28-4}{2^3})+\sqrt{\frac{100}{25}}=64-5(\frac{24}{8})+\sqrt{4}=64-5*3+2=64-15+2=51$. I hope this helps.
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.

9. Re: BEDMAS Question.

Originally Posted by MarkFL2
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.
I agree I was little rude to the original posters, you and Prove It, by posting the full solution, sorry, I will think more next time before I post after other's who are already helping. : )

10. Re: BEDMAS Question.

Thank you for your consideration and maturity.