anonymous
  • anonymous
Simplify the following expression: 8^2 /the square root of 16 + 5^2 - 8 forty-nine twenty-one thirty-three forty-three
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anything with a ^2 after it is supposed to be squared
myininaya
  • myininaya
Is that: \[\frac{8^2}{\sqrt{16+5^2-8}}\]
anonymous
  • anonymous
\[8\div \sqrt{16}+5^{2}-8\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

myininaya
  • myininaya
\[\frac{8}{\sqrt{16}}+5^2-8 \]
anonymous
  • anonymous
so its \[1+5^{2}-8\]?
myininaya
  • myininaya
\[\sqrt{16}=4\]
anonymous
  • anonymous
or actually its \[2+5^{2}-8?\]
myininaya
  • myininaya
Yes
anonymous
  • anonymous
then you simplify the 5 squared to \[2+25-8\]
myininaya
  • myininaya
yes!
anonymous
  • anonymous
so 2+25 = 27 27-8 = 19?
myininaya
  • myininaya
yes!
anonymous
  • anonymous
then what because 19 is not an answer and im lost after that.......
myininaya
  • myininaya
\[\frac{8}{\sqrt{16}}+5^2-8=\frac{8}{4}+5^2-8=2+25-8=27-8=19\] Yep it is the answer! Be sure that the problem I have looks exactly like the problem you have.
anonymous
  • anonymous
o i see thank you alot!

Looking for something else?

Not the answer you are looking for? Search for more explanations.