anonymous
  • anonymous
Multiply. (2-square root 2)(3 square root 6+1) Simplify your answer as much as possible.
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

precal
  • precal
|dw:1336880574692:dw| is this your problem?
anonymous
  • anonymous
yes it is
alexwee123
  • alexwee123
|dw:1336880633392:dw|

Looking for something else?

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

More answers

precal
  • precal
|dw:1336880612018:dw|
alexwee123
  • alexwee123
|dw:1336880664057:dw|
anonymous
  • anonymous
\[(2-\sqrt{2})(3 \sqrt{6}+1)\] \[6\sqrt{6}+2-3\sqrt{12}-\sqrt{2}\] \[6\sqrt{6}+2-6\sqrt{3}-\sqrt{2}\]
alexwee123
  • alexwee123
|dw:1336880788114:dw|
precal
  • precal
|dw:1336880688048:dw|
anonymous
  • anonymous
so whats the simplified answer>???? all these numbers and square roots are confusing me!!!
alexwee123
  • alexwee123
any of the answers are right :/
precal
  • precal
|dw:1336880938427:dw|
precal
  • precal
the numbers under the square root have to be the same to be considered like terms, since they are not, this is the solution (I put it in a box for you)
anonymous
  • anonymous
It just doesn't look simplified, but it is.
anonymous
  • anonymous
okay thanks so much!
precal
  • precal
anytime:)

Looking for something else?

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