anonymous
  • anonymous
...
Mathematics
katieb
  • katieb
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

anonymous
  • anonymous
\[-2+\frac{ 1 }{ 3 }+\frac{ 3 }{ 2} - \frac{ 2 }{ 3 }^{2} \]
anonymous
  • anonymous
then gives me \[-2+1\frac{ 1 }{ 6 }\]
anonymous
  • anonymous
After this don't know what to do

Looking for something else?

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

More answers

anonymous
  • anonymous
-2 + 7/6
anonymous
  • anonymous
common denominator and then add
anonymous
  • anonymous
ok
anonymous
  • anonymous
89/18
anonymous
  • anonymous
nathan917
  • nathan917
whats your question?
anonymous
  • anonymous
after this what do i do next \[-2+1\frac{ 1 }{ 6 }\]
anonymous
  • anonymous
nathan917
  • nathan917
Im kinda lost on this one ask @Mertsj or @radar
Mertsj
  • Mertsj
Is it (2/3)^2 or 2^2/3?
anonymous
  • anonymous
no
Mertsj
  • Mertsj
What do you mean no? It must be one or the other.
Mertsj
  • Mertsj
\[-2+\frac{1}{3}+\frac{3}{2}-\frac{4}{3}=-\frac{12}{6}+\frac{2}{6}+\frac{9}{6}-\frac{8}{6}=\]
Mertsj
  • Mertsj
\[\frac{-20}{6}+\frac{11}{6}=\frac{-9}{6}=\frac{-3}{2}\]

Looking for something else?

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