anonymous
  • anonymous
√(512)k^(2)
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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
i dont know
abb0t
  • abb0t
\[\sqrt{a^1} = a^\frac{ 1 }{ 2 }\] now, in your example, you have: \[\sqrt{a^2} = a^\frac{ 2 }{ 2 } \] and you know: \[\frac{ 2 }{ 2 } = 1\]. And when you have \[\sqrt{4} = \sqrt{ 2 \times 2} = 2\] Similarly: \[\sqrt{100} = \sqrt{20 \times 5 } = \sqrt{(5 \times 4) \times 5} = \sqrt{(5 \times (2 \times 2)) \times 5} = \sqrt{(5 \times 5) (2 \times 2)}\] hence: \[\sqrt{(5 \times 5) (2 \times 2)} = 5 \times 2 = 10\]
abb0t
  • abb0t
If you notice the patter, I am taking each pair of 5 and pair of 2 out of the radical. But only took out ONE PAIR of 5's and ONE PAIR of 2's. If you have more than one pair of 2's, then you would take out a 4. Because you know 2 x 2 = 4. and if you have two PAIRS of 5's you would take out a 25. Since you should know that 5 x 5 = 25.

Looking for something else?

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