Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

is V25 = ±5 always? why?

Mathematics
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.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
V25 = \[\sqrt{25}\]
actually it denotes positive square root of 25
so answer will be 5

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

@akash123 you're talking about principal square roots...
Because 5² = (=5)² = 25
yes it's... but if anyone use this symbol of sqrt then it implies positive sqrt only eg as we write sqrt(sin^2x) = I sinx I
(=5)^2? interesting
I think it's a convention only and moreover it's frequently used in calculus
So YES, the even root always yield 2 results :)
It's basic Pre Algebra!
Because 5² = (-5)² = 25
yup. but why? hm..
x^2= x* x
25= 5*5 and 25 =(-5)* (-5)
i know that. i mean... argh. i cant speak english very well. but i think that's enough thanks for your hardwork xD

Not the answer you are looking for?

Search for more explanations.

Ask your own question