LeoIsTheBest

See here: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:quadratic-formula-a1/a/quadratic-formula-proof-review

Note that i is additionally the index of the summation.

Binomial theorem expansion so we have \(\sum _{i=0}^4\binom{4}{i}\cdot \:1^{\left(4-i\right)}i^i=1+4i-6-4i+1=\boxed{-4}~~\blacksquare\)

Another way would be to evaluate \(|-2-2i\sqrt 3|^3\). We have: \(\sqrt{(-2)^2+(-2\sqrt 3)^2}=\sqrt{4+4\cdot 3}=\sqrt{16}=4\implies 4^3=\boxed{64}~~\blacksquare\)

Sorry that's wrong. It should like this: cos^2 0+cos^2 1+cos^2 2 ... cos^2 90 (in degrees). heureka was correct 

Thanks, CPhill! :)

If measurement of angle ABC=75 degrees, then how many segments will he draw until he reaches his starting point A.

It's supposed to say "If

Thank you :)

We can also complete the square. We can factor out a -1 from the given expression. We have -(t^2-8t+4).  We can work inside the parenthesis. We can try to make the quadratic a perfect square. We have: t^2-8t+4=(t-4)^2-16+4=(t-4)^2-12. Remember the negative ;) We now have: -(t^2-4)^2+12. Since the negative of a positive number (a square) is always negative, the maximum it can be is 0. Therefore, the only possible maximum is 12. Hope this gives helpful insight :)

Help, please?

Thank you so much, CPhill!

The links aren't working since the service was shut down.

Can I have help please?

Thanks so much!

I fixed it now.

That's incorrect, sorry.

Thank you so much! :)

"Can I have a non-trig solution?"