Convert to base 10, multiply, and convert back again. :)
Recall that an angle is obtuse if and only if it is between 90 degrees and 180 degrees. You should be able to get it from here, ask if you have any further questions.
Melody did get it right :) But I'd like to point out an easier method.
Multiply by \(\frac{\sqrt[16]{5}-1}{\sqrt[16]{5}-1}\), and the expression simplifies to \(\sqrt[16]{5}-1\). Our answer is hence \((\sqrt[16]{5})^{48} = \boxed{125}\).
Are you on AoPS? If so I would recommend asking there. I'm OlympusHero and you can get answered fast to this harder stuff there, though there are some really talented people here (melody, cphill, etc... I'm looking at you :P)
Give each child one cookie.
Now 10 children, 20 cookies.
So a+b+c+d+e+f+g+h+i+j=20
Result is (20+10-1) choose (10-1) = 29 C 9
= 10015005
Use the Binomial Theorem. What does this expansion look like in terms of $a$ and $n$? Can you set up some equations based on the given coefficients (-20 and 150 for x and x^2 respectively)?
Everyone, please do not answer this, it is an AoPS Algebra B writing problem. Ask on your message board. However if they are busy a helpful hint is that the sum of all coefficients is f(1) and the sum of the even coefficients minus the sum of odd coefficients is f(-1).
In both questions there are two angles congruent for the triangles that are in the question, so the desired results are each true by AA similarity.
Note that scaling the points up isn't going to change the area, so it's still 32.
I'm pretty sure there's a better way to do this. I've seen this question before. I have just forgotten how to approach it. Maybe let q(n)=p(2^n)-1/(2^n) for 0 <= n <=6 and then q(n)=n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)? Something like that? I could be wrong about the better way, though. It's just that 127/64 is a pretty nice number.
The roots can be (1,28) (2,14) (4,7) so that would mean 3.
