High hop

"Take control of a comical character and guide them to new heights in High Hop. With this engaging Android game, your goal is to help the hero reach the highest point possible by jumping on blocks. Equipped with a jet pack, your hero can soar with just a tap of your device's screen. Time your jumps perfectly to land on the approaching platforms and slow them down. Be careful to avoid collisions with moving objects and aim for a new record height to boost your ranking. Featuring block graphics, a humorous character, simple one-touch controls, and a high score system, High Hop is a fun and addictive game for all ages."<|endoftext|><|endoftext|> # 2015 AMC 10A Problems/Problem 1. ## Contents. 1 Problem 2 Solution 3 Video Solution 4 See Also ## Problem What is the value of \[2-4+6-8+10-12+14-16+18-20+22?\] $\textbf{(A)}\ -110\qquad\textbf{(B)}\ -44\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 44$ ## Solution. We can group the terms into pairs of two, and notice that each pair sums to $-2$. There are $11$ pairs, so the sum is $11 \cdot (-2) = \boxed{\textbf{(B)}\ -44}$. ## Video Solution. https://youtu.be/8SbJpYmzjzQ ~savannahsolver <|endoftext|>## Mathematical Forums ## Category: High School Olympiads ## Topic: Inequality ## Views: 389 ## [enter: math-user1, num_posts=697, num_likes_received=372] ## [math-user1, num_likes=1] Let $a,b,c>0$. Prove that $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+3\sqrt[3]{abc}\geq 2(a+b+c)$ ## [enter: math-user2, num_posts=545, num_likes_received=72] ## [math-user2, num_likes=0] By AM-GM we have $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+3\sqrt[3]{abc}\ge4\sqrt[4]{\frac{a^2}{b}\frac{b^2}{c}\frac{c^2}{a}3\sqrt[3]{abc}}=4\sqrt[4]{3abc}$ and by AM-GM again we have $4\sqrt[4]{3abc}\ge2(a+b+c)$ ## [math-user