Road crisis

"Manage Traffic on Busy City Crossroads - Prevent Road Crashes and Control Traffic Lights. Enhance Your Reflexes and Focus in this Thrilling Android Game. The Display Shows a Crossroads with Two Streets. Observe the Vehicles Carefully. Alter the Signals at the Perfect Time, When There is an Opening in Traffic, to Halt the Horizontal Cars and Allow the Vertical Cars to Pass. Earn Points and Unlock Different Cars. Game Highlights: Impressive Visuals Easy Gameplay Variety of Vehicles Top Scores"<|endoftext|><|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 $2+4+6+8+10+12+14+16+18+20$? $\textbf{(A)}\ 90\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 150$ ## Solution. We can pair the numbers up as follows: $(2+20)+(4+18)+(6+16)+(8+14)+(10+12)$. Each pair adds up to $22$, and there are $5$ pairs, so the sum is $22 \cdot 5 = \boxed{\textbf{(D) } 110}$. ## Video Solution. https://youtu.be/8WrdYLw9_ns ~savannahsolver <|endoftext|>## Mathematical Forums ## Category: High School Olympiads ## Topic: Inequality ## Views: 338 ## [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-user1, num_likes=1] [quote=math