Notice: Undefined index: HTTP_ACCEPT_LANGUAGE in /var/www/site/data/www/ezapk.net/main.php on line 27
Download Block puzzle pop APK for Android (FREE) - EZapk.net
  • Platform: Android
  • Updated: 24.01.2025
  • Android version: 5.0
  • Language: en fr de pl it es pt
  • Current version: 2021.6.30
  • Google Play: -
In the game "Block Puzzle Pop," you must strategically place blocks of various shapes, all composed of squares. Your goal is to create complete rows of blocks in order to eliminate them. Use your logical thinking and spatial imagination to conquer challenging levels in this Android game. At the bottom of the screen, you will see three blocks that will appear next. Simply drag these blocks onto the screen and position them in any available space. Take your time and carefully plan your moves, as there is no time limit in this game. Game features include minimalistic graphics, a day and night mode, leaderboards, and 15 achievements to unlock.<|endoftext|><|endoftext|><|endoftext|> # 2006 AMC 12B Problems/Problem 1. (Redirected from 2006 AMC 12B Problems / Problem 1) ## Problem. What is the value of $2^{-3}+3^{-3}$? $\text {(A) } \frac{5}{9}\qquad \text {(B) } \frac{17}{24}\qquad \text {(C) } \frac{19}{24}\qquad \text {(D) } \frac{7}{8}\qquad \text {(E) } \frac{11}{12}$ ## Solution. We can rewrite $2^{-3}$ and $3^{-3}$ as $\frac{1}{2^3}$ and $\frac{1}{3^3}$, respectively. Then, we can combine the fractions to get $\frac{1}{8}+\frac{1}{27}$. We can find a common denominator of $216$ to get $\frac{27}{216}+\frac{8}{216}=\frac{35}{216}$. We can simplify this fraction to get $\boxed{\textbf{(B) }\frac{17}{24}}$. <|endoftext|>## Mathematical Forums ## Category: High School Olympiads ## Topic: Inequality ## Views: 351 ## [enter: math-user1, num_posts=697, num_likes_received=372] ## [math-user1, num_likes=1] Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\geq 2(a^2+b^2+c^2)$ ## [enter: math-user2, num_posts=467, num_likes_received=180] ## [math-user2, num_likes=0] By AM-GM, $\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\ge3\sqrt[3]{\frac{a^2}{bc}}=\frac{3a}{\sqrt[3]{abc}}$. Summing cyclically gives $\frac{a}{b}+\frac{b}{c
  • Block puzzle pop
  • Block puzzle pop
  • Block puzzle pop
  • Block puzzle pop
Block puzzle pop

Download Block puzzle pop Android version for free

Main Menu