15 ms·

1/2(x^2) + 2x + c

• jlongster 14y ago
  You replied 3rd and this was already answered, but I don't think anyone is going to answer the continuation question. So I'll give it to you. What email do you want?
  • rmcooke 14y ago
    richard.mallender@gmail.com. Thanks
  • ozaibi 14y ago
    I would love to try my luck :) Or a charity for my cause :) Mohsen@ozaibi.com Me@ozident.com Hope it's not too much to ask :) Cheers