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Mad Lad / Madlad - The square root of child | /r/madlads

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Lily 7 Today at 9:22 PM "What is the square root of child?" @DJ Hammerhead DJ Hammerhead Today at 9:26 PM Alright so Lily Today at 9:26 PM NO NO BNO DJ Hammerhead Today at 9:26 PM First, we have to identify the value of each letter of child C is simple, it's equal to the speed of light h is a bit more complicated, so we're going to simplify it and say it's equal to the molar mass of hydrogen, H, which is equal to 1 We can say the same thing for I, which is the molar mass of lodine, which is 126.9 L and D are both far more complex than C, H, and I, since they don't have representative elements However, we can simplify L into l, which is the Azimuthal Quantum Number l ranges between O and (n - 1), which means it's simply a positive number less than infinity D neither has a representative element nor a representative symbol, which means we must keep it as a variable for now So we are trying to find the square root of chilD, which is equal to the square root of (299792458 m/s) * (1) * (126.9) * (0 to [n-1]) * D We can get rid of the h, since it only has a value of 1 Doing the multiplication, we now have, let's see here We can't do anything for the Azimuthal Quantum Number, since it's quite the complex variable So we're going to leave that out of the equation, for now Now all that's left is 299792458 m/s * 126.9 * D 299792458 * 126.9 is equal to 38043662920 Now we look at D, and we have 38043662920 * D We now know that the value of Child is 38043662920 * D We can now plug in the square root we've all been waiting for So now The square root of 38043662920 is equal to 195047.847771, and we also add on the square root of D for (195047.847771)*sqrt(D) Now we can go back to that I we've been missing Once we plug in the Azimuthal Quantum Number, we then have.. Between (195047.847771* sqrt[D]) * O m/s to (infinity-1)(195047.847771* sqrt[D]) (edited) Thus is the square root of child You're all welcome

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