Mad Lad / Madlad - Images
Madlass calls our her Uncle | /r/madlads
Mad Lad / Madlad
Oh dbrand you beautiful bastard | /r/madlads
Mad Lad / Madlad
The BTS of what the Ukranian farmers have been up to | /r/madlads
Mad Lad / Madlad
The Guy who gave the reward is a pure madlad | /r/madlads
Mad Lad / Madlad
madlad wishes someone a sad cakeday | /r/madlads
Mad Lad / Madlad
Hmm, I dunno. I think I'll pass on this one. | /r/madlads
Mad Lad / Madlad
Madlad wholeheartedly supports Ukraine by using the Finnish flag. | /r/madlads
Mad Lad / Madlad
Madlad surprises fans with 5 full novels written in 2 years. | /r/madlads
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Madlad sinks his boss’s multimillion dollar yacht | /r/madlads
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A YouTuber's satire video gets put on CNN believed to be a real video | /r/madlads
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Madlad commented the Ukrainian flag in all post from a Russian porn subreddit | /r/madlads
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That’s more than six hours as well | /r/madlads
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Hinata Is The Pedophile | /r/madlads
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He was told to change his name but roblox can’t contain him | /r/madlads
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Mad ancient lad has ancient cock ring | /r/madlads
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Mad lad celebrates cake day only every 4 years | /r/madlads
Mad Lad / Madlad
Well they couldn't have enforced stricter sanctions than this | /r/madlads
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this madlad just solved Russian censore | /r/madlads
Mad Lad / Madlad
Reddit made Ukraine's subreddit the first option when searching Russia | /r/madlads
Mad Lad / Madlad
The square root of child | /r/madlads
![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](https://i.kym-cdn.com/photos/images/masonry/002/320/018/954.png)
![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](https://i.kym-cdn.com/photos/images/original/002/320/018/954.png)
Mad Lad / Madlad
