i have love in my heart and i am bald

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DEAR READER
Sade Olutola

PR's Tumblrdome
Keni
Three Goblin Art
hello vonnie
Stranger Things

❣ Chile in a Photography ❣
occasionally subtle
Misplaced Lens Cap
he wasn't even looking at me and he found me
almost home
"I'm Dorothy Gale from Kansas"
d e v o n

#extradirty
we're not kids anymore.
PUT YOUR BEARD IN MY MOUTH
dirt enthusiast

Love Begins
seen from United States
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seen from Germany
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seen from China
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@seraphemmes
i have love in my heart and i am bald
does anyone know if it's okay to spend money you earned from working at your job on objects you need or want
Worry about it kitten
ok ♥️
fuck ♥️
nowhere left to go
SISYPHUS GETS A 9 TO 5
[spoilers for all of severance season 1. words by northernlion, brain worms by me]
If nobody got me I know drawing a little deer in ominous situations got me
Prints
Little begging dogfish!
Took some inspiration from spotted seabass for this one. sold! 🐟💙
average otherside sleepover
childhood
full
Chai tea bag + lil but of brown sugar + apple cider packet + 16 oz. mug of hot but not quite boiling water
it will not Fix You but like. maybe. maybe.
tags by @eridan-ampora
Update: this is the best post I've ever made because everyone is sharing their Warm Beverage recipes in the notes. Go check the notes for more Warm Beverages That Will Fix You.
i went to i love you island
i went there
i went to i love you island because i love you
I don't think this is possible????
Hello Ryan I am here to help. So the first step is pretty easy: Three cheeseburgers are worth 18, so each one is worth 6. If these are dollars, that's a steal!
From the second equation we get that cheeseburger plus fries-squared is five. Subtracting cheeseburger, which is six, from both sides, we get that fries-squared is negative-one. Math fans will know that there are two solutions to this; either fries are the "imaginary unit" 𝒾 or they are its negative, -𝒾. We'll do the rest of the problem with 𝒾, keeping in mind that at the end we should also take the complex conjugates as solutions.
Finally, we have that cup to the power of fries, minus cup, equals three. Replacing fries with 𝒾, and moving a cup to the other side, we get that cup-to-the-𝒾 is equal to cup-plus-three.
Now, the weird part about this is the cup-to-the-i. The problem with this is that complex exponentiation is technically not a thing. That is to say, there is no one function which is mathematically equal to "input-to-the-power-of-𝒾". In fact, there are infinitely many such functions.
Fortunately, due to reasons that take about six pages to explain (trust me I've done it), there is one particular function that many people have agreed is "the most reasonable one". This is not a mathematical notion, but a human preference. Seeing as this question was presumably written by a human, I am comfortable with using this function.
So, what function is this? Well, given a complex number r∠θ written in polar form (if you don't know what that means don't worry), where -π < θ ≤ π, then (r∠θ)^𝒾 = e^(-θ)∠ln(r).
Applying this to our problem a value r∠θ will be a possible solution for cup if e^(-θ)∠ln(r) = r∠θ + 3. Splitting this into real and imaginary parts, we get two equations: e^(-θ) cos(ln(r)) = r cos(θ) + 3 and e^(-θ) sin(ln(r)) = r sin(θ). We can graph these equations on Desmos:
The possible values of cup are the intersections between the red, green, and purple. There are infinitely many of these which have an angle of around -π/3, and there are two weirdos: One which is a complex number very close to -2.98, and one which is somewhere around -25. The possible values for cup are all of these infinitely many solutions, and also all of their complex conjugates.
They were right, 99% of people can't solve it.
@pyroclastic727 this seems right up your alley