funny title goes here

Comments

• OpenStars@startrek.website ⁨10⁩ ⁨months⁩ ago

  Hrm, it sounds like the teacher already knew the answer.

  Maybe stop wasting the time of the student then!?

  You’re welcome.

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• Dave@lemmy.nz ⁨10⁩ ⁨months⁩ ago

  Can someone explain this to me? Not the maths, the joke.

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  • fossilesque@mander.xyz ⁨10⁩ ⁨months⁩ ago

    knowyourmeme.com/memes/anti-memes

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    • Rolando@lemmy.world ⁨10⁩ ⁨months⁩ ago

      Calling !peterexplainsthejoke@lemmy.world…

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    • LinkOpensChest_wav@lemmy.blahaj.zone ⁨10⁩ ⁨months⁩ ago

      Old man voice: “Aunt eye memes? In my day, we only had regular memes, and I turned out fine!”

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  • CubbyTustard@reddthat.com ⁨10⁩ ⁨months⁩ ago

    at the core of humor is the twist, where you expected one thing but get another. In an anti joke or meme the twist is usually that there is no joke, but the format set you up to expect one.

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• driving_crooner@lemmy.eco.br ⁨10⁩ ⁨months⁩ ago

  The test:

  Solve e^{int_0^1 x2+x+1 dx}

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  • Usernamealreadyinuse@lemmy.world ⁨10⁩ ⁨months⁩ ago

    No, copiloot solve this test

    Answer:

    To solve the expression ( e^{\int_0^1 x^2+x+1 , dx} ), we first need to calculate the integral of ( x^2+x+1 ) from 0 to 1.

    The integral of ( x^2+x+1 ) is: [ \int x^2+x+1 , dx = \frac{x^3}{3} + \frac{x^2}{2} + x + C ]

    Evaluating this from 0 to 1 gives us: [ \left[ \frac{x^3}{3} + \frac{x^2}{2} + x \right]_0^1 = \left( \frac{1}{3} + \frac{1}{2} + 1 \right) - \left( 0 + 0 + 0 \right) = \frac{11}{6} ]

    Now, we raise ( e ) to the power of this result: [ e^{\frac{11}{6}} ]

    This is the solution to the given expression. If you need a numerical approximation, ( e^{\frac{11}{6}} ) is approximately 4.953.

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• AtomfriedMegaforce@lemmy.world ⁨10⁩ ⁨months⁩ ago

  …and this student was Albert Einstein

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