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fossilesque@mander.xyzM to Science Memes@mander.xyzEnglish · 1 year ago

funny title goes here

mander.xyz

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funny title goes here

mander.xyz

fossilesque@mander.xyzM to Science Memes@mander.xyzEnglish · 1 year ago
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  • OpenStars@startrek.website
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    1 year ago

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

    Maybe stop wasting the time of the student then!?

    You’re welcome.

  • Dave@lemmy.nz
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    1 year ago

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

    • fossilesque@mander.xyzOPM
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      1 year ago

      https://knowyourmeme.com/memes/anti-memes

      • Rolando@lemmy.world
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        1 year ago

        Calling [email protected]…

      • LinkOpensChest.wav@lemmy.blahaj.zone
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        1 year ago

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

        • Venator@lemmy.nz
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          1 year ago

          I prefer large memes

    • CubbyTustard@reddthat.com
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      1 year ago

      deleted by creator

  • driving_crooner@lemmy.eco.br
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    1 year ago

    The test:

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

    • Usernamealreadyinuse@lemmy.world
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      No, copilot solve this test

      Answer:

      To solve the expression ( e{\int_01 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.

  • Atomfried W. Megaforce@lemmy.world
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    1 year ago

    …and this student was Albert Einstein

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