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submitted 1 year ago by Toes@ani.social to c/196@lemmy.blahaj.zone

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[-] UnRelatedBurner@sh.itjust.works 1 points 8 months ago

Alright there buddy, I'd like to close this.

It's clear that your a troll. However, on the offchance that you didn't know, I'll tell you where you went wrong on the first one.

2(4)^2=(2x4)^2=8^2=64

You can't distribute into a bracket, that's raised to the power of anything other than 1, like this. To do this you need to raise distributed number to the bracket's power's inverse. in this case 1/2.

2(4)^2=(2^(1/2)*4)^2=(sqrt(2)*4)^2=2*4^2=2*16=32
or y'know 2*16=32

Maybe if we look at it with roots you'd get it. wolfram syntax

2(4)^2=2Surd[4,1/2]
2Surd[4,1/2]= Surd[4*2^(1/2),1/2]= (4*sqrt(2))^2= 4^2*2= 16*2= 32

I hope you don't get scared from this math, you're a teacher afterall. I have no Idea how you could have gotten a degree or not kicked from school on day 1. Unless.. you are trolling me, fuck you for that. If you respond with more bullshiting, I'll block you.

[-] SmartmanApps@programming.dev 1 points 8 months ago

2(4)^2=(2x4)^2=8^2=64

Yes, that's right. Brackets before Exponents, as per the order of operations rules.

You can’t distribute into a bracket

You know that's literally what The Distributive Law says you must do, right? Unless you have a source somewhere saying there's an exception?

Apparently you didn't bother reading any of the links I gave you, so here's one of the many textbooks which says you must distribute...

In case that's unclear, that means that 2x² and 2(x)² aren't the same thing (since 2(x)=(2x) by definition).

wolfram syntax

You know Wolfram disobeys The Distributive Law, right? I know I'm not the only one who knows this. Is that why you're insisting your way is right? Cos they're known to be wrong about this.