Pi is now 5 due to inflation
this post was submitted on 21 Apr 2024
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Having Pi equal five definitely does not help the world go round.
To be Devil's Advocate:
Given that the rest written in Comic Sans, it may be an early elementary school exercise, aimed at teaching kids to do multiplications. In this case, it's tolerable and/or defensible to find a simplification for pi.
That said, making pi equal to 3 would have been more accurate for that...
Unless the kid is even slightly above average and finds the idea that pi equaling 5 confusing.
...if they're above average, I think they'll figure out the explicitly defined variable. I think the instructor is trying to make sure this problem doesn't require a calculator and figured defining pi as 5 makes it clear that you can treat it as a whole number. 3 would be more accurate and just as easy, but meh idk that this is that great of a blunder.
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Or it's from an ME. They seldom can remember the rounded value of Pi, but they're pretty sure it's somewhere between 3 and 4. But you probably should use 5 just to be safe.............
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That's how you know math is OP when you can calculate volumes in parallel worlds where circles don't even looks like circles
They're just rounding up from "3."
And not that pussy rounding up where you go up by only 1, oh no, we’re rounding up 2 baby.
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Damn inflation is everywhere now, even the transcendentals :'(
This is how the wealthy calculate their tax exemptions.
With π=5 maths break down completely. If π=5, then e^(5i) = -1, meaning -1 = cos(5) + i * sin(5), or -1 ≈ 0.284 - 0.959 i
Intel®️ inside
Wow, that takes me back - you're referring to the floating point bug from ...98?
It's posts like these that makes me think we're all old here on Lemmy and then I get a response from someone who tells me they're 18...
665.999999657838 the floating point number of the beast
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I think you're overthinking it. The first thing you're told when you learn algebra is that a letter represents a number and you can say "let a equal (number), b equal (number)..." so you can let pi equal whatever you want for the purposes of one simple problem.
But the question is saying to find the volume of a cylinder. Which its clearly wrong.
Well, if we want to be pedantic, they never said that h is the height and r is the radius of the base circle. They could be just random numbers.
Also, since we never calculate with all the digits of pi, it is not any less weird to round to the nearest 5 and say that it's 5, than to the nearest 0.01 and saying it's 3.14. It just has a higher amount of rounding error.
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Idk, if you want to test people on how they understand formulae and order of operations without letting them just punch it into a calculator. The actual math isn't hard, but if you don't get substituting values into an equation then it's not trivial
Just let π be 3 ...
Or have them learn how to use a calculator at the same time.
Your life is easier and better if you can do this kind of simple math in your head.
Oh don't you try to sell me on the "you won't always have a calculator in your pocket" thing. I have fucking Excel in my pocket.
It's not just about haveing a calculator, it's also that it's faster and more convenient if you can do simple sums like this in your head. It also means you can sanity check the numbers your calculator gives you to make sure you didn't make a mistake entering the sum.
To your point below about products having their unit cost displayed, more than once I've seen that just be wrong, so I wouldn't rely on it. Make sure you can check it in your head.
Who really wants to use Excel to figure out if the 24-pack of Coca-Cola or the 3 12 packs is a better deal?
I don't need to, there's a legal requirement to print prices per liter or kg on every price tag here.
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When you want students to not use calculators but still rate a question for 10 second answer.
"One! Two! FIVE!" "Three, Sire!" "THREE!"
That just means the curvature of spacetime is negative.
Teaching them to to obey dumb instructions from incompetent bosses.
Very useful skill.
Assume the earth is a flat disc . . .
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π = 5 for very small values of r.
My favorite is when doing Order of Magnitude estimations, you are supposed to let pi = 1
I would be a smartass and leave Pi as a factor throughout and in the answer. I'm used to doing that in Calculus anyways.
V = πr^2^h
V = π⋅10^2^⋅10
V = π⋅100⋅10
V = π1000
BONUS SOLUTION:
V =∫~0~^10^ A⋅h dh
A = ∫~0~^10^ 2πr dr
V= ∫~0~^10^ ∫~0~^10^ h⋅2πr dr dh
h is a constant for A's integral so we can safely move it into V's integral
V= ∫~0~^10^ h⋅∫~0~^10^ 2πr dr dh
π is a constant so we can safely remove it from A's integral
A = π⋅∫~0~^10^ 2r dr
A = π⋅[r^2^]~0~^10^
A = π⋅( [10^2^] - [0^2^] )
A = π10^2^
A = π100
V = ∫~0~^10^ h⋅π100 dh
π100 is a constant so we can safely remove it from V's integral
V = π100⋅∫~0~^10^ h dh
V = π100⋅[h]~0~^10^
V = π100⋅([10] - [0])
V = π100⋅10
V = π1000
It goes a lot deeper but I'm not bored enough for that, yet.
EDIT: Hang on. I'm wrong with that height integral. Can somebody help remind me?
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Makes me miss the days of 22/7.
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