Americans are more fat so they need bigger Pi to keep geometry in touch with reality.
this post was submitted on 17 May 2025
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(13 replies)
Bye, bye, miss American PI.
Maybe Vader some day later, but now it's just about prime.
What kind of problem gives you the formula and all variable to replace? At this point, why not just write 5•10²•10=?
Intro to algebra type stuff to make sure you understand the concept of variables in the first place
Pi= 5 in this teachers reality. Circles must look wonky.
It makes it easy to do the math in your head without a calculator. But still , just tossing out pi=5 is not the way to go about creating these problems.
...fractal circumferences can be whatever length you want for any given mean radius...
Even then, I would want them to leave π in the problem itself. That would be much better for this exercise - teaching that you report “exact” values with π still in them.
Eg, if I rewrote this problem, I would expect an answer of 1000π.
One written in Comic fucking Sans
Cause reading comprehension is part of the test. Lots of kids will be able to solve that equation, but there's a bunch who can't understand it if it's presented this way.
Honestly here they should have done "round pi to two decimal places" or smth.
This was written by an engineer. They rounded up to 5 for the safety factor.
Ha Ha, non-Euclidian geometry go brr. :)
Assigning a value of 5 to pi, although ludicrous IRL, doesn't affect the problem. Plug the values into the equation and it will still give an answer that's correct in context.
For the benefit of doubt, maybe the test is from an alternate dimension that doesn't use euclidean space.
I've been there, I think, but it was really difficult to triangulate my location and confirm
That's because you were supposed to rhombusulate, not triangulate.
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I wish they would have used 22/7 for pi and 7 for the radius or height
If the goal is to avoid calculations with decimal places, why not just leave Pi in the result?
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It's official, the observable universe is ~3 times larger!
It’s clearly just saying that the surfaces on which the ends of the cylinder lie are metric spaces with distances defined using Chebyshev or Taxicab metrics based on pentagonal tilings of the parabolic plane so the ratio of a circle’s circumference to diameter is 5.
Since it’s a cylinder we assume the vertical dimension is Euclidean and voila the math checks out geometrically.
Calculator not allowed test probably
Even if so, the other factors are both 10. How hard can it be...
3.14 would be easy enough to solve this one. r^2*h resolves to 1000, so V would be 3140.
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This is how you develop trust issues.
Think of the Loch Ness Monster and use tree fiddy and you're much closer
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Ah yes, all those imperial units...
It went to far and Beck.
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