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.
What kind of problem gives you the formula and all variable to replace? At this point, why not just write 5•10²•10=?
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.Intro to algebra type stuff to make sure you understand the concept of variables in the first place
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.
Pi= 5 in this teachers reality. Circles must look wonky.
…fractal circumferences can be whatever length you want for any given mean radius…
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.
This question was written by an engineer
Nice try, physicists.
There’s good reasons they engineers over calculate, because they know things break, that people don’t do regular maintenance and that people will over stress the object. So engineers have to account for things like this when designing an object or a device so they don’t fail prematurely.
Engineer here, I always just use pi and a “safety factor” multiplier. Extra material is expensive, and I want the cheapest part (like a screw) to fail first. We don’t just oversimplify pi because half the time it’ll make your design weaker.
(If I just got whooshed I apologize)
This is the real answer, what engineer worth their salt is hiding margins in pi?
Clearly define that
Bye, bye, miss American PI.
Maybe Vader some day later, but now it’s just about prime.
It’s probably trying to teach kids algebra without using decimals. But it does look messed up. Everyone knows at least 3.14, except kids I guess
I don’t understand. Aren’t fractions better than decimals for algerba?
Like 22/7 is better than 3.14 when it comes to pi for example.
We always got taught to do everything as fractions and then convert to units at the last possible moment to reduce errors in rounding.
Kiddos would need to know how to divide for that though. I’m just trying to come up with a reason for it lol.
Man the Americans… everyone knows that π=-10
Nah, π = 3.2.
What in the idiocracy
the Speaker accepted another member’s recommendation to refer the bill to the Committee on Swamplands, where the bill could “find a deserved grave”
Lmao
Americans are more fat so they need bigger Pi to keep geometry in touch with reality.
I’m an American. I weigh about 140 lbs (63.5 kg). I’m about 5.75 ft (1.75 m) tall. Am I fat?
Fat? No, but dumb enough to understand my phrase as “each and every without a single exception”. Dumb as an American.
I’m not the one making dumb generalizations. Fuck off with your stereotypes.
You’re literally a walking stereotype, incredible!
Well, stereotypes are lazy humor, and I’m lazy & my life is a joke - so you’ve got me there.
Fair enough
Sick burn. Especially from a Belgian! Waddle waddle
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.
If the goal is to avoid calculations with decimal places, why not just leave Pi in the result?
That would work too.
Do cylinders even exist in metrics where pi = 5 ?
Technically no, because pi equals pi not 5. But you can approximate its value as 3 or 5 or whatever you want, knowing it’s not exact and that your result will only be an approximation. I mean you could also ask how long light takes to reach us from Alpha Centauri if the speed of light is 1000 mph. It’s not, but if you make that a condition of the problem you can do the calculation just fine.
I think that reason would make it “Technically Yes”, since False (pi = 5) implies False (cylinders exist) is (vacuously) True (“absurd premise”).
Yes. The 3d shape existence is not affected by changing pi values
Cause it’s just a (n-1)-dimensional ball extruded along the remaining axis, or do all 3d shapes exist on (nearly) all 3d metrics?
Mostly because the actual pi values can vary in between non/euclidean geometries. Within extremely strong gravitational fields, spacetime becomes highly non euclidean, affecting the C/d ratio of an actual circle, so I’d wager this would affect pi as well
For the benefit of doubt, maybe the test is from an alternate dimension that doesn’t use euclidean space.
Possible. I mean, electricity could actually be run by ghosts, but there’s no need for fanciful explanations when a mundane one is right there.
In America, numbers are just bigger.
Bigger numbers are better numbers!!! The best in the world!!!1!11!!!11!
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.
