this is a grade 12 exam question from 2001 from the provincial exam taken in Canada. can one of you math fags solve this. It's multiple choice but show proof please.
This is a grade 12 exam question from 2001 from the provincial exam taken in Canada...
Other urls found in this thread:
4cam.pro
twitter.com
I used to know how to do that shit
It looks like a conics problem user. They give you the inner lengths of the dome (43, 30). You will have to plug those numbers into the equation like in pic related, then solve for when x is 20 since it's centered on the origin.
Same. It's irrelevant now. Go fuck yourself OP.
20 m
none of the above
This is easy. First calculate the result for a semi circle
Then adjust for 30m height instead of 43m
either B or C
give me some time, i think i can do it in around 5 minutes
Assume a semi circle with radius 43.
Height at distance of 20 from the center would be sin(a) *43
calulate a from cos(a) = 20/43
Then adjust for the actual max height of 30
i.e. multiply the result by 30/43
No shit
Solved it, will post my solution, but i got D as my answer
its not a semi circle, its a semi elliptical shape
Why are all these noobs saying it's a semi circle? Conics is taught in grade 11, and it even says elliptical in the problem.
And so the answer is B
basically you to graph the ellipse, with the correct intercepts (86, and 30), then sub x = 20, and solve for y
You're not reading my whole story nigger
you doubled the full length of 86? When 20 m from the middle would be 43 - 20
The OP roof spans 86 meters. Your roof spans twice that.
Answer is B using the elliptical equation
fuck, nice catch
Then the answer is B, fuck, i'm gonna go now
my bad master whitey
Well now that we solved the question, Is this thread dead?
This is probably what the school wants you to do, but if you're too old to remember that the elliptical equation exists you can use the scaled trig method I outlined
want a new one?
Yes, scaling a semi circle can make it into an ellipse.. it's just rotating it in/out on the x axis.
But this reminds me of a bad memory of when a random ACT question used hyperbolas and i didnt know, which implies you should be 100% familiar with this crap to do well, for whatever reason.
Test your might!
Too fucking easy
what's your answer?
The problem of course is that it's not possible to draw a trapezoid with those dimensions
...
80, assuming the trapezoid is object a
It is not a trapezoid, that's the problem
you are all amazing ;_;
assuming that 5 is perpendicular to both lines measured 16 and 12. then the small segment of the triangle 8^2, 5^2 is root(39), and there on out it doesn't make sense, because 12 != 16-root(39) *2
Sup Forums is full of retards who can't do ma-
14
Yea it's not a trapezoid
too old
Its symetrical. So you just drop the left end on top of the right and its a square. Its 80.
It's not symetrical, you can't draw a symmetrical trapezoid with those dimensions, see
Can you guys help with this question? The answer is defs not B
I think the teacher just made random numbers up
Its D
Is this about Shannon limits?
enough math! go watch some porn!
4cam.pro
Its, about sending information down channels. I did a quick google of shannon limits and i don't remember it being said in class, but yet they look similar
How did you get D?
The whole course is based on information codes and ciphers
porn is the weak!
That slutty square. Taking it from the circle and wanting more. What a whore!
Thread is dead
Since there are some math fags around; Why do we define the closure of U, U being the subset of a metric space (X, d) to be the intersection of all closed sets containing A and then prove that U bar = bd(U) \cup U and not the other way around? Isn't it a lot simpler and neater this way?
Containing U. Damn I need sleep.
No, but if you wish
>[email protected]
I'm Not really but I'm too busy to be in the morning
i can't answer that for you, i'll have to pass