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Thank you. So it is Integration by parts, I guess, just expressed with dv as g(x). And u as f(x). By the way, how come when you take the derivative of something, the dx usually isn't written down, but when you go backwards for the integral, you always write down the dx. Shouldn't it be used in the answer of the derivative too?
It depends how far north you are in New England, and if you live in the mountains. But in general, in the six New England states you get a lot of snow, and very little rain in the winter. And usually not that windy either. I used to live in the northern and mountainous region of New Hampshire, and it was awesome for a half Swede like me. Now I live in southern NH, but it is still very nice. We've had two good snowfalls already, but then it did rain a little after Christmas, which is unusual.
The time delay for the speed of the Tower of Hanoi Solver is from 100 to 1000ms.
Could it be changed so you can choose 2000ms or at least 1500ms.
If you can only have 10 choices, maybe have choices 100ms, 200, 300, 400, 500, 750, 1000, 1250, 1500, 2000ms?
chromatic notes in musical octave = 8 - 1 (duplicate C = Doe a deer a female deer) + 5 (C#, Eb, F#, Ab = G#, Bb) = 12 notes
clock = 12 positions
C
B C#
Bb D
A . Eb
Ab E
G F
F#I just tried the knight move game, and got 83. Is that average or not so hot?
Also, I remember a game that 2 people play.
It has three groups of sticks of 3, 4, and 5.
One person crosses off a line. They go back and forth.
I can't remember who wins. It was kind of neat.
I found this formula online and wondered if anyone can tell me if it is true.
There was no explanation with it.
My turn.
II. white
The Braille dots should be 3 lines high because Braille characters are 2 dots wide and 3 dots high.
Hence there were other carriage returns that were lost somehow.
Here is a silly answer that uses operators that are probably disallowed in algebra in this fashion for some reason or other.
y=0(x<3) + (3 ≤ x<11)(1800/(11-3))(x-3) + (x ≥ 11)1800
Sounds good. I think with result from post #13 and your #14, you might be almost there, though it might be
a little messy. And I hope this reduction idea is all right. I think it is.
So the volume of the front slice to correct for the reduction would be:
(L-c)((s-s1)/s)h1b1
Okay, tack on the front a slice of thickness (s - s1)/s
The reason I say this is because if s1=8 and s=9, then the new proportion is 8/9,
so then the part we need to add to the reduced length is 1/9th.
So (9-1)/8, hence (s-s1)/s
I have a hunch if c > L/2, then when you reduce the object, the bottoms length gets too long.
Just a guess.
If c = L, then there is no bottom, just a point and the reduction should work fine, maybe.
If c is much smaller than L, then... who knows, it's all proportions somehow, I gotta think more...
I think that was all wrong thinking.
When I said the reduced image slides down the slant to hit the bottom, that is not really
correct. If you reduce the image holding the bottom C point at a fixed location, then
everything reduces right, but the length is just too short by (s1/s), so this shouldn't
be too hard...
I'm wrong about the one dimension ratio cubed because the length doesn't get shorter the same as everything else.
But if you could figure out how far the length is off, since it would be too short for the smaller similar shape of the
large one, then you could add on the prism slice to the front to make it the right length. But I don't know how yet.
I reread that, and it is unclear. If you reduce the big volume by the (s1/s)^3, then all the dimensions are right
except the length dimension, which is too short. But how far off is it??
As s1 gets smaller than s, c grows from zero upward, what is this equation?? Should be simple, but I am
not thinking of it. So when make the 3-d object reduced in size, length of object is reduced by (s1/s),
however, the object slides down the slant left and frontwards by c, so how is c related to s1?
Woops, I think that wouldn't work, because the length isn't changing proportionally.
I will use big L instead of little l, so you don't think it is a one.
Since we are told b=e, then the
Volume = he(L-c) + (2/3)hec
Then use pythagorean theorem perhaps to do
part 2, just guessing.
Sorry I'm doing the big volume with S, but I don't really
know what is the difference between the big one and
the small one.
Just multiply the big one's volume by (s1/s)^3 to get the small ones.
Hi fizzled, I figured out part of the problem, but not the whole thing.
If there was a vertical wall at point c (C-wall) from the back where the triangle slants and
hits the bottom, then the volume from the C-wall back to the slanted triangle
is exactly 2/3 rds the volume of going all the way back from the C-wall.
Yes, I sent them a second email telling of my mistake on problem #15, 1997.
mathsyperson is right, I'm in my late 30's.
I simply was trying out some of the problems I found on the internet at the
forementioned address in post #2 of this thread.
I just looked into old Cayley Contests and found that 1997 problem #15 had all four multiple choice answers incorrect, so I sent Waterloo an email.
http://www.cemc.uwaterloo.ca/english/contests/cayley.shtml
Woops, I made a mistake, I assumed the length of the integer was all zeros except for the
first digit, but actually their solution is right. Oops.
I actually did get a calculator, and also a beautiful sweater!
Is the back triangle slanted inward on the bottom, thus reducing the volume a little?
Is the front triangle side exactly vertical?
Yeah, I just read about that last week. Take the sine function and express it
with complex equation, then remove the imaginary i's and you get the
hyperbolic function. Something like that.
If you don't restrict t, it makes sawtooth waves.
Wow, different sizes!! Nice discovery!
'Course I only did the y=stuff equations.
I don't understand combining both together yet.
...
Oh I think I'm getting the idea of the parametric stuff.
Neat concept. I never had heard of it before!
...
So for like normal functions,
x = t and y = f(x) equation.
But now it's all in terms of t, wow, really flexible.
Never would have thought of that idea.
Cost of raw material ordering quantities is missing from the given data.
Assume if we maximize # units sold, we maximize profit, since we aren't
given raw material price or other overhead costs. Storage costs is not
specified if it is per unit or for all of each type. Also I am assuming
that no B's or C's are made, so you have to wait 1.5 hours before the
last step can be started in week 1, so this affects week 1 a little.
The results are guesswork below;I have no equations or methods:
WEEK 1:
making B 4men 160/.5=320units
making C 7men 312/1=312units
making A 1men 38.5/.25=154units
week 1 sold 154, 8 B's left, 158 C's left
WEEK 2:
making B 4men 320made + 8 units leftover = 328 units
making C 6men 240made + 158 units leftover = 398 units
making A 2men 320 units
week 2 sold 320, 88 B's left, 78 C's left
WEEK 3:
making B 4men 320made + 88 units leftover = 408 units
making C 6men 240made + 78 units leftover = 320 units
making A 2men 320 units
week 3 sold 320, 168 B's left, 0 C's left
WEEK 4: (ten workers on last week)
making B 1man 80made + 168 leftover = 248 units
making C 7men 248made with some wasted time
making A 2men 248made with much wasted time
week 4 sold 248, no product left over.
This analysis was just guesswork, so it's better than nothing.
Hope it helps.
What is krassi_holmz saying. I don't understand.
Did he solve the cube without learning some special moves??
When the rubik's cube first came out, I learned the solutions
and got pretty good at it. But that was quite some time ago,
like around 1980. I have a Rubik's cube now, and I relearned
the moves about a year ago and have again forgotten them,
but mostly I wondered if someone had some special math
up their sleeves when they figured out those amazing moves?
One move is called the rubik's manuever, so probably Erno Rubik
figured that one out himself, just guessing.