Pokemon Universe MMORPG
Off-Topic Discussions => In-Forum Games => Topic started by: Mr.Dodo on February 26, 2012, 03:16:20 PM
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So lets play Maths game :D
Rules are simple you have to solve maths task written on previous post
and post new maths task bellow.
If sombody post wrong number, straighten it in your post.
Simple?
Try not to use calculator.
So lets start:
2 + 8 x 2 = ?
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18
5+9x9=
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86
d( ln(sin(x))/x^2 )/dx =
Given x = pi
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^demolishes thread game^
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86
d( ln(sin(x))/x^2 )/dx =
Given x = pi
can we make it a rule that stuff this complicated is not allowed?
So I'm going to continue without answering this one.
51x283 (No Calculator)
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14,433?
(1-1/2)(1-1/3)(1-1/4)(1-1/5).......(1-1/20) =
Seems complicated, but it's not, actually.
EDIT: Read as= (one minus 1/2)times(one minus 1/3)times(one minus 1/4) until (1 minus 1/20)
There's no power thingy
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d( ln(sin(x))/x^2 )/dx =
Given x = pi
Sin(pi)=0, ln(0/pi^2)=ln(0)=impossible
No answer I think ;o
(1-1/2)(1-1/3)(1-1/4)(1-1/5).......(1-1/20) =
1/20?
4[squareroot(64)]+95x12.
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(1-1/2)(1-1/3)(1-1/4)(1-1/5).......(1-1/20) =
1/20?
4[squareroot(64)]+95x12.
Correct sir!
Hmm, 32+1140= 1172?
1+1=
(-__-") run out of question....
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(1-1/2)(1-1/3)(1-1/4)(1-1/5).......(1-1/20) =
1/20?
4[squareroot(64)]+95x12.
Correct sir!
Hmm, 32+1140= 1172?
1+1=
(-__-") run out of question....
I believe that The answer to Papi's question is 9,772 due to 32 + (95 x 12) where 95 x 12 = 9740
And 1+1=2
Now Find the common ratio in this geometric sequence, and determine the next term: 3, 12, 48, 192,...
or you can multiply 20 by 342 for a different question.
Edit: I honestly don't know what i was thinking when i said 95x12 was 9740, it was an extremely stupid moment i'm not proud of.
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95x12 is 1140.
768 is the next term as 4 is the ratio.
20 x 342 = 6840
And for my question
13/7 + 5/11 = ?
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d( ln(sin(x))/x^2 )/dx =
Given x = pi
Sin(pi)=0, ln(0/pi^2)=ln(0)=impossible
No answer I think ;o
Well, you should be applying the value of x later, but well, I can't blame you either :P There is no definite answer, right! ^-^
First use the quotient rule;
d( ln(sin(x))/x^2 )/dx = (x^2 cos(x)/sin(x) - 2x ln(sin(x)) / x^4)
Simplify; (x^2 cos(x)/sin(x) - 2x ln(sin(x)) / x^4) = (x cos(x)/sin(x) - 2 ln(sin(x)) )/ x^3)
Then put x = pi;
(pi cos(pi)/sin(pi) - 2 ln (sin(pi)) ) / pi^3
\infty - 2(-\infty) / pi^3 = indefinite solution
;D
13/7 + 5/11 = 1 (6/7) + 5/11 = 1 + (66+35)/77 = 1 + 101/77 = 2 24/77
Oh well. Another 'sequence';
3, 4, 6, 8, 12, 14, 18, 20, 24, 30, ?
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Is that the first 10 prime numbers +1, if so, then the next number would be 32.
Ummm. 5x2+3+7x0+12.5-1.
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5x2+3+7x0+12.5-1 = 10 + 3 + 0 + 12.5 - 1 = 24.5
Hmm, how about this one?
3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, ?
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5x2+3+7x0+12.5-1 = 10 + 3 + 0 + 12.5 - 1 = 24.5
Hmm, how about this one?
3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, ?
8.
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8=8?
You didn't give me a problem.
This shouldn't be too hard, just takes a little bit of time.
Domain of f(g(h(x))) given f(x)=(9^1/2)/(2x-2) g(x)=-5x^1/2 and f(x)=(128x^1/7)+3
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5x2+3+7x0+12.5-1 = 10 + 3 + 0 + 12.5 - 1 = 24.5
Hmm, how about this one?
3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, ?
8.
Haha, nice! ^_^
Though you forgot to put a question. Does that mean the sequence was that hard? :P
Domain of f(g(h(x))) given f(x)=(9^1/2)/(2x-2) g(x)=-5x^1/2 and f(x)=(128x^1/7)+3
Well, h is not defined, so... no solution ;D
I think you meant the last one was h, soooo:
f(g(h(x))) = f(g((128x^1/7)+3))
= f(-5((128x^1/7)+3)^1/2)
= (9^1/2)/(2(-5((128x^1/7)+3)^1/2)-2)
= -3/((100(128x^1/7)+300)^1/2+4)
Another sequence? ;D
1, 1, 2, 4, 7, 13, 24, 44, 81, ?
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1, 1, 2, 4, 7, 13, 24, 44, 81, ?
Similar to the Fibonacci sequence, 1+0+0=1 1+1+0=2 2+2+1=4 4+2+1=7, etc. which makes the next number 149.
(P.S I helped DarknighT ;D)
Hmm, In an Arithmatic sequence, where the first term is 3, and the common difference is 4 what is the tenth term?
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Yup! ^^
a = 3
d = 4
T(10) = 3 + (10 - 1)4 = 39
Hmmm, what could I find that's not too complicated...
You start with one bacteria. Each hour, one bacteria divides into two and grow to maturity. Thus, if you have 2 initially, you'll have 4 mature ones after one hour. Each 12 hours, two thirds the total number of bacteria die (don't ask why). You start with 1 bacteria. After how much time, to the nearest hour, would you reach 1 million bacteria?
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1 x 2 =
2 x 2 =
4 x 2 =
8 x 2 =
16 x 2 =
... this will take too long.
2^12 = 4096
2/3 of this would be about 2731 (Computer's calculator isn't very good)
After nine more hours, assuming they die on the 12th hour (you didn't specify), there would be 1398272 bacteria, the first hour the number is over 1 million.
It would take 21 hours. I think.
Assume rabbits, when they mate, produce an average of 8 offspring. It takes 3 months for each rabbit to reach the breeding age, and another 5 months for them to give birth. Half of each litter are males and half are females. Rabbits live to be 3 years old.
(Just made this up, not real facts)
If you start with a rabbit, how many months will it be before you have 5000 rabbits?
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Wait... so do you only start with 1 rabbit? What dies it breed with 0.o
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42.
1+1=
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1 x 2 =
2 x 2 =
4 x 2 =
8 x 2 =
16 x 2 =
... this will take too long.
2^12 = 4096
2/3 of this would be about 2731 (Computer's calculator isn't very good)
After nine more hours, assuming they die on the 12th hour (you didn't specify), there would be 1398272 bacteria, the first hour the number is over 1 million.
It would take 21 hours. I think.
Assume rabbits, when they mate, produce an average of 8 offspring. It takes 3 months for each rabbit to reach the breeding age, and another 5 months for them to give birth. Half of each litter are males and half are females. Rabbits live to be 3 years old.
(Just made this up, not real facts)
If you start with a rabbit, how many months will it be before you have 5000 rabbits?
Right! :)
And the question of yours doesn't have enough detail, such as consistency in the litter, if some are dying, or if some stop mating after a while, or after how much time some become mature to mate. That was why I posted a problem with bacteria instead XD
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Since the exam is getting near, I think it would be better to have some math exercises with you guys.
Here goes!
Say, I have thirteen sheets of money, the money I have only consists of one dollars and five dollars. How many one dollars and five dollars I have if the sum of the cash I have is fourty-one dollars?
Ah, also, did I write it correctly? (Is unfamiliar with dollar currency and English isn't my main language)
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Well, I'm not familiar with the dollars, but that's a simple simultaneous equation.
Anyone want some basic trigo question?
Prove that cos(2A) = 1- 2sin^2(A)
One of the identities you then use a lot afterwards.
Let o be the number of one dollar bill and f be the number of five dollar bill.
o + f = 13
o + 5f = 41
4f = 28
f = 7
o = 6
Number of one dollar bill = 6
Number of five dollar bill = 7
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cos2A
=cos(A+A)
=cosA.cosA - SinAsinA
=cos^2(A) - sin^2(A)
={1-sin^2(A)} - sin^2(A)
=1- 2sin^2(A)
oops forgot to add a questn... heres it...
How can you add eight 8's to get the number 1,000?
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no1...??
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That's because you edited a day later and there was no question before that :-\
888 + 88 + 8 + 8 + 8
Another pattern, I made up this one:
1, 2, 3, 7, 8, 17, 18, 27, ?
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That's because you edited a day later and there was no question before that
ya sry abt that... forgt to write it...
and is the ans 28 ?
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Right, and any explanation for it? ;)
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Right, and any explanation for it? ;)
i jus guessed it lol :P ...
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Well, let's push it further then:
1, 2, 3, 7, 8, 17, 18, 27, 28, 37, 38, ?