  # Topic, Math Whiz Quiz Game V2 Checkout our iPhone & iPad games!

Same for me, I'm not native English.
OK Simon, sorry for the misunderstanding. English is sometimes complicated for me ... much more than maths :)
I meant the second last point :)
@F12: no problem, we all understand that everybody has a real life. This topic is an entertainment, it must not become a constraint. It's a good idea to keep it as a maths discussion topic :)

@Simon: your explanation is much faster and less complicated than mine. Just a small precision: at the (n-2)th point, you still have 2 possibilities (2 points remaining). The result (n-1)! has to be divided by 2 because each polygon is counted twice (for example ABCDE is equal to AEDCB).
All scores=134

134/8=16,75 points :)))
Yeah sure, that's probably why people are solving this questions.
As you may have figured out, I will not be able to update this topic frequently enough for it to work as a game anymore. Sorry! But we could just keep it as a maths discussion topic, couldn't we? :)
I've drawn the solutions for 4 points, and I got 4 different point, the problem is one without symmetry. It could be that in certain situations symmetry 'deletes' a solution. I thought the solution was (n-1)!/2 too first, but then I found 4 for n = 4. I'll check it again.

Edit: I think I scrambled the n = 4 and n = 5. So I'm also guessing it's (n-1)!/2, but I don't have a conclusive proof.
My explanation for (n-1)!/2 is:
start at a random point, now you've n-1 options to draw a line to another point, at this point you've n-2 options, etc untill at the n-2th point you don't have any options anymore, so it becomes (n-1)!/2!=(n-2)!/2
Hello Simon! Your problem is very interesting and I will try to find an answer tomorrow. Just one remark: for n=4, I have found only 3 polygons.

Let A, B, C, D the four points. One polygon is ABCD, identical to BCDA, CDAB, DABC, DCBA, and others obtained by circular permutation and by putting the points in the reverse order.
Then, the 3 different polygons are ABCD, ACBD and ACDB. Do you agree?

edit: I think that the solution could be (n-1)!/2

- for n=3, there is one solution ABC
- for n=4, we obtain the new polygon by inserting the point D in each of the vertices of ABC: ADBC, ABDC, ABCD (3 solutions)
- for n=5, we do the same for each of the 3 previous polygon. for example, ADBC gives AEDBC, ADEBC, ADBEC, ADBCE (4 solutions multiplied by 3 = 12 polygons).
- for n, the number of polygons should be 3*4*5...(n-1) = (n-1)!/2
I was wondering if any of you would know the answer to this:
Suppose you've n points in the plane, where no 3 points are colinear. How many different polygons(n-gons) exist with the n points.(you have to use all points).
I found 4 polygons for n = 4.
The first time I'm answering here xD

(72+46+7+3+3+1+1):7=19

( all points in total : amount of players = average points )
Again, correct!
(2x^3/e^2x)'
= 2(x^3/e^2x)
= 2(3x^2*e^2x-x^3*2e^2x)/(e^2x)^2
= 2(3x^2-2x^3)/e^2x
= 2x^2(3-2x)/e^2x
Correct !
A cylinder with radius r and height 6r can hold 3 spheres with radius r.
Looks very good, but unfortunately I made a typo in the task again, so my answer is different, obviously. But of course you will get your points because that looks to me like it is the correct answer to the task I actually wrote.
line: y = -x+2
circle: (x-2)^2+(y-3)^2 = 25
subsituting x-2=-y ->
(-y)^2+(y-3)^2=25
y^2+y^2-6y+9=25
2y^2-6y-16=0
D=36-4*2*(-16)= 164
y=[6 +/- sqrt(164)]/4
y = 3/2 +/- sqrt(41)/2
Intersection points found with x = 2-y
[x = (1+sqrt(41))/2, y = (3-sqrt(41))/2] = [x = 3.70, y = -1.70]
and
[x = (1-sqrt(41))/2, y = (3+sqrt(41))/2] = [x = -2.70, y = 4.70]
You are of course orrect again :)
Let x = one chocolate bar, y = one bottle of Coca Cola
3x = y and 6x+y = 54
6x = 2y ==> 2y+y = 3y = 54 ==> y = 18
3x = y ==> x = 6
2y+x = 2*18+6 = 42kr
r(t) = [5t, 3t^3+2]
v(t) = [5, 9t^2]
a(t) = [0, 18t]
New questions!
It will be the 22.11.2013 (22th of November).
That's correct as well!
A=2B
5A+6B=80
10B+6B=80
16B=80
B=5 ==> A=10
Conclusion: He bought 10 apples and 5 bananas.
Correct !
Glass problem(3p) :)
V = (pi/3)*8cm*((2.3cm)^2+2.3cm*3.3cm +(3.3cm)^2)= 199.135 cm^3

Dice problem(2p)
Chance to have 2 dices with same number up is (1/6)*1/6)=1/36.
We have 6 sides so
chance is 6*(1/36)=1/6.
I need to look more closely to the newly commented topics list :/

And now I forgot to check how many point Simon was going to get. Was it four?
1) x/3+4=2x-3 Thus x+12=6x-9 Thus 5x=21. So x = 21/5

2) coordinates of centre of circle "x^2+y^2=25" is (0,0) and radius 5.
x=6x+8x-9-12
12x = 21
x = 21/12
x = 7/4

    

## First post of the topic

I thought I'd be fun to revive this little quiz-topic, orginally done by AK, who unfortunately is no longer active. The rules are more or less the same as in the old version, but if you are not familiar with them, here is how the game works:
- I post maths questions, and then it's up to you guys to find the answer(s), preferably with some kind of explanation.
- The first to post the correct answer, gets the points.
- The number of points any question gives, is stated along with the question itself. Few points → Easy question, and oposite.
- New questions will be added whenever the previous one(s) have been answered correctly.

CURRENT QUESTION(S):

What is the average score of Math Whiz Quiz Game V2? (1p) 