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3u Mathematics Marathon V 1.1 (1 Viewer)

haque

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By inspection
│x│≤√35 , pi/2≥y≥sininverse(1/6)
 
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.ben

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Riviet said:
Next Question:

How many different ways of painting 8 different colours on a cube?
You guys haven't answered this question!
 

followme

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.ben said:
8x7x6x5x4x3=20160

But basically isn't it the same because it just depends on perspective?
what's wrong with ur answer? i would've done the same...:confused:
The 1st face has 8 colours to pick from, 2nd 7 choices.... and 6th 3 choices. = 8x7x6x5x4x3
 

followme

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.ben said:
Next Question

A batsman hits a cricket ball 'off his toes' toward a fieldsman who is 65m away. The ball reaches a maxiumum height of 4.9m and the horizontal component of its velocity is 28m/s. Find the constant speed with which the fieldsman must run forward, starting at the instant the ball is hit, in order to catch the ball at a height of 1.3m above the ground. (g=9.8)
let Y'=0 t=usinα/g
4.9=-1/2 g(usinα/g)<sup>2</sup>+usinα(usinα/g)
u<sup>2</sup>sin<sup>2</sup>α/2g=4.9
usinα=9.8


let Y=1.3
1.3=-1/2gt<sup>2</sup>+usinαt
=-9.8/2t<sup>2</sup>+9.8t
ie 98t<sup>2</sup>-196t+26=0
t=0.143 or 1.857


X=utcosα
= 28t
=28x1.857=51.996m
65-51.996=13.004m


time of flight=1.857
therefore 13.004/1.857= 7m/sec forward
 
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haque

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the answer to the cube question is 8^6 as no restricitons are stated.
 

.ben

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@ haque can you explain please. thanks.
@ followme i did the projectile motion and got the same answer as you, but in the book the answer is 7m/s, surely they didn't round up did they?
 
P

pLuvia

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.ben said:
@ haque can you explain please. thanks.
It did not say that each colour could only be used once, hence all colours could be used more than once
 

haque

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yea what pluvia said- consider this- we can paint the first face in 8 ways, the second face in 8 ways and 3rd face in 8 ways and so on. there are six faces thus 8 times 8 times eight times and so on-get what i'm saying? i'm a bit unclear aren't i? sorry i'll explain again if u like.
 

.ben

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Nope that's perfect haque and pluvia. i get it now. it was just the word 'different' which confused me into thinking that one colour could only be used once.

thks
 

haque

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here's an easy one-how many diagonals may be constructed for an n sided polygon(regular).
 

haque

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slightly harder-what is the maximum number of points of intersection possible when m lines are drawn through n circles?
 

Yip

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label vertices of n-gon as A1,A2,A3....,An,
diagonal is a combination of 2 elements from this set ie A1A2 corresponds to a diagonal
hence there are nc2 diagonals, but the n-sides A1A2,A2A3,....are not diagonals, so there are nC2-n diagonals in total

Im not sure about the next problem, most likely wrong, but i'll take a stab...
Consider the separate cases of the maximum numbers of intersections of m-lines, maximum number of intersections of n-circles, and maximum numbers of m-lines with n-circles:
Case 1:
1 line has no intersection with any other line
2 lines has one intersection
3 lines has 1+2 intersections
.......
m-lines have 0+1+2+3+4+......+m=[m(m-1)]/2 intersections
Case 2:
Each circle can have a maximum of 2 intersections with another circle
hence n circles have a maximum of 2[(n-1)+(n-2)+(n-3)+...+2+1]
=2[n(n-1)-[1+2+...+(n-1)]]
=n(n-1)
Case 3:
A line cuts a circle at a maximum of 2 points, hence the maximum number of intersections of m-lines with n-circles is 2mn
Maximum number of points of intersection is: [m(m-1)]/2+n(n-1)+2mn

Most probably wrong......seems a bit iffy imo...
 
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Riviet

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haque said:
here's an easy one-how many diagonals may be constructed for an n sided polygon(regular).
There was a similar question that I posted earlier in this thread. :p
 

haque

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nah it's not iffy yip-the answerr's correct. oh sorry riviet i didn't see ur earlier posts which is why i posted that question
 

Yip

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Heres a quick one:
Prove, without resorting to induction, that the product of r continuous integers is divisible by r!
 

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Without resorting to Induction...............Whats wrong with induction lol

Hmm this question may hurt my brain for a bit
 

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