3u Mathematics Marathon V 1.1 (1 Viewer)
- Thread starter Riviet
- Start date
.ben said:
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
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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P
pLuvia
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It did not say that each colour could only be used once, hence all colours could be used more than once.ben said:
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.
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...
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...
Last edited:
Rax
Custom Me up Scotty
Without resorting to Induction...............Whats wrong with induction lol
Hmm this question may hurt my brain for a bit
Hmm this question may hurt my brain for a bit