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You made a slight typo here:
where a,b∈ R[sup]+[/sup]
If one of a and b is negative, the result breaks down. E.g. try a = 1.5, b = −0.5.
To prove the result for positive real a and b, observe that the function
is convex for x ∈ ℝ[sup]+[/sup]. Hence, by Jensens inequality,
QED.
Wait a minute! I believe Ive figured out how you did Level 8 in 26 moves (see my other thread). So Im down to 340 now.
Wow, I dont know what you did in Level 8, but I managed to knock six moves off in my latest try and get it down to 33 (and my overall score to 347).
Sorry to have to keep rewriting the record book, but Im now down to 39 moves for Level 8, 41 moves for Level 9 and 57 moves for Level 10. So 353 is the best overall total I can manage so far.
Btw, Jane, do you have a way of saving levels to avoid having to start again when you run out of credits?
When I want to do a level, I just start from the beginning and carry on to that level.
In the last few days, I have managed to find faster ways of completing most of the levels, with the result that I can now complete all ten levels in just 353 moves.
(12 → 11)
(53 → 48)
(27 → 23)
(39 → 31)
(44 → 40)
No change for Level 6 (44) and Level 7 (19).
(41 → 26)
(47 → 41)
(62 → 57)
Overall: 388 → 340
Ive brought my Level 9 score down to 43 now.
And my Level 5 score down to 40!
Ive also brought my Levels 1, 2 and 4 scores down to 11, 48 and 31 respectively, so my overall score is now 370.
I have now brought my fewest-move score down to 384.
My record is 388 moves.
Assuming no friction, t1=t2 for any position of P (as they would fall at the same rate, due to gravity).
That is not correct. The time taken for an object to fall a vertical distance h is
but if it slides down a frictionless slope inclined at angle θ to the horizontal, the time taken for it to fall the same vertical distance is
(as you can easily work out). So unless θ is 90°, it will take longer to slide down a slope than to fall freely under gravity.
If θ[sub]1[/sub] and θ[sub]2[/sub] are the angles at Q and R respectively and h is the , then
for i = 1, 2 (as can be deduced by application of elementary physics). As the area of the triangle is A, we have
You wanna minimize that?
Youre doing it free of charge?
1. Hint: consider making A a subgroup of B.
3. If
and , what is ?4. I will show you my own example so you can do yours. Consider
. Work from right to left. Under this permutation:So 1 maps to 3 and 3 maps back to 1, while 2 maps to 5 and 5 maps back to 2. Hence the permutation in disjoint cycles is
. You get the idea, hopefully.The order of a product of disjoint cycles is the LCM of the lengths of the cycles.
I did the test again and this time got them all in under five minutes 4 minutes 54 seconds to be precise.
I got 115 out of 118!
The ones I missed were hafnium (72), dubnium (105) and bohrium (107).
Is there any way we can use nPr to solve it.
The formula is
Your first
socks must be different colours and you can pick them in ways. Your th sock must then be one of the colours you have already picked.Note that this formula is only for when you have at least as many socks of each colour as there are colours.
No problem. In fact I misread the first question myself. I thought the answer was 8 until you pointed out the teeny words up to at the end.
No, 4 is the maximum!
We have
which is always non-negative for all
, i.e.Same for all
and ; adding up gives the required inequality.NB: This particular method of tackling inequalities is especially useful when there are no restrictions on how the variables vary with respect to each other and we can neatly separate out the variables into part sums.
There are no real solutions. Substituting y from the second equation into the first will yield a quartic with no real roots.
calculate:
Im looking over your problem again and cant help wondering if youve made a typo in the last bit there. Do you mean to calculate
instead? If this is the case, then the answer is just simply
.(both having the same sign).
NB: The precise value will depend on which quadrant you want
to be in.It follows that
.To complete the proof, show that