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What I mean to say is that some people after handshaking once will retire and do not participate further in handshaking. If too many people do so the total number of handshakes will fall short of 500.e.g. If 20 people take single handshake say for simplicity between them only they have exhausted 10 handshakes and the remaining 490 handshakes will have to be completed by remaining 30 people. But 30 people can have a maximum of only
handshakes. So the number of retirees has to be less than 20. An equation or rather type of equation has to solved.Rather than matrix it is a problem on vectors.
z-y=ai-bj
For x to be perpendicular to z-y dot product must be =0
here dot product=
When MB=MC and B.M and c are not collinear the locus is the circle passing through B,M and C in addition to the perpendicular bisector.
Now the original problem being over, what is the maximum number of people who can have single handshake, the total number of handshakes remaining 50?
In the equation t is measured from the time when the ball leaves the racket whereas h is height from the court. if you double the time it gives the time when the ball comes to the height of 0.5 m ,not the court level.
I think the summation has something to do with splitting the denominator say 1/(4*5) into (1/4-1/5) and combining first part with the previous term and the other with the later term.I can't figure out right now.
First 11 people who handshook with all others exhausted 484 handshakes. 12th man at best can shake hands with 16 more along with 11 he has already completed bringing his total to 27.What about the remaining? 16 people would have 12 handshakes and 22 would have 11 handshakes each. I think this competes the solution.
Since yesterday I am having trouble with \mbox {} It goes to submit mode.
for Q.2 Use variables separable method bringing all terms of y to left hand side and x terms to r.h.s. and integrate each side. Substitute
and solve l.h.s for r.h.s. use integration by parts. IR I thought integer regime.I am not sure but I have a hunch for 3.(b)
AM=A+M Premultiply by
One prisoner groups others according to the color of their hats. If there emerges any group one of the group will announce the color of the hat by looking at others in the group. If there is no group (all 4 having different colors) any other person will group the first person to the color of his hat. If now also group is not formed the first person can tell the color of his hat by missing color.
I think the prisoners are executed by now.
Complex roots must be in pairs
It must be cut into 3 pieces.First one L shaped with each side 8 blocks and width 4 blocks. Second one also L shaped 4x4x2 size. The green block will be in the third piece which is a square of 2x2.
I am sorry. I had not noticed that p is a number ending in 3.
PRIME NUMBERS WROTE
" p= any prime number ending with 3. Check p is not a square." This is redundant. No square will end with 3.
It is interesting to calculate some angles.
I've inserted spaces using space \ space. Bob Bundy.
Thank you.
Q6. either A=B or C is a right angle.
Q.7 A is aright angle.