... bare the fangs like a dog
laglerolarpok (Inuit) the gnashing of teeth
kashr (Persian) displaying the teeth in laughter
zhaghzhagh (Persian) the chattering of the teeth fromthe cold
or from ... rights reserved
THE LIBRARY OF CONGRESS HAS CATALOGED THE HARDCOVER EDITION AS FOLLOWS:
Jacot de Boinod, Adam.
The meaning of tingo and other extraordinary words from
around theworld / Adam ... word. In the Gilbert Islands of the Pacific,
arou pairi describes the process of rubbing noses in greeting. For
the Japanese, bowing is an important part of the process and a sign
16 The Meaning...
... APQ.
15. There are cards labeled from 1 to 2000. The cards are shuffled and placed in a pile. The
top card is placed on the table, then the next card at the bottom of the pile. Then the next ... ☺ The best problems fromaroundtheworld Cao Minh Quan
g
47
15. Two circles touch the x-axis and the line y = mx (m > 0). They meet at (9,6) and another
point and the product of their ... volume of the
top part (the small cone) divided by the volume of the bottom part (the frustrum) equals k and
painted area of the top part divided by the painted are of the bottom part also equals...
...
hotspot provider The Cloud. We discuss these and other service innovations that form a part of the
fixed-mobile convergence pattern in Parts 4 and 5. For now, we simply note that the European
experience ... and centers aroundthe world. I am deeply indebted to the many and diverse contributions
that each and every one of them made.
The project would not have been possible without the tremendous ... like
the U.S.; (2) The OECD data represent what companies tell their regulators, and what these regulators in
turn tell the OECD; the concern is that companies sometimes misreport to their...
... to the scien-
tic and learned societies, and he never was known to take
part in the sage deliberations of the Royal Institution or the
London Institution, the Artisan’s Association, or the ... how, from
the description you have, you will be able to recognise your
man, even if he is on board the Mongolia.’
‘A man rather feels the presence of these fellows, consul,
than recognises them. ... Englishmen on their travels, and the
hospitable eorts of the purser, the time passed quickly on
the Mongolia. e best of fare was spread upon the cabin
tables at breakfast, lunch, dinner, and the eight...
... Report
Conficker Worm Infects Millions AroundtheWorld
April 14, 2009
Introduction
The major news of the first quarter
was the rapid propagation of the
Conficker worm. Research indicates ... but the less
expected job search sites also made an appearance, albeit further down the list.
Criminal activity sites fell from first place last quarter to sixth place this quarter.
On the ... Commtouch Labs, the first
quarter saw an average turnover of 302,000 zombies each day. The graph below
shows the newly active zombies each day throughout the quarter; because the
Conficker...
... with UNAIDS, PEPFAR, the
countries suering the greatest burden
and other partners, virtually eliminate
the transmission of HIV from mother-to-
child by 2015 worldwide. The Business
Leadership ... loyalty.
They are also often owned by women.
New Public-Private Partnerships
The Alliance is working in partnership
with the H5 agencies (UNFPA, UNAIDS,
UNICEF, WHO, and theWorld Bank),
and the ...
diarrheal deaths in the two regions of the
world where they are most concentrated
— the northern states of India and Nigeria.
In Phase 2 this approach will be extended
to the other high child...
... associated
with participation in the SUUBI program. The orphans with a matched savings account had
more than twice the odds of rating their health as good or excellent than their counterparts
without ... biological father present in the household than did youths in the control group (Curley,
Ssewamala & Han, 2009; Ssewamala, Han, & Neilands, 2009).
11. Participants’ caregivers are the mother ... During the first year of the program, 2,743 parents and sponsors signed up. As of
December 2009, there were 6,929 Fundisa account owners. At the end of October 2009, the
total volume of the program...
... P in the interior of the triangle P QR, there exists
a tetrahedron ABCD such that P is the point of the face ABC at
the greatest distance (measured along the surface of the tetrahedron)
from ... |a
i
n
|. Note that either all of the terms are
positive, or they alternate in sign; in the latter case, the terms of
either sign form a geometric progression by themselves.
There cannot be three ... 1)
and the edges are parallel to the coordinate axes. If r is the radius
of the sphere, then (r, r, r) is its center, and (r, 1, 1) is the point of
tangency of one of the edges at B. Therefore...
... the parallelogram. It intersects the line AC in two
points; let the point farther from C be P
2
. then the locus is the
intersection of the segments AP
1
and CP
2
.
We begin by proving that the ... most 2d elements,
starting with the largest; i.e. we put the numbers from n/2 −
2d to n/2 − 1 into one package, then put the numbers from
n/2 − 4d to n/2 − 2d − 1 into another, and so forth, ... prefer B to C, but only the first prefers A
to C.
2. Let a, b, c be the sides, m
a
, m
b
, m
c
the lengths of the altitudes, and
d
a
, d
b
, d
c
the distances fromthe vertices to the orthocenter in...
... Ramsey’s Theorem, either three of them
(call them X, Y, Z) are all adjacent or all not adjacent. But then in
the group {V, X, Y, Z}, none of the players draws exactly once with
the other players, a ... contradiction.
On the other hand, 2
n
2
−1 days suffice. Suppose that n = 2t + 1
or 2t + 2; number the teams from 1 to n and the Sundays from 1 to
2t + 1. On the i-th Sunday, let team i either sit out ... whenever there is a 0 in the binary
representation of n − k, there is a 1 in the corresponding digit of k.
Then the corresponding
(n−k)
i
k
i
equals 0, and by Lucas’s Theorem
n−k
k
is even.
Therefore,...
... given
operations preserve the product of the values of the balls in the box.
This product is initially i
2000
= 1. If three balls were left in the box,
none of them green, then the product of their values ... green,
proving the claim in part (a). Furthermore, because no ball has value
1, the box must contain at least two balls at any time. Therefore, the
answer to the question in part (b) is “no.”
(To prove the ... position and then arranging the other n − m − 1
big numbe rs into the rest of chosen places. We then arrange all the
small numbers in the remaining m − 1 places. Hence, there are
x
m
=
n − 1
n...