1. It`s possible for a February to have ? Tuesdays, but not more. a. 3
1. It`s possible for a February to have ? Tuesdays, but not more. a. 3

Leap Frog Solutions 2015
Leap Frog Solutions 2015

... If b = 1, then we can take a = 1 (a = 2 is not possible since m < 45), giving m = 15. But if b = 0, we can satisfy both inequalities with a = 3, giving m = 27. This is thus the largest possible such m. ...
Problem of the Week Problem C and Solution One Step at a
Problem of the Week Problem C and Solution One Step at a

Week 12
Week 12

Problem 3.2 Apply nodal analysis to determine Vx in the circuit of Fig
Problem 3.2 Apply nodal analysis to determine Vx in the circuit of Fig

PROBLEM 16 : What angle between two edges of
PROBLEM 16 : What angle between two edges of

Geometry Scavenger Hunt (One page) Take pictures of the
Geometry Scavenger Hunt (One page) Take pictures of the

100 Degree Isosceles Triangle.pdf
100 Degree Isosceles Triangle.pdf

Answers
Answers

... 2. Since the three angles add to 180, 2b + b + 3b = 180 and b = 30. So 3b = 90 and we have a right triangle. 3. Trial and error shows the numbers are 45, 47, 49. 4. Since most of these cancel in pairs we just add 24 + 25 + 26 . . . + 31. These 8 integers average (24 + 31)/2 and so the total is 4 × 5 ...
Bellwork (Bonus ticket day!)
Bellwork (Bonus ticket day!)

45rd IMO 2004 Problem 1. Let ABC be an acute
45rd IMO 2004 Problem 1. Let ABC be an acute

Lesson 4.2
Lesson 4.2

1. Name an exterior angle of triangle XYZ. 2. Name two remote
1. Name an exterior angle of triangle XYZ. 2. Name two remote

COL100: Lab 13
COL100: Lab 13

Solution for Centroid of Equilateral Triangle Solution using Unit
Solution for Centroid of Equilateral Triangle Solution using Unit

Weber problem



In geometry, the Weber problem, named after Alfred Weber, is one of the most famous problems in location theory. It requires finding a point in the plane that minimizes the sum of the transportation costs from this point to n destination points, where different destination points are associated with different costs per unit distance.The Weber problem generalizes the geometric median, which assumes transportation costs per unit distance are the same for all destination points, and the problem of computing the Fermat point, the geometric median of three points. For this reason it is sometimes called the Fermat–Weber problem, although the same name has also been used for the unweighted geometric median problem. The Weber problem is in turn generalized by the attraction–repulsion problem, which allows some of the costs to be negative, so that greater distance from some points is better.