X - Maths, NUS
... Prove that FS is always a frisbee and that it is a a highflyer if and only if S is a singleton set. (b) For | X | show that F { A X : | X \ A | } is a freeflyer. (c) Use the Hausdorff Maximal Principle to prove that every frisbee is contained in a highflyer. (d) Show that a frisbee F is a ...
... Prove that FS is always a frisbee and that it is a a highflyer if and only if S is a singleton set. (b) For | X | show that F { A X : | X \ A | } is a freeflyer. (c) Use the Hausdorff Maximal Principle to prove that every frisbee is contained in a highflyer. (d) Show that a frisbee F is a ...
Geometry Chapter 6 REVIEW Problems 3/4/2015
... 8. This jewelry box has the shape of a regular pentagon. It is packaged in a rectangular box as shown here. The box uses two pairs of congruent right triangles made of foam to fill its four corners. Find the measure of the foam angle marked. ...
... 8. This jewelry box has the shape of a regular pentagon. It is packaged in a rectangular box as shown here. The box uses two pairs of congruent right triangles made of foam to fill its four corners. Find the measure of the foam angle marked. ...
HOMEWORK ASSIGNMENT #6 SOLUTIONS
... triangles ∆ABC and ∆JIC imply?) Assume that the Pythagorean Theorem does hold, then in ∆ABC we have AC 2 + BC 2 = AB 2 , and in ∆JIC we have, IC 2 + JC 2 = IJ 2 . Since I is the midpoint of AC and J is the midpoint of BC, the first equation gives us AB 2 = (2 IC ) 2 + (2 JC ) 2 = 4( IC 2 + JC 2 ) AB ...
... triangles ∆ABC and ∆JIC imply?) Assume that the Pythagorean Theorem does hold, then in ∆ABC we have AC 2 + BC 2 = AB 2 , and in ∆JIC we have, IC 2 + JC 2 = IJ 2 . Since I is the midpoint of AC and J is the midpoint of BC, the first equation gives us AB 2 = (2 IC ) 2 + (2 JC ) 2 = 4( IC 2 + JC 2 ) AB ...
Week 5 Lectures 13-15
... Then by the previous theorem δ is actually the minimum and hence is positive. Now let x ∈ K, and consider Bδ (x). If it is not contained in any of U1 , . . . , Uk , that would mean that the ball contains points from each of Fj which means that the distance of x from each Fj is strictly less that δ. ...
... Then by the previous theorem δ is actually the minimum and hence is positive. Now let x ∈ K, and consider Bδ (x). If it is not contained in any of U1 , . . . , Uk , that would mean that the ball contains points from each of Fj which means that the distance of x from each Fj is strictly less that δ. ...
2-1
... Determine if each conjecture is true. If false, give a counterexample. 3. The quotient of two negative numbers is a positive number. true 4. Every prime number is odd. false; 2 false; 90° and 90° 5. Two supplementary angles are not congruent. 6. The square of an odd integer is odd. true Holt McDouga ...
... Determine if each conjecture is true. If false, give a counterexample. 3. The quotient of two negative numbers is a positive number. true 4. Every prime number is odd. false; 2 false; 90° and 90° 5. Two supplementary angles are not congruent. 6. The square of an odd integer is odd. true Holt McDouga ...
3-3 Proving Lines Parallel
... 6. Vert.∠s Thm. 7. Substitution 8. Conv. of Same-Side Interior ∠s Post. ...
... 6. Vert.∠s Thm. 7. Substitution 8. Conv. of Same-Side Interior ∠s Post. ...
