PPT for Section 2.4, 2.5, 2.6, 2.7
... Locate the digit just to the right of the place you want to round off to If that digit is less than 5, then omit it and all other remaining numbers to its. The digit to its left remain as it is. If that digit is greater than 5 or equal to 5, then omit it and all other remaining numbers to its ...
... Locate the digit just to the right of the place you want to round off to If that digit is less than 5, then omit it and all other remaining numbers to its. The digit to its left remain as it is. If that digit is greater than 5 or equal to 5, then omit it and all other remaining numbers to its ...
Unit 1: Extending the Number System
... properties of circles, including central and inscribed angles, chords of a circle, and tangents of a circle. Then you build on this to explore polygons circumscribed and inscribed in a circle. You will then learn about the properties and construction of tangent lines. The measurement units of radian ...
... properties of circles, including central and inscribed angles, chords of a circle, and tangents of a circle. Then you build on this to explore polygons circumscribed and inscribed in a circle. You will then learn about the properties and construction of tangent lines. The measurement units of radian ...
teaching complex numbers in high school
... Suppose that i ϵ R. Then we know that i is greater than zero, equal to zero, or less than zero. If we take i to be greater than zero, then i2 = i ∙ i > 0 since the product of two positive numbers is positive. That is, -1 > 0 which is false. Therefore, i cannot greater than 0. Similar contradictions ...
... Suppose that i ϵ R. Then we know that i is greater than zero, equal to zero, or less than zero. If we take i to be greater than zero, then i2 = i ∙ i > 0 since the product of two positive numbers is positive. That is, -1 > 0 which is false. Therefore, i cannot greater than 0. Similar contradictions ...
Multiplying Fractions
... denominator of the second fraction by the GCF of them. Cross them out and put the quotient of each of them in their place. Then do the same with the denominator of the first fraction with the numerator of the second fraction. Finally multiply the fractions with the new numerators and denominators to ...
... denominator of the second fraction by the GCF of them. Cross them out and put the quotient of each of them in their place. Then do the same with the denominator of the first fraction with the numerator of the second fraction. Finally multiply the fractions with the new numerators and denominators to ...
4 - Sets of Real Numbers
... m and n are integers, then their sum m + n is also an integer. · Closed under Additive Inverses: If m ∈ Z, then there exists another integer (namely, −m ∈ Z) such that m + (−m) = 0. · Closed under Multiplication: If m, n ∈ Z, then m · n ∈ Z. In other words, if m and n are integers, then their produc ...
... m and n are integers, then their sum m + n is also an integer. · Closed under Additive Inverses: If m ∈ Z, then there exists another integer (namely, −m ∈ Z) such that m + (−m) = 0. · Closed under Multiplication: If m, n ∈ Z, then m · n ∈ Z. In other words, if m and n are integers, then their produc ...
2012 Gauss Contests - CEMC
... 20. If either Chris or Mark says, “Tomorrow, I will lie.” on a day that he tells a lie, then it actually means that tomorrow he will tell the truth (since he is lying). This can only occur when he lies and then tells the truth on consecutive days. For Chris, this only happens on Sunday, since he lie ...
... 20. If either Chris or Mark says, “Tomorrow, I will lie.” on a day that he tells a lie, then it actually means that tomorrow he will tell the truth (since he is lying). This can only occur when he lies and then tells the truth on consecutive days. For Chris, this only happens on Sunday, since he lie ...
36 it follows that x4 − x2 + 2 ̸= 0. 11. Proof. Consider the number
... 2. Proof. Assume, to the contrary, that 100 can be expressed as the sum of three odd integers, say x, y and z. Then x = 2a + 1, y = 2b + 1 and z = 2c + 1 for integers a, b and c. Thus 100 = x + y + z = (2a + 1) + (2b + 1) + (2c + 1) = 2(a + b + c + 1) + 1. Since a + b + c + 1 is an integer, 100 is o ...
... 2. Proof. Assume, to the contrary, that 100 can be expressed as the sum of three odd integers, say x, y and z. Then x = 2a + 1, y = 2b + 1 and z = 2c + 1 for integers a, b and c. Thus 100 = x + y + z = (2a + 1) + (2b + 1) + (2c + 1) = 2(a + b + c + 1) + 1. Since a + b + c + 1 is an integer, 100 is o ...
Theorem 1. Every subset of a countable set is countable.
... We draw attention to a simple principle, which can be used to prove many of the usual and important theorems on countabilty of sets. We formulate it as the Countability Lemma. Suppose to each element of the set A there is assigned, by some definite rule, a unique natural number in such a manner that ...
... We draw attention to a simple principle, which can be used to prove many of the usual and important theorems on countabilty of sets. We formulate it as the Countability Lemma. Suppose to each element of the set A there is assigned, by some definite rule, a unique natural number in such a manner that ...
hexadecimal-to-decimal conversion
... results in a carry beyond 8-bits in the binary representation and a carry beyond two digits in the hexadecimal representation. When doing arithmetic using fixed length numbers these carrys are potentially lost. ...
... results in a carry beyond 8-bits in the binary representation and a carry beyond two digits in the hexadecimal representation. When doing arithmetic using fixed length numbers these carrys are potentially lost. ...
Computing with Floating Point Numbers
... Problem 2.1.15. How many SP numbers are in each binade? Sketch enough of the positive real number line so that you can place marks at the endpoints of each of the SP binades 3, 2, 1, 0, 1, 2 and 3. What does this sketch suggest about the spacing between consecutive positive SP numbers? Problem 2.1.1 ...
... Problem 2.1.15. How many SP numbers are in each binade? Sketch enough of the positive real number line so that you can place marks at the endpoints of each of the SP binades 3, 2, 1, 0, 1, 2 and 3. What does this sketch suggest about the spacing between consecutive positive SP numbers? Problem 2.1.1 ...
