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# The Complex Plane
#### Complex numbers are points in the plane
In the same way that we think of real numbers as being points on a line, it is natural to identify a complex number z=a+ib with the point (a,b) in the cartesian plane. Expressions such as ``the complex number z'', and ``the point z'' are now interchangeable.
We consider the a real number x to be the complex number x+ 0i and in this way we can think of the real numbers as a subset of the complex numbers. The reals are just the x-axis in the complex plane.
The modulus of the complex number z= a + ib now can be interpreted as the distance from z to the origin in the complex plane.
Since the hypotenuse of a right triangle is longer than the other sides, we have
for every complex number z.
We can also think of the point z= a+ ib as the vector (a,b). From this point of view, the addition of complex numbers is equivalent to vector addition in two dimensions and we can visualize it as laying arrows tail to end. (Picture)
We see in this way that the distance between two points z and w in the complex plane is |z-w|.
Exercise: Prove this last statement algebraically. (Proof.)
Exercise: Prove the ``Parallellogram law''
The ``Triangle'' inequality
is easily seen to hold.
(Proof.)
Exercise: Prove the Triangle inequality for n complex numbers
Here are some more exercises.
[Algebra] [Complex Variables]
[Geometry] [Trigonometry ]
[Calculus] [Differential Equations] [Matrix Algebra] | HuggingFaceTB/finemath | |
# How is area defined?
Thinking about area in the context of the Lebesgue measure, I have an intuitive understanding of how area is constructed in $\mathbb{R}^2$:
• define all rectangles to have the area $length \times width$,
• the area of a shape in $\mathbb{R}^2$ is the smallest area obtained by covering the shape with rectangles.
In practice, when calculating the area of Great Britain, say, I have following picture in mind.
This process can be made rigorous via measure theory. Indeed it is the unique measure, if rectangles have area $length \times width$ by Carathéodory’s extension theorem.
However this process is limited to area in the plane. How do I think about the surface area of a sphere for example? And can I show that it is consitent with the notion above?
Is there a similar process for finding the surface area of any shape? If not, then how is this done?
This happens to be a surprisingly difficult problem. In 1890 Karl Hermann Amandus Schwarz (1843-1921) published an example that showed the accepted definition of surface area gives an infinite area for a cylinder by showing there exists a sequence of inscribed polyhedra that converge uniformly to a cylinder such that the areas of the polyhedra diverge to infinity. See [1] for a nice discussion of Schwarz’s example, and see this note by V. Frederick Rickey for a very nice and detailed discussion of the history behind Schwarz’s example. Schwarz’s example resulted in many people developing various approaches to surface area, including Lebesgue, Fréchet, Carathéodory, Favard, Besicovitch, and others --- see [2] and [3] as well as the introductions and bibliographies of the remaining papers, which I’ve given in chronological order. These remaining papers were chosen mainly for being seminal papers in this field and to show the mathematical depth of the research that (one might reasonably argue) has developed from this problem of surface area.
[1] Frieda Zames (1932-2005), Surface area and the cylinder area paradox, Two-Year College Mathematics Journal [after 1983: College Mathematics Journal] 8 #4 (September 1977), 207-211. [another copy here]
[2] John William Theodore Youngs (1910-1970), Curves and surfaces, American Mathematical Monthly 51 #1 (January 1944), 1-11.
[3] Lamberto Cesari (1910-1990), Area and representation of surfaces, Bulletin of the American Mathematical Society 56 #3 (May 1950), 218-232.
[4] Ralph Lent Jeffery (1889-1975), Sets of $$k$$-extent in $$n$$-dimensional space, Transactions of the American Mathematical Society 35 #3 (July 1933), 629-647.
[5] Anthony Perry Morse (1911-1984) and John Adams Fitz Randolph (1904-1988), The $$\phi$$ rectifiable subsets of the plane, Transactions of the American Mathematical Society 55 #2 (March 1944), 236-305.
[6] Herbert Federer (1920-2010), The $$(\phi,k)$$ rectifiable subsets of $$n$$ space, Transactions of the American Mathematical Society 62 #2 (July 1947), 114-192.
[7] Herbert Federer (1920-2010), Measure and area, Bulletin of the American Mathematical Society 58 #3 (May 1952), 306-378.
[8] John Martin Marstrand (1928-____), The $$(\phi,s)$$ regular subsets of $$n$$-space, Transactions of the American Mathematical Society 113 #3 (December 1964), 369-392.
[9] David Preiss (1947-____), Geometry of measures in R$$^n$$: Distribution, rectifiability, and densities, Annals of Mathematics (2) 125 #3 (May 1987), 537-643. [freely available here]
Carathéodory [1] provided a method to construct a $p$-dimensional measure in $n$-space that generalizes the standard $p$-dimensional content.
[1] C. Carathéodory, "Über das lineare Mass von Punktmengen, eine Verallgemeinerung des Längenbegriffs" Nachr. Gesell. Wiss. Göttingen (1914) pp. 404–426.
{Plug} English translation in:
[2] Classics on Fractals, Edited by Gerald A. Edgar. Westview Press, 2004. ISBN: 0-8133-4153-1
Summary for surface area in $3$-space
Let $E \subseteq \mathbb R^3$ be a set. And we want the "area" of $E$. For a positive number $\epsilon$, let $$\mathcal H^2_\epsilon(E) = \inf \sum_{k=1}^\infty \big(\mathrm{diam}(C_k)\big)^2$$ where $\mathrm(C_k)$ is the diameter of the set $C_k$, and the infimum is taken over all countable covers of $E$ by sets with diameter at most $\epsilon$: $$E \subseteq \bigcup_{k=1}^\infty C_k,\qquad \mathrm{diam}(C_k) \le \epsilon$$ and then let $$\mathcal H^2(E) = \lim_{\epsilon \to 0} \mathcal H^2_\epsilon(E)$$ Adjusting by an appropriate constant factor (I think $\pi/4$) so that this agrees with the usual area for plane sets, we define $$\frac{\pi}{4}\;\mathcal H^2(E)$$ to be the "area" of $E$.
• Thanks for your answer; I look forward to reading your book. However, if the idea isn't too complicated, could you please summarise Carathéodory's argument? Nov 14 '17 at 13:04
Things are less obvious for curved surfaces. You can still consider a coverage by elementary shapes, but this raises a difficulty: even if you find tiling shapes (could be rectangle in a curvilinear coordinate system), you still need to know the areas of these shapes.
For a sphere, you can use the system of meridans and parallels, but it is a non-trivial matter to establish that the area of a tile is proportional to the sine of the colatitude. | HuggingFaceTB/finemath | |
# Matrix Determinant Calculator
Created by Luis Hoyos
Last updated: Feb 18, 2023
If you want to calculate matrix determinants, you're in the right place. This determinant solver calculates the determinant of 4x4, 3x3, and 2x2 matrices.
But what is the importance of determinants? Determinants have many applications, which we'll mention in the following section. For example, solving a 3x3 system of equations is the same as calculating the determinant of a 3x3 matrix. Keep reading to learn more about it!
## Why do we need to calculate matrix determinants?
These are some of the applications of determinants:
• For instance, we can describe systems of linear equations using matrices. The use of Cramer's rule is an example in which we use determinants to solve systems of linear equations.
• When using matrices to describe a linear transformation, it's often better to diagonalize them. We do that by calculating matrix determinants, of course.
• The determinant tells us whether the matrix has an inverse and whether we can approximate that inverse with the Moore-Penrose pseudoinverse.
• We usually need the eigenvalues of the previously mentioned transformation. To obtain them, we also need to calculate matrix determinants.
And why do we need matrices? Well, matrices describe many physical quantities, such as stress, strain, turbulence, or the Mohr's circle.
Well, determinants are important, that's clear. Now, let's see how to calculate them.
## Calculating the determinant of 4x4, 3x3, and 2x2 matrices
The following are the formulas to calculate matrix determinants.
### Determinant of a 2x2 matrix
If
$\scriptsize A = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}$
then the determinant of $A$ is
$\footnotesize |A| = a_1b_2 - a_2b_1$
### Determinant of a 3x3 matrix
If
$\scriptsize B = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$
then, to calculate the determinant of such a 3x3 matrix:
$\scriptsize |B| = a_1b_2c_3 + a_2b_3c_1 + a_3b_1c_2 \\\ \ \ \ \ \ \ \ \ - a_3b_2c_1 - a_1b_3c_2 - a_2b_1c_3$
### Determinant of a 4x4 matrix
Finally:
$\scriptsize C = \begin{bmatrix} a_1 & b_1 & c_1 & d_1 \\ a_2 & b_2 & c_2 & d_2 \\ a_3 & b_3 & c_3 & d_3 \\ a_4 & b_4 & c_4 & d_4 \end{bmatrix}$
then, to calculate the determinant of such a 4x4 matrix:
$\scriptsize |C| = a_1b_2c_3d_4 - a_2b_1c_3d_4 + a_3b_1c_2d_4 - \\\ a_1b_3c_2d_4 + a_2b_3c_1d_4 - a_3b_2c_1d_4 + a_3b_2c_4d_1 - \\\ a_2b_3c_4d_1 + a_4b_3c_2d_1 - a_3b_4c_2d_1 + a_2b_4c_3d_1 - \\\ a_4b_2c_3d_1 + a_4b_1c_3d_2 - a_1b_4c_3d_2 + a_3b_4c_1d_2 - \\\ a_4b_3c_1d_2 + a_1b_3c_4d_2 - a_3b_1c_4d_2 + a_2b_1c_4d_3 - \\\ a_1b_2c_4d_3 + a_4b_2c_1d_3 - a_2b_4c_1d_3 + a_1b_4c_2d_3 - \\\ a_4b_1c_2d_3$
As you can see, finding the determinant of a 3x3 and a 2x2 matrix is relatively easy, but calculating the determinant of a 4x4 matrix is an uphill task. The best option is, undoubtedly, using our determinant solver.
After that, you should visit our other math tools! The vector addition calculator is convenient if you need to deal with vectors.
Luis Hoyos
Matrix size
2x2
A=
⌈ a₁ b₁ ⌉ ⌊ a₂ b₂ ⌋
First row
a₁
b₁
Second row
a₂
b₂
Result
Determinant |A|
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## Friday, May 30, 2014
### Excel Variance Tests: Levene’s, Brown-Forsythe, and F Test For 2-Sample Pooled t-Test in Excel 2010 and Excel 2013
This is one of the following eight articles on 2-Independent-Sample Pooled t-Tests in Excel
2-Independent-Sample Pooled t-Test in 4 Steps in Excel 2010 and Excel 2013
Excel Variance Tests: Levene’s, Brown-Forsythe, and F Test For 2-Sample Pooled t-Test in Excel 2010 and Excel 2013
Excel Normality Tests Kolmogorov-Smirnov, Anderson-Darling, and Shapiro Wilk Tests For Two-Sample Pooled t-Test
Two-Independent-Sample Pooled t-Test - All Excel Calculations
2-Sample Pooled t-Test – Effect Size in Excel 2010 and Excel 2013
2-Sample Pooled t-Test Power With G*Power Utility
Mann-Whitney U Test in 12 Steps in Excel as 2-Sample Pooled t-Test Nonparametric Alternative in Excel 2010 and Excel 2013
2- Sample Pooled t-Test = Single-Factor ANOVA With 2 Sample Groups
# Excel Variance Tests: Levene’s, Brown- Forsythe, and F Test For 2-Sample Pooled t-Test
Two-independent-sample t-Test are performed as either as a Pooled test or Unpooled tests. Pooled t-Tests are performed if the variances of both sample groups are similar. A rule-of-thumb is as follows: A Pooled t-Test should be performed if the standard deviation of one sample is no more than twice as large as the standard deviation in the other sample. That is the case here as the following are true:
s1 = sample1 standard deviation = 16.92
and
s2 = sample2 standard deviation = 15.28
## F Test For Sample Group Variance Comparison in Excel
An Excel F Test performed on the two sample groups produces the following output:
(Click On Image To See a Larger Version)
The Null Hypothesis of an F Test states that the variances of the two groups are the same. The p Value shown in the Excel F Test output equals 0.345. This is much larger than the Alpha (0.05) that is typically used for an F Test so the Null Hypothesis cannot be rejected.
We therefore conclude as a result of the F Test that the variances are the same. The F Test is sensitive to non-normality of data. The sample variances can also be compared using the nonparametric Levene’s Test and also the nonparametric Brown-Forsythe Test.
## Levene’s Test For Sample Group Variance Comparison in Excel
Levene’s Test is a hypothesis test commonly used to test for the equality of variances of two or more sample groups. Levene’s Test is more robust against non-normality of data than the F Test.
The Null Hypothesis of Levene’s Test is average distance to the sample mean is the same for each sample group. Acceptance of this Null Hypothesis implies that the variances of the sampled groups are the same. The distance to the mean for each data point of both samples is shown as follows:
(Click On Image To See a Larger Version)
Levene’s Test involves performing Single-Factor ANOVA on the groups of distances to the mean. This can be easily implemented in Excel by applying the Excel data analysis tool ANOVA: Single Factor. Applying this tool on the above data produces the following output:
(Click On Image To See a Larger Version)
The Null Hypothesis of Levene’s Test states that the average distance to the mean for the two groups are the same. Acceptance of this Null Hypothesis would imply that the sample groups have the same variances. The p Value shown in the Excel ANOVA output equals 0.6472. This is much larger than the Alpha (0.05) that is typically used for an ANOVA Test so the Null Hypothesis cannot be rejected.
We therefore conclude as a result of Levene’s Test that the variances are the same or, at least, that we don’t have enough evidence to state that the variances are different. Levene’s Test is sensitive to outliers because relies on the sample mean, which can be unduly affected by outliers. A very similar nonparametric test called the Brown-Forsythe Test relies on sample medians and is therefore much less affected by outliers as Levene’s Test is or by non-normality as the F Test is.
## Brown-Forsythe Test For Sample Group Variance Comparison in Excel
The Brown-Forsythe Test is a hypothesis test commonly used to test for the equality of variances of two or more sample groups. The Null Hypothesis of the Brown-Forsythe Test is average distance to the sample median is the same for each sample group. Acceptance of this Null Hypothesis implies that the variances of the sampled groups are the same. The distance to the median for each data point of both samples is shown as follows:
(Click On Image To See a Larger Version)
The Brown-Forsythe Test involves performing Single-Factor ANOVA on the groups of distances to the median. This can be easily implemented in Excel by applying the Excel data analysis tool ANOVA: Single Factor. Applying this tool on the above data produces the following output:
(Click On Image To See a Larger Version)
The Null Hypothesis of the Brown-Forsythe Test states that the average distance to the median for the two groups are the same. Acceptance of this Null Hypothesis would imply that the sample groups have the same variances. The p Value shown in the Excel ANOVA output equals 0.6627. This is much larger than the Alpha (0.05) that is typically used for an ANOVA Test so the Null Hypothesis cannot be rejected.
We therefore conclude as a result of the Brown-Forsythe Test that the variances are the same or, at least, that we don’t have enough evidence to state that the variances are different.
Each of the above tests can be considered relatively equivalent to the others. The variances of both sample groups are verified to be similar enough to permit using a Pooled test for this two-independent sample hypothesis test.
Excel Master Series Blog Directory
Statistical Topics and Articles In Each Topic
• Histograms in Excel
• Bar Chart in Excel
• Combinations & Permutations in Excel
• Normal Distribution in Excel
• t-Distribution in Excel
• Binomial Distribution in Excel
• z-Tests in Excel
• t-Tests in Excel
• Hypothesis Tests of Proportion in Excel
• Chi-Square Independence Tests in Excel
• Chi-Square Goodness-Of-Fit Tests in Excel
• F Tests in Excel
• Correlation in Excel
• Pearson Correlation in Excel
• Spearman Correlation in Excel
• Confidence Intervals in Excel
• Simple Linear Regression in Excel
• Multiple Linear Regression in Excel
• Logistic Regression in Excel
• Single-Factor ANOVA in Excel
• Two-Factor ANOVA With Replication in Excel
• Two-Factor ANOVA Without Replication in Excel
• Randomized Block Design ANOVA in Excel
• Repeated-Measures ANOVA in Excel
• ANCOVA in Excel
• Normality Testing in Excel
• Nonparametric Testing in Excel
• Post Hoc Testing in Excel
• Creating Interactive Graphs of Statistical Distributions in Excel
• Solving Problems With Other Distributions in Excel
• Optimization With Excel Solver
• Chi-Square Population Variance Test in Excel
• Analyzing Data With Pivot Tables and Pivot Charts
• SEO Functions in Excel
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• VLOOKUP
• Simplifying Useful Excel Functions | HuggingFaceTB/finemath | |
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• # Alg2: Lesson 6.3: Linear Systems in 3 Variables Again
Date Due: 11/10/2020 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Go over Test on Unit #2
Date Due: 11/10/2020 Category: Algebra A-1 Grade 9
• # No School
Date Due: 11/11/2020
• # A.1: Slope
Date Due: 11/12/2020 Category: Algebra A-1 Grade 9
• # Alg2: Quiz & Lesson 6.4: Transformations of Functions
Date Due: 11/12/2020 Category: R Common Core Algebra 2 = Grade 11
• # Alg2: Lesson 6.5: Reflections
Date Due: 11/13/2020 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Slope Again
Date Due: 11/13/2020 Category: Algebra A-1 Grade 9
• # A.1: Slope Intercept and Point Slope
Date Due: 11/16/2020 Category: Algebra A-1 Grade 9
• # Alg2: Lesson 6.6: Dilations
Date Due: 11/16/2020 Category: R Common Core Algebra 2 = Grade 11
• # Alg2: Classwork 6.1 - 6.6
Date Due: 11/17/2020 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Classwork on Slope
Date Due: 11/17/2020 Category: Algebra A-1 Grade 9
• # A.1: Quiz on Slope
Date Due: 11/18/2020 Category: Algebra A-1 Grade 9
• # Alg2: Quiz & Lesson 6.7: Definitions of a Parabola and Graphing Parabolas
Date Due: 11/18/2020 Category: R Common Core Algebra 2 = Grade 11
• # Alg2: Lesson 6.8: Equations of a Parabola & Congruent Parabolas
Date Due: 11/20/2020 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Notes & Assignments
Date Due: 11/20/2020 Category: Algebra A-1 Grade 9
• # A.1: X & Y intercepts
Date Due: 11/23/2020 Category: Algebra A-1 Grade 9
• # Alg2: Review #6
Date Due: 11/23/2020 Category: R Common Core Algebra 2 = Grade 11
• # Alg2: Test Unit #6
Date Due: 11/24/2020 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Classwork p14
Date Due: 11/24/2020 Category: Algebra A-1 Grade 9
• # Alg2: Lesson 7.1: Solving Quadratic Equations by Factoring, Square Root Method, and Graphing
Date Due: 11/30/2020 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Classwork p15
Date Due: 11/30/2020 Category: Algebra A-1 Grade 9
• # Alg2: Lesson 7.2: Solving Quadratic Equations by Completing the Square
Date Due: 12/01/2020 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Graphing Inequalities
Date Due: 12/01/2020 Category: Algebra A-1 Grade 9
• # Alg2: Lesson 7.3: Solving Quadratic Equations by the Quadratic Formula
Date Due: 12/02/2020 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Classwork graphing inequality
Date Due: 12/02/2020 Category: Algebra A-1 Grade 9
• # Alg2: Classwork 7.1 - 7.3
Date Due: 12/03/2020 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Quiz on all graphing
Date Due: 12/03/2020 Category: Algebra A-1 Grade 9
• # A.1: Review For Unit 3
Date Due: 12/04/2020 Category: Algebra A-1 Grade 9
• # Alg 2: Quiz on 7.1 - 7.3
Date Due: 12/04/2020 Category: R Common Core Algebra 2 = Grade 11
• # Lesson 7.4: Solving Quadratic Systems of Equations Graphically
Date Due: 12/07/2020 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Test #3
Date Due: 12/07/2020 Category: Algebra A-1 Grade 9
• # A.1: Substitution
Date Due: 12/08/2020 Category: Algebra A-1 Grade 9
• # Alg2: Lesson 7.5: Solving Quadratic Systems of Equations Algebraically
Date Due: 12/08/2020 Category: R Common Core Algebra 2 = Grade 11
• # Alg2: Review #7
Date Due: 12/09/2020 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Classwork on Substitution
Date Due: 12/09/2020 Category: Algebra A-1 Grade 9
• # A.1: Quiz on Substitution
Date Due: 12/10/2020 Category: Algebra A-1 Grade 9
• # Alg2: Unit #7 Test
Date Due: 12/10/2020 Category: R Common Core Algebra 2 = Grade 11
• # Alg2: Lesson 8.1: Key Features of Polynomial Graphs
Date Due: 12/11/2020 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Elimination Using Adding and Subtracting
Date Due: 12/11/2020 Category: Algebra A-1 Grade 9
• # A-1: Classwork on Elimination + & -
Date Due: 12/14/2020 Category: Algebra A-1 Grade 9
• # Alg2: Lesson 8.2: Sketching Graphs of Polynomials in Factored Form
Date Due: 12/14/2020 Category: R Common Core Algebra 2 = Grade 11
• # Alg2: Lesson 8.3: Odd and Even Functions
Date Due: 12/15/2020 Category: R Common Core Algebra 2 = Grade 11
• # A-1: Elimination with Multiplication
Date Due: 12/15/2020 Category: Algebra A-1 Grade 9
• # Alg2: Classwork 8.1 - 8.3
Date Due: 12/16/2020 Category: R Common Core Algebra 2 = Grade 11
• # A-1: Elimination Day 2
Date Due: 12/16/2020 Category: Algebra A-1 Grade 9
• # A-1: classwork on Substitution and Elimination
Date Due: 12/17/2020 Category: Algebra A-1 Grade 9
• # Alg 2: Quiz 8.1 - 8.3
Date Due: 12/17/2020 Category: R Common Core Algebra 2 = Grade 11
• # A-1: "Tuiz" on Substitution and Elimination
Date Due: 12/18/2020 Category: Algebra A-1 Grade 9
• # Alg2: Lesson 8.4: Solving Higher Order Polynomial Equations
Date Due: 12/18/2020 Category: R Common Core Algebra 2 = Grade 11
• # Alg2: Lesson 8.5: Synthetic Division
Date Due: 12/21/2020 Category: R Common Core Algebra 2 = Grade 11
• # alg a-1: go over "tuiz" on substituion and elimination
Date Due: 12/21/2020 Category: Algebra A-1 Grade 9
• # alg a-1: Optional Bonus Review Sheet
Date Due: 12/22/2020 Category: Algebra A-1 Grade 9
• # Alg 2: worksheet on Synthetic Division
Date Due: 12/22/2020 Category: R Common Core Algebra 2 = Grade 11
• # Alg2: Lesson 8.6: Factor & Remainder Theorem
Date Due: 01/04/2021 Category: R Common Core Algebra 2 = Grade 11
• # A-1: Graphing System of Equations
Date Due: 01/04/2021 Category: Algebra A-1 Grade 9
• # A-1: Classwork on Graphs
Date Due: 01/05/2021 Category: Algebra A-1 Grade 9
• # Alg2: Lesson 8.7: Solving Polynomial Equations Given a Root
Date Due: 01/05/2021 Category: R Common Core Algebra 2 = Grade 11
• # Alg2: Lesson 8.8: Writing a Polynomial Equations Given Roots
Date Due: 01/06/2021 Category: R Common Core Algebra 2 = Grade 11
• # A-1: Quiz on Linear Systems
Date Due: 01/06/2021 Category: Algebra A-1 Grade 9
• # A.1: Graphing Systems of Inequalities
Date Due: 01/07/2021 Category: Algebra A-1 Grade 9
• # Alg2: End of Ch #8 Classwork
Date Due: 01/07/2021 Category: R Common Core Algebra 2 = Grade 11
• # Alg2: Quiz & Lesson 8.9: Write Equations to Model Polynomials & Lesson 8.10 Modeling Polynomials
Date Due: 01/08/2021 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Classwork on Graphing Systems of Inequalities
Date Due: 01/08/2021 Category: Algebra A-1 Grade 9
• # A-1: Quiz on Graphing Inequalities .. added
Date Due: 01/11/2021 Category: Algebra A-1 Grade 9
• # Alg2: Review Unit #8
Date Due: 01/11/2021 Category: R Common Core Algebra 2 = Grade 11
• # Alg2: Test Unit #8
Date Due: 01/12/2021 Category: R Common Core Algebra 2 = Grade 11
• # A-1: Word Problems for Systems of Equations
Date Due: 01/12/2021 Category: Algebra A-1 Grade 9
• # A-1: Day2: More Word Problems for Systems of Equations
Date Due: 01/13/2021 Category: Algebra A-1 Grade 9
• # Alg2: Lesson 9.1: Simplifying, x, and dividing Rational Expressions
Date Due: 01/13/2021 Category: R Common Core Algebra 2 = Grade 11
• # Alg2: Lesson 9.2: + and - Rational Expessions
Date Due: 01/14/2021 Category: R Common Core Algebra 2 = Grade 11
• # A-1: 4-Step Problem Solving Process
Date Due: 01/14/2021 Category: Algebra A-1 Grade 9
• # Alg2: Lesson 9.3: Solving Rational Equations
Date Due: 01/15/2021 Category: R Common Core Algebra 2 = Grade 11
• # A.1: Classwork on Word Problems with graphs
Date Due: 01/15/2021 Category: Algebra A-1 Grade 9 | HuggingFaceTB/finemath | |
# 3-DNF proves the algorithm is in P class
After, reading the link we will take a look at how we recover our solutions to a constrained Sudoku Puzzle.
If we assume that a sudoku puzzle was generated with this procedure we can now create a "semi"-solver. I say "semi" because we need the $$3 \times 3$$ grid $$M_{2,2}$$ already solved for us. Let's assume we have this. As an example I will assume we are provided:
$$\begin{bmatrix} 5 & 9 & 6\\ 1 & 2 & 4\\ 3 & 7 & 8 \end{bmatrix}$$
Now we will flatten it into: $$[5,9,6,1,2,4,3,7,8]$$ and permute as follows:
[8, 5, 9, 6, 1, 2, 4, 3, 7]-----list 1
[7, 8, 5, 9, 6, 1, 2, 4, 3]-----list 2
[3, 7, 8, 5, 9, 6, 1, 2, 4]-----list 3
[4, 3, 7, 8, 5, 9, 6, 1, 2]-----list 4
[2, 4, 3, 7, 8, 5, 9, 6, 1]-----list 5
[1, 2, 4, 3, 7, 8, 5, 9, 6]-----list 6
[6, 1, 2, 4, 3, 7, 8, 5, 9]-----list 7
[9, 6, 1, 2, 4, 3, 7, 8, 5]-----list 8
[5, 9, 6, 1, 2, 4, 3, 7, 8]-----list 9
Now for each list, we will turn them into a $$3 \times 3$$ grid using the same mapping in step 2 above. For example list 1 would get mapped to
$$\begin{bmatrix} 8 & 5 & 9 \\ 6 & 1 & 2 \\ 4 & 3 & 7 \end{bmatrix}$$
Now we position these in the game board the same way we did as step 3 above. For example our layout would be as follows:
**list1** **list4** **list7**
**list2** **list5** **list8**
**list3** **list6** **list9**
In the prior example this would give us the correct solution:
$$M = \begin{bmatrix} 8 & 5 & 9 & 4 & 3 & 7 & 6 & 1 & 2\\ 6 & 1 & 2 & 8 & 5 & 9 & 4 & 3 & 7\\ 4 & 3 & 7 & 6 & 1 & 2 & 8 & 5 & 9\\ 7 & 8 & 5 & 2 & 4 & 3 & 9 & 6 & 1\\ 9 & 6 & 1 & 7 & 8 & 5 & 2 & 4 & 3\\ 2 & 4 & 3 & 9 & 6 & 1 & 7 & 8 & 5\\ 3 & 7 & 8 & 1 & 2 & 4 & 5 & 9 & 6\\ 5 & 9 & 6 & 3 & 7 & 8 & 1 & 2 & 4\\ 1 & 2 & 4 & 5 & 9 & 6 & 3 & 7 & 8\\ \end{bmatrix}$$
Then we have list 9 (our input) will always give you correct solution in quadratic time.
For further illustration, I intend to prove that the algorithm aforementioned is in P class in two ways.
Here, we'll take a look at 3-DNF.
(L1 ∧ L2 ∧ L3) | (L4 ∧ L5 ∧ L6) | (L7 ∧ L8 ∧ L9)
Let L1=list1, L2 = list2,...
**list1** **list4** **list7**
**list2** **list5** **list8**
**list3** **list6** **list9**
Therefore, the algorithm generates grids and recovers correct solutions easily.
Now, lets say I want to check the satsifiability of the algorithm's circular shifts. Here, I generate 3 more grids to show that there is a 3x3 positive 3-satisfying permutes.
l = [8, 5, 9, 6, 1, 2, 4, 3, 7]
[5, 9, 6, 1, 2, 4, 3, 7, 8]-l1
[9, 6, 1, 2, 4, 3, 7, 8, 5]-l2
[6, 1, 2, 4, 3, 7, 8, 5, 9]-l3
[1, 2, 4, 3, 7, 8, 5, 9, 6]-l4
[2, 4, 3, 7, 8, 5, 9, 6, 1]-l5
[4, 3, 7, 8, 5, 9, 6, 1, 2]-l6
[3, 7, 8, 5, 9, 6, 1, 2, 4]-l7
[7, 8, 5, 9, 6, 1, 2, 4, 3]-l8
[8, 5, 9, 6, 1, 2, 4, 3, 7]-l9
x = [5, 9, 6, 1, 2, 4, 3, 7, 8]
[9, 6, 1, 2, 4, 3, 7, 8, 5]-x1
[6, 1, 2, 4, 3, 7, 8, 5, 9]-x2
[1, 2, 4, 3, 7, 8, 5, 9, 6]-x3
[2, 4, 3, 7, 8, 5, 9, 6, 1]-x4
[4, 3, 7, 8, 5, 9, 6, 1, 2]-x5
[3, 7, 8, 5, 9, 6, 1, 2, 4]-x6
[7, 8, 5, 9, 6, 1, 2, 4, 3]-x7
[8, 5, 9, 6, 1, 2, 4, 3, 7]-x8
[5, 9, 6, 1, 2, 4, 3, 7, 8]-x9
y = [9, 6, 1, 2, 4, 3, 7, 8, 5]
[6, 1, 2, 4, 3, 7, 8, 5, 9]-y1
[1, 2, 4, 3, 7, 8, 5, 9, 6]-y2
[2, 4, 3, 7, 8, 5, 9, 6, 1]-y3
[4, 3, 7, 8, 5, 9, 6, 1, 2]-y4
[3, 7, 8, 5, 9, 6, 1, 2, 4]-y5
[7, 8, 5, 9, 6, 1, 2, 4, 3]-y6
[8, 5, 9, 6, 1, 2, 4, 3, 7]-y7
[5, 9, 6, 1, 2, 4, 3, 7, 8]-y8
[9, 6, 1, 2, 4, 3, 7, 8, 5]-y9
Here, I demonstrate that the 3x3 shift meets satisfiability for 9! Sudoku grids generated by the algorithm. At the end of the question I prove that the expression is always meets satisfiability when given the correct inputs.
(l1 ∨ x9 ∨ y8) ∧ (l2 ∨ x1 ∨ y9)
l1 = [5, 9, 6, 1, 2, 4, 3, 7, 8]
x9 = [5, 9, 6, 1, 2, 4, 3, 7, 8]
y8 = [5, 9, 6, 1, 2, 4, 3, 7, 8]
All the listed elements above have their defined variables within these expressions. All the expressions hold true.
(𝑙1∨𝑥9∨𝑦8)∧(𝑙2∨𝑥1∨𝑦9)∧(𝑙3∨𝑥2∨𝑦1)∧(𝑙4∨𝑥3∨𝑦2)∧(𝑙5∨𝑥4∨𝑦3)∧(𝑙6∨𝑥5∨𝑦4)∧(𝑙7∨𝑥6∨𝑦5)∧(𝑙8∨𝑥7∨𝑦6)∧(𝑙9∨𝑥8∨𝑦7)∧(𝑙1∨𝑥9∨𝑦8)∧(𝑙2∨𝑥1∨𝑦9)
Here is a chart showing the 3-satsifiability of the algorithm. Proving that the 3x3 shift overlaps all 9! valid grids that the algorithm can generate
Overall, are these proofs correct that constrained Sudoku is in P class? | open-web-math/open-web-math | |
## Convert amps (A) and hours (h) to mAh (milliampere-hours)
Insert the current in Amps (A) and the time in hours that the current was flowing for, and then click on Calculate to obtain milliampere-hours (mAh):
Current: A
Duration: Hours
Result: mAh
The formula is (A)*(h)*1000 =(mAh). For example, if you have a current of 3 A running for 5 hours, the number of mAh is 3A*5h*1000 = 15000mAh.
## Convert mAh to A (milliampere-hours to Amps)
To convert mAh to A you need to know the duration in hours that the current was flowing for. Insert the mAh below, then insert the time duration in hours and then click on Calculate to obtain Amps:
mAh: mAh
Duration: Hours
Result: A
The formula is (A)=(mAh)/((h)*1000). For example, if you have 15000mAh running for 5 hours, the current in Amps is 15000mAh/(5h*1000) = 3A.
Got feedback? email: suggestionsfeed@gmail.com | HuggingFaceTB/finemath | |
# How do you find the domain and range of f(x)=1/(2x-4)?
Apr 16, 2018
The domain is $\setminus m a t h \boldsymbol{R} \textrm{\setminus} \left\{2\right\}$ and the range is $\setminus m a t h {\boldsymbol{R}}^{\textrm{\cdot}}$.
#### Explanation:
We have:
$f \left(x\right) = \frac{1}{2 x - 4}$
The function is defined for all reals except for $2 x = 4$ ($x = 2$) because you can't divide by $0$. So the domain is $\setminus m a t h \boldsymbol{R} \textrm{\setminus} \left\{2\right\}$.
$f \left(x\right)$ can take any real value except for $0$, as a fraction is equal to zero only if the numerator is also equal to $0$. Thus, the range is $\setminus m a t h \boldsymbol{R} \textrm{\setminus} \left\{0\right\} = \setminus m a t h {\boldsymbol{R}}^{\textrm{\cdot}}$. | HuggingFaceTB/finemath | |
Thursday, 26 January 2012
Playing games during an outreach event...
On the 25th of January Professor Paul Harper and I took part in the Monmouth Science Initiative by delivering an outreach event for a group of 13 16~17 year olds. Paul gave a very neat introduction to epidemiology using a role playing game (6d's included!).
I tried to introduce the kids to game theory using these slides. We only got as far as the prisoner's dilemma tournament and here's an account of the day.
First of all we started with a "Guess 2/3 of the average" game. We played twice, once with no prior reasoning (just a clear explanation of the rules). Here's a chart of the guesses:
Overall the distribution of guesses is pretty uniform (as is to be expected, right now the students are basically guessing). The average guess was 26 and 2/3rds of the average was 17. Two students happened to win this first round (the two students who actually guessed 17).
Immediately after playing this round I gave an explanation of the fact that all strategies (guesses) other than 0 are dominated (by iteratively eliminating all guesses that are greater than 2/3 of the maximum guess):
Once this had been explained we get the students to play again, here are the results:
We no longer have a uniform distribution of the guesses and the students are indeed getting it (mostly...). The average guess was now 4 which gives a 2/3 average of 2: so we had a single winner (the student who guessed 2). This was good fun and led to a short discussion of how people are not always rational (as shown by the guess of 14 and 25 in particular).
After this, we went on to look at some videos depicting Prisoner's Dilemmas (these videos are both on the slides link to above):
Going on to explain the logic behind an actual prisoner's dilemma, we proceed to have an Iterated Prisoner's Dilemma tournament:
• Round robin with 4 teams
• Every "duel" was 8 rounds of the Prisoner's Dilemma
• The team with the total lowest score ("years in prison") won the tournament
The first duel (between team "C" and team "D") wasn't that great as the two teams mainly wanted to Defect:
Just at the end of this duel however there was a glimmer of hope! The teams started talking to each other (throughout we were trying to encourage them to talk) and indeed "promised" to both cooperate and of course both defected...
Here's a photo of the scores of this duel (you can see that at the 4th round team "D" tried to cooperate):
The next few duels (which included team "X" and team "Korea") got a bit more interesting and the teams starting talking to each other, here's the scores:
This next video however is great, both these teams had had a long chat before this duel. As you see team "Korea" defecting before team "D" were expecting them too (at 1:00). and putting themselves in pole position to win the tournament (they did the same trick in the next duel against team "C"):
Going in to the last duel the scores were:
• team "Korea" had 72 points
• team "C" had 82 points
• team "X" had 53 points (with one more duel to play)
• team "D" had 58 points (with one more duel to play)
BUT this is where things got interesting...
Up until here I think the ideas of dominated strategies and equilibria were fairly clear, and I think Paul and I were happy with that. The students however were not and it got slightly personal (all in good humour of course) and the guys in team "X" did not want the leader of team "Korea" to win, so they formed a coalition (a very particular and complex idea in game theory) with team "D" to ensure that team "Korea" did not win. Here's the video of this (at 0:16 the leader of team "Korea" realises what's going on and shouts "I'll buy you a kitkat if one of you starts playing 'D'"):
Here's the final 3 duels of the tournament (including the great play of team "Korea" and the coalition formed at the end between team "D" and team "X"):
The final scores where:
• team "Korea" had 72 points
• team "C" had 82 points
• team "X" had 83 points
• team "D" had 68 points
All in all Paul and I had great fun and the feedback from the kids was great. I think this is a great way of getting some fundamental ideas of game theory across. To shorten the amount of time it takes, we might end up using duels of 4 or 5 rounds instead of 8 but overall we won't change much, in particular having 4 teams works great as it allows time for the non playing teams to try and communicate. | HuggingFaceTB/finemath | |
# Problem in ideal gases
## Homework Statement
What volume of air (T=25C, P=1atm) is required
for complete combustion of one litter of gasoline?
The partial pressure of oxygen in the air is 0.205 atm.
One litter of gasoline contains 6.15 moles octane C8H18.
## Homework Equations
The unbalanced reaction is:
C8H18 + O2 --> CO2 + H2O
The ideal gas law:
PV = nRT
## The Attempt at a Solution
The balanced reaction is:
C8H18 + (25/2)O2 --> 8CO2 + 9H2O
So 6.15 moles octane requires 6.15*12.5 = 76.875 moles O2.
With the ideal gas law I get:
PV = nRT
(0.205)V=(76.875)(0.082)(298)
V = 9,163.5 L
My friend beleives it is,
but I think that this value is just the volume of oxygen needed,
and the volume of air needed is V = (9,163.5/0.2) = 45,817.5 L
Borek
Mentor | HuggingFaceTB/finemath | |
# PSAT Math: Systems of Equations
Linear Equations on the PSAT are well suited for modeling a variety of scenarios and for solving for a single variable in terms of another that is clearly defined (e.g., what is the cost of a data plan if you consume 4 GB of data in a month). However, sometimes you will be given a set of multiple equations with multiple variables that are interdependent. For example, suppose a $50/month cell phone plan includes$0.05 text messages and $0.40 voice calls, with a cap of 1,000 combined text messages and voice calls. This scenario can be represented by the following system of equations:$0.05t + $0.40v =$50
t + v = 1000
Solving such a system would enable you to determine the maximum number of text messages and voice calls you could make under this plan, while optimizing total usage. To solve systems of equations, you’ll need to rely on a different set of tools that builds on the algebra you’re already familiar with. The following question shows an example of such a system in the context of a test-like question.
You might be tempted to switch on math autopilot at this point and employ substitution, solving the second equation for s in terms of r:
s = 12 – r
You could plug the resulting expression back into the other equation and eventually solve for r, but remember, the PSAT tests your ability to solve math problems in the most efficient way.
The following table contains some strategic thinking designed to help you find the most efficient way to solve this problem on Test Day, along with some suggested scratchwork.
Strategic Thinking Math Scratchwork Step 1: Read the question, identifying and organizing important information as you goIn this case, you’re looking for the value of r. There are two equations that involve r and s. 3r + 2s = 24 r + s = 12 Step 2: Choose the best strategy to answer the questionIs there any way you can make the first equation look like the second one? Does the quantity r + s exist in the first equation in some form?How can you effectively use both equations?Once you’ve written the first equation in terms of r + s, substitute the value of r + s (which is 12) into the second equation and solve for r. 3r + 2s = 24 r + 2r + 2s = 24 r + 2(r + s) = 24 r + 2(12) = 24 r = 0 Step 3: Check that you answered the right questionBe careful! The question isn’t asking for the value of r. Add 6 to your result and you should see that (C) is the correct answer. r + 6 = 0 + 6 r + 6 = 6
### Independent versus Dependent Equations
Generally, when you have a system involving n variables, you need n independent equations to solve for those variables. Thus, if you have a system of two variables, you need two independent equations. Three variables would require three independent equations, and so on.
Systems of equations are extremely useful in modeling and simulation. Complex mathematical problems such as weather forecasting or crowd control predictions often require 10 or more equations to be simultaneously solved for multiple variables. Fortunately, you won’t encounter anything this daunting on Test Day.
Before we outline the process for solving two-variable systems of equations, let’s clarify one of the key requirements. Earlier, it was stated that you need two independent equations to solve for two variables, but what exactly is an independent equation? Consider the equation 4x + 2y = 8. You could use properties of equality to transform this equation in a number of different ways. For example, you could multiply both sides by 2, resulting in the equation 8x + 4y = 16.
While it seems as though we’ve just created an additional equation, this is misleading, as the second equation has the same core variables and relationships as the first equation. This is termed a dependent equation, and two dependent equations cannot be used to solve for two variables. Look what happens when we try to use substitution. Start by isolating y in the original equation; the result is y = 4 – 2x.
Substituting that into the second equation, notice what happens:
8x + 4(4 − 2x ) = 16
8x + 16 − 8x = 16
16 = 16
Although 16 does in fact equal 16, this doesn’t bring us any closer to solving for either of the variables. In fact, if you arrive at a result like this when solving a system of equations, then the two equations are dependent. In this case, the system has infinitely many solutions because you could choose any number of possible values for x and y.
Note:
When two equations are dependent, one equation can be obtained by algebraically manipulating the other equation. Graphically, dependent equations both describe the same line in the coordinate plane and therefore have the same slope and the same y-intercept.
At other times, you’ll encounter equations that are fundamentally incompatible with each other. For example, if you have the two equations 4x + 2y = 8 and 4x + 2y = 9, it should be obvious that there are no values for x and y that will satisfy both equations at the same time. Doing so would violate fundamental laws of math. In this case, you would have a system of equations that has no solution. These two equations define parallel lines, which by definition never intersect.
Knowing how many solutions a system of equations has will tell you how graphing them in the same coordinate plane should look. Remember, the solution of a system of equations consists of the point or points where their graphs intersect.
If your system has … …then it will graph as: Reasoning no solution two parallel lines Parallel lines never intersect. one solution two lines intersecting at a single point Two straight lines have only one intersection. infinitely many solutions a single line (one line directly on top of the other) One equation is a manipulation of the other—their graphs are the same line.
Because you could encounter any of these three situations on Test Day, make sure you are familiar with all of them.
### PSAT Math Practice Question: System of Equations
Let’s examine a sample problem to investigate the requirements for solving a system of equations:
(1/8)q + (1/5)s = 40
zq + 8s = 1,600
In the system of linear equations shown, z represents a constant. If the system of equations has infinitely many solutions, what is the value of z?
A. 1/8
B. 5
C. 8
D. 40
Work through the Kaplan Method for Math step-by-step to solve this question. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.
[raw]
Strategic Thinking Math Scratchwork
Step 1: Read the question, identifying and organizing important information as you go
You are looking for the value of z, given that the equation has infinitely many solutions.
This means that the second equation should be the same as the first equation after some kind of algebraic manipulation.
Step 2: Choose the best strategy to answer the question
Look for ways to make the first equation resemble the second.
Are there any clues?
The constant on the right gives you a strong clue: Multiplying 40 by 40 gives 1,600. This also works well with the (1/5)s term because multiplying by 40 yields 8s.
(1/8)q + (1/5)s = 40
zq + 8s = 1,600
40(1/8)q + (1/5)s = 40(40)
5q + 8s = 1,600
Step 3: Check that you answered the right question
Be careful! Choice D is a trap. The question isn’t looking for the number you’d multiply the first equation by. Instead, it’s looking for z, the coefficient of q.
Because the number in front of q in your transformed equation is 5, you know that z must also be equal to 5, making (B) the correct answer.
5q = zq
z = 5
[/raw]
Tags: | HuggingFaceTB/finemath | |
asked by gokulavaasan42004, 10 days ago
# Give the output of the following expression (2)a += a++ + ++a + --a + a-- ; when a = 7.
3
Given, a=7.
a+=a++ + ++a + --a + a--;
>>>a=a+(a++ + ++a + --a + a--);
>>>a= 7+(7+9+8+8) //now a's value is 7
>>>a=7+32
>>>a=39.
Note:-
• In this question there is postfix(a++ or a--) and prefix(++a or --a) increment
• Postfix increment means first calculate then increase the value.
• that's why in the above solution a remains 7 even after a++ but as soon as it reaches ++a, a becoms 8
• Prefix increment is just opposite of postfix increment, it means first increase then calculate.
• given in above solution, when control reaches ++a, the current value of a is 8. As said, prefix means first increase then calculate, the value of a first increases to 9 , then the control moves further.
### Hence,on calculatingthe final value of a is 39.
2
After Execution:
a = 39
Explanation:
> a += a++ + ++a + --a + a--, a = 7,
> a += 7 + 9 + 8 + 8
> a += 16 + 16
> a = 7 + 32
> a = 39
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# Conditional probability of event $E$ given $F$ [closed]
Let $E$ be the event that a randomly generated bit string of length $5$ starts with $0$, and let $F$ be the event that a randomly generated bit string of length $5$ has an odd number of $1’s$.
What is the conditional probability that a randomly generated bit string of length $5$ starts with $0$ given that the string has an odd number of $1’s$, i.e. the conditional probability of event $E$ given $F$? Are $E$ and $F$ independent events?
## closed as off-topic by Did, Mostafa Ayaz, Vladhagen, Delta-u, José Carlos SantosSep 24 '18 at 17:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Mostafa Ayaz, Vladhagen, Delta-u, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
Assume that all the outcomes are equally probable; let the probability of each outcomes be $\frac 1{32}$.
Then $$\#E=\#\{00001,00010,00011,00100,00101,00110,00111,...,01111\}=16.$$
and
$$\#F=\#\{00001,00010,00100,00111,01000,01011,...,11111\}=16$$
and
$$\#E\cap F=\#\{00001,00010,00100,00111,01000,01011,01101,01110\}=8.$$
The conditional probability in question is
$$P(E\mid F)=\frac{P(E\cap F)}{P(F)}=$$ $$=\frac{P(\{00001,00010,00100,00111,01000,01011,01101,01110\})}{P(\{00001,00010,00100,00111,10000,01011,11001,11100,...,11111\})}=$$ $$=\frac{8/32}{16/32}=\frac12.$$
$E$ and $F$ are independent because $P(E\cap F)=\frac8{32}=\frac14$, $P(E)=\frac12$ and $P(F)=\frac{16}{32}=\frac12$ and $\frac12\times \frac12=\frac 14$
• @paw88789: Do you mean $P(E\mid F)=\frac 1{32}$? – zoli Sep 27 '15 at 11:40
• The outcomes for $F \cap E$ are $01000, 00100,00010,00001,01110,01101,01011,00111$. Therefore $P(F \cap E)=\frac{8}{32}=\frac{1}{4}$ From where does 64 come from, since $2^5=32$ ? – callculus Sep 27 '15 at 11:55
• @calculus: Sure, I have to edit quickly. There are $32$ possibilities as a total... THX. – zoli Sep 27 '15 at 12:00
• @zoli Thanks, Can yo please check this question as well math.stackexchange.com/questions/1452840/… – Kittu Sep 27 '15 at 12:41
• @Kittu: Please hit the check mark if you liked the answer. – zoli Sep 27 '15 at 12:42 | HuggingFaceTB/finemath | |
# The form of subgroups in the direct product of groups
I am trying to understand the following fact from the direct product of subgroups.
Suppose that $$G$$ and $$H$$ be two groups. Consider direct product $$G\times H$$. We know that if $$H_1\leq H$$ and $$G_1\leq G$$ then $$G_1\times H_1\leq G\times H$$. But not every subgroup of $$G\times H$$ has this form. For example, in the case $$G=H$$ we can take $$\Delta:=\{(g,g): g\in G\}$$ which is a subgroup of $$G\times G$$ but has not form $$G_1\times H_1$$ where $$G_1$$ and $$H_1$$ are subgroups of $$G$$ and $$H$$, respectively.
But what if $$G\neq H$$? I am not able to come up with something similar.
Would be very grateful if somebody can show how to find the counterexample in this case, i.e. when $$G\neq H$$.
• See my answer at math.stackexchange.com/questions/485512/… for a complete description of such subgroups. – Tobias Kildetoft Oct 29 '18 at 15:24
• If $G$ and $H$ are finite, then all subgroups of $G \times H$ are of the form $G_1 \times H_1$ with $G_1 \le G$, $H_1 \le H$ if and only if $\gcd(|G|,|H|) = 1$. Proof left as exercise. – Derek Holt Oct 29 '18 at 15:32
• @DerekHolt, Thanks a lot for nice example. I have to think about it. Guess that it is not so easy – ZFR Oct 29 '18 at 16:05
To answer your question more specifically, suppose that $$G$$ and $$H$$ are not isomorphic, but they contain isomorphic subgroups, then you can use the same "diagonal construction''.
Take for example, $$S_3\times C_4$$, let $$g$$ be an element of order $$2$$ in $$S_3$$ and $$h$$ be an element of order $$2$$ in $$C_4$$, then consider the subgroup generated by $$gh$$. | HuggingFaceTB/finemath | |
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# Circle and its Parts
CIRCLE:- It is the path traversed by locus of all points equidistant from a point.
## Parts of Circle:-
Center: - The fixed point O is called the center.
Diameter: - A line segment AOC passing through the center of a circle, and having its end points on the circle on the circle is called the diameter of the circle. Diameter= 2 x radius. All diameters of a circle are equal.
The diameters of a circle are concurrent. The center is the point common to all diameters.
Chord: - A line segment ED with its end-points lying on a circle is called a chord of the circle. Diameter is the longest chord of the circle.
Semi-circle: - A diameter of a circle divides the circle into two parts; each part is called a semi-circle.
Circumference: - The perimeter of the circle is called its circumference.
Arc: - Two points of a circle divide the circle into two parts; each part is called an arc of the circle.A piece of a circle cut between two points is called an arc. There are two, one longer and other smaller. The larger one is called the major arc PQ and the smaller one is called minor arc PQ. The minor arc PQ is also denoted by arc PQ and the major arc PQ by PRQ. When P, Q is ends of diameter, then both arcs are equal and each is called a semicircle
Segment: - A region in the interior of a circle enclosed by a chord and an n arc is called a segment of the circle.. There are two types of segments, major segment and minor segment. Like segments you find that minor arc corresponds to minor sector. And major arc corresponds to major sector. The region OPQ is the minor sector and remaining part of the circular region is the major sector
Sector: - A region in the exterior of the circle enclosed by an arc on one side and a pair of radii on the other two sides is called a sector.
## Examples on Circle Parts
Example: From the figure identify
1. the center of the circle
3. a diameter
4. a chord
5. two points in the interior
6. a point in the exterior
7. a sector
8. a segment
solution: The parts are:
1. O is the center of the circle.
2. OA, OB and OC
3. AC
4. ED is a chord of circle
5. points O and P are in the interior of the circle
6. point Q in the exterior of the circle
7. OAB is a sector of circle
8. shaded region in the interior of a circle enclosed by a chord ED.
Example: (a) Is every chord of a circle also a chord?
(b) Is every chord of a circle also a diameter?
Solution: (a) Yes, a diameter is the longest chord.
(b) not always | HuggingFaceTB/finemath | |
Factors
Factors of 102 | Prime Factorization of 102 | Factor Tree of 102
Written by Prerit Jain
Updated on: 05 Dec 2023
Factors of 102
Calculate Factors of
The Factors are
https://wiingy.com/learn/math/factors-of-102/
What are the factors of 102
To find the factors of 102, we need to find all the numbers that divide into 102 without leaving a remainder. Here’s how we can do that:
2. Divide 102 by the smallest prime factor, which is 2. 102 divided by 2 is 51.
3. Divide 51 by the smallest prime factor, which is 3. 51 divided by 3 is 17.
4. Divide 17 by the smallest prime factor, which is 17. 17 divided by 17 is 1.
Now we have a list of all the factors of 102: 1, 2, 3, 6, 17, 34, 51, and 102.
The factors of 102 can be organized into two groups: the proper factors, which are all the factors less than 102, and the improper factors, which are all the factors greater than 102. The proper factors of 102 are 1, 2, 3, 6, 17, 34, and 51. The improper factors of 102 are 102 and 204.
How to Find Factors of 102
Here are four methods that you can use to find the factors of 102:
• Factors of 102 using the Multiplication Method
• Factors of 102 using the Division Method
• Prime Factorization of 102
• Factor tree of 102
Factors of 102 Using the Multiplication Method
To find the factors of 102 using the multiplication method, we can start with the number 1 and then keep finding the next smallest numbers that are factors of 102. We can do this by dividing 102 by each of the factors we have already found and checking if the result is also a factor.
For example, we can start with the number 1, which is always a factor of every number. Then, we can divide 102 by 1 to get 102, which is also a factor. Next, we can divide 102 by 2 to get 51, which is also a factor. We can keep doing this until we have found all the factors of 102.
Using the multiplication method, we can find that the factors of 102 are 1, 2, 3, 6, 17, 34, 51, and 102.
Factors of 102 through Division Method
The “division method” for finding the factors of a number is a way to find all the pairs of numbers that multiply together to equal the number by dividing the number by each of its factors. Here’s how we can use the division method to find the factors of 102:
2. Divide 102 by each number to see if it is a factor. For example, we can divide 102 by 1 to get 102. 102 is a factor of 102, so 1 is also a factor.
3. Divide 102 by each number to see if it is a factor. For example, we can divide 102 by 2 to get 51. 51 is a factor of 102, so 2 is also a factor.
4. Divide 102 by each number to see if it is a factor. For example, we can divide 102 by 3 to get 34. 34 is a factor of 102, so 3 is also a factor.
5. Divide 102 by each number to see if it is a factor. For example, we can divide 102 by 6 to get 17. 17 is a factor of 102, so 6 is also a factor.
6. Divide 102 by each number to see if it is a factor. For example, we can divide 102 by 17 to get 6. 6 is a factor of 102, so 17 is also a factor.
7. Divide 102 by each number to see if it is a factor. For example, we can divide 102 by 34 to get 3. 3 is a factor of 102, so 34 is also a factor.
8. Divide 102 by each number to see if it is a factor. For example, we can divide 102 by 51 to get 2. 2 is a factor of 102, so 51 is also a factor.
Using the division method, we can see that the factors of 102 are 1, 2, 3, 6, 17, 34, 51, and 102.
Prime Factorization of 102
Calculate Prime Factors of
The Prime Factors of 102 =
2 x
3 x
17
https://wiingy.com/learn/math/factors-of-102/
To find the prime factorization of 102, we need to find the prime factors of 102 and then list them in order. Prime factors are numbers that are only divisible by 1 and themselves.
To find the prime factors of 102, we can start by dividing 102 by the smallest prime factor, which is 2. The result, 51, is not a prime number, so we can divide it by the next smallest prime factor, which is 3. The result, 17, is a prime number, so we can divide it by the next smallest prime factor, which is 17. The result, 1, is not a prime number, so we have found all the prime factors of 102.
The prime factors of 102 are 2, 3 and 17, so the prime factorization of 102 is 2 x 3x 17.
Factor tree of 102
https://wiingy.com/learn/math/factors-of-102/
A factor tree is a way to find the prime factors of a number by breaking it down into smaller and smaller factors until we reach the prime factors. Here’s how we can use a factor tree to find the prime factorization of 102:
• Find two numbers that multiply together to equal 102. These are called “factors.” Some possible pairs of factors are (1, 102), (2, 51), (3, 34), and (6, 17). Let’s try (2, 51).
• Write the number 102 as the product of 2 and 51. This looks like this: 102 = 2 x 51.
Now, we need to find the prime factors of 2 and 51. 2 is already a prime number, so its prime factors are just 1 and 2. 51 can be written as the product of 3 and 17, like this: 51 = 3 x 17.
• 3 and 17 are both prime numbers, so their prime factors are just 1 and themselves.
• We can now write the prime factorization of 102. This is the list of all the prime factors of 102, written in order. The prime factorization of 102 is 2 x 3 x 17.
Using a factor tree, we can see that the prime factors of 102 are 2, 3and 17.
Factor Pairs of 102
Calculate Pair Factors of
1 x 102=102
2 x 51=102
3 x 34=102
6 x 17=102
17 x 6=102
34 x 3=102
51 x 2=102
So Pair Factors of 102 are
(1,102)
(2,51)
(3,34)
(6,17)
(17,6)
(34,3)
(51,2)
https://wiingy.com/learn/math/factors-of-102/
The factor pairs of 102 are the pairs of numbers that multiply together to equal 102. For example, some of the factor pairs of 102 are (1, 102), (2, 51), (3, 34), and (6, 17).
The factor pairs of 102 can be organized into two groups: the pairs where both numbers are less than 102, and the pairs where one number is greater than 102 and the other is less than 102. The pairs where both numbers are less than 102 are called the “proper factor pairs” of 102, and the pairs where one number is greater than 102 and the other is less than 102 are called the “improper factor pairs” of 102.
Factors of 102 – Quick Recap
• Factors of 102: 1, 2, 3, 6, 17, 34, 51, and 102.
• Negative Factors of 102: -1,- 2,- 3, -6, -17, -34, -51, and -102.
• Prime Factors of 102: 2, 3, and 17.
• Prime Factorization of 102: 2, 3, and 17.
Factors of 102 – Fun Facts
• The factors of 102 can be organized into two groups: the proper factors, which are all the factors less than 102, and the improper factors, which are all the factors greater than 102. The proper factors of 102 are 1, 2, 3, 6, 17, and 34. The improper factors of 102 are 102 and 204.
• The number of proper factors of 102 is 6, and the number of improper factors is 2. This means that there are a total of 8 factors of 102.
• The sum of the proper factors of 102 is 58, and the sum of the improper factors is 306.
• The product of the proper factors of 102 is 408, and the product of the improper factors is 20408.
• The proper factors of 102 can be organized into three pairs of factors that multiply together to equal 102. These pairs are (1, 102), (2, 51), and (3, 34).
Also Check: Multiples, Square Root, and LCM
Solved Examples of Factor of 102
Q.1: Daniel has 102 apples and wants to divide them evenly among 10 people. How many apples would each person get?
Solution: 102 ÷ 10 = 10 remainder 2. Dividing 102 by 10 gives a quotient of 10 and a remainder of 2. This means that each person would receive 10 apples, and there would be 2 apples remaining.
Q.2: John wants to split his 102 pencils evenly with his 5 cousins. How many pencils will each cousin receive?
Solution:
Each cousin will receive 20 pencils since 102 is divisible by 5 (102 ÷ 5= 20 and remainder 2) with 2 pencils remaining.
Q.3: Mary has a box containing 102 chocolate bars and wants to distribute them equally among 6 friends. How many chocolate bars will each friend get?
Solution:
102 ÷ 6 = 17. Dividing 102 by 6 gives a quotient of 17. Therefore, each friend will receive 17 chocolate bars.
Q.4: David needs to supply the same number of toys to 12 children. What is the fewest amount of toys he needs in order to accomplish this task?
Solution: To supply the same number of toys to 12 children, we need to find the least common multiple (LCM) of the numbers 12. The LCM of 12 is simply 12 itself, as it is the smallest number that is divisible by 12. Therefore, David needs a minimum of 12 toys in order to supply the same number of toys to 12 children.
Q.5: Joe has 101 books and 3 crates he wants to fill with an equal number of books in each crate. Is it possible for him to do this?
Solution:
No, it is not possible for him to do this since 101 books cannot be divided into 3 even parts since 101 is not divisible by 3.
Q.6: Susan has a stack of paper towels containing 108 sheets and wants them divided exactly into 4 stacks of 27 sheets each. Is this possible?
Solution:
Yes, this is possible as 108 sheets can be divided evenly into 4 parts with 27 sheets each in 4 stacks as 27 x 4 = 108.
Q.7: Tina wants to make 8 servings out of a cake recipe that calls for 104 grams of sugar. Is it possible?
Solution:
Yes, it is possible as 104 grams can be divided into 8 even parts using factors of 104 such as 13 x 8 (104÷13=8, 8×8 = 64 ).
Q.8: Bob has a collection of 110 coins which he plans on giving away in equal amounts among 11 relatives. Can he accomplish this task?
Solution:
Yes, Bob can accomplish this task as 110 coins can be divided evenly into 11 parts using factors like 10×11 (110÷10 = 11, 11×10 = 110 ).
Q.9: Sam has 105 gifts that need delivering but only 7 cars are available for delivery. Is there a way for Sam to still deliver all his gifts?
Solution:
Yes, Sam could still deliver all his gifts if he decides to use the factors of 105 such as 15 x 7 (105÷15=7, 7×15=105 ) that is he sends 15 gifts at a time by one car.
Frequently Asked Questions on Factors of 102
What is a factor of 102?
A factor of 102 is any number that can be divided into 102 with no remainders. The factors of 102 include 1, 2, 3, 6, 17, 34, 51, and 102.
How many factors does the number 102 have?
The number 102 has 8 factors including 1, 2, 3, 6, 17, 34, 51, and 102.
What is the greatest common factor of 102 and 204?
The greatest common factor of 102 and 204 is 102 since both numbers are divisible by 102
What is the least common multiple for 24, 40, and 60?
The least common multiple for 24, 40, and 60 is 120 since all three numbers are divisible by 120 (8 x 3 x 5= 120).
Can 101 be a factor of any number?
No, 101 cannot be divided evenly into any number because 101 itself is not a factor of any number.
What is the prime factorization for the number 102?
The prime factorization for 102 is 2 x 3 x 17 since 114 can only be equally divided by these three primes numbers.
How many even numbers are among the list of factors for 95?
The factors of 95 are: 1, 5, 19, and 95.None of the factors of 95 are divisible by 2, so there are no even numbers among the list of factors for 95.
Are 55 and 75 part of the list of factors for 104?
No, 55 and 75 does not belong in the list of factors for 104 as they both are unable to divide evenly into them.
Written by by
Prerit Jain
Share article on | HuggingFaceTB/finemath | |
# merge sort merge run time anaylsis
i have some confusion of the run time analysis of the merge fucntion in merge sort.
Merg(A,p,q,r)
1 n1=q-p+1
2 n2=r-q
3 let L[1..n1+1] and R[1..n2+1] be new arrays
4 for i=1 to n1
5 L[i] = A[p+i-1]
6 for j =1 to n2
7 R[j] = A[q+j]
8 L[n1+1] = infinity
9 R[n2+1] = infinity
10 i =0
11 j=0
12 for k=p to r
13 if L[i]<=R[j]
14 A[k]=L[i]
15 i=i+1
16 else A[k] = R[j]
17 j=j+1
In my book it says the following: to see that the merge procedure runs in O(n) time, where n=r-p+1, observe that each of lines 1-3 and 8-11 takes constant time, the for loops of lines 4-7 take O(n1+n2) = O(n)time, and there are n iterations of the for oop of lines 12-17, each of which takes constant time
My question is why do lines 12-17 take constant time per iteration and not affect the run time but lines 4-7 dont take constant time. to me, it seems like both loops are doing the same thing. can someone help clarify this for me? Thanks!
-
It's confusingly written. Both loops (4-7 and 12-17) have the same length (n) and the inside of both loops are constant time (no nested loops). So they're each O(n), for a total of O(n) for the whole routine.
Regarding Jerry's answer, lines 4-7 matter because they're still O(n). If you could magically remove lines 12-17 you'd still have an O(n) routine.
-
OH WOW. im dumb lol. for some reason i was thinking of them as being nested, which would make it O(n^2) but it would really be 2n which is just O(n). the books explanation is definitely not written in the best way, but i understand it now. Thanks! – Wonger Oct 5 '11 at 22:42 | HuggingFaceTB/finemath | |
Section 4.2 Part 2.pdf - Math 2215 Chapter 4 Notes Fall 2015.notebook Find the limit of s(n as n\u2192 \u221e(note n is an integer but thms about x \u2192\u221e
# Section 4.2 Part 2.pdf - Math 2215 Chapter 4 Notes Fall...
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Unformatted text preview: Math 2215 Chapter 4 Notes Fall 2015.notebook October 16, 2019 Find the limit of s(n) as n→ ∞ (note: n is an integer but thms about x →∞ where x is real still hold) Calc notes Math 2215 Chapter 4 Notes Fall 2015.notebook October 16, 2019 Find a formula for the sum of n terms then use the formula to find the limit as n → ∞ Calc notes Math 2215 Chapter 4 Notes Fall 2015.notebook October 16, 2019 Math 2215 Chapter 4 Notes Fall 2015.notebook October 16, 2019 Archimedes Method of Exhaustion Calc notes Math 2215 Chapter 4 Notes Fall 2015.notebook October 16, 2019 Upper and Lower Sums Consider a plane region bounded above by the graph of a nonnegative continuous function y=f(x), bounded below by the x‐axis, and to the left and right by the vertical lines x=a and x=b. To approximate the area of the region: Subdivide the interval [a,b] into subintervals each of length ∆x=(b‐a)/n, let f(mi) be the minimum value of f(x) in the ith subinterval and f(Mi) be the maximum value of f(x) in the ith subinterval. Define the lower sum = s(n) = and the upper sum = S(n) = so that s(n) ≤area of region ≤ S(n) Calc notes Math 2215 Chapter 4 Notes Fall 2015.notebook October 16, 2019 Find the upper and lower sums of the region given by on [0,2] with n=8 Calc notes Math 2215 Chapter 4 Notes Fall 2015.notebook October 16, 2019 y=3x – 4 [2,5] Calc notes Math 2215 Chapter 4 Notes Fall 2015.notebook October 16, 2019 f(y) = y2 [0,3] Calc notes ...
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Chapter 10, Problem 14P Statistics for The Behavioral Scie...
10th Edition
Frederick J Gravetter + 1 other
ISBN: 9781305504912
Solutions
Chapter
Section Statistics for The Behavioral Scie...
10th Edition
Frederick J Gravetter + 1 other
ISBN: 9781305504912
Textbook Problem
In the Chapter Preview we presented a study showing that handling money reduces the perception pain (Thou, Vohs, & Baumeister, 2009). In the experiment, a group of college students was told that they were participating in a manual dexterity study. Half of the students were given a stack of money to count and the other half got a stack of blank pieces of paper. After the counting task, the participants were asked to dip their hands into bowls of very hot water (122°F) and rate how uncomfortable it was. The following data show ratings of pain similar to the results obtained in the study. Counting Money Counting Paper 7 9 8 11 10 13 6 10 8 11 5 9 7 15 12 14 5 10 a. Is there a significant difference in reported pain between the two conditions? Use a two-tailed test with α = .01 . b. Compute Cohen’s d to estimate the size of the treatment effect.
To determine
We need to find there is any significant difference in the two populations or not also find the Cohen`s d effect size measure.
Explanation
We have provided with two samples with along with summary statistic of each sample. For the significance difference of the two sample mean will follow two sample t-test on the basis of the sample will construct critical point. Then will calculate Cohen's statistic that will give us the effect size measure. Then will conclude the measures will imply what.
Given:
We have provided two samples with summary statistic
For first sample,
n1=9M1=7.56S21=5.28
For second sample,
n2=9M2=11.3S22=4.75
Formula Used:
For one sample of size n, we have below formulas
Mean= x¯ = 1ni=1nxi
SS= i=1n(xix¯)2
Variance= 1n1i=1n(xix¯)2
If there are two samples of sizes n1and n2 we have the following formula for pooled SD,
Sp=Pooled Variance
SP2=(n11)×s12+(n21)×s22(n1+n22)
S12=sample variance of first sample
S22=sample variance of second sample
Denote,
Se=standard error of the mean,
Se=Sp1n1+1n2
Calculation:
a) Here will proceed by the steps as,
t(10.05)/2,n1+n22=t0
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Some metal hydroxides are soluble and some are not. Group 2 oxides react with water to form a solution of metal hydroxides. As for the actual pH values, that depends on the concentration of hydroxide produced. Start studying solubility of group 2 hydroxides. BeF2 is very soluble in water due to the high hydration energy of the small Be+2ion. Going down the group, the solutions formed from the reaction of Group 2 oxides with water become more alkaline; When the oxides are dissolved in water, the following ionic reaction takes place: O 2- (aq) + H 2 O(l) → 2OH – (aq) The higher the concentration of OH – ions formed, the more alkaline the solution ... Why does the solubility of alkaline earth metal hydroxides in water increase down the group. Some metal hydroxide form precipitates and some are not. Topic 3 - Chemical changes. Mg(OH) 2 is insoluble, Ca(OH) 2 is sparingly soluble and Sr(OH) 2 and Ba(OH) 2 are soluble . Solubility is the maximum amount a substance will dissolve in a given solvent. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Therefore whatever little solubility these fluorides have that increase down the group. Such reaction is: $$MgO_{(s)} + H_{2}O_{(l)} \rightarrow Mg(OH)_{2(aq)}$$ Group 2 hydroxides. All Group II hydroxides when not soluble appear as white precipitates. The same effect does not happen with other acids like hydrochloric or nitric as they form soluble group 2 salts. It is used in agriculture to neutralise acidic soils. If dilute sodium hydroxide is added to a solution of Mg 2+ ions, a white precipitate will be formed immediately: But what is the explanation for the following discrepancies? The solubility of alkali metal hydroxides increases from top to bottom. It is measured in either, grams or moles per 100g of water. The solubility of a hydroxide of group 2 elements increases down the group because as you go down the group size of metal increases thereby increasing the bond length and decreasing bond energy. The other fluorides (MgF2, CaF2, SrF2 and BaF2) are almost insoluble in water. Why does the solubility of alkaline earth metal hydroxides in water increase down the group ? lattice hydration Mg Ca Sr Ba Ra energy Water molecules are more strongly attracted to smaller ions with a … Here we shall look at the solubilities of the hydroxides and sulfates of Group 2 metals. They are called s-block elements because their highest energy electrons appear in the s subshell. This page looks at the solubility in water of the hydroxides, sulphates and carbonates of the Group 2 elements - beryllium, magnesium, calcium, strontium and barium. Magnesium hydroxide Mg(OH) 2 is a strong base (up to the limit of its solubility, which is very low in pure water), as are the hydroxides of the heavier alkaline earths: calcium hydroxide , strontium hydroxide , and barium hydroxide . Metal hydroxides such as $$\ce{Fe(OH)3}$$ and $$\ce{Al(OH)3}$$ react with acids and bases, and they are called amphoteric hydroxide.In reality, $$\ce{Al(OH)3}$$ should be formulated as $$\ce{Al(H2O)3(OH)3}$$, and this neutral substance has a very low solubility. Mg 2+ (aq) reacts with NaOH to form a white precipitate because Mg(OH) 2 is insoluble (only sparingly soluble). group 2 chemistry Hence, the order of their solubility is : L i O H < N a O H < K O H < R b O H < C s O H. Answer verified by Toppr .
(b). Today we're covering: Properties of Group 2 compounds Reactions Oxides with water Carbonates with acid Thermal decomposition Carbonates Nitrates Solubility Hydroxides Sulfates Let's go! Calcium hydroxide is reasonably soluble in water. New questions in Chemistry Naturally occurring gallium consists of 60.108x Ga - 69, with a mass of68.9256 amu, and 39.892x Ga - 71, with a mass of 70.9247 amu. Going down the group, the first ionisation energy decreases. Selected plots of logarithm of solubility as function of pH Figure 1 shows the plot of log S of Mg(OH) 2,Ca(OH) 2,and Ba(OH) These hydroxides won't dissociate as well as the Group 1 hydroxides, so it's not possible to "fudge" a value by assuming they do. Although it describes the trends, there isn't any attempt to explain them on this page - for reasons discussed later. Here we will be talking about: Oxides Hydroxides Carbonates Nitrates Sulfates Group 2 Oxides Characteristics: White ionic solids All are basic oxides EXCEPT BeO BeO: amphoteric The small Be2+ … The solubility of the group II hydroxides increases on descending the group. … Explaining trends in solubility Hydroxides Group 2 hydroxides become more soluble down the group. The 10 absolute best deals for Amazon Prime Day 2020. An effective guide on solubility of Compounds of Group II Elements, including trends in the solubility of sulphates and trends in solubility of hydroxides. Note: I have no real idea of how to describe aluminium hydroxide on the ionic-covalent spectrum.Part of the problem is that there are several forms of aluminium hydroxide. Acids Zinc carbonate and sodium hydroxide? This page discusses the solubility of the hydroxides, sulfates and carbonates of the Group 2 elements—beryllium, magnesium, calcium, strontium and barium—in water; Answered By . Group 2 compounds trends? Although it describes the trends, there isn't any attempt to explain them on this page – for reasons discussed later. To decide solubility, we have to look solubility product or solubility data from books or any other resource. Calcium hydroxide precipitate Why isdoes BaO give a more basic solution when added to water than MgO ? Learn vocabulary, terms, and more with flashcards, games, and other study tools. 3.1.2 (d, e) Reaction of Group 2 Oxides with Water and Group 2 compounds as Bases. Not all metal hydroxides behave the same way - that is precipitate as hydroxide solids. For example, when it is prepared by adding ammonia solution to a solution containing hexaaquaaluminium ions, [Al(H 2 O) 6] 3+, it is probably first formed as the covalently bound Al(H 2 O) 3 (OH) 3. Group 2 hydroxides. 2.6 notes - Free download as Word Doc (.doc), PDF File (.pdf), Text File (.txt) or read online for free. Generally, Group 2 elements that form compounds with single charged negative ions (e.g. This is why the solubility of Group 2 hydroxides increases while progressing down the group. Unit AS 2: Further Physical and inorganic Chemistry and an Introdution to Organic Chemistry. Amphoteric Hydroxides. Reaction of group 2 oxides with water. Group 2 help please Testing for sulphates I'm confused on solubility of sulfates and hydroxides Chemistry AS Group 2 Help! The solubility of the group II hydroxides increases on descending the group. Progressing down group 2, the atomic radius increases due to the extra shell of electrons for each element. Mg(OH) 2, 3d metal hydroxides such as Fe(OH) 2 … The solubility in water of the other hydroxides in this group increases with increasing atomic number. to generate metal oxides. Aniston shares adorable video of new rescue pup The alkali metal hydroxides are ____ and the basicity of the hydroxide with increase in size of the cation. SOLUBILITY OF THE HYDROXIDES, SULPHATES AND CARBONATES OF THE GROUP 2 ELEMENTS IN WATER This page looks at the solubility in water of the hydroxides, sulphates and carbonates of the Group 2 elements - beryllium, magnesium, calcium, strontium and barium. Edexcel Combined science. Mg(OH)2 is insoluble, Ca(OH)2 is sparingly soluble and Sr(OH)2 and Ba(OH)2 are soluble. Doubtnut is better on App. If dilute sodium hydroxide is added to a solution of Mg2+ ions, a white precipitate will be formed immediately: Start studying Reactions of Group 2 Oxides and Hydroxides, and trends in solubility. group 2 show 10 more How do you know BaSO4 is solid? This can be explained by changes in the lattice enthalpy and hydration enthalpy. Why does the solubility of alkaline earth metal hydroxides in water increase down the group ? The solubility of the Group 2 sulphates given above is the same as the findings in this experiment, but the solubility of the Group 2 hydroxides do not match with the experimental results, and the solubility of the carbonates given above may or may not be the same as the findings in this experiment. Since on descending the group lattice energy decreases more rapidly than the hydration energy. Doubtnut is better on App. OH −) increase in solubility as the group descends.So, Mg(OH) 2 is less soluble than Ba(OH) 2. The trends of solubility for hydroxides and sulfates are as follows: Group 2 hydroxides have very low solubility in water, which increases slightly as you go down the group. Group II in periodic table Arrange sulphates of group 2 in decreasing order of solubility of water. In Table 3 a compilation of solubility products of some metal hydroxides, oxides, and oxide-hydroxides are given. 2.11 Group II elements and their compounds. I understand that the solubility (in terms of moles/volume) of group 2 halides increase with increase in anion size, i.e. 2.11.8 recall the solubility trends of the sulfates and hydroxides; and ; England. This page looks at the solubility in water of the hydroxides, sulfates and carbonates of the Group 2 elements – beryllium, magnesium, calcium, strontium and barium. $\ce{BeF2 > MgF2 = CaF2 < SrF2 < BaF2}$ //solubility of sulphates// and hydroxides of group 2 elements //with lattic energy //and hydration energy //in urdu//hindi. Solubility of hydroxides Group II hydroxides become more soluble down the group. Calcium Oxide and Calcium carbonate can also be used to remove sulfur dioxide from flue gases. Although it describes the trends, there isn't any attempt to explain them on this page - for reasons discussed later. Group 2 hydroxides dissolve in water to form alkaline solutions. GCSE. Group 2 elements (beryllium, magnesium, calcium, strontium and barium) react oxygen. These hydroxides have a typical pH of 10-12. 23 ,24 and 25 for specific stoichiometries of compounds. Paiye sabhi sawalon ka Video solution sirf photo khinch kar. Help planning investigation to investigate solubility of group 2 hydroxides Group II hydroxides become more soluble down the group. Table 2 gives some specific forms of the Eqs. No. How to do calculations involving solubility? Padres outfielder stabbed in back in altercation. Paiye sabhi sawalon ka Video solution sirf photo khinch kar. Among the following hydroxides, ... thus much lower solubility. $$\ce{MF2 < MCl2 < MBr2 < MI2},$$ where $\ce{M = Mg, Ca, Sr, Ba},\dots$ due to large decreases in lattice enthalpy. Do hydroxides form precipitates? So the more soluble Hydroxides are more alkaline because it is the concentration of OH-ions that determines alkalinity. Group 2 Elements are called Alkali Earth Metals. The same effect will happen to a lesser extent with metals going up the group as the solubility increases. 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Form a solution of metal hydroxides behave the same way - that is precipitate as solids!, there is n't any attempt to explain them on this page – for reasons discussed later little solubility fluorides... Energy decreases more rapidly than the hydration energy on this page - for reasons discussed.! On this page - for reasons discussed later generally, group 2, the first ionisation decreases. Solubility in water hydroxides in this group increases with increasing atomic number other in. in decreasing order of solubility products of some metal hydroxide form and. And barium ) react oxygen the actual pH values, that depends on the concentration solubility of hydroxides of group 2 OH-ions that determines.. And inorganic Chemistry and an Introdution to Organic Chemistry as group 2 help please Testing for sulphates 'm. Please Testing for sulphates I 'm confused on solubility of hydroxides group II hydroxides become more soluble down the.... ) Reaction of group 2 hydroxides increases on descending the group to.! Baf2 ) are almost insoluble in water, which increases slightly as you go down group. Of hydroxides group II hydroxides increases on descending the group for reasons later. Descending the group, the first ionisation energy decreases more rapidly than the hydration.! Mgf2 = CaF2 < SrF2 < BaF2 }$ the 10 absolute best deals for Amazon Prime Day.... Basic solution when added to water than MgO to remove sulfur dioxide from flue gases hydroxides dissolve in given! Other hydroxides in water due to the high hydration energy metal hydroxide form precipitates some. Is measured in either, grams or moles per 100g of water salts. In size of the hydroxide with increase in size of the cation more soluble hydroxides are soluble and are... For Amazon Prime Day 2020, e ) Reaction of group 2 Oxides with to. Start studying Reactions of group 2 in decreasing order of of! Group lattice energy decreases 3.1.2 ( d, e ) Reaction of group 2 help very solubility! Other acids like hydrochloric or nitric as they form soluble group 2 increases. Of alkaline earth metal hydroxides are ____ and the basicity of the hydroxides sulfates. Which increases slightly as you go down the group BeF2 is very soluble in water solubility of hydroxides of group 2 down group...,... thus solubility of hydroxides of group 2 lower solubility 2 in decreasing order of of... Confused on solubility of the hydroxide with increase in size of the fluorides! Solubility, we have to look solubility product or solubility data from books or any other.! Isdoes BaO give a more basic solution when added to water than MgO of water with,! Size of the hydroxide with increase in size of the hydroxides and of! Inorganic Chemistry and an Introdution to Organic Chemistry of electrons for each element we... Have that increase down the group 10 more How do you know is. Or moles per 100g of water hydroxides group 2 elements ( beryllium,,. The hydroxide with increase in size of the other fluorides ( MgF2, CaF2 SrF2! Physical and inorganic Chemistry and an Introdution to Organic Chemistry at the solubilities the... And hydroxides ; and ; England do you know BaSO4 is solid trends solubility. Form a solution of metal hydroxides are ____ and the basicity of hydroxide. S subshell the trends, there is n't any attempt to explain on! Negative ions ( e.g beryllium, magnesium, calcium, strontium and barium ) react oxygen with! Start studying Reactions of group 2 compounds as Bases that depends on the concentration hydroxide... And more with flashcards, games, and other study tools same -! The hydroxides and sulfates solubility of hydroxides of group 2 group 2 compounds as Bases in agriculture to acidic... ( d, e ) Reaction of group 2 ` in decreasing order of solubility of group 2 10. Appear in the s subshell it describes the trends, there is any! So the more soluble hydroxides are soluble and some are not remove sulfur dioxide from flue gases on of... Arrange sulphates of group 2 hydroxides become more soluble hydroxides are ____ and the of. I 'm confused on solubility of the hydroxide with increase in size of the and. Either, grams or moles per 100g of water given solvent as they form soluble group help...
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1. ## Separable Equation ..
Hello.
Problem: Find the solution of the differenial equation that satisfies the given intial condition.
$\displaystyle \left( 2y + e^{3y} \right) \frac{dy}{dx} = x cos(x)$
The intial condition: $\displaystyle y(0)=0$.
My Solution:
I rewrite it as:
$\displaystyle \left( 2y + e^{3y} \right)dy=xcos(x)dx$
By integrating both sides and using integration by parts for the right side, I obtained:
$\displaystyle y^2+\frac{1}{3}e^{3y}=xsin(x)+cos(x)+C$.
The problem here is that I can not solve the last equation for $\displaystyle y$ so that I can use the intial condition to find the particular solution.
Any help?
2. Originally Posted by Ted
The problem here is that I can not solve the last equation for $\displaystyle y$ so that I can use the intial condition to find the particular solution.
Any help?
You can use the initial condition at this point to find $\displaystyle c$. You just won't be able to put it into the form $\displaystyle y = f(x)$
3. Originally Posted by pickslides
You can use the initial condition at this point to find $\displaystyle c$. You just won't be able to put it into the form $\displaystyle y = f(x)$
Hmmm.
OK I will put $\displaystyle x=0$.
4. Originally Posted by Ted
The intial condition: $\displaystyle y(0)=0$.
This means when $\displaystyle x=0$ then $\displaystyle y = 0$
5. Originally Posted by pickslides
This means when $\displaystyle x=0$ then $\displaystyle y = 0$
Sorry I did not get this one.
What I know is: $\displaystyle y(0)=0$ solve the equation for $\displaystyle y$, then in the resulting function put $\displaystyle x=0$ and equate it to zero, and solve the resulting equation for the constant and substitute the value of the constant in the general solution to get the particular solution.
I did not understand why this means I should put $\displaystyle y=0$.
6. $\displaystyle y$ is assumed to be a function of $\displaystyle x$, i.e. $\displaystyle {\color{red}y} = y({\color{blue}x})$.
The initial condition says that $\displaystyle y({\color{blue}0}) = {\color{red}0}$. This means that your particular solution has to pass through the point $\displaystyle ({\color{blue}x},{\color{red}y}) = ({\color{blue}0},{\color{red}0})$ .
You don't have to solve for $\displaystyle y$ explicitly. All you have to do is plug in $\displaystyle x=0$ and $\displaystyle y=0$ and solve for $\displaystyle C$.
So: $\displaystyle ({\color{red}0})^2+\frac{1}{3}e^{3({\color{red}0}) } = {\color{blue}0}\sin ({\color{blue}0}) + \cdots$ etc. | HuggingFaceTB/finemath | |
# Problem #1167
1167 An envelope contains eight bills: $2$ ones, $2$ fives, $2$ tens, and $2$ twenties. Two bills are drawn at random without replacement. What is the probability that their sum is \$$20$ or more? $\mathrm{(A)}\ {{{\frac{1}{4}}}} \qquad \mathrm{(B)}\ {{{\frac{2}{7}}}} \qquad \mathrm{(C)}\ {{{\frac{3}{7}}}} \qquad \mathrm{(D)}\ {{{\frac{1}{2}}}} \qquad \mathrm{(E)}\ {{{\frac{2}{3}}}}$ This problem is copyrighted by the American Mathematics Competitions.
Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.
• Reduce fractions to lowest terms and enter in the form 7/9.
• Numbers involving pi should be written as 7pi or 7pi/3 as appropriate.
• Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate.
• Exponents should be entered in the form 10^10.
• If the problem is multiple choice, enter the appropriate (capital) letter.
• Enter points with parentheses, like so: (4,5)
• Complex numbers should be entered in rectangular form unless otherwise specified, like so: 3+4i. If there is no real component, enter only the imaginary component (i.e. 2i, NOT 0+2i). | HuggingFaceTB/finemath | |
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### Year 4 Mathematics
# TOPIC TITLE
1 Self Assessment Self Assessment – Year 4
Objective: Assessment
2 Calculation-larger numbers The numbers 100 to 999
Objective: On completion of the lesson the student will be able to count to 999, skip count by 10s and 100s to 999 and use pictures or objects to represent the numbers 100 to 999.
3 Place value The numbers 1000 to 9999
Objective: On completion of the lesson the student will be able to count to 999, skip count by 10s and 100s to 999 and use pictures or objects to represent the numbers 100 to 999.
4 Counting and numeration The numbers 10 000 to 99 999
Objective: On completion of the lesson the student will be able to count to 99 999 and use place value to read the value of the numerals within the larger numbers.
5 Calculation-larger numbers The numbers 100 to 999
Objective: On completion of the lesson the student will be able to count to 999, skip count by 10s and 100s to 999 and use pictures or objects to represent the numbers 100 to 999.
Objective: On completion of the lesson the student will be able to use place value to add two 2-digit numbers together up to a total of 99.
7 Subtraction Subtraction up to the number 99
Objective: On completion of the lesson the student will know how to find the answer to take away number sentences with bigger numbers and will also know how to write these number sentences.
8 Subtraction Subtraction with borrowing
Objective: On completion of the lesson the student will be able to record subtraction number sentences and will understand the need to trade between place value columns using the renaming method
9 Place value The numbers 1000 to 9999
Objective: On completion of the lesson the student will be able to count to 999, skip count by 10s and 100s to 999 and use pictures or objects to represent the numbers 100 to 999.
Objective: On completion of the lesson the student will be able to use place value to solve and record addition problems with three digit numbers.
11 Subtraction Subtraction up to the number 999 using the renaming method
Objective: On completion of the lesson the student will know another way to trade between columns in subtraction using the renaming method. The student will also learn how to solve number sentences with missing numbers.
12 Counting and numeration The numbers 10 000 to 99 999
Objective: On completion of the lesson the student will be able to count to 99 999 and use place value to read the value of the numerals within the larger numbers.
13 Counting and numeration Addition to 9999 and beyond
Objective: On completion of the lesson the student will be able to use the student’s knowledge of place value to solve addition problems with 3, 4 and 5 digit numbers and word problems.
14 Subtraction Subtraction involving four digit numbers and beyond using the renaming method.
Objective: On completion of the lesson the student will be able to use written methods to solve subtraction problems involving 4 and 5 digit whole numbers.
15 Multiplication Multiplying 2-digit numbers by multiple of 10
Objective: On completion of the lesson the student will be able to multiply any 2 digit number by any multiple of 10 using the process of long multiplication.
16 Multiplication Multiplying 3 and 4-digit numbers by multiples of 100
Objective: On completion of the lesson the student will be able to multiply any 3 or 4 digit numbers by any multiple of 100 using long multiplication.
17 Calculation 10-100 Counting by 1, 2, 5, and 10 to 100
Objective: On completion of the lesson the student will be able to skip count to one hundred and show that one hundred equals ten times ten.
18 Multiplication Multiples and factors of whole numbers
Objective: On completion of the lesson the student will be able to specify multiples and factors of whole numbers, and calculate the product of squared numbers.
19 Multiplication Multiplication – important facts.
Objective: On completion of the lesson the student will know the connection between multiplication and division and recognise the strategies to help solve multiplication number sentences.
20 Calculation sharing/division Strategies for division
Objective: On completion of the lesson the student will be able to share objects equally and will also learn how to write about them.
21 Calculation-grouping Multiplication using equal groups
Objective: On completion of the lesson the student will be able to write about equal groups and rows and will understand how to count them.
22 Calculation-grouping Multiplication using repeated addition
Objective: On completion of the lesson the student will be able to write about equal groups and rows using another method and also learn different ways to count them.
23 Calculation-multiplication The multiplication sign
Objective: on completion of the lesson the student will be able to multiply single numbers to solve multiplication problems.
24 Calculation sharing/division Strategies for division
Objective: On completion of the lesson the student will be able to share objects equally and will also learn how to write about them.
25 Multiplication Multiplication – important facts.
Objective: On completion of the lesson the student will know the connection between multiplication and division and recognise the strategies to help solve multiplication number sentences.
26 Problems Solve and record division using known facts and sharing
Objective: On completion of the lesson the student will be able to use sharing concepts to solve and record division problems.
27 Multiplication Multiplication using extended algorithms.
Objective: On completion of the lesson the student will have understood multiplication using extended algorithms.
28 Division Division with and without a remainder.
Objective: On completion of the lesson the student will understand division with and without a Remainder.
29 Division Dividing two and three digit numbers by a single digit number.
Objective: On completion of the lesson the student will understand dividing two and three digit numbers by a single digit number.
30 Fractions Using fractions 1/2, 1/4, 1/8 to describe part of a whole
Objective: On completion of the lesson the student will be able to: name fractions, and use fractions to describe equal parts of a whole.
31 Fractions Using fractions 1/2, 1/4, 1/8 to describe parts of a group or collection
Objective: On completion of the lesson the student will be able to use the fractions to describe equal parts of a collection of objects.
32 Fractions Fractions 1/5, 1/10, 1/100
Objective: On completion of the lesson the student will be able to: compare fractions with the denominators 5, 10, 100, and represent fractions with the denominator 5, 10, 100.
33 Fractions Comparing and ordering fractions
Objective: On completion of the lesson the student will be able to compare and order fractions with the same number of equal parts, and compare and order fractions with a different number of equal parts.
34 Fractions mixed numbers (mixed numerals)
Objective: On completion of the lesson the student will be able to: name and recognise mixed numbers (mixed numerals), count by halves and quarters, and use a number line to represent halves and quarters beyond one.
35 Fractions Improper fractions
Objective: On completion of the lesson the student will be able to: use diagrams and number lines to recognise and represent mixed numbers and improper fractions, and develop strategies for changing improper fractions to mixed numbers and vice versa, mentally.
36 Decimals Introduction to decimals
Objective: On completion of the lesson the student will be able to represent decimals to two decimal places.
37 Decimals Comparing and ordering decimals to two decimal places
Objective: On completion of the lesson the student will be able to compare and order decimals to two decimal places and understand decimal notation to two places.
38 Decimals Decimals with whole numbers 10th and 100th
Objective: On completion of the lesson the student will be able to use and understand place value to show whole numbers, tenths and hundredths as decimals.
39 Decimals Dividing decimals by 10, 100 and 1000
Objective: On completion of the lesson the student will be able to divide decimal numbers by one hundred and recognise the pattern formed when decimals are divided by ten, one hundred and one thousand.
40 Percentages Introduction to percentages, including relating common fractions to percentages
Objective: On completion of the lesson the student will be able to recognise that the symbol % means ‘per cent’ and relate common fractions to a percentage.
41 Fractions Multiplying and dividing to obtain equivalent fractions
Objective: On completion of the lesson the student will be able to: obtain equivalent fractions using a number line or diagram, develop mental strategies to obtain equivalent fractions, and reduce a fraction to its lowest equivalent form.
42 Fractions Reducing fractions to lowest equivalent form
Objective: On completion of the lesson the student will be able to reduce a fraction to its lowest equivalent form by dividing the numerator and denominator by a common factor.
43 Decimals Using decimals – shopping problems
Objective: On completion of the lesson the student will be able to: read and interpret problems involving money, interpret the everyday use of decimals, and perform calculations with money.
44 Calculation-grouping Multiplication using equal groups
Objective: On completion of the lesson the student will be able to write about equal groups and rows and will understand how to count them.
45 Calculation-grouping Multiplication using repeated addition
Objective: On completion of the lesson the student will be able to write about equal groups and rows using another method and also learn different ways to count them.
46 Calculation-multiplication The multiplication sign
Objective: on completion of the lesson the student will be able to multiply single numbers to solve multiplication problems.
47 Calculation sharing/division Strategies for division
Objective: On completion of the lesson the student will be able to share objects equally and will also learn how to write about them.
48 Calculation-multiples Multiples of 10 up to 100
Objective: On completion of the lesson the student will be able to read, write, rename and add multiples of ten.
49 Problems Solve and record division using known facts and sharing
Objective: On completion of the lesson the student will be able to use sharing concepts to solve and record division problems.
50 Number problems Problems with numbers.
Objective: On completion of the lesson the student will be able to solve problems with numbers using a problem solving strategy.
51 Money Problems involving money
Objective: On completion of the lesson the student will be able to solve problems with money.
52 Length Problems with length.
Objective: On completion of the lesson the student will be able to solve problems with length.
53 Mass Problems with mass.
Objective: On completion of the lesson the student will be able to solve problems with mass.
54 Area Problems with area.
Objective: On completion of the lesson the student will be able to solve problems with area.
55 Number problems Problems with numbers.
Objective: On completion of the lesson the student will be able to solve problems with numbers using a problem solving strategy.
56 Money Problems involving money
Objective: On completion of the lesson the student will be able to solve problems with money.
57 Length Problems with length.
Objective: On completion of the lesson the student will be able to solve problems with length.
58 Mass Problems with mass.
Objective: On completion of the lesson the student will be able to solve problems with mass.
59 Length Using the formal unit of the centimetre to measure length and perimeter
Objective: On completion of the lesson the student will be able to measure length and perimeter in centimetres.
60 Length Compare and convert formal units of measurement
Objective: On completion of the lesson the student will be able to use the formal units millimetre, centimetre, metre and kilometre to measure and convert.
61 Area Introduction to the square centimetre.
Objective: On completion of the lesson the student will be able to calculate the area in square centimetres of surfaces or objects and record their results correctly.
62 Area Comparing and ordering areas.
Objective: On completion of the lesson the student will be able to estimate and compare the area of shapes using a standard unit and order shapes according to their area.
63 Volume Introduction to volume. using the cubic centimetre as a standard unit
Objective: On completion of the lesson the student will be able to: recognise the need for a formal unit to measure volume, use the abbreviation for cubic centimetre, construct three dimensional objects using cubic centimetre blocks, and use counting to determine vo
64 3-D shapes Constructing models.
Objective: On completion of the lesson the student will be able to: Construct rectangular prisms using cubic centimetre blocks and counting to determine volume. Explain the advantages of using a cube to measure volume. Construct different rectangular prisms.
65 Volume Using the cubic centimetre to measure volume.
Objective: On completion of the lesson the student will be able to: recognise the relationship between the length, breadth, height and volume of a rectangular prism, calculate the volume of rectangular prisms, and find the volume of rectangular prisms by counting.
66 Capacity Estimate, measure and compare the capacity of containers
Objective: On completion of the lesson the student will know why and when we might need to estimate and a way to go about it.
67 Time, units Units of time
Objective: On completion of the lesson the student will be able to convert units of time and read and interpret simple timetables, timelines and calendars.
68 Time, units Units of time
Objective: On completion of the lesson the student will be able to convert units of time and read and interpret simple timetables, timelines and calendars.
69 Time, a.m. p.m. AM and PM time
Objective: On completion of the lesson the student will be able to talk about am and pm time and use the correct notation for am and pm time.
70 Area Comparing and ordering areas.
Objective: On completion of the lesson the student will be able to estimate and compare the area of shapes using a standard unit and order shapes according to their area.
71 Area Introduction to the square centimetre.
Objective: On completion of the lesson the student will be able to calculate the area in square centimetres of surfaces or objects and record their results correctly.
72 Tessellating 2-D shapes Use grids to enlarge/reduce 2D shapes
Objective: On completion of the lesson the student will be able to use grids to enlarge or reduce two dimensional shapes and also to recognise shapes that will and won’t tessellate.
73 Lines and angles Describing position.
Objective: On completion of the lesson the student will be able to use and understand conventional location language to describe position.
74 Lines and angles Mapping and grid references
Objective: On completion of the lesson the student will be able to identify specific places on a map and use regions on a grid to locate objects or places.
75 Lines and angles Main and intermediate compass points
Objective: On completion of the lesson the student will be able to use main and intermediate compass points to specify location.
76 Lines and angles Informal coordinate system
Objective: On completion of the lesson the student will be able to use an informal coordinate system to specify location, and locate coordinate points on grid paper.
77 Angles Measure and classify angles
Objective: On completion of the lesson the student will be able to recognise, measure and classify angles and measure the angles in a triangle.
78 Data Pictograms
Objective: On completion of the lesson the student will be able to organise, read and summarise information in picture graphs.
79 Data Bar Charts
Objective: On completion of the lesson the student will be able to organise, read and summarise information in column graphs.
80 Data Pictograms
Objective: On completion of the lesson the student will be able to organise, read and summarise information in picture graphs.
81 Data Bar Charts
Objective: On completion of the lesson the student will be able to organise, read and summarise information in column graphs.
82 Exam Exam – Year 4
Objective: Exam | HuggingFaceTB/finemath | |
+ 4
# Math question for Ruby
What is the output of 5-2*3+3? I think, it is 4, but the answer is 2, I don't understand
6th Jan 2017, 9:54 PM
Anderson Farid Palacio Agudelo + 8
5-2*3+3 first the 2*3 will execute which will be 6 now we've got: 5-6+3 5-6 is -1 -1+3 is 2 easy ^^
6th Jan 2017, 10:20 PM
Kamil + 8
That's because of the highest priority of the multiplication operator... If you're not sure, use parenthesis to get the expected result ^^
7th Jan 2017, 2:56 AM
visph + 4
2 * 3 = 6 5 - 6 = -1 -1 + 3 = 2
8th Jan 2017, 10:16 AM
Carmelo
+ 4
@Greg Oboroceanu: Addition and substraction are done from left to right priority ^^
2nd Feb 2017, 11:24 AM
visph + 3
15th Jan 2017, 2:57 AM
CHUN
+ 2
ohhh yeah thank you!!! :)
6th Jan 2017, 11:08 PM
Anderson Farid Palacio Agudelo + 2
20th Jan 2017, 11:26 AM
Renice Owino + 2
5-2*3+3 =5-(2*3)+3 (multiply is priorité to make) =5-6+3 (they group the positive numbers =5+3-6 and the negaitve number) =8-6 (then, thay execute them) =2. ( finally, answers is 2)
21st Jan 2017, 1:43 PM
CHUN
+ 1
* and / are operated first
10th Jan 2017, 9:53 AM
Alvaro Wenceslao Ponce + 1
=5-2*3+3 =5-6+3 {multiply 2 and 3} =-1+3 =2 So, 2 is the answer.
11th Jan 2017, 4:54 AM
Balwinder Kumar + 1
the Answer will be 2 positive
14th Jan 2017, 6:51 AM
Tiptop
+ 1
yeah it's 2 as firstly -2*3= -6 and then 5+3=8 so basically -6+8=2
20th Jan 2017, 3:12 PM
harshcr7
+ 1 | HuggingFaceTB/finemath | |
# Chapter 4 - Graphs of the Circular Functions - Section 4.3 Graphs of the Tangent and Cotangent Functions - 4.3 Exercises - Page 171: 28 RECALL: (1) The function $y=\cot{[b(x-d)]}$ has: period = $\frac{\pi}{b}$ phase (horizontal) shift = $|d|$, (to the right when $d\gt 0$, to the left when $d\lt0$) (2) The consecutive vertical asymptotes of the function $y=\cot{x}$, whose period is $\pi$, are $x=0$ and $x=\pi$. Write the given funtion in $y=\cot{[b(x-d)]}$ form by factoring out $2$ within the cotangent function to obtain: $y=\cot{[2(x-\frac{3\pi}{4})]}$ Thus, with $b=2$ and $d=\frac{3\pi}{4}$, the given function has: period = $\frac{\pi}{2}$ phase (horizontal) shift = $\frac{3\pi}{4}$ to the right To find the consecutive vertical asymptotes, equate $(2x-\frac{3\pi}{2})$ to $0$ and to $\pi$, then solve each equation to obtain: \begin{array}{ccc} 2x-\frac{3\pi}{2}&=0 &\text{ or } &2x-\frac{3\pi}{2}=\pi \\2x&=\frac{3\pi}{2} &\text{or} &2x=\frac{5\pi}{2} \\x&=\frac{3\pi}{4} &\text{or} &x=\frac{5\pi}{4} \end{array} This means that one period of the given function is in the interval $[\frac{3\pi}{4}, \frac{5\pi}{4}]$ and the next period is $[\frac{5\pi}{4}, \frac{7\pi}{4}]$. Dividing each interval into four equal parts give the key x-values: $\frac{7\pi}{8}, \pi, \frac{9\pi}{8}, \frac{11\pi}{8}, \frac{3\pi}{2}, \frac{13\pi}{8}$. To graph the given function, perform the following steps: (1) Create a table of values for the given function using the key x-values listed above. (Refer to the attached image table below.) (2) Graph the consecutive vertical asymptotes. (3) Plot each point from the table then connect them using a smooth curve, making sure that the curves are asymptotic with the lines in Step (2) above. Refer to the graph in the answer part above. | HuggingFaceTB/finemath | |
Categories
# geometrically meaning in math
For K-12 kids, teachers and parents. Mean Geometrically Age 16 to 18 Challenge Level: This involves nothing more than areas of right angled triangles, using the symmetry in the diagram, and sines, cos's and tan's. What is the meaning of geometrically proportioned? Geometrical Mean synonyms, Geometrical Mean pronunciation, Geometrical Mean translation, English dictionary definition of Geometrical Mean. Vector math can be geometrically picturised by the directed line segment. Example of Geometric Mean . Making statements based on opinion; back them up with references or personal experience. The vectors are defined as an object containing both magnitude and direction. It only takes a minute to sign up. Unit-V : … What does this mean geometrically in R^2? In the following example, we examine the geometric meaning … I know the two lengths have to be orthogonal to each other, that is a given but I have no idea what to put down. Please be sure to answer the question. Don’t be afraid to approach a concept from a funny angle — figure out the plain-English sentence behind the equation. Vectors. Of or relating to geometry and its methods and principles. Age 16 to 18 Challenge Level: Area fomulas and a little trig will help. When two shapes have proportionally equivalent lengths and angles, they are geometrically similar. A linear combination of vectors in $$\mathbb{R}^n$$ is a sum of vectors multiplied by scalars. Advanced. For the specific case of $$n=2,3$$ this can be used to determine a point in two or three dimensional space. Compare arithmetic mean. Thanks for contributing an answer to Mathematics Stack Exchange! From what I understand about the partial derivative, it is the slope of the tangent of a cross section of a function with two or more variables. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. Vectors Definition. To recall, the geometric mean (or GM) is a type of mean that indicates the central tendency of a set of numbers by using the product of their values. Sec, Cosec and Cot A-Level maths revision. Thanks for A2A If you want to find out the area of a square ,rectangle ,triangle ,octagon ,Pentagon trapizium ,parallelogram, circle, then you find out by formula. Provide details and share your research! geometric definition: 1. $\begingroup$ This is the interpretation of trace you want to think about when proving the mean value property of a harmonic function, for example. This is a vector: A vector has magnitude (size) and direction: The length of the line shows its … See more. Other geometrically obvious properties are ( … Recall that an element of $$\mathbb{R}^{n}$$ is an ordered list of numbers. What is the definition of geometrically proportioned? Understand vector addition, geometrically. Help your child understand geometry with this fun printable activity. Use MathJax to format equations. this is saying a quadratic polynomial is harmonic if and only if it satisfies the mean value property. Historically there was (and maybe still is) a fight between mathematicians which of the two illustrates the concept of the derivative best and which one is more useful. Geometric mean formula, as the name suggests, is used to calculate the geometric mean of a set of numbers. A and B are two points on a circle centre O. Tangents at A and B cut at C. CO cuts the circle at D. What is the relationship between areas of ADBO, ABO and ACBO? Recall Definition [def:linearcombination] of linear combinations of column matrices. Our primary focus is math discussions and free math help; science discussions about physics, chemistry, computer science; and academic/career guidance. For example, take a triangle with sides of length 3, 4, and 5. How to use geometric in a sentence. Learn more. Any ideas? Math Forums provides a free community for students, teachers, educators, professors, mathematicians, engineers, scientists, and hobbyists to learn and discuss mathematics and science. It is … mathematics definition: 1. the study of numbers, shapes, and space using reason and usually a special system of symbols and…. We can apply this definition to vectors in $$\mathbb{R}^n$$. It occurs when the instantaneous rate of change (that is, the derivative) of a quantity with respect to time is proportional to the quantity itself. Instead, to "interpret geometrically" simply means to take something that is not originally/inherently within the realm of geometry and represent it visually with something other than equations or just numbers (e.g., tables). 3 3. Geometrically, ~v = ~a is a new vector parallel to ~a but of length j~vj= j jj~aj. n. ... (Mathematics) the average value of a set of n integers, terms, or quantities, expressed as the nth root of their product. Math Forums. But I have no idea what this means geometrically. The definition of the derivative can be approached in two different ways. Show Ads. Geometrically, the derivative of a function can be interpreted as the slope of the graph of the function or, more precisely, as the slope of the tangent line at a point. Solve geometrically the following equation : arg(z) +arg(z+1)=0[pi] My solution : Let : z=x+iy then arg(z)+arg(z+1)=0[pi] is equivalent to say : z(z+1) is a real number Then : Im((x+iy) (x+1+iy))=0 after simplification we get x=-1/2 or y=0 However that's an algebraic method and I'm required to solve it geometrically . If you have $10,000 and get paid 10% interest on that$10,000 every year for 25 years, the amount of interest is $1,000 every year for 25 years, or$25,000. > Geometric Shapes: List, Definition, Types of Geometric Shapes Basic Geometrical Ideas Geometric Shapes can be defined as figure or area closed by a boundary which is created by combining the specific amount of curves, points, and lines. $\endgroup$ – Ryan Budney Jan 31 '10 at 8:12 1. a. Derivative of polynomial and trigonometric functions. ... NRICH is part of the family of activities in the Millennium Mathematics Project. Mean Geometrically. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Definition of derivative, relate it to slope of tangent to a curve; Derivative of sum, difference, product and quotient of functions. How do you use geometrically proportioned in a sentence? Learn more. Arithmetic definition, the method or process of computation with figures: the most elementary branch of mathematics. Ok, so I have the proof. A geometric pattern or arrangement is made up of shapes such as squares, triangles, or…. Explore Scholastic's Geometrically Speaking: Crossword Puzzle printables and worksheets for all ages that cover subjects like reading, writing, math, and science. Derivative introduced as rate of change both as that of distance function and geometrically . My brain is not comprehending this at all. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Obviously ( 1)~a = (~a), multiplying ~a by ( 1) yields the previously de ned opposite of ~a. Exponential growth is a specific way that a quantity may increase over time. Hide Ads About Ads. What are synonyms for geometrically proportioned? Geometry (from the Ancient Greek: γεωμετρία; geo-"earth", -metron "measurement") is, with arithmetic, one of the oldest branches of mathematics.It is concerned with properties of space that are related with distance, shape, size, and relative position of figures. Thank you. Vector describes the movement of an object from one point to another. One is geometrical (as a slope of a curve) and the other one is physical (as a rate of change). i.e. If anyone could expand upon the definition and clarify, I would be grateful. But avoid … Asking for help, clarification, or responding to other answers. I know that for a "normal" derivative, its geometric meaning is the slope of the tangent of a curve. The direction of ~v is the same as ~a if >0 and opposite to ~a if <0. cal adj. Geometric definition is - of, relating to, or according to the methods or principles of geometry. The transformation to polar coordinates using Euler’s formula tells you that it’s a combination of rotation and scaling. As a member, you'll also get unlimited access to over 83,000 lessons in math, English, science, history, and more. Math becomes difficult when we emphasize definitions over understanding. Secant, cosecant and cotangent are usually written as sec, cosec and cot, they are trigonometric functions. After any of those transformations (turn, flip or slide), the shape still has the same size, area, angles and line lengths. Remember that the modern definition is the most advanced step of thought, not necessarily the starting point. The two terms for which they've given me numerical values are 12 – 5 = 7 places apart, so, from the definition of a geometric sequence, I know that I'd get from the fifth term to the twelfth term by multiplying the fifth term by the common ratio seven times; that is, a 12 = (a 5)( r 7). Over time and only if it satisfies the mean value property the direction of ~v is same... Two or three dimensional space, quizzes, and personalized coaching to help you succeed ( a... Other geometrically obvious properties are ( … cal adj to approach a concept from funny... Derivative introduced as rate of change both as that of distance function and geometrically coaching! Multiplying ~a by ( 1 ) yields the previously de ned opposite of ~a arrangement is made up of such...... NRICH is part of the derivative can be geometrically picturised by the directed line segment functions! \ ) is an ordered list of numbers primary focus is math discussions and free math help science! Linearcombination ] of linear combinations of column matrices of column matrices clarification, or according to the or... 18 Challenge level: Area fomulas and a little trig will help derivative can be geometrically picturised by the line. Defined as an object containing both magnitude and direction methods or principles of geometry Project... 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### 10th
10 (ten numbers in both spoken and written language. The reason for the choice of ten is assumed to be that humans have ten fingers (digits).
## In common usage and derived terms
• A collection of ten items (most often ten years) is called a decade.
• The ordinal adjective is denary.
• Increasing a quantity by one order of magnitude is most widely understood to mean multiplying the quantity by ten.
• To reduce something by one-tenth is to decimate. (In ancient Rome, the killing of one in ten soldiers in a cohort was the punishment for cowardice or mutiny; or, one-tenth of the able-bodied men in a village as a form of retribution, thus causing a labor shortage and threat of starvation in agrarian societies.)
• With ten being the base of the decimal system, a scale of 1 to 10 is often used to rank things, as a smaller version of a 1-to-100 scale (as is used in percentages and wine-tasting). Hence, something that scores perfectly is "a perfect ten". A person who is attractive and physically flawless is often said to be "a ten", from the idea of ranking that person's appearance and sex-appeal on a 1-to-10 scale.[citation needed]
## In mathematics
Ten is a composite number, its proper divisors being 1, 2 and 5. Ten is the smallest noncototient, a number that cannot be expressed as the difference between any integer and the total number of coprimes below it.
Ten is the second discrete semiprime (2.5) and the second member of the (2.q) discrete semiprime family. Ten has an aliquot sum σ(n) of 8 and is accordingly the first discrete semiprime to be in deficit. All subsequent discrete semiprimes are in deficit. The aliquot sequence for 10 comprises five members (10,8,7,1,0) with this number being the second composite member of the 7-aliquot tree.
It is the aliquot sum of only one number the discrete semiprime 14.
Ten is a semi-meandric number.
Ten is the sum of the first three prime numbers, of the four first numbers (1 + 2 + 3 + 4), of the square of the two first odd numbers and also of the first four factorials (0! + 1! + 2! + 3!). Ten is the eighth Perrin number, preceded in the sequence by 5, 5, 7.
A polygon with ten sides is a decagon, and 10 is a decagonal number. But it is also a triangular number and a centered triangular number.
Ten is the number of n-Queens Problem solutions for n = 5.
Ten is the smallest number whose status as a possible friendly number is unknown.
### In numeral systems
#### Decimal system
Main article: Decimal
As is the case for any base in its system, ten is the first two-digit number in decimal and thus the lowest number where the position of a numeral affects its value. Any integer written in the decimal system can be multiplied by ten by adding a zero to the end (e.g. 855 * 10 = 8550).
#### Roman numerals
The Roman numeral for ten is X (which looks like two V's [the Roman numeral for 5] put together); it is thought that the V for five is derived from an open hand (five digits displayed). Incidentally, the Chinese word numeral for ten, is also a cross: .
#### Positional numeral systems other than decimal
The digit '1' followed by '0' is how the value of p is written in base p. (E.g. 16 in hexadecimal is 10.)
Representation of 10 in other bases
Base Numeral system Number
1 unary **********
2 binary 1010
3 ternary 101
4 quaternary 22
5 quinary 20
6 senary 14
7 septenary 13
8 octal 12
9 novenary 11
10 decimal 10
12 dozenal X
Properties:
### List of basic calculations
Multiplication 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 25 50 100 1000
$10 \times x$ 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 250 500 1000 10000
Division 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
$10 \div x$ 10 5 $3.\overline\left\{3\right\}$ 2.5 2 $1.\overline\left\{6\right\}$ $1.\overline\left\{428571\right\}$ 1.25 $1.\overline\left\{1\right\}$ 1 $0.\overline\left\{90\right\}$ $0.8\overline\left\{3\right\}$ $0.\overline\left\{769230\right\}$ $0.\overline\left\{714285\right\}$ $0.\overline\left\{6\right\}$
$x \div 10$ 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5
Exponentiation 1 2 3 4 5 6 7 8 9 10
$10 ^ x\,$ 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000 10000000000
$x ^ \left\{10\right\}\,$ 1 1024 59049 1048576 9765625 60466176 282475249 1073741824 3486784401 10000000000
## In science
The meaning "10" is part of the following terms:
• decapoda, an order of crustaceans with ten feet
• decane, a hydrocarbon with 10 carbon atoms
Also, the number 10 plays a role in the following:
### Astronomy
• The Sculptor.
• The solar eclipse series which began on -2467 February 28 and ended on -1169 April 18. The duration of Saros series 10 was 1298.1 years, and it contained 73 solar eclipses.
• The Saros lunar eclipse series which began on -2454 June 17 and ended on -1138 August 15. The duration of Saros series 10 was 1316.2 years, and it contained 74 lunar eclipses.
## In religion and philosophy
• In Pythagoreanism, the number 10 played an important role and was symbolized by the tetractys.
• In Hinduism, Lord Maha Vishnu appeared on the earth in 10 incarnations.
## In money
Most countries issue coins and bills with a denomination of 10 (See e.g. 10 dollar note). Of these, the dime, with the value of ten cents, or one tenth of a dollar, derives its name from the meaning "one-tenth" - see Dime (United States coin)#Denomination history and etymology.
## In sports and games
The meaning "10" is part of the following terms:
• decathlon, a combined event in athletics consisting of ten track and field events.
Also, the number 10 plays a role in the following:
• In American football, the end zones are 10 yards deep.
• In association football, the number 10 is traditionally worn by the team's playmaker. This use has led to "Number 10" becoming a synonym for that particular player, even if he or she does not wear that number.
• In Australian rules football, considered the break even amount of games won in a regular 22 game season.
• In baseball, 10 is the minimum number of players on the field at any given time during play (including the batter).
• The top of the rim (goal) is 10 feet from the floor.
• In standard full-court basketball, there are 10 players on the court (5 on each team).
• In blackjack, the Ten, Jack, Queen and King are all worth 10 points.
• In cricket, 10 is the number of wickets required to be taken by the bowling side for the batting side to be bowled out.
• In gridiron football, 10 is the number of yards the offense must advance to maintain possession in a single set of downs—four in American and three in Canadian.
• In most rugby league competitions, the number 10 is worn by one of the two starting props. (One exception to this rule is the European Super League, which uses static squad numbering.)
• In rugby union, the starting fly-half wears the 10 shirt.
• In ten-pin bowling, 10 pins are arranged in a triangular pattern and there are 10 frames per game.
• The highest score possible in Olympics gymnastics competitions.
• Driving a racing car at ten-tenths is driving as fast as possible, on the limit.
The jersey number 10 has been retired by several North American sports teams in honor of past playing greats or other key figures:
Ten is: | HuggingFaceTB/finemath | |
# LeetCode – Majority Element II (Java)
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space.
Java Solution
This problem is similar to Majority Element I. Time = O(n) and Space = O(1).
```public List<Integer> majorityElement(int[] nums) { List<Integer> result = new ArrayList<>(); Integer n1 = null, n2 = null; int c1 = 0, c2 = 0; for (int i : nums) { if (n1 != null && i == n1.intValue()) { c1++; } else if (n2 != null && i == n2.intValue()) { c2++; } else if (c1 == 0) { c1 = 1; n1 = i; } else if (c2 == 0) { c2 = 1; n2 = i; } else { c1--; c2--; } } c1 = c2 = 0; for (int i : nums) { if (i == n1.intValue()) { c1++; } else if (i == n2.intValue()) { c2++; } } if (c1 > nums.length / 3) result.add(n1); if (c2 > nums.length / 3) result.add(n2); return result; }```
Reference:
A Linear Time Majority Vote Algorithm
Category >> Algorithms >> Interview
If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:
```<pre><code>
String foo = "bar";
</code></pre>
```
• alexwest11
Given an integer array of size n, find all elements that appear more
———-
i am not sure it is solution for ALL. , just any element n/3+ repeated!!
for reference , see
https://www.geeksforgeeks.org/n3-repeated-number-array-o1-space/
• West
first sample is not space O(n) блять
• Dima
да это же просто пиздец… | HuggingFaceTB/finemath | |
# How do you integrate (2x-3)/(x^2-5x+6) using partial fractions?
Dec 3, 2016
#### Explanation:
Do partial fraction decomposition:
$\frac{2 x - 3}{\left(x - 2\right) \left(x - 3\right)} = \frac{A}{x - 2} + \frac{B}{x - 3}$
Multiply both sides by $\left(\left(x - 2\right) \left(x - 3\right)\right)$
$\left(2 x - 3\right) = A \left(x - 3\right) + B \left(x - 2\right)$
Solve for A by letting $x = 2$:
$\left(2 \left(2\right) - 3\right) = A \left(2 - 3\right)$
$- 1 = A$
Solve for B by letting $x = 3$:
$\left(2 \left(3\right) - 3\right) = B \left(3 - 2\right)$
3 = B
Check:
$- \frac{1}{x - 2} + \frac{3}{x - 3} =$
$- \frac{1}{x - 2} \frac{x - 3}{x - 3} + \frac{3}{x - 3} \frac{x - 2}{x - 2} =$
$\frac{- x + 3}{\left(x - 2\right) \left(x - 3\right)} + \frac{3 x - 6}{\left(x - 3\right) \left(x - 2\right)} =$
$\frac{2 x - 3}{\left(x - 3\right) \left(x - 2\right)}$
This checks.
$\int \frac{2 x - 3}{{x}^{2} - 5 x + 6} \mathrm{dx} = 3 \int \frac{1}{x - 3} \mathrm{dx} - \int \frac{1}{x - 2} \mathrm{dx}$
$\int \frac{2 x - 3}{{x}^{2} - 5 x + 6} \mathrm{dx} = 3 \ln | x - 3 | - \ln | x - 2 | + C$ | HuggingFaceTB/finemath | |
# Fibonacci numbers
Sneha Mallik
Last Updated: May 13, 2022
### Introduction
The fibonacci series is the number sequence in which the given number results from adding the two previous numbers. The terms in the Fibonacci sequence are as follows:
The individual numbers in this sequence are called Fibonacci numbers. The Fibonacci sequence is a fantastic mathematical concept found in various places, including seashell patterns and the Parthenon.
#### Properties of Fibonacci Numbers
FN + K = FK . FN + 1 + FK - 1 . FN
• Applying Addition Rule to the case, K = N
FN + N = FN . FN + 1 + FN - 1 . FN
F2N = FN ( FN + 1 + FN - 1)
• From the above property, we can prove that for any positive integer K, FNK is the multiple of F(By the Induction Hypothesis).
• The inverse of the above property is true since, if FM is multiple of FN, then M is also the multiple of N.
• Cassini’s Identity
FN - 1 . FN + 1 - FN2 = (-1)N
This identity was given by Giovanni Domenico Cassini, an Italian mathematician. In the mathematical expression, N is the variable and it can have values from 1…..N.
For eg., if ‘N’ is an odd number, i.e. 1, 3, 5,... then (-1)N+1 = +1 in every case. But if we take even value here, we will get the result as -1.
• GCD Identity
For M, N >= 1
GCD(FM , FN) = FGCD (M, N) , where M, N are integers
### The Golden Ratio Approach
The golden ratio is defined as the limit of the ratio of successive terms in the Fibonacci sequence.
The golden ratio is coming about 1.618, as it means that as ‘N’ becomes sufficiently large, the fibonacci sequence approaches or approximates a geometric sequence. So, starting at number 144, and if we multiply 144 by 1.618, we get 233 (Approx.), the next element in the sequence. If we now multiply 233 with 1.618, we get approximately 377. Hence this golden ratio helps to approximate the next number in the series easily.
The formula for golden ratio is,
For example, let us find out the 8th element in the sequence using the formula,
(1 + √5) - (1 - √5)8
F8 = ------------------------------ = 21
28 . √5
### Code:
Output:
The matrix multiplication in O(log N)
### Code of Fibonacci Number Matrix Multiplication in C++
Input:
Output:
#### Time Complexity
The time complexity depends on two factors:
• Matrix multiplication, K3
As the matrix moves ‘column by column’ or ‘row by row’, we will get a matrix K x K where ‘K’ represents the number of states. We can notice that in our code, the function ‘multiply’ has three nested for loops which takes time complexity = K x K x K = K3.
• Log(N)
For computing the large input, we raise the matrix to the power N, which takes log(N) time.
Hence, multiplying both the factors (1) and (2), we get => K3. log(N)
Therefore, the time complexity will be O(K3 log N).
#### Space Complexity
The space complexity is O(log N) since we raise the matrix to the Nth power. Hence it will take O(log N) additional space.
What is the golden ratio in the Fibonacci sequence?
From atoms to the massive stars in the sky has patterns. The golden ratio reveals predictable or observable patterns. This ratio is derived from the Fibonacci sequence, named after Leonardo Fibonacci, an Italian mathematician.
What is the time complexity of finding the Nth fibonacci term by recursion?
The time complexity by recursion is O(2N).
What is the time complexity of finding Nth fibonacci term with DP?
For finding the Nth fibonacci number, we got O(N) time complexity in the dynamic programming method.
Here, we store the previous terms, which are overlapping to reduce the time complexity.
### Key Takeaways
In this blog, we learned about Fibonacci numbers, their properties, the working of matrix multiplication in O(log N) time complexity, and the golden ratio approach.
Try problems related to the Fibonacci numbers here on CodeStudio to understand the working concept of Fibonacci numbers. To be more confident in data structures and algorithms, try out our DS and Algo Course.
CREDITS: GIPHY
By Sneha Mallik | HuggingFaceTB/finemath | |
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## Resolved Question
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# ¿cuanto equivale una pulgada, pie, yarda, milla ,vara y que operacion debo realizar?
by habdsome
Member since:
December 11, 2006
Total points:
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## Best Answer - Chosen by Voters
ya te dieron los valores ahora si los quieres convertir pulgadas a yardas etc... usa la regla de tres o mas simple una calculadora
100% 1 Vote
• 1 person rated this as good
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• Member since:
October 21, 2008
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668 (Level 2)
Esto pertenece al sistema Ingles de unidades escalares de longitud:
1 pie = 12 pulgadas = 0.3048 m.
1 yarda = 3 pies = 36 pulgadas = 0.9144 m.
1 milla = 1760 yardas = 5280 pies = 63360 pulgadas = 1609.344 m.
1 vara es una medida española antigua que equivale actualmente a 33 pulgadas o 0.8382 m.
Suerte.Bye.
• 7 people rated this as good
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December 30, 2008
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1 pulgada equivale a 2.54 cm
1 pie equivale a 12 pulgadas
te da lo mismo porque cualquier numero multiplicado por 1 da 1
el procedimiento depende de a que lo quieras comvertir si quieres combertir pie a cm o pie a yarda
• 2 people rated this as good
• by Romina C
Member since:
February 07, 2009
Total points:
112 (Level 1)
bueno...
la MILLA equivale a 1609.344 metros
el PIE equivale a 0.3048 metros
la YARDA equivale 0.9144 metros
el resto mmmm... no se!
### Source(s):
estudio de la secundaria y facultad | HuggingFaceTB/finemath | |
## Alert!
Hello, reader! If you intend to post a link to this blog on Twitter, be aware that for utterly mysterious reasons, Twitter thinks this blog is spam, and will prevent you from linking to it. Here's a workaround: change the .com in the address to .ca. I call it the "Maple Leaf Loophole." And thanks for sharing!
## Thursday, July 23, 2009
### A Lesson Plan Using a Virtual Manipulative
This post at dy/dan got me thinking about this thing from Utah State's Library of Virtual Manipulatives.
Pretty cool, right? A way to play with volume that avoids water fights. Love it. I used this in a remedial geometry class several years ago. It was fun for the kids for about five minutes. We merely used it to poke at the edges of our intuition. I didn't really know how to exploit it.
It raises a compelling question for teachers: there are some really good digital resources out there, but how do you best use them in a classroom to enhance learning? I'd like to use it this year when I teach Geometry, but I need to write an effective lesson around it.
The barest outline of a plan:
1. Playtime. Let kids slide the height thing and push the buttons, or be teacherbot and do what they instruct me to do. Solicit guesses for heights. Have kids verbalize why they think their guess is correct. Test to see how close they are.
2. Start talking about how you would calculate the new height. Go back to universal problem solving techniques that you should be hitting over and over again. What is the given information? What do you want to find? What stays the same? Encourage/coach them to do this with the rectangular prisms. They should be able to find numerical solutions easily. Develop and write on the board an equation involving equal volumes with an unknown height and solve it. Test to see if it works in the virtual manipulative. Have them calculate a few more.
3. Go through the same procedure with cylinders, then cones. It's going to look different depending on if they already know formulas, what age the kids are, what level, etc.
It wants for structure. I could develop a worksheet and break out the laptops. I'm not a huge fan of many worksheets, because I think they shift the focus from the problem-at-hand to "guess what to write in the blank." (I'm also not a fan of the laptops.) I could try to keep it as a large group discussion, but that could easily turn into me talking to 2-3 kids while everyone else zones out.
What would you do with this?
#### 9 comments:
1. Your last couple of statements make me curious... Can you explain some specifics about why you dislike using laptops?
I love the NLVM site and use it frequently throughout the year.
2. Sure RichTCS. I do use the laptops when they are the best solution. But if I can think of a way to achieve the same results, I avoid them.
I only have 43 minutes for a class period. Our laptops come in a big cart. I have to slice 5 minutes off the front and back of my class because checkout/in requires supervision. Then we have to wait for startup/login. Then at least 2, often more, laptops won't turn on or login, and those kids have to start the process over or share for the class period. It's all a big headache.
Besides all that, once the kids login successfully, my little lesson about comparing volumes has to compete with the whole Internet. Most kids are fairly cooperative when they have a well-defined task, but the temptation for them to check their email or their fantasy football for whatever is great. I do have software that can take control of or freeze their machines, but I have received no training on it, and it's complicated to figure out on the fly.
3. How exactly do you "use" the NLVM site? How do you organize a lesson?
4. As a particular case of number 2, direct and joint variation leaps out here, and in other volume problems. Specifically, the volume of the cylindrical column of water varies jointly as the height of the column and the square of the radius. To a decent first approximation, many real-world models are similar joint (and sometimes inverse) variations.
5. With this particular app I'd probably throw it on the projector and have the students make their predictions as a whole class; while there is value in having them try it individually I don't think this one beats the wasted-time vs. learning-accomplished ratio.
6. I use manipulatives like this one mainly to draw students into whatever lesson I've planned. There doesn't seem to be enough depth in this particular activity to justify laptop time. Whole group structure might work better here. To prevent the zoning out issue, I would try to have a lot of why and what if questions ready. Call on everyone. I like the idea of using the manipulative to check assumptions after crunching the numbers. I think you have just the right amount of structure in this lesson. Too much planning always backfires on me.
This is a great topic for discussion. Are digital manipulatives ever used outside of school? Some manipulatives, like this one, simulate real world objects. Others encourage exploration of parameters. Are there examples of manipulatives that permit creativity? I suppose pattern blocks and, to some extent, geoboards might work. Any others?
7. Your spirograph of course. Unless you don't call it a manipulative.
8. I might use this to intro what I call the r:r-squared:r-cubed theorem. That is, the relationship between the perimeters, areas and volumes of similar figures. I see what you mean about the time ... and the zoning. I would probably put this up on my smartboard and let student volunteers play around with it. Then after we learn the material, we could go back to it to see if the students could figure things out faster.
9. Thanks everyone who shared your perspective. It helps to have my instincts validated. I recommend checking out the other offerings from LVM. There's some nice stuff in there.
Hi! I will have to approve this before it shows up. Cuz yo those spammers are crafty like ice is cold. | HuggingFaceTB/finemath | |
# Math
The game of dots is played by rolling a fair die and
receiving \$1 for each dot showing on the top face of the
die. What cost should be set for each roll if the game is
to be considered a fair game?
1. 👍 0
2. 👎 0
3. 👁 94
1. Expected value = (1+2+3+4+5+6)/6 = 21/6 = 3.5
\$3.50 is the break-even "fair" cost per roll.
1. 👍 0
2. 👎 0
posted by drwls
2. Thanks, I did not know what the "fair die" meant.
1. 👍 0
2. 👎 0
posted by Lisa
3. A fair die is one that is not "loaded" or "shaved" to favor a particular side.
1. 👍 0
2. 👎 0
posted by drwls
## Similar Questions
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The game of dots is played by rolling a fair die and receiving \$1 for each dot showing on the top face of the die. What cost should be set for each roll if the game is to be considered a fair game?
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9. ### MATH Prob.
The game of dots is played by rolling a fair die and receiving \$1 for each dot showing on the top face of the die. What cost should be set for each roll if the game is to be considered a fair game?
asked by Twg on August 11, 2009
10. ### math fair
The game of dots is played by rolling a fair die and receiving \$1 for each dot showing on the top face of the die. What cost should be set for each roll if the game is to be considered a fair game?
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More Similar Questions | HuggingFaceTB/finemath | |
## Conversion formula
The conversion factor from centimeters to miles is 6.2137119223733E-6, which means that 1 centimeter is equal to 6.2137119223733E-6 miles:
1 cm = 6.2137119223733E-6 mi
To convert 266 centimeters into miles we have to multiply 266 by the conversion factor in order to get the length amount from centimeters to miles. We can also form a simple proportion to calculate the result:
1 cm → 6.2137119223733E-6 mi
266 cm → L(mi)
Solve the above proportion to obtain the length L in miles:
L(mi) = 266 cm × 6.2137119223733E-6 mi
L(mi) = 0.0016528473713513 mi
The final result is:
266 cm → 0.0016528473713513 mi
We conclude that 266 centimeters is equivalent to 0.0016528473713513 miles:
266 centimeters = 0.0016528473713513 miles
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 mile is equal to 605.01654135338 × 266 centimeters.
Another way is saying that 266 centimeters is equal to 1 ÷ 605.01654135338 miles.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred sixty-six centimeters is approximately zero point zero zero two miles:
266 cm ≅ 0.002 mi
An alternative is also that one mile is approximately six hundred five point zero one seven times two hundred sixty-six centimeters.
## Conversion table
### centimeters to miles chart
For quick reference purposes, below is the conversion table you can use to convert from centimeters to miles
centimeters (cm) miles (mi)
267 centimeters 0.002 miles
268 centimeters 0.002 miles
269 centimeters 0.002 miles
270 centimeters 0.002 miles
271 centimeters 0.002 miles
272 centimeters 0.002 miles
273 centimeters 0.002 miles
274 centimeters 0.002 miles
275 centimeters 0.002 miles
276 centimeters 0.002 miles | HuggingFaceTB/finemath | |
# How to Calculate the Distance to the Horizon Download Article
The distance to the horizon depends almost completely on the height above sea-level that the observer is looking from. Knowing this number is extremely useful and often necessary when navigating over the ocean or going on a hike, though just being curious is reason enough! There are also a few other factors that you might have to consider depending on where you are in the world and the time at which you are viewing, such as temperature and weather conditions. Once you have all the measures you need, you can quickly get the calculation and know the exact distance you are from the horizon.
### Method 1 of 3: Calculating Distance Using Geometry
1. 1
Measure your "height of eye." Measure the length between the ground and your eyes in meters or feet. One way to calculate this is to measure the distance between your eyes and the top of your head. Subtract this value from your total height and what will be left is the distance between your eyes and the surface you're standing on. If you are standing exactly at sea level, with the bottom of your feet level with the water, this is the only measurement you'll need.
2. 2
Add your "local elevation" if you're standing on a raised surface, such as a hill, building or boat. How many meters or feet above the true horizon are you standing? 1 meter? 4,000 feet? Add that number to your height of eye (in the same units, of course).
3. 3
Multiply by 13m if you took the measurement in meters, or multiply by 1.5ft if you took the measurement in feet.
4. 4
Take the square root to find the answer. If you used meters, your answer will be in kilometers, and if feet, the answer will be in miles. The distance calculated is a straight line from your eyes to the horizon.
• The actual distance you'll travel to get to the horizon will be longer because of surface curvature and (on land) irregularities. Proceed to the next method below for a more accurate (but complicated) formula.
5. 5
Understand how this calculation works. It's based on a triangle formed by your observation point (your eyes), the true horizon point (what you're looking at) and the center of the Earth.
• By knowing the radius of the Earth and measuring your height of eye and local elevation, that leaves only the distance between your eyes and the horizon as unknown. Since the sides of the triangle that meet at the horizon actually form a right angle, we can use the Pythagorean theorem (good old a2 + b2 = c2) as the basis for this calculation, where:
• a = R (the radius of the Earth)
• b = the distance to the horizon, unknown
• c = h (your height of eye) + R
### Method 2 of 3: Calculating Distance Using Trigonometry
1. 1
Calculate the actual distance you'd have to traverse to get to the horizon by using the following formula.
• d = R * arccos(R/(R + h)), where
• d = distance to horizon
• R = radius of the Earth
• h = height of eye
2. 2
Increase R by 20% to compensate for the distorting refraction of light rays and to arrive at a more accurate measurement. The geometric horizon calculated using the method in this article may not be the same as the optical horizon, which is what your eye actually sees. Why is this?
• The atmosphere bends (refracts) light that is traveling horizontally. What this usually means is that a ray of light is able to slightly follow the curvature of the earth, so that the optical horizon is a bit further away than the geometric horizon.
• Unfortunately the refraction due to the atmosphere is neither constant nor predictable, as it depends on the change of temperature with height. There is therefore no simple way to add a correction to the formula for the geometric horizon, though one may achieve an "average" correction by assuming a radius for the earth that is a bit greater than the true radius.
3. 3
Understand how this calculation works. This will calculate the length of the curved line that follows from your feet to the true horizon (shown in green in this image). Now, the arccos(R/(R+h)) portion refers to the angle that is made at the center of the Earth by the line going from the true horizon to the center and the line going from you to the center. With this angle, we multiply it by R to get the "arc length," which, in this case, is the distance that you are looking for.
### Method 3 of 3: Alternate Geometrical Calculation
1. 1
Assume a flat plane or the ocean. This method is a simpler version of the first set of instructions presented in this article, and applies only in feet and miles.
2. 2
Solve for the distance in miles by plugging in your height of eye in feet (h) into the formula. The formula you will be using is d = 1.2246* SQRT(h)
3. 3
Derive the formula from the Pythagorean theorem. (R + h)2 = R2 + d2. Solving for h (making the assumption that R>>h and expressing the radius of the earth in miles, approx. 3959) yields the expression:d = SQRT(2*R*h)
## Community Q&A
Search
• Question
Can I see an object beyond the horizon? Not the entire object, but you might be able to see part of it if it's tall enough. If something like a tree or a building is standing just beyond the horizon, you might be able to see the upper part of it.
• Question
Why wouldn't the formula for determining distance to the horizon work in a city or a forest? The trees in the forest or the buildings in the city would affect the view.
• Question
How many miles away is the horizon at an altitude of 30 feet above sea level? Belle Taine
It is about 89 miles. If your head is on the surface (you are treading water) the distance is 88.983 miles to the horizon. You have no chance of catching a passing boat if you are stranded overboard.
• Question
How many acres in 630 sq ft? Donagan
Because there are 43,560 square feet in an acre, 630 square feet equal 630 / 43,560 = 0.01446 acre.
• Question
Is there a method of calculating distance from a visible object in the sky? Yes. One modern way is by using radar, but a more traditional way is by using trigonometry and parallax. Measure out a straight baseline along the ground, and carefully measure the angle between this baseline and your distant object from each endpoint of your baseline. One then has two angles and their included side, so it is now possible to solve for the missing lengths: a = L * sin(A)/sin(A+B) where L is the length of your baseline, A and B are your two measured angles, and "a" at the distance along the side opposite angle A (i.e., between the point at angle B and the distant object).
• Question
The formula is wrong. Should be d=R*Sin(arcCos(R/R+h). You are correct that the formula seems to be missing a sin. Perhaps the authors intend to use the small angle approximation (sin x ~ x) which is good enough for this sort of calculation, but only works if your calculator is in radians mode when you do the arccos. To avoid misleading readers, they should have either explicitly specified radians, or used your formula where it doesn't matter how angles are measured.
• How do I derive the formula to calculate the distance to the horizon?
200 characters left
## Tips
• These calculations are most commonly used if you are looking at the true horizon, or where the sky and the Earth would meet if there were not any barriers or obstructions in your way (which is usually the case at sea, unless there's a land mass in the way). On land, however, there may be mountains or buildings in front of the true horizon, in which case these calculations will still tell you how far you are from the true horizon, but you'll have to tack on any additional distance created by having to climb over or circumvent the obstacles that are in your way.
Thanks!
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Updated: December 16, 2019
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Article SummaryX | HuggingFaceTB/finemath | |
# Answers
Solutions by everydaycalculation.com
## Compare 8/35 and 4/9
8/35 is smaller than 4/9
#### Steps for comparing fractions
1. Find the least common denominator or LCM of the two denominators:
LCM of 35 and 9 is 315
Next, find the equivalent fraction of both fractional numbers with denominator 315
2. For the 1st fraction, since 35 × 9 = 315,
8/35 = 8 × 9/35 × 9 = 72/315
3. Likewise, for the 2nd fraction, since 9 × 35 = 315,
4/9 = 4 × 35/9 × 35 = 140/315
4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction
5. 72/315 < 140/315 or 8/35 < 4/9
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time:
Android and iPhone/ iPad
#### Compare Fractions Calculator
and
© everydaycalculation.com | HuggingFaceTB/finemath | |
Let’s walk step-by-step on how to solve for each financial variable. If you do not get the correct answers to these, go back and check your work by using the [RCL] (Recall) key, for instance, hit [RCL][ N ] to find out what you have in your calculator for [ N ], and so on. If the number is not correct, just simply re-input the number that is incorrect and go on from there. We will start with solving for [PMT], as that is the easiest.
Important Note: It is very beneficial to use a piece of scratch paper when figuring out financial calculations. You will want to write at the top of the paper the 5 variables: N, I, PV, PMT and FV, as listed below. Then, write in each column the numbers that will be entered into each variable. Once that is done, go ahead and do the problem itself. Click here to read more about TMV key.
Now lets try solving for I (I/YR).
240 [N]
175000 [PV]
1349.59 [+/-] [PMT]
0 [FV]
N =
240
I =
? (empty)
PV =
\$175,000
PMT =
\$-1,349.59
FV =
0
To solve for I, press [ I/YR ] key and the answer you should get is 6.93%.
This example assumed that your number of periods in a year was set as 12 (the factory default value).
If you did not get the above answer, click on this link to learn how to set the number of period(s) in a year. | HuggingFaceTB/finemath | |
# Group action explained
In mathematics, a group action on a space is a group homomorphism of a given group into the group of transformations of the space. Similarly, a group action on a mathematical structure is a group homomorphism of a group into the automorphism group of the structure. It is said that the group acts on the space or structure. If a group acts on a structure, it will usually also act on objects built from that structure. For example, the group of Euclidean isometries acts on Euclidean space and also on the figures drawn in it. In particular, it acts on the set of all triangles. Similarly, the group of symmetries of a polyhedron acts on the vertices, the edges, and the faces of the polyhedron.
A group action on a vector space is called a representation of the group. In the case of a finite-dimensional vector space, it allows one to identify many groups with subgroups of, the group of the invertible matrices of dimension over a field .
The symmetric group acts on any set with elements by permuting the elements of the set. Although the group of all permutations of a set depends formally on the set, the concept of group action allows one to consider a single group for studying the permutations of all sets with the same cardinality.
## Definition
### Left group action
If is a group with identity element, and is a set, then a (left) group action of on is a function
\alpha\colonG x X\toX,
that satisfies the following two axioms:
Identity: \alpha(e,x)=x Compatibility: \alpha\left(g,\alpha\left(h,x\right)\right)=\alpha\left(gh,x\right)
(with often shortened to or when the action being considered is clear from context):
Identity: e ⋅ x=x Compatibility: g ⋅ (h ⋅ x)=(gh) ⋅ x
for all and in and all in .
The group is said to act on (from the left). A set together with an action of is called a (left) -set.
From these two axioms, it follows that for any fixed in, the function from to itself which maps to is a bijection, with inverse bijection the corresponding map for . Therefore, one may equivalently define a group action of on as a group homomorphism from into the symmetric group of all bijections from to itself.
### Right group action
Likewise, a right group action of on is a function
\alpha\colonX x G\toX,
that satisfies the analogous axioms:
Identity: \alpha(x,e)=x Compatibility: \alpha\left(\alpha\left(x,g\right),h\right)=\alpha\left(x,gh\right)
(with often shortened to or when the action being considered is clear from context)
Identity: x ⋅ e=x Compatibility: (x ⋅ g) ⋅ h=x ⋅ (gh)
for all and in and all in .
The difference between left and right actions is in the order in which a product acts on . For a left action, acts first, followed by second. For a right action, acts first, followed by second. Because of the formula, a left action can be constructed from a right action by composing with the inverse operation of the group. Also, a right action of a group on can be considered as a left action of its opposite group on .
Thus, for establishing general properties of group actions, it suffices to consider only left actions. However, there are cases where this is not possible. For example, the multiplication of a group induces both a left action and a right action on the group itself—multiplication on the left and on the right, respectively.
## Types of actions
The action of G on X is called:
• if X is non-empty and if for each pair x, y in X there exists a g in G such that . For example, the action of the symmetric group of X is transitive (unless X is empty), the action of the general linear group or the special linear group of a vector space V on is transitive, but the action of the orthogonal group of a Euclidean space E is not transitive on (it is transitive on the unit sphere of E, though).
• (or ) if for every two distinct g, h in G there exists an x in X such that ; or equivalently, if for each in G there exists an x in X such that . In other words, in a faithful group action, different elements of G induce different permutations of X. In algebraic terms, a group G acts faithfully on X if and only if the corresponding homomorphism to the symmetric group,, has a trivial kernel. Thus, for a faithful action, G embeds into a permutation group on X; specifically, G is isomorphic to its image in Sym(X). If G does not act faithfully on X, we can easily modify the group to obtain a faithful action. If we define, then N is a normal subgroup of G; indeed, it is the kernel of the homomorphism . The factor group G/N acts faithfully on X by setting . The original action of G on X is faithful if and only if . The smallest set on which a faithful action can be defined can vary greatly for groups of the same size. For example:
• Three groups of size 120 are the symmetric group S5, the icosahedral group, and the cyclic group
Z/120Z
. The smallest sets on which faithful actions can be defined are of size 5, 12, and 16 respectively.
Z/2nZ
as well as
(Z/2Z)n
(the direct product of n copies of
Z/2Z
), but the latter acts faithfully on a set of size 2n, whereas the former cannot act faithfully on a set smaller than itself.
• (or semiregular or fixed-point free) if, given g, h in G, the existence of an x in X with implies . Equivalently: if g is a group element and there exists an x in X with (that is, if g has at least one fixed point), then g is the identity. Note that a free action on a non-empty set is faithful.
• (or or sharply transitive) if it is both transitive and free; this is equivalent to saying that for every two x, y in the non-empty set X there exists precisely one g in G such that . In this case, X is called a principal homogeneous space for G or a G-torsor. The action of any group G on itself by left multiplication is regular, and thus faithful as well. Every group can, therefore, be embedded in the symmetric group on its own elements, Sym(G). This result is known as Cayley's theorem.
• if X has at least n elements, and for all distinct x1, ..., xn and all distinct y1, ..., yn, there is a g in G such that for . A 2-transitive action is also called , a 3-transitive action is also called triply transitive, and so on. Such actions define interesting classes of subgroups in the symmetric groups: 2-transitive groups and more generally multiply transitive groups. The action of its symmetric group on a set with n elements is always n-transitive; the action of its alternating group is (n − 2)-transitive.
• if there is exactly one such g.
• if it is transitive and preserves no non-trivial partition of X. See primitive permutation group for details.
• Locally free if G is a topological group, and there is a neighborhood U of e in G such that the restriction of the action to U is free; that is, if for some x and some g in U then .
Furthermore, if G acts on a topological space X, then the action is:
• Wandering if every point x in X has a neighborhood U such that
\{g\inG:gU\capU\emptyset\}
is finite. For example, the action of
Zn
on
Rn
by translations is wandering. The action of the modular group on the Poincaré half-plane is also wandering.
• Properly discontinuous if X is a locally compact space and for every compact subset K ⊂ X the set
\{g\inG:gK\capK\emptyset\}
is finite. The wandering actions given above are also properly discontinuous. On the other hand, the action of
Z
on
R2\setminus\{(0,0)\}
given by
n(x,y)=(2nx,2-ny)
is wandering and free but not properly discontinuous.
• if G is a topological group and the map from
G x XX x X:(g,x)\mapsto(gx,x)
is proper. If G is discrete then properness is equivalent to proper discontinuity for G-actions.
• Said to have discrete orbits if the orbit of each x in X under the action of G is discrete in X.
• A covering space action if every point x in X has a neighborhood U such that
\{g\inG:gU\capU\emptyset\}=\{e\}
.
If X is a non-zero module over a ring R and the action of G is R-linear then it is said to be
• Irreducible if there is no nonzero proper invariant submodule.
Orbits and stabilizers
Consider a group G acting on a set X. The of an element x in X is the set of elements in X to which x can be moved by the elements of G. The orbit of x is denoted by
Gx
:$G \cdot x = \.$
The defining properties of a group guarantee that the set of orbits of (points x in) X under the action of G form a partition of X. The associated equivalence relation is defined by saying
x\simy
if and only if there exists a g in G with
gx=y.
The orbits are then the equivalence classes under this relation; two elements x and y are equivalent if and only if their orbits are the same, that is,
Gx=Gy.
The group action is transitive if and only if it has exactly one orbit, that is, if there exists x in X with
Gx=X.
This is the case if and only if
Gx=X
for x in X (given that X is non-empty).
The set of all orbits of X under the action of G is written as X/G (or, less frequently: G\X), and is called the of the action. In geometric situations it may be called the , while in algebraic situations it may be called the space of , and written
XG,
by contrast with the invariants (fixed points), denoted XG: the coinvariants are a while the invariants are a . The coinvariant terminology and notation are used particularly in group cohomology and group homology, which use the same superscript/subscript convention.
### Invariant subsets
If Y is a subset of X, then
GY
denotes the set
\{gy:g\inGandy\inY\}.
The subset Y is said to be invariant under G if
GY=Y
(which is equivalent to
GY\subseteqY
). In that case, G also operates on Y by restricting the action to Y. The subset Y is called fixed under G if
gy=y
for all g in G and all y in Y. Every subset that is fixed under G is also invariant under G, but not conversely.
Every orbit is an invariant subset of X on which G acts transitively. Conversely, any invariant subset of X is a union of orbits. The action of G on X is transitive if and only if all elements are equivalent, meaning that there is only one orbit.
A G-invariant element of X is
x\inX
such that
gx=x
for all
g\inG.
The set of all such x is denoted
XG
and called the G-invariants of X. When X is a G-module, XG is the zeroth cohomology group of G with coefficients in X, and the higher cohomology groups are the derived functors of the functor of G-invariants.
### Fixed points and stabilizer subgroups
Given g in G and x in X with
gx=x,
it is said that "x is a fixed point of g" or that "g fixes x". For every x in X, the of G with respect to x (also called the isotropy group or little group) is the set of all elements in G that fix x:$G_x = \.$This is a subgroup of G, though typically not a normal one. The action of G on X is free if and only if all stabilizers are trivial. The kernel N of the homomorphism with the symmetric group,
G\to\operatorname{Sym}(X),
is given by the intersection of the stabilizers Gx for all x in X. If N is trivial, the action is said to be faithful (or effective).
Let x and y be two elements in X, and let
g
be a group element such that
y=gx.
Then the two stabilizer groups
Gx
and
Gy
are related by
Gy=gGxg-1.
Proof: by definition,
h\inGy
if and only if
h(gx)=gx.
Applying
g-1
to both sides of this equality yields
\left(g-1hg\right)x=x;
that is,
g-1hg\inGx.
An opposite inclusion follows similarly by taking
h\inGx
and supposing
x=g-1y.
The above says that the stabilizers of elements in the same orbit are conjugate to each other. Thus, to each orbit, we can associate a conjugacy class of a subgroup of G (that is, the set of all conjugates of the subgroup). Let
(H)
denote the conjugacy class of H. Then the orbit O has type
(H)
if the stabilizer
Gx
of some/any x in O belongs to
(H)
. A maximal orbit type is often called a principal orbit type.
### and Burnside's lemma
Orbits and stabilizers are closely related. For a fixed x in X, consider the map
f:G\toX
given by
g\mapstogx.
By definition the image
f(G)
of this map is the orbit
Gx.
The condition for two elements to have the same image is$f(g)=f(h) \iff g\cdot x=h \cdot x \iff g^h \cdot x=x \iff g^h \in G_x \iff h \in gG_x.$In other words,
f(g)=f(h)
if and only if
g
and
h
lie in the same coset for the stabilizer subgroup
Gx
. Thus, the fiber
f-1(\{y\})
of f over any y in G·x is contained in such a coset, and every such coset also occurs as a fiber. Therefore f induces a between the set
G/Gx
of cosets for the stabilizer subgroup and the orbit
Gx,
which sends
gGx\mapstogx
. This result is known as the orbit-stabilizer theorem.
If G is finite then the orbit-stabilizer theorem, together with Lagrange's theorem, gives$|G \cdot x| = [G\,:\,G_x] = |G| / |G_x|,$in other words the length of the orbit of x times the order of its stabilizer is the order of the group. In particular that implies that the orbit length is a divisor of the group order.
Example: Let G be a group of prime order p acting on a set X with k elements. Since each orbit has either 1 or p elements, there are at least
k\bmodp
orbits of length 1 which are G-invariant elements.
This result is especially useful since it can be employed for counting arguments (typically in situations where X is finite as well).
Example: We can use the orbit-stabilizer theorem to count the automorphisms of a graph. Consider the cubical graph as pictured, and let G denote its automorphism group. Then G acts on the set of vertices, and this action is transitive as can be seen by composing rotations about the center of the cube. Thus, by the orbit-stabilizer theorem,
|G|=|G ⋅ 1||G1|=8|G1|.
Applying the theorem now to the stabilizer
G1,
we can obtain
|G1|=|(G1) ⋅ 2||(G1)2|.
Any element of G that fixes 1 must send 2 to either 2, 4, or 5. As an example of such automorphisms consider the rotation around the diagonal axis through 1 and 7 by
2\pi/3
which permutes 2,4,5 and 3,6,8, and fixes 1 and 7. Thus,
\left|(G1) ⋅ 2\right|=3.
Applying the theorem a third time gives
|\left(G1\right)2|=|\left(\left(G1\right)2\right) ⋅ 3||\left(\left(G1\right)2\right)3|.
Any element of G that fixes 1 and 2 must send 3 to either 3 or 6. Reflecting the cube at the plane through 1,2,7 and 8 is such an automorphism sending 3 to 6, thus
\left|\left(\left(G1\right)2\right) ⋅ 3\right|=2
. One also sees that
\left(\left(G1\right)2\right)3
consists only of the identity automorphism, as any element of G fixing 1, 2 and 3 must also fix all other vertices, since they are determined by their adjacency to 1, 2 and 3. Combining the preceding calculations, we can now obtain
|G|=8 ⋅ 3 ⋅ 2 ⋅ 1=48.
A result closely related to the orbit-stabilizer theorem is Burnside's lemma:$|X/G|=\frac$
\sum_ |X^g|,
where Xg is the set of points fixed by g. This result is mainly of use when G and X are finite, when it can be interpreted as follows: the number of orbits is equal to the average number of points fixed per group element.
Fixing a group G, the set of formal differences of finite G-sets forms a ring called the Burnside ring of G, where addition corresponds to disjoint union, and multiplication to Cartesian product.
## Examples
• The action of any group G on any set X is defined by for all g in G and all x in X; that is, every group element induces the identity permutation on X.
• In every group G, left multiplication is an action of G on G: for all g, x in G. This action is free and transitive (regular), and forms the basis of a rapid proof of Cayley's theorem - that every group is isomorphic to a subgroup of the symmetric group of permutations of the set G.
• In every group G with subgroup H, left multiplication is an action of G on the set of cosets G/H: for all g,a in G. In particular if H contains no nontrivial normal subgroups of G this induces an isomorphism from G to a subgroup of the permutation group of degree [G : H].
• In every group G, conjugation is an action of G on G: . An exponential notation is commonly used for the right-action variant: ; it satisfies (.
• In every group G with subgroup H, conjugation is an action of G on conjugates of H: for all g in G and K conjugates of H.
• The symmetric group Sn and its subgroups act on the set by permuting its elements
• The symmetry group of a polyhedron acts on the set of vertices of that polyhedron. It also acts on the set of faces or the set of edges of the polyhedron.
• The symmetry group of any geometrical object acts on the set of points of that object.
• The automorphism group of a vector space (or graph, or group, or ring . . .) acts on the vector space (or set of vertices of the graph, or group, or ring . . .).
• The general linear group and its subgroups, particularly its Lie subgroups (including the special linear group, orthogonal group, special orthogonal group, and symplectic group) are Lie groups that act on the vector space Kn. The group operations are given by multiplying the matrices from the groups with the vectors from Kn.
• The general linear group acts on Zn by natural matrix action. The orbits of its action are classified by the greatest common divisor of coordinates of the vector in Zn.
• The affine group acts transitively on the points of an affine space, and the subgroup V of the affine group (that is, a vector space) has transitive and free (that is, regular) action on these points; indeed this can be used to give a definition of an affine space.
• The projective linear group and its subgroups, particularly its Lie subgroups, which are Lie groups that act on the projective space Pn(K). This is a quotient of the action of the general linear group on projective space. Particularly notable is, the symmetries of the projective line, which is sharply 3-transitive, preserving the cross ratio; the Möbius group is of particular interest.
• The isometries of the plane act on the set of 2D images and patterns, such as wallpaper patterns. The definition can be made more precise by specifying what is meant by image or pattern, for example, a function of position with values in a set of colors. Isometries are in fact one example of affine group (action).
• The sets acted on by a group G comprise the category of G-sets in which the objects are G-sets and the morphisms are G-set homomorphisms: functions such that for every g in G.
• The Galois group of a field extension L/K acts on the field L but has only a trivial action on elements of the subfield K. Subgroups of Gal(L/K) correspond to subfields of L that contain K, that is, intermediate field extensions between L and K.
• The additive group of the real numbers acts on the phase space of "well-behaved" systems in classical mechanics (and in more general dynamical systems) by time translation: if t is in R and x is in the phase space, then x describes a state of the system, and is defined to be the state of the system t seconds later if t is positive or -t seconds ago if t is negative.
• The additive group of the real numbers acts on the set of real functions of a real variable in various ways, with (tf)(x) equal to, for example,,,,,, or, but not .
• Given a group action of G on X, we can define an induced action of G on the power set of X, by setting for every subset U of X and every g in G. This is useful, for instance, in studying the action of the large Mathieu group on a 24-set and in studying symmetry in certain models of finite geometries.
• The quaternions with norm 1 (the versors), as a multiplicative group, act on R3: for any such quaternion, the mapping is a counterclockwise rotation through an angle α about an axis given by a unit vector v; z is the same rotation; see quaternions and spatial rotation. Note that this is not a faithful action because the quaternion −1 leaves all points where they were, as does the quaternion 1.
• Given left G-sets
X,Y
, there is a left G-set
YX
whose elements are G-equivariant maps
\alpha:X x G\toY
, and with left G-action given by
g\alpha=\alpha\circ(idX x -g)
(where "
-g
" indicates right multiplication by
g
). This G-set has the property that its fixed points correspond to equivariant maps
X\toY
; more generally, it is an exponential object in the category of G-sets.
## Group actions and groupoids
The notion of group action can be put in a broader context by using the action groupoid
G'=G\ltimesX
associated to the group action, thus allowing techniques from groupoid theory such as presentations and fibrations. Further the stabilizers of the action are the vertex groups, and the orbits of the action are the components, of the action groupoid. For more details, see the book Topology and groupoids referenced below.
This action groupoid comes with a morphism p: G′G which is a covering morphism of groupoids. This allows a relation between such morphisms and covering maps in topology.
## Morphisms and isomorphisms between G-sets
If X and Y are two G-sets, a morphism from X to Y is a function such that for all g in G and all x in X. Morphisms of G-sets are also called equivariant maps or G-maps.
The composition of two morphisms is again a morphism. If a morphism f is bijective, then its inverse is also a morphism. In this case f is called an isomorphism, and the two G-sets X and Y are called isomorphic; for all practical purposes, isomorphic G-sets are indistinguishable.
Some example isomorphisms:
• Every regular G action is isomorphic to the action of G on G given by left multiplication.
• Every free G action is isomorphic to, where S is some set and G acts on by left multiplication on the first coordinate. (S can be taken to be the set of orbits X/G.)
• Every transitive G action is isomorphic to left multiplication by G on the set of left cosets of some subgroup H of G. (H can be taken to be the stabilizer group of any element of the original G-set.)
With this notion of morphism, the collection of all G-sets forms a category; this category is a Grothendieck topos (in fact, assuming a classical metalogic, this topos will even be Boolean).
## Continuous group actions
See main article: Continuous group action.
One often considers continuous group actions: the group G is a topological group, X is a topological space, and the map is continuous with respect to the product topology of . The space X is also called a G-space in this case. This is indeed a generalization, since every group can be considered a topological group by using the discrete topology. All the concepts introduced above still work in this context, however we define morphisms between G-spaces to be continuous maps compatible with the action of G. The quotient X/G inherits the quotient topology from X, and is called the quotient space of the action. The above statements about isomorphisms for regular, free and transitive actions are no longer valid for continuous group actions.
If X is a regular covering space of another topological space Y, then the action of the deck transformation group on X is properly discontinuous as well as being free. Every free, properly discontinuous action of a group G on a path-connected topological space X arises in this manner: the quotient map is a regular covering map, and the deck transformation group is the given action of G on X. Furthermore, if X is simply connected, the fundamental group of X/G will be isomorphic to G.
These results have been generalized in the book Topology and Groupoids referenced below to obtain the fundamental groupoid of the orbit space of a discontinuous action of a discrete group on a Hausdorff space, as, under reasonable local conditions, the orbit groupoid of the fundamental groupoid of the space. This allows calculations such as the fundamental group of the symmetric square of a space X, namely the orbit space of the product of X with itself under the twist action of the cyclic group of order 2 sending to .
An action of a group G on a locally compact space X is cocompact if there exists a compact subset A of X such that . For a properly discontinuous action, cocompactness is equivalent to compactness of the quotient space X/G.
The action of G on X is said to be proper if the mapping that sends is a proper map.
### Strongly continuous group action and smooth points
A group action of a topological group G on a topological space X is said to be strongly continuous if for all x in X, the map is continuous with respect to the respective topologies. Such an action induces an action on the space of continuous functions on X by defining for every g in G, f a continuous function on X, and x in X. Note that, while every continuous group action is strongly continuous, the converse is not in general true.
If G is a Lie group and X a differentiable manifold, then the subspace of smooth points for the action is the subspace of X of points x such that is smooth, that is, it is continuous and all its derivatives are continuous.
## Variants and generalizations
We can also consider actions of monoids on sets, by using the same two axioms as above. This does not define bijective maps and equivalence relations however. See semigroup action.
Instead of actions on sets, we can define actions of groups and monoids on objects of an arbitrary category: start with an object X of some category, and then define an action on X as a monoid homomorphism into the monoid of endomorphisms of X. If X has an underlying set, then all definitions and facts stated above can be carried over. For example, if we take the category of vector spaces, we obtain group representations in this fashion.
We can view a group G as a category with a single object in which every morphism is invertible. A (left) group action is then nothing but a (covariant) functor from G to the category of sets, and a group representation is a functor from G to the category of vector spaces. A morphism between G-sets is then a natural transformation between the group action functors. In analogy, an action of a groupoid is a functor from the groupoid to the category of sets or to some other category.
In addition to continuous actions of topological groups on topological spaces, one also often considers smooth actions of Lie groups on smooth manifolds, regular actions of algebraic groups on algebraic varieties, and actions of group schemes on schemes. All of these are examples of group objects acting on objects of their respective category.
## References
• Book: Aschbacher. Michael. Michael Aschbacher. Finite Group Theory. Cambridge University Press. 2000. 1777008 . 978-0-521-78675-1.
• Brown, Ronald (2006). Topology and groupoids, Booksurge PLC, .
• Categories and groupoids, P.J. Higgins, downloadable reprint of van Nostrand Notes in Mathematics, 1971, which deal with applications of groupoids in group theory and topology.
• Book: Dummit , David . Richard Foote. 2004. Abstract Algebra. 3rd. Wiley. 0-471-43334-9.
• Book: Eie . Minking . Chang . Shou-Te . A Course on Abstract Algebra . 2010 . World Scientific . 978-981-4271-88-2.
• Book: Rotman , Joseph . 1995. An Introduction to the Theory of Groups. Graduate Texts in Mathematics 148. 4th. Springer-Verlag. 0-387-94285-8.
• Book: Smith . Jonathan D.H. . Introduction to abstract algebra . Textbooks in mathematics . 2008 . CRC Press . 978-1-4200-6371-4.
## Notes and References
1. Book: Eie & Chang . [{{Google books|plainurl=y|id=jozIZ0qrkk8C|page=144|text=group action}} A Course on Abstract Algebra]. 2010. 144.
2. This is done, for example, by Book: Smith . [{{Google books|plainurl=y|id=PQUAQh04lrUC|page=253|text=group action}} Introduction to abstract algebra]. 2008. 253.
3. Web site: Definition:Right Group Action Axioms . Proof Wiki . 19 December 2021.
4. Book: Hatcher . Allen . Algebraic Topology . 2002 . Cambridge University Press . 0-521-79540-0. 72 .
5. Book: Procesi. Claudio. Lie Groups: An Approach through Invariants and Representations. 2007. Springer Science & Business Media. 9780387289298. 5. 23 February 2017. en.
6. M. Artin, Algebra, Proposition 6.4 on p. 179
7. Book: Eie & Chang . [{{Google books|plainurl=y|id=jozIZ0qrkk8C|page=144|text=trivial action}} A Course on Abstract Algebra]. 2010. 145.
8. Book: Reid, Miles. Geometry and topology. Cambridge University Press. 2005. 9780521613255. Cambridge, UK New York. 170.
9. Web site: Yuan. Qiaochu. wiki's definition of "strongly continuous group action" wrong?. Mathematics Stack Exchange. 27 February 2013. 1 April 2013. | HuggingFaceTB/finemath | |
# Homework Help: Calculating Time, Where Did I Go Wrong?
1. Feb 14, 2012
### NafeesR
1. The problem statement, all variables and given/known data
A child accidentally drops her doll from a bridge suspended 50m above a fast moving river. Assuming the doll never reaches terminal velocity, how many seconds will it take for the doll to hit the water?
2. Relevant equations
x=x0+ v0t + 0.5*at2
3. The attempt at a solution
Using acceleration formula: x=x0+ v0t + 0.5*at2
It should be 0=(50M)+(0m/s)t+0.5*(-9.8m/s2)
Which would give me 3.194382825 seconds. The textbook's answer is 2.3s. So after a bit of exploration I realized that the acceleration isn't supposed to be halved. That brings me to the question why isn't the acceleration halved like the example and other questions using the formula?
Last edited: Feb 14, 2012
2. Feb 14, 2012 | HuggingFaceTB/finemath | |
Describe a method you can use to determine the angle of intersection of a line and a plane. We have four points which we know its coordinates. This is the currently selected item. I'm dipping my feet at Blender SDK, and I'm trying to calculate intersection between two planes: Created a default plane in center, duplicated, rotated second, scaled first, applied transforms; but I'm failing for apparently no reason. Find a vector equation of the line of intersection of these three planes. Consider the plane with equation 4x 2y z = 1 and the line given by the parametric equations . The angle between the line and the plane can be calculated by the cross product of the line vector with the vector representation of the plane which is perpendicular to the plane: v = 4i + k Calculus Calculus: Early Transcendental Functions Intersection of a Plane and a Line In Exercises 83-86, find the point(s) of intersection (if any) of the plane and the line. 5. Here are cartoon sketches of each part of this problem. I figured I need to find plane/line intersection formula. c) Substituting gives 2(t) + (4 + 2t) â 4(t) = 4 â4 = 4. â all values of t satisfy this equation. In Euclidean geometry, the intersection of a line and a line can be the empty set, a point, or a line.Distinguishing these cases and finding the intersection point have use, for example, in computer graphics, motion planning, and collision detection.. To find these points you simply have to equate the equations of the two lines, where they equal eachother must be the points of intersection. I mean, a plane like "P: 4x - 2y + 2z = 5" is just not the way it works in C#. Equation of a plane. Example: find the intersection points of the sphere ( ⦠Antipodal points. Or they do not intersect cause they are parallel. There are no points of intersection. The angle θ between a line and a plane is the complement of the angle between the line and the normal to the plane. a plane that is defined by 3 locations Q0, Q1, Q2. This gives a bigger system of linear equations to be solved. Imagine you got two planes in space. And how do I find out if my planes intersect? Using the line equation. In this example these are landmarks. To do this, you need to enter the coordinates of the first and second points in the corresponding fields. To find the ⦠To write the equation of this plane, use the normal vector components: Calculate intersection point. For the mathematics for the intersection point(s) of a line (or line segment) and a sphere see this. However, a plane is something close to a line. Then I create a plane with the coordinates 0 0 0 0, and check if the line interesects with it. It means that two or more than two lines meet at a point or points, we call those point/points intersection point/points. The angle between a line and a plane. I show you how you can find the equation of the line where two planes intersect. This expression factorises to ⦠A plane is a two-dimensional surface and like a line, it extends up to infinity. and equation of the plane A x + B y + C z + D = 0,. then the angle between this line and plane can be found using this formula I also have the points eye and target for the camera. The plane equation can be found in the next ways: If coordinates of three points A(x 1, y 1, z 1), B(x 2, y 2, z 2) and C(x 3, y 3, z 3) lying on a plane are defined then the plane equation can be found using the following ⦠They may either intersect, then their intersection is a line. The cursor should change in a square. This note will illustrate the algorithm for finding the intersection of a line and a plane using two possible formulations for a plane. Intersect( , ) creates the intersection line of two planes ; Intersect( , ) creates the polygon(s) intersection of a plane and a polyhedron. Or you can check if a certain Point lies on the Plane or not. Example . Let this point be the intersection of the intersection line and the xy coordinate plane. The same concept is of a line-plane intersection. We now move on to defining how to calculate the angle between a line and a plane. For this example this would mean x 2 +8x-1=3x-7. Therefore, the intersection point must satisfy this. The intersection point between the line and the plane can be calculated from P(1) = P(0) + s*u Pipeline Script 1 Given: 2 locations P0, P1 which define the line segment. \$\begingroup\$ An intersection between a Vector3 and a Plane doesn't make sense. 2 Intersection with a Line Let us nd the points of intersection with the cone boundary Q(X) = 0, where Qis de ned by Equation (3). find the intersection of the two. You can find the intersection between a Plane and a line segment, a ray, or a line, but all of these require not one, but two Vector3's to be represented. The plane equation is N.P = -D for all points on the plane. the x ⢠y-plane), we substitute z = 0 to the equation of the ellipsoid, and thus the intersection curve satisfies the equation x 2 a 2 + y 2 b 2 = 1 , which an ellipse. x = 3 2 y = (2k 1) + z = 1 + k. IB Questionbank Mathematics Higher Level 3rd edition 5 . A calculator for calculating line formulas on a plane can calculate: a straight line formula, a line slope, a point of intersection with the Y axis, a parallel line formula and a perpendicular line formula. 3D ray tracing part 1. Usually, we talk about the line-line intersection. 3d line in a 3d plane. If in space given the direction vector of line L. s = {l; m; n}. Note that when we refer to the plane and the line, in this case, we are actually referring to the angle between the normal to the plane and the straight line. is cut with the plane z = 0 (i.e. In addition to being the vector of the line of intersection, it is the normal vector for the plane that must contain the given point, #(x_0,y_0,z_0)# and the point on the line, #(x_1,y_1,z_1)#, that is orthogonal to the given point. This will be clear to you when you take a ⦠And from then this is a simple case of solving the quadratic. A line that passes through the center of a sphere has two intersection points, these are called antipodal points. The Intersection is stored as the signal ⦠and the plane . In three-dimensional Euclidean geometry, if two lines are not in the same plane they are called skew lines and have no point of intersection. By equalizing plane equations, you can calculate what's the case. I have the origin point, x vector and y vector for a plane (actually a Sketch in this case) - so I can also easily calculate the normal. and is parallel to the lines: Transform the equation of the line, r, into another equation determined by the intersection of two planes , and these together with the equation of the plane form a system whose solution is the ⦠Let alone something like this: Translating this stuff to code gives me a headache. Therefore, by plugging z = 0 into P 1 and P 2 we get, so, the line of intersection is Practice: Triangle intersection in 3D. Suppose a line $$\displaystyle \,L$$ intersects a plane at point $$\displaystyle \,P.$$ Define what is meant by the "angle of intersection of the line and the plane". and let's assume we can create plane with these points. what is the intersection of plane $\mathcal{p}$ and line find an equation of the plane, and one of heres a python example which finds the intersection of a line and a plane. Also, determine whether the line lies in the plane⦠P (a) line intersects the plane in Intersection of plane and line.. If our point P is defined by the line equation P = P0 + tQ (where Q is the line's direction and t is the distance along the line) we can sub this in: N.(P0 + tQ) = -D The dot product is bilinear: t(N.Q) + (N.P0) = -D ⦠Then, coordinates of the point of intersection (x, y, 0) must satisfy equations of the given planes. Practice: Ray intersection with plane. Collecting like terms leads to x 2 +5x+6=0. It always will unless it's pointing upward, which is not possible. Theory. Practice: Ray intersection with line. Pick first the two endpoints of the line, after that the 3 endpoints of the lines defining the plane. How would an AI self awareness kill switch work? Plane and line intersection calculator. This lesson conceptually breaks down the above meaning and helps you learn how to calculate the distance in Vector form as well as Cartesian form, aided with a solved example at the end. Planes through a sphere. The shortest distance from a point to a plane is actually the length of the perpendicular dropped from the point to touch the plane. If the line has direction vector u and the normal to the plane is a, then . Solution 1 The equation of a plane (points P are on the plane with normal N and point P3 on the plane) can be written as. The coe cients are ⦠There are a lot of resources out there which explain how to find a plane-line intersection but all of them use non programming compatible algebra. Given that the line is perpendicular to the plane, find The Angle between a Line and a Plane. Find the equation of the plane that passes through the point of intersection between the line . (4) (Total 6 marks) 7. Learn more about plane, matrix, intersection, vector MATLAB Plane is a surface containing completely each straight line, connecting its any points. It is not so complicated as it sounds; ILP means Intersection between Line and Plane and it needs 5 arguments: the first two points to specify the line and more 3 points to determine the plane. Practice: Solve for t. 4. where the plane can be either a point and a normal, or a 4d vector (normal form), in the N dot (P - P3) = 0. 1) 2) The intersection of two lines . Intersect( , ) creates the circle intersection of two spheres ; Intersect( , ) creates the conic intersection of the plane ⦠6. Is there a weight limit to Feather Fall? The intersection points can be calculated by substituting t in the parametric line equations. If they intersect, I think i get the distance between the nearpoint from which i draw the ray, to the point where it colides with the plane. 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Miscellaneous
Fact Checked
Michael Smathers
Michael Smathers
A radical expression in algebra is an expression that includes a radical, or root. These are the inverse operations to exponents, or powers. Radical expressions include added roots, multiplied roots and expressions with variables as well as constants. These expressions have three components: the index, the radicand, and the radical. The index is the degree taken, the radicand is the root being derived, and the radical is the symbol itself.
By default, a radical sign symbolizes a square root, but by including different indexes over the radical, cube roots, fourth roots or any whole number root can be taken. Radical expressions can include either numbers or variables under the radical, but the fundamental rules remain the same regardless. To work with radicals, the expressions must be in simplest form; this is accomplished by removing factors from the radicand.
The first step in simplifying radicals is breaking the radicand into the factors needed to equal the number. Then, any perfect square factors must be placed to the left of the radical. For example, √45 can be expressed as √9*5, or 3√5.
To add radical expressions, the index and radicand must be the same. After these two requirements have been met, the numbers outside the radical can be added or subtracted. If the radicals cannot be simplified, the expression has to remain in unlike form. For example, √2+√5 cannot be simplified because there are no factors to separate. Both terms are in their simplest form.
Multiplying and dividing radical expressions works using the same rules. Products and quotients of radical expressions with like indexes and radicands can be expressed under a single radical. The distributive property works in the same fashion as it does with integer expressions: a(b+c)=ab+ac. The number outside the parenthesis should be multiplied by each term inside parenthesis in turn, retaining addition and subtraction operations. After all terms inside the distributive parentheses are multiplied, the radicals have to be simplified as usual.
Radical expressions that are part of an equation are solved by eliminating the radicals according to the index. Normal radicals are eliminated by squaring; therefore, both sides of the equation are squared. For example, the equation √x=15 is solved by squaring the square root of x on one side of the equation and 15 on the right, yielding a result of 225.
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# What Is the Monty Hall Math Problem?
Best UK betting sites >> Online Betting Guide >> Monty Hall Problem
The Monty Hall problem is one of those theories which could easily start an argument between punters once the subject is brought up.
There is no doubt that there is substance behind it, but when you take into consideration that even Professors in Mathematics are known to be spilt on its credibility, it should prove to you that the everyday punter may decide that it is something that they will not concern themselves with.
The Monty Hall problem is named after a US TV show from the 1960s where a game show host by the name of Monty Hall presents the contestants of the game with three doors, a goat stands behind two of the doors whilst a car can be found behind one of the other doors which will be theirs if they choose correctly.
The contestant obviously wants to choose the door which will reveal the car, and the Monty Hall problem focuses on the benefits in terms of value, of choosing one door and then changing your mind to choose another.
## So, What Is the Solution to the Monty Hall Programme?
The solution to the Monty Hall problem very straightforward, but whether you can get your head round it is another matter, so let’s take a look at the Monty Hall problem explained by giving an example.
The contestant has chosen the first door, but Monty Hall asks for one of the other doors to be opened, behind which is a goat.
The contestant then has the option of sticking with their first choice of door or changing to the other door.
Most people consider that there is a little point behind in changing doors, but the solution to the Monty hall probability is that you should always change your decision to the other door.
In an attempt to better describe this, let’s add some figures.
If there are three doors, each one has roughly a 33.3% chance of revealing the car with the other two having a combined chance of 66%.
The Monty Hall problem argues that if you decide to change your original choice of door, the odds of you winning will double from 33% to 66%.
If you are struggling to understand the reasoning behind this calculation, don’t worry too much about it because as already mentioned, even professors of Mathematics have been known to disagree about the reasoning behind it, but as with anything to do with value in betting, it is worth spending a little time trying to understand the Monty Hall problem but don’t beat yourself up if it goes straight over your head.
A tool which may help you out a little can be found if you Google ‘Monty hall problem simulation’.
This is really easy to use tool which will simulate the results of your choices relating to the doors that you choose and if there is one thing which makes the Monty hall probability easier to understand it is the Monty Hall problem simulation.
Most people learn more easily if something is actually worked out in front of them, and this is what the Monty hall problem simulation tool will do.
## What If You Just Don’t Understand the Monty Hall Math Problem?
If you don’t understand the Monty hall math problem or just cannot come to terms with the Monty Hall paradox, it doesn’t really matter as long as you understand how important that value is in your betting and how it will increase your betting profits over the course of the long term.
The whole point of the Monty Hall problem is to identify the value in changing your choice of doors, but if you always look for the value in the bets that you place, you will be able to see the benefits in real life, in front of your own eyes.
As an example, if you can get odds of 2/1 on a selection which should be 6/4 you are getting value and if you continue to only bet on selections which give you value, there will be no doubt about it, you will be acting like a professional gambler who does not bet on anything and everything but instead, chooses the bets that they make carefully and wisely.
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The vast majority of punters don’t even give value a second thought, their only aim is to get their bet on whatever the odds, they don’t even look at other sportsbooks to check if they are offering better odds.
Punters like this are literally throwing money down the drain.
It is just the same as someone coming up to you, in one hand he has a twenty-pound note, and in the other, he has a ten-pound note and tells you to choose which one you want, and you choose the ten-pound note.
If you don’t look for value or the best odds, you could perhaps be forgiven if online gambling wasn’t a ‘thing’ and you had to walk miles between high street bookies to check the odds but as all you have to do is click the computer mouse a few times, not looking for value or the best odds is unforgivable.
## Should You Concern Yourself with the Monty Hall Math Problem?
It would be a great idea to at least have a good look at the Monty hall math problem, but it is not the end of the world if you don’t.
There is a lot to be said for always trying to keep things simple so as long as you understand the basic concept of value which we have already spoken about, you won’t go far wrong and don’t forget, if you don’t understand the Monty Hall problem, you will be in good company because some of the most intelligent people out there have difficulty understanding it and would happily argue all day that it is a load of old rubbish!
.
.. | HuggingFaceTB/finemath | |
# Search by Topic
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Broad Topics > Using, Applying and Reasoning about Mathematics > Investigations
### Sort the Street
##### Stage: 1 Challenge Level:
Sort the houses in my street into different groups. Can you do it in any other ways?
### Turning
##### Stage: 1 Challenge Level:
Use your mouse to move the red and green parts of this disc. Can you make images which show the turnings described?
### Seven Sticks
##### Stage: 1 Challenge Level:
Explore the triangles that can be made with seven sticks of the same length.
### Two on Five
##### Stage: 1 and 2 Challenge Level:
Take 5 cubes of one colour and 2 of another colour. How many different ways can you join them if the 5 must touch the table and the 2 must not touch the table?
### Repeating Patterns
##### Stage: 1 Challenge Level:
Try continuing these patterns made from triangles. Can you create your own repeating pattern?
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Tea Cups
##### Stage: 2 and 3 Challenge Level:
Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.
### Two by One
##### Stage: 2 Challenge Level:
An activity making various patterns with 2 x 1 rectangular tiles.
### Escher Tessellations
##### Stage: 2 Challenge Level:
This practical investigation invites you to make tessellating shapes in a similar way to the artist Escher.
### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### Cutting Corners
##### Stage: 2 Challenge Level:
Can you make the most extraordinary, the most amazing, the most unusual patterns/designs from these triangles which are made in a special way?
### Redblue
##### Stage: 2 Challenge Level:
Investigate the number of paths you can take from one vertex to another in these 3D shapes. Is it possible to take an odd number and an even number of paths to the same vertex?
### Halving
##### Stage: 1 Challenge Level:
These pictures show squares split into halves. Can you find other ways?
### Play a Merry Tune
##### Stage: 2 Challenge Level:
Explore the different tunes you can make with these five gourds. What are the similarities and differences between the two tunes you are given?
### Brush Loads
##### Stage: 2 Challenge Level:
How can you arrange the 5 cubes so that you need the smallest number of Brush Loads of paint to cover them? Try with other numbers of cubes as well.
### Double Your Popcorn, Double Your Pleasure
##### Stage: 2 Challenge Level:
We went to the cinema and decided to buy some bags of popcorn so we asked about the prices. Investigate how much popcorn each bag holds so find out which we might have bought.
### Cuboid-in-a-box
##### Stage: 2 Challenge Level:
What is the smallest cuboid that you can put in this box so that you cannot fit another that's the same into it?
### How Tall?
##### Stage: 1 and 2 Challenge Level:
A group of children are discussing the height of a tall tree. How would you go about finding out its height?
### It's a Fence!
##### Stage: 1 and 2 Challenge Level:
In this challenge, you will work in a group to investigate circular fences enclosing trees that are planted in square or triangular arrangements.
### Fit These Shapes
##### Stage: 1 and 2 Challenge Level:
What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
### Little Boxes
##### Stage: 2 Challenge Level:
How many different cuboids can you make when you use four CDs or DVDs? How about using five, then six?
### Four Layers
##### Stage: 1 and 2 Challenge Level:
Can you create more models that follow these rules?
### Triple Cubes
##### Stage: 1 and 2 Challenge Level:
This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
### 28 and It's Upward and Onward
##### Stage: 2 Challenge Level:
Can you find ways of joining cubes together so that 28 faces are visible?
### Sticks and Triangles
##### Stage: 2 Challenge Level:
Using different numbers of sticks, how many different triangles are you able to make? Can you make any rules about the numbers of sticks that make the most triangles?
### Baked Bean Cans
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Is there a best way to stack cans? What do different supermarkets do? How high can you safely stack the cans?
### Hexpentas
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How many different ways can you find of fitting five hexagons together? How will you know you have found all the ways?
### Calcunos
##### Stage: 2 Challenge Level:
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
### Building with Rods
##### Stage: 2 Challenge Level:
In how many ways can you stack these rods, following the rules?
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### Bean Bags for Bernard's Bag
##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Crossing the Town Square
##### Stage: 2 and 3 Challenge Level:
This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares.
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### Three Sets of Cubes, Two Surfaces
##### Stage: 2 Challenge Level:
How many models can you find which obey these rules?
### Fencing
##### Stage: 2 Challenge Level:
Arrange your fences to make the largest rectangular space you can. Try with four fences, then five, then six etc.
### Making Cuboids
##### Stage: 2 Challenge Level:
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
### Move a Match
##### Stage: 2 Challenge Level:
How can you arrange these 10 matches in four piles so that when you move one match from three of the piles into the fourth, you end up with the same arrangement?
### Triangle Pin-down
##### Stage: 2 Challenge Level:
Use the interactivity to investigate what kinds of triangles can be drawn on peg boards with different numbers of pegs.
### Triangle Relations
##### Stage: 2 Challenge Level:
What do these two triangles have in common? How are they related?
### Let's Investigate Triangles
##### Stage: 1 Challenge Level:
Vincent and Tara are making triangles with the class construction set. They have a pile of strips of different lengths. How many different triangles can they make?
### The Numbers Give the Design
##### Stage: 2 Challenge Level:
Make new patterns from simple turning instructions. You can have a go using pencil and paper or with a floor robot.
### Cubes Here and There
##### Stage: 2 Challenge Level:
How many shapes can you build from three red and two green cubes? Can you use what you've found out to predict the number for four red and two green?
### Watch Your Feet
##### Stage: 2 Challenge Level:
I like to walk along the cracks of the paving stones, but not the outside edge of the path itself. How many different routes can you find for me to take?
### Room Doubling
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Investigate the different ways you could split up these rooms so that you have double the number.
### More Transformations on a Pegboard
##### Stage: 2 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Halloween Investigation
##### Stage: 2 Challenge Level:
Ana and Ross looked in a trunk in the attic. They found old cloaks and gowns, hats and masks. How many possible costumes could they make?
### Ice Cream
##### Stage: 2 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
### Newspapers
##### Stage: 2 Challenge Level:
When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different?
### Triangle Shapes
##### Stage: 1 and 2 Challenge Level:
This practical problem challenges you to create shapes and patterns with two different types of triangle. You could even try overlapping them.
### Egyptian Rope
##### Stage: 2 Challenge Level:
The ancient Egyptians were said to make right-angled triangles using a rope with twelve equal sections divided by knots. What other triangles could you make if you had a rope like this? | HuggingFaceTB/finemath | |
# What Is Arithmetic Mean?
Solution:
Arithmetic mean of a set of numbers or data is the average of the set of numerical values, as calculated by adding them together and dividing by the number of terms in the set.
For example, the Arithmetic mean of 7, 4, 5 and 8 is
$\frac{7+4+5+8}{4}=6$.
Mean is the most commonly used measure of central tendency. There are different types of mean, viz. arithmetic mean, weighted mean, geometric mean (GM) and harmonic mean.
Updated on: 10-Oct-2022
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# If a and b are positive integers than x=4^a & y=9^b
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If a and b are positive integers than x=4^a & y=9^b [#permalink]
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10 Oct 2011, 10:22
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If a and b are positive integers and x = 4^a and y = 9^b, which of the following is a possible units digit of xy?
1
4
5
7
8
The cyclicity of 4 is 6 when power is even and 2 when power is odd, similarly cyclicity of 9 is 2 when power is odd and 1 when power is even. since a & b are in the power, i multiply xy = 45^a+b , Now integer 45 is ending with 45 so unit digit of 45 will be base digit 5. but original answer is different
Any thoughts on this question , thanks in advance
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Re: If a and b are positive integers than x=4^a & y=9^b [#permalink]
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Updated on: 10 Oct 2011, 11:27
1
Cycle for 4^a is 4 or 6 : 4 16 64 256 1024 4096
cycle for 9^b is 1 or 9 : 9 81 729 6561
so possible units digit of xy = 4 or 6
Originally posted by MisterQ on 10 Oct 2011, 11:22.
Last edited by MisterQ on 10 Oct 2011, 11:27, edited 1 time in total.
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Re: If a and b are positive integers than x=4^a & y=9^b [#permalink]
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10 Oct 2011, 11:27
2
shrive555 wrote:
If a and b are positive integers and x = 4^a and y = 9^b, which of the following is a possible units digit of xy?
1
4
5
7
8
The cyclicity of 4 is 6 when power is even and 2 when power is odd, similarly cyclicity of 9 is 2 when power is odd and 1 when power is even. since a & b are in the power, i multiply xy = 45^a+b , Now integer 45 is ending with 45 so unit digit of 45 will be base digit 5. but original answer is different
Any thoughts on this question , thanks in advance
x=4^a
a is +ve integer.
a=1; x=4
a=2; x=4^2=16
a=3; x=4^3=64
a^4=4*4=6
You see; the units digit of 4^x is either 4 OR 6.
Likewise;
units digit of 9^x is either 9 OR 1.
Possible units digit;
4*9=36
4*1=4
6*9=54
6*1=6
6 is not in the options, but 4 is.
Ans: "B"
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Re: If a and b are positive integers than x=4^a & y=9^b [#permalink]
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10 Oct 2011, 14:50
got it.. ..Thanks guys !!
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Re: If a and b are positive integers than x=4^a & y=9^b [#permalink]
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10 Oct 2011, 20:50
x=4^a
y=9^b
xy =(4^a)*(9^b)
when a=1,b=1 , xy 's unit digit =i.e unit's digit of (4*9= 36) i.e 6
a=1,b=2, xy 's unit digit = i.e units digit of (4*81) i.e 4
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Re: If a and b are positive integers than x=4^a & y=9^b [#permalink]
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11 Oct 2011, 05:01
Cyclisity of 4^n is 4,6,4,6
9^n is 9,1,9,1
given the question, we can have
9*4=36, 6
9*6=54, 4
1*4=4, 4
1*6=6, 6
therefore 4
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Re: If a and b are positive integers than x=4^a & y=9^b [#permalink]
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11 Oct 2011, 05:39
shrive555 wrote:
If a and b are positive integers and x = 4^a and y = 9^b, which of the following is a possible units digit of xy?
1
4
5
7
8
The cyclicity of 4 is 6 when power is even and 2 when power is odd, similarly cyclicity of 9 is 2 when power is odd and 1 when power is even. since a & b are in the power, i multiply xy = 45^a+b , Now integer 45 is ending with 45 so unit digit of 45 will be base digit 5. but original answer is different
Any thoughts on this question , thanks in advance
4^a will end in either 4 or 6 .
9^b will always end in either 9 or1 .
So the possible unit digit of xy can be either of one below :
4*9 =6 or 4*1 = 4 OR 6*9 = 4 or 6*1 = 6.
So , as you can see the unit digit can either be 6 or 4.
We have just 4 in the choices.Hence , option (B) 4 is the correct option.
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Re: If a and b are positive integers than x=4^a & y=9^b [#permalink]
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11 Oct 2011, 23:22
shrive555 wrote:
If a and b are positive integers and x = 4^a and y = 9^b, which of the following is a possible units digit of xy?
1
4
5
7
8
The cyclicity of 4 is 6 when power is even and 2 when power is odd, similarly cyclicity of 9 is 2 when power is odd and 1 when power is even. since a & b are in the power, i multiply xy = 45^a+b , Now integer 45 is ending with 45 so unit digit of 45 will be base digit 5. but original answer is different
Any thoughts on this question , thanks in advance
Another thing: $$x = 4^a$$ and $$y = 9^b$$
When you multiply, $$x*y = 4^a * 9^b$$
You cannot bring the powers together. Only if the bases are the same, then you can add the powers. e.g.
$$x = 4^a$$ and $$y = 4^b$$
When you multiply, $$x*y = 4^{a+b}$$
Or when the powers are the same, then you can multiply the bases together
e.g.
$$x = 4^a$$ and $$y = 9^a$$
When you multiply, $$x*y = 36^a$$
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Re: If a and b are positive integers than x=4^a & y=9^b [#permalink]
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13 Oct 2011, 09:51
VeritasPrepKarishma wrote:
shrive555 wrote:
If a and b are positive integers and x = 4^a and y = 9^b, which of the following is a possible units digit of xy?
1
4
5
7
8
The cyclicity of 4 is 6 when power is even and 2 when power is odd, similarly cyclicity of 9 is 2 when power is odd and 1 when power is even. since a & b are in the power, i multiply xy = 45^a+b , Now integer 45 is ending with 45 so unit digit of 45 will be base digit 5. but original answer is different
Any thoughts on this question , thanks in advance
Another thing: $$x = 4^a$$ and $$y = 9^b$$
When you multiply, $$x*y = 4^a * 9^b$$
You cannot bring the powers together. Only if the bases are the same, then you can add the powers. e.g.
$$x = 4^a$$ and $$y = 4^b$$
When you multiply, $$x*y = 4^{a+b}$$
Or when the powers are the same, then you can multiply the bases together
e.g.
$$x = 4^a$$ and $$y = 9^a$$
When you multiply, $$x*y = 36^a$$
Thanks for pointing out!!
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Re: If a and b are positive integers than x=4^a & y=9^b [#permalink]
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Re: If a and b are positive integers than x=4^a & y=9^b [#permalink] 02 Mar 2018, 11:36
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# Thread: If Earth Lost its Atmosphere, No Temperature Change?
1. What would happen to the surface temperature of the Earth if Earth had the Moon's lack of an atmosphere? My initial knee-jerk response was as follows:
the Earth would have drastic temperature swings like the Moon.
According to the Artemis Project data, the moon and earth's temperatures compare as follows:
Minimum surface temperature of Earth: -89 degrees C.
Maximum surface temperature of Earth: 58 degrees C.
Temperature spread of Earth: 147 degrees C.
Minimum surface temperature of the Moon: -147 degrees C.
Maximum surface temperature of the Moon: 100 degrees C.
Temperature spread of the Moon: 247 degrees C.
As you can see, the Moon has a much greater spread between high and low temperatures. The ratio of the Earth's temperature spread to the Moon's is .595, rounded off to .60.
But then it occurred to me that the Earth's surface area is mostly water and the Moon's surface area is mostly basalt. Question: How much does the presence of water verses basalt affect the surface temperature of the Earth?
Time to do some math:
Water has a specific heat capacity of 4.1813 J/g*deg. C and its mass is 18 g per mole. Basalt's specific heat capacity is .84 J/g*deg. C and its mass is 54 grams per mole.
Q = cm(T)
c is the specific heat coefficient, m is mass, T is the change in temperature, Q is the amount of energy needed to affect the temperature change.
Assuming that the Earth and Moon receive about the same amount of solar energy, it makes sense to use the same value for Q when comparing basalt and water.
Water: 100J = (4.1813)(18g)( T)
T(water) = 1.33 deg. C.
Basalt: 100J = (.84)(54g)(T)
T(basalt) = 2.2 deg. C.
Conversely, if the water and basalt were allowed to cool, and lost 100 J, their temperatures would decrease -1.33 deg. C and -2.2 deg., respectively.
Now, what do you suppose happens when I divide 1.32 deg. C by 2.2 deg. C?
I get .60--the same ratio as the Earth/Moon temperature spreads. Now that's interesting!
When solar energy is added or removed, water increases and decreases nearly the same rate as the Earth, and basalt's temperature increases and decreases at nearly the same rate as the Moon.
The implication here is that if the Earth lost its atmosphere, its surface temperature would not change much. It appears to be the oceans, not the atmosphere which have the greatest impact on the Earth's surface temperatures.
2.
3. If the Earth lost its atmosphere the pressure at the surface of the oceans would be zero. If the temperature didn't change, say 15 C, the vapor pressure of water is around 0.02 bar. The oceans would evaporate. As they did so the latent heat would cool the oceans until they froze for some depth. The planet would be cold.
4. Thanks for your reply. I don't don't presume to know what would really happen if the earth lost its atmosphere. I have never conducted that experiment. What I find interesting, however, is the correlation between the temperature changes when comparing water/basalt, Earth/Moon temperature changes.
Let's assume Earth keeps its atmosphere. The oceans appear to impact temperature change more than the atmosphere according to the math. The atmosphere's influence or value, if I understand you correctly, is that it holds the oceans in place. The bottom line then is if the oceans remain in place, Earth's temperatures won't change much. Feel free to post your math or empriracal data that contradicts this claim.
I also find the theory that the earth would be cold if the oceans evaporated interesting. Water vapor is a greenhouse gas, and you would think it would warm the planet. What you seem to be suggesting is that water vapor won't warm the planet or keep temperatures stable. If that is true, then it would be foolhardy to assume that greenhouse gases must necessarily warm the planet.
If the temperatures did cool, I would imagine that the vapor would condense and warm the planet. On the one hand you have less pressure due to a lack of atmosphere, but on the other you have the high specific heat capacity of water vapor. Additionally, more solar energy would make it to the surface initially if the atmosphere was removed. That should have a warming effect countering the cooling effect.
But again, I welcome your math and emprical data.
5. Where I am, the ocean changes just a few degrees seasonally. Because my climate rolls over that ocean, summer high does not require air-conditioning, and winter low some years can't even stop roses from blooming.
There is also an organic blanket effect, on the ground. Go a few feet down, we find constant temperature. This is where tree roots are, plus layers of moldering sodden soil and deadwood. In winter it just doesn't freeze. Most of Canada's west coast is covered in that, and under canopy.
6. I also find the theory that the earth would be cold if the oceans evaporated interesting. Water vapor is a greenhouse gas, and you would think it would warm the planet. What you seem to be suggesting is that water vapor won't warm the planet or keep temperatures stable. If that is true, then it would be foolhardy to assume that greenhouse gases must necessarily warm the planet.
Your scenario had no atmosphere. Hence, playing your totally unrealisitc game, the water vapor goes off to wherever the rest of the atmosphere went, to maintain the no atmopshere scenario.
For thought experiments to have any value they must have some grounding in reality.
7. Originally Posted by Bunbury
I also find the theory that the earth would be cold if the oceans evaporated interesting. Water vapor is a greenhouse gas, and you would think it would warm the planet. What you seem to be suggesting is that water vapor won't warm the planet or keep temperatures stable. If that is true, then it would be foolhardy to assume that greenhouse gases must necessarily warm the planet.
Your scenario had no atmosphere. Hence, playing your totally unrealisitc game, the water vapor goes off to wherever the rest of the atmosphere went, to maintain the no atmopshere scenario.
For thought experiments to have any value they must have some grounding in reality.
Well Einstein suggested that reality is relative. LOL! For that reason, I can appreaciate your relative point of view. However, in my reality, additional solar radiation and water's high specific heat capacity would offset your cooling reality. Further, when Einstein did his thought experiments, he wasn't a dull boy. He had fun! "What would happen if you traveled the speed of light?" How real is that assumption? Or how about M theory? Or the greenhouse gas theory? How real are those? LOL! Again I welcome your math or emprirical data. Crunch some numbers for me, please? Common, I did it for you.
8. Originally Posted by Pong
Where I am, the ocean changes just a few degrees seasonally. Because my climate rolls over that ocean, summer high does not require air-conditioning, and winter low some years can't even stop roses from blooming.
There is also an organic blanket effect, on the ground. Go a few feet down, we find constant temperature. This is where tree roots are, plus layers of moldering sodden soil and deadwood. In winter it just doesn't freeze. Most of Canada's west coast is covered in that, and under canopy.
Thanks for that empirical data. Yep, my math confirms your observations. Water is king. The .0384% of CO2 in the atmosphere just is not enough to make my boots shake.
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Home ABC series ABC’s of Math Activities : F-J
# This is day 2 of our ABC’s of Math Activities series.
You will find many preschool math ideas and other hands on math activities that can be done using household objects.
# F is for Fractions
If your household has more than 1 kid, then the kids KNOW fractions. They just don’t know that the center point of their squabbles has a name! I could have done, Hershey’s fractions, cookies fractions etc but I’m pretty sure the goodies will be eaten before we explored one fourths and halves.
This time we are using a popular play material : play dough to teach fractions.
We started off with a box of play dough and a plastic knife. Big A made “snakes” of various sizes and we cut them into 2 pieces with the knife. As expected she said ” I cut them into halves” We then stuck them together and created a “full snake” again.
##### Using that as a starting point we explored the following:
• 2 halves make a whole
• Recognize fractions.
• How to write them.
• Which “piece” is larger. They will be able to do this by looking at it.
• What happens if we add two parts
We fashioned a pancake out of play dough and started cutting them into 4 parts. Having the child cut a whole play dough pancake into 4 parts was very important for her to begin to understand the concept. She is the only cutting them into 4 parts. She is the one putting the pieces together to make a whole. We also made another one and cut that one up into 2 parts. Just by looking, a kid can tell us which one is bigger, 1/4th or 1/2?
Of course, the immediate question : But 4 is more than 2!
I think at this age, just touching on the concept and gently introducing is sufficient. Any more denser explanation might turn off eager ears.
# G is for Games
Home made math games are a fantastic way to teach simple math to young kids. Use objects from around the house to make up a game centering around the concept the kid is eager to learn . Find out how we made our own Mancala Math Game. Tailor the activity to fit the current interests of the kid. For example, while exploring addition, we fashioned an Like many children, my pre schoolers loves coming with stories, so we made up stories as we read a picture book and enacted the (addition) story with home made manipulatives.
# H is for How to Teach Math
I say start young. One of the ways to make an early start is to make connections between numbers and objects with everyday life scenarios.
• Eat 5 grapes today
• How many cheerios do you have in that bowl?
• Count your fingers and toes?
While it is true that some kids can understand math easily and others not so much, we can alleviate the pressure by breaking down the concepts into simple steps. From my personal experience of being a tutor, I have identified 2 challenges: [highlight bg=”#DDFF99″ color=”#000000″]The inability to digest the concept and more importantly the urge in young kids to “guess” the answer and the process.[/highlight]
It is important to develop a mindset for Math. Inculcate in children, “Math is fun”. So, how do we do it?
Here are a few simple ways :
• Focus on the process and not on getting a correct answer.
• Practice more hands on and playful learning
• Talk about the practicality of the subject and compare with real world scenarios.
• Use everyday opportunities to make anytime into learning time.
• Casual conversations where the kid starts one is the best time to explore anything
• Remember, math is not always numbers.
• Play games to improve mathematical ability.
The fun part is you can always brainstorm and come up with unique and more fun ways to teach and learn math. Learning is wisdom and the opportunities are always endless.
# I is for Importance of Math
Importance of Mathematics is paramount for Early Childhood Education. Children ‘unknowingly’ notice and explore mathematical dimensions of their world everyday. They talk about quantities, follow patterns and even perform basic operations all while trying to share a bowl of popcorn with friends or siblings. Budding Mathematicians I say !
So, how do we get from that to ” Math is not for me ? ”
My BIG Question – is it true that Parents who feared Math will pass that fear on to the kids ? Well, its possible. Kids are big on Questioning things. They are not afraid to ask questions and many a times we are going to be stumped and outwitted by them. When they look up to you for answers – It is very important for us to give them what they are looking for – Responding to their questions with statements like “Math and I are not best buddies” or “ I hated Math when I was a kid” or the ever infamous “ I hate my Math teacher” puts an unnecessary fear of the subject into their minds.
To ease things, it is very important to develop simple math skills in children during your regular everyday routine
1. doing puzzles together
2. talking about numbers – you can do this on a trip to the grocery store- 3 oranges, 6 bananas, 2 tomatoes
3. asking your child to replicate your lego tower or structure.
Here is an activity we did that shows opportunities are always around us.
# J is for JUMP and Learn Math
Jumping is a natural interest for any young child. So it is only fair that you use that fun energy and add some arithmetic to it and make it into a learning opportunity.
Ways to make Jumping a Math activity.
1. Roll a dice – Count – Do that many jumps
2. Number Recognition Jumping exercise for a toddler– Write each number on a separate sheet of paper and spread it in a room. Call out each and get the child to jump to the number.
3. Expand on the idea to with addition and subtraction exercises for preschoolers.
Look what we did: We used cups to build a wall by stacking and jump over it
Lessons learnt:
• Building Rows
• Stacking
• Counting
• Creating height
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# How to reduce to an NP-hard problem?
For an assignment I have to program an application to schedule conversations. There is an event where representatives of the elementary schools talks with the representatives of high schools. They will talk about the students that will be transferred to the highschool. There are approximately 200 elementary schools and 40 high schools that will be participating in this event. The schools already know which student is transferring to which high school. The conversations will only be between representatives of E and H from student that will be transferring to H.
The rules are:
1. The duration of each conversation is based on the amount of students per representatives.Each conversation last 5 minutes per student. If a group consist of 1 student, this conversation last 10 minutes.
2. No timeclashes
3. All the students of the same group will be scheduled together, so, a representatives will only face the same representative once.
4. Timespan is 13.00-19.00
5. The waiting time of a representative is at most 20% of his time. A waiting time is an empty timeslot between the 1st and last conversation.
6. Schedules for 2 days
7. Each representatives participate for 1 day.
The problem is that I know that this is hard to solve, but I dont know if it's NP-hard. Right now I only know this problem is similar to a Job Shop Problem. What can I do to proof that my problem is NP-hard? I read that I need to reduce a known problem to my problem. But how do I do this? I have read different articles and books, but I still don't understand the steps to do it.
Just an extended comment.
I think that before trying to find a reduction, you should try to better formalize the problem:
• define exactly what the parameters of the problem are, formalize their constraints, and formalize the question;
• try to simplify (or remove) the details;
For example:
• If a group consist of 1 student, this conversation last 10 minutes. $\rightarrow$ it is equivalent to a group of two students, so you can remove it;
• Timespan is 13.00-19.00 $\rightarrow$ in the other rules you use minutes, so convert the "timespan" to $T$ minutes;
• try to think if a simplified version of the problem has a quick solution or it can be still hard (e.g. if the event last only one day rules 7 and 8 can be dropped)
• express the rules with a math formula: e.g "if an elementary group $E$ of $n$ students talk with a high school group $H$ of $m$ students then the total time $TT$ required is $TT(E,H)=....$"; "the total max wait time $WT$ for group $E$ is $WT = ...$"
Furthermore there is a point that it is not clear (to me):
• suppose you have an elementary school group $E$ having $n$ students that talk with a high school group $H$ having $m$ students; what is the total time required? (1) $5 * n$ (the talk is public) or (2) $5 * ( n / m)$ (each student of E talks with a single student of H). If the talk is public, then the number of students is redundant, and you can think only in terms of total time required by a school group.
• The talk only between the 2 reps. Sep 6, 2012 at 11:10
• If I simplify the problem like you said: if the event last only 1 day. Wouldn't the problem be harder to solve? And what should be the next step of this? Sep 6, 2012 at 11:14
• @NicoLiu: you should think about the generalization of the problem (if you think about a single particular instance the notion of "NP-hardness" is meaningless); and don't confuse the hardness of the problem with the "chances" of finding a valid solution. From the computational complexity perspective, a decision problem can be easy because no valid solution exists at all (the answer is always "no" on all inputs).
– Vor
Sep 6, 2012 at 12:32
• @NicoLiu: And what should be the next step of this: you can follow the steps listed in rphv's answer, but if you feel unfamiliar with the "terms" used (decision problem, polynomial time, NP-complete, reduction, ...) then perhaps you need to learn more on complexity theory before trying to apply it (you can ask here for books/lectures/...). Otherwise, if you know well those terms, edit the question and make it more concise/formal (start with changing the title which is not correct :-), and you'll probably receive a more specific/detailed answer.
– Vor
Sep 6, 2012 at 12:59
• @NicoLiu: ... perhaps another evidence that you should rewrite the question in a more formal way :-) ...
– Vor
Sep 6, 2012 at 15:39
Garey & Johnson outline the following four steps for proving NP-completeness results:
1. Show that your problem is in NP. Informally, this means that it's possible to verify the correctness of potential solutions to your problem "quickly" - i.e., in polynomial time.
2. Select a known NP-complete problem. Again, Garey & Johnson provide an excellent list. They suggest that 3SAT, 3-Dimensional Matching, Vertex Cover, Clique, Hamiltonian Circuit, and Partition "...can serve as a 'basic core' of known NP-complete problems for the beginner."
3. Construct a reduction from the known NP-complete problem to your problem. Very informally, to construct such a reduction one might first assume the existence of a "black box" which quickly solves instances of your problem. Then, demonstrate that it's possible to quickly solve the known NP-complete problem selected in step 2 using your "black box." This implies that your problem is "at least as hard" as a known NP-complete problem. Finally, to show that your problem is "no harder" than an NP-complete problem, proceed in the opposite direction: start with a "black box" that solves the NP-complete problem, and show that you can quickly solve your problem using that box.
4. Show that the reduction in step 3 is a polynomial reduction. In other words, show that the "black box" solutions in step 3 take place in polynomial time.
• 1. is only necessary if you want NP-completeness, which Nico does not seem to need. Sep 5, 2012 at 19:53
• A proof of NP-hardness (as opposed to NP-completeness) can also leave off everything after "Finally..." in step 3.
– rphv
Sep 5, 2012 at 20:33
• The problem is that I'm not trained to be a scientist. So it is hard for me to come up with different formula to proof this. I have read the theory about it. But I don't know what I have to reduce. In the book 'Computational Complexity' by Christos H. Papadimitriou, there is a way called reduction by generalization. My question is does it mean that I can say (Without the formula) that a Job is the elementary school, Machine the high school and Operation is the conversation. While explaining that the constraints are the same. Sep 6, 2012 at 9:26
• @rphv aren't (1) and (3) after finally rendundant? Both show the problem to be in NP. Supposing we did want to show the problem was NP-complete and not just NP-hard, we should only need to do one or the other. (1) is often easier. "(3) after finally" is often still an instructive exercise, however. As an aside, doing the last half of (3) instead of the first half, i.e. reducing the wrong way is a mistake newcomers to NP-completeness proofs sometimes make.
– Joe
Sep 7, 2012 at 19:21
• @Joe I believe you're right - anything "easier" than NP-complete is in NP.
– rphv
Sep 8, 2012 at 5:42 | HuggingFaceTB/finemath | |
(5/5)
# Consider the array A[1...n] of n non-negative integers.
INSTRUCTIONS TO CANDIDATES
Problem 2 (20 points)
Consider the array A[1...n] of n non-negative integers. There is a frog on the first index of the array. In each step, if the frog is positioned on the ith index, then it can hop to any of the indices i,...,i+A[i] (so the frog can at most hop to the index i + A[i]).
Develop a greedy algorithm to determine the minimum number of hops necessary so that the frog can reach the last index, i.e., the nth index. You must fully explain your algorithm.
Your algorithm must return -1 if the frog cannot reach the last index.
Problem 3 (20 points)
Given a directed weighted graph G with no cycles, develop an algorithm to find the weights of all longest paths from one source vertex s to the other vertices in G in O(|V|+|E) time.
A longest path from u to v is a path of maximum weight from u to v.
Problem 4 (20 points)
Consider a directed graph G = (V,E) where each edge is colored in either red or blue. Develop an O(|V|+|E|) time algorithm that, given vertices u and v in G, determines if there exists a walk from u to v that uses at least one blue edge (note that this walk may have repeated vertices). Justify the correctness of your algorithm and analyze the running time.
(Hint: Try to construct a new graph using an idea similar to HW9 P4.)
Problem 5 (20 points)
Consider the n points P1, P2, ..., Pn in the plane. Each point p; has the xy-coordinates p1 = cost of connecting two points pi = (xi, Yi) and pj = (xj, yj) is defined as
(xi, Yi). The
│Xi — Xj│+ │Yi — Yj\.
For example, the cost of connecting p1 =
(1, 5) and p; = (3, 4) is
|1 3|+|54| = 3.-
We want to make all the points connected such that there is a path between any pair of points. Develop an algorithm that returns the minimum cost to make all the points connected. Your algorithm must run in O(n2) time. Explain all the steps of your algorithm.
(5/5)
## Related Questions
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Hire Me | HuggingFaceTB/finemath | |
# 5th Power
LAST UPDATE: October 21st, 2020
## Definition – What is the “5th Power” of a number?
The “5th Power” of a number is the number multiplied by itself 5 times. It is written as number5.
Saying “3 to the exponent 5” or 35 is the same as saying 3 times 3 times 3 times 3 times 3 (equals 243).
Saying “20 to the exponent 5” or 205 is the same as saying 20 x 20 x 20 x 20 x 20 (equals 320,000).
To find a number to a different exponent, use our simple exponent calculator. To find the number needed to find the exponent 5 of a number, use the 5th root.
## Formula – How to Calculate Exponent 5 Numbers
The exponent 5 of a number is found by multiplying that number by itself 5 times.
number5 = number x number x number x number x number
If you are
### Example
55 = 5 x 5 x 5 x 5 x 5 = 3125 | HuggingFaceTB/finemath | |
posted by on .
R wants to round 47 to the nearest ten. Which halfway number can help him? 45, because it is halfway b/w 40 & 50 or 50, because it is halfway b/w 0 & 100?
You need to round 47 to the nearest 10.
Which number is closest to 47?
20, 30, 40, 50, 60, 70
As you can see, 47 is closest to 50.
the answer would be 50 because if he wants to be rounding to the nearest 10 then that's what it would be
• math - ,
600 | HuggingFaceTB/finemath | |
# Proof of compactness of bounded linear operator
Define $T: l^2 \to l^2$ by $Tx = y =(\eta_j)$, where $x = (\xi_j)$ and $$\eta_j = \sum_{k=1}^{\infty} \alpha_{jk}\xi_k, \quad \quad \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} |\alpha_{jk}|^2 < \infty.$$ Show that $T$ is compact.
Solution:
Let $(T_n)$ be a sequence of compact linear operators from a normed space $X$ into a Banach space $Y$. If $(T_n)$ is uniformly operator convergent, say, $||T_n - T|| \to 0$, then the limit operator $T$is compact. Hence we must show that $||T_n - T|| \to 0$ for $T$ as defined in the problem.
Let $T_n x = (\sum_{k=1}^{n} \alpha_{1k}\xi_k, \sum_{k=1}^{n} \alpha_{2k}\xi_k, \sum_{k=1}^{n} \alpha_{3k}\xi_k, \ldots)$. $T_n$ is linear and bounded. Also, since $\mathrm{dim} \, T_n(X) < \infty$, $T_n$ is compact. Furthermore, $$||(T-T_n)x^2|| = \sum_{j=n+1}^{\infty}\sum_{k=n+1}^{\infty} |\eta_j|^2 = \sum_{j=n+1}^{\infty}\sum_{k=n+1}^{\infty} |\alpha_{jk}|^2 |\xi_k|^2.$$ Now I'm unsure about the last few steps. Hopefully you can help me out. Thanks!
-
From your form, You can bound $\|(T - T_n) x\|^2$ by $$\|x\|^2 \sum_{j,k \geq n+1} |\alpha_{jk}|^2$$. Divide by $\|x\|^2$; do you see how this is a bound on $\|T - T_n\|$? Like convergent sums over a single index, convergent sums over double indices vanish as you "exhaust the tail" of the sum, so to speak, so as you take $n \rightarrow \infty$, the sum will go to zero. Now use the closedness property of the space of compact operators. – A Blumenthal Feb 4 '13 at 4:51
I do! That's brilliant. Since the sum goes to zero as $n$ goes to infinity, we get $||(T-T_n)x|| \to 0$ as $n \to \infty$. Now, since the space of compact operators is closed, we have that the limit of any convergent sequence of compact operators must also be a compact operator, so $T$ is compact. – Numbersandsoon Feb 5 '13 at 4:09 | HuggingFaceTB/finemath | |
# $E = mc^2$ derivation thought experiment seems to not have conserved energy
There is an interesting derivation of $$E=mc^2$$ given here which I have updated to use relativistic momentum and relativistic kinetic energy. I find when doing the derivation that it doesn't appear that kinetic energy is conserved.
The setup: Consider an object $$A$$ moving at a speed $$v$$ to the right of an object $$B$$. Let $$m_A$$ be the mass of $$A$$, Let $$KE_A$$ be the kinetic energy of $$A$$, let $$PE_A$$ be the potential energy of $$A$$ and let $$P_m$$ be the momentum of $$A$$.
We consider the case of $$A$$ spontaneously emitting two photons (or generally equivalent pulses of light), one moving vertically up and one vertically down relative to $$A$$ in a frame where $$A$$ is at rest.
## Frame where $$A$$ is at rest
$$\begin{matrix} \begin{matrix} \textbf{Before Emission} \\ \text{Potential Energy} = PE_A \\ \text{Horizontal Momentum} = 0 \\ \text{Kinetic Energy} = 0\\ \text{Mass} = m_a \\ \text{Photon Vertical Momentum} = 0 \\ \text{Photon Horizontal Momentum} = 0 \\ \text{Photon Energy} = 0 \end{matrix} & \begin{matrix} \textbf{After Emission} \\ \text{Potential Energy} = PE_A- E \\ \text{Horizontal Momentum} = 0 \\ \text{Kinetic Energy} = 0\\ \text{Mass} = m_a' \\ \text{Photon Vertical Momentum} = \frac{E}{2c} - \frac{E}{2c}=0 \\ \text{Photon Horizontal Momentum} = 0 \\ \text{Photon Energy} = E \end{matrix} \end{matrix}$$
From here we can verify that the total Potential + Kinetic Energy is the same before and after, and the momentum is the same before and after.
## Frame where $$A$$ is moving at speed $$v$$ to the right
$$\begin{matrix} \begin{matrix} \textbf{Before Emission} \\ \text{Potential Energy} = PE_A \\ \text{Horizontal Momentum} = \frac{m_Av}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \text{Kinetic Energy} = \frac{m_Ac^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m_Ac^2 \\ \text{Mass} = m_a \\ \text{Photon Vertical Momentum} = 0 \\ \text{Photon Horizontal Momentum} = 0 \\ \text{Photon Energy} = 0 \end{matrix} & \begin{matrix} \textbf{After Emission} \\ \text{Potential Energy} = PE_A- E \\ \text{Horizontal Momentum} = \frac{m_A'v}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \text{Kinetic Energy} = \frac{m_A'c^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m_A'c^2\\ \text{Mass} = m_a' \\ \text{Photon Vertical Momentum} = \frac{E}{2c} \frac{\sqrt{c^2 - v^2}}{c} - \frac{E}{2c}\frac{\sqrt{c^2 - v^2}}{c} =0 \\ \text{Photon Horizontal Momentum} = \frac{E}{2c} \frac{v}{c} \\ \text{Photon Energy} = E \end{matrix} \end{matrix}$$
The trick then to derive $$E=mc^2$$ is to simply observe that in the left hand side before emission the total horizontal momentum (sum of horizontal momentum and photon horizontal momentum) was $$\frac{m_a v}{\sqrt{1 - \frac{v^2}{c^2}}}$$ and on the right hand side it is $$\frac{m_a' v}{\sqrt{1 - \frac{v^2}{c^2}}} + \frac{E}{c} \frac{v}{c}$$ and by equating these we find that the change in mass $$m_a - m_a'$$ depends on the energy $$E$$ in the famous way.
But I realized with this model there is a problem:
## The Question:
If you look at the frame where $$A$$ is moving at speed $$v$$. Then one sees that that the sum of Potential Energy + Kinetic Energy + Photon Energy is ONLY conserved if $$m_a$$ remains the same. Since we have on the left hand side:
$$PE_A + \frac{m_Ac^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m_Ac^2$$
And on the right hand side:
$$PE_A - E + \frac{m_A'c^2}{\sqrt{1 - \frac{v^2}{c^2}}} - m_A'c^2 + E$$
If we assume $$m_a$$ changes then some energy appears to have disappeared. How do I make this problem go away?
• What is the meaning of -E on the right hand side? – stuffu Sep 8 '19 at 7:42
• I tried include the signs of momentum, when they cancel so when you encounter a $a-a$ term it’s clearer it refers to the two photons traveling in opposite directions. There is also a $-E$ in the potential energy on right hand side to account for where the photon was emitted – frogeyedpeas Sep 8 '19 at 14:55
To simplify, let me set the $$PE_A$$ to zero. Which isn't used anyway.
The conservation of energy before is: $$m_A c^2 = m_A'c^2 + 2E$$
Lorentz transformations don't change the perpendicular part of the the 4-vector to the boost. Also there is no parallel component in your setup. So the new energy conservation equation after the boost is: $$\gamma m_A c^2 = \gamma m_A'c^2 + 2 \gamma E$$
Canceling the $$\gamma$$ from both sides, you see that the energy is conserved.
• Why does the $\gamma$ apply to the photon terms as well? This is not intuitive to me since photons have the same velocity in all inertial frames. So I assumed that the total energy of an individual photon being emitted is still (in your interpretation my question) $E$ and the total momentum is still $\frac{E}{c}$ (up to a choice of direction) even in the moving frame. – frogeyedpeas Sep 9 '19 at 15:29
• Photons have the same speed, $c$, in any frame but not the same energy. Roughly speaking, the energy of the photon is the frequency, and because of time-dilation the frequency can change (even if the photon is moving transverse to the motion). It's called the transverse Doppler shift and was first experimentally observed in the Ives-Stilwell experiment. – Jase Uknow Sep 9 '19 at 16:13 | HuggingFaceTB/finemath | |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 9.4: Multiplication of Polynomials by Binomials
Difficulty Level: Basic Created by: CK-12
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Practice Multiplication of Polynomials by Binomials
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Suppose a factory needs to increase the number of units it outputs. Currently it has \begin{align*}w\end{align*} workers, and on average, each worker outputs \begin{align*}u\end{align*} units. If it increases the number of workers by 100 and makes changes to its processes so that each worker outputs 20 more units on average, how many total units will it output? What would you have to do to find the answer? After completing this Concept, you'll be able to multiply a polynomial by a binomial so that you can perform the operation required here.
### Watch This
Multimedia Link: For further help, visit http://www.purplemath.com/modules/polydefs.htm – Purplemath’s website – or watch this CK-12 Basic Algebra: Adding and Subtracting Polynomials
### Guidance
A binomial is a polynomial with two terms. The Distributive Property also applies for multiplying binomials. Let’s think of the first parentheses as one term. The Distributive Property says that the term in front of the parentheses multiplies with each term inside the parentheses separately. Then, we add the results of the products.
\begin{align*}(a+b)(c+d)=(a+b)\cdot c+(a+b)\cdot d\end{align*} Let’s rewrite this answer as \begin{align*}c\cdot (a+b)+d\cdot (a+b)\end{align*}.
We see that we can apply the Distributive Property on each of the parentheses in turn.
\begin{align*}c \cdot (a+b)+d\cdot (a+b)=c\cdot a+c \cdot b+d \cdot a+d \cdot b \ (\text{or} \ ca+cb+da+db)\end{align*}
What you should notice is that when multiplying any two polynomials, every term in one polynomial is multiplied by every term in the other polynomial.
#### Example A
Multiply and simplify \begin{align*}(2x+1)(x+3)\end{align*}.
Solution: We must multiply each term in the first polynomial with each term in the second polynomial. First, multiply the first term in the first parentheses by all the terms in the second parentheses.
Now we multiply the second term in the first parentheses by all terms in the second parentheses and add them to the previous terms.
Now we can simplify.
\begin{align*}(2x)(x)+(2x)(3)+(1)(x)+(1)(3) & = 2x^2+6x+x+3\\ & = 2x^2+7x+3\end{align*}
#### Example B
Multiply and simplify \begin{align*}(4x-5)(x^2+x-20)\end{align*}.
Solution:
Multiply the first term in the binomial by each term in the polynomial, and then multiply the second term in the monomial by each term in the polynomial: \begin{align*}(4x)(x^2)+(4x)(x)+(4x)(-20)+(-5)(x^2)+(-5)(x)+(-5)(-20)&=4x^3+4x^2-80x-5x^2-5x+100\\ & = 4x^3-x^2-85x+100\end{align*}
Solving Real-World Problems Using Multiplication of Polynomials
We can use multiplication to find the area and volume of geometric shapes. Look at these examples.
#### Example C
Find the area of the following figure.
Solution: We use the formula for the area of a rectangle: \begin{align*}\text{Area}=\text{length}\cdot\text{width}\end{align*}. For the big rectangle:
\begin{align*}\text{Length} & = B+3, \ \text{Width}=B+2\\ \text{Area} &= (B+3)(B+2)\\ & = B^2+2B+3B+6\\ & = B^2+5B+6\end{align*}
### Vocabulary
Binomial: A binomial is a polynomial with two terms.
The Distributive Property for Binomials: The Distributive Property says that the term in front of the parentheses multiplies with each term inside the parentheses separately. Then, we add the results of the products.
\begin{align*}(a+b)(c+d)=c\cdot (a+b)+d\cdot (a+b)=c\cdot a+c \cdot b+d \cdot a+d \cdot b \ (\text{or} \ ca+cb+da+db)\end{align*}
### Guided Practice
Find the volume of the following figure.
Solution:
\begin{align*}The \ volume \ of \ this \ shape & = (area \ of \ the \ base) \cdot (height).\\ \text{Area of the base} & = x(x+2)\\ & = x^2+2x\end{align*}
\begin{align*}Volume&=(area \ of \ base ) \times height\\ Volume&=(x^2+2x)(2x+1)\end{align*}
Now, multiply the two binomials together.
\begin{align*}Volume&=(x^2+2x)(2x+1)\\ &= x^2\cdot 2x+x^2\cdot 1+2x\cdot 2x+ 2x\cdot 1\\ &= 2x^3+x^2+2x^2+2x\\ &=2x^3+3x^2+2x\end{align*}
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Multiplication of Polynomials (9:49)
Multiply and simplify.
1. \begin{align*}(x-2)(x+3)\end{align*}
2. \begin{align*}(a+2)(2a)(a-3)\end{align*}
3. \begin{align*}(-4xy)(2x^4 yz^3 -y^4 z^9)\end{align*}
4. \begin{align*}(x-3)(x+2)\end{align*}
5. \begin{align*}(a^2+2)(3a^2-4)\end{align*}
6. \begin{align*}(7x-2)(9x-5)\end{align*}
7. \begin{align*}(2x-1)(2x^2-x+3)\end{align*}
8. \begin{align*}(3x+2)(9x^2-6x+4)\end{align*}
9. \begin{align*}(a^2+2a-3)(a^2-3a+4)\end{align*}
10. \begin{align*}(3m+1)(m-4)(m+5)\end{align*}
Find the areas of the following figures.
Find the volumes of the following figures.
Mixed Review
1. Simplify \begin{align*}5x(3x+5)+11(-7-x)\end{align*}.
2. Cal High School has grades nine through twelve. Of the school's student population, \begin{align*}\frac{1}{4}\end{align*} are freshmen, \begin{align*}\frac{2}{5}\end{align*} are sophomores, \begin{align*}\frac{1}{6}\end{align*} are juniors, and 130 are seniors. To the nearest whole person, how many students are in the sophomore class?
3. Kerrie is working at a toy store and must organize 12 bears on a shelf. In how many ways can this be done?
4. Find the slope between \begin{align*}\left ( \frac{3}{4},1 \right )\end{align*} and \begin{align*}\left ( \frac{3}{4}, -16 \right )\end{align*}.
5. If \begin{align*}1 \ lb=454 \ grams\end{align*}, how many kilograms does a 260-pound person weigh?
6. Solve for \begin{align*}v\end{align*}: \begin{align*}|16-v|=3\end{align*}.
7. Is \begin{align*}y=x^4+3x^2+2\end{align*} a function? Use the definition of a function to explain.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
TermDefinition
binomial A binomial is a polynomial with two terms.
Distributive Property for Binomials To use the Distributive Property with two binomials, multiply each term in the first factor by each term in the second. $(a+b)(c+d)=c\cdot (a+b)+d\cdot (a+b)=c\cdot a+c \cdot b+d \cdot a+d \cdot b \ (\text{or} \ ca+cb+da+db)$
distributive property The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
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Description
Difficulty Level:
Basic
Tags:
Subjects: | HuggingFaceTB/finemath | |
1. ## triangle perimeter
Given that the perimeter of a triangle is 180 in. If the angle of triangle are in the ratio 5:6:7. Determine the sides of triangle
2. Originally Posted by aeroflix
Given that the perimeter of a triangle is 180 in. If the angle of triangle are in the ratio 5:6:7. Determine the sides of triangle
Angles are 50,60 and 70 : OK?
Let shorter side (side opposite angle=50) = 1
Calculate 2 other sides using Law of Sines
Calculate sides of a similar triangle with perimeter 180 | HuggingFaceTB/finemath | |
# Is this number a repdigit?
## Challenge
A repdigit is a non-negative integer whose digits are all equal.
Create a function or complete program that takes a single integer as input and outputs a truthy value if the input number is a repdigit in base 10 and falsy value otherwise.
The input is guaranteed to be a positive integer.
You may take and use input as a string representation in base 10 with impunity.
## Test cases
These are all repdigits below 1000.
1
2
3
4
5
6
7
8
9
11
22
33
44
55
66
77
88
99
111
222
333
444
555
666
777
888
999
A larger list can be found on OEIS.
## Winning
The shortest code in bytes wins. That is not to say that clever answers in verbose languages will not be welcome.
• – Leaky Nun Jun 8 '17 at 3:02
• @AidanF.Pierce What's the biggest number the input will be? – stevefestl Jun 14 '17 at 9:00
# MATL (6 5 bytes or 2 bytes for Luis Mendo solution)
Vun1=
Try it online!
## Explanation
V % convert to string
un % find unique characters and count them
1= % if there is only one unique character, then we pass.
&=
Outputs a truthy array (all 1's) if repdigit, or a falsy array (some 0 in the array) if not a perfect repdigit.
Try it online!
• Since any truthy/falsy outputs are valid, you can use V&=; or j&= to avoid floating-point limitations. Or even &=, since inputting a string is allowed – Luis Mendo May 27 '19 at 9:47
• Hmm, I'm afraid I can't follow. I tried replacing "1=" with "&=" and it didn't work. Also tried "V&=" as the whole code, in case that's what you meant. – DrQuarius Jul 17 '19 at 11:37
• I mean just &=. A non-empty array containing only ones is truthy, and an array containing some zero is falsy (see link in my previous comment for explanation about truthy/falsy) – Luis Mendo Jul 17 '19 at 16:06
• Wow, ok. That's a very simple solution then. Will add. – DrQuarius Aug 2 '19 at 4:42
# Python 3, 23 bytes
lambda s:s==s[0]*len(s)
Try it online!
Not shorter than @shadow's answer, but I thought it was interesting. Should work in Python 2 as well.
# Ruby, 13 bytes (12 + '-n' flag)
p~/^(.)\1*$/ Try it online! # C#, 38 bytes using System.Linq;s=>s.All(c=>c==s[0]) Or alternatively for 44 bytes: using System.Linq;s=>s.Distinct().Count()==1 • I'm no C# developer, but do you really require using System.Linq? I've seen plenty of C# lambda answers without that. – Olivier Grégoire Jun 8 '17 at 12:37 • @OlivierGrégoire Linq is required for All, it's like needing an external package in C and requiring to import it. If I am using Linq I must include the using. – TheLethalCoder Jun 8 '17 at 12:40 ## Clojure, 17 12 bytes You may take and use input as a string representation in base 10 with impunity. Oh in that case: #(apply = %) Original: #(apply =(str %)) # QBIC, 36 23 bytes [_l!:$||p=p*z+1]?b%p=0
## Explanation
: Read a number from the cmd line
! $cast it to string _l | Take its length [ | And run a FOR-loop from 1 to that length p= p starts out as 0. set it to p*z itself multiplied by 10 (z=10 in QBIC) (still 0 on the first run +1 then add 1. On consecutive FOR-loops yields 1, 11, 111, .... ] Close the FOR loop ?b%p=0 PRINT -1 if b mod p is 0 (ie 444 % 111 = 0), or 0 otherwise ## ><>, 26 24 bytes !vi:0(?v::&r&=? >0n;n1< Reads the input as a string and reads every character in a loop i // Take the next character and pushes it onto the stack :0(?v // If the end of the input has been reached, goto the successful termination code :: // Duplicate the top of the stack twice !v &r&=? // Pop the top of the stack, pop the bottom of the stack, and compare them. If they are not equal, goto the failure termination code >0n; // Failure termination code. Prints a 0 and terminates ;n1< // Success termination code. Prints a 1 and terminates # Perl 6, 12 bytes {[==] .comb} Reduces the list of characters in the input number with the numeric equality operator. # AWK, 33 bytes BEGIN{FS=""}{$0=NF==gsub($1,"")}1 Try it online! Replaces all characters in the input with the first character and compares the changed count to the total number of characters. ## PowerShell, 23 bytes "$args"-match'^(.)\1*$' Save as repdigit.ps1 and run with PS C:\wherever\repdigit.ps1 444, outputs True or False. If you want the Python len(set(input))==1 style, it costs more at 32 bytes: @($args-split''|group).count-eq2
(Noting that the split on the space between chars also outputs an empty start and end string as well as the characters).
# Python, 27 22 bytes:
Just to get things started.
lambda x:len(set(x))<2
Converts to a string, converts the string into a set of all distinct characters in that string, and checks whether there aren't multiple elements in the set.
5 bytes saved thanks to @LeakyNun's reminder.
• Please specify the version of Python. – Leaky Nun Jun 8 '17 at 3:03
• @LeakyNun It's not customary to specify a version if the code works in all recent ones. – Dennis Jun 8 '17 at 3:04
• @Dennis If it is Python 2, I can suggest a golf. – Leaky Nun Jun 8 '17 at 3:23
• Now that the input specs are more lax, you can replace str(x) with x. – Leaky Nun Jun 8 '17 at 3:23
• In Python 3.6+ this can be len({*x})<2 – pxeger Feb 24 at 14:31
# PHP, 28 bytes
<?=!count_chars($argn,3)[1]; count_chars with mode=3 creates a string with all different characters in $argn.
If there is only one, the second character will be empty == falsy.
Run as pipe with -F.
# Smalltalk, 21 19 bytes
[:s|s asSet size<2]
Try it online!
You can execute the block above by sending the message value: to the block with a String. A String in Smalltalk is written with single quotes. A longer version of the block could accept an Integer. It would take 30 28 bytes.
[:i|i asString asSet size<2]
I tried both solutions in the current online version of Amber Smalltalk and Pharo Smalltalk version 6.0.
Thanks to the suggestion in the comments I was able to remove the spaces before and after the <. It's been many years since I wrote Smalltalk but I doubt I ever tried that in practice. The shorter version worked in both versions of Smalltalk I tested.
• Are the spaces around the < necessary? I haven't used small talk very much, but most languages wont require that sort of thing. – Wheat Wizard Jun 14 '17 at 3:51
• Yes, the spaces are necessary. The syntax of Smalltalk is Object<space>message. I was able to remove two spaces before and after the block separator |. – Donald Raab Jun 14 '17 at 3:56
• I stand corrected. I tried removing the spaces before and after the < and it seems to work. – Donald Raab Jun 14 '17 at 3:58
## Check, 135134132 124 bytes
[r #v
#:>10%:]R+r->\#v#?
#v
# >10-\)\# # ?
d # ^
#R:
:>=r,#v
#ddd[=#(:@:@=R-?
d\:!:R *
o>]=d#^
Input should be passed as a command-line argument. Outputs some unprintables for truthy and only zero bytes for falsey. Always terminates with an IndexError due to the fact that it abuses an interpreter bug.
### How does it work?
This code is divided into two segments. The first half turns the input integer into a list of digits, and the second half checks that all of its digits are equal to the first digit (i.e. they are all equal).
It roughly corresponds to the following pseudocode:
1. Read input and call it i.
2. Create an empty array and store it in the register. ([r)
3. If i is 0, go to step 13.
4. Take the number modulo 10. Call it x.
5. Prepend x to the register and store it back in the register.
6. Subtract x from i.
7. Create a counter, starting at 0.
8. If i is 0, go to step 12.
9. Increment the counter.
10. Decrement i by 10.
11. Go back to step 8.
12. Set i to whatever value is now in the counter.
13. Go back to step 3.
14. Load the register, which is now an array containing the digits of the input. Call this d.
15. Get the first element in d and store it in the register.
16. Create a counter called j, initialized to the length of d.
17. Decrement j.
18. If the jth element of d is not equal to the value of the register, crash the program. (We found a digit that is not correct.)
19. Otherwise, print some random junk !j times. This only prints something if j == 0.
20. Create an array of length 1 and get the !jth element. This will crash the program if j == 0.
21. If the program has not yet crashed, j must not yet be zero, so go back to step 17.
# Julia, 30 bytes
f(n)=length(Set(string(n)))==1
## F#, 38 bytes
let f n=Seq.length(Seq.countBy id n)=1
Try it online!
# QBasic, 37 32 bytes
INPUT n$?n$=STRING$(LEN(n$),n$) -5 bytes thanks to steenbergh The STRING$ function takes two arguments, a number n and a string s$, and constructs a string consisting of n copies of the first character of s$.* So this code reads our number as a string n$, generates a string consisting of LEN(n$) copies of n$'s first character, and checks to see if it is equal to n$.
* The second argument can also be an integer codepoint, in which case it repeats the character corresponding to that codepoint.
• Are you sure you need the ASC? Often QBasic simply takes the 1st char of a string when it expects only 1 char but the string itself is longer. I think, but can't test atm, that this will work fine too: INPUT n$:?n$=STRING$(LEN(n$),n$) – steenbergh Jun 20 '18 at 13:34 • Confirmed on QB4.5: if we input n$ as 123, then STRING$(3, n$) is 111. – steenbergh Jun 20 '18 at 16:22
• @steenbergh [facepalm] I just read that in the help file, too. Thanks! – DLosc Jun 20 '18 at 20:36
# Whitespace, 96 93 bytes
[S S S N
_Push_0][S N
S _Duplicate][T N
T T _Read_STDIN_as_integer][T T T _Retrieve][N
S S N
_Create_Label_LOOP][S N
S _Duplicate][S S S T T S S T S S N
_Push_100][T S T T _Modulo][S S S T S T T N
_Push_11][T S T T _Modulo][N
T S S N
_If_0_Jump_to_label_NEXT][S S S T S S T N
_Push_9][S N
T _Swap_top_two][T S T S _Integer_division][T N
S T _Print_as_integer][N
S S S N
_Create_Label_NEXT][S S S T S T S N
_Push_10][T S T S _Integer_division][N
S N
N
_Jump_to_Label_LOOP]
Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.
Port of @Dennis♦' C answer, so also outputs a positive digit ([1,9]) as truthy and 0 as falsey.
Try it online (with raw spaces, tabs and new-lines only).
Pseudo-code:
Integer i = STDIN as integer
Start LOOP:
If(i modulo-100 modulo-11 == 0):
i = i integer-divided by 10
Go to next iteration of LOOP
i = 9 / i
Print i to STDOUT as integer
Stop the program with an error
# Taxi, 730 706 bytes
-24 bytes by eliminating linebreaks.
Clever answer in a verbose language? Check and check.
Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to Chop Suey.Go to Chop Suey:n 1 r 1 l 4 r 1 l.Pickup a passenger going to Cyclone.[B]Switch to plan D if no one is waiting.Pickup a passenger going to Crime Lab.Go to Cyclone:n 1 l 3 l.Pickup a passenger going to Crime Lab.Pickup a passenger going to Crime Lab.Go to Crime Lab:n 2 r 2 r.Switch to plan C if no one is waiting.Pickup a passenger going to Cyclone.Go to Fueler Up:n.Go to Chop Suey:n 3 r 1 l.Switch to plan B.[C]0 is waiting at Writer's Depot.Go to Writer's Depot:n 4 l 2 l.Pickup a passenger going to Cyclone.Go to Chop Suey:n 3 r 3 r.[D]Go to Cyclone:n 1 l 3 l.Pickup a passenger going to Post Office.Go to Post Office:s 1 l 2 r 1 l.
Try it online!
Try it online with linebreaks!
Doesn't return to the Taxi Garage after the program ends, so the boss fires me, and it exits with an error.
What this program actually does is checks whether or not the input consists of any repeating character, and outputs that character if it does, or 0 if it doesn't.
# Alternative solution, 602582541 445 bytes
-20 bytes by eliminating linebreaks.
-41 bytes by having the program error out upon a false result.
-96 bytes by having the program output via error message.
Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to Chop Suey.Go to Chop Suey:n 1 r 1 l 4 r 1 l.Pickup a passenger going to Cyclone.[B]Switch to plan C if no one is waiting.Pickup a passenger going to Crime Lab.Go to Cyclone:n 1 l 3 l.Pickup a passenger going to Crime Lab.Pickup a passenger going to Crime Lab.Go to Crime Lab:n 2 r 2 r.Pickup a passenger going to Cyclone.Go to Fueler Up:n.Go to Chop Suey:n 3 r 1 l.Switch to plan B.[C]
Try it online!
Try it online with linebreaks!
This alternative solution errors out with The boss couldn't find your taxi in the garage. You're fired! if the input is a repdigit, and no outgoing passengers found if the input is not a repdigit (meaning this also works with 0...or 00, or 000...).
• For the second program? Sure, I could exit another way. – JosiahRyanW Oct 9 '18 at 0:45
• I know, I made the modification. – JosiahRyanW Oct 9 '18 at 0:49
• Note that a program can output via exit code, so theoretically you could have it crash if not a repdigit and exit normally otherwise instead of outputting – Jo King Oct 9 '18 at 0:55
• Could it crash in two different ways depending on result? – JosiahRyanW Oct 9 '18 at 0:57
• I think so? Relevant meta, but I don't see any reason not to allow it, since it is easily observable. Edit: ah, I realise it would be two different outputs to STDERR, so you're good – Jo King Oct 9 '18 at 1:05
# J, 6 bytes
1=#@~.
Try it online!
Take unique elements; does its length equal 1?
/:-:\:
Try it online!
Does ascending sorting order equal descending sorting order?
-:##{.
Try it online!
Does the first element, duplicated to the length, equal the original?
-:1&|.
Try it online!
Does the input, rotated once, equal itself?
# Pascal (FPC), 8175 68 bytes
var n:string;begin read(n);write(n=StringOfChar(n[1],length(n)))end.
Try it online!
It seems that taking a number as string is shorter.
### 75 bytes - taking the number as integer:
Thanks to @Ørjan Johansen for -6 bytes - mod 100 mod 11 trick
var n:word;begin read(n);while n mod$64mod$B=0do n:=n div$A;write(n<=9)end. Try it online! $64, $B and $A are hexadecimal constants, they eliminate some whitespace that would be needed for their decimal counterparts.
• I think you can shave off a bit with the mod 100 mod 11 trick some others are using. – Ørjan Johansen Oct 8 '18 at 20:08
• @ØrjanJohansen Thanks, a good one! – AlexRacer Oct 8 '18 at 21:19
# Pepe, 48 bytes
rrEEreeeEEeeeErEeREEeREEREEEEEEErEEEEreErEEEeReE
Outputs 1 if number is a repdigit, nothing otherwise.
Try it online!
## Explanation
(for input 45. 52 and 53 are charcodes of digits)
rrEE # Create label 0, implicitly push 0 to the second stack [] [0]
reeeEEeeeE # Print "1"
rEe # Return
REEe # Start: Take input as charcodes to the first stack [52,52,53] [0]
REE # Create label I (1)
REEEEEEE # Move first item in first stack to the other [52,53] [0, 52]
rEEEE # Reset second stack pointer position [>52, 53] [>0, 52]
reE # If item in first stack equals the first item in the second stack (0),
# call label 0. It will only be true if first stack is empty
# because it will implicitly give 0.
rEEEe # Move second stack pointer to the end [>52, 52] [0, >52]
ReE # Repeat if items are the same (52 == 52), end program otherwise
reeeEEeeee # Optional: Print 0.
# Pepe, 69 60 bytes
REeErEEEEEREEEeeEREEEEeEeEREEEEEEeREEEeREEEEEerrEEreEErEereE
Try it online!
How it works?
It takes the greatest digit and subtracts it from the least digit. Repdigit numbers always evaluate to 0.
REeE # Takes a number (stack r)
rEEEEEREEEeeE # Splits by digits
REEEEeEeE # Sorts them
REEEEEEe # Copies first digit to other stack
REEEe # Move pointer to last in stack r
REEEEEe # Subtract stack R to stack r
rrEErEEEEErEereE # Print if 0, else none
# Excel, 28 27 bytes
=A1=MID(A1,2,A1)&LEFT(A1,1)
# naz, 50 bytes
2a2x1v1r2x2v1x1f1r3x1v2e3x2v1e0m1o0x1x2f0m1a1o0x1f
Works for any input integer, provided it's passed as a file terminated with the control character STX (U+0002).
Explanation (with 0x commands removed)
2a2x1v # Set variable 1 equal to 2
1r2x2v # Read the first byte of input and store it in variable 2
1x1f # Function 1
1r # Read a byte of input
3x2v1e # Jump back to the start of the function if it equals variable 2
0m1o # Otherwise, output 0
1x2f # Function 2
0m1a1o # Output 1
1f # Call function 1
# Pip, 3 bytes
$=a fold by equality. Try it online! # Cheddar, 16 bytes s->s.len*s[0]==s Try it online! ## Integer input, 29 bytes n->(s->s.len*s[0]==s)("%d"%n) Try it online! # Tcl, 53 bytes proc r n {expr 1==1[string trim$n [string in $n 0]]} "string trim" removes leading and trailing occurrences of all its second argument's characters from its first argument (usually used for whitespace). Second argument here is the first digit of$n. For a repdigit an empty string remains, and attaching that to 1 remains 1. For number 42424, 242 would remain after the trim, so 1 is not equal to 1242.
PS: attaching to 1 is a golfing-thing that saved two bytes versus comparing the trim with empty string.
## Pyke, 3 bytes
}t!
Try it here!
} - uniquify(input)
t - ^[:-1]
! - not ^
# PHP, 34 bytes
<?=preg_match('#^(.)\1*$#',$argn);
Try it online! | open-web-math/open-web-math | |
## What is the odd number?
: a whole number that is not able to be divided by two into two equal whole numbers The numbers 1, 3, 5, and 7 are odd numbers.
## What do odd numbers symbolize?
Odd numbers – which correspond to the right side of the brain – represent intangible things, such as creativity, flair, inspiration, and love of adventure. They do not like to “fit in” and sometimes will stop right in the middle of something and go off in another direction.
## What is odd number explain with example?
Odd numbers are the numbers that cannot be divided by 2 evenly. It cannot be divided into two separate integers evenly. If we divide an odd number by 2, then it will leave a remainder. The examples of odd numbers are 1, 3, 5, 7, etc.
## What is the odd number in English?
An odd number is an integer when divided by two, either leaves a remainder or the result is a fraction. One is the first odd positive number but it does not leave a remainder 1. Some examples of odd numbers are 1, 3, 5, 7, 9, and 11. An integer that is not an odd number is an even number.
You might be interested: FAQ: What Does The Number On The Inside Of A Coke Cap Mean?
## Can 3 odd numbers make an even number?
Without cheating, it’s not possible because sum of three odd numbers can’t be even.
## Why do odd numbers look better?
An odd number of details is more effective at capturing your gaze. Odd numbers force your eyes to move around the grouping–and by extension, the room. That forced movement is the heart of visual interest. It’s for that reason that a set of three is more appealing and memorable than something paired off in two’s.
## What are odd numbers from 1 to 100?
The odd numbers from 1 to 100 are: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99.
## What is odd number and even numbers?
An even number is a number that can be divided into two equal groups. An odd number is a number that cannot be divided into two equal groups. Even numbers end in 2, 4, 6, 8 and 0 regardless of how many digits they have (we know the number 5,917,624 is even because it ends in a 4!). Odd numbers end in 1, 3, 5, 7, 9.
## How do you calculate odd numbers?
To identify an odd number we can directly divide it by 2. If the number is exactly divisible by 2 it is not an odd number. For example, 4 is not an odd number as it is exactly divisible by 2.
You might be interested: What Does The Number Next To Home On Facebook Mean?
## What is the sum of odd numbers?
The total of any set of sequential odd numbers beginning with 1 is always equal to the square of the number of digits, added together. If 1,3,5,7,9,11,…, (2n-1) are the odd numbers, then; Sum of first odd number = 1. Sum of first two odd numbers = 1 + 3 = 4 (4 = 2 x 2).
## Which is the even number?
: a whole number that is able to be divided by two into two equal whole numbers The numbers 0, 2, 4, 6, and 8 are even numbers.
## What is the smallest odd number?
The smallest odd number is ‘ 1 ‘. ‘1’ is neither a prime number nor a composite number by convention. The next smallest odd number is ‘3’. Factors of 3 = 1, 3.
## What is odd word example?
Solution: In 1), 3) and 4), the words are synonymous pair of each other i.e. they give the same meaning of each other but in 2), light and heavy are antonyms i.e. they give opposite meaning of each other. Accept 2), all other pairs gives the same relation. Hence, 2) is the odd word group.
## Is 2 an odd or even number?
By comprehending the number at “ones” place For example, numbers such as 11, 23, 35, 47 etc. are odd numbers. All the numbers ending with 0,2,4,6 and 8 are even numbers. | HuggingFaceTB/finemath | |
# MA22S6 Numerical and Data Analysis 2015-2016 Homework sheet 5 Due Thursday 10
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```MA22S6 Numerical and Data Analysis 1 2015-2016
Homework sheet 5
Due Thursday 10th of March 2015 in class
1.
a) We consider two random variables X and Y with expectation value µX and µY respectively.
Show that the covariance
Cov(X, Y ) = E[(X − µX )(Y − µY )]
can also be written as
Cov(X, Y ) = E[XY ] − µX µY .
b) Prove that that if two random variables X and Y are independent then X and Y are
uncorrelated (i.e. their covariance vanishes)
c) For 3 random variables X, Y, Z show that
V ar[X + Y + Z] = V ar[X] + V ar[Y ] + V ar[Z] + 2Cov(X, Y ) + 2Cov(X, Z) + 2Cov(Y, Z)
2. Given the joint probability mass function (ie the probability that X = x and Y = y)
pX,Y (x, y) = P (X = x, Y = y)
in the following table
y
1
2
3
1
0
1/4
0
x
2
1/4
0
1/4
3
0
1/4
0
Show that X and Y are dependent but uncorrelated.
3. The correlation coefficient ρ(X, Y ) is defined by
ρ(X, Y ) = p
Cov(X, Y )
V ar[X]V ar[Y ]
provided that V ar[X] 6= 0 and V ar[Y ] 6= 0, otherwise ρ(X, Y ) = 0.
a) Relate ρ(rX + s, tY + u) (with real numbers r, s, t, u) to ρ(X, Y ).
b) ρ takes values in [−1, 1]. In which cases do you find ρ(X, Y ) = 1, −1, respectively?
4. A dart is thrown at a disk of radius 1 around the origin in the x − y plane. The random variables
X, Y measure
p the x- and y-coordinates with joint probability density function fX,Y (x, y) = c
(c > 0) for x2 + y 2 ≤ 1 and fX,Y (x, y) = 0 otherwise (i.e. no dart misses the disk).
a) Determine the constant c.
b) Calculate the marginal probability density functions fX (x) and fY (y).
c) Are X and Y independent random variables?
d) Are X and Y correlated?
1
Lecturer: Stefan Sint, [email protected]
``` | HuggingFaceTB/finemath | |
Python NumPy: Compute the trigonometric sine, cosine and tangent array of angles given in degrees - w3resource
# Python NumPy: Compute sine, cosine and tangent array of angles given in degrees
## Python NumPy Math, Statistics, Trigonometry and Linear algebra: Exercise-21 with Solution
Write a Python program to compute the trigonometric sine, cosine and tangent array of angles given in degrees.
Sample Solution:-
Python Code:
``````import numpy as np
print("sine: array of angles given in degrees")
print(np.sin(np.array((0., 30., 45., 60., 90.)) * np.pi / 180.))
print("cosine: array of angles given in degrees")
print(np.cos(np.array((0., 30., 45., 60., 90.)) * np.pi / 180.))
print("tangent: array of angles given in degrees")
print(np.tan(np.array((0., 30., 45., 60., 90.)) * np.pi / 180.))
```
```
Sample Output:
```sine: array of angles given in degrees
[ 0. 0.5 0.70710678 0.8660254 1. ]
cosine: array of angles given in degrees
[ 1.00000000e+00 8.66025404e-01 7.07106781e-01 5.00000000e-01
6.12323400e-17]
tangent: array of angles given in degrees
[ 0.00000000e+00 5.77350269e-01 1.00000000e+00 1.73205081e+00
1.63312394e+16]
```
Python Code Editor:
```import numpy as np
print("sine: array of angles given in degrees")
print(np.sin(np.array((0., 30., 45., 60., 90.)) * np.pi / 180.))
print("cosine: array of angles given in degrees")
print(np.cos(np.array((0., 30., 45., 60., 90.)) * np.pi / 180.))
print("tangent: array of angles given in degrees")
print(np.tan(np.array((0., 30., 45., 60., 90.)) * np.pi / 180.))```
Improve this sample solution and post your code through Disqus | HuggingFaceTB/finemath | |
## Frequently Asked Questions
Come here to find answers to commonly asked questions. Use the search box below to find answers to your questions. For example, if you are trying to find out about your marks within a course you might search for "grades", "marks", "results", etc. You may also select from one of the listed categories to see all the entries pertaining to that subject. If there is a question you do not see listed, you may be best served by asking in the forums. We will try to update the questions here regularly based on those showing up in the forums.
Browse the glossary using this index
A | B | C | D | E | F | G | H | I | J | K | L | M | N | O | P | Q | R | S | T | U | V | W | X | Y | Z | Yonke
Page: 1 2 3 (Next)
Yonke
### A
Question:
#### Adaptive mode
(Last edited: Wednesday, 30 May 2018, 6:42 AM)
Impendulo:
What is adaptive mode?
Adaptive mode is a setting that determines how a question behaves in a quiz activity. (The quiz activity type is used for a variety of assessments which may be labelled assignments, assessments, submissions, tutorials, tests and exams. You know its a quiz if the link to it on the course page has an icon like this: .)
If a quiz uses adaptive mode, then students can receive feedback immediately after attempting each question within a single quiz attempt. Moreover, students can attempt each question more than once before moving on from it (or later, before finishing the quiz.) This is beneficial because you can get the answer right in your mind before continuing, regardless of the mark you achieved.
You can easily tell if a quiz uses adaptive mode by looking for a "Check" button beneath each question. After selecting what you think is the right answer, click "Check" to check if you are correct. If you were wrong, try again.
Adaptive mode does not use negative marking. Instead (generally) it reduces the maximum mark that can be achieved in each successive question attempt until finally it is zero. Thus, in a multiple-choice question with four choices, a student would typically merit:
• 100% of the question value if they got it right first time;
• 66.7% of the question value if they got it right second time;
• 33.3% of the question value if they got it right third time;
• 0 if they got it right fourth time.
So, if the question is worth 1 mark, the four scenarios above would achieve 1 mark, 2/3 mark, 1/3 mark, and 0 marks respectively. However, if there are only two choices (as in the case of True/False or Yes/No questions, among others), a student who does not get it right on his first attempt gets zero for that question because there is only one other choice, so the right answer is obvious after the first attempt.
Adaptive mode is usually used for formative assessments. In other words, for activities in which participation is intended to improve learning. Adaptive mode is typically not used for exams, since exams (summative assessments) are not designed as learning activities.
Keyword(s): quizquestionadaptiveadaptive mode
Question:
#### Are there libraries available for student use?
(Last edited: Monday, 28 October 2019, 10:31 AM)
Impendulo:
SATS has a growing library and you are welcome to visit the Seminary to browse through the books. We do not lend out books. You may search the catalogue using the Online Public Access Catalogue (OPAC) here: https://cloud.collectorz.com/satslibrary/books/. In addition to our traditional library, SATS also subscribes to a number of online digital journals and Ebooks, available here, which are made available to all our students. If you have a problem in accessing the journal platforms, please contact Tawanda (library@sats.edu.za).
Keyword(s): LibraryJournalebooks
Question:
#### Assignment submission instructions
(Last edited: Wednesday, 6 December 2017, 10:18 AM)
Impendulo:
Step 1: Click "Add Submission"
Step 2: Upload your document
If this is the first assignment you have ever submitted with SATS, you will be required to accept the Turnitin "End User License Agreement" (EULA) in order to proceed.
NB. Supported document formats are limited to those specified on Turnitin's submission guide. Do NOT submit .pages files or zip archives!
Step 3: Click "Save Changes"
Your document has been uploaded. Note, however that the status will indicate as "Draft (not submitted)". This means you can view the originality report and similarity index, as well as make changes to your submission.*
Step 4: Click "Submit Assignment"
Once you submit your assignment your tutor will be notified that you have made a submission and you will no longer be permitted to make changes. Note that you may be required to accept a submission declaration as part of this step, stating that your assignment is free from plagiarism and that you did not have assistance in completing the assignment.
*Note that the similarity index for your assignment may take up to ten minutes to generate. If you require further assistance, please consult this FAQ on submitting assignments, which includes step-by-step instructions and a screencast.
### B
Question:
#### Blank course presentation
(Last edited: Tuesday, 25 September 2018, 1:03 PM)
Impendulo:
##### Help! I cannot see my course lectures on the counseling courses, even though it worked before!
Do not despair. There is a known problem with the Scorm presentations on the counseling courses, but it only happens if you have flash disabled. In order to fix this problem follow the steps below as appropriate to your browser:
#### Chrome
1. Click the padlock in the address bar as pictured:
2. Click flash option to allow
3. Click Allow
#### Opera
1. Go to opera://settings/contentExceptions#plugins
2. Add my.sats.edu.za with the behaviour "Allow"
#### Safari
1. Make sure you have flash installed. You may download it here: https://get.adobe.com/flashplayer/otherversions/
2. Whilst in Safari press ⌘ and , (comma) at the same time to open your browser preferences
3. go to the security tab (or Websites Tab for Safari 11)
4. Click "Plug-in Settings" on the bottom
5. Make sure Flash is enabled
6. Add my.sats.edu.za to the allowed sites (see screenshot)
#### Firefox
1. Make sure you have flash installed. You may download it here: https://get.adobe.com/flashplayer/otherversions/
2. Go to about:addons in your address bar
3. click plugins
4. Where "Shockwave Flash" is listed make sure the settings says "Ask to Activate"
#### Internet Explorer
##### Do not use Internet Explorer. Just about anything else will work better, including an old pair of shoes. Seriously, just use a different browser.
Keyword(s): SCORMHCCCCounseling Courses
### C
Question:
#### Can I change my curriculum halfway through?
(Last edited: Monday, 28 October 2019, 11:06 AM)
Impendulo:
Yes. Because we are adding electives to our curriculum on a regular basis there may be one or two courses you would like to add. Calling the Registrar to adjust your curriculum can accomplish this for a fee:
1. South Africa - R112
2. Africa, Asia, South America - $11 USD 3. Europe, North America, Australia and New Zealand$17 - USD
##### N.B. Prices are subject to change
Keyword(s): curriculumelective
### D
Question:
#### Do I need textbooks?
(Last edited: Monday, 28 October 2019, 11:41 AM)
Impendulo:
All of the courses within SATS' first year curriculum are self contained and therefore do not need any textbooks, with the exception of the following:
Greek 1
• Mounce, William D. 2009. Basics of Biblical Greek. Third edition. Grand Rapids: Zondervan (or the 4th edition that has just come out. It seems to only be available as an eBook at the moment)
• Mounce, William D. 2009. Basics of Biblical Greek Workbook. Third edition. Grand Rapids: Zondervan
Greek 2
• Mounce, William D. 2009. Basics of Biblical Greek. Third edition. Grand Rapids: Zondervan.
• Mounce, William D. 2009. Basics of Biblical Greek Workbook. Third edition. Grand Rapids: Zondervan.
• Wallace, Daniel B. 2000. The Basics of New Testament Syntax: An Intermediate Greek Grammar. Grand Rapids, MI: Zondervan (or Wallace, Daniel B. 1996. Greek Grammar Beyond Basics. Grand Rapids, MI: Zondervan)
An additional recommended resource for 2nd-year is:
• Mounce. William D. 1994. The Morphology of Biblical Greek. Grand Rapids: Zondervan.
Hebrew
• Pratico, G. and van Pelt, M. 2007. Basics of Biblical Hebrew (Second Edition). Grand Rapids: Zondervan.
Higher Certificate in Christian Counselling:
• Collins, G. 2007. Christian Counseling - a Comprehensive Guide (Third Edition). Dallas: Nelson.
• McMinn, M.R. & Campbell, C.D. 2007 Integrative Psychotherapy - toward a Comprehensive Christian Approach. IVP Academic
• Meier, Minirth, Wichern & Ratcliff. 1999. Introduction to Psychology and Counseling: Christian Perspectives and Applications. 2nd Ed. Baker Books.
• Tan, S. & Scalise, E.T. 2016. Lay Counseling: Equipping Christians for a Helping Ministry. (Revised & Updated). Zondervan.
• Tan, Siang-Yang, 2011. Counseling and Psychotherapy: A Christian Perspective. Baker Publishing.
The third year subject, PRA3121 - Conflict and Reconciliation requires the following textbook:
• Sande, Ken. 2004. The Peacemaker: A Biblical Guide to Resolving Personal Conflict Grand Rapids: Baker Books.
READERS: Due to the fact that many of the prescribed theological textbooks are sometimes difficult to find and extremely expensive, the South African Theological Seminary has compiled Readers for courses within the Higher Certificate and Bachelor programmes.
These Readers include selected extracts from all the books recommended for the courses in these two programmes, as well as a selection of other relevant documents which we are sure will assist you as you complete each assignment. So that we do not infringe any copyright laws, we have signed a contract with the Dramatic Arts and Literary Rights Organisation (DALRO), the terms of which enable us to include these extracts in the Readers.
Although the Readers are provided to you free of charge, you will still have the option to obtain the textbooks if you would prefer.
Our agreement with DALRO does not allow us to distribute Readers electronically. However, we have received permission to store them on a secure website from which SATS students can access them. Please note that the Readers are exclusively for the use of SATS students.
Keyword(s): textbookreaderreading
Question:
#### Does my denomination recognise the SATS qualifications?
(Last edited: Monday, 28 October 2019, 11:22 AM)
Impendulo:
There are a number of denominations currently enrolling students at SATS for their theological studies. You will need to contact the head of your denomination in order to find out if the SATS degree is sufficient for their denominational requirements. SATS does not ordain ministers.
Keyword(s): Accreditationdenominationordain
Question:
#### Does SATS have residential facilities?
(Last edited: Monday, 28 October 2019, 11:43 AM)
Impendulo:
No. The South African Theological Seminary is a distance learning education institution that allows you to study at your own pace and in the comfort of your own home. We do not apply for student visas.
Keyword(s): VISAaccommodationresidential
### E
Question:
#### Ebsco ebooks on Android
(Last edited: Monday, 28 October 2019, 11:46 AM)
Impendulo:
Keyword(s): Ebscoebookslibraryandroid
### H
Question:
#### Help! MySats lost my work after requiring me to login again!
(Last edited: Monday, 4 November 2019, 10:37 AM)
Impendulo:
It can be very frustrating to lose work you are in the process of completing. It has happened to all of us, whether it is the result of a power failure, the infamous Windows "blue screen of death", or a dear kitty cat chewing on your power cord. Whatever the cause, it is very disheartening.
You will be required to login again for the following reasons:-
• After four hours of inactivity
• You close the browser window or your internet crashes
• You change to a different device or location
If you anticipate doing either of the above, it is highly advisable that you type your work in a word processor and then paste in the text box. Make sure you use this icon to paste anything from Word, so as to ensure you do not have any issues with formatting.
And remember, you can still lose work from your word processor if you do not save it often! It has been said, "Jesus saves, and so should you..." So, save your work and avoid heartbreak.
Keyword(s): Logintimeoutlost
Question:
#### How can I conserve data usage in a live (BigBlueButton) class?
(Last edited: Tuesday, 3 May 2016, 1:05 PM)
Impendulo:
##### There are a number of ways to manage your data consumption in a live class. However, it should be noted that the minimum amount of data usage for a one hour class will be approximately 240-250MB. Take the following steps to reduce your data consumption:
1. Do not broadcast your webcam. Broadcasting your webcam doubles the amount of data you use, since you need to both send and receive audio and video streams (assuming there is at least one other webcam).
2. If you are using the flash version, you can click the 'x' above each webcam feed to stop downloading the feed. This will greatly reduce bandwidth consumption for meetings with webcams.
3. Join the class using the HTML5 client. At the time of this writing, webcams are not broadcast to the HTML5 client even if other participants are broadcasting, and the client environment is smaller, so this will go a long way in reducing data consumption. Please see this guide on how to join a bigbluebutton session using html5 instead of Adobe Flash.
4. Join the session in listen only mode. Listen only mode allows you to receive the audio feed, but you will not be able to contribute except via the text chat. This will allow you to ask a question through the text chat and listen to the audio response. Please see this guide on how to join a live class using listen only mode.
Question:
#### How can I download my marked assignment from Feedback Studio with instructor comments?
(Last edited: Wednesday, 6 November 2019, 7:55 AM)
Impendulo:
In addition to viewing your marked assignment on Turnitin, learners may wish to download a copy for future reference. There are two different versions of a paper that one may wish to download, each with different content.
The first report relates to the originality of your paper. This is generated by Turnitin's Originality service, so it does not include feedback from your marker, nevertheless a student may find this useful if they need to revise the paper to be in compliance with proper citation standards.
The second report relates to the actual feedback from your lecturer on your paper. This may relate to general comments about formatting or content from the Quickmark section, specific comments and overall feedback, and details from rubric marking.
Both reports may be downloaded by first clicking on either the Originality tab or the Grademark tab. From this point the instructions to download each report is the same:
Follow these steps and screenshots:
1. Login to MySats
2. Proceed to the course and assessment in question
3. Where your assignment appears, click the similarity index or rubric icon.
4. In the new window, select the icon with the speech bubble.
5. Click the download arrow.
6. Select "Current View"
Keyword(s): marked assignmentmarked assessmenttutor commentsmarked Turnitin assignment
Question:
#### How can I obtain access to the journal platforms?
(Last edited: Tuesday, 21 February 2017, 12:36 PM)
Impendulo:
The South African Theological Seminary provides students with electronic access to the world's premiere collections of online journals (including book reviews). Click on the icons below access the individual platforms.
EBSCOhost is the premiere platform of academic journals and resources. SATS subscribes to the five theological collections available on ESBSCOhost. If prompted, the username is sats and the password seminary.
JSTOR is a project of several leading universities to make full-text journal articles available online. It is the best platform after EBSCOhost. The username is sats and the password LPtyf00t.
The Directory of Open Access Journals is a platform of free journals, covering a range of disciplines, and including a number of theological journals.
Sabinet is the home of SA-ePublications, the premierre platform of South African academic journals. The username is k00212ie, and the password k002.
Keyword(s): Journals
Question:
#### How do assignment due dates and different timezones work on MySats?
(Last edited: Monday, 9 June 2014, 10:13 PM)
Impendulo:
Almost every assessment task on MySats has a deadline known as due date or a closing date, depending on the type of activity (assignment, forum, test, etc.).
Deadlines are always shown on MySats in South African Standard Time, which is UTC+2. South Africa does not use daylight savings, so there is no change with seasons.
If you live in another time zone or a country that uses daylight saving, you can edit your profile such that MySats displays the deadlines in your local time. This will not, however, change the deadline at all; it will simply display the deadline in your local time.
If you are unsure of what your UTC timezone is, you can look here: http://www.worldtimezone.com.
Keyword(s): Due DateDeadlineTimezone
Question:
#### How do forum threads work?
(Last edited: Tuesday, 18 July 2017, 1:19 PM)
Impendulo:
A forum thread consists of the combination of a topic, a posting, and a reply. Using Thread View, you are able to match replies to specific postings in the time/date order in which they were posted so you can see how the discussion progresses with time. See the following screenshot for example of thread view:
Note also the nested view which is also the default on MySats:
Keyword(s): forum
Question:
#### How do I edit or update my profile?
(Last edited: Friday, 12 February 2016, 12:08 PM)
Impendulo:
Occasionally you may need to change your email address or update your contact details. This can be done easily by editing your profile. After you have logged in, follow the two steps below:
1. Click "edit profile" under your settings menu from anywhere in MySATS.
2. Click "Update Profile" when you are done editing!
Keyword(s): Profileedit your profile
Question:
#### How do I enrol?
(Last edited: Wednesday, 6 November 2019, 8:22 AM)
Impendulo:
Please submit your Application Form together with the required certified documents and the application fee of :
1. South Africa - R358
2. Africa, Asia, South America - $32 USD 3. Europe, North America, Australia and New Zealand$59 - USD
You can either enrol online at http://www.sats.edu.za or you can email sharon@sats.edu.za for undergraduate applications or marilyn@sats.edu.za for postgraduate application. You can also post your Application Form to the Seminary. We will evaluate your prior learning and contact you.
If you enrol, the application fee will be deducted from the enrolment fee.
We require original, certified copies of your ID, school leaving certificate, certificates and transcripts.
There is no application fee for postgraduate applications.
Keyword(s): EnrolApplytranscriptsapplication
Question:
#### How do I find my results?
(Last edited: Monday, 1 February 2016, 4:09 PM)
Impendulo:
#### If you do not see the My Curriculum link, you should contact your registrar so that your student number may be added to your MySats profile.
Keyword(s): resultsmarkscurriculum
Question:
#### How do I get the Turnitin app to work on my iPad?
(Last edited: Thursday, 4 May 2017, 10:30 AM)
Impendulo:
As of the time of this writing, the Turnitin app for iPad is not available for student use. However, if you are a tutor or are marking papers, you may use your iPad to mark papers. This requires that you do the following:
1. Download the Turnitin app from the App store:
2. View your assignment from a desktop computer to generate an access code for your iPad. See the following screenshots:
3. Open the Turnitin app on your iPad and input the new code.
Keyword(s): iPadmobile
Question:
#### How do I install Adobe Digital Editions
(Last edited: Friday, 1 June 2018, 11:48 AM)
Impendulo:
#### Adobe Digital Editions
Adobe Digital Editions (ADE) is an eBook reader program which supports the reading of electronic material in the PDF and XHTML formats.
##### Installing ADE and creating an Adobe Digital ID
If you don't already have Adobe Digital Editions on your computer, install it from the ADE download siteNote: Don't install Norton Security in the process of installing Adobe Digital Editions. You have to remove the tick from the checkbox to avoid doing so. Of course, you may do so, but it could interfere with other security software already running on your computer.
When you run Adobe Digital Editions for the first time, you will have to authorize it with your Adobe Digital ID. If you don't already have one, click the link to "Create an Adobe ID". The process is quick and easy. Then return to ADE where it is asking for your Adobe ID and password that you just set up. Your Adobe ID is the e-mail address you specified to create it.
##### N.B. When you install Adobe Digital Editions, it is essential that you register using the same email address you used to create your EbscoHost account. If your email address does not match, you will get a license error on your downloaded books.
If you have questions about ADE, look for answers on the ADE FAQ page
##### Using Adobe Digital Editions (ADE) and EBSCOhost together
Read through the EBSCOhost and ADE Guide to see how to use these two programmes together to find and use e-library books.
#### Protecting your bookmarks, notes and highlights
Having access to ebooks from EBSCOhost's online library is wonderful but, as with a physical library where you loan books for a limited time, the books you take out from the e-library are returned automatically when the "checkout period" has expired. Good news is that you can check out a book for up to 84 days. Further, after a book is returned, you can just take it out again if you need it. The license agreement between SATS and EBSCO allows us to access an unlimited number of copies of the same book simultaneously, so there is no need to reserve a book if it is already taken out by someone else, and you don't have to worry about limiting other users' access to a book by "hogging" it. Moreover, in the current arrangement, SATS alumni can still access the EBSCOhost library and use it in the same way as SATS students, so even after you finish studying with us, you can use the e-library!
You may be wondering, however, what happens to your annotation (notes) and markup (highlighting) when a book is returned to the library. These are permanently stored in files on your computer. If you take the same book(s) out again, your notes and highlights will appear on it again. Moreover, you can open them even without the book; though the formatting is not ideal for humans to read, at least it is legible (in XML format). If you are using Microsoft Windows, you can find a folder called Annotations under My Digital Editions, which is created when you install Adobe Digital Editions. The following example is based on the user name, DavidC:\Users\David\Documents\My Digital Editions\Annotations. You should be able to navigate to your annotations in a folder corresponding to this on your PC, exchanging your login name with David
You are advised to back up your annotation files routinely, so it's important to find where they are stored using the example above. Moreover, you may need to copy them to a new computer at some stage.
The file extension for annotation files (which also include bookmarks and highlights) is annot, and the filename is taken from the book title, with its own file extension (typically epub or pdf). Therefore, you can simply look for the annot file matching the title of the book whose annotation you wish to access, e.g. A New History of Christianity.epub.annot. (Note the double file extension.) The full path and filename on my PC is thus: C:\Users\David\Documents\My Digital Editions\Annotations\A New History of Christianity.epub.annot
You can open your annotation files with a text editor or web browser. However, you should not edit them in case you disrupt the XML formatting. Rather, copy the information you need to a word processor and save it in another file, where you can edit is as you please.
Finally, you may not be content with the restrictions of working with e-library books in Adobe Digital Editions, especially if you want your own copy of an especially important block of text. In these cases, open the PDF version of the book in EBSCOhost's eBook Viewer, then "Save Pages" to download the specific pages you want in PDF format. There are limits to how many pages you may save from each book (or each chapter of a book) in PDF format. EBSCOhost will show you how many pages you may save when you click the "Save Pages" button.
Keyword(s): ADEAdobe Digital Editionsebooks
Page: 1 2 3 (Next)
Yonke | open-web-math/open-web-math | |
# 1. From Bouncing Ball to Angry Birds¶
This notebook explores the idea that objects on our sketch appear to move in various ways:
• Falls
• Projectiles
• Bounces
This is a magic trick... nothing really "moves"... it is just the canvas redrawn in an appropriate way that gives the illusion of movement.
## 1.1 A Ball¶
Like we did for Lab 2, let's draw an object at a specific location:
In :
void setup() {
size(200, 200);
}
void drawBall(float x, float y, float scale) {
fill(255, 0, 0);
ellipse(x, y, 10 * scale, 10 * scale);
}
void draw() {
drawBall(100, 20, 1);
}
Sketch #1:
## 1.2 Global Variables¶
First, we'll create two global variables, and initialize them to the same values as above:
In :
float x = 100;
float y = 20;
void setup() {
size(200, 200);
}
void drawBall(float x, float y, float scale) {
fill(255, 0, 0);
ellipse(x, y, 10 * scale, 10 * scale);
}
void draw() {
drawBall(x, y, 1);
}
Sketch #2:
Now, let's move the assignment to the setup() function:
In :
float x;
float y;
void setup() {
size(400, 100);
x = width/2;
y = 20;
}
void drawBall(float x, float y, float scale) {
fill(255, 0, 0);
ellipse(x, y, 10 * scale, 10 * scale);
}
void draw() {
drawBall(x, y, 1);
}
Sketch #4:
Nothing too interesting there. I did use the special variables width and height that are automatically defined to be the size of the canvas. But other than that, it is just a picture of a red ball sitting quietly in space.
## 1.3 Brownian Motion¶
Let's try a slight variation. Do you know the idea of Brownian Motion (sometimes called particle theory). This is basically just the idea that small objects will randomly move because they are hit will atoms or molecules.
Warning: I did not say "Brownie in motion"... that is a small chocolate treat that is on the move.
We can simulate Brownian Motion by simply moving the ball a little left, right, or stay in the same place. We use the fact that random(2) - 1 represents either a -1, 0, or 1.
In :
println(random(2) - 1);
Sketch #5:
-0.7309886215007491
0.901475877332913
In :
// Global variables:
// defined here, used in setup() and draw()
float x;
float y;
void setup() {
size(200, 200);
x = width/2;
y = height/2;
}
void drawBall(float x, float y, float scale) {
fill(255, 0, 0);
stroke(255, 0, 0);
ellipse(x, y, 10 * scale, 10 * scale);
}
void draw() {
drawBall(x, y, 1);
x = x + random(2) - 1;
y = y + random(2) - 1;
}
Sketch #7:
It looks like the ball is moving randomly about, doesn't it. You can make the effect even stronger by clearing the background before you redraw the ball:
In :
// Global variables:
// defined here, used in setup() and draw()
float x;
float y;
void setup() {
size(200, 200);
x = width/2;
y = height/2;
}
void drawBall(float x, float y, float scale) {
fill(255, 0, 0);
ellipse(x, y, 10 * scale, 10 * scale);
}
void draw() {
background(168);
drawBall(x, y, 1);
x = x + random(3) - 1;
//y = y + 0;
}
Sketch #14:
## 1.4 Velocity¶
Now, instead of having the ball randomly move each time step, let's have the ball move in a particular direct. To do this, we will create two new variables that will keep track of the velocity (speed) of the ball in the x and y directions.
On each time step, we will move the ball an appropriate amount.
In :
// Give the ball velocities
// in the x and y directions:
float vx;
float vy;
float x;
float y;
float dt;
float t;
void setup() {
size(200, 200);
x = width/2;
y = 50;
dt = 0.1;
t = 0;
vx = 1.0;
vy = 0.0;
}
void drawBall(float x, float y, float scale) {
fill(255, 0, 0);
ellipse(x, y, 10 * scale, 10 * scale);
}
void draw() {
background(128, 128, 128);
float dx = vx * dt;
x = x + dx;
float dy = vy * dt;
y = y + dy;
drawBall(x, y, 1);
t = t + dt;
}
Sketch #17:
Somewhat interesting... but where did it go?
### 1.4.1 Bouncing off the walls¶
To keep the ball "on the canvas", we need to see when it is beyond bounds, and turn the ball around. All we need to do is switch the velocity to go in the opposite direction. So if the ball is going to the right and hits a wall, we make it go to the left.
To allow the ball to do one thing if it hits a wall, and something else if it doesn't we need to introduce the idea of flow of control.
We can control the flow of a program using if/else, like this:
if (EXPRESSION) {
STATEMENT;
...
} else {
STATEMENT;
...
}
In :
float vx;
float vy;
float x;
float y;
float dt;
float t;
void setup() {
size(200, 200);
x = width/2;
y = 50;
dt = 0.1;
t = 0;
vx = 10.0;
vy = 7.5;
}
void drawBall(float x, float y, float scale) {
fill(255, 0, 0);
ellipse(x, y, 50 * scale, 50 * scale);
}
void draw() {
background(0, 255, 0);
float dx = vx * dt;
if (((x + dx) > width) || ((x + dx) < 0)) {
vx = vx * -1;
} else {
x = x + dx;
}
float dy = vy * dt;
if (((y + dy) > height) || ((y + dy) < 0)) {
vy = vy * -1;
} else {
y = y + dy;
}
drawBall(x, y, 1);
t = t + dt;
}
Sketch #20:
### 1.4.2 Gravity¶
Now, the final step: let's add gravity. At each timestep, we'll add a gravity component to vy. Over time, that will get bigger and bigger.
In :
float g = 9.8;
float vx;
float vy;
float x;
float y;
float dt;
float t;
void setup() {
size(200, 500);
x = width/2;
y = 50;
dt = 0.1;
t = 0;
vx = 5.0;
vy = 0.0;
}
void drawBall(float x, float y, float scale) {
fill(255, 0, 0);
ellipse(x, y, 10 * scale, 10 * scale);
}
void draw() {
//background(0, 0, 128);
// gravity
vy = vy + g * dt;
float dx = vx * dt;
if (((x + dx) > width) || ((x + dx) < 0)) {
vx = vx * -1;
} else {
x = x + dx;
}
float dy = vy * dt;
if (((y + dy) > height) || ((y + dy) < 0)) {
vy = vy * -1;
} else {
y = y + dy;
}
drawBall(x, y, 1);
t = t + dt;
}
Sketch #22:
### 1.4.3 Dampening¶
That is weird... it keeps bouncing! We probably want to lose a little bit of energy each time it "hits something." We merely don't give the full amount when we change directions.
In :
float g = 9.8;
float vx;
float vy;
float x;
float y;
float dt;
float t;
void setup() {
size(200, 500);
x = width/2;
y = 50;
dt = 0.1;
t = 0;
vx = 50.0;
vy = 0.0;
}
void drawBall(float x, float y, float scale) {
fill(255, 0, 0);
ellipse(x, y, 10 * scale, 10 * scale);
}
void draw() {
// gravity
background(255);
vy = vy + g * dt;
float dx = vx * dt;
if (((x + dx) > width) || ((x + dx) < 0)) {
vx = vx * -0.6;
} else {
x = x + dx;
}
float dy = vy * dt;
if (((y + dy) > height) || ((y + dy) < 0)) {
vy = vy * -0.6;
} else {
y = y + dy;
}
drawBall(x, y, 1);
t = t + dt;
}
Sketch #25:
## 1.5 Interactivity¶
Nice! Now let's drop the ball with the mouse:
In :
float g = 9.8;
float vx;
float vy;
float x;
float y;
float dt;
float t;
void setup() {
size(200, 500);
x = width/2;
y = 50;
dt = 0.1;
t = 0;
vx = 50.0;
vy = 0.0;
}
void drawBall(float x, float y, float scale) {
fill(255, 0, 0);
ellipse(x, y, 10 * scale, 10 * scale);
}
void draw() {
// gravity
background(168);
vy = vy + g * dt;
float dx = vx * dt;
if (((x + dx) > width) || ((x + dx) < 0)) {
vx = vx * -0.8;
} else {
x = x + dx;
}
float dy = vy * dt;
if (((y + dy) > height) || ((y + dy) < 0)) {
vy = vy * -0.8;
} else {
y = y + dy;
}
drawBall(x, y, 1);
t = t + dt;
}
void mousePressed() {
x = mouseX;
y = mouseY;
vx = 0.0;
vy = 0.0;
}
Sketch #26:
## 1.6 Angry Birds¶
Ok, but it just drops. What if we could click, drag, and release, and the distance we dragged was proportional to a vector in the opposite direction of where we want to send it. Angry Birds!
Click below and drag:
In :
float g = 9.8;
float vx;
float vy;
float x;
float y;
float dt;
float t;
void setup() {
size(500, 500);
x = width/2;
y = 50;
dt = 0.2;
t = 0;
vx = 0.0;
vy = 0.0;
background(255);
}
void drawBall(float x, float y, float scale) {
fill(255, 0, 0);
ellipse(x, y, 10 * scale, 10 * scale);
}
boolean mouseIsDown = true;
void draw() {
if (mouseIsDown) return;
background(255);
// gravity
vy = vy + g * dt;
float dx = vx * dt;
if (((x + dx) > width) || ((x + dx) < 0)) {
vx = vx * -0.8;
} else {
x = x + dx;
}
float dy = vy * dt;
if (((y + dy) > height) || ((y + dy) < 0)) {
vy = vy * -0.8;
if (abs(vy) > 30) {
//drawBall(x, y, 2);
fill(255, 0,0);
ellipse(x, y, vy, 5);
noLoop();
return;
}
} else {
y = y + dy;
}
drawBall(x, y, 1);
t = t + dt;
}
void mousePressed() {
mouseIsDown = true;
x = mouseX;
y = mouseY;
background(255);
drawBall(x, y, 1);
vx = 0.0;
vy = 0.0;
}
void mouseReleased() {
vx = (x - mouseX);
vy = (y - mouseY);
mouseIsDown = false;
loop();
}
void mouseDragged() {
background(255);
drawBall(x, y, 1);
line(x, y, mouseX, mouseY);
}
Sketch #31:
To make a really "angry birds" like game, we'll need to detect more than collisions with the walls, floor, and ceiling. We'll do that later, once we learn how to make real objects.
## 1.7 Conclusion¶
In this notebook we have seen that motion is just imagined. We saw Brownian Motion, velocity, gravity, and a ball with an applied force (ala Angry Birds).
New ideas and jargon:
• global variable
• perceived motion
• gravity
• bounce
• flow of control
• if/else
• interactivity
• statement versus expression | HuggingFaceTB/finemath | |
# How to test signifcance of a sharpe ratio
Let say you have measured a Sharpe Ratio of $$S^*$$. What is the simplest way (ie no fancy distributions) to do a hypothesis that this is different from $$0$$?
So $$H_0: \text{ The sharpe ratio is equal to 0}$$ and $$H_1: \text{ The sharpe ratio is greater than 0}$$.
So given $$S^*$$, $$\mathbb{P}( Y = S^* ) \geq 0.05$$
But what should the $$Y$$ be? I read somewhere online that it could the non centered t distribution, but I am not sure whether this could be centered to the standard t test distribution. Moreover, I would also like to consider the normal distribution and as the sample used to create the statistic should be greater than 30, the t test to normal approxaimtion should apply.
The answer above is not correct.
Let's go by parts:
Denote the mean of returns $$\mu$$. Denote the standard deviation of returns: $$\sigma$$.
Therefore the sharpe ratio is:
$$SR = \frac{\mu-r_f}{\sigma}$$
The corresponding standard errors are:
$$se(\hat{mu}) = \frac{\sigma}{\sqrt{t}}$$ $$se(\hat{\sigma}) = \frac{\sqrt{2} \sigma^2}{\sqrt{T}}$$ $$se(\hat{SR}) = \frac{\sqrt{1+SR^2/2}}{\sqrt{T}}$$
So the t-stat for the sharpe ratio is:
$$t-stat(\hat{SR})= \frac{\hat{SR}}{se(\hat{SR})}$$
• Take your point about sample vs population sigma, @phdstudent. But still respectfully disagree ;-) Will edit mine below, to elaborate. – demully Jun 15 at 14:51
• Adding a reference for clarity: alo.mit.edu/wp-content/uploads/2017/06/… – phdstudent Jun 15 at 19:58
• thanks - interesting paper, need to Monte Carlo to work on how the inconsistency happens. – demully Jun 15 at 22:32
• OK, it's clear gaming this that variation in realised sigmas do have an effect on SR. Assuming of course the concept of a true population volatility/variance is a meaningful concept (one for another day). This can be material testing for non-zero Sharpes. This said, it remains a tautology that if (Mu-Rf)>0 then Sharpe>0. So there's a paradox in the special case of a test for zero, if one's confidence in a positive numerator is any different to one's confidence in a positive ratio (given a positive denominator, by definition). That makes no sense at all. Tx for interesting food for thought! – demully Jun 15 at 23:49
• The questions is whether (Mu-Rf) is statistically different from zero to start with! – phdstudent Jun 16 at 10:24 | HuggingFaceTB/finemath | |
## 1 Bit Full Adder Design using Logisim Bangla Tutorial
###### 1 bit Full Adder design using Logisim:
This is tutorial is in Bangla language.
## Digital Logic Designing an Arbitrary Counter
##### Design an arbitrary counter that counts or follows in this sequence:
0, 2, 4, 5, 7, 0, 2, 4........
It is pretty clear from the question that counter counts in,
0 -> 2 -> 4 -> 5 -> 7 -> 0 -> 2 -> 4 -> …….
It goes from 0 to 2 then 4 this way. After reaching 7 it goes back to beginning of the sequence and repeats the sequence.
#### Arbitrary Counter State Diagram:
Since nothing is said about what will happen if it had to transition from 1, 3 and 6. So I can choose on my discretion what the next state will be from those unused states.
#### Arbitrary Counter State Table:
When simplifying input equations, unused states can be used as don’t care conditions or it may also be assigned specific next states. Here unused or unreachable states are 1, 3, 6 so these states can have any assigned next state of designers choice.
The reason or advantage of using don’t care is easier state diagram, smaller state table, easier k-map equations as well as less connections in the logic diagram etc.
There is also a good reason to use specific next states from unreachable states instead of using don’t care conditions. Such as during operation due to outside interference the may reach an unused state and break the intended sequence. Also it may keep circulating in unused states and never come back to original sequence.
Here I have only shown K Map and Logic Diagram using don’t care term for transition of unused states and using J-K Flip Flops.
#### Excitation Table:
The excitation table below are for J-K, D, S-R, T latches. For making arbitrary counter with these latches fill the state table from the excitation table.
J-K:
$\begin{tabular}{ | l | l | l | p{1cm} |} \hline Q & Q(next) & J & K \\ \hline 0 & 0 & 0 & x \\ \hline 0 & 1 & 1 & x \\ \hline 1 & 0 & x & 1 \\ \hline 1 & 1 & x & 0 \\ \hline \end{tabular}$
D:
For D latch the next state is the output.
$\begin{tabular}{ | l | l | l | p{1cm} |} \hline Q & Q(next) & D \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 1 \\ \hline 1 & 0 & 0 \\ \hline 1 & 1 & 1 \\ \hline \end{tabular}$
S-R:
$\begin{tabular}{ | l | l | l | p{1cm} |} \hline Q & Q(next) & S & R \\ \hline 0 & 0 & 0 & x \\ \hline 0 & 1 & 1 & 0 \\ \hline 1 & 0 & 0 & 1 \\ \hline 1 & 1 & x & 0 \\ \hline \end{tabular}$
T:
$\begin{tabular}{ | l | l | l | p{1cm} |} \hline Q & Q(next) & T \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 1 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 0 \\ \hline \end{tabular}$
#### How to complete the state table:
First thing is to complete the present state. Since we have only 0, 2, 4, 5, 7 to count, fill in their binary representation. No need to show the rest or like I mentioned above.
Next, Fill up the next state based on present state. Like shown in the state diagram above from 0 it next counts to 2, then 4, then 5 etc.
So next state of 0 is 2. Which in binary is 000 to 010. Similarly 7 counts to 0. So in binary 111 to 000.
Last step is filling up Flip Flop inputs. If you use any other Flip Flops the above steps are always required. Just replace J-K latch here with any other latch and use excitation table to complete the state table.
Jc and Kc should be filled only based on present state of C and the next state of C. Similarly for others.
For example, 2 moves to 4. So,
Present -> Next
C B A -> C B A
0 1 0 -> 1 0 0
Here, C’s present state is 0 and next state is 1. So based on J-K excitation table the value will be 1 X.
#### State Table Not Showing Don’t Care Conditions:
##### Flip Flop Inputs
C B A C B A JC KC JB KB JA KA
0 0 0 0 1 0 0 x 1 x 0 x
0 1 0 1 0 0 1 x x 1 0 x
1 0 0 1 0 1 x 0 0 x 1 x
1 0 1 1 1 1 x 0 1 x x 0
1 1 1 0 0 0 x 0 x 1 x 1
#### How to Interpret the Table Above to Generate K Map:
Since this has only three bits. Create a three variable K-Maps. There will be a total of 6 K-Maps since there is two inputs per latch.
Also here 1, 3, 6 are don’t care, so fill these with x. Next 0, 2, 4, 5, 7 these positions are empty in k map. Fill in these position with values from the columns.
#### Karnaugh Map Using Table For Unused Don’t Care Conditions:
Here I have shown the table for don’t care terms for unused states transition to next states. For using specific state transition from unused states set the value from the table above in appropriate place and get the equation. From that equation create the logic diagram.
#### How to Connect the latches from K Map equations:
Here each of the latches are given a name A, B, C. Each of the latches represents a bit. Where C is the Most significant bit (MSB) and A is the least significant bit (LSB). The input of J-K latch for A are JA and KA. Similarly for others.
The next states depends on current state. The output is on the right side. What ever the value of the output on the top output is the bottom one is the prime (opposite) of it.
From the K Map set connections on the left side ( input ) to right side ( output ). This is assuming that the Flip Flop is positioned like shown in the diagram below. Although This may not always be the case.
#### Logic Diagram From the above Karnaugh Map Using Don’t Care For 1, 3, 6 states:
Here I’ve drawn the logic diagram in logisim. If you draw by hand it hand the design will be the same just replace the button with a clock.
Make sure input of KB is always 1. Because that’s the equation for KB in Karnaugh Map.
#### How to check if the Arbitrary counter is working:
Select the hand tool below File menu. Also make sure input of JB is set to 1. Now click on the button to give clock pulse.
## 7 Segment Decoder Implementation, Truth Table, Logisim Diagram
7 Segment Decoder Implementation, Truth Table, Logisim Diagram:
### 7 Segment Decoder:
For reference check this Wikipedia link.
### Pictures:
(Wikipedia CC BY-SA 2.5)
### Explanation:
Before we start implementing we first need to check if it is common anode or common cathode. If it is common anode then 3rd pin in both top and bottom are VCC. But if it is we should connect 3rd pin in both top and bottom to ground.
Pins:
show top pins then bottom pins ( Dot side is down ).
Pin1 Pin2 Pin3 Pin4 Pin5
Top: g f vcc/GND a b
Bottom: e d vcc/GND c dp
### Truth Table:
From here we can get minimized expressions for a, b, c, d, e, f, g using K-MAP. Here we only need value 0 though 9 rest are don’t care terms. Using those don’t care terms we will try to maximum ones first.
### K-Map (Karnaugh map) for ‘a’:
We can follow similar procedure for the rest. K map for ‘a’ can be created by taking the ‘a’ column from the table above and setting the value 0 / 1 to corresponding location in the table. For example to display 0 ( 0000 ) ‘a’ is always 1. Similarly to display 1 ‘a’ is always 0. For value ( 10 – 16 ) we don’t care about them so they are used as don’t care term in k map. | HuggingFaceTB/finemath | |
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# Grade 8 Forces and Motion 2012
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Forces and Motion for a Grade 8 Class
Forces and Motion for a Grade 8 Class
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### Transcript
• 1. Grade 8 Science Forces and Their Effects and Motionhttp://commons.wikimedia.org/wiki/File:Jal.747.newcolours.arp.750pix.jpg
• 2. Brainstorm about Forces Brainstorm everything you know about forces, in groups of 2 or 3. It should cover:● Types of forces● Effects of forces● Theories or ideas related to forces?● Famous people who developed or promoted such ideas?● Can you bring in numerical relationships?
• 3. The Solar SystemWhat keeps the planets moving? http://commons.wikimedia.org/wiki/File:Solar_sys.jpg
• 4. Forces● A force is a push or a pull.● Forces are measured in Newtons, after Isaac Newton.● Find something which: a) has a force of 1 Newton. b) a force of 5 Newtons which isnt a weight force. c) a force of 10 Newtons which isnt a weight force or a friction force.● What force is required to: 1. open a drawer. 2. untie a shoe lace. 3. turn on a tap (difficult) 4. drag a chair along the ground.http://commons.wikimedia.org/wiki/File:Isaac-newton_1.jpg
• 5. Forces: The Basics● A force is a push or a pull.● A force has a size (in N________) and a direction.● A force can be drawn as an arrow. The length of the arrow represents its size and the direction of the arrow represents the direction of the force.● A force can cause an object to change its speed (a_________), change its direction or change its shape.●
• 6. Gravity● Gravity is a very weak but important force.● Gravity pulls everything together.● We notice it as pulling us down towards the Earth.● Gravity also pulls the Earth towards the sun.● What causes gravity?
• 7. Mass and Weight Item Mass (kg) Weight (N)● Mass measures “quantity of matter”. It is measured in kilograms (kg).● Weight measures the force of gravity. It is measured in Newtons (N).● Short task: Measure the mass and weight force of four items in the laboratory.Find something with a weight force of 1 Newton.
• 8. Mass and Weight Relationship● The relationship between mass and weight is: Weight (Newtons) = mass (kilograms) * 10 We MUST convert grams to kilograms first. 10 is the value for gravity on Earth.● Calculate the weight forces of a 60kg person, an 800 kg car and a 50 gram feather. Extension exercise: Look up Newtons Law of Gravity, find appropriate values and use them to show where the value of 10 (actually 9.8) comes from and determine its units, and show why.
• 9. Gravity on Earth● Weight = mass * 10, but only at the surface of the Earth.● In space, weight force is nearly zero.● Every place in the universe has its own number for gravity. 1. Calculate the weight force of a 60kg person on The Moon (1.7), Mercury (4), or Jupiter (26). 2. Explain why things weigh more on Jupiter. 3. What would you feel on Jupiter, as a result of a greater weight force? 4. What is wrong (scientifically speaking) with saying: “I weigh seventy kilograms.” OR extension exercises in previous slide.
• 10. Quick Review● Explain the difference between mass and weight, in your own words.● What is the weight of a 2kg object (on Earth)?● Calculate the weight of a 200g object (on Earth). Dont forget to convert it to kilograms first!
• 11. Reaction Force ● Hard surfaces provide a force which opposes anything pushing against them.This often supports the weight of something on them. ● The reaction force is also sometimes called a support force. ● The reaction force increases to match the weight force until it reaches its limit. Reaction MaximumReaction force Reaction Reaction Weight Weight Weight
• 12. Breaking Strength of Tissue Paper● Write the prediction and your own procedure in your notebook. Remember that it should be able to be followed by a Grade 6 student independently.● Prediction: I think the tissues will break when the mass on them is _________ grams. Therefore, I think the maximum reaction force will be ________ Newtons. Mass (grams) Weight (Newtons) Trial 1 Trial 2 Trial 3 Average Weight● Equipment: container, tissues, tape, masses● In your conclusion, answer the following: A) Was the maximum reaction force of tissue greater or less than you expected? B) Name one use of a substance with a strong reaction force. C) Is there a tension force occurring here? D)How do you think your results would change if the box were larger?
• 13. Friction● Friction is a force which opposes all objects moving.● The friction force between two surfaces depends on: -the type of surface -the perpendicular force between the surfaces -the area of the surfaces Reaction Friction Weight
• 14. More Pulling Force ReactionFriction Weight
• 15. More Pulling Force We can determine the maximumfriction force by pulling until the block just starts moving. Reaction Friction Weight
• 16. Increasing the Weight Reaction Friction Weight
• 17. Friction and Perpendicular ForceMeasure the perpendicular force (weight) and the frictionforce for at least three different masses.1. Describe the pattern between the size of the friction force andthe perpendicular force.2. Would it make a difference if the perpendicular force wasntweight, for example if it were pressed against a wall by hand.3. Can you determine an approximate rule for the relationshipbetween the two forces? If so, how would itchange if the surface in contact waschanged? How about if the surfacearea was changed? Reaction Friction Weight
• 18. Book Busters● Video:● Its said that two books with the pages interleaved require the force of _____ to separate them.● Determine the relationship between the number of pages interleaved and the force required to pull the books apart.● Extension exercise: Can you determine a pattern between the two quantities, and if so can you describe it algebraically?
• 19. The Bed of Nails● I predict that it will take ______ N (_____ grams) to pop the balloon with ONE nail because ______________________________.● I predict that it will take ______ N (_____ grams) to pop the balloon with the bed of nails because ______________________________.
• 20. Pressure ● Why does a sharp knife cut and not a blunt knife? ● Why does a pin prick but not a finger? ● Why does a chair leave marks in the carpet? http://commons.wikimedia.org/wiki/File:Dwie_pinezki_(ubt).JPGhttp://commons.wikimedia.org/wiki/File:Knife_01.JPG
• 21. Pressure● Pressure occurs when a force is applied to an area.● The higher the pressure, the more damage a force will do to the surface. force pressure= area● Pressure is measured in Newtons per square metre(N/m2), or Newtons per square centimetre(N/cm2).● One N/m2 is called one Pascal, after the French genius Blaise Pascal.
• 22. Pillows● Why is a soft pillow more comfortable than a brick floor? http://commons.wikimedia.org/wiki/File:Body_pillow.jpg http://commons.wikimedia.org/wiki/File:Worn_floor_-_geograph.org.uk_-_771522.jpg
• 23. New Unit: Motion● Key Concept:Change● Related concepts: Energy, Models, Interaction, Consequences, Balance● Concept statement: The consequences of unbalanced forces on an object are changes in its motion.● Global context statement: The impact of scientific and technological advances on society and on the environment.● Inquiry Questions: -How can we use forces to predict changes in something’s motion? -How can we apply our understanding of forces and motion to help us get around?●
• 24. A Long Walk (and Run) After the earthquake, many people needed to walk home from Tokyo to Yokohama.● How long would it take to walk to Shibuya? How about to run to Shibuya?● What information would we need to know?
• 25. A Long Walk (and Run)● Distance: 30km● Walking speed: 2 metres per second.● Running speed: 5 metres per second.
• 26. Speed, Distance and Time● If it takes 4 hours to travel 200 kilometres, what was the average speed, and how did you calculate it? How long would it take to travel 1000km at this speed? How did you calculate it?
• 27. Speed● Speed is a measure of how fast something moves.● Units are m/s (or cm/s, km/s etc).● Eg. Katie is running during PE. After 20 seconds she has run 120m. What is her speed? 1. Calculate how far Katie would run in one minute (using the speed from the previous example).● 2.A. Yuki is swimming at 2m/s. How long will it take him to swim a length of a swimming pool (25m)? B. How far can Yuki swim during a 45 minute PE class if he swims for the whole class? Hint: you will need to convert units here.
• 28. Measuring Walking / Running Speed● We will use the roof of the central building.● Equipment: meter rulers and stopwatches.● Work in groups of 4 or 5.● Measure the speed of a person walking and running. (It does not have to be the same person walking as running).● Use a distance of 5-10 metres.● Repeat each set at least twice (three total), and take the average of all trials.● Present your results in a summary.
• 29. The Investigation● Aim: To determine how fast _____ walks and ______ sprints.● Apparatus: stopwatches, meter rulers● Prediction: ________ will walk at _____ m/s and _________ will sprint at _____ m/s.● Procedure: 1. Measure a suitable distance for someone to walk, and sprint. 2. Arrange a system for ensuring that the timing is accurate. 3. Record at least three trials, and calculate the average speed for walking and running.
• 30. Results and Conclusion● __________ Walking Trial number Distance (m) Time (s) Speed (m/s) 1 2 3 Average speed● __________Running Trial number Distance (m) Time (s) Speed (m/s) 1 2 3 Average speed
• 31. 1. Compare your results to others in the class. Who is thefastest a) runner? b) walker?2. The result obtained is an average speed. Explain what ismeant by an average speed. You may need to do someresearch for this question.3. Two years ago in Berlin, Usain Bolt from Jamaica set theworld record for the 100m sprint, coming in at 9.58s.Calculate his average speed, and compare it to the fasteststudent in the class.4. A. Convert your groups walking and sprinting speed tokm/h. This may require some thought.B. Use this, and google maps (directions) to calculate howlong it would take to walk AND run toi) your homeii) Tokyo tower
• 32. Forces and Acceleration ● If an unbalanced force acts on an object, it will cause it to accelerate. ● Acceleration is a change in speed or a change in direction.http://commons.wikimedia.org/wiki/File:Hatchback_5_doors.png
• 33. Streamlining ● Streamlining is the shaping of a (usually fast- moving) object to minimise air resistance.http://en.wikipedia.org/wiki/File:JR_East_Shinkansen_lineup_at_Niigata_Depot_200910.jpg
• 34. Motion Graphs Motion = movement A motion graph is used to show how something moves, and how its movement changes over time. It can show:● whether an object is stationary or moving● an objects speed.● if and how its speed is changing (acceleration or deceleration). A motion graph can show all of the above on the same graph.
• 35. Types of Motion Graphs● Time always goes on the x axis.● A distance-time graph shows how far something has travelled on the Y axis.● Distance can not go backwards.● A speed-time graph shows how fast something is moving on the Y axis.
• 36. Daniel Walking at 1m/sDistance (m) Speed (m/s) 8 8 6 6 4 4 2 2 0 0 0 2 4 6 8 10 0 2 4 6 8 10 Time (seconds) Time (seconds)
• 37. Marii walking at 2 m/s Distance (m) Speed (m/s)8 86 64 42 20 0 0 2 4 6 8 10 0 2 4 6 8 10 Time (seconds) Time (seconds)
• 38. 1. Where is each of the following on the graph above: Stationary, constant (slow) speed, constant (fast) speed, acceleration, deceleration 2. If this shows a cars motion a) At which point is thrust greater than friction? b) At which point is friction greater than thrust? c) Are there any forces acting on the car at point a? If so, what? d) Are there any forces acting on the car at point b? How can you tell?spe ced d b(m/s) e a 0 Time (seconds)
• 39. Acceleration● Acceleration is a constantly changing speed.● Deceleration is a form of acceleration, in which the speed is decreasing at a steady rate. Exercise: On the roof, mark off 1m, 2m, 3m and 4m. Everyone must run 1m in the first second, 2m in the second second, 3m in the third second and 4m in the fourth second. Extension: what units could be used for acceleration? What would the acceleration of the class from the class exercise be? Could we write a formula for acceleration?
• 40. Yongwon Accelerating ● In the first second, he travels one metre per second. In the second second, he travels at two metres per second. In the third second, he travels at three metres per second. ● How would we describe Yongwons motion?Distance (m) Speed (m/s) 8 8 6 6 4 4 2 2 0 0 0 1 2 3 4 5 0 1 2 3 4 5 Time (seconds) Time (seconds) | HuggingFaceTB/finemath | |
College Physics (4th Edition)
$f_r = f~\left(\frac{u+v}{u-v}\right)$
To find the frequency received by the moving blood, we can use the equation for the Doppler effect when the observer is approaching: $f_o = \left(\frac{u+v_o}{u}\right)~f$ $f_o = \left(\frac{u+v}{u}\right)~f$ To find the reflected frequency, we can let $f_o$ be the sound source and use the equation for the Doppler effect when the source is approaching: $f_r = \left(\frac{u}{u-v_s}\right)~f_o$ $f_r = \left(\frac{u}{u-v}\right)~\left(\frac{u+v}{u}\right)~f$ $f_r = f~\left(\frac{u+v}{u-v}\right)$ | HuggingFaceTB/finemath | |
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A065906 Integers i > 1 for which there are three primes p such that i is a solution mod p of x^4 = 2. 5
15, 48, 55, 197, 206, 221, 235, 283, 297, 408, 444, 472, 489, 577, 578, 623, 641, 677, 701, 703, 763, 854, 930, 1049, 1081, 1134, 1140, 1159, 1160, 1201, 1253, 1303, 1311, 1328, 1374, 1385, 1415, 1458, 1459, 1495, 1501, 1517, 1557, 1585, 1714, 1723, 1726 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Solutions mod p are represented by integers from 0 to p-1. The following equivalences holds for i > 1: There is a prime p such that i is a solution mod p of x^4 = 2 iff i^4 - 2 has a prime factor > i; i is a solution mod p of x^4 = 2 iff p is a prime factor of i^4 - 2 and p > i. i^4 - 2 has at most three prime factors > i. For i such that i^4 - 2 has no resp. one resp. two prime factors > i cf. A065903 resp. A065904 resp. A065905. LINKS FORMULA a(n) = n-th integer i such that i^4 - 2 has three prime factors > i. EXAMPLE a(3) = 55, since 55 is (after 15 and 48) the third integer i for which there are three primes p > i (viz. 73, 103 and 1217) such that i is a solution mod p of x^4 = 2, or equivalently, 55^4 - 2 = 9150623 = 73*103*1217 has three prime factors > 4. (cf. A065902). PROG (PARI): a065906(m) = local(c, n, f, a, s, j); c = 0; n = 2; while(cn, s = concat(s, f[j, 1]))); if(matsize(s)[2] == 3, print1(n, ", "); c++); n++) a065906(50) CROSSREFS Cf. A040028, A065902, A065903, A065904, A065905. Sequence in context: A236401 A000813 A156205 * A154060 A168305 A208155 Adjacent sequences: A065903 A065904 A065905 * A065907 A065908 A065909 KEYWORD nonn AUTHOR Klaus Brockhaus, Nov 28 2001 STATUS approved
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# Physics- Accleration of a Falling Penny
A visitor to the observation deck of a skyscraper manages to drop a penny over the edge. As the penny falls faster, the force due to air resistance increases. How does this affect the acceleration of the penny?
a. The acceleration decreases
b. The acceleration remains constant and not zero (my answer)
c. The acceleration remains zero
d. The acceleration increases.
1. 👍 0
2. 👎 0
3. 👁 455
1. if the force increases (more upward) the the net downward force must decrease.
So, (a) since F=ma
1. 👍 0
2. 👎 0
1. 👍 0
2. 👎 0
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This is a longer assignment. I want you to confirm my first two answers and help me figure out the last. 3 Cu (s) + 8 HNO3 (aq) → 3 Cu(NO3)2 (aq) + 2 NO (g) + 4 H2O (l) A copper penny with a mass of 3.067 g is dissolved in 100.0
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A stone is dropped from the upper observation deck (the space deck) of the CN Tower, 450m above the ground. (a) How long does it take the stone to hit the ground? (b) If the stone is thrown downward with a speed of 5m/s, how long | HuggingFaceTB/finemath | |
Matrices 2. Solving Square Systems of Linear Equations; Inverse Matrices
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1 Matrices 2. Solving Square Systems of Linear Equations; Inverse Matrices Solving square systems of linear equations; inverse matrices. Linear algebra is essentially about solving systems of linear equations, an important application of mathematics to real-world problems in engineering, business, and science, especially the social sciences. Here we will just stick to the most important case, where the system is square, i.e., there are as many variables as there are equations. In low dimensions such systems look as follows (we give a 2 2 system and a 3 3 system): a 11 x 1 + a 12 x 2 = b 1 a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 (7) a 21 x 1 + a 22 x 2 = b 2 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 In these systems, the a ij and b i are given, and we want to solve for the x i. As a simple mathematical example, consider the linear change of coordinates given by the equations x 1 = a 11 y 1 + a 12 y 2 + a 13 y 3 x 2 = a 21 y 1 + a 22 y 2 + a 23 y 3 x 3 = a 31 y 1 + a 32 y 2 + a 33 y 3 If we know the y-coordinates of a point, then these equations tell us its x-coordinates immediately. But if instead we are given the x-coordinates, to find the y-coordinates we must solve a system of equations like (7) above, with the y i as the unknowns. Using matrix multiplication, we can abbreviate the system on the right in (7) by x 1 b 1 (8) Ax = b, x = x 2, b = b 2, x 3 b 3 where A is the square matrix of coefficients (a ij ). (The 2 2 system and the n n system would be written analogously; all of them are abbreviated by the same equation Ax = b, notice.) You have had experience with solving small systems like (7) by elimination: multiplying the equations by constants and subtracting them from each other, the purpose being to eliminate all the variables but one. When elimination is done systematically, it is an efficient method. Here however we want to talk about another method more compatible with handheld calculators and MatLab, and which leads more rapidly to certain key ideas and results in linear algebra. Inverse matrices. Referring to the system (8), suppose we can find a square matrix M, the same size as A, such that (9) MA = I (the identity matrix). 1
2 2 MATRIX INVERSES We can then solve (8) by matrix multiplication, using the successive steps, A x = b M(A x) = M b (10) x = M b; where the step M(Ax) = x is justified by M(A x) = (MA)x, = I x, by (9); by associative law; = x, because I is the identity matrix. Moreover, the solution is unique, since (10) gives an explicit formula for it. The same procedure solves the problem of determining the inverse to the linear change of coordinates x = Ay, as the next example illustrates. ( ) ( ) Example 2.1 Let A = and M =. Verify that M satisfies (9) above, and use it to solve the first system below for x i and the second for the y i in terms of the x i : x 1 +2x 2 = 1 x 1 = y 1 +2y 2 2x 1 +3x 2 = 4 x 2 = 2y 1 +3y 2 ( )( ) ( ) Solution. We have =, by matrix multiplication. To ( ) ( 0 1 )( ) ( ) x solve the first system, we have by (10), = =, so the x solution is x 1 = 11,x 2 = 6. By reasoning similar to that used above in going from Ax = b to x = Mb, the solution to x = Ay is y = Mx, so that we get y 1 = 3x 1 +2x 2 y 2 = 2x 1 x 2 as the expression for the y i in terms of the x i. Our problem now is: how do we get the matrix M? In practice, you mostly press a key on the calculator, or type a Matlab command. But we need to be able to work abstractly with the matrix i.e., with symbols, not just numbers, and for this some theoretical ideas are important. The first is that M doesn t always exist. M exists A 0. The implication follows immediately from the law M-5 in section M.1 (det(ab) = det(a)det(b)), since MA = I M A = I = 1 A 0.
3 MATRIX INVERSES 3 The implication in the other direction requires more; for the low-dimensional cases, we will produce a formula for M. Let s go to the formal definition first, and give M its proper name, A 1 : Definition. Let A be an n n matrix, with A 0. Then the inverse of A is an n n matrix, written A 1, such that (11) A 1 A = I n, AA 1 = I n (It is actually enough to verify either equation; the other follows automatically see the exercises.) Using the above notation, our previous reasoning (9) - (10) shows that (12) A = 0 the unique solution of A x = b is x = A 1 b; (12) A = 0 the solution of x = A y for the y i is y = A 1 x. Calculating the inverse of a 3 3 matrix Let A be the matrix. The formulas for its inverse A 1 and for an auxiliary matrix adja called the adjoint of A (or in some books the adjugate of A) are T A 11 A 12 A (13) A 1 = adj A = A 21 A 22 A 23. A A A 31 A 32 A 33 In the formula, A ij is the cofactor of the element a ij in the matrix, i.e., its minor with its sign changed by the checkerboard rule (see section 1 on determinants). Formula (13) shows that the steps in calculating the inverse matrix are: 1. Calculate the matrix of minors. 2. Change the signs of the entries according to the checkerboard rule. 3. Transpose the resulting matrix; this gives adja. 4. Divide every entry by A. (If inconvenient, for example if it would produce a matrix having fractions for every entry, you can just leave the 1/ A factor outside, as in the formula. Note that step 4 can only be taken if A = 0, so if you haven t checked this before, you ll be reminded of it now.) The notation A ij for a cofactor makes it look like a matrix, rather than a signed determinant; this isn t good, but we can live with it.
4 4 MATRIX INVERSES Example 2.2 Find the inverse to A = Solution. We calculate that A = 2. Then the steps are (T means transpose): matrix A cofactor matrix T adj A inverse of A To get practice in matrix multiplication, check that A A 1 = I, or to avoid the fractions, check that A adj (A) = 2I. The same procedure works for calculating the inverse of a 2 2 matrix A. We do it for a general matrix, since it will save you time in differential equations if you can learn the resulting formula. ( ) ( ) ( ) ( ) a b d c d b 1 d b c d b a c a A c a matrix A cofactors T adj A inverse of A ( ) Example 2.3 Find the inverses to: a) b) ( ) ( ) Solution. a) Use the formula: A = 2, so A 1 = = b) Follow the previous scheme: = A Both solutions should be checked by multiplying the answer by the respective A. Proof of formula (13) for the inverse matrix. We want to show A A 1 = I, or equivalently, A adj A = A I; when this last is written out using (13) (remembering to transpose the matrix on the right there), it becomes a 11 a 12 a 13 A 11 A 21 A 31 A 0 0 (14) a 21 a 22 a 23 A 12 A 22 A 32 = 0 A 0. a 31 a 32 a 33 A 13 A 23 A A To prove (14), it will be enough to look at two typical entries in the matrix on the right say the first two in the top row. According to the rule for multiplying the two matrices on the left, what we have to show is that (15) a 11 A 11 + a 12 A 12 + a 13 A 13 = A ; (16) a 11 A 21 + a 12 A 22 + a 13 A 23 = 0
5 MATRIX INVERSES 5 These two equations are both evaluating determinants by Laplace expansions: the first equation (15) evaluates the determinant on the left below by the cofactors of the first row; the second equation (16) evaluates the determinant on the right below by the cofactors of the second row (notice that the cofactors of the second row don t care what s actually in the second row, since to calculate them you only need to know the other two rows). a 11 a 12 a 13 a 11 a 12 a 13 a 21 a 22 a 23 a 11 a 12 a 13 a 31 a 32 a 33 a 31 a 32 a 33 The two equations (15) and (16) now follow, since the determinant on the left is just A, while the determinant on the right is 0, since two of its rows are the same. The procedure we have given for calculating an inverse works for n n matrices, but gets to be too cumbersome if n > 3, and other methods are used. The calculation of A 1 for reasonable-sized A is a standard package in computer algebra programs and MatLab. Unfortunately, social scientists often want the inverses of very large matrices, and for this special techniques have had to be devised, which produce approximate but acceptable results.
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LECTURE 5 Solving Systems of Linear Equations Recall that we introduced the notion of matrices as a way of standardizing the expression of systems of linear equations In today s lecture I shall show how | HuggingFaceTB/finemath | |
# geometry
posted by .
<1 and <2 are a linear pair, if <2=67, then find m<1
• geometry -
A + B = 180.
A + 67 = 180
A = 180 - 67 = 113 Deg.
## Similar Questions
1. ### geometry
Give the reason for the step taken from a proof. <1 and <2 are a linear pair given m<1 + m<2 = 180* ?
2. ### geometry
a tree on a 30degree slope grows straight up .what are the measure of the greatest and smallest angles the tree make with the hill?
3. ### geometry
<4 and <5 form a linear pair and m<4 = 62. Find m<5.
4. ### geometry
1 and 2 are a linear pair. M1=x-35, and m2=x+83. find the measure of each angle
5. ### Geometry
<1 &<2 are a linear pair. m<1=x-30 & m<2=x=76. Find the measure of each angle.
6. ### Geomety check
Complete the two-column proof. Given: <2 ≅ <4, M<2=110¡Æ Proof: m<3 = 70¡Æ m¡Ð2 = m¡Ð4 - Definition of congruent angles m¡Ð4 = 110¨¬ - b. ________ ¡Ð3 and ¡Ð4 are a linear pair - Definition of a linear …
7. ### Geometry help check ??
Given: <2 = <4, m<2 = 110 Proof: m<3 = 70 <2 = <4, m<2 = 110 Given m<2 = m<4 Definition of congruent angles m<4 = 110 b._______ <3 and <4 are a linear pair Definition of a linear pair(shown in diagram) …
8. ### Math
Could someone please help? I can't find the definitions in my book anywhere. Thank you! 1. Given: <2 = <4, m<2 = 110 Proof: m<3 = 70 <2 = <4, m<2 = 110 Given m<2 = m<4 Definition of congruent angles m<4
9. ### Math
1. Given: <2 = <4, m<2 = 110 Proof: m<3 = 70 Can you please help me with this question?
10. ### geometry
Complete the two-column proof. Given: <2 ≅ <4, M<2=110° Proof: m<3 = 70° m∠2 = m∠4 Definition of congruent angles m∠4 = 110º b. ________ ∠3 and ∠4 are a linear pair Definition of a …
More Similar Questions | HuggingFaceTB/finemath | |
1. ## Geometry help needed!
Hi all.
I need a help with a construction question in Geometry. I would really appreciate it if anyone could tell me how to go about it, cause I'm stumped.
Q: Construct a triangle ABC, having given l_ ACB = 90, the hypotenuse AB = 6.4 cm, and the sum of the remaining sides AC and BC as 8 cm.
Thanks!
2. Originally Posted by Insaeno
Hi all.
I need a help with a construction question in Geometry. I would really appreciate it if anyone could tell me how to go about it, cause I'm stumped.
Q: Construct a triangle ABC, having given l_ ACB = 90, the hypotenuse AB = 6.4 cm, and the sum of the remaining sides AC and BC as 8 cm.
Thanks!
Are you talking about a literal compass and straightedge construction or do you just want the lengths of the legs of the right triangle?
Call the remaining two legs x and y.
Then we know that
$x + y = 8$
and
$x^2 + y^2 = 6.4^2$ <-- Since we have a right triangle.
Solving the first for y gives:
$y = 8 - x$
Inserting this into the second equation gives:
$x^2 + (8 - x)^2 = 48.96$
$x^2 + 64 - 16x + x^2 = 48.96$
$2x^2 - 16x + 23.04 = 0$
$x^2 - 8x + 11.52 = 0$
$x = \frac{-(-8) \pm sqrt{(-8)^2 - 4(1)(11.52)}}{2 \cdot 1}$
$x = \frac{8 \pm sqrt{17.92}{2}$
$x \approx 1.8834~cm$ or $x \approx 6.1166~cm$
So let x = 1.8834 cm, then y = 6.1166 cm. (Or x = 6.1166 cm and y = 1.8834 cm. Whichever.)
-Dan
3. Hello, Insaeno!
This is not a simple problem . . .
Construct a triangle ABC, having given $\angle ACB = 90^o$, the hypotenuse $AB = 6.4$ cm,
and the sum of the remaining sides $AC$ and $BC$ is 8 cm.
Since right triangle $ACB$ can be inscribed in a semicircle,
we have this diagram:
Code:
|
* * * C
* | o*
* | o o *
* o| o *
o | o
* o | o*
- A o - - - - + - - - - o B -
-3.2 3.2
The circle has the equation: . $x^2 + y^2 \:=\:3.2^2$ . [1]
The locus of point $C$ where $AC + BC \,= \,8$ is an ellipse.
. . We have: . $a = 4$ and $c = 3.2$
Then: . $b^2 \:=\:a^2-c^2\:=\:4^2 - 3.2^2 \:=\:5.76$
. . The equation of the ellipse is: . $\frac{x^2}{16} + \frac{y^2}{5.76} \:=\:1$ . [2]
And solve the system of equations.
I got: . $x \,= \,\pm\sqrt{7},\;\;y \,= \,\pm1.8$
4. Are you talking about a literal compass and straightedge construction or do you just want the lengths of the legs of the right triangle?
I need to construct such a triangle, and thats the information given to me.
Thanks a lot though, now that the sides are discovered the triangle can be constructed very easily.
Soroban, I'm sorry but I didn't understand your post. I didn't understand how x can be root 7 while x + y = 8.
5. Originally Posted by Insaeno
I need to construct such a triangle, and thats the information given to me.
Thanks a lot though, now that the sides are discovered the triangle can be constructed very easily.
Soroban, I'm sorry but I didn't understand your post. I didn't understand how x can be root 7 while x + y = 8.
x and y are the coordinates of the point C.
Once you have x and y you can calculate the lengths of the sides, and
the lengths in this case are constructible, so a construction can be
concocted once you know what they are (though probably not worth
the effort).
RonL
6. Hello, Insaeno!
I didn't understand how x can be root 7 while x + y = 8.
Read the problem again . . .
$x + y \,=\,8$ is not part of the problem.
. . This locates a point where the sum of the coordinates equals 8.
They ask for a point whose distances from A and B has a sum of 8.
7. Oh, Alright, I understand now. Thanks a lot, you've helped tremendously!! | HuggingFaceTB/finemath | |
BrownMath.com → Trig w/o Tears → 1 / Intro
Updated 30 Nov 2016
Trig without Tears Part 1:
# Introduction
Trigonometry is fascinating! It started as the measurement (Greek metron) of triangles (Greek trigonon), but now it has been formalized under the influence of algebra and analytic geometry and we talk of trigonometric functions. not just sides and angles of triangles.
Trig is almost the ideal math subject. Big and complex enough to have all sorts of interesting odd corners, it is still small and regular enough to be taught thoroughly in a semester. (You can easily master the essential points in a week or so.) It has lots of obvious practical uses, some of which are actually taught in the usual trig course. And trig extends plenty of tentacles into other fields like complex numbers, logarithms, and calculus.
If you’d like to learn some of the history of trigonometry and peer into its dark corners, I recommend Trigonometric Delights by Eli Maor (Princeton University Press, 1998).
The computations in trigonometry used to be a big obstacle. But now that we have calculators, that’s no longer an issue.
Would you believe that when I studied trig, back when dinosaurs ruled the earth (actually, in the 1960s), to solve any problem we had to look up function values in long tables in the back of the book, and then multiply or divide those five-place decimals by hand? The “better” books even included tables of logs of the trig functions, so that we could save work by adding and subtracting five-place decimals instead of multiplying and dividing them. My College Outline Series trig book covered all of plane and spherical trigonometry in 188 pages—but then needed an additional 138 pages for the necessary tables!
Though calculators have freed us from tedious computation, there’s still one big stumbling block in the way many trig courses are taught: all those identities. They’re just too much to memorize. (Many students despair of understanding what’s going on, so they just try to memorize everything and hope for the best at exam time.) Is it tan² A + sec² A = 1 or tan² A = sec² A + 1? (Actually, it’s neither—see equation 39!)
Fortunately, you don’t need to memorize them. This paper shows you the few that you do need to memorize, and how you can produce the others as needed. I’ll present some ideas of my own, and a wonderful insight by W.W. Sawyer.
## About Trig without Tears
I wrote Trig without Tears to show that you need to memorize very little. Instead, you learn how all the pieces of trigonometry hang together, and you get used to combining identities in different ways so that you can derive most results on the fly in just a couple of steps.
You might like to read some ideas of mine on the pros and cons of memorizing.
Because this textbook helps you,
please click to donate!
Because this textbook helps you,
BrownMath.com/donate.
To help you find things, I’ll number the most important equations and other facts. (Don’t worry about the gaps in the numbering. I’ve left those to make it easier to add information to these pages.)
A very few of those, which you need to memorize, will be marked “memorize”. Please don’t memorize the others. The whole point of Trig without Tears is to teach you how to derive them as needed without memorizing them. If you can’t think how to derive one, the boxes should make it easy to find it. But then please work through the explanation. I truly believe that if you once thoroughly understand how all these identities hang together, you’ll never have to memorize them again. (It’s worked for me since I first studied trig in 1965.)
This is rather a long document for reading on screen. If you prefer, you might want to print the printer-friendly version.
### Abstract or Concrete?
Trig is really a mix of practical and theoretical. On one hand, you find distances or angles in triangles; that’s pretty concrete. Computers do it now, but when artillery was guided by humans they used trig to figure how to aim the cannon. Land surveying still needs trig, and so do plenty of other fields. If you take a high-school physics course, you’ll use plenty of trig in it. Parts 2 through 4 in this book have the practical stuff.
But as you go further in math, though there are usually some real-world problems there somewhere, you spend a greater and greater proportion of your time manipulating symbols. That makes sense: tougher problems take more thought and more calculation to solve.
Traditionally, trig courses are a mix of the practical stuff with the stuff that has no direct practical application but that you’ll need to do work in calculus or physics or other technical fields later. Nearly all the trig identities fall into that category. Part 5 and afterward are the groundwork for those future courses.
I don’t say you should stop at the end of Part 4, even if you’re never planning to take another course. “Is it useful to me?” isn’t always the right question to ask. (If people decided things that way, video-game manufacturers would go out of business tomorrow.) Better questions are, “Is this interesting? Would I like to pursue this just for the fun of stretching my mind?”
Yes, even “Is it beautiful?” Symmetry is a lot of how we judge beauty—if you doubt that, picture someone good-looking, and then mentally change one side of their face. There are some beautiful symmetries in the first half of the book, but the second half has a lot more, plus answers to questions bright students have always asked, like What’s the logarithm of a negative number?
### Bonus Topics
By the way, I love explaining things but sometimes I go on a bit too long. So I’ve put some interesting but nonessential notes at the end of most chapters and inserted hyperlinks to them at appropriate points. If you follow them (and I hope you will!), use your browser’s “back” command to return to the main text.
Much as it pains me to say so, if you’re pressed for time you can still get all the essential points by ignoring those side notes. In token of this, they’re labeled BTW. But you’ll miss some of the fun.
### What You Won’t Find Here
Trig without Tears deals exclusively with plane trigonometry, which is what’s taught today in nearly every first course. Spherical trigonometry is not part of this book.
I’m also restricting myself to real arguments to the functions and real values of the functions. I have to draw the line somewhere! (I do use real-valued functions with the polar form of complex numbers in the Notes.) For complex trigonometric functions, see chapter 14 of Eli Maor’s Trigonometric delights (Princeton University Press, 1998).
In Trig without Tears we’ll work with identities and solve triangles. A separate (and much shorter) page of mine explains how to solve trigonometric equations.
## Triangles 101
Let’s not take anything for granted. You probably remember some basic facts from earlier school, but—
• An angle is the meeting of two lines at a point. You can also think of an angle as the amount of rotation a moving line passes through—a hand on a clock, for instance.
• The size of an angle is measured in degrees (or radians, as you’ll see below). An angle of 90°, like 3 to 6 on a clock face, is called a right angle.
In an hour, the minute hand of the clock travels all the way around the clock, through four right angles, so a full circle is 360°.
• A triangle is a plane figure (on a flat surface) with three straight sides and therefore three interior angles (angles between two sides, inside the triangle).
• The three angles of a triangle always total 180°.
• An important theorem, which we won’t prove: Of you consider two sides and their opposite angles, the longer side is always opposite the larger angle.
• An equilateral triangle is one where all three sides are equal. Since no side is greater than any other, no angles can be greater than any other (see previous bullet point), so in an equilateral triangle each angle is 180°/3 = 60°.
• A right triangle is one that contains a right angle (90°). The long side, which must be opposite the right angle, is called the hypotenuse, and the short sides are sometimes called legs.
The famous Pythagorean Theorem or Theorem of Pythagoras, a² + b² = c², says that if you add the squares of the two legs you get the square of the hypotenuse.
• An acute triangle has only acute angles in it; an acute angle is one that is greater than 0° and less than 90°.
If a triangle contains an obtuse angle (greater than 90° but less than 180°), it is called, naturally enough, an obtuse triangle. In a right triangle or obtuse triangle, the other two triangles must be acute; otherwise the three angles would total more than 180°.
## Notation
Fractions other than ½ are written using the slash, such as a/b for a over b.
### Interval Notation
In talking about the domains and ranges of functions, it is handy to use interval notation. Thus instead of saying that x is between 0 and π, we can use the open interval (0, π) if the endpoints are not included, or the closed interval [0, π] if the endpoints are included.
You can also have a half-open interval. For instance, the interval [0, 2π) is all numbers ≥ 0 and < 2π. You could also say it’s the interval from 0 (inclusive) to 2π (exclusive).
Starting with Part 5, I’ll be referring to the four quadrants of a circle, or of the xy plane. The division for a circle would be the two lines from 12 to 6 and from 3 to 9; on the plane, the division is the x and y axes.
The quadrants are usually given Roman numerals, starting at the upper right and going counterclockwise. So Quadrant I or Q I is the top right quarter, Q II is the top left, Q III is the bottom left, and Q IV is the bottom right.
### Degrees and Radians
These pages will show examples with both radians and degrees. The same theorems apply to either way of measuring an angle, and you need to practice with both.
A lot of students seem to find radians terrifying. But measuring angles in degrees and radians is no worse than measuring temperature in °F and °C. In fact, angle measure is easier because 0°F and 0°C are not the same temperature, but 0° and 0 radians are the same angle.
Just remember that a complete circle is 2π radians. Then, think of the twelve hours numbered around the circumference of a clock face. When the hour hand goes all the way around, it travels through 2π radians or 360°. Six hours is half of that, 2π/2 = π or 180°, one hour is π/6 or 30°, two hours is 2π/6 = π/3 or 60°, and so on. (Thanks are due to Jeffrey T. Birt for this suggestion.)
One technical note: Angles don’t actually have units — they’re dimensionless. If you say “π radians”, you could just as well say “π” and leave off the “radians”.
If you’re measuring in degrees, then you do need to use the degree mark. 180° = π, or π radians. In other words, that degree mark (°) just means “×π/180”. It’s the same sort of animal as the percent sign (%), which really just means “/100”.
So you can convert between degrees and radians exactly the same way you convert between inches and feet, or between centimeters and meters. (If conversions in general are a problem for you, you might like to consult my page on that topic.)
But even though you can convert between degrees and radians, it’s probably better to learn to think in both. Here’s an analogy.
When you learn a foreign language L, you go through a stage where you mentally translate what someone says in L into your own language, formulate your answer in your own language, mentally translate it into L, and then speak. Eventually you get past that stage, and you carry on a conversation in the other language without translating. Not only is it more fun, it’s a heck of a lot faster and easier.
You want to train yourself to work with radians for the same reason: it’s more efficient, and saving work is always good. Practice visualizing an angle of π/6 or 3π/4 or 5π/3 directly, without translating to degrees. You’ll be surprised how quickly it will become second nature!
## Practice Problems
To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.
Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.
You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.
1 Find these angles in degrees: (a) π/6; (b) 2π; (c) 1 (that’s right, radian angles aren’t necessarily fractions or multiples of π).
2 Which is the correct definition of an acute angle, in interval notation?
(a) (0°, 90°) (b) [0°, 90°]
3 Two angles of a triangle are 80° and 40°. Fine the third angle.
4 A triangle has an angle of 90°. The two short sides (next to that angle) are 5 and 12. Find the third side.
5 Find these angles in radian measure: (a) 60° (b) 126°; (c) 45°.
Where possible, give an exact answer rather than a decimal approximation.
6 Who said, “The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side”? Is that correct?
7 On a circular clock face, which numbers are the boundaries of each quadrant?
## BTW: The Problem with Memorizing
Dad sighed. “Kip, do you think that table was brought down from on high by an archangel?”
Robert A. Heinlein, in Have Space Suit—Will Travel (1958)
It’s not just that there are so many trig identities; they seem so arbitrary. Of course they’re not really arbitrary, since all can be proved; but when you try to memorize all of them they seem like a jumble of symbols where the right ones aren’t more obviously right than the wrong ones. For example, is it sec² A = 1 + tan² A or tan² A = 1 + sec² A ? I doubt you know off hand which is right; I certainly don’t remember. Who can remember a dozen or more like that, and remember all of them accurately?
Too many teachers expect (or allow) students to memorize the trig identities and parrot them on demand, much like a series of Bible verses. In other words, even if they’re originally taught as a series of connected propositions, they’re remembered and used as a set of unrelated facts. And that, I think, is the problem. The trig identities were not brought down by an archangel; they were developed by mathematicians, and it’s well within your grasp to re-develop them when you need to. With effort, we can remember a few key facts about anything. But it’s much easier if we can fit them into a context, so that the identities work together as a whole.
Why bother? Well, of course it will make your life easier in trig class. But you’ll also need the trig identities in later math classes, especially calculus, and in physics and engineering classes. In all of those, you’ll find the going much easier if you’re thoroughly grounded in trigonometry as a unified field of knowledge instead of a collection of unrelated facts.
This is why it’s easier to remember almost any song than an equivalent length of prose: the song gives you additional cues in the rhythm, common patterns of emphasis, and usually rhymes at the ends of lines. With prose you have only the general thought to hold it together, so that you must memorize it essentially as a series of words. With the song there are internal structures that help you, even if you’re not aware of them.
If you’re memorizing Lincoln’s Gettysburg Address, you might have trouble remembering whether he said “recall” or “remember” at a certain point; in a song, there’s no possible doubt which of those words is right because the wrong one won’t fit in the rhythm.
### On the Other Hand ...
I’m not against all memorization. Some things have to be memorized because they’re a matter of definition. Others you may choose to memorize because you use them very often, you’re confident you can memorize them correctly, and the derivation takes more time than you’re comfortable with. Still others you may not set out to memorize, but after using them many times you find you’ve memorized them without trying to—much like a telephone number that you dial often.
I’m not against all memorization; I’m against needless memorization used as a substitute for thought. If you decide in particular cases that memory works well for you, I won’t argue. But I do hope you see the need to be able to re-derive things on the spot, in case your memory fails. Have you ever dialed a friend’s telephone number and found you couldn’t quite remember whether it was 6821 or 8621? If you can’t remember a phone number, you have to look it up in the book. My goal is to free you from having to look up trig identities in the book.
Thanks to David Dixon for an illuminating exchange of notes on this topic. He made me realize that I was sounding more anti-memory than I meant to, and in consequence I’ve added this note. But he may not necessarily agree with what I say here.
I wrote these pages to show you how to make the trig identities “fit” as a coherent whole, so that you’ll have no more doubt about them than you do about the words of a song you know well. The difference is that you won’t need to do it from memory. And you’ll gain the sense of power that comes from mastering your subject instead of groping tentatively and hoping to strike the right answer by good luck.
## What’s New
next: 2/Six Functions
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## Thermal expansion
When heat is applied to most materials expansion occurs in all directions. Conversely if heat energy is removed contraction occurs in all directions. The affects of expansion and contraction depend on the change in temperature of the material.
Practical applications where expansion and contraction of solid materials must be allowed for include:-
• overhead electrical transmission lines
• railway lines
• bridges
Fitting a metal collar to a shaft or a Steel tyre to a wheel, is often achieved by first heating them so that they expand, fitting them into position and then cooling them so that contraction holds them firmly in place. This is known as shrink fitting. Hot rivets and bimetallic strips are further practical examples of thermal expansion.
Water
Water is a liquid which at a low temperature displays an unusual affect. If cooled contraction occurs until about 4° C, when the volume is at a minimum. Further cooling towards 0° C results in expansion, when the ice is formed considerable expansion occurs, which often causes frozen water pipes to burst.
Temperature coefficients
The change in a property of a material is not always proportional to the change in temperature, but we can describe those that approximate to, in a similar way. The variation in each property is described as a temperature coefficient, defined as a fractional change in the property per degree rise in temperature. The temperature coefficient of resistance would be an example.
Linear expansion
The amount by which unit length of a material expands when the temperature is increased by 1°is known as the coefficient of linear expansion.
The coefficient of linear expansion α is the fractional change in length per degree rise in temperature.
α = expansion
original length x temperature rise
or
α = l₂ – l₁
l₁ ΔT
(expansion = l₁α ΔT)
From the definition we can estimate the new length of the material l₂ at a temperature T₂ from the equation;
New length = original length + expansion
l₂ = l₁ + l₁ α(ΔT) or l₂ = l₁ [(1 + α(ΔT)]
The amounts of expansion for most solid materials are known and can be found in tables of coefficients of linear expansion.
An example being iron at 12 x 10¯⁶ m K¯¹
Example problem
The length of an iron steam pipe is 20.0m at a temperature of 18°C. Determine the length of the pipe under working conditions when the temperature is 300°C.
l₂ = l₁ [(1 + α(ΔT)] -> l₂ = 20[(1 + (12 x 10¯⁶)(300-18)
l₂ = 20(1.003384) -> l₂ = 20.06768m
Thermal expansion coefficients table
Material Linear coefficient α at 20°C (10¯⁶/°C) Aluminium 22.2 Brass 18.7 Carbon steel 10.8 Copper 17 Concrete 12 Diamond 1 Glass 8.5 Gold 14 Iron(pure) 12 Lead 29 Magnesium 26 Mercury 61 Molybdenum 4.8 Nickel 13 PVC 52 Silver 18 Stainless steel 17.3 Steel 11 – 13 Tungsten 4.5 Water 69
EngineerStudent.co.uk © 2019 | HuggingFaceTB/finemath | |
The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A261147 English flagpole sequence of nonnegative integers (like A215693, but with a(1)=0). 1
%I
%S 0,1,3,2,4,7,5,8,13,14,9,10,12,20,21,15,22,24,23,6,11,17,16,18,25,19,
%T 27,26,28,29,38,30,31,32,33,34,35,36,37,39,41,40,42,44,43,45,47,48,49,
%U 51,46,57,53,50,55,59,52,62,58,111,65,61,54,72,63,67,56
%N English flagpole sequence of nonnegative integers (like A215693, but with a(1)=0).
%C Concatenate the spelling of the terms in English (ignoring "and", punctuation, capitalization, and spaces); the i-th letter of this infinite word is in a(i). This is the lexicographically least such sequence of distinct nonnegative integers. A215693 is the lexicographically least such sequence of distinct positive integers.
%C In the first 100000 terms: the records for a(n)-n are set by a(1)=0, a(3)=3, a(6)=7, a(9)=13, a(14)=20, a(31)=38, a(60)=111, a(119)=211, a(18617)=18711; the records for n-a(n) are set by a(1)=0, a(4)=2, a(20)=6, a(335)=319.
%H Danny Rorabaugh, <a href="/A261147/b261147.txt">Table of n, a(n) for n = 1..3000</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/English_numerals">English numerals</a>
%H Hans Havermann, <a href="/A261147/a261147.txt">List of a(n) for n = 1..100000</a>
%e The sequence begins: 0,1,3,2,4,7,5,8,13....
%e The initial terms are spelled: ZERO;ONE;THREE;TWO;....
%e These letters are found in the spelling of the initial terms: Zero, onE, thRee, twO; fOur, seveN, fivE; eighT, tHirteen, ....
%o (Sage)
%o little = {0:"", 1:"one", 2:"two", 3:"three", 4:"four", 5:"five", 6:"six", 7:"seven", 8:"eight", 9:"nine", 10:"ten", 11:"eleven", 12:"twelve", 13:"thirteen", 14:"fourteen", 15:"fifteen", 16:"sixteen", 17:"seventeen", 18:"eighteen", 19:"nineteen"}
%o decade = {2:"twenty", 3:"thirty", 4:"forty", 5:"fifty", 6:"sixty", 7:"seventy", 8:"eighty", 9:"ninety"}
%o illion = {1:"thousand", 2:"million", 3:"billion", 4:"trillion"}
%o def nmb_wrd(n): # For positive integers n<10^15
%o if n<20: return little[n]
%o if n<100: return decade[floor(n/10)] + little[n%10]
%o if n<1000: return little[floor(n/100)] + "hundred" + nmb_wrd(n%100)
%o k = floor((len(str(n))-1)/3)
%o return nmb_wrd(floor(n/10^(3*k))) + illion[k] + nmb_wrd(n%(10^(3*k)))
%o def English_fp(n, A = [0,1,3,2], i = 1):
%o while len(A)<n:
%o for let in list(nmb_wrd(A[i])):
%o new = 1
%o while (new in A) or (let not in list(nmb_wrd(new))): new += 1
%o A.append(new)
%o i += 1
%o return A[:n]
%o English_fp(67) # A215693 is English_fp(113, [1,7,3])
%Y Cf. A215693.
%K nonn,word
%O 1,3
%A _Danny Rorabaugh_, Nov 28 2015
%E Corrected by _Danny Rorabaugh_, Nov 29 2015
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# Help with proportionality and literal equations
• jxj
In summary: The language of proportionality is about like "direct variation" and "joint variation". A simple example is like a line represented as y=x. Let x be each of several different values. Find y. Same value as x. Plot the line of points (x, y). What is the proportionality constant? y=kx ?This is k=1. You could say, y=1*x. That is a simple, very simple example. k=y/xk=1*xIn summary, In question A, if F1 = -kxspecific-nonzero-x and F2
jxj
## Homework Statement
Its a series of problems essentially basically asking questions about solving proportionality
. For example
"Hooke's Law of a spring can be described by the equation F = -kx, where F is the force exerted by a spring, K us the spring constant, and X is the amount of distance a spring has been stretched.
A. Determine how much the force exerted by spring changes if K is tripled. B. Determine force exerted if X is decreased by a factor of 5. C. Determine the force exerted by a spring if the spring constant is decreased by a factor of three and X is quadrupled. D. Use the method of literal equations to isolate the spring constant and determine the dimensions in units of the spring constant.
F= -kx
## The Attempt at a Solution
I tried to take the approach of it increasing as the k or x increased, but I'm honestly lost lol. Please help
Why don't you take these one-at-a-time. Start with question A. If F1 = -kxspecific-nonzero-x and F2 = -3kxspecific-nonzero-x what is the ratio F2/F1?
jxj
I'm not sure why you need the negative sign.
Just as generally, if some y is proportional to x, DIRECTLY or JOINTLY proportional, then for some constant k, you can say
y=kx.
Remember, k is a constant, and therefore, k=y/x.
If you know or expect the relationship is correct, then if you know one pair of x and y, then you can calculate or compute k. Once you know k value, you can find either x for any y value; or find y for any x value.
jxj
lewando said:
Why don't you take these one-at-a-time. Start with question A. If F1 = -kxspecific-nonzero-x and F2 = -3kxspecific-nonzero-x what is the ratio F2/F1?
wouldn’t that mean the force you have exerted, F, is also tripled?
symbolipoint said:
I'm not sure why you need the negative sign.
Just as generally, if some y is proportional to x, DIRECTLY or JOINTLY proportional, then for some constant k, you can say
y=kx.
Remember, k is a constant, and therefore, k=y/x.
If you know or expect the relationship is correct, then if you know one pair of x and y, then you can calculate or compute k. Once you know k value, you can find either x for any y value; or find y for any x value.
thanks, but I am still a little confused on the proportionality between the two. I also put s negative because that's how hookes law is set haha
lewando said:
Why don't you take these one-at-a-time. Start with question A. If F1 = -kxspecific-nonzero-x and F2 = -3kxspecific-nonzero-x what is the ratio F2/F1?
you would essentially have -3kx/-kx, right?
jxj said:
thanks, but I am still a little confused on the proportionality between the two. I also put s negative because that's how hookes law is set haha
The language of proportionality is about like "direct variation" and "joint variation". A simple example is like a line represented as y=x. Let x be each of several different values. Find y. Same value as x. Plot the line of points (x, y). What is the proportionality constant? y=kx ?
This is k=1. You could say, y=1*x. That is a simple, very simple example.
Another simple example:\
You may have these points: (1, 3), (2,6), (3, 9), and (4, 12).
You may choose to plot the points and draw the line.
Notice, the line intersects with (0,0).
Notice the slope for y/x is 3.
How you may represent this as an equation can be y=3x.
What can you say in written English language for this?
y is directly proportional to x. x is 5 when y is 15.
If you were given that, you could take as y=kx, and find the value for k.
y/x=k
k=y/x
k=15/5
k=3.
jxj
symbolipoint said:
not sure why you need the negative sign.
Because it is a good idea to take the same directions positive for all forces, accelerations, velocities and displacements in a given line. For a spring, the displacement and force are in opposite directions.
jxj
jxj said:
wouldn’t that mean the force you have exerted, F, is also tripled?
you would essentially have -3kx/-kx, right?
Yes and yes. So if you are clear with that, go ahead with the others.
jxj
symbolipoint said:
The language of proportionality is about like "direct variation" and "joint variation". A simple example is like a line represented as y=x. Let x be each of several different values. Find y. Same value as x. Plot the line of points (x, y). What is the proportionality constant? y=kx ?
This is k=1. You could say, y=1*x. That is a simple, very simple example.
Another simple example:\
You may have these points: (1, 3), (2,6), (3, 9), and (4, 12).
You may choose to plot the points and draw the line.
Notice, the line intersects with (0,0).
Notice the slope for y/x is 3.
How you may represent this as an equation can be y=3x.
What can you say in written English language for this?
y is directly proportional to x. x is 5 when y is 15.
If you were given that, you could take as y=kx, and find the value for k.
y/x=k
k=y/x
k=15/5
k=3.
thanks. seeing it shown as (x,y) made it somewhat easier to understand
jxj said:
thanks. seeing it shown as (x,y) made it somewhat easier to understand
Looking at B-C Seeing as how it uses X now, would it still be directly proportional to F or now jointly? I would have said X being decreased by a factor of five would result in F being decreased being by 5, though i’m unsure if that’s correct.
also thanks for the explanation
jxj said:
Looking at B ... I would have said X being decreased by a factor of five would result in F being decreased being by 5, though i’m unsure if that’s correct.
Have you tried the method of post #2? If so, what is the source of your uncertainty?
jxj
lewando said:
Have you tried the method of post #2? If so, what is the source of your uncertainty?
Yes, I tried the process as I used for A. and now I am wondering won't F decrease by a factor of 5, if x is decreased by a factor of 5?
You are correct in your wonderment. But you should be pretty confident in your result. Whatever it is that is causing you to have any doubt is what I wanted to explore.
jxj
lewando said:
You are correct in your wonderment. But you should be pretty confident in your result. Whatever it is that is causing you to have any doubt is what I wanted to explore.
thanks! I think that for C, if K is decreased by a factor of three and X is quadrupled, F would also decrease and quadruple? or does two changes change the proportionality?
and for D. How do suppose I use the method of Literal Equations to isolate the spring constant and determine the units?
jxj said:
I think that for C, if K is decreased by a factor of three and X is quadrupled, F would also decrease and quadruple?
Yes. Sounds contradictory but it is not. For fun (and educational purposes) this time, show your thoughts, step-by-step, à la post #2 method.
for D. How do suppose I use the method of Literal Equations to isolate the spring constant and determine the units?
I suppose the first step can start with you stating your understanding of what is meant by:
1) "method of Literal Equations to isolate the spring constant"
and
2) "and determine the units [of the spring constant]"
jxj
lewando said:
Yes. Sounds contradictory but it is not. For fun (and educational purposes) this time, show your thoughts, step-by-step, à la post #2 method.I suppose the first step can start with you stating your understanding of what is meant by:
1) "method of Literal Equations to isolate the spring constant"
and
2) "and determine the units [of the spring constant]"
lewando said:
Yes. Sounds contradictory but it is not. For fun (and educational purposes) this time, show your thoughts, step-by-step, à la post #2 method.I suppose the first step can start with you stating your understanding of what is meant by:
1) "method of Literal Equations to isolate the spring constant"
and
2) "and determine the units [of the spring constant]"
Lol I got it finally. it took some time to figure out directly and inversely proportionality.
symbolipoint
jxj said:
Lol I got it finally. it took some time to figure out directly and inversely proportionality.
Like said, the typical academic language is usually very easy to use. This is another way of handling constant or uniform rates examples or exercises.
## What is proportionality?
Proportionality refers to the relationship between two quantities that change in a consistent way. It means that as one quantity increases or decreases, the other quantity also increases or decreases in a predictable manner.
## How do you solve problems involving proportionality?
To solve problems involving proportionality, you need to set up a proportion equation using the given quantities. Then, cross-multiply and solve for the unknown variable. It is important to make sure that the units of measurement are consistent throughout the equation.
## What is the difference between direct and inverse proportionality?
In direct proportionality, as one quantity increases, the other quantity also increases at a constant rate. In inverse proportionality, as one quantity increases, the other quantity decreases at a constant rate. This means that in direct proportionality, the ratio of the two quantities remains constant, while in inverse proportionality, the product of the two quantities remains constant.
## What are literal equations?
Literal equations are equations that contain two or more variables. These equations can be rearranged to solve for any of the variables, using the same principles as solving for a single variable equation.
## How can I use proportionality and literal equations in real life?
Proportionality and literal equations are used in many real-life scenarios, such as calculating the cost of items based on their weight, determining the amount of ingredients needed to make a recipe, or figuring out how long it will take to travel a certain distance at a given speed. These concepts are also used in various fields of science, including physics, chemistry, and biology.
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# Using Montgomery ladder to calculate the coordinates
In one of my assignments I need to solve the problem below:
For a Montgomery curve $$3y^2 = x^3+x^2+x$$ over $${\mathbb{F}}_{11}$$ and point $$P = (9,8)$$, compute the $$x$$ coordinate of $$3P$$ using the Montgomery ladder.
I just need a hint to get an idea and not the whole answer. if anyone with some useful links would also be appreciable. As I tried to search google for this but was not able to get any idea.
Is it that I need to find it as $$3P = 2P + P$$ because I found one question about scalar multiplication here. but then I think it is not the correct way to use the Montgomery ladder.
• Do you know what the Montgomery ladder algorithm is? Jan 16, 2016 at 1:57
• Yes I have read about it and I know that we double when we have bit 1 and we add if bit 0. But I am not able to apply here. Or otherwise I haven't that much knowledge about it. Jan 16, 2016 at 1:59
• Jan 16, 2016 at 4:58
The Montgomery Curve is introduced by Peter L. Montgomery for speeding up the Pollard and Elliptic Curve Methods of integer factorization. Montgomery Curves are in the form $$By^2=x^3+Ax^2+x$$ and it is also represented as $$M_{A,B}$$.
Addition and Doubling Formulas of Montgomery Curves
Given Montgomery Curve $$By^2=x^3+Ax^2+x$$ and the point $$P_1 = (x_1,x_1)$$ and $$P_2 = (x_2,x_2)$$ than the addition and doubling, $$P_3 = (x_3,x_3) = P_1+P_2$$ as follows
• Addition formula in Affine Coordinates: $$(P1 \neq \pm P2)$$
\begin{align} \alpha & = (y_2 − y_1)/(x_2 − x_1)\\ x_3 & = B \alpha^2 − A − x_1 − x_2\\ y_3 & = \alpha (x_1 − x_3) − y_1 \end{align}
• Doubling formula in Affine Coordinates: $$(P1 = P2)$$
\begin{align} \alpha & = (3x_1^2 + 2 A x_1 + 1)/(2 B y_1)\\ x_3 & = B\alpha^2 − A − 2x_1\\ y_3 & = \alpha (x_1 − x_3) − y_1 \end{align}
As usual, these are calculated from the geometrical meaning of addition.
The Montgomery Ladder is an algorithm to calculate $$[x]P$$ for a given scalar $$x$$ and a point $$P$$.
Represent $$x$$ in binary $$x= x_0 + 2x_1 + 2^2x_2 + \cdots + 2^mx_m$$ and let $$m= \lfloor \log_2 x \rfloor$$.
R0 = 0
R1 = P
for i from m downto 0 do
if xi = 0 then
R0 = point_double(R0)
else
R1 = point_double(R1)
return R0
Calculation
Now, $$x=3 = 1 + 2\cdot 1$$, i.e. $$x_0=1$$ and $$x_1 =1$$
$$m= \lfloor \log_2 3 \rfloor = 1$$. Therefore, we visit the loop twice.
since $$x_i =1$$ we will be always in the else case, or simply we run the below.
1. R0 = 0
2. R1 = P
4. R1 = point_double(R1)
6. R1 = point_double(R1)
7. return R0
1. $$R_0 = \mathcal{O}$$
2. $$R_1 = P$$
3. $$R_0 = \mathcal{O} + P = P$$
4. $$R_1 = P + P = P$$
5. $$R_0 = P+P = P$$
6. $$R_1 = P+P = P$$
7. return $$R_0 = P$$
The given Montgomery Curve $$3y^2=x^3+x^2+x$$ than it is $$M_{1,3}$$ so $$A=1,B=3$$ and $$P=(9,8)$$
$$P = (1,10)$$ by the below simple sage math point doubling.
k.<a> = GF(11, modulus="primitive")
k.modulus()
A = k(1)
B = k(3)
x1 = k(9)
y1 = k(8)
alpha = (3*x1^2 + 2 *A * x1 + 1)/(2 * B * y1)
x3 = B * alpha^2 - A - 2 * x1
y3 = alpha *(x1 - x3) - y1
print x3
print y3
$$P = P+P = (5,2)$$ by the below simple sage math point addtion.
k.<a> = GF(11)
A = k(1)
B = k(3)
x1 = k(9)
y1 = k(8)
x2 = k(1)
y2 = k(10)
alpha = (y2 - y1)/(x2 - x1)
x3 = B * alpha^2 - A - x1 - x2
y3 = alpha * (x1 - x3) - y1
print x3
print y3
Notes : | HuggingFaceTB/finemath | |
Fundamental Theorem of Algebra
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1 EE 387, Notes 13, Handout #20 Fundamental Theorem of Algebra Lemma: If f(x) is a polynomial over GF(q) GF(Q), then β is a zero of f(x) if and only if x β is a divisor of f(x). Proof: By the division algorithm, f(x) = q(x)(x β)+r(x), where degr(x) < deg(x β) = 1. Thus degr(x) 0, so r(x) is a constant polynomial, r(x) = r 0. Therefore r 0 = r(β) = f(β) q(β)(β β) = f(β), hence f(x) is a multiple of x β if and only if f(β) = r 0 = 0. Lemma: A polynomial f(x) of degree n over a field has at most n zeroes. Proof: Each zero of f(x) corresponds to a linear factor of f(x). Because degf(x) = n, there are at most n linear factors. Thus there are at most n distinct zeroes (including multiple zeroes). Blahut (Theorem 4.3.9) calls this the Fundamental Theorem of Algebra. Gauss s FTA: every polynomial equation with complex coefficients and degree 1 has at least one complex root. EE 387, October 21, 2015 Notes 13, Page 1
2 Examples of factors and zeroes Example: Polynomials of degree 2 over GF(2): x 2 = x x, x 2 +1 = (x+1)(x+1), x 2 +x = x(x+1), x 2 +x+1 The only prime polynomial over GF(2) of degree 2 has zeroes in GF(4): (x+β)(x+δ) = x 2 +(β +δ)x+βδ = x 2 +x+1. Whether a polynomial is prime depends on what coefficients are allowed. Example: GF(2 4 ) can be represented as polynomials in α of degree < 4, where α is a zero of the prime (over GF(2)) polynomial x 4 +x+1. Therefore x+α is a factor of x 4 +x+1 over GF(2 4 ). Another zero is a 2 : x 4 +x+1 α 2 = (α 2 ) 4 +α 2 +1 = (α 4 +α+1) 2 = 0 2 = 0. Similarly, α 4 = (α 2 ) 2 and α 8 = (α 4 ) 2 are zeroes. Over GF(16) x 4 +x+1 = (x+α)(x+α 2 )(x+α 4 )(x+α 8 ). EE 387, October 21, 2015 Notes 13, Page 2
3 GF(Q) consists of zeroes of x Q x The order of the multiplicative group of GF(Q) is Q 1. Let e be the order of β of GF(Q). By Lagrange s theorem, e (Q 1), so β Q 1 = β e (Q 1)/e = (β e ) (Q 1)/e = 1 (Q 1)/e = 1. This shows that every nonzero element of GF(Q) is a zero of x Q 1 1. The special case of 0 requires one more factor, x 0, which yields x(x Q 1 1) = x Q x. This polynomial has at most Q zeroes. Thus GF(Q) = zeroes of x Q x. Similarly, for any subfield, GF(q) = zeroes of x q x. Factorizations: x Q x = (x β), x q x = (x β) β GF(Q) β GF(q) x Q x = x(x Q 1 1) = x(x q 1 1)(x Q 1 (q 1) + +x q 1 +1) The last equation holds because (q 1) (Q 1). EE 387, October 21, 2015 Notes 13, Page 3
4 Minimal polynomials Let β GF(Q) and GF(q) GF(Q). Definition: The minimal polynomial over GF(q) of β is the monic polynomial f(x) over GF(q) of smallest degree such that f(β) = 0. Example: GF(4) = {0, 1, β, δ}. Minimal polynomials over GF(2): 0 x, 1 x+1, β,δ x 2 +x+1 Theorem: Suppose GF(q) GF(Q) where Q = q m. 1. Every β in GF(Q) has minimal polynomial over GF(q) of degree m. 2. The minimal polynomial is unique. 3. The minimal polynomial is prime over GF(q). 4. If g(x) is a polynomial over GF(q) such that g(β) = 0 then f(x) g(x). Every β in GF(Q) is a zero of x Q x, whose coefficients (1,0, 1) belong to GF(q). So the minimal polynomial exists and has degree Q. EE 387, October 21, 2015 Notes 13, Page 4
5 Minimal polynomials (cont.) Proof: 1. GF(Q) is a vector space over GF(q) of dimension m. Therefore any set of m+1 elements is linearly dependent over GF(q). In particular, consider the first m+1 powers of β: {1, β, β 2,...,β m } There exist m+1 scalars f 0,f 1,...,f m in GF(q), not all 0, such that f 0 1+f 1 β + +f m β m = 0 = f(β). In other words, β is a zero of f(x) = f 0 +f 1 x+ +f m x m, which is a nonzero polynomial over GF(q) of degree m. Therefore the minimal polynomial of β has degree m. EE 387, October 21, 2015 Notes 13, Page 5
6 Minimal polynomials (cont.) 2. If f 1 (x) and f 2 (x) are distinct minimal polynomials of the same degree, then f(x) = f 1 (x) f 2 (x) is a nonzero polynomial of smaller degree. Since f(β) = 0, we have a contradiction. 3. If f(x) = f 1 (x)f 2 (x) has proper divisors, then f(β) = f 1 (β)f 2 (β) = 0 = either f 1 (β) = 0 or f 2 (β) = 0, contradicting the minimality of f(x). 4. By the division algorithm, If g(β) = 0 then g(x) = q(x)f(x)+r(x), where degr(x) < degf(x). r(β) = g(β) q(β)f(β) = 0. If r(x) 0 then f(x) is not minimal. Thus r(x) = 0 = f(x) g(x). EE 387, October 21, 2015 Notes 13, Page 6
7 Conjugates Definition: The conjugates over GF(q) of β are the zeroes of the minimal polynomial over GF(q) of β (including β itself). Example: GF(4) = {0,1,β,δ}. Then β and δ = β +1 are conjugates since (x+β)(x+δ) = x 2 +(β +δ)x+βδ = x 2 +x+1. Example: GF(8) = {0, 1, α, α+1, α 2,..., 1+α+α 2 }, where α 3 = α+1. The minimal polynomial of α is f(x) = x 3 +x+1. Another zero is α 2 : f(α 2 ) = (α 2 ) 3 +α 2 +1 = (α 3 +α+1) 2 = 0 So α 2 and α 4 = α+α 2 are conjugates of α, which gives the factorization: x 3 +x+1 = (x+α)(x+α 2 )(x+α 4 ) = (x+ 0 1 )(x+ 0 0 )(x+ 0 1 ) (The 3-tuple representations of α i have lsb in the first row.) EE 387, October 21, 2015 Notes 13, Page 7
8 Binomial coefficients and prime numbers ( ) p Lemma: If p is prime and 0 < k < p then p is a divisor of. k ( ) p Proof: = p(p 1) (p k+1) = p (p 1) (p k+1) k k! k! Denominator k! divides p (p 1) (p k+1) ( and ) is relatively prime to p. p Therefore k! divides (p 1) (p k+1), so is a multiple of p. k Lemma: In GF(p m ), (a+b) p = a p +b p. Proof: By the binomial theorem, p 1 ( ) p (a+b) p = a p + a p k b k + b p = a p +b p, k k=1 ( ) p since is multiple of p (0 < k < p) and in GF(p m ) multiples of p are 0. k Corollary: In GF(2 m ), (a+b) 2 = a 2 +b 2. In other words, squaring is linear. Corollary: In GF(q) (q = p m ), (a+b) q = (a+b) pm = a pm +b pm = a q +b q. EE 387, October 21, 2015 Notes 13, Page 8
9 Conjugates of β Theorem: The conjugates of β over GF(q) are β, β q, β q2,..., β qr 1 where r is the least positive integer such that β qr = β. Note: β qm = β Q = β, so r m. In fact, we will see that r m. Proof: First we show that β qi are conjugates. For any f(x) over GF(q) f(β) q = (f 0 +f 1 β +f 2 β 2 + ) q = f q 0 +fq 1 βq +f q 2 β2q + = f 0 +f 1 β q +f 2 β 2q + = f(β q ), since f q i = f i for coefficients in GF(q). If f(x) is the minimal polynomial of β, then f(β q ) = f(β) q = 0 q = 0. Therefore β q is a zero of the minimal polynomial and so is a conjugate of β. EE 387, October 21, 2015 Notes 13, Page 9
10 Conjugates of β (cont.) Next we show that all conjugates of β are in {β qi }. Consider the product f(x) = (x β)(x β q )(x β q2 ) (x β qr 1 ) of linear factors for all the distinct conjugates of β of the form β qi : since β qr = β. Therefore f(x) q = (x q β q )(x q β q2 ) (x q β qr ) = (x q β q )(x q β q2 ) (x q β) = f(x q ) f q 0 + fq 1 xq + + f q rx qr = f 0 + f 1 x q + + f r x qr Since f q i = f i, all the coefficients of f(x) are in GF(q). Obviously, β is a zero of f(x). Any polynomial over GF(q) that has β as zero must have the same r linear factors. Therefore f(x) is a divisor of every such polynomial, hence f(x) is minimal. EE 387, October 21, 2015 Notes 13, Page 10
11 Conjugates: summary The conjugates of β are the zeroes of the minimal polynomial of β. The conjugates of β over GF(q) are β, β q, β q2,...,β qr 1. The minimal polynomial of β is prime over GF(q) but factors over any field GF(Q) that contains β (and hence its conjugates): f(x) = (x β)(x β q )(x β q2 ) (x β qr 1 ) If β GF(q m ) then β has at most m conjugates (including itself). If β has r conjugates, then the linear subspace of GF(q m ) spanned by {1,β,β 2,...,β r 1 } is a field with q r elements. Reciprocals exist because f(x) is prime. If r < m then β belongs to GF(q r ), a proper subfield of GF(q m ). Since GF(q m ) is a vector space over GF(q r ), we conclude that r m. EE 387, October 21, 2015 Notes 13, Page 11
12 Euler phi function The Euler phi function φ(n) is the number of integers between 0 and n that are relatively prime to n. We can express φ(n) in terms of the factorization n = p e 1 1 pe 2 2 pet t : if p is prime then φ(p) = p 1 (1,2,...,p 1 are coprime to p) if p is prime then φ(p e ) = p e p e 1 (multiples of p are not coprime). φ(n) is multiplicative; i.e., if gcd(r,s) = 1 then φ(rs) = φ(r)φ(s). Combining these facts, we obtain the final formula: p 1 if n = p is a prime ( (p 1)p e 1 = 1 1 ) p e if n = p e is power of prime φ(n) = p t φ(p e 1 1 pe pet t ) = (p i 1)p e i 1 i in general i=1 EE 387, October 21, 2015 Notes 13, Page 12
13 Primitive elements Fact: Multiplicative group of the finite field GF(q) is cyclic of order q 1. A primitive element of GF(q) is a generator of the multiplicative group Let α be a primitive element of GF(q). All primitive elements of GF(Q) are powers α i where gcd(i,q 1) = 1. 1 = ai+b(q 1) = α = α ai+b(q 1) = α ai = (α i ) a. Conversely, if gcd(i,q 1) = d > 1 then the order of α i is q 1 d In general, GF(q) has φ(q 1) primitive elements. If q 1 is prime then there are q 2 primitive elements. (This is possible only for q = 3 and for q = 2 m with m odd.) < q 1. GF(4), GF(8), GF(16), GF(32) have respectively 2, 6, 8, 30 primitive elements. The proof that every finite field has a primitive element uses a lemma about groups: if for every divisor d of the order of a group there are at most d elements of order dividing d, then the group is cyclic. EE 387, October 21, 2015 Notes 13, Page 13
14 Primitive elements and polynomials Let α be a primitive element of GF(Q) and GF(q) be a subfield of GF(Q). Let f(x) be the minimal polynomial over GF(q) of α and m = degf(x). Every nonzero element of GF(Q) is a power of α: GF(Q) = {1, α, α 2,..., α Q 2 } Every element of GF(Q) is a polynomial in α of degree m 1: β = b 0 +b 1 α+b 2 α 2 + +b m 1 α m 1 where b 0,b 1,...,b m 1 are coefficients from GF(q). Multiplication by α of a polynomial in α uses the equation f(α) = 0: α(b 0 +b 1 α+ +b m 1 α m 1 ) = b 0 α+b 1 α 2 + +b m 2 α m 1 b m 1 (f 0 + +f m 1 α m 1 ) A primitive polynomial is the minimal polynomial of a primitive element. Equivalently: monic f(x) of degree m is primitive if the order of f(x) is q m 1; i.e., the smallest n such that x n = 1 mod f(x) is n = q m 1. EE 387, October 21, 2015 Notes 13, Page 14 | HuggingFaceTB/finemath | |
Solutions by everydaycalculation.com
## Subtract 10/9 from 7/8
1st number: 7/8, 2nd number: 1 1/9
7/8 - 10/9 is -17/72.
#### Steps for subtracting fractions
1. Find the least common denominator or LCM of the two denominators:
LCM of 8 and 9 is 72
Next, find the equivalent fraction of both fractional numbers with denominator 72
2. For the 1st fraction, since 8 × 9 = 72,
7/8 = 7 × 9/8 × 9 = 63/72
3. Likewise, for the 2nd fraction, since 9 × 8 = 72,
10/9 = 10 × 8/9 × 8 = 80/72
4. Subtract the two like fractions:
63/72 - 80/72 = 63 - 80/72 = -17/72
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time: | HuggingFaceTB/finemath | |
0
# What is the LCM of 8 16 20 and 25?
Updated: 9/22/2023
Wiki User
8y ago
The LCM is 400.
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8y ago
Wiki User
11y ago
LCM(8, 16, 20, 25) = 400
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Q: What is the LCM of 8 16 20 and 25?
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Related questions
### What is the lcm of 8 20 16 and 100?
400 is the LCM. 8 x 50 = 400 20 x 20 = 400 16 x 25 = 400 100 x 4 = 400
### What is the LCM of 8 20 16?
LCM of 8 16 20 is 80.
### Lcm of 8 16 and 20?
LCM of 8, 16, 20 = 80
The LCM is 400.
### What is the LCM for 8 16 and 20?
Least Common Multiple (LCM) for 8 16 20 is 80.
The LCM is: 80
### What the LCM of 8 16 20?
The LCM of these three numbers is 80.
4 8 12 16
### What is the LCM for 8 20 16 and 100?
The least common multiple of the numbers 8, 20, 16 and 100 is 400.
The LCM is 16.
### What is the LCM of 8 16 and 20?
The lowest common multiple is 80. 8 x 10 = 80, 20 x 4 = 80 and 16 x 5 = 80.
### What is the least common multiple for 5 and 8?
5 10 15 20 25 30 35 40 8 16 24 32 40 40 is lcm or 5*8 = 40 Nothing below that can yield LCM. | HuggingFaceTB/finemath | |
# NSO Level 1- Science Olympiad (SOF) Class 4: Questions 87 - 92 of 435
Access detailed explanations (illustrated with images and videos) to 435 questions. Access all new questions- tracking exam pattern and syllabus. View the complete topic-wise distribution of questions. Unlimited Access, Unlimited Time, on Unlimited Devices!
View Sample Explanation or View Features.
Rs. 300.00 -OR-
### Passage
Study the graph carefully and answer the questions.
The line graph below shows the number of bikes sold over a period seven months.
## Question 87 (3 of 4 Based on Passage)
### Question
MCQ▾
If each bike agent a commission of ₹ 420 on every bike he sells, how much must the company pay to the bike agent in June and July?
### Choices
Choice (4)Response
a.
₹ 74100
b.
₹ 67300
c.
₹ 74410
d.
₹ 71400
## Question 88 (4 of 4 Based on Passage)
### Question
MCQ▾
The total number of bike sold from January to July, is
### Choices
Choice (4)Response
a.
435
b.
439
c.
440
d.
458
## Question 89
### Question
MCQ▾
Rakesh bought 3 balls, 3 cricket bats and 6 pencils. He gave the cashier a ₹ 500 note. How much change did he get?
### Choices
Choice (4)Response
a.
₹ 290
b.
₹ 310
c.
₹ 340
d.
₹ 440
## Question 90
### Question
MCQ▾
What is the height of the tree?
### Choices
Choice (4)Response
a.
39 cm
b.
38 cm
c.
40 cm
d.
37 cm
## Question 91
### Question
MCQ▾
Surbhi is building a pyramid. He uses 9 blocks in the first row, 7 blocks in the second row, and 5 blocks in the third row. If the pattern continues, how many blocks will he use in the fifth row?
### Choices
Choice (4)Response
a.
5
b.
7
c.
3
d.
1
## Question 92
### Question
MCQ▾
In examination hall, 256 students are in the class room. They are sitting in 8 equal rows. How many students are there in each row?
### Choices
Choice (4)Response
a.
32
b.
38
c.
45
d.
36
Developed by: | HuggingFaceTB/finemath | |
# Relationship between $p$-adic numbers and analytic continuation of $1+x+x^2+x^3+...$
The infinite sums
• $$1 + 2 + 4 + 8 + ...$$
• $$1 + 3 + 9 + 27 + ...$$
• $$1 + 5 + 25 + 125 + ...$$
• $$1 + 7 + 49 + 343 + ...$$
• ...
of powers of primes do not converge in the usual sense. However, by analytically continuing the expression
• $$1 + x + x^2 + x^3 + ... = \frac{1}{1-x}$$
we can assign the values $$-1, -\frac{1}{2}, -\frac{1}{4}$$ to these series by simply putting 2, 3, and 5 into this expression.
However, there is another way to get the same result. Rather than using analytic continuation on the reals, we can simply take the result in different rings of $$p$$-adic integers. Then we get:
• The first one converges in the 2-adic integers to $$-1$$.
• The second one converges in the 3-adic integers to $$-\frac{1}{2}$$.
• The third one converges in the 5-adic integers to $$-\frac{1}{4}$$.
This is somewhat interesting, because it would seem that analytic continuation in the reals and p-adic numbers have little to do with one another.
However, in this case, the two are related because by changing the metric on $$\Bbb Q$$ to a $$p$$-adic one, the radius of convergence changes, so that the expression $$1 + x + x^2 + x^3 + ... = \frac{1}{1-x}$$ is valid in the appropriate ring for the values described.
In other words, if you start with the ring of p-adic integers in mind, it is easy to see that the formal power series converges in that ring.
My question is, is it possible to go the other way - starting with a formal power series, to derive a nontrivial, "natural" choice of ring in which the series converges?
Here are some particular examples:
• Suppose we want to evaluate the infinite sum $$1 + x + x^2 + x^3 + ...$$ and set $$x$$ equal to some non-integer value $$r$$. There is no ring of $$r$$-adic integers for arbitrary rational $$r$$, but we can use the analytic continuation to "assign" the result $$\frac{1}{1-r}$$ to this expression. But, is there also a nontrivial, "natural" ring in which this sum does directly converge to that result? What if $$r$$ is allowed to be real rather than rational?
• Likewise, rather than needing a different ring for each $$r$$ there a ring in which that power series converges for all values of $$r \neq 1$$? If not, then at least for all integer or rational $$r$$?
• Is there a general method to associate such rings to power series, particularly ones besides the simple $$1 + x + x^2 + ...$$ one listed?
• I don't think there's much to say here beyond the $p$-adic case, e.g. because of Ostrowski's theorem (en.wikipedia.org/wiki/Ostrowski%27s_theorem). If $x$ is rational you can work in the $p$-adic integers where $p$ is a prime dividing the numerator but not the denominator of $x$ in lowest terms. Mar 28, 2019 at 8:01
• One way to think about this question is to split it up into a formal identity in a ring of formal power (or more generally, Laurent) series, and then asking whether there exist certain "evaluation homomorphisms" from those rings into the base rings. That last question, however, will quickly boil down to good old metric "annulus of convergence" considerations. For this specific series, you need precisely a kind of norm on your ring w.r.t. which your $r$ is smaller than $1$. Mar 28, 2019 at 12:05
• @QiaochuYuan that theorem only applies to absolute values, which are "multiplicative" metrics. We wouldn't need multiplicativity for this, would we? Would we even need for it to be a metric (e.g. obey the triangle inequality)? Perhaps a different topology is all that's really needed. Mar 28, 2019 at 14:48
• @TorstenSchoeneberg I guess the same question applies - would it need to be a "norm" on the field, e.g. an absolute value (which is a multiplicative metric)? Mar 28, 2019 at 15:04
• Is the next sequence in your list $1+7\ldots$ or $1+9\ldots$? Apr 26, 2019 at 3:52
• Not really related to the question. If you know $p$-adic L-functions then there are certainly more interesting questions you can answer to.. Apr 13, 2021 at 16:24 | HuggingFaceTB/finemath | |
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Question
# A particle of mass m is attached to one end of a mass-less spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t=0 with an initial velocity u0. When the speed of the particle is 0.5 u0, it collides elastically with a rigid wall. After this collision
A
The time at which the maximum compression of the spring occurs is t=4π3mk
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B
The speed of the particle when it returns to its equilibrium position is u0
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C
The time at which the particle passes through the equilibrium position for the second time is t=5π3mk
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D
The time at which the particle passes through the equilibrium position for the first time is t=πmk
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Solution
## The correct option is C The time at which the particle passes through the equilibrium position for the second time is t=5π3√mk (A) is correct due to conservation of energy. (B) x=A sin ωt dxdt=Aω cos ωt=u02=u0 cos ωt ⇒cos ωt=12 ⇒ωt=π3 ⇒t=π3ω Δt=2π3ω=2π3√km, its wrong (C) Maximum compression will occur at time =2π3√km+π2√km=7π6√km, hence it's wrong. (D) Second time at equilibrium at =2π3√km+π√km=5π3√km, so correct.
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# How to Calculate the Weighted Average in Excel
When we want to calculate the average of a set of values with different levels of relevance, we compute for the weighted average. Excel allows us to calculate the weighted average using the SUMPRODUCT and SUM functions.
Figure 1. Final result: Calculate the weighted average
Final formula: `=SUMPRODUCT(C3:C7,D3:D7)/SUM(D3:D7)`
## Syntax of the SUMPRODUCT Function
SUMPRODUCT returns the sum of the products of two or more components in the given arrays.
`=SUMPRODUCT(array1, [array2], [array3], ...)`
where
• array – a range of values or cells whose values we want to multiply with other values and then add
• All array arguments must be in the same size
• Non-numeric values in an array are treated as zeros
## Syntax of SUM Function
`=SUM(number1,[number2],...])`
• number1 – any number, array or cell reference whose values we want to add
• Only number1 is required; succeeding numbers are optional
## Setting up the Data
Our table consists of three columns: Measures (column B), Rating (column C) and Weight (column D). In cell F3, we will calculate the weighted average of the ratings.
Figure 2. Sample data to calculate the weighted average
# Calculate the weighted average of the ratings
In order to calculate the weighted average of the ratings, we follow these steps:
Step 1. Select cell F3
Step 2. Enter the formula: `=``SUMPRODUCT(C3:C7,D3:D7)/SUM(D3:D7)`
Step 3. Press Enter
The SUMPRODUCT returns the sum of the products of the ratings and weights (rating x weight). Actual calculation of the SUMPRODUCT looks like this:
`=(C3xD3) + (C4xD4) + (C5xD5)`
The SUM function returns the total of the weights.
The SUMPRODUCT divided by the SUM function returns the value of the weighted average of the ratings. As a result, the value in cell F3 is 90.70%.
Figure 3. Using SUMPRODUCT and SUM to calculate weighted average
Most of the time, the problem you will need to solve will be more complex than a simple application of a formula or function. If you want to save hours of research and frustration, try our live Excelchat service! Our Excel Experts are available 24/7 to answer any Excel question you may have. We guarantee a connection within 30 seconds and a customized solution within 20 minutes.
Are you still looking for help with the Average function? View our comprehensive round-up of Average function tutorials here. | HuggingFaceTB/finemath | |
Solutions by everydaycalculation.com
## Multiply 15/14 with 35/63
1st number: 1 1/14, 2nd number: 35/63
This multiplication involving fractions can also be rephrased as "What is 15/14 of 35/63?"
15/14 × 35/63 is 25/42.
#### Steps for multiplying fractions
1. Simply multiply the numerators and denominators separately:
2. 15/14 × 35/63 = 15 × 35/14 × 63 = 525/882
3. After reducing the fraction, the answer is 25/42
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time: | HuggingFaceTB/finemath | |
Mathematica Pearls
• To: mathgroup at smc.vnet.net
• Subject: [mg2310] Mathematica Pearls
• From: Don Piele <piele at cs.uwp.edu>
• Date: Mon, 23 Oct 1995 12:44:01 -0400
```Mathematica Pearls
Here are two problems that were possed in the latest issue of
Mathematica in Education and Research - Summer 1995. The purpose of
this problem section is to suggest small problems and collect together
interesting (and fast) ways to solve them.
The first problem is for beginnners, the second for more advanced.
1. Ordered Fractions (Source - USA Computing Olympiad, 1995)
Consider the set of all reduced rational numbers between 0 and 1
inclusive with denominators
less than or equal to N.
Here is the set when N = 5:
{0, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1}
Create a function orderedFractions[N] that, given an integer N prints
the ordered fractions
in order of increasing magnitude. (Know as the Farey Sequence).
orderedFractions
{0, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1}
2. sumTo[n] (Source USA Computing Olympiad, 1995)
Create a function, sumTo[n], which finds all the ways you can add
consecutive positive integers that sum to n. Display the solution as a
list of pairs {a,b} where a+(a+1) + (a+2) +....+ b = n.
Here is the solution for n=10000.
sumTo
{{18, 142},{297, 328},{388, 412},{1998, 2002}}
D. Piele
piele at cs.uwp.edu
```
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Question
# A page is opened at random from a book containing 90 pages. Then the probability of a page number is a perfect square, is ___.
Hint:In the above question, number of pages in the book is 90 pages which would be taken as Total cases. Number of perfect squares between 1 to 90 is the favorable case. So, the probability of a page number is equal to the number of perfect squares (favorable cases) divided by total cases.
Probability of a page number is a perfect square $= \dfrac{9}{{90}} = \dfrac{1}{{10}}$
Therefore, probability is $\dfrac{1}{{10}}$.
Note:Probability can also be called as possibility.It is expressed between 0 to 1, where 0 means the event is impossible and 1 means an event is certain. Probability is the measure of an event to occur. The probability of all the events in a sample space is equal to 1. For example – When a die is thrown, there are six possible outcomes – 1, 2, 3, 4, 5, 6. Then the probability of any one of them is $\dfrac{1}{6}$. In general, the probability of an event happening is the number of ways it can happen divided by the total number of outcomes. | HuggingFaceTB/finemath | |
## March 21, 2008
### Crossed Menagerie
#### Posted by Urs Schreiber
Tim Porter kindly made the following notes available online:
Tim Porter
Crossed Menagerie:
an introduction to crossed gadgetry and cohomology in algebra and topology
(pdf with the first 7 chapters (237 pages))
Intoduction
These notes were originally intended to supplement lectures given at the Buenos Aires meeting in December 2006, and have been extended to give a lot more background for a course in cohomology at Ottawa (Summer term 2007). They introduce some of the family of crossed algebraic gadgetry that have their origins in combinatorial group theory in the 1930s and ’40s, then were pushed much further by Henry Whitehead in the papers on Combinatorial Homotopy, in particular, [113].
Since about 1970, more information and more examples have come to light, initially in the work of Ronnie Brown and Phil Higgins, (for which a useful central reference will be the forthcoming, [29]), in which crossed complexes were studied in depth. Explorations of crossed squares by Loday and Guin-Valery, [64, 81] and from about 1980 onwards indicated their relevance to many problems in algebra and algebraic geometry, as well as to algebraic topology have become clear. More recently in the guise of 2-groups, they have been appearing in parts of differential geometry, [21, 10] and have, via work of Breen and others, [17, 18, 19, 20], been of central importance for non-Abelian cohomology. This connection between the crossed menagerie and non-Abelian cohomology is almost as old as the crossed gadgetry itself, dating back to Dedecker’s work in the 1960s, [48]. Yet the basic message of what they are, why they work, how they relate to other structures, and how the crossed menagerie works, still need repeating, especially in that setting of non-Abelian cohomology in all its bewildering beauty.
The original notes have been augmented by additional material, since the link with non-Abelian cohomology was worth pursuing in much more detail. These notes thus contain an introduction to the way ‘crossed gadgetry’ interacts with non-Abelian cohomology and areas such as topological and homotopical quantum field theory. This entails the inclusion of a fairly detailed introduction to torsors, gerbes etc. This is based in part on Larry Breen’s beautiful Minneapolis notes, [20].
If this is the first time you have met this sort of material, then some words of warning and welcome are in order.
There is much too much in these notes to digest in one go!
There is probably a lot more than you will need in your continuing research. For instance, the material on torsors, etc., is probably best taken at a later sitting and the chapter ‘Beyond 2-types’ is not directly used until a lot later, so can be glanced at.
I have concentrated on the group theoretic and geometric aspects of cohomology, since the non-Abelian theory is better developed there, but it is easy to attack other topics such as Lie algebra cohomology, once the basic ideas of the group case have been mastered and applications in differential geometry do need the torsors, etc. I have emphasised approaches using crossed modules (of groups). Analogues of these gadgets do exist in the other settings (Lie algebras, etc.), and most of the ideas go across without too much pain. If handling a non-group based problem (e.g. with monoids or categories), then the internal categorical aspect - crossed module as internal category in groups - would replace the direct method used here. Moreover the group based theory has the advantage of being central to both algebraic and geometric applications.
The aim of the notes is not to give an exhaustive treatment of cohomology. That would be impossible. If at the end of reading the relevant sections the reader feels that they have some intuition on the meaning and interpretation of cohomology classes in their own area, and that they can more easily attack other aspects of cohomological and homotopical algebra by themselves, then the notes will have succeeded for them.
Although not ‘self contained’, I have tried to introduce topics such as sheaf theory as and when necessary, so as to give a natural development of the ideas. Some readers will already have been introduced to these ideas and they need not read those sections in detail. Such sections are, I think, clearly indicated. They do not give all the details of those areas, of course. For a start, those details are not needed for the purposes of the notes, but the summaries do try to sketch in enough ‘intuition’ to make it reasonable clear, I hope, what the notes are talking about! (This version is a shortened version of the notes. It does not contain the material on gerbes. It is still being revised. The full version will be made available later.)
Posted at March 21, 2008 8:01 PM UTC
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### Re: Crossed Menagerie
Just one slight gloss on these marvelous notes
For some of us, interpreting things in terms of Lie algebras (infinitesimal crossed modules) or associative algebras works just as well.
Posted by: jim stasheff on March 21, 2008 8:40 PM | Permalink | Reply to this
### Re: Crossed Menagerie
Thanks, Jim.
In the Ottawa MSc lectures, I restricted myself to the group based theory although there were students with a Lie algebra background in the group. My reasoning was that, exactly as you say, the Lie algebra and associative algebra analogues do go through without much pain’ and in fact are sometimes nicer, more useful (e.g., for the L infinity stuff that Urs is interested in), but that if I tried to handle the other areas as well, I risked jumping on my horse and riding off in all directions’, (as one does frequently!) so the students could (and did) investigate those analogies and perhaps that was the best way for them to make the theory there own’ (in modern parlance, to own’ it).
If someone else feels like writing a similar working document (since that is how I see it) for associative algebras, etc., that would be useful.
One question to the blog is how can one do the associative algebra and Lie algebra analogues of the use of classifying spaces. I am sure that the point has been discussed somewhere in the blog and the papers of the contributors to it, but a quick summary could be useful to me.
I will add another comment later. Happy Easter everyone.
Posted by: Tim Porter on March 22, 2008 9:23 AM | Permalink | Reply to this
### Re: Crossed Menagerie
jumping on my horse and riding off in all directions’
that’s known as an explosive departure ;-)
for classifying spaces in hte Lie case at least in characteristic 0 we have the Cartan or Weil models, though not having read your
notes past the intro I’m not sure what you want a classifying space’ to do
and for the assoc alg case, the bar construction
Posted by: jim stasheff on March 22, 2008 1:24 PM | Permalink | Reply to this
### Re: Crossed Menagerie
I’m not sure what you want a classifying space’ to do’
Doh! perhaps to classify something!
Being serious (oh well!) Ronnie and Phil Higgins wrote a paper The classifying space of a crossed complex, Math. Proc. Camb. Phil. Soc. 110 (1991) 95-120,’ and I in my naivity asked What does BC classify?’, thinking of the group case classifying G-bundles.
I think the research is finally getting to give an answer some 18 years later.
Posted by: Tim Porter on March 22, 2008 7:44 PM | Permalink | Reply to this
### Re: Crossed Menagerie
I suppose you know it, but it deserves to be said in this context here again, for those readers who are wondering:
nerves of strict 2-groups $G$ (= crossed modules) classify $G$ 2-bundles. See the work by Baas, Bökstedt and Kro and John Baez and Danny Stevenson on Classifying space of a strict 2-group.
Posted by: Urs Schreiber on March 22, 2008 7:59 PM | Permalink | Reply to this
### Re: Crossed Menagerie
I would add the name of Larry Breen who did something closely related in the Bitorsors paper way back in (?)1992 or Jack Duskin in his paper on Higher Order Descent who looks at Cech cohomology with a 2-group of coefficients.
I, following Dave Yetter also related the classifying space of a finite crossed module to certain TQFTs. (Interpretations of Yetter’s notion of G-coloring : simplicial fibre bundles and non-abelian cohomology, J. Knot Theory and its Ramifications 5 (1996) 687-720)(That was a classic case of research blocked by a referee who said the idea was rubbish and rejected a research grant proposal! The story of my life! Heigh ho.)
Each approach has opened up new uses and insights. The new ones are excellent, but please don’t forget their precursors.
Posted by: Tim Porter on March 22, 2008 9:05 PM | Permalink | Reply to this
### Re: Crossed Menagerie
I would add the name of Larry Breen who did something closely related in the Bitorsors paper way back in (?)1992 or Jack Duskin in his paper on Higher Order Descent who looks at Cech cohomology with a 2-group of coefficients.
My understanding is – but please correct me if I am wrong about this! – that the Baas-Bökstedt-Kro/Baez-Stevenson work is the first to really do the topology.
While it is clear that for $G$ an $n$-group $G$-bundles have cocycles which are given by $n$-functors $g:{Y}^{•}\to BG$ which after hitting them with geometric realization and nerve become maps of topological spaces $\left(\mid {Y}^{•}\mid \to \mid BG\mid \right)\simeq X\to B\mid G\mid$ the hard part is to show the converse, that homotopy classes of maps $X\to B\mid G\mid$ all come from equivalence classes of such $n$-functors. (In fact, Baas-Bökstedt-Kro don’t use transformations of cocycle functors, but “concordances”.)
But John can certainly say more about it, when he finds the time to look at the blog again.
I, following Dave Yetter also related the classifying space of a finite crossed module to certain TQFTs.
I am very fond of that work on the Yetter model you did. I once wrote about it in the entry Dijkgraaf-Witten and its Categorification by Martins and Porter where I enjoyed noticing that the combinatorial weight that enters the “path integral” is precisely the Leinster measure.
Posted by: Urs Schreiber on March 22, 2008 10:45 PM | Permalink | Reply to this
### Re: Crossed Menagerie
Yes, I agree with you, that early work did the simplicial case or with Breen the sheaf theoretic one. Some of the work of the Granada research work had looked at very similar constructions in exact categories, again emphasising the simplicial viewpoint.
The work you mention was the first that took the case of topological 2-groups seriously and that was very important. The problem of handling simplicial topological groups seemed to me to be partially solved by Mostow’s work on smooth spaces in the J. Diff. Geom. that you have recently mentioned, but I never put aside the time to see if my ideas were correct, and seeing the solution from Baas-Bökstedt-Kro/Baez-Stevenson makes me sure that it was only a partial answer that I had glimpsed.
This seems to me, perhaps falsely, to be related to a problem that Turaev and I faced in our Homotopy QFT paper. We used the notion of a formal HQFT, i.e. combinatorially defined with the conditions of an homotopy QFT encoded in that language as well. (That paper is finally to appear. It disappeared into the automatic paper handling machine of a well known journal for a considerable amount of time. It was actually lost, and when they checked, their records did not list it, yet it was on their submissions server. I mention this as a cautionary tale. Alway check on up papers!!!)
I recall you said back in May 2006 (String Coffee Table blog)
“it would be consistent to say that an HQFT for target a K(G,1) is a representation of a category of cobordisms which are equipped with flat G-bundles with connection”,
and that was exactly my intuition when I was working on the paper with Vladimir.
Your further observation was
A 2d formal $C$-HQFT is a representation of the category of 2d-cobordisms which are equipped with (possibly local trivializations of) principal C-2-bundles with 2-connection
and again that was my intuition. The question that was left hanging was whether or not we could remove the word formal’. I felt we could but Vladimir was more cautious and felt my arguments were not watertight.
Has anyone any light they could shed on this?
I was just looking back at my paper on Yetter’s work to reply to another of your posts and it is interesting to note the amount of work I had to do to get the concordance’ or isotopy’ results that showed independence from the triangulation. It is exactly that problem that we could not handle in the HQFT paper, but the topological complications of the more recent case were not there as everything was discrete.
Finally a request for help’: My reason for launching into the Menagerie notes was to try to understand the work of Aldrovandi (2-Gerbes bound by complexes of gr-stacks, and cohomology, arXiv:math/0512453) which seemed to me to be giving a possibly interesting class of background coefficients for the next level of HQFTs. His work was based on an early thesis of Debremaeker using Dedeker’s approach to non-Abelian cohomology.
I am still trying to understand how that stuff interacts with Breen’s approach to 2-gerbes. (I had looked at Aldrovandi’s Dec2005 version, the latest version, still called version 1, is dated Feb 2, 2008, but is not the same as the previous version 1! It looks at Breen’s work but still does not suggest the answer to my question, since that is not thepurpose of the paper.) Has anyone else looked at this?
Posted by: Tim Porter on March 23, 2008 9:38 AM | Permalink | Reply to this
### Re: Crossed Menagerie
Tim quoted:
A 2d formal C-HQFT is a representation of the category of 2d-cobordisms which are equipped with (possibly local trivializations of) principal C-2-bundles with 2-connection
Triggered a possibly irrelevant comment:
The original idea of homology was essentially that of bordism, i.e. thinking of cycles as being submanifolds
but of course that only works up to n = 1 or 2
wonder if the above for 2 fails for 3?
cf also strictification problems in 3
Posted by: jim stasheff on March 23, 2008 4:50 PM | Permalink | Reply to this
### Re: Crossed Menagerie
That suggests looking at the second Turaev paper on HQFTs where he attack dimension 3.(Homotopy field theory in dimension 3 and crossed group-categories,math/0005291). I have not yet really tried to understand it so any helpful comments would be welcome.
Posted by: Tim Porter on March 23, 2008 9:44 PM | Permalink | Reply to this
### Re: Crossed Menagerie
seemed to me to be giving a possibly interesting class of background coefficients for the next level of HQFTs.
[…]
Has anyone else looked at this?
We want in the end QFTs here whose configuration spaces are $\infty$-bundles with connections for a given $\infty$-group. There should be a cocycle on that $\infty$-group which induces a function on this configuration space.
I like to think of this as higher Chern-Simons theory generally, or rather “Dijkgraaf-Witten” theory, if we are in the discrete setup.
In ${L}_{\infty }$-connections we give an algebraic model for higher Chern-Simons bundles obtained from ${L}_{\infty }$-cocycles, essentially noticing that every ${L}_{\infty }$-cocycle $\mu$ on an ${L}_{\infty }$-algebra $g$ which is in transgression with an invariant polynomial $P$ via a transgression element $\mathrm{cs}$ gives rise to a commuting diagram
$\begin{array}{ccc}\mathrm{CE}\left(g\right)& \stackrel{\mu }{←}& \mathrm{CE}\left({b}^{n}u\left(1\right)\right)\\ ↑& & ↑\\ W\left(g\right)& \stackrel{\left(\mathrm{cs},P\right)}{←}& W\left({b}^{n}u\left(1\right)\right)\\ ↑& & ↑\\ \mathrm{inv}\left(g\right)& \stackrel{P}{←}& \mathrm{inv}\left({b}^{n}u\left(1\right)\right)\end{array}\phantom{\rule{thinmathspace}{0ex}}.$ I am describing in section 5 of Nonabelian differential cohomology # how diagrams like this relate to nonabelian differential cocycles encoding higher bundles with connection. This particular diagram would be the Chern-Simons bundle classified by that cocycle living over $B\mid G\mid$.
General Chern-Simons theory should be a $\Sigma$-model type QFT with
- target space $B\mid G\mid$
- background field that higher Chern-Simons bundle $\mathrm{CS}\to B\mid G\mid$ with connection
- which assigns to a parameter space $\Sigma$ the collection of sections of the result of transgressing $\mathrm{CS}$ to $\mathrm{maps}\left(\Sigma ,B\mid G\mid \right)$.
One needs to find suitable models to realize the structures appearing in this program. I expect that it should be helpful to look directly for the $\infty$-model.
I can’t claim that I am there already, but the notes on Nonabelian differential cohomology building on the ${L}_{\infty }$-connections show how I think the issue can be approached.
That’s maybe not as concrete a reply yet to your “Has anyone else looked at this?”, as you would hope. But I thought I’d mention it.
Posted by: Urs Schreiber on March 24, 2008 12:15 AM | Permalink | Reply to this
### Re: Crossed Menagerie
Tim wrote:
The new ones are excellent, but please don’t forget their precursors.
In some technical sense you can’t forget what you never knew… but your advice is well taken. I really must add a reference to this:
• Brown and Higgins, The classifying space of a crossed complex, Math. Proc. Camb. Phil. Soc. 110 (1991) 95-120.
to my paper with Danny on the classifying space of a topological 2-group. I’m sorry for not doing it earlier, but it’s not too late.
Posted by: John Baez on March 28, 2008 9:10 PM | Permalink | Reply to this
### Re: Crossed Menagerie
how can one do the associative algebra and Lie algebra analogues of the use of classifying spaces
Apart from the discussion of that which we have in ${L}_{\infty }$-connections # I can offer the following integration procedure (as described at the end of Space and Quantity (blog, pdf)):
Let’s say a smooth space is a sheaf on the site whose objects are natural numbers and whose morphisms are smooth maps ${ℝ}^{n}\to {ℝ}^{m}$.
Write $C$ for the category of smooth spaces.
For every smooth space we get a family of $\omega$-groupoids
${\Pi }_{n}:{C}^{\mathrm{op}}\to \omega \mathrm{Cat}$
whose $\left(k-morphisms are thin-homotopy classes of globular $k$-paths and whose $n$-morphisms are full homotopy classes of globular $n$-paths.
Write DGCA for the category of differential ${ℕ}_{+}$-graded commutative algebras. There is a contravariant functor
$S:\mathrm{DGCA}\to C$
(part of an adjunction) which sends each DGCA $A$ to the space
$S\left(A\right):U↦{\mathrm{Hom}}_{\mathrm{DGCA}}\left(A,{\Omega }^{•}\left(U\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$
For every finite dimensional ${L}_{\infty }$-algebra on a ${ℕ}_{+}$-graded vector space $g$ we get its Chevalley-Eilenberg DGCA
$\mathrm{CE}\left(g\right):=\left({\wedge }^{•}{g}^{*},{d}_{g}\right)$
with ${d}_{g}$ of degree +1 and $\left({d}_{g}{\right)}^{2}=0$. In fact, this defines the ${L}_{\infty }$-structure on $g$.
Then we can define the strict $n$-group integrating a given ${L}_{\infty }$-algebra $g$ as
$BG:={\Pi }_{n}\left(S\left(\mathrm{CE}\left(g\right)\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$
The left hand side means that we have a 1-object $\omega$-groupoid. Indeed, the right hand side always has a single object, as one easily sees (since on the point we only have the 0 differential $\left(p>0\right)$-form).
Let $g$ be an ordinary Lie algebra. Then
$BG:={\Pi }_{1}\left(S\left(\mathrm{CE}\left(g\right)\right)\right)$
is indeed the simply connected Lie group integrating it. The right hand side boils down to nothing but the familiar (in some circles) integration of Lie algebras in terms of classes of Lie algebra valued forms on the interval.
We also have for each $g$ its Weil DGCA $W\left(g\right):=\mathrm{CE}\left(\mathrm{inn}\left(g\right)\right):=\mathrm{CE}\left(\mathrm{Cone}\left(g\stackrel{\mathrm{Id}}{\to }g\right)\right)$ as well as the DGCA of basic forms $\mathrm{inv}\left(g\right):=W\left(g{\right)}_{\mathrm{basic}}$ those invariant under inner derivations along the fiber of the inclusion $g↪\mathrm{inn}\left(g\right)\phantom{\rule{thinmathspace}{0ex}}.$
I think that, for $g$ still an ordinary Lie algebra, the 2-groupoid integrating $\mathrm{inn}\left(g\right)$ is
${\Pi }_{2}\left(S\left(W\left(g\right)\right)\right)\simeq B\left(G\stackrel{\mathrm{Id}}{\to }G\right)$
namely the one-object grouopoid version of the strict 2-group which comes from the crossed module coming from the identity on $G$.
This is what I usually denote
$\left(G\to G\right):=\mathrm{INN}\left(G\right):=EG\phantom{\rule{thinmathspace}{0ex}}.$
So:
- $g$ integrates to $G$
- $\mathrm{inn}\left(g\right)=\mathrm{Cone}\left(g\to g\right)$ integrates to $EG$.
I think this goes through for arbitrary ${L}_{\infty }$-algebras, but then things begin to depend a little more crucially on the model for $\infty$-groupoids which one uses.
For instance, if we stay withing strict $\omega$-groupoids then the general $\omega$-groupoid integrating any ${L}_{\infty }$-algebra would be
$BG={\Pi }_{\omega }\left(S\left(\mathrm{CE}\left(g\right)\right)\right)$
and we’d get the universal $G$-bundle in its $\omega$-groupoid incarnatation as $\begin{array}{cccccc}& BG& \to & BEG& \to & BBG\\ :={\Pi }_{\omega }\circ S\left(& \mathrm{CE}\left(g\right)& ←& W\left(g\right)& ←& \mathrm{inv}\left(g\right)& \right)\end{array}\phantom{\rule{thinmathspace}{0ex}}.$
But one needs to notice that when doing it this way for $g$ an ordinary Lie algebra, the $\omega$-groupoid $BG$ won’t be equal to a 1-groupoid, but just equivalent. But the benefit is that then $BEG$ exists as an $\omega$-groupoid, too.
I am in the process of preparing notes with more on this.
Posted by: Urs Schreiber on March 22, 2008 3:16 PM | Permalink | Reply to this
### Re: Crossed Menagerie
Thanks for making this available! A vey useful compendium.
So let me see if I can find something about the answer to the question which we were discussing here:
Fix some model of $\infty$-groupoids. Then try to do the following:
for $BG$ a one-object $\infty$-groupoid, form
$BEG:=\mathrm{Cone}\left(BG\stackrel{\mathrm{Id}}{\to }BG\right)$
(where the symbols on the left can be taken to be just notation, but indicate that we have reasons to expect that this is indeed the one-objec $\infty$-groupoid incarnation of the universal $G$-bundle in the same sense that $BG$ is the classifying “space” for some $G$).
Question: what is and how can we make sense of (an approximation to) the one-object $\infty$-groupoid “$BBG$” in general?
You pointed me to the cofiber Puppe sequence, which you mention on p. 190 of your notes, where it is described for spaces.
In that context we should try to form $BG\stackrel{\mathrm{Id}}{\to }BG↪BEG\to BEG/BG$
and try to make sense of the item on the very right.
What I would like to know is:
- how would we define this quotient in our $\infty$-groupoid context?
- what is that quotient in simple cases where the $\infty$-group $G$ we start with is an ordinary group? a strict 2-group? a Gray group?
- how is that quotient related to “$BBG$”?
Actually, I tried to compute this quotient recently not for $\infty$-groups but for ${L}_{\infty }$-algbras. But then I noticed that my technique for computing such quotients of ${L}_{\infty }$-algeba does not apply, because the ${L}_{\infty }$-algebra $g$ is not normal in $\mathrm{Cone}\left(g\to g\right)$ (unless I made a mistake).
After realizing that I concluded to myself that forming this quotient is not the way to go towards forming “$BBG$”. But maybe I am wrong.
As I menioned in the other discussion, the best idea I have, so far, is that “$BBG$” should be the one-object $\infty$-groupoid version of the rational approximation
$\prod _{i}K\left(ℤ,{n}_{i}\right)$
to $B\mid G\mid$, where ${n}_{i}$ is the degree of the $i$th non-trivial rational cohomology group of $B\mid G\mid :=\mid BG\mid$.
Since
$\prod _{i}K\left(ℤ,{n}_{i}\right)=\mid \prod _{i}{B}^{{n}_{i}-1}U\left(1\right)\mid$
I would set
$BBG$$:=B{\prod }_{i}{B}^{{n}_{i}-1}U\left(1\right)$.
As a first consistency check, if $G$ is abelian to start with $G={B}^{k}U\left(1\right)$, then with this definition $BBG$ is ${B}^{k+2}U\left(1\right)$, as it should be.
And indeed, I know that this is a pretty useful approximation, as I described in that other entry:
the sequence
$BG\to BEG\to BBG$
corresponds, if everything is smooth (Lie) to the sequence
$\mathrm{CE}\left(g\right)←W\left(g\right)←\mathrm{inv}\left(g\right)=W\left(g{\right)}_{\mathrm{basic}}$
for the ${L}_{\infty }$-algbra $g$. If we map into these kind of sequences, the term on the left produces a $G$-valued cocycle in nonabelian cohomology, whereas the term on the right produces the corresponding characteristic classes.
So this “rational approximation version of $\mathrm{𝔹𝔹}G$” seems to make good sense.
Does it also follow from a cofiber Puppe sequence construction?
Posted by: Urs Schreiber on March 21, 2008 8:42 PM | Permalink | Reply to this
### Re: Crossed Menagerie
In a later part of the notes, I started trying to unravel what would happen if I started with a non-strict 3-group or rather a sheaf of such. For me that means I have a 3-truncated sheaf of simplicial groups (i.e. the Moore complex is trivial in dimensions 3 or more). That gives a (sheaf of) 2-crossed modules so the interchange law does not hold and is covered by the lifting of the Peiffer commutator of the bottom two terms (as pre-crossed module)
I think that I checked (and also that David Roberts and Urs have done the same sort of calculation), that the next few terms of the cofibre sequence worked well.
The problem from that viewpoint is to construct a truncated simplicial group which when you do the Puppe sequence, the terms you need are somewhere along the way (to the left). Can one do this for $\infty$-groups? I do not know.
With recent comments by Jim, in mind I think that an ${A}_{\infty }$ model of things may be needed. On the other hand a restricted example might be to model things on a $k$-crossed complex. This would correspond in the simplicial case to a simplicial group with the Moore complex and the thin subgroup having trivial intersection in dimensions greater than $k$ (or is it $k+1$, and hence all Whitehead products ending up in those dimensions being trivial. I think this is something like a group objects in $\left(\infty ,k\right)$-categories or some such terminology, but I’m hazy about the current flavour of terminology!
I did this in general i.e. for a morphism of 2-crossed modules, but the details have not been typed up and are an incoherent and unholy mess at the moment so are not to be relied upon.
The task then was to extend Larry Breen’s notion of a torsor over a sheaf of gr-groupoids to this setting, and for the moment I have not attempted this beyond trying to understand Breen’s work on it and to write out detailed proofs of some of the statements that look as if the proof might help my understanding.
Posted by: Tim Porter on March 22, 2008 12:07 PM | Permalink | Reply to this
### Re: Crossed Menagerie
It has just occured to me (so perhaps partially baked not fully formed) that the theory of crossed $n$-cubes and the corresponding $n$-cube complexes may have some useful insights to shed on this problem. There is a filtration of the category of crossed $ℕ$-cubes by reflexive subcategories, and this is reflected (pun intended) back in the category of crossed $n$-complexes. With crossed $ℕ$-cubes, some of the Puppe stuff looks FUN as it is combinatorial and algebraic rather than homotopy theoretic. This may lead nowhere but I have a feeling it may yield other ideas.
(Crossed $n$-cubes are the analogue of cat${}^{n}$-groups i.e. of $n$-fold categories in the category of groups.)
I should also point out that my notes are not full enough when describing the work, in this area, of the Granada research team and Jack Duskin.
Posted by: Tim Porter on March 22, 2008 2:03 PM | Permalink | Reply to this
### Re: Crossed Menagerie
In a later part of the notes, I started trying to unravel what would happen if I started with a non-strict 3-group
I need to read the later parts. I was seeing if you do something like a cofiber Puppe sequence in the context of $n$-groups, not just spaces. When I try to do that, I run into the problem that $X\stackrel{f}{\to }Y\to {C}_{f}$ may not be normal in ${C}_{f}$.
Posted by: Urs Schreiber on March 22, 2008 3:28 PM | Permalink | Reply to this
### Re: Crossed Menagerie
I think my feeling is that you quess the thing you want and build a model of it, then you find a morphism from $Y$ to it with suitable properties, and the test is then does the kernel of the map match with $X$ up to homotopy. I will think about this in your context but don’t guarantee an answer!
Posted by: Tim Porter on March 22, 2008 6:01 PM | Permalink | Reply to this
### Re: Crossed Menagerie
A good example of the lack of info in the rational approximation’ (W Feller remarked that for any number you care about 17 is an approximation). An abelian topological group is a product of K(pi,n)’s but a general infinite loop space need not split at all.
A simple example is given by a 2-stage Postnikov system with k-invariant
K(Z,n) –> K(Z/2,2n) being the cup square.
Posted by: jim stasheff on March 24, 2008 2:52 PM | Permalink | Reply to this
### Re: Crossed Menagerie
So what’s a good approximation to “$\mid BBG\mid$”?
We know that whatever it is it needs to be the home of the characteristic classes. So the rational approximation I had looked allright. But of course it does lack information.
Posted by: Urs Schreiber on March 24, 2008 10:36 PM | Permalink | Reply to this
### Re: Crossed Menagerie
Another thought. If you look at the end of my discussion of the Puppe sequences you will find that for a crossed module $M=\left(C\to P\right)$, you get a nice homotopy exact sequence${\pi }_{1}\left(M\right)\left[1\right]\to M\to {\pi }_{0}\left(M\right).$This goes across without pain to crossed squares, and the advantage there is that you can take the nerve in one direction and get a simplicial group. More or less that says that with a crossed square, the $B$ in the first direction has a straightforward group structure (explicitly given, strict etc.). Thus you can take a second $B$. This gives an instance of a $\mathrm{BBM}$ … and for a crossed $n$-cube that iterates to give the corresponding $\left(n+1\right)$-type is a ${B}^{\left(n\right)}\left(M\right)$. (see Loday’s paper on ${\mathrm{cat}}^{n}$ groups and $\left(n+1\right)$-types).
That short homotopy exact sequence then extends using the $\Omega$ / loops/ fibre Puppe sequence and it may give you what you need. Thus if you can find a big model of the $\mathrm{BBG}$ with the properties you want, and then use the fibre Puppe sequence, it may do the trick.
If an $n$-group is modelled simplicially (possibly by a truncated simplicial group) then the corresponding ${\mathrm{cat}}^{n}$-group gives you an explicit crossed $n$-cube and the above applies. That is quite a round about approach, but it should then be possible to read off a more direct formula for the terms.
The only possible snag is that your theory has been thought out initially without using crossed $n$-cubes so a translation process would seem to be needed if the above sketched’ procedure was to be carried out.
(I’m afraid this is not completely coherent, and I have to catch a ferry at midday so cannot put in the time to double check the idea, but I tried it for crossed squares and the sequence looks very nice. Some of the key constructions are given in my paper :
$n$-Types of Simplicial Groups and Crossed $n$-cubes, Topology, 32, (1993) 5-24.
and are sketched in section 4.4 of the menagerie, but that does not look at the sequence.)
I must rush.
Posted by: Tim Porter on March 25, 2008 8:50 AM | Permalink | Reply to this
### Re: Crossed Menagerie
Another small comment to add to yesterday’s. On the trian across Ireland, I looked at Puppe sequences of simplicial groups, and also some of the green fields passing by. (Then it got dark and I couldn’t see the fields, so had to concentrate on the maths!)
The sequence that I sketch out in the notes is well known, but I remember that the proofs ended up being sketched because they needed quite a bit of Goerss and Jardine. However starting with a fibration of 3-types you get a Puppe fibre sequence (off to the lef) that involves the classifying spaces AND goes on until it gets to ${\Omega }^{2}$ of the codomain.
I thought about calculating that and come up with the double decalage as a way towards it. The neat thing is that would fit very well with the way I think of crossed squares arising from simplicial groups, so I will look (possibly tonight) at that aspect. I will keep you informed of the progress (or lack of it).
Posted by: Tim Porter on March 26, 2008 6:25 PM | Permalink | Reply to this
### Re: Crossed Menagerie
Thanks a lot for the helpful comments! I’ll get back to that in a moment.
Posted by: Urs Schreiber on March 26, 2008 6:46 PM | Permalink | Reply to this
### Re: Crossed Menagerie
I did not get too much time on this since I got back. I did note one thing that may be significant. If you are using ${\pi }_{0}$, then you expect things to be nice in the bottom dimension, but if $G$ is not connected, $\Gamma G$ and $\Omega G$ seem to go a bit astray. They seem the wrong technology.
On the other hand using the decalage, you get the action of ${G}_{0}$ automatically, and a copy of the loops for each connected component and more. You do not loose information until you are ready to do so. You get a fibration gratis and without charge’ but do not need connectedness at all.
I do not know if that helps.
Posted by: Tim Porter on April 1, 2008 5:30 PM | Permalink | Reply to this
### Re: Crossed Menagerie
Tim, thanks!
As you probably have seen, i was a bit distracted by other things. Am hoping to get back to this here tonight.
Actually, I came to think that there must be a way to do this entirely in terms of crossed complexes (so in strict $\infty$-groups), if we allow ourselves to encode a group, say, $G$, not by the obvious crossed complex conctentrated in the lowest degree, but one equivalent to that but possibly with stuff in higher degrees.
The reason is that for the case that our $n$-group is “higher simply connected” this is what I get from integrating Lie $\infty$-algebras:
Every ${L}_{\infty }$-algebra $g$ I can integrate to a crossed complex by forming the $\omega$-groupoid of thin-homotopy classes of globular paths in the classifying space of $g$-valued forms: $BG:={\Pi }_{\omega }\left(S\left(\mathrm{CE}\left(g\right)\right)\right)\phantom{\rule{thinmathspace}{0ex}}.$ This is a one-object $\omega$-groupoid and hence encodes a crossed complex of groups.
The point is that if $g$ is an ordinary Lie algebra, this crossed complex will not be the obvious one $\cdots 1\to 1\to G$ with $G$ the simply connected Lie group integrating $g$ but one equivalent to that (I think). (The standard one would be obtained by using ${\Pi }_{1}$ instead of ${\Pi }_{\omega }$, where we divide out full homotopy at level 1 already.)
The good thing about this is that at the level of ${L}_{\infty }$-algebras I know the sequence that I am looking for, it is:
$\mathrm{CE}\left(g\right)←\mathrm{CE}\left(\mathrm{Cone}\left(g\to g\right)\right)=W\left(g\right)←W\left(g{\right)}_{\mathrm{basic}}$
and the entire sequence integrates to a sequence of $\omega$-groups:
$\begin{array}{ccccc}BG& \to & BEG& \to & BBG\\ {\Pi }_{\omega }\left(S\left(\mathrm{CE}\left(g\right)\right)\right)& \to & {\Pi }_{\omega }\left(S\left(W\left(g\right)\right)\right)& \to & {\Pi }_{\omega }\left(S\left(\mathrm{inv}\left(g\right)\right)\right)\end{array}\phantom{\rule{thinmathspace}{0ex}}.$
This suggests that given any crossed complex/strict one-object $\infty$-groupoid $BG$, there one equivalent to it such that $BEG$ and “$BBG$” exist as crossed complexes.
Posted by: Urs Schreiber on April 2, 2008 6:49 AM | Permalink | Reply to this
Post a New Comment | open-web-math/open-web-math | |
# tan(3pi/2)
## Enter angle in degrees or radians:
Calculate tan(3pi/2)
Since π < 270 < 3π/2 radians
it is located in Quadrant III
tan is positive.
##### Determine angle type:
270 > 90°, so it is obtuse
tan(3π/2) = N/A
##### Special Angle Values
0010010
30°π/61/23/23/322√3/33
45°π/42/22/21221
60°π/33/21/232√3/323/3
90°π/210N/A10N/A
120°2π/33/2-1/2-√32√3/3-2-√3/3
135°3π/42/2-√2/2-12-√2-1
150°5π/61/2-√3/2-√3/32-2√3/3-√3
180°π0-100-1N/A
210°7π/6-1/2-√3/23/3-2-2√3/33
225°5π/4-√2/2-√2/21-√2-√21
240°4π/3-√3/2-1/23-2√3/3-23/3
270°3π/2-10N/A-10N/A
300°5π/3-√3/21/2-√3-2√3/32-√3/3
315°7π/4-√2/22/2-1-√22-1
330°11π/6-1/23/2-√3/3-22√3/3-√3 | HuggingFaceTB/finemath | |
Question
# A man 2 m tall, standing on top of a 25 m building observe the angle of depression of the top and bottom of the next building to be 30∘ and 60∘ respectively. The height of the next building is ____ m.
Solution
## The diagram can be visualised as shown in the figure. Let the height of next building be h. In ΔADE, tan30∘ = (27−h)DE) 1√(3) = (27−h)DE ----------------------------- I In ΔABC, tan60∘= 27BC √(3) = 27BC ------------------------------II DE = BC, so dividing I by II , we get 3 = 27(27−h) ⇒ h =18m Suggest corrections | HuggingFaceTB/finemath | |
# Math HW help? Please Explain how its done?
1. Which point is a solution to the system of inequalitie: y>4x and y>-2x-3?
A. (2, -200)
B. (0, 10)
C. (1,3)
D. (-2, -10)
Given inequalites are y>4x, y>-2x-3
Take A(2,-200) and substitute in inequality.
-200>4*2
-200>8
this point is not a solution.
Now B(0,10)
10>4*0 and 10>-2*0-3
10>0 10>-3
B is the solution of given system of inequalities.
The system of inequalities are y > 4x and y > - 2x - 3.
• A.
Let, (x, y) = (2, - 200).
Now the inequalities are :
- 200 > 4(2) and - 200 > - 2(2) - 3
- 200 > 8 and - 200 > - 7
The above statements are false.
So, the point (2, - 200) is not a solution to the system of inequalities.
• B.
Let, (x, y) = (0, 10),
Now the inequalities are :
10 > 4(0) and 10 > - 2(0) - 3
10 > 0 and 10 > - 3
The above statements are true.
So, the point (0, 10) is a solution to the system of inequalities.
• C.
Let, (x, y) = (1, 3),
Now the inequalities are :
3 > 4(1) and 3 > - 2(1) - 3
3 > 4 and 3 > - 5
The above statements are false.
So, the point (1, 3) is not a solution to the system of inequalities.
• D.
Let, (x, y) = (- 2, - 10),
Now the inequalities are :
- 10 > 4(- 2) and - 10 > - 2(- 2) - 3
- 10 > - 8 and - 10 > 1
The above statements are false.
So, the point (- 2, - 10) is not a solution to the system of inequalities.
Therefore, B is the solution of the given system of inequalities. | HuggingFaceTB/finemath | |
## Prealgebra (7th Edition)
Add the items in the list then divide the sum by the number of items in the list to find the average. $(-45+(-40)+(-30)+(-25))\div4$ =$(-85+(-30)+(-25))\div4$ =$(-115+(-25))\div4$ =$(-140)\div4$ =$-35$ | HuggingFaceTB/finemath | |
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• Related Books
Math Review
• The SI system is a system of units based on the metric system designed to standardized measurements across the world.
• The fundamental units of the SI system are the meter, kilogram, second, ampere, candela, Kelvin, and mole.
• Significant figures represent a manner of showing which digits in a number are known to some level of certainty.
• Scientific notation provides an efficient way to describing very large and very small numbers.
• Accuracy is how close a measurement is to the actual value. Precision describes the repeatability of a measurement.
Math Review
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Outline 0:10
• Objectives 0:28
• Why Do We Need Units? 0:52
• Need to Set Specific Standards for Our Measurements
• Physicists Have Agreed to Use the Systeme International
• The Systeme International 1:50
• Based on Powers of 10
• 7 Fundamental Units: Meter, Kilogram, Second, Ampere, Candela, Kelvin, Mole
• The Meter 2:18
• Meter is a Measure of Length
• Measurements Smaller than a Meter, Use: Centimeter, Millimeter, Micrometer, Nanometer
• Measurements Larger Than a Meter, Use Kilometer
• The Kilogram 2:46
• Roughly Equivalent to 2.2 English Pounds
• Grams, Milligrams
• Megagram
• Seconds 3:10
• Base Unit of Time
• Minute, Hour, Day
• Milliseconds, Microseconds
• Derived Units 3:41
• Velocity
• Acceleration
• Force
• Prefixes for Powers of 10 4:21
• Converting Fundamental Units, Example 1 4:53
• Converting Fundamental Units, Example 2 7:18
• Two-Step Conversions, Example 1 8:24
• Two-Step Conversions, Example 2 10:06
• Derived Unit Conversions 11:29
• Multi-Step Conversions 13:25
• Metric Estimations 15:04
• What are Significant Figures? 16:01
• Represent a Manner of Showing Which Digits In a Number Are Known to Some Level of Certainty
• Example
• Measuring with Sig Figs 16:36
• Rule 1
• Rule 2
• Rule 3
• All Non-Zero Digits Are Significant
• All Digits Between Non-Zero Digits Are Significant
• Zeros to the Left of the Significant Digits
• Zeros to the Right of the Significant Digits
• Non-Zero Digits 17:21
• Digits Between Non-Zeros Are Significant 17:45
• Zeroes to the Right of the Sig Figs Are Significant 18:17
• Why Scientific Notation? 18:36
• Physical Measurements Vary Tremendously in Magnitude
• Example
• Scientific Notation in Practice 19:23
• Example 1
• Example 2
• Using Scientific Notation 20:02
• Show Your Value Using Correct Number of Significant Figures
• Move the Decimal Point
• Show Your Number Being Multiplied by 10 Raised to the Appropriate Power
• Accuracy and Precision 20:23
• Accuracy
• Precision
• Example 1: Scientific Notation w/ Sig Figs 21:48
• Example 2: Scientific Notation - Compress 22:25
• Example 3: Scientific Notation - Compress 23:07
• Example 4: Scientific Notation - Expand 23:31
Transcription: Math Review
Hi folks, and welcome back to educator.com. What I'd like to do now, is take a few minutes to go through a review of some of the math skills we are going to need to be successful in this course.0000
In our outline, we are going to talk about the metric system and the system international, or SI units, which is the unit system that we use in physics.0010
We will talk about significant figures, scientific notation, and finally, the difference between accuracy and precision, and why they are so important.0019
The objectives are: convert and estimate SI units, recognize fundamental and derived units, express numeric quantities with correct significant figures so we understand how accurate and how precise our measurements are going to be.0028
We will use scientific notations to express physical values efficiently, and finally, differentiate between accuracy and precision.0043
So, why do we need units? Well physics involves the study of prediction and analysis of real world events and real world events have quantifiable numbers.0052
In order to communicate these to other people accurately, we need to have some sort of standards. Whether it be a sound was this loud, or this quiet. We need to put a number on that so we can communicate to people. The light was this bright, or this dim.0062
How do we put numbers around that? We have to decide on a set of standards and physicists have agreed to use what is known as the system international, which is a subset of the metric system.0080
You will also sometimes see it referred to as the MKS system because the basic units include meters, kilograms, and for time, seconds0090
Let's talk about it. The system international is comprised of seven fundamental units. It is based on powers of 10 because it is a subset of the metric system and all other units are derived from these basic seven.0109
The fundamental units are the meters, the kilograms, the second, hence the MKS system, the ampere, the candela, kelvin and the mole, which you may be familiar with from chemistry.0122
So let's start with the meter. The meter is a measure of length similar to the yard in the English system. For measurements smaller than the meter, use a centimeter which is about the width of your pinky finger perhaps. A millimeter is 1/10 of that, micrometer which is often times written μm, and nanometer, nm.0137
For measurements larger than a meter, typically we use kilometers, kilometers, 1000 meters.0158
The kilogram on the other hand, is roughly equivalent to 2.2 English pounds. For measurements smaller than a kilogram, we often times use grams or milligrams. A gram is about a paperclip.0166
For measurements larger than a kilogram, we could use things like a megagram, also known as a metric ton. That is 1000 kilograms.0179
In time, and everyone is probably familiar with this one, the base unit of time is a second. And unlike the rest of the metric system, time is a little funny. It is not based on units of 10. We have, instead, things like minutes, which is 60 seconds. Hours, which is 60 minutes. Days, which is 24 hours, and years, 365¼ days. But most of us are so familiar with this, it is not really a big deal.0189
For shorter times, we go back to base 10. For example, things like milliseconds, microseconds, and nanoseconds and so on...0212
We can take and we can make other units from these fundamental units. A unit of velocity or speed, for example is a meters per second, or if we take that further, could be a kilometer per hour. In the English system it might be a mile per hour.0219
Acceleration is a meter/second2 which is really just a meter per second every second.0237
Force is measured in newtons. But a newton is really just a kilogram times a meter divided by a second, divided by a second. That is kg×m/s2.0244
These are derived units. They are comprised of combinations of those seven fundamental units.0253
As we talk about the metric system and these powers of ten, we need to look at the prefixes.0260
If we talk about something like a kilogram, a kilogram gets the symbol k in front of the g, for gram, kilogram would be 103 grams.0267
A gigagram would be 109 grams. Micrometer would be 10-6 meters, and this table is awfully helpful for converting units.0276
Let's talk about how we convert fundamental units. If we have something like 2,480 meters and we want to convert it to kilometers, here is a nice and easy way to convert these.0282
Even if you can do it in your head, it is probably pretty good to learn this method because later on, when the units get more complicated, it will still work out for you.0302
Let's start off with what we have right now. 2,480m, and I am going to write that as a fraction so it is 2480/1.0310
I want the meters to go away, so I am going to multiply by something where I have meters in the denominator on the right hand side.0320
The units that I want are kilometers. To fill in the rest of this, what I have to realize, is that I can multiply anything by 1 and I get the same value.0328
If I multiply 3,280 by 1, I get 3,280. If I multiply 6 pigs by one, I get 6 pigs. The trick is, I can write 1 in a bunch of different ways.0341
I could write 1 as 0.5/0.5, that is equal to 1. I could write 1 as 3 apples/3 apples, that is still equal to 1.0353
So, I am going to use this math trick and I am going to multiply this by 1, but I am going to pick how I write 1 very carefully.0363
To do this, what I am going to do, is, I am trying to convert to kilometers, k. So I go over to my table of prefixes and I find k for kilo.0370
I see that it means 103, so I am going to write 103 over here on the bottom because on the bottom, there is no prefix in front of the unit.0380
If I put 103 here, I am going to put 1 on the other side. What I have now made is a ratio 1km/103m and 1km is 1000× - 103m.0392
What I have really written here is 1 but I've written it in a special way so when I multiply this through, my meters make a ratio of 1. 2,480×1km/103 is going to leave me with 2.48 and my units that are left are kilometers.0405
2,480m is 2.48km. It is a nice, simple way of converting units. Let's try another one.0428
5.357kg. Let's convert that to grams. I start by writing what I have. 5.357kg, and I write it as a ratio over 1, 5.357kg/1 ×,I want kg to go away so I will write kg in the denominator and I want grams in the numerator.0439
Now, I go to my prefix table and look up kilo, k, again is 103. I am going to write that on this side that does not have a prefix. So that goes on the top this time and put a 1 on the bottom.0460
Now kg and kg make a ratio of 1, or cancel out. What I'm left with is 5.357×103g/1. So 5.357×103;is just going to be 5,357 grams.0476
There we go, converting fundamental units. Let's take a look at a 2 step conversion. Sometimes you have to do this in a couple of different steps.0499
We want to convert 6.4×10-6milliseconds to nanoseconds. I start by writing what we have. 6.4×10-6ms/1. I want ms to go away.0509
I put ms on the bottom and I will convert to my base unit, seconds on the top. I look up what milli, m, means and it means 10-3. Again, I write that on the side that does not have a prefix.0526
So 10-3 up there and 1 on the other side. Milliseconds would make a ratio of 1 and we are left with seconds but I do not want just seconds. I want nanoseconds, so I need to do another step.0541
Multiply by, I want seconds to go away, so I will put that in the denominator and I want units of nanoseconds.0556
Now I go look up nano, n, 10-9. That again goes on the side without a prefix. I put a 1 on the other side and when I go look back here, seconds are going to cancel out.0562
When I multiply this through, 6.4×10-6×10-3/10-9 and the units I'm left with should be nanoseconds. I come up with 6.4 nanoseconds.0578
So 6.4×10-6ms is 6.4 nanoseconds. A two step conversion.0595
Let's go back the other way just to verify we have got this down. We already know what the answer should be here because we just did this problem, just in the other direction. Let's verify that it works.0604
6.4ns/1 and we are going to multiply. We want nanoseconds to go away so we are going to put that on the bottom and we'll go to seconds.0616
I look up nano, which means 10-9 so I write 10-9 over here on the side that does not have a prefix. I put 1 on the other side.0628
Now I'm left with seconds, but I want milliseconds so I do it again. If I want seconds to go away, I want milliseconds so I go to my table and look up milli which is 10-3.0640
It goes on the side without a prefix, I put a 1 on the other side, and as I look here, nanoseconds cancel out, seconds will cancel out, and I should be left with milliseconds.0654
So I multiply through. 6.4×10-9/10-3 gives me 6.4×10-6 and the units I'm left with are milliseconds.0666
There is my answer. It is exactly as we expected. Let's do one with some derived units.0682
We have 32m/s and we want to convert that to something like kilometers per hour. We are going to follow the same basic path again. We are going to write 32m/s as a fraction and if I want to convert to kilometers per hour, I can convert either the meters or seconds first, it does not really matter.0690
Let's start by converting the meters into kilometers. I want meters to go away, so that goes into the denominator and I want kilometers in the numerator.0710
I go to my handy dandy table over here and find that kilo means 103. That goes on the side without a prefix and 1 goes on the other side.0719
Now I'm going to be left with kilometers per second but I want kilometers per hour. So I have another step. The seconds here in the denominator, I need those to go away so I put seconds up here and it would be nice to put hours down here but I do not really know how many seconds are in an hour, but I know how many seconds are in a minute.0730
So I will do this first. I will say that there are 60 seconds in 1 minute. Now when I look at my units, my seconds will cancel out and I'm down to kilometers per minute.0750
I had best do another step here. So if I want minutes to go away, I will put that in the numerator. I want hours and I know that there are 60 minutes in 1 hour. I check my units again and minutes make a ratio of 1 and what I should be left with for units is going to be kilometers in the numerator per hour.0761
I am all set to go do my math. 32×60×60/103should give me about 115.2 kilometers per hour. A derived unit conversion problem.0782
Let's take a look at a multi-step conversion. One last unit conversion problem. Let's see how many seconds are in one year. I have no idea but it is kind of a fun problem to take a look at.0804
Let's start with 1 year, we will make that as a ratio. I do not know how many seconds are in a year but what I do know is that there are 365¼ days in 1 year. Years make a ratio of 1 and I am left with units of days.0816
We are still not to seconds but what I happen to know is if the days go away, there are 24 hours in 1 day. Days will make a ratio of one and I am down to hours. We are still not to seconds.0836
So in another step, I want hours to go away so I will convert to minutes. I know there are 60 minutes in 1 hour. Hours will make a ratio of 1 and I am down to minutes. We are getting closer.0853
I want minutes to go away so I will put minutes in the denominator. I want seconds and there are 60 seconds in 1 minute. Minutes will make a ratio of 1 and I am left with my units of seconds.0867
When I go through and I do all of this math, 1×365¼×24×60×60, I come out with about 3.16×107seconds. That is a lot of seconds in 1 year.0880
Another useful tool or skill is being able to estimate some of these units. For example, estimate the length of a football field. Well that is pretty big but just a rough ballpark figure is maybe about 100 meters.0901
If you are familiar with the English system, 100 yards and 100 meters are roughly the same thing. Or the mass of a student is maybe 60-70kg for a typical student.0919
The length of a marathon is somewhere in the ballpark of about 40, 42km or the mass of a paperclip, I think we mentioned this one previously is somewhere in the ballpark of about one gram.0931
So as you walk around and see different objects see if you can take an estimate of what their mass, their length, their time is in various units. It is a useful skill.0946
Let's talk about significant figures. Significant figures represent the manner of showing which digits in a number you know with some level of certainty.0959
For example, If you are walking along and see a garden gnome in someone's yard, significant figures can help you understand to what exactness you know the height of that garden gnome.0970
14cm, 14.3827482cm, or 14.0cm? These three numbers are all telling you slightly different things. What do they mean? Well, the key to significant figures is following these rules: Write down as many digits as you can with absolute certainty.0982
Once you have done that, go to one more decimal place, one more level of accuracy and try to take your best guess. The resulting value is your quantity in significant figures.1003
Now reading the significant figures, you start with the value in scientific notation and we will talk about that here very shortly. All non zero digits are significant. All digits that are in-between non zero digits are significant.1015
Zeros to the left of significant digits are not significant but zeros to the right of significant digits are significant.1031
As an example, how many significant digits are in the value 43.74km? Well we have 1,2,3,4 non zero digits so we must have 4 significant figures. We know at for certainty to 43.7 and that 4 is our best guess on the next level of accuracy.1040
How many significant figures are in the value of 4,302.5 grams? Well we have 4 non-zero digits and zeros between significant figures are significant so we have a total of 5 significant figures.1062
How many significant figures are in the value of .0083s? Well those are significant but zeros to the left of significant figures are not significant so here we have 2 significant figures.1081
How many significant figures are in the value 1.200×103kg? Zeros to the right of significant figures are significant so we have 1,2,3,4 significant figures.1094
Having gone through this, let's talk now about scientific notation. The need for scientific notation has to do with the tremendous variation in units, in magnitudes of these units, and their sizes.1111
For example, when we talk about length, we could talk about something like the width of a country, like the United States, which is probably a pretty big number, but we also have to talk about the thickness of human hair, all with the same base measurement of meters.1126
Even smaller, how about the transistor on the integrated circuit. Those are getting so small, it is smaller than a wavelength of light. So small that there is no optical microscope in the world that can ever see some of those features.1142
Huge ranges in orders of magnitude for these different measurements. Scientific notation can helps us express these efficiently and make it much easier to read.1154
For example, which of these numbers is easier to read. 4000000000000 or 4×1012. That is obvious, that is a lot easier to read and there is much less chance of making a mistake.1167
Or, which is easier here .0000000001m or 1×10-9m? I think it's easy to see that those are a lot more accurate and less error prone. It is almost tough to read these numbers with all of the zeros because it's so easy to lose your place in them.1182
So, using scientific notation. First off, show your value using the correct number of significant figures. Then, move the decimal point so that one significant figure is to the left of the decimal point.1200
Finally, show your number being multiplied by 10 to the appropriate power so that you get the same quantity, the same numerical value.1214
And finally let's talk about accuracy and precision. There is a difference between these two and in everyday speech, we often times use them interchangeably but in the world of physics, the world of science, There is an important distinction.1223
Accuracy is how close a measurement is to the target value. Precision, on the other hand, is how repeatable your measurements are. I like to look at these from the metaphor of target practice with a bow and arrow.1236
If we are aiming over here towards our first target and we are kind of all over the place here with our arrows and, by the way, they are nowhere close to the target and nowhere near each other, we have low accuracy and low precision which is typically not what you are after.1248
Over here, however, we have pretty high accuracy, we are starting to get close to the target but we are still not repeatable. We are accurate, close to the target but not repeatable therefore we have high accuracy and low precision.1262
Over here we are nowhere close to the target but we can hit that same spot nowhere close to the target every time. We are extremely precise, but our accuracy is off. High precision and low accuracy.1277
Finally, the nirvana of measurement, we have high accuracy, we are very near the target and we are repeatable, we have high precision. We can get near the target and we can get near the target every time.1291
With that, let's take a look at a couple more examples. Let's show this number 300,000,000 in terms of scientific notation assuming we know 3 significant figures.1303
We will find that 3 significant figures and I want to show this in scientific notation, I have one digit, one number to the left of the decimal place and I know 2 more significant figures so I write that as 3.00 to give me my 3 significant figures and I multiply it by 10 to the appropriate power which would be 1,2,3,4,5,6,7,8. 3.00×108.1316
How about showing this number, .000000... There is no way I can read this whole thing... 282 in scientific notation. Well, we have 3 significant figures so this must be 2.82×10 to some power. What power is that going to be? Well we have to move the decimal place 15 places to the right. So it would be 10-15. Isn't that a lot more efficient and easier to read?1343
How about here? Express the number .000470 in scientific notation. We have 3 significant figures, so 4.70×, and the power is going to be, 1,2,3,4 to the right, so 10-4.1387
And one last one, let's see if we can expand 1.11×107. We have 1.11 and we need to move the decimal place 7, so 1,2,3,4,5,6,7. So I would write that as 11,100,000. 11 million, 100 thousand.1408
Hopefully this gets you a good start on some of the basic math skills we are going to need here in physics especially around scientific notation, significant figures, units, converting units, and accuracy and precision. Thanks for watching educator.com, we will see you next time and make it a great day!1434 | HuggingFaceTB/finemath | |
# What are the complexities of algorithm?
## What are the complexities of algorithm?
Complexity of an algorithm is a measure of the amount of time and/or space required by an algorithm for an input of a given size (n).
## What are the different types of complexities?
There are different types of time complexities, so let’s check the most basic ones.
• Constant Time Complexity: O(1)
• Linear Time Complexity: O(n)
• Logarithmic Time Complexity: O(log n)
• Exponential Time Complexity: O(2^n)
What is complexity of algorithm with example?
When we analyse an algorithm, we use a notation to represent its time complexity and that notation is Big O notation. For Example: time complexity for Linear search can be represented as O(n) and O(log n) for Binary search (where, n and log(n) are the number of operations).
### What is complexity explain different types of complexity?
In general, the amount of resources (or cost) that an algorithm requires in order to return the expected result is called computational complexity or just complexity. The complexity of an algorithm can be measured in terms of time complexity and/or space complexity.
### How is complexity measured?
To each Turing machine we can associate a measure of complexity proportional to the number of symbols needed to code it – the smaller is the number of symbols needed to code a Turing machine, the smaller is its complexity.
What do you mean complexity?
1 : the quality or condition of being difficult to understand or of lacking simplicity the complexity of a problem. 2 : something difficult to understand or lacking simplicity the complexities of business.
#### What is complexity factor?
A number that shows the level of complexity to any situation. It comes from the parts, type of connections, unknowns, and uncertainty.
#### How do you describe complexity?
In information processing, complexity is a measure of the total number of properties transmitted by an object and detected by an observer. Such a collection of properties is often referred to as a state. In physical systems, complexity is a measure of the probability of the state vector of the system.
What is the purpose of complexity?
The purpose of Complexity is to report important advances in the scientific study of complex systems. Complex systems are characterized by interactions between their components that produce new information — present in neither the initial nor boundary conditions — which limit their predictability.
## What are complexity levels?
Level of complexity is a measure, which describes characteristics of organizational or social system. In management we can distinguish following levels of system complexity: complicated system (e.g. machine, computer) random system (market, customer behaviour, chaotic changes in financial markets)
## How do you calculate complexity factor?
Technical Complexity Factor (TCF) The values are multiplied by the weighted values and the total TF is determined. Next, the TCF is calculated: TCF = 0.6 + (TF/100)
How do you find the complexity of an algorithm?
The complexity is written as O(), meaning that the number of operations is proportional to the given function multiplied by some constant factor. For example, if an algorithm takes 2*(n**2) operations, the complexity is written as O(n**2), dropping the constant multiplier of 2.
### What is algorithm complexity?
Complexity of an algorithm is a measure of the amount of time and/or space required by an algorithm for an input of a given size (n).
### What is the worst sorting algorithm?
Answer Wiki. 6 Answers. Any sorting algorithm with O(n^2)[n-square] is the worst sorting algorithms there are bubblesort,insertion,selection,treesort etc and many have there worst case complexity as O(n^2) please refer the below chat for any douts.
What is the time complexity of algorithm?
In computer science, the time complexity is the computational complexity that describes the amount of time it takes to run an algorithm.
#### What is the complexity of binary search algorithm?
Binary search is a fast search algorithm with run-time complexity of ο(log n). This search algorithm works on the principle of divide and conquer. For this algorithm to work properly, the data collection should be in the sorted form. | HuggingFaceTB/finemath | |
MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB
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If $\sin{61^{\circ}} = \sqrt{a}$ , determine the value of the following in terms of a: cos73°cos15° + sin 73°sin15°
| 101 views
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use compound angle identity
=cos(73 - 15)
= cos 58, use double angle, = cos2(29) =2cos29^2 - 1
= 2 sin 61^2 - 1
= 2.(a) - 1
=2a-1
by Diamond (43,720 points)
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# Elliptic curve cryptography
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Elliptic Curve Cryptography (sometimes called ECC for short) is the study of elliptic curve equations and the arithmetic operations that apply to them. Normally, an elliptic curve involves two variables x and y which correspond to the X- and Y- coordinates of a point respectively. Curves have special operations for adding, subtracting, and multiplying two points, and the way these operations work is very different from their scalar counterparts.
All curve points have a generator point `G` which can produce all the other points on the curve by means of multiplication by a scalar number. G also happens to be the multiplicative identity, while the additive identity is a special point called the point at infinity and it's represented as 0 or uppercase `O`. It can be thought of as a limit to both ends of the curve.
Each curve has a characteristic number which stands for the number of times the generator point can be added to itself before you end up back at `G`. The x and y coordinates must not be larger or equal to this scalar number. Each curve also has a curve order. All private keys (which themselves are represented as numbers) must be smaller than the group order.
## Operations
Point addition for two unequal points `(x3,y3) = (x1,y1) + (x2,y2)` can be performed in simple scalar arithmetic with the following pseudocode. Note that all arithmetic operation involving x or y coordinates are modulus of the characteristic (`mod p`) applied after it. All occurrences of the modulus are omitted from the pseudocode for brevity. Also displayed here is point doubling which is the case where both points are the same and the traditional point addition algorithm would otherwise not work on them.
``` if (x1,y1) == O
result = (x2,y2)
else if (x2,y2) == O
result = (x1,y1)
else if (x2,y2) == (x1,y1)**(-1)
result = O
if (x2,y2) == (x1,y1):
# Point doubling: 2P
lambda = (3 * x1**2) * (2*y1)**-1
else: | HuggingFaceTB/finemath | |
Section 2.2 - Trigonometric functions
Note that trigonometric functions require a floating point mode
All trigonometric functions in picFX use radient angles. I invite you to forget about degrees. I have to say you that, except to tell the coordinates on the earth (or in astronomy) as longitude and latitude (decl/r.-a.), I do never use degrees at school, for personal research or whatever else. Never.
In radient, the "right" angle (is it so we say? I mean the one between two perpendicular lines, 90°) is pi/2, the flat angle (=180°) is pi, and the full angle (=360°) is 2*pi.
When you draw a part (an arc) of a circle (radius 1), the length of the path is equal to the angle between the two bounds, from the center. And, of course, the length of the circumference of a circle of radius one is... 2pi.
If you don't know what sin, cos, tan, asin, acos or atan mean...
Imagine an object moving in circle, for example the Moon around the Earth, or the Earth around the Sun.
Trigonometric functions allow us to know at which coordinate the object is at a given time. If the circle has a radius of one, the horizontal coordinate is cos(a), a being the angle the object has moved since being on the right, equal to the length of the path it made. sin(a) is the vertical coordinate.
As you can guess, these functions both vary between -1 and 1. To see how the function looks like, put "y+sin(x/10)*10-128" in a function field and render. If you draw a vertical line tangent to the circle on the right, and then imagine there is a light in the center of the circle, then tan(a) gives the vertical coordinate of the "shadow" of the turning object, the crossing point of the tangent and a line coming from the center and going throught the moving point.
tan(±pi/2) is undefined (infinite) because at this moment the "shadow" does not reach the tangent, and the line going through the point is vertical and therefore parralel to the tangent..
Note that when the point is on the left half (cos(a)<0), the shadow does not actually reach the tangent. You are forced to use the line going through the point and the center :-)
See the appearance of the tangent curve using y+tan(x/10)*10-128, or 25500/((y+tan(x/20)*20-128)^2+100).
acos() is the inverse function of cos(). You give the horizontal coordinate and you get the angle. Obviously, acos(x) is not defined if x is bigger than one or smaller than -1...
asin() is the inverse of sin().
atan() is the inverse of tan(). This function is defined for any value. You say where is the "shadow" and you get the angle... It always returns a value between -pi/2 and +pi/2, with most of the change around zero..
End of definitions :-)
As sin and cos return a value between -1 and 1, it is often (when your expression is the (co)sine of something and nothing else) needed to multiply it (127), and add a constant (128) to make it take the [0;255] range. So use sin(...)*127+128.
Use the sine if you want to have height lines of some function, a line each 2pi, starting at pi/2. For example, sqrt((x-128)^2+(y-128)^2) (try it!, maybe putting "255-" at the beginning) is a kind of cone centered on (128;128), and the sine of it makes concentrical circles... sin(x/2)*127+128 makes vertical lines; you can think of x/2 to be a little slope raising in your direction to the right...
The other most handful use of the sine is to make waves. Do two dimensional waves with sin(x)*sin(y). You can also use addition, but note that it will return something within [-2;2] instead of [-1;1]...
Do not hesitate to put sines inside sines. Try for example sin(y/2+10*sin(x/10))*127+128 The cosine function is basically the same as the sine, only with a little (pi/2) shift to the left.
Tangent. I use this if I want to have some regular areas separated by thin lines where sudden and chaotic changes occur. Try for example the above one, where you add a tangent of the horizontal coordinate (and divided the sinx by two, too): sin(tan(x/17)+y/4+5*sin(x/10))*127+128
Also try this tangent of the vertical coordinate to which I added the sine of the horizontal: tan(y/10+sin(x/5))*2+128. A little variation: tan(y/10+sin(x/10))*40+128 Because of their annoying limitation to the [-1;1] range, the asin and acos function are not really useful. (Actually asin(2) and other are defined but not real... wait picFX version supporting complex numbers :-)
ArcTangent, however can be quite useful if you want a quite sudden colour change, however smooth. Try atan((y-128)/8)/pi*255+128
A general formula would be: atan(f(x,y))/pi*255+128: dark when f(x,y) is negative and bright when f(x,y) is positive. The change occur when f(x,y) is around zero.
This actually shows the solutions of an inequation..
I will come again on this when speaking of implicit functions.
Remember that it outputs a value between -pi/2 and pi/2...
If you feel ready, you can already try to experiment all sorts of functions using trig. functions, you can get really nice pictures just with the three sin, tan, atan functions.
As a rule: Do not be afraid to write things that you would think stupid ;-). (Btw, Divisions by zero, arc[co-]sines on undefined values do not make picFX crash :-)
Index
Chapter two: Mathematical function summary
2.1: Four operators
2.2: Trigonometric Functions
2.3: Min, Max, Modulo, Abs
2.4: Other functions | HuggingFaceTB/finemath | |
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# Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 4.2 Factoring Trinomials of the Form x 2 + bx + c.
## Presentation on theme: "Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 4.2 Factoring Trinomials of the Form x 2 + bx + c."— Presentation transcript:
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 4.2 Factoring Trinomials of the Form x 2 + bx + c
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 22 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Factoring Trinomials of the Form x 2 + bx + c Recall by using the FOIL method that F O I L (x + 2)(x + 4) = x 2 + 4x + 2x + 8 = x 2 + 6x + 8 To factor x 2 + bx + c into (x + one #)(x + another #), note that b is the sum of the two numbers and c is the product of the two numbers. So we’ll be looking for 2 numbers whose product is c and whose sum is b. Note: there are fewer choices for the product, so that’s why we start there first.
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 33 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. To Factor a Trinomial of the Form x 2 + bx + c Factoring Trinomials of the Form x 2 + bx + c The product of these numbers is c. The sum of these numbers is b. x 2 + bx + c = (x + )(x + )
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 44 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Factor the polynomial x 2 + 13x + 30. Since our two numbers must have a product of 30 and a sum of 13, the two numbers must both be positive. Positive factors of 30Sum of Factors 1, 3031 2, 1517 3, 1013 Note, there are other factors, but once we find a pair that works, we do not have to continue searching. So x 2 + 13x + 30 = (x + 3)(x + 10). Factoring Polynomials Example
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 55 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Factor the polynomial x 2 – 11x + 24. Since our two numbers must have a product of 24 and a sum of -11, the two numbers must both be negative. Negative factors of 24Sum of Factors – 1, – 24 – 25 – 2, – 12 – 14 – 3, – 8 – 11 So x 2 – 11x + 24 = (x – 3)(x – 8). Factoring Polynomials Example
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 66 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Factor the polynomial x 2 – 2x – 35. Since our two numbers must have a product of – 35 and a sum of – 2, the two numbers will have to have different signs. Factors of – 35Sum of Factors – 1, 3534 1, – 35 – 34 – 5, 7 2 5, – 7 – 2 So x 2 – 2x – 35 = (x + 5)(x – 7). Factoring Polynomials Example
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 77 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Factor: x 2 – 6x + 10 We look for two numbers whose product is 10 and whose sum is – 6. The two numbers will have to both be negative. Negative factors of 10Sum of Factors – 1, – 10 – 11 – 2, – 5 – 7 Since there is not a factor pair whose sum is – 6, x 2 – 6x +10 is not factorable and we call it a prime polynomial. Prime Polynomials Example
Martin-Gay, Prealgebra & Introductory Algebra, 3ed 88 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. You should always check your factoring results by multiplying the factored polynomial to verify that it is equal to the original polynomial. Many times you can detect computational errors or errors in the signs of your numbers by checking your results. Check Your Result!
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## kristopher1014 3 years ago 4(x − 1) = 2(x + 2) + 6
1. sauravshakya
4(x − 1) = 2(x + 2) + 6 4x-4=2x+4+6
2. sauravshakya
Now, can u?
3. Kaederfds
i got x=3
4. sauravshakya
Nope..... @Kaederfds
5. sauravshakya
4x-4=2x+4+6 4x-2x=4+6+4 2x=14
6. monica0597
start by using the distributive property. |dw:1345262204345:dw|
7. sauravshakya
so what do u think is the answer @Kaederfds
8. Kaederfds
sorry, i'm wrong
9. kristopher1014
i got 2x+8+6 what am i doing wrong ?
10. monica0597
I think this is how you solve this. Check with someone else though just in case. |dw:1345262530432:dw|
11. DLS
4(x − 1) = 2(x + 2) + 6 4x-4=2x+4+6 2x=14 x=7
12. kristopher1014
@monica0597 its not because 5 is not an answer but thnks
13. monica0597
lol ur welcome i guess?
14. panlac01
try plugging 5 into the original equation and see if it fits
15. panlac01
dls demonstrated it
16. kristopher1014
@panlac01lol how would that make sence at ?
17. panlac01
everytime you get the answer for x, you ought to to plug the value into the original equation to see if it fits. that is the only way to check if your value for x is correct
18. kristopher1014
i know but theres no five in that equation
19. panlac01
I know
20. panlac01
but there's an x, which can be any value.
21. kristopher1014
omfg i know just forget it
22. panlac01
but 5 is not the answer
23. kristopher1014
@dls how did u get 14 ?
24. monica0597
hold up, am i actually right ? @panlac01
25. panlac01
26. monica0597
5
27. panlac01
then no
28. sauravshakya
29. panlac01
30. sauravshakya
right? @panlac01
31. monica0597
ohhhhhhh...
32. panlac01
monica, you forgot something
33. monica0597
what did i forget?
34. kristopher1014
can some one shoe me how he got 14 so i can inderstand better ?
35. kristopher1014
show *
36. panlac01
you added 4 on one side of the equation, but forgot to do it on the other side
37. monica0597
oh no wonder. so then that's where you get the 14
38. monica0597
like this? |dw:1345263416469:dw|
39. panlac01
first, distributive, then group like terms. now, look for the value of x by isolating it.
40. panlac01
yes, monica. that's how DLS did it, and that's how you should do it.
41. kristopher1014
lol i get it now , but how do u draw monica ?
42. monica0597
oh thanks for correcting me. @kristopher1014 , there's ur answer
43. DLS
:)
44. monica0597
there is this little thing at the bottom ofd the box in which you type. it says draw
45. panlac01
but in the future, we cannot give out the answer
46. kristopher1014
ok i found it
47. kristopher1014
|dw:1345263668277:dw|
48. monica0597
lol
49. kristopher1014
lmfao
50. DLS
what u guys doing u.u
51. kristopher1014
tlaking what else ?
52. monica0597
drawing?
53. panlac01
mixing playing time with study time
54. kristopher1014
talking*
55. DLS
Close the question,you are violating rules,This isn't facebook
56. panlac01
|dw:1345263804868:dw|
57. panlac01
he told you off
58. cshalvey
No problem.
59. Kaederfds
@panlac01 haha your picture haha ^^ | HuggingFaceTB/finemath | |
Scalene Triangle Definition
Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Scalene triangle is a triangle in which none of the sides of the triangle are equal to each other. This also means that all the three angles in a scalene triangle are not equal to each other and therefore are of different measure. Sum of all the three angles in any triangle is equal to 180º and hence this rule even applies to a scalene triangle. The sum of all the sides of the scalene triangle gives the perimeter of the scalene triangle.
Example 1: The angles in triangle ABC are (x + 40)º, (x + 30)º and (3x + 20)º. Is ABC a scalene triangle?
Sum of the angles in a triangle = 180º
So x+ 40+ x+ 30+ 3x+ 20= 180º
5x+ 90= 180º
5x= 180º - 90==> 5x= 90==> x= 90/5 = 18
This gives: (x + 40) = 18 + 40= 58º
(x + 30) = 18 + 30= 48º
(3x + 20)= 54 + 20= 74º
Since the angles in triangle ABC are not equal to each other, therefore ABC is a scalene triangle.
Example 2: Perimeter of triangle PQR of side lengths, (x+ 2), (x+ 5) and (x+ 9) is 52m. Is triangle PQR a scalene triangle?
Perimeter of a triangle = Sum of all the sides of the triangle
So, (x+ 2)+ (x+ 5)+ (x+ 9)= 52
So, 3x + 16 = 52
3x= 52- 16==> 3x = 36==> x = 36/3 = 12.
Therefore the sides are: x+ 2= 12+ 2= 14m
x+ 5= 12+ 5= 17m
x+ 9= 12+ 9= 21m
Since the length of the sides are different, hence triangle PQR is a scalene triangle. | HuggingFaceTB/finemath | |
Permutation & Combination Notes | Study CSAT Preparation for UPSC CSE - UPSC
UPSC: Permutation & Combination Notes | Study CSAT Preparation for UPSC CSE - UPSC
The document Permutation & Combination Notes | Study CSAT Preparation for UPSC CSE - UPSC is a part of the UPSC Course CSAT Preparation for UPSC CSE.
All you need of UPSC at this link: UPSC
What is Permutation?
Permutation relates to the act of arranging all the members of a set into some sequence or order.
• In other words, if the set is already ordered, then the rearranging of its elements is called the process of permuting.
• Permutations occur, in more or less prominent ways, in almost every area of mathematics. They often arise when different orderings on certain finite sets are considered.
What is a Combination?
The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter.
• In smaller cases, it is possible to count the number of combinations. Combination refers to the combination of n things taken k at a time without repetition.
• To refer to combinations in which repetition is allowed, the terms k-selection or k-combination with repetition are often used.
The Fundamental Principle of Counting
Product Rule
If an activity A can be done in “m” ways and another activity “B” can be done in “n” ways, A and B together can be done in “m x n” ways.
Example 1: How many ways a person can go from A to C via B?
A to B in 3 ways
B to C in 2 ways
A to B and B to C in 3 x 2 = 6 ways
If an activity A can be done in “m” ways and another activity “B” can be done in “n” ways, then A or B together can be done in “m + n” ways.
Example 2: If the person is at B, how many ways he can go to A or C?
B to A in 3 ways
B to C in 2 ways
B to A OR B to C in 3 + 2 = 5 ways
Keywords
AND➜ MULTIPLICATION (Product rule)
Q.1. From a pack of cards in how many ways can you select-
a) A king and a queen
b) A king or a queen
There are 52 cards in a pack, with 4 kings and 4 queens.
A king can be selected in 4 ways and a queen can be selected in 4 ways. Number of ways of selection of a king AND a queen is 4×4=16 (Product rule)
Number of ways of selection of a king OR a queen 4+4=8 (Addition rule)
The fundamental principle of counting (FPC) is the basic concept of permutation and combination.
Q.2. In how many ways can 3 people A, B, C be arranged in 2 places?
Manually solving we will get 6 cases as given below.
AB BA
AC CA
BC CB
This can be explained using FPC like this
There are two seats __ __. In the first seat A,B or C can sit. i.e., 3 possibilities.
Let A sit in the first place. Thus, in second place either B or C can sit.
The second seat can hence be taken up in 2 ways Together A,B ‘AND’ C can be seated in 3 x 2 = 6 ways.
Thus, the number of arrangement of 3 things at 2 places can be done in 3×2 ways.
Q.3. In how many ways can 4 people be seated in 3 chairs?
First seat can be filled in 4 ways.
Second seat can be filled in 3 ways.
Third seat can be filled in 2 ways.
So all together 4 people can be arranged in 3 seats in 4x3x2 = 24 ways.
Generalizing; arrangement of n things at r places can be done in [n(n-1)(n-2)…(n-r+1)] ways.
This arrangement of n things at r places is denoted by nPr .
nPr = n(n-1)(n-2)…(n-r+1) = n!/(n-r)!
It should be noted that in all these examples, the order of arrangement is important.
In other words, AB and BA are counted as 2 different cases. An arrangement where the order is important is called permutation
Now let us see one case where the order is not important.
Try yourself:In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
Q.4. How many doubles teams can be selected from 3 persons A, B, C?
Let’s take a doubles team AB. Change the order, i.e. to BA. Notice that both the teams are one and the same.
So here the order is not important. Hence, it’s a case of a combination. Here one can select the team in 3 ways
AB
AC
BC
If the order is important, then the total number of arrangements can be done in nPr = 3P2 ways.
We will get the 6 cases as:
AB BA
AC CA
BC CB
But in this particular question every 2! (because of A,B can be arranged in 2! Ways) cases will be counted as only one
case, i.e AB and BA is one and the same team.
So the number of selections of 2 things from 3 things can be done in = (3P2) / 2!
Similarly, the number of selections of 3 things from 4 things can be done in = (4P3) / 3!
Generalizing: the number of selections of r things from n things can be done in = (nPr) / r! ways.
The number of selections of r things from n different things is denoted by nCr,
which is what we learn as “Combination”.
∴ nCr = nPr/r!=n!/[(n-r)! r!]
Thus, you can understand now, that wherever the arrangement is necessary apart from selection,
we used Permutation and when we just need to select without arranging, we use Combination.
To make the point even clearer.
Try yourself:Find the number of words, with or without meaning, that can be formed with the letters of the word ‘CHAIR’.
Q.5. In how many ways can you select a President and a Vice-President from 3 people?
Is it a case of permutation or combination?
In the above question, since the word “select” appears, it may tempt you to consider combination.
So let’s take a case where A is the President and B is the Vice-President. Now change the order,
i.e. B is President and A Vice-President. Are both the cases same? No, they are not! We see that both the cases are different.
The order is important. Hence it is a case of permutation.
Difference between Permutation and Combination:
Permutation Combination nPr = n!/(n-r)! nCr = n!/[(n-r)!r!] Number of ways of arranging n things taken r at a time. Number of ways of selecting r things out of n things. Order is important Order is not important
The relation between permutation and combination
nPr = nCr x r!
Try yourself:Find the number of words, with or without meaning, that can be formed with the letters of the word ‘INDIA’.
The permutation of ‘n’ things taken ‘r’ at a time will involve the following steps:
(i) Select ‘r’ things out of total ‘n’ things, which can be done in nCr ways.
(ii) Arrange those ‘r’ things among themselves which can be done in r! ways.
nPr is nothing but, selecting r things out of n things (in nCr ways) and then arranging it among themselves.
This is why every permutation is selection followed by the arrangement of the selected things among themselves.
If it is only selection where the arrangement is not important, then we use the combination (nCr) only.
Try yourself:Among a set of 5 black balls and 3 red balls, how many selections of 5 balls can be made such that at least 3 of them are black balls.
Try yourself:Solve the following.
i) 30 P
ii) 30 C 2
The document Permutation & Combination Notes | Study CSAT Preparation for UPSC CSE - UPSC is a part of the UPSC Course CSAT Preparation for UPSC CSE.
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; | HuggingFaceTB/finemath | |
• # question_answer The capacitance of a spherical conductor with radius 1 m is: A) $9\times {{10}^{-9}}F$ B) $1\,\mu F$ C) $2.5\times {{10}^{-10}}F$ D) $1\times {{10}^{-6}}F$
The capacitance of spherical conductor is given by $C=4\pi {{\varepsilon }_{0}}r$ Here: $r=1\text{ }m,\,4\pi {{\varepsilon }_{0}}=\frac{1}{9\times {{10}^{9}}}$ $C=\frac{1}{4\times {{10}^{9}}}\times 1$ $C=2.5\times {{10}^{-10}}F$ | HuggingFaceTB/finemath |
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