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Video: EG19M2-Statistics-Q11
If π₯ and π¦ are two variables such that β π₯ = 15, β π¦ = 25, β π₯π¦ = 75, and π = 5, find the linear correlation coefficient, π, between π₯ and π¦.
02:05
Video Transcript
If π₯ and π¦ are two variables such that the summation of π₯ equals 15, the summation of π¦ equals 25, the summation of π₯ times π¦ equals 75, and π equals five, find the linear correlation coefficient, π, between π₯ and π¦.
What do we know about the linear correlation coefficient? And this coefficient, π, can be found by multiplying π times the summation of π₯ times π¦ minus the summation of π₯ times the summation of π¦. All over the square root of π times the summation of π₯ squared minus the summation of π₯ squared times π times the summation of π¦ squared minus the summation of π¦ squared.
And while itβs a long formula, in this case, thereβs something we can notice. Our π equals five. And the summation of π₯ times π¦ equals 75. We need to subtract that from 15 times 25. This will all be over the square root of some value in the denominator.
However, letβs examine the numerator a little bit more closely. I can rewrite 75 as three times 25. Five times 75 equals five times three times 25. If we regroup one more time, five times three equals 15. We rewrite five times 75 as 15 times 25. And then we see that our numerator is 15 times 25 minus 15 times 25. Our numerator equals zero. And that means the linear correlation coefficient, π, equals zero over some value. And we donβt need to know whatβs in the denominator. π equals zero.
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# Complex Modulus/Examples
## Examples of Complex Modulus
### Complex Modulus of $i$
$\cmod i = \cmod {-i} = 1$
### Complex Modulus of $-5$
$\left\vert{-5}\right\vert = 5$
### Complex Modulus of $1 + i$
$\cmod {1 + i} = \sqrt 2$
### Complex Modulus of $4 + 3 i$
$\cmod {4 + 3 i} = 5$
### Complex Modulus of $-4 + 2 i$
$\cmod {-4 + 2 i} = 2 \sqrt 5$
### Complex Modulus of $3iz - z^2$
Let:
$w = 3 i z - z^2$
where $z = x + i y$.
Then:
$\cmod w^2 = x^4 + y^4 + 2 x^2 y^2 - 6 x^2 y - 6 y^3 + 9 x^2 + 9 y^2$
### Complex Modulus of $\tan \theta + i$
$\left\vert{\tan \theta + i}\right\vert = \left\vert{\sec \theta}\right\vert$
where:
$\theta \in \R$ is a real number
$\tan \theta$ denotes the tangent function
$\sec \theta$ denotes the secant function.
### Complex Modulus of $\dfrac {1 + 2 i t - t^2} {1 + t^2}$
$\cmod {\dfrac {1 + 2 i t - t^2} {1 + t^2} } = 1$
where:
$t \in \R$ is a real number.
Let $z_1 = 4 - 3 i$ and $z_2 = -1 + 2 i$.
### Complex Modulus of $z_1 + z_2$
$\cmod {z_1 + z_2} = \sqrt {10}$
### Complex Modulus of $z_1 - z_2$
$\cmod {z_1 - z_2} = 5 \sqrt 2$
### Complex Modulus of $2 \overline z_1 - 3 \overline z_2 - 2$
$\cmod {2 \overline {z_1} - 3 \overline {z_2} - 2} = 15$
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Confirm
Archive
Special Issues
Profit Maximization in a Product Mix Bakery Using Linear Programming Technique
Journal of Investment and Management
Volume 9, Issue 1, March 2020, Pages: 27-30
Received: May 10, 2019; Accepted: Jan. 13, 2020; Published: Mar. 6, 2020
Authors
Kayode Olakunle Oluwaseyi, Department of Business Administration, University of Benin, Benin City, Nigeria
Atsegameh Elizabeth, Department of Business Administration, University of Benin, Benin City, Nigeria
Omole Ezekiel Olaoluwa, Department of Mathematics & Statistics, Joseph Ayo Babalola University, Ikeji-Arakeji, Osun State, Nigeria
Article Tools
Abstract
Linear Programming is one of the optimization techniques in finding solutions to managerial decisions making. Linear Programming is a widely used mathematical modelling technique designed to help managers in planning and decisions making relative to resource allocation. This study applied linear programming technique to decision making problem in university of Benin Bakery, Benin city, Edo state, Nigeria, and intended to determine the quantity of Bread that the firm should produce in a day to maximize profit, subject to constraints in the production process. Data on quantity of major raw material used in production of large, medium and small size bread per day were collected from the extract of the financial record of the bakery. The problem was formulated in mathematical term and solved using computer software known as Linear Programming Solver (LIPS). The solution obtained from a single iteration showed that 667 units of extra- large bread should be produced daily for the firm to achieve a maximum daily profit of #100,000. It is therefore recommended that the firm should concentrate more on production of extra- large bread to obtain maximum profit of #100,000 per day.
Keywords
Linear Programming, Optimization Problem, Simplex Method, Linear Programming Solver
Kayode Olakunle Oluwaseyi, Atsegameh Elizabeth, Omole Ezekiel Olaoluwa, Profit Maximization in a Product Mix Bakery Using Linear Programming Technique, Journal of Investment and Management. Vol. 9, No. 1, 2020, pp. 27-30. doi: 10.11648/j.jim.20200901.14
References
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Agarana, M., Anake, T., & Adeleke, O. (2014). Application of Linear Programming Model To Unsecured Loans and Bad Debt Risk Control In Banks. International Journal of Management, Information Technology and Engineering, 2 (7), 93-102.
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Akiniyi, J. (2008). Allocating Available Resources With The Aid of Linear Programming: A Roadmap To Economic Recovery. Multidisciplinary Journal of Research Development, 5, 113-119.
[3]
Anieting, A., Ezugwu, V., & Ologun, S. (2013). Application of Linear Programming Technique In The Determination of Optimum Production Capacity: IOSR. Journal of Mathematics (IOSR-JM), 5 (6), 62-65.
[4]
Balogun, O., Jolayemi, E., Akingbade, T., & Muazu. (2012). Use Linear Programming for Optimal Production In A Production Line In Coca-cola Bottling Company, Ilorin. International Journal of Engineering Research and Application, 2 (5), 2004-2007.
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Balogun, O., Role, M., Akingbade, T., & Akinrefon, A. (2015). An Optimisation Procedure in a Production Line Sokat Soap Industry. Marsland Press (Researcher), 5 (10), 50-54.
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Fagoyinbo, I., & Ajibode, I. (2010). Application of Linear Programming Techniques In The Effective Use of Resources For Staff Trainning. Journal of Emerging Trends In Engineering and Applied Sciences, 1 (2), 127-132.
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Fagoyinbo, I., Akinbo, R., & Ajibode, I. (2011). Maximisation of Profits In Manufacturing Industries Using Linear Programming Technique. Meditteranean Journal of Social Science, 7 (3), 97-105.
[8]
Felix, M., Judith, M., Jonathan, M., & Munashe, S. (2013). Modelling a Small Farm Livelihood System Using Linear Programming In Bindura, Zimbabwe. Research Journal of Management Science, 2 (5), 20-23.
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Hazar, J., & Render, J. (2004). Operation Management: Process and Value Chain. New Jersey: Prentice Hall.
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Kourosh, R., Farhang, K., & Reza, J. (2013). Using Linear Programming In Solving The Problem of Services Company's Cost. Singaporean Journal of Business Economics and Management, 4 (2), 93-104.
[11]
Lee, S. (1972). Goal Programming for Decision Analysis. Philadephia: Auerback Publishers.
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Tien, J., & Kamiyama, A. (1982). On Manpower Scheduling Algorithm. SIAM Review, 24 (3), 257-287.
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Yahya, W. (2004). Determination of Optimum Product Mix at Minimum Raw Material Cost Using Linear Programming. Nigeria Journal of Pure and Applied Sciences, 19 (2), 1712-1721. ******
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Raimi O. A, & Adedayo, O. C. (2017). Application of linear programming techniques on bread production optimization in Rufus Giwa Polytechnic Ondo State. American Journal of Operations Management and Information Systems, 2 (1), 32 – 36.
[15]
Ibitoye O, Atoyebi K. O., Genevieve K., & Kadiri K. (2015). Entrepreneur decision making process and application of linear programming techniques. European Journal of Business, Economics and Accounting, 3 (5), 1 – 5.
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Adebiyi S. O., Amole B. B., & Soils I. O. (2014). Linear Optimization techniques for production mix of paints production in Nigeria. AUDCE, 10 (1), 181 – 190.
[17]
Oladejo, N. K, Abolarinwa, A, Salary, S. O & Lukman, A, F. (2019). Optimization principle and its application in optimizing landmark university bakery production using linear programming. International Journal of Civil Engineering and Technology, 10 (2), 183 – 190.
[18]
Molina, M. G. (2018). Product mix optimization at minimum supply cost of online clothing store using linear programming. International Journal of Applied Mathematics, Electronics and Computers, 6 (3), 33 – 38.
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Ailobhio, T. D. & Suleiman, A. I. (2018). Optimizing profit in lace baking industry Latia with linear programming model. International Journal of Statistics and Application, 8 (1), 18 – 22. Doi: 10.59231statistics.20180801.03.
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## 130. Rational Functions.
After integrating polynomials it is natural to turn our attention next to rational functions. Let us suppose $$R(x)$$ to be any rational function expressed in the standard form of, viz. as the sum of a polynomial $$\Pi(x)$$ and a number of terms of the form $$A/(x – \alpha)^{p}$$.
We can at once write down the integrals of the polynomial and of all the other terms except those for which $$p = 1$$, since $\int \frac{A}{(x – \alpha)^{p}}\, dx = -\frac{A}{p – 1}\, \frac{1}{(x – \alpha)^{p-1}},$ whether $$\alpha$$ be real or complex ( § 117).
The terms for which $$p = 1$$ present rather more difficulty. It follows immediately from Theorem (6) of § 113 that $\begin{equation*} \int F’\{f(x)\}\, f'(x)\, dx = F\{f(x)\}. \tag{3} \end{equation*}$ In particular, if we take $$f(x) = ax + b$$, where $$a$$ and $$b$$ are real, and write $$\phi(x)$$ for $$F(x)$$ and $$\psi(x)$$ for $$F'(x)$$, so that $$\phi(x)$$ is an integral of $$\psi(x)$$, we obtain $\begin{equation*} \int \psi(ax + b)\, dx = \frac{1}{a}\phi(ax + b). \tag{4} \end{equation*}$
Thus, for example, $\int \frac{dx}{ax + b} = \frac{1}{a} \log|ax + b|,$ and in particular, if $$\alpha$$ is real, $\int \frac{dx}{x – \alpha} = \log|x – \alpha|.$ We can therefore write down the integrals of all the terms in $$R(x)$$ for which $$p = 1$$ and $$\alpha$$ is real. There remain the terms for which $$p = 1$$ and $$\alpha$$ is complex.
In order to deal with these we shall introduce a restrictive hypothesis, viz. that all the coefficients in $$R(x)$$ are real. Then if $$\alpha = \gamma + \delta i$$ is a root of $$Q(x) = 0$$, of multiplicity $$m$$, so is its conjugate $$\bar{\alpha} = \gamma – \delta i$$; and if a partial fraction $$A_{p}/(x – \alpha)^{p}$$ occurs in the expression of $$R(x)$$, so does $$\bar{A}_{p}/(x – \bar{\alpha})^{p}$$, where $$\bar{A}_{p}$$ is conjugate to $$A_{p}$$. This follows from the nature of the algebraical processes by means of which the partial fractions can be found, and which are explained at length in treatises on Algebra.1
Thus, if a term $$(\lambda + \mu i)/(x – \gamma – \delta i)$$ occurs in the expression of $$R(x)$$ in partial fractions, so will a term $$(\lambda – \mu i)/(x – \gamma + \delta i)$$; and the sum of these two terms is $\frac{2\{\lambda(x – \gamma) – \mu\delta\}}{(x – \gamma)^{2} + \delta^{2}}.$ This fraction is in reality the most general fraction of the form $\frac{Ax + B}{ax^{2} + 2bx + c},$ where $$b^{2} < ac$$. The reader will easily verify the equivalence of the two forms, the formulae which express $$\lambda$$, $$\mu$$, $$\gamma$$, $$\delta$$ in terms of $$A$$, $$B$$, $$a$$, $$b$$, $$c$$ being $\lambda = A/2a,\quad \mu = -D/(2a\sqrt{\Delta}),\quad \gamma = -b/a,\quad \delta = \sqrt{\Delta}/a,$ where $$\Delta = ac – b^{2}$$, and $$D = aB – bA$$.
If in (3) we suppose $$F\{f(x)\}$$ to be $$\log |f(x)|$$, we obtain $\begin{equation*} \int \frac{f'(x)}{f(x)}\, dx = \log |f(x)|; \tag{5} \end{equation*}$ and if we further suppose that $$f(x) = (x – \lambda)^{2} + \mu^{2}$$, we obtain $\int \frac{2(x – \lambda)}{(x – \lambda)^{2} + \mu^{2}}\, dx = \log\{(x – \lambda)^{2} + \mu^{2}\}.$ And, in virtue of the equations (6) of § 128 and (4) above, we have $\int \frac{-2\delta\mu}{(x – \lambda)^{2} + \mu^{2}}\, dx = -2\delta \arctan \left(\frac{x – \lambda}{\mu}\right).$
These two formulae enable us to integrate the sum of the two terms which we have been considering in the expression of $$R(x)$$; and we are thus enabled to write down the integral of any real rational function, if all the factors of its denominator can be determined. The integral of any such function is composed of
the sum of a polynomial, a number of rational functions of the type $-\frac{A}{p – 1}\, \frac{1}{(x – \alpha)^{p-1}},$ a number of logarithmic functions, and a number of inverse tangents.
It only remains to add that if $$\alpha$$ is complex then the rational function just written always occurs in conjunction with another in which $$A$$ and $$\alpha$$ are replaced by the complex numbers conjugate to them, and that the sum of the two functions is a real rational function.
Example XLVIII
1. Prove that $\int \frac{Ax + B}{ax^{2} + 2bx + c}\, dx = \frac{A}{2a} \log |X| + \frac{D}{2a \sqrt{-\Delta}} \log \left|\frac{ax + b – \sqrt{-\Delta}}{ax + b + \sqrt{-\Delta}}\right|$ (where $$X = ax^{2} + bx + c$$) if $$\Delta < 0$$, and $\int \frac{Ax + B}{ax^{2} + 2bx + c}\, dx = \frac{A}{2a} \log |X| + \frac{D}{2a \sqrt{\Delta}} \arctan \left(\frac{ax + b}{\sqrt{\Delta}}\right)$ if $$\Delta > 0$$, $$\Delta$$ and $$D$$ having the same meanings as above.
2. In the particular case in which $$ac = b^{2}$$ the integral is $-\frac{D}{a(ax + b)} + \frac{A}{a} \log |ax + b|.$
3. Show that if the roots of $$Q(x) = 0$$ are all real and distinct, and $$P(x)$$ is of lower degree than $$Q(x)$$, then $\int R(x)\, dx = \sum \frac{P(\alpha)}{Q'(\alpha)} \log |x – \alpha|,$ the summation applying to all the roots $$\alpha$$ of $$Q(x) = 0$$.
[The form of the fraction corresponding to $$\alpha$$ may be deduced from the facts that $\frac{Q(x)}{x – \alpha} \to Q'(\alpha),\quad (x – \alpha) R(x) \to \frac{P(\alpha)}{Q'(\alpha)},$ as $$x \to \alpha$$.]
4. If all the roots of $$Q(x)$$ are real and $$\alpha$$ is a double root, the other roots being simple roots, and $$P(x)$$ is of lower degree than $$Q(x)$$, then the integral is $$A/(x – \alpha) + A’\log |x – \alpha| + \sum B\log |x – \beta|$$, where $A = -\frac{2P(\alpha)}{Q”(\alpha)},\quad A’ = \frac{2\{3P'(\alpha) Q”(\alpha) – P(a) Q”'(\alpha)\}} {3\{Q”(\alpha)\}^{2}},\quad B = \frac{P(\beta)}{Q'(\beta)},$ and the summation applies to all roots $$\beta$$ of $$Q(x) = 0$$ other than $$\alpha$$.
5. Calculate $\int \frac{dx}{\{(x – 1) (x^{2} + 1)\}^{2}}.$
[The expression in partial fractions is $\frac{1}{4(x – 1)^{2}} – \frac{1}{2(x – 1)} – \frac{i}{8(x – i)^{2}} + \frac{2 – i}{8(x – i)} + \frac{i}{8(x + i)^{2}} + \frac{2 + i}{8(x + i)},$ and the integral is $-\frac{1}{4(x – 1)} – \frac{1}{4(x^{2} + 1)} – \tfrac{1}{2} \log |x – 1| + \tfrac{1}{4} \log (x^{2} + 1) + \tfrac{1}{4} \arctan x.]$
6. Integrate $\begin{gathered} \frac{x}{(x – a)(x – b)(x – c)},\quad \frac{x}{(x – a)^{2}(x – b)},\quad \frac{x}{(x – a)^{2} (x – b)^{2}},\quad \frac{x}{(x – a)^{3}},\\ \frac{x}{(x^{2} + a^{2}) (x^{2} + b^{2})},\quad \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b)^{2}},\quad \frac{x^{2} – a^{2}}{x^{2}(x^{2} + a^{2})},\quad \frac{x^{2} – a^{2}}{x(x^{2} + a^{2})^{2}}.\end{gathered}$
7. Prove the formulae: \begin{aligned}\int \frac{dx}{1 + x^{4}} = \frac{1}{4\sqrt{2}} \biggl\{ &\log \biggl(\frac{1 + x\sqrt{2} + x^{2}}{1 – x\sqrt{2} + x^{2}}\biggr) + 2\arctan \biggl(\frac{x\sqrt{2}}{1 – x^{2}}\biggr)\biggr\},\\ \int \frac{x^{2}\, dx}{1 + x^{4}} = \frac{1}{4\sqrt{2}} \biggl\{&-\log \biggl(\frac{1 + x\sqrt{2} + x^{2}}{1 – x\sqrt{2} + x^{2}}\biggr) + 2\arctan \biggl(\frac{x\sqrt{2}}{1 – x^{2}}\biggr)\biggr\},\\ \int \frac{dx}{1 + x^{2} + x^{4}} = \frac{1}{4\sqrt{3}}\biggl\{ &\sqrt{3}\log \biggl(\frac{1 + x + x^{2}}{1 – x + x^{2}}\biggr) + 2\arctan \biggl(\frac{x\sqrt{3}}{1 – x^{2}}\biggr)\biggr\}.\end{aligned}
## 131. Note on the practical integration of rational functions.
The analysis of § 130 gives us a general method by which we can find the integral of any real rational function $$R(x)$$, provided we can solve the equation $$Q(x) = 0$$. In simple cases (as in Ex. 5 above) the application of the method is fairly simple. In more complicated cases the labour involved is sometimes prohibitive, and other devices have to be used. It is not part of the purpose of this book to go into practical problems of integration in detail. The reader who desires fuller information may be referred to Goursat’s Cours d’Analyse, second ed., vol. i, pp. 246 et seq., Bertrand’s Calcul Intégral, and Dr Bromwich’s tract Elementary Integrals (Bowes and Bowes, 1911).
If the equation $$Q(x) = 0$$ cannot be solved algebraically, then the method of partial fractions naturally fails and recourse must be had to other methods.2
1. See, for example, Chrystal’s Algebra, vol. i, pp. 151–9.↩︎
2. See the author’s tract “The integration of functions of a single variable” (Cambridge Tracts in Mathematics, No. 2, second edition, 1915). This does not often happen in practice.↩︎
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# Divide Unit Fractions and Whole Numbers
## Fractions
As you all know part of a whole is known as a fraction. The top number is called the numerator and the bottom one is called the denominator.
For example, 2/3 is a fraction with 2 as the numerator and 3 as the denominator. A unit fraction is a fraction whose numerator is 1. It can also be considered as a special type of proper fraction.
### Arithmetic way of dividing a unit fraction by a whole number
To divide a unit fraction by a whole number, first, we take the reciprocal of the whole number and simultaneously change the division operation to the multiplication. After that, we multiply both the numerators of the given fractions together and write the answer as the numerator of the resulting fraction. Similarly, we multiply both the denominators of the given fractions together and write the answer as the denominator of the resulting fraction.
#### Visual Representation of Division of a Unit Fraction by a Whole Number
We have learnt the arithmetic way of dividing a unit fraction by a whole number. Now, let’s
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https://www.vedantu.com/formula/rectangle-formula
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# Rectangle Formula
View Notes
## Area of Rectangle Formula
The area of a rectangle is the region encompassed by the rectangle in a two-dimensional closed plane figure. In simple words, the area of a rectangle is the total space covered by the figure. On the other hand, the space within the bounds of the perimeter of the rectangle is also the area of a rectangle. Here, you will also get to know why the area of a rectangle is the product of its two sides and also the units of measurement.
### What is a Rectangle?
A rectangle is a 2 dimensional geometric shape consisting of four sides, four angles and four vertices. Opposite sides of the rectangle are equivalent and parallel to each other in a rectangle. Remember that all the interior four angles of the rectangle are right angles. Pointing out, a parallelogram also has its opposite sides equal and parallel to each other but the angles do not make 90 degrees of measurement.
### Important Rectangle Formulas
The formula of the area and perimeter of the rectangle is given below.
Given that ‘l’ represents the length and ‘b’ represents the breadth of the rectangle, then the
Element Formula Perimeter of rectangle formula 2(l + b) units Length of the rectangle formula P/2 - b units Breadth of the rectangle P/2 - l units Area of rectangle formula l × b sq. units Length of the rectangle A/b units Breadth of the rectangle A/l units Diagonal of the rectangle √(l²+b²) units
### Derive and Calculate the Area of a Rectangle
To derive the area of a rectangle, we use the unit squares. Divide the rectangle MNOP into unit squares. The area of a rectangle MNOP is the total number of unit squares in it.
Now, finding the area of a rectangle
Rectangle MNOP in unit squares each with 1 sq. inch
Therefore, the total area of the rectangle MNOP is equal to 48 sq. inch.
Also, following the perspective, we discovered that the area of a rectangle is mandatorily the product of its two sides. Here, the length of MN is 6 inches and the breadth of NO is 8 inches. The area of MNOP is the product of 6 and 8, which is 48.
Instead, formulas to find the area of a rectangle is derived by dividing the figure into two equal-sized right triangles. For example, in the given rectangle MNOP, a diagonal from the vertex M is drawn to O.
The diagonal MO divides the rectangle into 2 equivalent right-angle triangles.
Therefore, the area of MNOP will be:
  ⇒ Area (MNOP) = Area (MNO) + Area (MPO)
  ⇒ Area (MNOP) = 2 × Area (MNO)
  ⇒ Area (MNO) = ½ × base × height
  ⇒ Area (MNOP) = 2 × (½ × b × h)
   ⇒ Area (MNOP) = b × h
### Application of Rectangle Formula
The early documentation script of Babylonian culture represents the use of geometric objects with lengths, width, angles, and areas for construction and astronomy. The skill and understanding of stone cutting in standard shapes such as squares, triangles and rectangles along with principles having reference to area and perimeter helped ancient Egyptians in constructing giant structures like pyramids. In modern times, these rectangle volume formula concepts are greatly helpful in object modelling, land surveying and others.
### Examples of Rectangle
Some examples of rectangular figures are:
• Parks and agricultural fields,
• Painting Canvas
• Tiles and walkway with rectangular tiles
• Daily life objects such as a table, serving tray, glass, etc.
### Solved Examples
Here, we will perform practical problems using step-by-step solutions in order to find the area of a rectangle as well as other measurements.
Example:
Calculate the area and the perimeter of the rectangle box that measures 15 cm in length and 8cm in breadth.
Solution:
Given:
Length = 15 cm,
Now applying the area of rectangle formula i.e. = length × breadth
= (15 × 8) cm²
= 120 cm²
Using the formula for Perimeter of rectangle = 2 (length + breadth)
 = 2 (15 + 8) cm
= 2 × 23 cmÂ
= 46 cm
Example:
Evaluate the breadth of the Television set rectangular in shape whose area is 320 m2 and whose length is 40 m. Find its perimeter.
Solution:
We know that the breadth of the rectangular television set = Area/length
= 320m/40m
= 8 m
Thus, the perimeter of the rectangular television set = 2 (length + breadth)
= 2(40 + 8) m
 = 2 × 48 m
= 96 m
### Fun Facts
• In a rectangle, both the diagonals are equal in length.
• We also have a rectangular figure that we call a cyclic rectangle. A circle that contains a rectangle with its entire vertex touching the circumference is a cyclic rectangle.
• This concept of area calculation is also very useful in land surveying, map designing, modelling, and others.
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https://stackoverflow.com/questions/5323818/condensed-matrix-function-to-find-pairs/14839010
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# Condensed matrix function to find pairs
For a set of observations:
``````[a1,a2,a3,a4,a5]
``````
their pairwise distances
``````d=[[0,a12,a13,a14,a15]
[a21,0,a23,a24,a25]
[a31,a32,0,a34,a35]
[a41,a42,a43,0,a45]
[a51,a52,a53,a54,0]]
``````
Are given in a condensed matrix form (upper triangular of the above, calculated from `scipy.spatial.distance.pdist` ):
``````c=[a12,a13,a14,a15,a23,a24,a25,a34,a35,a45]
``````
The question is, given that I have the index in the condensed matrix is there a function (in python preferably) f to quickly give which two observations were used to calculate them?
``````f(c,0)=(1,2)
f(c,5)=(2,4)
f(c,9)=(4,5)
...
``````
I have tried some solutions but none worth mentioning :(
The formula for an index of the condensed matrix is
``````index = d * (d - 1) / 2 - (d - i) * (d - i - 1) / 2 + j - i - 1
``````
Where `i` is the row index, `j` is the column index, and `d` is the row length of the original (d X d) upper triangular matrix.
Consider the case when the index refers to the leftmost, non-zero entry of some row in the original matrix. For all the leftmost indices,
``````j == i + 1
``````
so
``````index = d * (d - 1) / 2 - (d - i) * (d - i - 1) / 2 + i + 1 - i - 1
index = d * (d - 1) / 2 - (d - i) * (d - i - 1) / 2
``````
With some algebra, we can rewrite this as
``````i ** 2 + (1 - (2 * d)) * i + 2 * index == 0
``````
Then we can use the quadratic formula to find the roots of the equation, and we only are going to care about the positive root.
If this index does correspond to leftmost, non-zero cell, then we get a positive integer as a solution that corresponds to the row number. Then, finding the column number is just arithmetic.
``````j = index - d * (d - 1) / 2 + (d - i) * (d - i - 1)/ 2 + i + 1
``````
If the index does not correspond to the leftmost, non-zero cell, then we will not find an integer root, but we can take the floor of the positive root as the row number.
``````def row_col_from_condensed_index(d,index):
b = 1 - (2 * d)
i = (-b - math.sqrt(b ** 2 - 8 * index)) // 2
j = index + i * (b + i + 2) // 2 + 1
return (i,j)
``````
If you don't know `d`, you can figure it from the length of the condensed matrix.
``````((d - 1) * d) / 2 == len(condensed_matrix)
d = (1 + math.sqrt(1 + 8 * len(condensed_matrix))) // 2
``````
• I had to search a long time to find this. Your answer deserves more attention. PS: if you swap out `math` for `numpy`, your solution is actually vectorized. Oct 31, 2014 at 1:52
• I think the problem may that the question title is not so clear. Do you have a suggestion for a better title? Oct 31, 2014 at 15:48
You may find triu_indices useful. Like,
``````In []: ti= triu_indices(5, 1)
In []: r, c= ti[0][5], ti[1][5]
In []: r, c
Out[]: (1, 3)
``````
Just notice that indices starts from 0. You may adjust it as you like, for example:
``````In []: def f(n, c):
..: n= ceil(sqrt(2* n))
..: ti= triu_indices(n, 1)
..: return ti[0][c]+ 1, ti[1][c]+ 1
..:
In []: f(len(c), 5)
Out[]: (2, 4)
``````
• This works, although it won't scale up. More than 10k of 2 dimensional observations will fill up the memory Mar 16, 2011 at 11:11
• @Ηλίας: Care to elaborate more, assuming your condensed matrix data type is double, then triu_indices consume same amount of memory.
– eat
Mar 16, 2011 at 12:01
• @eat `from scipy.spatial.distance import pdist`, the `pdist` would happily crunch up to 10k of data. And your function would go up to 10.000.000 size. So I take back my comment! The problem was on pdist Mar 16, 2011 at 14:53
• @Ηλίας: You may describe on a separate question what you are aiming for. Is it absolutely necessary to calculate all pairwise distances? Thanks
– eat
Mar 16, 2011 at 15:51
• No doubt that this solution is inefficient for even moderate 'n' sizes. Dec 26, 2012 at 10:40
Cleary, the function f you are searching for, needs a second argument: the dimension of the matrix - in your case: 5
First Try:
``````def f(dim,i):
d = dim-1 ; s = d
while i<s:
s+=d ; d-=1
return (dim-d, i-s+d)
``````
• True, the function would should have a reference to the condensed matrix. But it should be able to deduce the dimension from the length of the condensed matrix. Mar 16, 2011 at 10:27
• Unfortunately this goes into inf loop Mar 16, 2011 at 11:17
• dim can be found by solving for n in `n*(n-1)=len(condensed matrix)` (or just keep a lookup table of the likely/supported sizes) Mar 16, 2011 at 14:14
• f( 5, 1 ) gives (11,-4). Not sure I can follow what goes on in there Mar 16, 2011 at 14:45
• It's actually `(n*(n-1))/2` Nov 17, 2015 at 15:42
To complete the list of answers to this question: A fast, vectorized version of fgreggs answer (as suggested by David Marx) could look like this:
``````def vec_row_col(d,i):
i = np.array(i)
b = 1 - 2 * d
x = np.floor((-b - np.sqrt(b**2 - 8*i))/2).astype(int)
y = (i + x*(b + x + 2)/2 + 1).astype(int)
if i.shape:
return zip(x,y)
else:
return (x,y)
``````
I needed to do these calculations for huge arrays, and the speedup as compared to the un-vectorized version (https://stackoverflow.com/a/14839010/3631440) is (as usual) quite impressive (using IPython %timeit):
``````import numpy as np
from scipy.spatial import distance
test = np.random.rand(1000,1000)
condense = distance.pdist(test)
sample = np.random.randint(0,len(condense), 1000)
%timeit res = vec_row_col(1000, sample)
10000 loops, best of 3: 156 µs per loop
res = []
%timeit for i in sample: res.append(row_col_from_condensed_index(1000, i))
100 loops, best of 3: 5.87 ms per loop
``````
That's about 37 times faster in this example!
• There's a syntax error with an extra `(`. Also why would `i` have a shape? The condensed distance matrix is always a 1d array. Jun 6, 2018 at 6:20
• AMAZING ANSWER. thanks. I only modified it to `return zip(x,y)` so that I get the output in a list Jun 18, 2020 at 9:26
This is in addition to the answer provided by phynfo and your comment. It does not feel like a clean design to me to infer the dimension of the matrix from the length of the compressed matrix. That said, here is how you can compute it:
``````from math import sqrt, ceil
for i in range(1,10):
thelen = (i * (i+1)) / 2
thedim = sqrt(2*thelen + ceil(sqrt(2*thelen)))
print "compressed array of length %d has dimension %d" % (thelen, thedim)
``````
The argument to the outer square root should always be a square integer, but sqrt returns a floating point number, so some care is needed when using this.
• Doesn't `n= ceil(sqrt(2* len(c)))' just be sufficient?
– eat
Mar 16, 2011 at 12:03
• @eat: yes, absolutely. Above is overly contrived. Mar 16, 2011 at 12:42
Here's another solution:
``````import numpy as np
def f(c,n):
tt = np.zeros_like(c)
tt[n] = 1
return tuple(np.nonzero(squareform(tt))[0])
``````
To improve the efficiency using `numpy.triu_indices`
use this:
``````def PdistIndices(n,I):
'''idx = {} indices for pdist results'''
idx = numpy.array(numpy.triu_indices(n,1)).T[I]
return idx
``````
So `I` is an array of indices.
However a better solution is to implement an optimized Brute-force search, say, in `Fortran`:
``````function PdistIndices(n,indices,m) result(IJ)
!IJ = {} indices for pdist[python] selected results[indices]
implicit none
integer:: i,j,m,n,k,w,indices(0:m-1),IJ(0:m-1,2)
logical:: finished
k = 0; w = 0; finished = .false.
do i=0,n-2
do j=i+1,n-1
if (k==indices(w)) then
IJ(w,:) = [i,j]
w = w+1
if (w==m) then
finished = .true.
exit
endif
endif
k = k+1
enddo
if (finished) then
exit
endif
enddo
end function
``````
then compile using `F2PY` and enjoy unbeatable performance. ;)
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https://tradingqna.com/t/how-to-calculate-for-aluminium-and-zinc-parity-and-disparity-between-mcx-price-and-international-price/52418/4
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How to calculate for Aluminium and Zinc, parity and disparity between MCX price and International price?
As per SEBI guidelines, Aluminium contracts from the month of March 2019 and Zinc contracts from the April 2019 will be physically settled similar to Gold and Silver. Just saw this SEBI/MCX circular.
I can also see that MCX is now showing up a separate spot market price for these physically settled contracts.
Now that these are physically settled, does someone have a calculation on how to calculate the parity/disparity between international price and local price of Zinc and Aluminium?
Aluminium
Spot in \$ (per MT) \$1,850.00
Add CIF* in \$ (per MT) \$50.00
RBI Reference rate* as of Jan 3,2019(USDINR) ₹70.36
Convert Spot +Premium(INR per MT) ₹133,689.13
Convert Spot + Premium(INR per KG) ₹133.69
Landing Cost(INR per KG) ₹135.03
Add Basic Customs Duty + Surcharge(8.25%)* ₹11.14
Landed Cost ₹146.17
Add Port Clearing & Handling charges(Rs 2/KG) 2.00
Final Landed Cost 149.17
Similarly for Zinc,
Zinc
Spot in \$ (per MT) \$2,475.00
Add CIF* in \$ (per MT) \$200.00
RBI Reference rate* as of Jan 3,2019(USDINR) ₹70.36
Convert Spot +Premium(INR per MT) ₹188,220.22
Convert Spot + Premium(INR per KG) ₹188.22
Landing Cost(INR per KG) ₹190.10
Add Basic Customs Duty + Surcharge(5.5%)* ₹10.46
Landed Cost ₹200.56
Add Port Clearing & Handling charges(Rs 2/KG) 2.00
Final Landed Cost 203.56
*CIF - Cost of Insurance & Freight)
*RBI Reference Rates
*Surcharge or SWS stands Social Welfare Surcharge charged by the Govt.
*Delivery overheads will vary if Warehouse location is different
Here is a link to the above table.
1 Like
Now that Lead and Nickel contracts are also physically settled, can you provide the calculations for the same
It’s really appreciable, that you keep on solving the problems and this one is really a big problem as even 1 rs. change cost us rs. 5000.
I have been using this sheet for last 7-10 days for calculation basis, but never get the right values which matches with MCX, initially the difference was somewhere 30 paise, now it has gone upto rs. 3, which is really big for the trader like me.
This is the theoretical calculation, the actual price difference is based on demand supply on the exchange. It can be less or more.
are the surcharges and CIF same even today? where do we find the updated charges ?
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This week in Pre-Calc, I had some trouble understanding how to put equations into piecewise form, and how to solve absolute value equations. So, today I will explain how you do them.
When you are putting an equation into piecewise form, you first take the equation out of the absolute value signs. Once you have done this, you make the equation equal to zero. After this, you put all the numbers (without variables) on the other side of the equation. Next, you isolate “x” by diving by the number that is next to it (coefficient) on both sides, and then you will have your answer for what “x’ equals. Once you have solved for “x”, you put the point or points on a number line, and choose two (one point) or three (two points) on the number line. Based on which point is negative, you will then write the formula in piecewise form. To do this, you will write the original formula as it is, and then write “x” is < or > depending on where your point is positive and negative for all your points (one or two). Next, you will write out the original equation again, this time for the negative point, and since it is a negative, you will have to multiply in a negitive sign in the front of the original equaiton to make it positive. Once you have writtent that, you will then write “x” is < or > the point or points that are negative on your number line.
To solve an equation, you seperate it into two different equations. One will be the original equation (without the absolute value symbols) and the other will be the equation with the negative symbole multiplying in (to make the negative part of the number line positive). To solve your first side (the original one), you will move any numbers without variables to the other side. Once you have done this, you divide both sides by the number in front of “x” (coefficient), and this will give you your answer. For the second side (the original one with the negative multiplied in), you will start by multiplying the negative symbole in. Once you have done this, you move all the numbers without variables to the other side. Next, you will divide both sides by the number in front of “x” (coefficient), and this will give you the other answer.
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# [<"12-14. The dragster starts from rest and has a velocity described by the graph.
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If You Have an Enemy, Give Him Information
# If You Have an Enemy, Give Him Information
Information Bias
In his short story “Del rigor en la ciencia,” which consists of just a single paragraph, Jorge Luis Borges describes a special country. In this country, the science of cartography is so sophisticated that only the most detailed of maps will do—that is, a map with a scale of 1:1, as large as the country itself. Their citizens soon realize that such a map does not provide any insight, since it merely duplicates what they already know. Borges’s map is the extreme case of the information bias, the delusion that more information guarantees better decisions.
Searching for a hotel in Miami a little while ago, I drew up a short list of five good offers. Right away, one jumped out at me, but I wanted to make sure I had found the best deal and decided to keep researching. I plowed my way through dozens of customer reviews and blog posts and clicked through countless photos and videos. Two hours later, I could say for sure which the best hotel was: the one I had liked at the start. The mountain of additional information did not lead to a better decision. On the contrary, if time is money, then I might as well have taken up residence at the Four Seasons.
Jonathan Baron from the University of Pennsylvania asked physicians the following question: A patient presents symptoms that indicate with a probability of 80 percent that he is suffering from disease A. If this is not the case, the patient has either disease X or Y. Each of these diseases is equally bad, and each treatment results in similar side effects. As a doctor, what treatment would you suggest? Logically, you would opt for disease A and recommend the relevant therapy. Now suppose there is a diagnostic test that flashes “positive” when disease X is present and “negative” when disease Y is detected. However, if the patient really does have disease A, the test results will be positive in 50 percent of the cases and negative in the other 50 percent. Would you recommend conducting the test? Most doctors said yes, even though the results would be irrelevant. Assuming that the test result is positive, the probability of disease A is still much greater than that of disease X. The additional information contributes nothing of value to the decision.
Doctors are not the only professionals with a penchant for surplus information. Managers and investors are almost addicted to it. How often are studies commissioned one after the other, even though the critical facts are readily available? Additional information not only wastes time and money, it can also put you at a disadvantage. Consider this question: Which city has more inhabitants, San Diego or San Antonio? Gerd Gigerenzer of the Max Planck Institute in Germany put this question to students in the University of Chicago and the University of Munich. Sixty-two percent of Chicago students guessed right: San Diego has more. But, astonishingly, every single German student answered correctly. The reason: All of them had heard of San Diego but not necessarily of San Antonio, so they opted for the more familiar city. For the Chicagoans, however, both cities were household names. They had more information, and it misled them.
Or consider the hundreds of thousands of economists—in service of banks, think tanks, hedge funds, and governments—and all the white papers they have published from 2005 to 2007: The vast library of research reports and mathematical models. The formidable reams of comments. The polished PowerPoint presentations. The terabytes of information on Bloomberg and Reuters news services. The bacchanal dance to worship the god of information. It was all hot air. The financial crisis touched down and upended global markets, rendering the countless forecasts and comments worthless.
Forget trying to amass all the data. Do your best to get by with the bare facts. It will help you make better decisions. Superfluous knowledge is worthless, whether you know it or not. The historian Daniel J. Boorstin put it right: “The greatest obstacle to discovery is not ignorance—it is the illusion of knowledge.” And next time you are confronted by a rival, consider killing him—not with kindness but with reams of data and analysis.
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https://www.numbersaplenty.com/15271
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Search a number
15271 is a prime number
BaseRepresentation
bin11101110100111
3202221121
43232213
5442041
6154411
762344
oct35647
922847
1015271
1110523
128a07
136c49
1457cb
1547d1
hex3ba7
15271 has 2 divisors, whose sum is σ = 15272. Its totient is φ = 15270.
The previous prime is 15269. The next prime is 15277. The reversal of 15271 is 17251.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 15271 - 21 = 15269 is a prime.
Together with 15269, it forms a pair of twin primes.
15271 is a lucky number.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (15277) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7635 + 7636.
It is an arithmetic number, because the mean of its divisors is an integer number (7636).
215271 is an apocalyptic number.
15271 is a deficient number, since it is larger than the sum of its proper divisors (1).
15271 is an equidigital number, since it uses as much as digits as its factorization.
15271 is an evil number, because the sum of its binary digits is even.
The product of its digits is 70, while the sum is 16.
The square root of 15271 is about 123.5758876157. The cubic root of 15271 is about 24.8097559605.
The spelling of 15271 in words is "fifteen thousand, two hundred seventy-one".
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# Georgia Standards of Excellence Curriculum Map. Mathematics. Accelerated GSE Geometry B / Algebra II
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1 Georgia Standards of Excellence Curriculum Map Mathematics Accelerated GSE Geometry B / Algebra II These materials are for nonprofit educational purposes only. Any other use may constitute copyright infringement.
2 Georgia Department of Education Accelerated GSE Geometry B/Algebra II Curriculum Map 1 st Semester 2 nd Semester Unit 1 (5 6 weeks) Circles and Volume MGSE9-12.G.C.1 MGSE9-12.G.C.2 MGSE9-12.G.C.3 MGSE9-12.G.C.4 MGSE9-12.G.C.5 MGSE9-12.G.GMD.1 MGSE9-12.G.GMD.2 MGSE9-12.G.GMD.3 MGSE9-12.G.GMD.4 Unit 2 Geometric and Algebraic Connections MGSE9-12.G.GPE.1 MGSE9-12.G.GPE.4 MGSE9-12.G.GPE.5 MGSE9-12.G.GPE.6 MGSE9-12.G.GPE.7 MGSE9-12.G.MG.1 MGSE9-12.G.MG.2 MGSE9-12.G.MG.3 Unit 3 Applications of Probability MGSE9-12.S.CP.1 MGSE9-12.S.CP.2 MGSE9-12.S.CP.3 MGSE9-12.S.CP.4 MGSE9-12.S.CP.5 MGSE9-12.S.CP.6 MGSE9-12.S.CP.7 Unit 4 (2 3 weeks) Quadratics Revisited MGSE9-12.N.CN.1 MGSE9-12.N.CN.2 MGSE9-12.N.CN.3 MGSE9-12.N.CN.7 MGSE9-12.N.CN.8 MGSE9-12.A.REI.4 MGSE9-12.A.REI.4b MGSE9-12.N.RN.1 MGSE9-12.N.RN.2 Unit 5 (2 3 weeks) Operations With Polynomials MGSE9-12.A.APR.1 MGSE9-12.A.APR.5 MGSE9-12.A.APR.6 MGSE9-12.F.BF.1 MGSE9-12.F.BF.1b MGSE9-12.F.BF.1c MGSE9-12.F.BF.4 MGSE9-12.F.BF.4a MGSE9-12.F.BF.4b MGSE9-12.F.BF.4c Unit 6 Polynomial Functions MGSE9-12.N.CN.9 MGSE9-12.A.SSE.1 MGSE9-12.A.SSE.1a MGSE9-12.A.SSE.1b MGSE9-12.A.SSE.2 MGSE9-12.A.APR.2 MGSE9-12.A.APR.3 MGSE9-12.A.APR.4 MGSE9-12.F.IF.4 MGSE9-12.F.IF.7 MGSE9-12.F.IF.7c Unit 7 (4 5 weeks) Rational & Radical Relationships MGSE9-12.A.APR.7 MGSE9-12.A.CED.1 MGSE9-12.A.CED.2 MGSE9-12.A.REI.2 MGSE9-12.F.IF.4 MGSE9-12.F.IF.5 MGSE9-12.F.IF.7 MGSE9-12.F.IF.7b MGSE9-12.F.IF.7d Unit 8 Exponential & Logarithms MGSE9-12.A.SSE.3 MGSE9-12.A.SSE.3c MGSE9-12.F.IF.7 MGSE9-12.F.IF.7e MGSE9-12.F.IF.8 MGSE9-12.F.IF.8b MGSE9-12.F.BF.5 MGSE9-12.F.LE.4 Unit 9 Mathematical Modeling MGSE9-12.A.SSE.4 MGSE9-12.A.CED.1 MGSE9-12.A.CED.2 MGSE9-12.A.CED.3 MGSE9-12.A.CED.4 MGSE9-12.A.REI.11 MGSE9-12.F.IF.6 MGSE9-12.F.IF.9 MGSE9-12.F.BF.3 These units were written to build upon concepts from prior units, so later units contain tasks that depend upon the concepts addressed in earlier units. All units will include the Mathematical Practices and indicate skills to maintain. NOTE: Mathematical standards are interwoven and should be addressed throughout the year in as many different units and tasks as possible in order to stress the natural connections that exist among mathematical topics. Grade 9-12 Key: Number and Quantity Strand: RN = The Real Number System, Q = Quantities, CN = Complex Number System, VM = Vector and Matrix Quantities Algebra Strand: SSE = Seeing Structure in Expressions, APR = Arithmetic with Polynomial and Rational Expressions, CED = Creating Equations, REI = Reasoning with Equations and Inequalities Functions Strand: IF = Interpreting Functions, LE = Linear and Exponential Models, BF = Building Functions, TF = Trigonometric Functions Geometry Strand: CO = Congruence, SRT = Similarity, Right Triangles, and Trigonometry, C = Circles, GPE = Expressing Geometric Properties with Equations, GMD = Geometric Measurement and Dimension, MG = Modeling with Geometry Statistics and Probability Strand: ID = Interpreting Categorical and Quantitative Data, IC = Making Inferences and Justifying Conclusions, CP = Conditional Probability and the Rules of Probability, MD = Using Probability to Make Decisions July 2016 Page 1 of 5
3 Georgia Department of Education Accelerated GSE Geometry B/Algebra II Expanded Curriculum Map 1 st Semester 1 Make sense of problems and persevere in solving them. 2 Reason abstractly and quantitatively. 3 Construct viable arguments and critique the reasoning of others. 4 Model with mathematics. Standards for Mathematical Practice 5 Use appropriate tools strategically. 6 Attend to precision. 7 Look for and make use of structure. 8 Look for and express regularity in repeated reasoning. 1 st Semester Unit 1 Unit 2 Unit 3 Unit 4 Circles and Volume Geometric and Algebraic Applications of Probability Quadratics Revisited Connections Understand and apply theorems about circles Translate between the geometric description and Understand independence and conditional Perform arithmetic operations with complex MGSE9-12.G.C.1 Understand that all circles are the equation for a conic section probability and use them to interpret data numbers. similar. MGSE9-12.G.GPE.1 Derive the equation of a MGSE9-12.S.CP.1 Describe categories of events as MGSE9-12.N.CN.1 Understand there is a complex MGSE9-12.G.C.2 Identify and describe circle of given center and radius using the subsets of a sample space using unions, number i such that i 2 = 1, and every complex relationships among inscribed angles, radii, chords, Pythagorean Theorem; complete the square to find intersections, or complements of other events (or, number has the form a + bi where a and b are real tangents, and secants. Include the relationship the center and radius of a circle given by an and, not). numbers. between central, inscribed, and circumscribed equation. MGSE9-12.S.CP.2 Understand that if two events A MGSE9-12.N.CN.2 Use the relation i 2 = 1 and the angles; inscribed angles on a diameter are right Use coordinates to prove simple geometric and B are independent, the probability of A and B commutative, associative, and distributive properties angles; the radius of a circle is perpendicular to the theorems algebraically occurring together is the product of their to add, subtract, and multiply complex numbers. tangent where the radius intersects the circle. MGSE9-12.G.GPE.4 Use coordinates to prove probabilities, and that if the probability of two MGSE9-12.N.CN.3 Find the conjugate of a MGSE9-12.G.C.3 Construct the inscribed and simple geometric theorems algebraically. For events A and B occurring together is the product of complex number; use the conjugate to find circumscribed circles of a triangle, and prove example, prove or disprove that a figure defined by their probabilities, the two events are independent. the absolute value (modulus) and quotient of properties of angles for a quadrilateral inscribed in a four given points in the coordinate plane is a MGSE9-12.S.CP.3 Understand the conditional complex numbers. circle. rectangle; prove or disprove that the point (1, 3) probability of A given B as P (A and B)/P(B). Use complex numbers in polynomial identities MGSE9-12.G.C.4 Construct a tangent line from a lies on the circle centered at the origin and Interpret independence of A and B in terms of and equations. point outside a given circle to the circle. containing the point (0,2). conditional probability; that is the conditional MGSE9-12.N.CN.7 Solve quadratic equations with Find arc lengths and areas of sectors of circles (Focus on quadrilaterals, right triangles, and circles.) probability of A given B is the same as the real coefficients that have complex solutions by (but MGSE9-12.G.C.5 Derive using similarity the fact MGSE9-12.G.GPE.5 Prove the slope criteria for probability of A and the conditional probability of B not limited to) square roots, completing the square, that the length of the arc intercepted by an angle is parallel and perpendicular lines and use them to given A is the same as the probability of B. and the quadratic formula. proportional to the radius, and define the radian solve geometric problems (e.g., find the equation of MGSE9-12.S.CP.4 Construct and interpret two-way MGSE9-12.N.CN.8 Extend polynomial identities to measure of the angle as the constant of a line parallel or perpendicular to a given line that frequency tables of data when two categories are include factoring with complex numbers. For proportionality; derive the formula for the area of a passes through a given point). associated with each object being classified. Use the example, rewrite x as (x + 2i)(x 2i). sector. MGSE9-12.G.GPE.6 Find the point on a directed two-way table as a sample space to decide if events Solve equations and inequalities in one variable Explain volume formulas and use them to solve line segment between two given points that are independent and to approximate conditional MGSE9-12.A.REI.4 Solve quadratic equations in problems partitions the segment in a given ratio. probabilities. For example, use collected data from one variable. MGSE9-12.G.GMD.1 Give informal arguments for MGSE9-12.G.GPE.7 Use coordinates to compute a random sample of students in your school on their MGSE9-12.A.REI.4b Solve quadratic equations by geometric formulas. perimeters of polygons and areas of triangles and favorite subject among math, science, and English. inspection (e.g., for x 2 = 49), taking square roots, a. Give informal arguments for the formulas of rectangles, e.g., using the distance formula. Estimate the probability that a randomly selected factoring, completing the square, and the quadratic the circumference of a circle and area of a Apply geometric concepts in modeling situations student from your school will favor science given formula, as appropriate to the initial form of the circle using dissection arguments and informal MGSE9-12.G.MG.1 Use geometric shapes, their that the student is in tenth grade. Do the same for equation (limit to real number solutions). limit arguments. measures, and their properties to describe objects other subjects and compare the results. Extend the properties of exponents to rational b. Give informal arguments for the formula of (e.g., modeling a tree trunk or a human torso as a MGSE9-12.S.CP.5 Recognize and explain the exponents. the volume of a cylinder, pyramid, and cone cylinder). concepts of conditional probability and MGSE9-12.N.RN.1 Explain how the meaning of using Cavalieri s principle. MGSE9-12.G.MG.2 Apply concepts of density independence in everyday language and everyday rational exponents follows from extending the MGSE9-12.G.GMD.2 Give an informal argument based on area and volume in modeling situations situations. For example, compare the chance of properties of integer exponents to rational numbers, using Cavalieri s principle for the formulas for the (e.g., persons per square mile, BTUs per cubic foot). having lung cancer if you are a smoker with the allowing for a notation for radicals in terms of volume of a sphere and other solid figures. MGSE9-12.G.MG.3 Apply geometric methods to chance of being a smoker if you have lung cancer. rational exponents. For example, we define 5 (1/3) to MGSE9-12.G.GMD.3 Use volume formulas for solve design problems (e.g., designing an object or Use the rules of probability to compute be the cube root of 5 because we want [5 (1/3) ] 3 = cylinders, pyramids, cones, and spheres to solve structure to satisfy physical constraints or minimize probabilities of compound events in a uniform 5 [(1/3) x 3] to hold, so [5 (1/3) ] 3 must equal 5. problems. cost; working with typographic grid systems based probability model MGSE9-12.N.RN.2 Rewrite expressions involving Visualize relationships between two-dimensional on ratios). MGSE9-12.S.CP.6 Find the conditional probability radicals and rational exponents using the properties and three-dimensional objects of A given B as the fraction of B s outcomes that of exponents. July 2016 Page 2 of 5
4 MGSE9-12.G.GMD.4 Identify the shapes of twodimensional cross-sections of three-dimensional objects, and identify three-dimensional objects generated by rotations of two-dimensional objects. Georgia Department of Education also belong to A, and interpret the answer in context. MGSE9-12.S.CP.7 Apply the Addition Rule, P(A or B) = P(A) + P(B) P(A and B), and interpret the answers in context. July 2016 Page 3 of 5
5 Georgia Department of Education Accelerated GSE Geometry B/Algebra II Expanded Curriculum Map 2 nd Semester 1 Make sense of problems and persevere in solving them. 2 Reason abstractly and quantitatively. 3 Construct viable arguments and critique the reasoning of others. 4 Model with mathematics. Standards for Mathematical Practice 5 Use appropriate tools strategically. 6 Attend to precision. 7 Look for and make use of structure. 8 Look for and express regularity in repeated reasoning. 2 nd Semester Unit 5 Unit 6 Unit 7 Unit 8 Unit 9 Operations With Polynomials Polynomial Functions Rational & Radical Exponential & Logarithms Mathematical Modeling Relationships Perform arithmetic operations on polynomials MGSE9-12.A.APR.1 Add, subtract, and multiply polynomials; understand that polynomials form a system analogous to the integers in that they are closed under these operations. MGSE9-12.A.APR.5 Know and apply that the Binomial Theorem gives the expansion of (x + y) n in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal s Triangle. Rewrite rational expressions MGSE9-12.A.APR.6 Rewrite simple rational expressions in different forms using inspection, long division, or a computer algebra system; write a(x)/b(x) in the form q(x) + r(x)/b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x). Build a function that models a MGSE9-12.N.CN.9 Use the Fundamental Theorem of Algebra to find all roots of a polynomial equation Interpret the structure of expressions MGSE9-12.A.SSE.1 Interpret expressions that represent a quantity in terms of its context. MGSE9-12.A.SSE.1a Interpret parts of an expression, such as terms, factors, and coefficients, in context. MGSE9-12.A.SSE.1b Given situations which utilize formulas or expressions with multiple terms and/or factors, interpret the meaning (in context) of individual terms or factors. MGSE9-12.A.SSE.2 Use the structure of an expression to rewrite it in different equivalent forms. For example, see x 4 y 4 as (x 2 ) 2 - (y 2 ) 2, thus recognizing it as a difference of squares that can be factored as (x 2 y 2 ) (x 2 + y 2 ). Understand the relationship between zeros and factors of polynomials MGSE9-12.A.APR.2 Know and apply Rewrite rational expressions MGSE9-12.A.APR.7 Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions. Create equations that describe numbers or relationships MGSE9-12.A.CED.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear, quadratic, simple rational, and exponential functions (integer inputs only). MGSE9-12.A.CED.2 Create linear, quadratic, and exponential equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. (Limit to rational and radical functions. The phrase in two or more variables refers to formulas like the Write expressions in equivalent forms to solve problems MGSE9-12.A.SSE.3 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. MGSE9-12.A.SSE.3c Use the properties of exponents to transform expressions for exponential functions. For example, the expression 1.15 t, where t is in years, can be rewritten as [1.15 (1/12) ] (12t) (12t) to reveal the approximate equivalent monthly interest rate is 15%. Analyze functions using different representations MGSE9-12.F.IF.7 Graph functions expressed algebraically and show key features of the graph both by hand and by using technology. MGSE9-12.F.IF.7e Graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, Write expressions in equivalent forms to solve problems MGSE9-12.A.SSE.4 Derive the formula for the sum of a finite geometric series (when the common ratio is not 1), and use the formula to solve problems. For example, calculate mortgage payments. MGSE9-12.A.CED.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear, quadratic, simple rational, and exponential functions (integer inputs only). MGSE9-12.A.CED.2 Create linear, quadratic, and exponential equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. (The phrase in two or more variables refers to formulas like the compound interest formula, in which A = P(1 + r/n) nt has multiple variables.) MGSE9-12.A.CED.3 Represent constraints by equations or inequalities, relationship between two quantities the Remainder Theorem: For a compound interest formula, in which A = midline, and amplitude. and by systems of equation and/or MGSE9-12.F.BF.1 Write a function that describes a relationship between two quantities. MGSE9-12.F.BF.1b Combine standard function types using arithmetic operations in contextual situations (Adding, subtracting, and multiplying functions of different types). MGSE9-12.F.BF.1c Compose functions. For example, if T(y) is the temperature in the atmosphere as a function of height, and h(t) is the height of a weather polynomial p(x) and a number a, the remainder on division by x a is p(a), so p(a) = 0 if and only if (x a) is a factor of p(x). MGSE9-12.A.APR.3 Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial. Use polynomial identities to solve problems MGSE9-12.A.APR.4 Prove P(1 + r/n) nt has multiple variables.) Understand solving equations as a process of reasoning and explain the reasoning MGSE9-12.A.REI.2 Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise. MGSE9-12.F.IF.4 Using tables, graphs, and verbal descriptions, interpret the key characteristics of a function which models the relationship between two MGSE9-12.F.IF.8 Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. MGSE9-12.F.IF.8b Use the properties of exponents to interpret expressions for exponential functions. For example, identify percent rate of change in functions such as y = (1.02) t, y = (0.97) t, y = (1.01) (12t), y = (1.2) (t/10), and classify them as representing exponential growth and decay. inequalities, and interpret data points as possible (i.e. a solution) or not possible (i.e. a non-solution) under the established constraints. MGSE9-12.A.CED.4 Rearrange formulas to highlight a quantity of interest using the same reasoning as in solving equations. Examples: Rearrange Ohm s law V = IR to highlight resistance R; Rearrange area of a circle formula A = π r 2 to highlight the radius r. Represent and solve equations and balloon as a function of time, then T(h(t)) polynomial identities and use them quantities. Sketch a graph showing key inequalities graphically is the temperature at the location of the to describe numerical relationships. features including: intercepts; interval Build new functions from existing MGSE9-12.A.REI.11 Using graphs, weather balloon as a function of time. where the function is increasing, functions tables, or successive approximations, For example, the polynomial Build new functions from existing decreasing, positive, or negative; relative MGSE9-12.F.BF.5 Understand the show that the solution to the equation July 2016 Page 4 of 5
6 functions MGSE9-12.F.BF.4 Find inverse functions. MGSE9-12.F.BF.4a Solve an equation of the form f(x) = c for a simple function f that has an inverse and write an expression for the inverse. For example, f(x) =2(x 3 ) or f(x) = (x+1)/(x-1) for x 1. MGSE9-12.F.BF.4b Verify by composition that one function is the inverse of another. MGSE9-12.F.BF.4c Read values of an inverse function from a graph or a table, given that the function has an inverse. identity (x 2 + y 2 ) 2 = (x 2 y 2 ) 2 + (2xy) 2 can be used to generate Pythagorean triples. Interpret functions that arise in applications in terms of the context MGSE9-12.F.IF.4 Using tables, graphs, and verbal descriptions, interpret the key characteristics of a function which models the relationship between two quantities. Sketch a graph showing key features including: intercepts; interval where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity. Analyze functions using different representations MGSE9-12.F.IF.7 Graph functions expressed algebraically and show key features of the graph both by hand and by using technology. MGSE9-12.F.IF.7c Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior. Georgia Department of Education maximums and minimums; symmetries; end behavior; and periodicity. Interpret functions that arise in applications in terms of the context MGSE9-12.F.IF.5 Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. For example, if the function h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function. Analyze functions using different representations MGSE9-12.F.IF.7 Graph functions expressed algebraically and show key features of the graph both by hand and by using technology. MGSE9-12.F.IF.7b Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. MGSE9-12.F.IF.7d Graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing end behavior. inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents. Construct and compare linear, quadratic, and exponential models and solve problems MGSE9-12.F.LE.4 For exponential models, express as a logarithm the solution to ab (ct) = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology. f(x) = g(x) is the x-value where the y- values of f(x) and g(x) are the same. Interpret functions that arise in applications in terms of the context MGSE9-12.F.IF.6 Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph. MGSE9-12.F.IF.9 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a graph of one function and an algebraic expression for another, say which has the larger maximum. Build new functions from existing functions MGSE9-12.F.BF.3 Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them. July 2016 Page 5 of 5
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# 3 Ways to Solve a 750+ Level GMAT Question About Irregular Polygons
We have examined how to deal with polygons when you encounter them on a GMAT question in a previous post. Today, we will look at a relatively difficult polygon question, however we would like to remind you here that the concepts being tested in this question are still very simple (although we won’t give away exactly which concepts they are yet). First, take a look at the question itself:
The hexagon above has interior angles whose measures are all equal. As shown, only five of the six side lengths are known: 10, 15, 4, 18, and 7. What is the unknown side length?
(A) 7
(B)10
(C) 12
(D) 15
(E) 16
There are various ways to solve this question, but each takes a bit of effort. Note that the polygon we are given is not a regular polygon, since the side lengths are not all equal. The angles, however, are all equal. Let’s first find the measure of each one of those angles using the formula discussed in this previous post.
(n – 2)*180 = sum of all interior angles
(6 – 2)*180 = 720
Each of the 6 angles = 720/6 = 120 degrees
Though we would like to point out here that if you see a question such as this one on the actual GMAT exam, you should already know that if each angle of a hexagon is equal, each angle must be 120 degrees, so performing the above calculation would not be necessary.
Method 1: Visualization
This is a very valid approach to obtaining the correct answer on this GMAT question since we don’t need to explain the reasoning or show our steps, however it may be hard to comprehend for the beginners. We will try to explain it anyway, since it requires virtually no work and will help build your math instinct.
Note that in the given hexagon, each angle is 120 degrees – this means that each pair of opposite sides are parallel. Think of it this way: Side 4 turns on Side 18 by 120 degrees. Then Side 15 turns on Side 4 by another 120 degrees. And finally, Side 10 turns on Side 15 by another 120 degrees. So Side 10 has, in effect, turned by 360 degrees on Side 18.
This means Side 10 is parallel to Side 18.
Now, think of the 120 degree angle between Side 4 and Side 15 – it has to be kept constant. Plus, the angles of the legs must also stay constant at 120 degrees with Sides 10 and 18. Since the slopes of each leg of that angle are negatives of each other (√3 and -√3), when one leg gets shorter, the other gets longer by the same length (use the image below as a visual of what we’re talking about).
Hence, the sum of the sides will always be 15 + 4 = 19. This means 7 + Unknown = 19, so Unknown = 12. Our answer is C.
If you struggled to understand the approach above, you’re not alone. This method involves a lot of intuition, and struggling to figure it out may not be the best use of your time on the GMAT, so let’s examine a couple of more tangible solutions!
Method 2: Using Right Triangles
As we saw in Method 1 above, AB and DE are parallel lines. Since each of the angles A, B, C, D, E and F are 120 degrees, the four triangles we have made are all 30-60-90 triangles. The sides of a 30-60-90 triangle can be written using the ratio 1:√(3):2.
AT = 7.5*√3 and ME = 2*√3, so the distance between the sides of length 10 and 18 is 9.5*√3. We know that DN = 3.5*√3, so BP = (9.5*√3) – (3.5*√3) = 6*√3.
Since the ratios of our sides should be 1:√(3):2, side BC = 2*6 = 12. Again, the answer is C. Let’s look at our third and final method for solving this problem:
Method 3: Using Equilateral Triangles
First, extend the sides of the hexagon as shown to form a triangle:
Since each internal angle of the hexagon is 120 degrees, each external angle will be 60 degrees. In that case, each angle between the dotted lines will become 60 degrees too, and hence, triangle PAB becomes an equilateral triangle. This means PA = PB = 10. Triangle QFE and triangle RDC also become equilateral triangles, so QF = QE = 4, and RD = RC = 7.
Now note that since angles P, Q, and R are all 60 degrees, triangle PQR is also equilateral, and hence, PQ = PR.
PQ = 10 + 15 + 4 = 29
PR = 10 + BC + 7 = 29
BC = 12 (again, answer choice C)
Note the geometry concepts that we used to solve this problem: regular polygon, parallel lines, angles, 30-60-90 right triangles, and equilateral triangles. We know all of these concepts very well individually, but applying them to a GMAT question can take some ingenuity!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# Using Parallel Lines and Transversals to Your Advantage on the GMAT
Today, we will look at a Geometry concept involving parallel lines and transversals (a line that cuts through two parallel lines). This is the property:
The ratios of the intercepts of two transversals on parallel lines is the same.
Consider the diagram below:
Here, we can see that:
• “a” is the intercept of the first transversal between L1 and L2.
• “b” is the intercept of the first transversal between L2 and L3.
• “c” is the intercept of the second transversal between L1 and L2.
• “d” is the intercept of the second transversal between L2 and L3.
Therefore, the ratios of a/b = c/d. Let’s see how knowing this property could be useful to us on a GMAT question. Take a look at the following example problem:
In triangle ABC below, D is the mid-point of BC and E is the mid-point of AD. BF passes through E. What is the ratio of AF:FC ?
(A) 1:1
(B) 1:2
(C) 1:3
(D) 2:3
(E) 3:4
Here, the given triangle is neither a right triangle, nor is it an equilateral triangle. We don’t really know many properties of such triangles, so that will probably not help us. We do know, however, that AD is the median and E is its mid-point, but again, we don’t know any properties of mid-points of medians.
Instead, we need to think outside the box – parallel lines will come to our rescue. Let’s draw lines parallel to BF passing through the points A, D, and C, as shown in the diagram below:
Now we have four lines parallel to each other and two transversals, AD and AC, passing through them.
Consider the three parallel lines, “line passing through A”, “BF”, and “line passing through D”. The ratio of the intercepts of the two transversals on them will be the same.
AE/ED = AF/FP
We know that AE = ED since E is the mid point of AD. Hence, AE/ED = 1/1. This means we can say:
AE/ED = 1/1 = AF/FP
AF = FP
Now consider these three parallel lines: “BF”, “line passing through D”, and “line passing through C”. The ratio of the intercepts of the two transversals on them will also be the same.
BD/DC = FP/PC
We know that BD = DC since D is the mid point of BC. Hence, BD/DC = 1/1. This means we can also say:
BD/DC = 1/1 = FP/PC
FP = PC
From these two calculations, we will get AF = FP = PC, and hence, AF:FC = 1:(1+1) = 1:2.
Therefore, the answer is B. We hope you see that Geometry questions on the GMAT can be easily resolved once we bring in parallel lines.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# The Pythagorean Triples Properties You’ll See on the GMAT
Today, let’s discuss a few useful properties of primitive Pythagorean triples. A primitive Pythagorean triple is one in which a, b and c (the length of the two legs and the hypotenuse, respectively) are co-prime. So, for example, (3, 4, 5) is a primitive Pythagorean triple while its multiple, (6, 8, 10), is not.
Now, without further ado, here are the properties of primitive Pythagorean triples that you’ll probably encounter on the GMAT:
I. One of a and b is odd and the other is even.
II. From property I, we can then say that c is odd.
III. Exactly one of a, b is divisible by 3.
IV. Exactly one of a, b is divisible by 4.
V. Exactly one of a, b, c is divisible by 5.
If you keep in mind the first primitive Pythagorean triple that we used as an example (3, 4, 5), it is very easy to remember all these properties.
If we look at some other examples:
(3, 4, 5), (5, 12, 13), (8, 15, 17) (7, 24, 25) (20, 21, 29) (12, 35, 37) (9, 40, 41) (28, 45, 53) (11, 60, 61) (16, 63, 65) (33, 56, 65) (48, 55, 73), etc.
we will see that these properties hold for all primitive Pythagorean triples.
Now, let’s take a look at an example GMAT question which can be easily solved if we know these properties:
The three sides of a triangle have lengths p, q and r, each an integer. Is this triangle a right triangle?
Statement 1: The perimeter of the triangle is an odd integer.
Statement 2: If the triangle’s area is doubled, the result is not an integer.
We know that the three sides of the triangle are all integers. So if the triangle is a right triangle, the three sides will represent a Pythagorean triple. Given that p, q and r are all integers, let’s use the properties of primitive Pythagorean triples to break down each of the statements.
Statement 1: The perimeter of the triangle is an odd integer.
Looking at the properties above, we know that a primitive Pythagorean triple can be represented as:
(Odd, Even, Odd) (The first two are interchangeable.)
Non-primitive triples are made by multiplying each member of the primitive triple by an integer n greater than 1. Depending on whether n is odd or even, the three sides can be represented as:
(Odd*Odd, Even*Odd, Odd*Odd) = (Odd, Even, Odd)
or
(Odd*Even, Even*Even, Odd*Even) = (Even, Even, Even)
However, the perimeter of a right triangle can never be odd because:
Odd + Even + Odd = Even
Even + Even + Even = Even
Hence, the perimeter will be even in all cases. (If the perimeter of the given triangle is odd, we can say for sure that it is not a right triangle.) This statement alone is sufficient.
Statement 2: If the triangle’s area is doubled, the result is not an integer.
If p, q and r are the sides of a right triangle such that r is the hypotenuse (the hypotenuse could actually be either p, q, or r but for the sake of this example, let’s say it’s r), we can say that:
The area of this triangle = (1/2)*p*q
and
Double of area of this triangle = p*q
Double the area of the triangle has to be an integer because we are given that both p and q are integers, but this statement tells us that this is not an integer. In that case, this triangle cannot be a right triangle. If the triangle is not a right triangle, double the area would be the base * the altitude, and the altitude would not be an integer in this case.
This statement alone is sufficient, too. Therefore, our answer is D.
As you can see, understanding the special properties of primitive Pythagorean triples can come in handy on the GMAT – especially in tackling complicated geometry questions.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# How to Find the Maximum Distance Between Points on a 3D Object
How do we find the the two farthest points on a 3D object? For example, we know that on a circle, any two points that are diametrically opposite will be the farthest from each other than from any other points on the circle. Which two points will be the farthest from each other on a square? The diagonally opposite vertices. Now here is a trickier question – which two points are farthest from each other on a rectangular solid? Again, they will be diagonally opposite, but the question is, which diagonal?
A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box?
(A) 15
(B) 20
(C) 25
(D) 10 * √(2)
(E) 10 * √(3)
There are various different diagonals in a rectangular solid. Look at the given figure:
BE is a diagonal, BG is a diagonal, GE is a diagonal, and BH is a diagonal. So which two points are farthest from each other? B and E, G and E, B and G, or B and H?
The inside diagonal BH can be seen as the hypotenuse of the right triangle BEH. So both BE and EH will be shorter in length than BH.
The inside diagonal BH can also be seen as the hypotenuse of the right triangle BHG. So both HG and BG will also be shorter in length than BH.
The inside diagonal BH can also be seen as the hypotenuse of the right triangle BDH. So both BD and DH will also be shorter in length than BH.
Thus, we see that BH will be longer than all other diagonals, meaning B and H are the points that are the farthest from each other. Solving for the exact value of BH then should not be difficult.
In our question we know that:
l = 10 inches
w = 10 inches
h = 5 inches
Let’s consider the right triangle DHB. DH is the length, so it is 10 inches.
DB is the diagonal of the right triangle DBC. If DC = w = 10 and BC = h = 5, then we can solve for DB^2 using the Pythagorian Theorem:
DB^2 = DC^2 + BC^2
DB^2 = 10^2 + 5^2 = 125
Going back to triangle DHB, we can now say that:
BH^2 = HD^2 + DB^2
BH^2 = 10^2 + 125
BH = √(225) = 15
Thus, our answer to this question is A.
Similarly, which two points on a cylinder will be the farthest from each other? Let’s examine the following practice GMAT question to find out:
The radius of cylinder C is 5 inches, and the height of cylinder C is 5 inches. What is the greatest possible straight line distance, in inches, between any two points on a cylinder C?
(A) 5 * √2
(B) 5 * √3
(C) 5 * √5
(D) 10
(E) 15
Look at where the farthest points will lie – diametrically opposite from each other and also at the opposite sides of the length of the cylinder:
The diameter, the height and the distance between the points forms a right triangle. Using the given measurements, we can now solve for the distance between the two points:
Diameter^2 + Height^2 = Distance^2
10^2 + 5^2 = Distance^2
Distance = 5 * √5
In both cases, if we start from one extreme point and traverse every length once, we reach the farthest point. For example, in case of the rectangular solid, say we start from H, cover length l and reach D – from D, we cover length w and reach C, and from C, we cover length h and reach B. These two are the farthest points.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# Quarter Wit, Quarter Wisdom: Beware of Sneaky Answer Choices on the GMAT!
Test-takers often ask for tips and short cuts to cut down the amount of work necessary to solve a GMAT problem. As such, the Testmaker might want to award the test-taker who pays attention to detail and puts in the required effort.
Today, we will look at an example of this concept – if it seems to be too easy, it is a trap!
In the figure given above, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?
(A) 8√(2)
(B) 24√(3)
(C) 72√(2)
(D) 144√(2)
(E) 384
The first thing I notice about this question is that we have an equilateral triangle. So I am thinking, the area = s^2 * √(3)/4 and/or the altitude = s*√(3)/2.
The irrational number in play is √(3). There is only one answer choice with √(3) in it, so will this be the answer?
Now, it actually makes me uncomfortable that there is only one option with √(3). At first glance, it seems that the answer has been served to us on a silver plate. But the question format doesn’t seem very easy – it links two geometrical figures together. So I doubt very much that the correct answer would be that obvious.
The next step will be to think a bit harder:
The area of the triangle has √(3) in it, so the side would be a further square root of √(3). This means the actual irrational number would be the fourth root of 3, but we don’t have any answer choice that has the fourth root of 3 in it.
Let’s go deeper now and actually solve the question.
The area of the equilateral triangle = Side^2 * (√(3)/4) = 48
Side^2 = 48*4/√(3)
Side^2 = 4*4*4*3/√(3)
Side = 8*FourthRoot(3)
Now note that the side of the equilateral triangle is the same length as the sides of the squares, too. Hence, all sides of the three squares will be of length 8*FourthRoot(3).
All nine sides of the figure are the sides of squares. Hence:
The perimeter of the nine sided figure = 9*8*FourthRoot(3)
The perimeter of the nine sided figure =72*FourthRoot(3)
Now look at the answer choices. We have an option that is 72√(2). The other answer choices are either much smaller or much greater than that.
Think about it – the fourth root of 3 = √(√(3)) = √(1.732), which is actually very similar to √(2). Number properties will help you figure this out. Squares of smaller numbers (that are still greater than 1) are only a bit larger than the numbers themselves. For example:
(1.1)^2 = 1.21
(1.2)^2 = 1.44
(1.3)^2 = 1.69
(1.414)^2 = 2
Since 1.732 is close to 1.69, the √(1.732) will be close to the √(1.69), i.e. 1.3. Also, √(2) = 1.414. The two values are quite close, therefore, the perimeter is approximately 72√(2). This is the reason the question specifically requests the “approximate” perimeter.
We hope you see how the Testmaker could sneak in a tempting answer choice – beware the “easiest” option!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# How to Use the Pythagorean Theorem With a Circle
It does not surprise anyone when they learn that the properties of circles are tested on the GMAT. Most test-takers will nod and rattle off the relevant equations by rote: Area = Π*radius^2; Circumference = 2Π* radius; etc. However, many of my students are caught off guard to learn that the equation for a circle on the coordinate plane is our good friend the Pythagorean theorem. Why on earth would an equation for a right triangle describe a circle?
Remember: the GMAT loves to test shapes in combination: a circle inscribed in a square, for example, or the diagonal of a rectangle dividing it into two right triangles. So you should expect that triangles will appear just about anywhere – including in circles. Especially in coordinate geometry questions, where the coordinate grid allows for right angles everywhere, you should bring the Pythagorean Theorem with you to just about every GMAT geometry problem you see, even if the triangle isn’t immediately apparent. Let’s talk about how the Pythagorean Theorem can present itself in circle problems – “Pythagorean circle problems” if you will. (And note that the Pythagorean Theorem doesn’t have to “announce itself” by telling you you’re dealing with a right triangle! Very often it’s on you to determine that it applies.)
Take a look at the following diagram in which a circle is centered on the origin (0,0) in the coordinate plane:
Designate a random point on the circle (x,y). If we draw a line from the center of the circle to x,y, that line is a radius of the circle. Call it r. If we drop a line down from (x,y) to the x-axis, we’ll have a right triangle (and an opportunity to therefore apply the Pythagorean Theorem to this circle):
Note that the base of the triangle is x, and the height of the triangle is y. So now we have our Pythagorean Theorem equation: x^2 + y^2 = r^2. This is also the equation for a circle centered on the origin on the coordinate plane. [The more general equation for a circle with a center (a,b) is (x-a)^2 + (y-b)^2 = r^2. When a circle is centered on the origin, (a,b) is simply (0,0.)]
This Pythagorean equation of a circle ends up being an immensely useful tool to use on the GMAT. Take the following Data Sufficiency question, for example:
A certain circle in the xy-plane has its center at the origin. If P is a point on the circle, what is the sum of the squares of the coordinates of P?
(1) The radius of the circle is 4
(2) The sum of the coordinates of P is 0
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient
C. Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
D. EACH statement ALONE is sufficient
E. Statements (1) and (2) TOGETHER are NOT sufficient
So let’s draw this, designating P as (x,y):
Now we draw our trust right triangle by dropping a line down from P to the x-axis, which will give us this:
We’re looking for x^2 + y^2. Hopefully, at this point, you notice what the question is going for – because we have a right triangle, x^2 + y^2 = r^2, meaning that all we need is the radius!
Statement 1 is pretty straightforward – if r = 4, we can insert this into our equation of x^2 + y^2 = r^2 to get x^2 + y^2 = 4^2. So x^2 + y^2 = 16. Clearly, this is sufficient.
Now look at Statement 2. If the sum of x and y is 0, we can say x = 1 and y = -1 or x = 2 and y = -2 or x = 100 and y = -100, etc. Each of these will yield a different value for x^2 + y^2, so this statement alone is clearly not sufficient. Our answer is A.
Takeaway: any shape can appear on the coordinate plane, and given the right angles galore in the coordinate grid you should be on the lookout for right triangles, specifically. If the shape in question is a circle, remember to use the Pythagorean theorem as your equation for the circle, and what would have been a challenging question becomes a tasty piece of baklava. (We are talking about principles elucidated by the ancient Greeks, after all.)
And a larger takeaway: it’s easy to memorize formulas for each shape, so what does the GMAT like to do? See if you can apply knowledge about one shape to a problem about another (for example, applying Pythagorean Theorem to a circle). For this reason it’s important to know the “usual suspects” of how shapes get tested together. Triangles and circles work well together, for example:
-If a triangle is formed with two radii of a circle, that triangle is therefore isosceles since those radii necessarily have the same measure.
-If a triangle is formed by the diameter of a circle and two chords connecting to a point on the circle, that triangle is a right triangle with the diameter as the hypotenuse (another way that the GMAT can combine Pythagorean Theorem with a circle).
-When a circle appears in the coordinate plane, you can use Pythagorean Theorem with that circle to find the length of the radius (which then opens you up to diameter, circumference, and area).
In general, whenever you’re stuck on a geometry problem on the GMAT a great next step is to look for (or draw) a diagonal line that you can use to form a right triangle, and then that triangle lets you use Pythagorean Theorem. Whether you’re dealing wit a rectangle, square, triangle, or yes circle, Pythagorean Theorem has a way of proving extremely useful on almost any GMAT geometry problem, so be ready to apply it even to situations that didn’t seem to call for Pythagorean Theorem in the first place.
By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.
# Improve Your Speed on the ACT Math Section Using Math Fluidity
Speed is key on the Math Section of the ACT – you have only 60 minutes to complete 60 questions. However, this doesn’t mean you should spend one minute on each question, as not every question on in this section is created equal. Many questions (particularly Questions 1-30) are problems that you can solve in under one minute. In fact, you should aim to solve Questions 1-30 in less than 30 minutes – around 25 minutes is the goal.
That’s because some of the later questions, particularly the questions from Questions 40-60, will require more than a minute. Basically, you want to put aside extra time for the tricky questions at end of the section by completing the easier, earlier questions as quickly as possible. If you do Questions 1-30 in 25 minutes, then you have 35 minutes to do Questions 31-60.
One way to improve your speed on the Math Section is to develop what I call “math fluidity.” That means recognizing how common patterns, formulas and special rules can help you solve any particular problem. To illustrate, take a look at the following triangle problem:
Triangle ABC (below) is an equilateral triangle with side of length 4. What is the area of triangle ABC?
The first step to any geometry problem is writing down what relevant common formula you’ll need to solve the problem; i.e. whenever I’m asked the area of a triangle, at the top of my work space I’ll write:
A = (b*h)/2
Having the formula in front of you will be helpful because right away, it’s clear that although we have some information, we don’t have all the information we need to solve this problem – we have the base of the triangle (4), but not the height. Since the height of an equilateral triangle always goes from one angle to the opposite side, where it forms two 90-degree angles, drawing the height of an equilateral triangle creates two identical triangles, as shown below:
Many students would now conclude that they need the Pythagorean theorem to solve for the height (that line bisecting the equilateral triangle). This is where math fluidity comes in. Although you could use the Pythagorean theorem, it’s much faster to instead recognize what type of triangle you are dealing with.
Whenever you split an equilateral triangle in half, you create two 30-60-90 triangles. These are also called “special right triangles” because they always follow the rule that the shortest side is always “x,” the side opposite the 60-degree angle is always x√3, and the hypotenuse is always 2x. See the triangle below:
So, rather than spend any time solving for the height of the our triangle by using the Pythagorean Theorem, recognize that because the hypotenuse is 4 and the base is 2 (of either of the smaller triangles), and because the triangle is a right triangle, the height must be 2√3. Therefore, the area of the larger triangle is (2√3)(4)(1/2), which equals 4√3.
Instantly recognizing that the two smaller triangles are 30-60-90 triangles only saves a little bit of time – if you can regularly shave off 20 seconds on question after question by recognizing special rules or how best to apply formulas, you’ll accrue saved time that can later be spent on harder math questions. Speaking of which, math fluidity also applies to tricky questions – similar to what we previously saw, recognition will break down hard questions into easier, faster steps.
So, let’s take a look at a more difficult question. Note, this next example is especially relevant for students shooting for 99th percentile or perfect scores. Although many students can solve the following question if given enough time, few students can solve it quickly enough to get it correct on the ACT. Here’s the problem:
In triangle ABC below, angle BAE measures 30 degrees. What is the value of angle AED minus angle ABE?
A) 30
B) 60
C) 90
D) 120
E) 150
Although there are several ways to solve this problem, math fluidity will help with whatever approach you choose. As I mentioned earlier, it is always best to start by writing down a relevant formula, as it will include what information you have and what information you need. In this case, I’m looking for AED-ABE. Because I’ve also been given the measure of angle BAE, I’ll write down:
BAE = 30 and BAE + ABE = AED
Here’s where math fluidity comes in; the second formula is based off a theorem that you probably learned (and then forgot!) in your geometry class. I do recommend (re)memorizing it for the ACT as follows: a measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles.
Are you drawing a blank? If so, take a moment to think about why that statement is true. If the smaller two angles of a right-angle triangle, as shown at left, are 40 and 50, then if we extend a line as shown to form the adjacent exterior angle x, then x + 50 = 180, so x = 130.
Also, 40 + 50 + 90 = 180, since the sum of interior angles of a triangle always add up to 180. So, if x + 50 = 180, and 40 + 50 + 90 = 180, then x+ 50 = 40 + 50 + 90.
Removing the 50 from both sides, we can conclude that x = 40 + 90, or x (the adjacent exterior angle of one interior angle) is equal to the sum of the other two interior angles.
Now, returning to our original problem:
If BAE = 30 and BAE + ABE = AED, then:
30 + ABE = AED
AED – ABE = 30
Therefore, our answer is A, 30.
Still need to take the ACT? We run a free online ACT prep seminar every few weeks. And be sure to find us on Facebook, YouTube, Google+ and!
# GMAT Tip of the Week
Approach Geometry Questions from the Right (tri)Angle
(This is one of a series of GMAT tips that we offer on our blog.)
As the writers of the GMAT create difficult quantitative questions, one favorite technique is to require test-takers to use skills that would not, on the surface, seem relevant. As examinees struggle to find a foothold on the question with the tools that seem more obvious by comparison, they lose time, make calculation errors, and suffer not only from an incorrect answer but also from a decreased level of confidence that carries on to future questions.
One such seemingly-irrelevant skill that appears on a majority of geometry-based questions is the right triangle, a shape for which you should train yourself to look whenever a geometry question appears. Every shape that the GMAT tests could contain a right triangle:
• The area of a non-right triangle is created by taking a perpendicular line from the base to the angle opposite it; this process divides the original triangle in to two right triangles.
• The diagonal of a square or rectangle is a right triangle.
• The area of a parallelogram or trapezoid is found by taking a perpendicular line between the bases; doing so can create a right triangle.
• Even in a circle, connecting the diameter to a point on the edge of the circle, using two chords to do so, will create a right triangle with the diameter as the hypotenuse.
• In 3-D figures, the longest difference between two points on the item is always a hypotenuse of a right triangle that must be drawn in space.
By training yourself to look for opportunities to use your knowledge of right triangles, even if it doesn’t appear to be the obvious skill to apply, you will save time and build confidence on most geometry-based questions.
For more help on the GMAT, take a look and see what Veritas Prep has to offer.
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Question
# Corn: In a random sample of 88 ears of corn, farmer Carl finds that 15 of...
Corn: In a random sample of 88 ears of corn, farmer Carl finds that 15 of them have worms. He wants to find the 99%confidence interval for the proportion of all his corn that has worms.
(a) What is the point estimate for the proportion of all of Carl's corn that has worms? Round your answer to 3 decimal places.
(b) What is the critical value of z (denoted zα/2) for a 99% confidence interval? Use the value from the table or, if using software, round to 2 decimal places.
zα/2 =
(c) What is the margin of error (E) for a 99% confidence interval? Round your answer to 3 decimal places.
E =
(d) Construct the 99% confidence interval for the proportion of all of Carl's corn that has worms. Round your answers to 3 decimal places.
?? < p < ??
(e) Based on your answer to part (d), are you 99% confident that less than 30% of Carl's corn has worms?
No, because 0.30 is above the upper limit of the confidence interval.
No, because 0.30 is below the upper limit of the confidence interval.
Yes, because 0.30 is below the upper limit of the confidence interval.
Yes, because 0.30 is above the upper limit of the confidence interval.
a)
the point estimate for the proportion of all of Carl's corn that has worms = 15/88 = 0.170
b)
What is the critical value of z (denoted zα/2) for a 99% confidence interval
Here then
c)
Margin of error is
M.E = 0.103
d)
e)
No, because 0.30 is above the upper limit of the confidence interval.
#### Earn Coins
Coins can be redeemed for fabulous gifts.
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Calculation of probability and statistical analysis
lmat1271 2023-2024 Louvain-la-Neuve
Calculation of probability and statistical analysis
6.00 credits
30.0 h + 30.0 h
Q2
Teacher(s)
Language
Prerequisites
Courses LMAT1121 and LMAT1122 (real analysis/calculus, in particular bivariate integration).
Main themes
The general aim of the course consists in giving an introduction into the thinking and the tools of probability theory and statistical analysis, with a view towards applications. The addressed topics cover the basic notions of probability (and conditional probability) and the main distributions of random vectors. The course treats the concepts of independence and correlation, and some aspects of large sample properties. For the statistical analysis, priority is given to the parametric approach (estimation of the parameters of a probability distribution) and to methods of statistical inference (hypothesis testing and confidence intervals). The statistical concepts are applied to the specific problems of analysis of variance (ANOVA) and of (simple) linear regression.
Learning outcomes
At the end of this learning unit, the student is able to : 1 Contribution of the course to learning outcomes in the Bachelor in Mathematics programme.By the end of this activity, students will be able to :Recognise and understand a basic foundation of mathematics.Choose and use the basic tools of calculation to solve mathematical problems.Recognise the fundamental concepts of important current mathematical theories.Establish the main connections between these theories, analyse them and explain them through the use of examples.Show evidence of abstract thinking and of a critical spirit.Argue within the context of the axiomatic method. Recognise the key arguments and the structure of a proof.Distinguish between the intuition and the validity of a result and the different levels of rigorous understanding of this same result.Learning outcomes specific to the course.The general goal of the course is to introduce the student to the notion and the tools of probability theory and statistical analysis, with a view towards applications.By the end of the course, students will be able to :- Use the basic notions of probabilistic modelling, being able to worki with random variables :- Apply the most frequently used techniques of probability theory (conditional probabilities and expectation, normal, Poisson and exponential laws) in various fields of application- Explore structured data sets by the methods of statistical inference- Apply the techniques of confidence intervals and hypothesis testing
Content
• Recalling the concepts of random variables, generalised to random vectors, of conditional probabilities and conditional moments and of the transformation of random variables (in particular the concept of the moment-generating function)
• Derivation of the random sampling distributions (Chi squared, F, Student, …), necessary in order to derive the properties of the most common statistics (estimators)
• Derivation and applications of asymptotic laws (Chebycheff inequality, law of large numbers, central limit theorem, ...)
• Point estimation: method of moments, maximum likelihood method, least-squares method; theoretical properties of estimators (bias, variance, mean-square error, consistency, asymptotic normality, efficiency)
• Confidance intervals (exact and asymptotical; based on the maximum likelihood estimator)
• Statistical hypothesis testing (for the mean, variance, and proportion of one or two normal or binomial populations, respectively): method of pivots; likelihood ratio test,..
• Linear regresion (simple and multiple), general notion of model fitting
• (Time permitting) Introduction into modern data analysis techniques (resampling techniques; principal component analysis; clustering/classification)
Teaching methods
This second introductory course in probability and statistics will consist of :
• lectures that will present the subject on the basis of examples and the development of mathematical reasoning,
• exercise sessions aiming at systematically putting into practice the different notions seen in the course on well-targeted cases and with the help of a specialized software,
• projects that will give the student the opportunity to integrate the different tools in the fields of application of mathematics and physics.
The pedagogical approach used will privilege the active learning of the students and will try to respect the pedagogical orientations proposed by the Faculty.
Evaluation methods
• During the semester :
Continued evaluation via the preparation of a data-analysis project (in groups) using some statistical software and/or via some occasional tests (in groups) during the tutorials.
• During the exam sessions :
Assessment is based on a written examination that focuses on theory and on exercises.
The examination tests knowledge and understanding of fundamental concepts and results, ability to construct and write a coherent argument, mastery of the techniques of calculation and, above all, the applicability of the methods covered in the course to problems in the statistical analysis of data.
Online resources
Site Moodle
https://moodleucl.uclouvain.be/course/view.php?id=8921
On the website can be found : copies of transparencies, exercice problems and their solutions, a list of formulas and statistical tables, the help file for using the statistical software, a copy of a recent exam and the detailed table of contents of the course.
Bibliography
D. Wackerly, W. Mendenhall, R. Scheaffer : "Mathematical Statistics with Applications" (7th ed.) 2008, Brooks/Cole.
Teaching materials
• matériel sur moodle
Faculty or entity
Programmes / formations proposant cette unité d'enseignement (UE)
Title of the programme
Sigle
Credits
Prerequisites
Learning outcomes
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# Unit 4 Multiple Choice Test
34 Questions
Unit 4 Multiple Choice Test (Chapters 5 and 8) ~Please enter the letter of the best choice in the space available.
• 1.
Is it possible for an object moving with a constant speed to accelerate? Explain. A) No, if the speed is constant then the acceleration is equal to zero. B) No, an object can accelerate only if there is a net force acting on it. C) Yes, although the speed is constant, the direction of the velocity can be changing. D) Yes, if an object is moving it can experience acceleration
• A.
A
• B.
B
• C.
C
• D.
D
• 2.
Consider a particle moving with constant speed such that its acceleration of constant magnitude is always perpendicular to its velocity. A) It is moving in a straight line. B) It is moving in a circle. C) It is moving in a parabola. D) None of the above is definitely true all of the time.
• A.
A
• B.
B
• C.
C
• D.
D
• 3.
When an object experiences uniform circular motion, the direction of the acceleration is A) in the same direction as the velocity vector. B) in the opposite direction of the velocity vector. C) is directed toward the center of the circular path. D) is directed away from the center of the circular path.
• A.
A
• B.
B
• C.
C
• D.
D
• 4.
Consider a particle moving with constant speed such that its acceleration of constant magnitude is always perpendicular to its velocity. A) It is moving in a straight line. B) It is moving in a circle. C) It is moving in a parabola. D) None of the above is definitely true all of the time.
• A.
A
• B.
B
• C.
C
• D.
D
• 5.
What type of acceleration does an object moving with constant speed in a circular path experience? A) free fall B) constant acceleration C) linear acceleration D) centripetal acceleration
• A.
A
• B.
B
• C.
C
• D.
D
• 6.
What force is needed to make an object move in a circle? A) kinetic friction B) static friction C) centripetal force D) weight
• A.
A
• B.
B
• C.
C
• D.
D
• 7.
When an object experiences uniform circular motion, the direction of the net force is A) in the same direction as the motion of the object. B) in the opposite direction of the motion of the object. C) is directed toward the center of the circular path. D) is directed away from the center of the circular path.
• A.
A
• B.
B
• C.
C
• D.
D
• 8.
A roller coaster car is on a track that forms a circular loop in the vertical plane. If the car is to just maintain contact with track at the top of the loop, what is the minimum value for its centripetal acceleration at this point? A) g downward B) 0.5g downward C) g upward D) 2g upward
• A.
A
• B.
B
• C.
C
• D.
D
• 9.
A roller coaster car (mass = M) is on a track that forms a circular loop (radius = r) in the vertical plane. If the car is to just maintain contact with the track at the top of the loop, what is the minimum value for its speed at that point? A) rg B) (rg)1/2 C) (2rg)1/2 D) (0.5rg)1/2
• A.
A
• B.
B
• C.
C
• D.
D
• 10.
A pilot executes a vertical dive then follows a semi-circular arc until it is going straight up. Just as the plane is at its lowest point, the force on him is A) less than mg, and pointing up. B) less than mg, and pointing down. C) more than mg, and pointing up. D) more than mg, and pointing down.
• A.
A
• B.
B
• C.
C
• D.
D
• 11.
A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What is the minimum coefficient of static friction between the turntable and the coin if the coin is not to slip? A) (4π2f2r)/g B) (4π2fr2)/g C) (4πf2r)/g D) (4πfr2)/g
• A.
A
• B.
B
• C.
C
• D.
D
• 12.
A car goes around a curve of radius r at a constant speed v. What is the direction of the net force on the car? A) toward the curve's center B) away from the curve's center C) toward the front of the car D) toward the back of the car
• A.
A
• B.
B
• C.
C
• D.
D
• 13.
A car goes around a curve of radius r at a constant speed v. Then it goes around the same curve at half of the original speed. What is the centripetal force on the car as it goes around the curve for the second time, compared to the first time? A) twice as big B) four times as big C) half as big D) one-fourth as big
• A.
A
• B.
B
• C.
C
• D.
D
• 14.
A car goes around a curve of radius r at a constant speed v. Then it goes around a curve of radius 2r at speed 2v. What is the centripetal force on the car as it goes around the second curve, compared to the first? A) four times as big B) twice as big C) one-half as big D) one-fourth as big
• A.
A
• B.
B
• C.
C
• D.
D
• 15.
The gravitational force between two objects is proportional to A) the distance between the two objects. B) the square of the distance between the two objects. C) the product of the two objects. D) the square of the product of the two objects.
• A.
A
• B.
B
• C.
C
• D.
D
• 16.
The gravitational force between two objects is inversely proportional to A) the distance between the two objects. B) the square of the distance between the two objects. C) the product of the two objects. D) the square of the product of the two objects.
• A.
A
• B.
B
• C.
C
• D.
D
• 17.
Two objects attract each other gravitationally. If the distance between their centers is cut in half, the gravitational force A) is cut to one fourth. B) is cut in half. C) doubles. D) quadruples
• A.
A
• B.
B
• C.
C
• D.
D
• 18.
Two objects, with masses m1 and m2, are originally a distance r apart. The gravitational force between them has magnitude F. The second object has its mass changed to 2m2, and the distance is changed to r/4. What is the magnitude of the new gravitational force? A) F/32 B) F/16 C) 16F D) 32F
• A.
A
• B.
B
• C.
C
• D.
D
• 19.
Two objects, with masses m1 and m2, are originally a distance r apart. The magnitude of the gravitational force between them is F. The masses are changed to 2m1 and 2m2, and the distance is changed to 4r. What is the magnitude of the new gravitational force? A) F/16 B) F/4 C) 16F D) 4F
• A.
A
• B.
B
• C.
C
• D.
D
• 20.
Compared to its mass on the Earth, the mass of an object on the Moon is A) less. B) more. C) the same. D) half as much.
• A.
A
• B.
B
• C.
C
• D.
D
• 21.
The acceleration of gravity on the Moon is one-sixth what it is on Earth. An object of mass 72 kg is taken to the Moon. What is its mass there? A) 12 kg B) 72 kg C) 72 N D) 12 N
• A.
A
• B.
B
• C.
C
• D.
D
• 22.
As a rocket moves away from the Earth's surface, the rocket's weight A) increases. B) decreases. C) remains the same. D) depends on how fast it is moving.
• A.
A
• B.
B
• C.
C
• D.
D
• 23.
A spaceship is traveling to the Moon. At what point is it beyond the pull of Earth's gravity? A) when it gets above the atmosphere B) when it is half-way there C) when it is closer to the Moon than it is to Earth D) It is never beyond the pull of Earth's gravity.
• A.
A
• B.
B
• C.
C
• D.
D
• 24.
Suppose a satellite were orbiting the Earth just above the surface. What is its centripetal acceleration? A) smaller than g B) equal to g C) larger than g D) Impossible to say without knowing the mass.
• A.
A
• B.
B
• C.
C
• D.
D
• 25.
A hypothetical planet has a mass of half that of the Earth and a radius of twice that of the Earth. What is the acceleration due to gravity on the planet in terms of g, the acceleration due to gravity at the Earth? A) g B) g/2 C) g/4 D) g/8
• A.
A
• B.
B
• C.
C
• D.
D
• 26.
The acceleration of gravity on the Moon is one-sixth what it is on Earth. The radius of the Moon is one-fourth that of the Earth. What is the Moon's mass compared to the Earth's? A) 1/6 B) 1/16 C) 1/24 D) 1/96
• A.
A
• B.
B
• C.
C
• D.
D
• 27.
Two planets have the same surface gravity, but planet B has twice the radius of planet A. If planet A has mass m, what is the mass of planet B? A) 0.707m B) m C) 1.41m D) 4m
• A.
A
• B.
B
• C.
C
• D.
D
• 28.
Two planets have the same surface gravity, but planet B has twice the mass of planet A. If planet A has radius r, what is the radius of planet B? A) 0.707r B) r C) 1.41r D) 4r
• A.
A
• B.
B
• C.
C
• D.
D
• 29.
Consider a small satellite moving in a circular orbit (radius r) about a spherical planet (mass M). Which expression gives this satellite's orbital velocity? A) v = GM/r B) (GM/r)1/2 C) GM/r2 D) (GM/r2)1/2
• A.
A
• B.
B
• C.
C
• D.
D
• 30.
Satellite A has twice the mass of satellite B, and rotates in the same orbit. Compare the two satellite's speeds. A) The speed of B is twice the speed of A. B) The speed of B is half the speed of A. C) The speed of B is one-fourth the speed of A. D) The speed of B is equal to the speed of A.
• A.
A
• B.
B
• C.
C
• D.
D
• 31.
A person is standing on a scale in an elevator accelerating downward. Compare the reading on the scale to the person's true weight. A) greater than their true weight B) equal to their true weight C) less than their true weight D) zero
• A.
A
• B.
B
• C.
C
• D.
D
• 32.
Who was the first person to realize that the planets move in elliptical paths around the Sun? A) Kepler B) Brahe C) Einstein D) Copernicus
• A.
A
• B.
B
• C.
C
• D.
D
• 33.
The speed of Halley's Comet, while traveling in its elliptical orbit around the Sun, A) is constant. B) increases as it nears the Sun. C) decreases as it nears the Sun. D) is zero at two points in the orbit.
• A.
A
• B.
B
• C.
C
• D.
D
• 34.
The average distance from the Earth to the Sun is defined as one "astronomical unit" (AU). An asteroid orbits the Sun in one-third of a year. What is the asteroid's average distance from the Sun? A) 0.19 AU B) 0.48 AU C) 2.1 AU D) 5.2 AU
• A.
A
• B.
B
• C.
C
• D.
D
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# MATLAB Questions and Answers – Convolution – 2
This set of MATLAB Multiple Choice Questions & Answers (MCQs) focuses on ” Convolution – 2″.
1. After what instant in time will the signal output become zero?
`p=conv([ones(1,100)],[ones(1,50),zeros(1,50)]);`
a) 150
b) 149
c) 151
d) Error
Explanation: The graphical convolution gives an intuitive method to realize such cases. We observe that the second vector completely overlaps the first vector at an instant of 100 units in time. Now since we’ve flipped the second vector, we observe that the pulse of ones leaves the first vector at the instant of 149 units in time and we’re left with zeros. From here on, the output becomes 0 and hence, the correct option is 149.
2. For a causal L.T.I. system, the impulse response is 0 for _________
a) t<0
b) t=0
c) t>0
d) Always
Explanation: Since the system is causal, the impulse response won’t exist for t<0. This is because the output of the system should not depend on future inputs. Thus only option t<0 is correct.
3. The convolution of a discrete signal with itself is _________
a) Squaring the signal
b) Doubling the signal
d) is not possible
Explanation: This is proved by the fact that since discrete signals can be thought of as a one variable polynomial with the coefficients, along with the order, representing the amplitude, at an instant equal to the order of the variable, of the signal- they are simply multiplied during convolution.
4. The convolution of a function with an impulse function delayed to an instant 3 in time results in ____________
a) An advance in the function by 3 units in time
b) The function itself
c) A delay in the function by 3 units in time
d) Cannot be determined
Explanation: The convolution of an impulse function with a function results in the function itself. But if the impulse function is delayed, the output will also get delayed by an equal amount. This is because
-∞ f(k).δ(t-m-k)dk=f(t-m).
5. What is the output of the following code?
```a=con([1 2],[1 2]);
b=cconv([1 2],[1 2]);```
a) a=b
b) a<b
c) a>b
d) a!=b
Explanation: Circular and linear convolution produce the same equivalent results in MATLAB and hence the correct option is a=b. This can also be checked from the tabular method for linear and circular convolution.
6. A continuous signal can be represented as the product of an impulse function and the signal itself.
a) True
b) False
Explanation: The continuous signal can be represented as an integral of impulses. This representation buries it down to the form of convolution of two signal where one signal is the impulse function while the other is the continuous signal. Hence, the above statement is true.
7. What is the output of the following code?
`cconv([1 2],[0 1 0])`
a) [0 1 2 0]
b) [1 2 0 0]
c) [1 2 0 0]
d) [0 0 1 2]
Explanation: The signal is getting circularly convolved with a n impulse function which is delayed by 1 unit in time. Hence, the output of the above code will be the original function which gets delayed by 1 unit in time.
8. What is the Scope value if the signal generator has a frequency of 2 Hz only?
a) An attenuated signal of same frequency
b) The entire signal at the same frequency
c) The entire signal at reduced frequency
d) An attenuated signal at reduced frequency
Explanation: The above transfer function is that of a high pass filter. The cut-off frequency for allowing signals is 3Hz but the given signal frequency is that of 2Hz only. If the frequency of the signal generator was more than 2Hz- the output would’ve been entire signal at the same frequency but for >>3Hz. Now, the convolution in time domain is multiplication in frequency domain and the output of the transfer function block is the product of the transfer function and the laplace transform of the sinusoid. It can be checked in MATLAB that the poles of the resultant function will consist of one pole at s=-2+(-).707i which equivalently suggest that there is a 3 db decade if the input frequency becomes less than 2hz.
9. What is the output of the following code?
```P=tf([1 2],[3 4]);
Q=tf([1 2],[3 4]);
Z=P.Q;```
a) Z is the response of the system whose pole is at s=-4/3
b) Z is the response of the system whose pole is at s=4/3
c) Z is the response of the system whose poles are at s=-4/3 & s=+4/3
d) Z is the response of the system whose zeros are at s=-4/3
Explanation: Z is the multiplication of laplace transform of two sets of two transfer functions. So if they get multiplied, the output can be the response of a system- provided one of the transfer function is the impulse response of a system while the other is just a representation of some signal. Now, this means that the poles of each function is provided in the second row vector of tf() command. This means the pole can possibly be at s=-4/3 since the elements indicate increasing power of s from left to right.
10. What is the inverse laplace transform of Z from the following?
```P=tf([1],[1 0 0]);
Q=tf([1],[1 0]);
Z=P.Q;```
a) t2
b) t2/2
c) t3
d) Error
Explanation: Z is the convolution of P and Q in time domain which has been done in MATLAB by converting P and Q into the frequency domain and multiplying them. Observing that P and Q are the laplace transform of ramp and step functions respectively, Z is 1/s3. The inverse laplace transform of Z is simply option t2 then.
11. What is the output of the following code?
```P=tf([1 2],[3 4]);
Q=tf([1 2],[3 4]);
Z=P.Q;
ilpalace(Z)```
a) Error
b) t
c) t2
d) Cannot be determined
Explanation: Z is in the tf domain. So, the ilaplace command wont be able to take such variables. We need to write the entire Z as an input to the ilpalace command and then it will give the inverse lapalce transform.
12. What is the output of the following code?
`conv[1 2]`
a) Error in input
b) Error in []
c) 2
d) 0
Explanation: The first error MATLAB notices is that the input to the conv command is given within [] but not () and this is the error it’ll return. The next error is the fact that there is no ‘ , ‘ between 1 and 2 so the conv command doesn’t understand the input.
Sanfoundry Global Education & Learning Series – MATLAB.
To practice all areas of MATLAB, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
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# What is 59 Oil Barrels in Cubic Inches?
## Convert 59 Oil Barrels to Cubic Inches
To calculate 59 Oil Barrels to the corresponding value in Cubic Inches, multiply the quantity in Oil Barrels by 9701.9999999787 (conversion factor). In this case we should multiply 59 Oil Barrels by 9701.9999999787 to get the equivalent result in Cubic Inches:
59 Oil Barrels x 9701.9999999787 = 572417.99999874 Cubic Inches
59 Oil Barrels is equivalent to 572417.99999874 Cubic Inches.
## How to convert from Oil Barrels to Cubic Inches
The conversion factor from Oil Barrels to Cubic Inches is 9701.9999999787. To find out how many Oil Barrels in Cubic Inches, multiply by the conversion factor or use the Volume converter above. Fifty-nine Oil Barrels is equivalent to five hundred seventy-two thousand four hundred eighteen Cubic Inches.
## Definition of Oil Barrel
A barrel is one of several units of volume included fluid barrels (UK beer barrel, US beer barrel), dry barrels, oil barrel, etc. Since medieval times, the size of barrel has been used with different meanings around Europe, from about 100 liters to above 1000 in special cases. Modern barrels are made of aluminum, stainless steel, and different types of plastic, such as HDPE. Now in most countries, the barrels are replaced by SI units. However, prices per barrel in USD are commonly used.
## Definition of Cubic Inch
The cubic inch is a unit of measurement for volume in the Imperial units and United States customary units systems. It is the volume of a cube with each of its three dimensions (length, width, and depth) being one inch long. The cubic inch and the cubic foot are still used as units of volume in the United States, although the common SI units of volume, the liter, milliliter, and cubic meter, are also used, especially in manufacturing and high technology. One cubic foot is equal to exactly 1,728 cubic inches because 123 = 1,728.
## Using the Oil Barrels to Cubic Inches converter you can get answers to questions like the following:
• How many Cubic Inches are in 59 Oil Barrels?
• 59 Oil Barrels is equal to how many Cubic Inches?
• How to convert 59 Oil Barrels to Cubic Inches?
• How many is 59 Oil Barrels in Cubic Inches?
• What is 59 Oil Barrels in Cubic Inches?
• How much is 59 Oil Barrels in Cubic Inches?
• How many in3 are in 59 bbl?
• 59 bbl is equal to how many in3?
• How to convert 59 bbl to in3?
• How many is 59 bbl in in3?
• What is 59 bbl in in3?
• How much is 59 bbl in in3?
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# constructing an explicit homotopy
I can see that the paths $(\cos(\pi s), \sin(\pi s))$ and $(\cos(\pi s), -\sin(\pi s))$ in $\mathbb{R}^2 \setminus \{0\}$ are 'homotopic' But can't construct an explicit homotopy between them.
Could anyone suggest me an explicit homotoy function?
Here I mean just homotopy, not path-homotopy
• What do you mean when you say "just homotopy"? Not endpoint-preserving? Also, what's the range of $s$? $[0,1]$, $[0,2]$...? – Najib Idrissi Oct 7 '14 at 13:58
• (Please look at how I edited your post to get an idea of how to properly format your questions. You can also read this.) – Najib Idrissi Oct 7 '14 at 14:02
• oh sorry I have not been specific enough. I mean not endpoint-preserving and the range of s is [0,1] – Keith Oct 7 '14 at 16:17
You can explicitly construct a "retraction" of the first path to the constant path based at $(1,0)$ ($s=0$), by putting $\gamma_t(s) = (\cos(\pi s t), \sin(\pi s t))$. You can do the same for the second path. Now do the first "retraction", then the second one in reverse direction.
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2015-03-25T23:18:09-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
The change in x is equal to the difference in x-coordinates (also called run), and the change in y is equal to the difference in y coordinates (also called rise).
m=y2−y1/x2−x1
Substitute in the values of x and y into the equation to find the slope.
m=−3−(2)/2−(−1)
Simplify.
m=−5/3
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https://www.teacherspayteachers.com/Product/Area-of-Circles-Coloring-Book-Math-2514944
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# Area of Circles Coloring Book Math
Subject
Grade Levels
Resource Type
Common Core Standards
Product Rating
4.0
3 Ratings
File Type
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
548 KB|4 pages
Share
Product Description
This Scooby Doo "Coloring Book Math" worksheet includes a worksheet on finding the area of circles given their radius or diameter, a coloring sheet, worksheet key, and coloring page key! Have fun finding the area of circles with Scooby Doo!!!
“Why??? Coloring is for kindergarten.”
My motto has always been: “If you’re bored, your students are definitely bored!” After grading homework sheet after workbook page… I realized that grading wasn’t super fun, and I’m positive doing the homework wasn’t very fun either. Then I remembered how much I loved kindergarten and the lower grades because the teachers made EVERYTHING fun. That is when I decided to make my assignments a little more ‘interesting.’ I didn’t know how my students would react, but surprisingly they LOVE IT and grading is a lot more exciting as well! Who doesn’t love coloring? (Ok… I do have one student who chooses not to color but 1 out of 40 isn’t too bad :)
“How/when would I use these activities?”
Whenever you want! Use them as homework, review, extra credit, etc.
“Why are the colors so crazy? Don’t you know that Mario's pants aren't blue?
After being a middle school student myself and teaching middle school for the past few years, I know how clever they can be. If the colors were accurate, it would be simple to find the problem that is “blue” and match it up with the answer on his pants which are supposed to be “blue.” However, if they are random colors, they have to do the work!
“Do I grade them on their coloring?”
I don’t. I simply remind them that the ANSWERS are on the back! It would be silly not to look at them! (Remind them that this is not cheating since you are giving it to them.) I give them an extra credit point for coloring.
“What are the benefits?”
-Students actually check their work. If the answer isn’t on the coloring page, they know they did something wrong and can redo the problem.
-Easier to grade (just look at their coloring page). TIP: I usually at least glance through their work to make sure they didn’t simply copy someone else’s coloring.
-Students actually get excited about homework!
“Do you have a coloring book page for ___________”
Message me and let me know what topic/problems you would like to have a coloring page for and what cartoon/picture you would prefer and I will make you one if I do not already have one!
Total Pages
4 pages
Answer Key
Included
Teaching Duration
N/A
Report this Resource
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https://www.cryptoground.com/amis-profit-calculator
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• Market Cap: \$286b
• Feb 20, 2020
AMIS Profit Calculator or you can say AMIS ROI Calculator is a simple tool to calculate how much profit you would have made if you had invested in AMIS (AMIS) in past. This helps you measure the return on investment (ROI) of AMIS (AMIS) .
If you are looking for mining calc check it here: AMIS Mining Calculator
What my profit would be If I have invested
\$
in AMIS on the date
?
## How does AMIS Profit Calculator Works?
This AMIS Profit Calculator uses a simple mathematical principal to calculate the ROI of AMIS. It fetches the historical AMIS price from the database and compares with current AMIS Price and calculate the profit or loss made on it.
It does this simple calculation get the amount AMIS you would have got by investing x\$'s on that day (\$x/price of AMIS). Now it calculates the current price of that amount in USD (current AMIS price * amount of AMIS purchased in past). Now the return on investment (ROI) is calculated by dividing amount in USD today by amount invested and multiplying it by 100.
Mathmetical logic behind the same:
\$invested_USD = USD invested in past date;
\$historical_AMIS_price = Price of AMIS in past date;
\$quantity_AMIS = Quantity of AMIS in past = \$amount invested / \$price_on_that_day;
\$price_AMIS = Current price of AMIS;
\$USD_today = (\$price_AMIS * \$quantity_AMIS) - \$invested_USD;
\$ROI = (\$USD_today/\$invested_USD)*100;
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If you have any query regarding the above calculator you can comment it in comment box below.
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https://www.cbse24.com/2020/10/sign-convention-for-spherical-mirror.html
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# Sign Convention for spherical Mirror and Mirror Formula class 10 | cbse24
## Sign Convention for Spherical Mirror:
1:-All the distance are measured from pole of the mirror as origin.
2:-Distance measured in the same direction as that of incident light are taken as positive.
3:-Distance measured against the direction of incident light are taken as negative.
4:-Distance measured upward and perpendicular to the principal axis are taken as positive.
5:-Distance measured downward and perpendicular to the principal axis are taken as negative
## Mirror Formula:-
$\frac{1}{Image-distance}+\frac{1}{Object-distance}=\frac{1}{Focal Length}$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
so
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
Where
v=distance of image from mirror
u=distance of object from mirror
f=focal length of the mirror
### Linear Magnification(m):-
The ratio of the height of image to the height of object is known as linear magnification.
$m=\frac{height -of- image}{height- of -object}=\frac{-v}{u}$
$m=\frac{h_{i}}{h_{o}}=\frac{-v}{u}$
Note:
1:- ho always will be positive
2:- If hi is virtual than it will be positive and hi is real than it will be negative
3:-If the magnification has a minus sign, therefore image is real and inverted.
4:-If the magnification has a positive sign, therefore image is virtual and erect.
*
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http://www.docstoc.com/docs/25265051/Interruptible-Transit-Service-Contract
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# Interruptible Transit Service Contract
Document Sample
``` Interruptible Transportation Service Contract
between
Baumgarten-Oberkappel Gasleitungsgesellschaft m.b.H.
Floridsdorfer Hauptstrasse 1, floridotower
1210 Vienna, Austria
(hereinafter referred to as Carrier)
and
xxxxxxxxxxxxxx
xxxxxxxxxx
Postal Code, City
(hereinafter referred to as Shipper)
Interruptible Transit Service Contract Page 1 of 11
- -2-
Content
Article 1 Object 3
Article 2 Inlet Point and Outlet Point 4
Article 3 Capacities 5
Article 4 Tariff 6
Article 5 Interruption 8
Article 6 Arbitration 9
Article 7 Term 10
Article 8 General Terms and Conditions 11
Interruptible Transit Service Contract Page 2 of 11
- -3-
Article 1
Object
This Contract sets out the terms and conditions for the right of Shipper to
have transported by Carrier via the West Austria Gasleitung (WAG) on an
interruptible basis quantities of Natural Gas corresponding to the Capacity
as specified in Article 3.
Interruptible Transit Service Contract Page 3 of 11
- -4-
Article 2
Inlet Point and Outlet Point
2.1 The Inlet Point is the ____________________________________________.
2.2 The Outlet Point is the _________________________________________.
2.3 The Transportation Distance is (TD) is equal to: ____________________ km.
Interruptible Transit Service Contract Page 4 of 11
- -5-
Article 3
Capacities
Shipper commits to reserve in the WAG for interruptible transportation of
Natural Gas from the Inlet Point to the Outlet Point a Capacity for a maximum
hourly flow-rate (RS) of ________ Nm³/h in the period from ___________, 6:00
a.m., to _____________, 6:00 a.m.
Interruptible Transit Service Contract Page 5 of 11
- -6-
Article 4
Tariff
4.1 For each Month “m” from ______________ until ______________ Shipper
shall pay to Carrier for the services provided for in this Contract an
amount Em in EUR calculated on the 1st of October each Year - for the
first time on DD.MM.YYYY - as follows:
RS * EuK n RS * TD * EaK n
E m
= +
12 12
where
⎛ G L ⎞
EuK n = EuK o * ⎜ 0,70 + 0,15 * m + 0,15 * m ⎟
⎜
⎝ G0 L0 ⎟
⎠
⎛ G L ⎞
EaK n = EaK o * ⎜ 0,70 + 0,15 * m + 0,15 * m ⎟
⎜
⎝ G0 L0 ⎟
⎠
EuK0 (distance-independent component) = _________ EUR/(m³/h) per Year
EaK0 (distance-dependent component) = _____ EUR/((m³/h) * km) per
Year
RS = maximum hourly flow-rate in Nm³/h according to Article 3
TD = ______ km, as specified in Article 2
Em = __________ EUR/month
Gm value of the Austrian wholesale price index without seasonal effect
(“Großhandelspreisindex – Nichtsaisonwaren,”) as published in the
“Statistische Nachrichten” issued by the “Statistik Austria” valid at
the time of recalculation. Gm shall be rounded to one (1) decimal place
behind the decimal point.
G0 Gm valid for the month of [to be filled in (Month Year)]; i.e. _____
Lm the arithmetic average of all Monthly minimum wages of workers of
the Austrian oil producing industry according to the “Kollektivvertrag
für Arbeiter der erdöl- und erdgasgewinnenden Industrie Österreichs”,
expressed in EUR per Month, in force on the last Day of the Month of
February immediately preceding each revision of Lm. Lm shall be
rounded to four (4) decimal places behind the decimal point.
L0 Lm valid for the month of [to be filled in (Month Year)]; i.e. _____ EUR
Interruptible Transit Service Contract Page 6 of 11
- -7-
4.2 Any payment under this Article 4 is exclusive of any taxes, duties or
levies of a similar nature. Carrier is entitled to add to the amount due by
Shipper all taxes, duties or levies of similar nature imposed on Carrier by
any authority concerning the transportation services under this Contract.
At present, except VAT, no such charges are imposed.
4.3 Carrier shall have the right to interrupt partly or totally the Capacity as
specified in Article 3. In case of such interruption an amount Rm, as sum
of Ri, shall be refunded by Carrier to Shipper. This amount Rm, to be
calculated for the Month in which such interruption is effected, shall be
deducted from the amount Em as per the invoice issued by Carrier to
Shipper for the following Month.
Ri and Rm shall be calculated as follows:
Em *1,5 * IRS
Ri =
RS
and
1 Hi
Rm = ∑ Ri ≤ Em
H m i =1 where Rm shall not exceed Em and
where
Rm = Monthly amount to be refunded by Carrier to Shipper for interruptions of
Capacity.
Em = Monthly amount to be paid by Shipper as defined in Article 4.1 excluding
any refund for interruption.
Hi Total number of hours Carrier has interrupted Shipper’s capacity
Hm Total number of hours in the month Shipper’s capacity has been
interrupted by Carrier.
IRS Interrupted hourly flow rate in Nm³/h.
Interruptible Transit Service Contract Page 7 of 11
- -8-
Article 5
Interruption
5.1 The exclusive justification for an interruption is the lack of transportation
capacity necessary to fulfil the contracted transportation service from the
Inlet-Point to the Outlet-Point and vice versa. Transportation services on a
firm basis have precedence over transportation services on an interruptible
basis.
5.2 Notwithstanding Article 2.5 of the General Terms and Conditions for
Transportation of Natural Gas, Carrier shall inform Shipper as soon as
possible, in any case at least two (2) hours in advance of such
interruptions.
Interruptible Transit Service Contract Page 8 of 11
- -9-
Article 6
Arbitration
5.1 All disputes arising out of or in connection with the Contract including all
amendments, if any, shall be exclusively and finally settled under the Rules of
Arbitration of the International Chamber of Commerce, Paris.
5.2 The arbitral tribunal shall be composed of three arbitrators. Unless otherwise
agreed each Party shall appoint one (1) arbitrator and the third arbitrator will be
selected by the first two arbitrators within thirty (30) days after the appointment
of the second arbitrator. The third arbitrator shall act as chairman of the board
of arbitration and shall be fully educated and trained as a lawyer.
5.3 The seat of the arbitration tribunal and the proceedings shall be in Vienna.
Notwithstanding Article 16.2 of the General Terms and Conditions, the
language of the proceedings shall be German.
Interruptible Transit Service Contract Page 9 of 11
- - 10 -
Article 7
Term
This Contract shall come into full force and effect upon signature and shall
remain in force until ______________, 6.00 a.m. Transportation services under
this Contract shall commence as laid down in Article 3.
Interruptible Transit Service Contract Page 10 of 11
- - 11 -
Article 8
General Terms and Conditions
8.1 The English version of the General Terms and Conditions for
Transportation of Natural Gas (“General Terms and Conditions”) is an
integral part of this Contract. Shipper acknowledges the General Terms
and Conditions valid at the time of signing this Contract and as published
at the time of signing on the Internet (http://www.bog-gmbh.at). Shipper
is fully aware of the provisions of the General Terms and Conditions and
accepts them. In case of any contradictions/discrepancies between the
General Terms and Conditions and the terms of this Contract, the latter
shall prevail.
8.2 For the purpose of Article 15.4 of the General Terms and Conditions the
address given on the first page of this Contract shall be the relevant one.
IN WITNESS WHEREOF, this Contract has been signed in two original copies by
duly authorised officers of the Parties on the day and year written below.
Vienna, ...................
Baumgarten-Oberkappel <Shipper>
Gasleitungsgesellschaft m.b.H. <xxxxxxxxxxx>
Interruptible Transit Service Contract Page 11 of 11
```
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Edit Article
# wikiHow to Play Go Fish
If you're new to playing games with cards, Go Fish is a great place to start. This classic children's card game may be played with 2 to 6 players, and all you need is a standard 52-card deck. Learn the rules of the game and a few variations.
### Method 1 Understanding the Rules
1. 1
Know the objective. The goal in Go Fish is to collect as many "books," or sets of four cards of the same rank, as possible. The person with the most books at the end of the game is the winner.
• An example of a book would be having all four queens in a deck: the queen of hearts, the queen of spades, the queen of clubs, and the queen of diamonds.
• A book doesn't have to be composed of face cards. You could have a book of nines: the nine of hearts, the nine of spades, the nine of clubs, and the nine of diamonds.
2. 2
Know how to build a book. Players collect complete books by taking turns asking each other for a card they need to make a set of cards complete. For example, if a player is dealt a two of clubs and a two of hearts, she will ask another player if he has a two. In this way she adds cards to the book until it is complete.
3. 3
Know what it means to "fish." If a player is asked for a card she has in her hand, she is obligated to hand over all cards of that rank. If she does not have the card, she replies, "Go fish." The player who requested the card then "fishes" a card from the pile of extra cards called the "draw pile." This gives her an extra chance to obtain a card for one of the books she is building.
• If a player receives the card she asked for or draws it from the pile, she gets another turn.
• If a player doesn't end up with the card she asked for, her turn is over.
4. 4
Understand how the game ends. Players continue to go around the circle, asking for cards, drawing cards and creating books, until either someone has no cards left in his or her hands or the draw pile runs out. The person with the most books is the winner.
### Method 2 Shuffling and Dealing
1. 1
Nominate a dealer. In Go Fish, one person starts out as the dealer, the person who distributes the first hand of cards and gets the game started. The person who had the idea to play the game usually plays the role of the dealer. The other players arrange themselves in a circle extending from both sides of the dealer.
• Some people like to follow certain rules to figure out who the dealer should be. For example, the dealer could be the youngest or oldest person present, or the person whose birthday is coming up the soonest.
• If you end up playing more than one game of Go Fish, the dealer of the second game is usually the person who won the first game.
2. 2
Shuffle the deck. Any time you're about to start a card game, shuffle the deck to redistribute the cards from the last game. This ensures the cards aren't organized in a predictable pattern and shows the other players that no cheating is involved.
3. 3
Deal 5 cards to each player. Start with the deck of cards face down, so they cannot be seen by any of the players. Give the topmost card to the first player on the left, the next card to the next player in the circle, and so on. Continue dealing cards one at a time around the table until everyone has five cards.
• If you are playing with just two players, deal 7 cards to each player instead of 5.
4. 4
Make a draw pile, or "pool." Place the remaining cards face down in the center of the circle or table so that they are within everyone's reach. They don't have to be orderly, but they should all be face down. This is the pool from which everyone "fishes."
### Method 3 Playing the Game
1. 1
Examine your cards. Hold the cards in a fan so that the other players can't see them, and take a look at what you were dealt. If you have two or more cards of the same rank, you may want to plan to pursue more cards of that suit to build a book. If you don't have any cards of the same rank, you may choose to pursue any of the cards in your hand.
2. 2
Start gameplay with the player to the left of the dealer. This player chooses someone else, it doesn't matter who, to ask whether he or she has a card of a specific rank. For example, the player might say, "Moirin, do you have any threes?"
• If Moirin has threes, she is obligated to hand them over, and the player gets another turn.
• If Moirin has no threes, she says "Go fish." The player then draws a card from the draw pile. If it's the card he or she asked for, another turn is granted. If not, game play moves to the next person on the left.
3. 3
Make complete books. As play moves around the circle, players start collecting enough cards to make complete books. When a book is complete, the player shows the other players the book, then lays the cards in the book face down.
• As players ask each other for cards, try to remember which cards they asked for. When it's your turn, you'll have the advantage of knowing what's in their hand. For example, if you hear a player ask someone else for an eight, and you're hoping to collect a book of eights yourself, remember to ask that player for an eight next time it's your turn.
4. 4
Finish the game. Eventually the draw pile will dwindle and the cards will run out. When this happens, each player counts his or her books. The person with the most books is the winner.
### Method 4 Using a Variation
1. 1
Ask for specific cards. Rather than asking for a card in the same rank, ask for a specific card. For example, if you have a jack of hearts, ask another player for a jack of diamonds, rather than just asking for a jack. This variation makes the game more difficult, and it tends to last longer this way.
2. 2
Play with pairs instead of books. When you form a pair of two cards of the same rank and color, show the other players and lay the pair down. An easier variation would be to form pairs of two cards of the same rank, even if they aren't the same color.
3. 3
Disqualify players who run out of cards. In a typical game of Go Fish, the game is over when a player runs out of cards. Play a variation in which the game continues among the players who still have cards.
## Community Q&A
Search
• If I ask for nines and the opponent has 3 nines, does she give me all 3 nines?
wikiHow Contributor
Yes. But if you're playing pairs, she should only have 1.
200 characters left
If this question (or a similar one) is answered twice in this section, please click here to let us know.
## Tips
• Do not use the jokers.
## Things You'll Need
• Deck of playing cards
• 2-6 players
## Sources and Citations
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• An online midpoint calculator helps to find the distance and midpoint of a line segment and shows you the step-by-step calculations. The midpoint typically useful in geometry and our midpoint coordinate calculator uses the simple midpoint formula geometry to find the missing midpoint coordinates between the points.
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• Class 10 mathematics Coordinate Geometry - Distance Formula - Collinear Points - Non-collinear points - Concept and application In this video we use distanc...
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• coordinate system has been chosen: a point O, the origin, and two perpendicular lines through the origin, the x- andy-axes. A vectorV is determined by its length, j V and its direction, which we can describe by the angle θthat V makes with the horizontal (see figure 13.4). In this figure, we have realized V as the vector OP ~ from the origin ...
• Oct 19, 2019 · Students can solve NCERT Class 10 Maths Coordinate Geometry MCQs with Answers to know their preparation level. Class 10 Maths MCQs Chapter 7 Coordinate Geometry. 1. The distance of the point P(2, 3) from the x-axis is (a) 2 (b) 3 (c) 1 (d) 5. Answer/ Explanation. Answer: b Explaination: Reason: The distance from x-axis is equal to its ordinate ...
• Formulas used in these situations almost always involve r, theta, and phi and not x, y, and z. It should be noted that formulas for objects like the gradient, curl, and divergence have a different form in different coordinate systems. Always be careful to use the proper formula when dealing with these objects.
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• From a series of Coordinate Geometry Worksheets – By HelpingWithMath.com Similar to the above listing, the resources below are aligned to related standards in the Common Core For Mathematics that together support the following learning outcome:
• Using Cartesian coordinates on the plane, the distance between two points (x 1, y 1) and (x 2, y 2) is defined by the formula, which can be viewed as a version of the Pythagorean Theorem. Similarly, the angle that a line makes with the horizontal can be defined by the formula θ = tan-1(m), where m is the slope of the line. Midpoint formula
• All Formulas of Coordinate Geometry; General Form of a Line: Ax + By + C = 0: Slope Intercept Form of a Line: y = mx + c: Point-Slope Form: y − y 1 = m(x − x 1) The slope of a Line Using Coordinates: m = Δy/Δx = (y 2 − y 1)/(x 2 − x 1) The slope of a Line Using General Equation: m = −(A/B) Intercept-Intercept Form: x/a + y/b = 1: Distance Formula
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Shows work with distance formula and graph. Enter 2 sets of coordinates in the 3 dimensional Cartesian coordinate system, (X1, Y1, Z1) and (X2, Y2, Z2), to get the distance formula calculation for the 2 points and calculate distance between the 2 points.The definition extends to any object in n-dimensional space: its centroid is the mean position of all the points in all of the coordinate directions. [2] While in geometry the word barycenter is a synonym for centroid , in astrophysics and astronomy , the barycenter is the center of mass of two or more bodies that orbit each other.
Jul 08, 2016 · " 2B: Derive and use the distance, slope, and midpoint formulas to verify geometric relationships, including congruence of segments and parallelism or perpendicularity of pairs of lines." Students will write down the formula on the front of their foldable.
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Sep 23, 2020 · You can use the Distance Formula to find the length of such a line. This formula is basically the Pythagorean Theorem, which you can see if you imagine the given line segment as the hypotenuse of a right triangle. By using a basic geometric formula, measuring lines on a coordinate path becomes a relatively easy task.
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• Jan 23, 2020 · Check below the solved MCQs from Class 10 Maths Chapter 7 Coordinate Geometry: 1. If the distance between the points (2, –2) and (–1, x) is 5, one of the values of x is
• Improve your math knowledge with free questions in "Distance formula" and thousands of other math skills. Distance formula. 59F. Share skill.
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• Determine by distance formula, whether the given points are collinear : (-1, 2), (5, 0) and (2, 1). Let the given points be A (-1, 2), B (5, 0) and C (2, 1)then Therefore, the points A, B and C are collinear.
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# Ppt on distance formula coordinate geometry
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Geometry Notes Perimeter and Area Page 2 of 57 We are going to start our study of geometry with two-dimensional figures. We will look at the one-dimensional distance around the figure and the two-dimensional space covered by the figure. The perimeter of a shape is defined as the distance around the shape. Since
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# Calculating the Acceleration of Spacecraft Orbiting Earth
Since 1957 when the first artificial satellite was launched, thousands of spacecraft have orbited Earth. While some are placed into polar orbits for communications or weather purposes, others such as communications satellites orbit in geostationary orbits for longer-term performance.
Like throwing a ball into the air creates an unpredictable path towards its landing point, spacecraft are also designed to navigate Earth in an orderly manner by striking an equilibrium between forward momentum and gravitational pull.
## Orbits
Spacecraft are capable of staying in orbit by matching their velocity with Earth’s gravitational pull, and by expending enough energy to combat it. As speed increases, more energy must be expended to counter this force; most satellites usually only need fuel for two purposes: altering orbit or avoiding debris collisions.
A satellite’s orbit depends on several factors, including its height, eccentricity and inclination relative to Earth’s equatorial plane; as well as what view of our planet it offers.
Sun-synchronous (sun-SIN-kron-ous) satellites, which orbit at approximately the same time every day, are known as sun-synchronous satellites (sun-SIN-kron-ous). Their unique polar orbit enables them to observe Earth from all corners without being obscured by atmospheric obscurity; such orbits are popularly used for navigation satellites and remote sensing purposes. At present there are 56 geosynchronous orbit satellites primarily utilized for communications, satellite radio and remote sensing activities.
## Distances
Distance from Earth for any spacecraft depends on its orbit and can be affected by factors like its shape and size as well as initial speed from launch affecting final velocity and distance from planet.
Satellites in lower-Earth orbit (LEO) orbit closer to Earth than their higher counterparts and are frequently utilized for satellite imaging due to higher resolution capabilities. Furthermore, LEO orbit serves as home for the International Space Station where astronauts spend months at a time living and working aboard its facilities.
Middle orbit satellites (GEO and MEO) cover more of Earth as they pass over both poles. Their elliptical orbits allow them to track changes on the ground as Earth rotates, such as weather patterns.
Spacecraft traveling at extreme velocities are typically far removed from Earth, due to velocities exceeding its gravitational force. Still, they must know where they are in space using “dead reckoning,” which measures the round trip time it takes for radio signals from their spacecraft back to Earth and back again.
## Velocity
Each orbit a satellite undertakes requires it to maintain its trajectory at a steady velocity. Both its altitude and travel speed must be taken into consideration to calculate this velocity.
Spacecraft in low Earth orbit (LEO) must regain their speed four times each year to counter atmospheric drag, as satellites traverse thicker air during solar maxima which produces greater drag.
Gravity keeps satellites moving constantly around Earth, but too much slowdown or speedup may cause them to spiral away from its center rather than staying on track with its circle path. To prevent this from occurring, orbital velocity must remain stable – as illustrated below for geostationary orbits such as those used by television and communication satellites that remain over one point on its surface.
## Acceleration
Acceleration for satellites orbiting Earth depends on their mass and their distance from its center of gravity; to calculate this acceleration use the formula (m / r) x (2 pi / orbital period).
As an illustration of gravitation at work, let’s imagine we climb to the peak of a mountain and throw an outgoing cricket ball horizontally outward from there – gravity will draw it down in an arc rather than straight. Satellites orbiting their appropriate orbits are subject to similar effects of gravitation, whereby its force exactly balances their forward momentum to keep them flying around circular paths.
As satellites approach Earth, their orbit becomes increasingly affected by atmospheric drag that reduces their trajectory over time and brings them back closer to planet. That is why satellites like International Space Station reside in geostationary orbits rather than polar ones.
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# Dynamic occupancy models
This R Markdown Notebook accompanies a SEEC Toolbox Seminars given by Res Altwegg and David Maphisa at the University of Cape Town on 28 November 2019. Use this notebook together with the slides and the recorded seminar.
Occupancy is the state variable describing where a species occurs and is a key quantity in ecology and conservation. The problem with occupancy is that it is usually not directly observable because non-detection of a species at a site does not establish absence of that species at this site. Two earlier SEEC Toolbox seminars showed how occupancy models can be used to deal with this problem by explicitly accounting for the detection probability. Occupancy models assume closure, i.e. that a site is either occupied or not for the entire duration of the study. However, unoccupied sites do get colonised and species can go extinct from formerly occupied sites. Such changes in occupancy are often the main study focus, i.e. interest is in the dynamics rather than just static patterns. In this Toolbox Seminar, we show how dynamic occupancy models can be used to examine changes in occupancy, by estimating colonsation and local extinction.
### Simulation
We demonstrate the basic ideas using a simulated data set. This code is adapted from the unmarked Vignette on dynamic occupancy models, written by Marc Kéry and Richard Chandler. We simulate data for 250 sites that were visited twice per year for 5 years. Initial occupancy $$\Psi_1$$ was 0.6. We drew values for the $$\phi_t$$ (1 minus the extinction probability, $$\epsilon$$) from a uniform distribution U(0.5, 0.7), and the $$\gamma_t$$ (colonisation probabilities) from U(0.1, 0.2). Extinction was thus quite a bit higher than colonisation. We chose some arbitrary values for the detection probabilities, $$p_t$$.
M <- 250 # Number of sites
J <- 2 # num secondary sample periods
T <- 5 # num primary sample periods
psi <- rep(NA, T) # Occupancy probability
muZ <- z <- array(dim = c(M, T)) # Expected and realized occurrence
y <- array(NA, dim = c(M, J, T)) # Detection histories
set.seed(13973)
psi[1] <- 0.6 # Initial occupancy probability
p <- c(0.3,0.1,0.5,0.5,0.1)
phi <- runif(n=T-1, min=0.5, max=0.7) # Survival probability (1-epsilon)
gamma <- runif(n=T-1, min=0.1, max=0.2) # Colonization probability
# Generate latent states of occurrence
# First year
z[,1] <- rbinom(M, 1, psi[1])
# Initial occupancy state
# Later years
for(i in 1:M){
# Loop over sites
for(k in 2:T){
# Loop over years
muZ[k] <- z[i, k-1]*phi[k-1] + (1-z[i, k-1])*gamma[k-1]
z[i,k] <- rbinom(1, 1, muZ[k])
} }
# Generate detection/non-detection data
for(i in 1:M){ for(k in 1:T){ prob <- z[i,k] * p[k]
for(j in 1:J){ y[i,j,k] <- rbinom(1, 1, prob)
} } }
# Compute annual population occupancy
for (k in 2:T){ psi[k] <- psi[k-1]*phi[k-1] + (1-psi[k-1])*gamma[k-1] }
Then, we need to get the data into the correct format for the colext function in unmarked. The unmarked package has a really useful helper function for that, unmarkedMultFrame.
## Loading required package: lattice
## Loading required package: parallel
## Loading required package: Rcpp
## Loading required package: reshape2
yy <- matrix(y, M, J*T)
year <- matrix(c(' 01' ,' 02' ,' 03' ,' 04' ,' 05'), nrow(yy), T, byrow=TRUE)
head(yy)
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
## [1,] 0 0 0 0 0 0 0 0 0 0
## [2,] 0 1 0 0 0 0 0 1 0 0
## [3,] 0 0 0 0 0 0 0 0 0 0
## [4,] 1 0 0 0 0 0 0 0 0 0
## [5,] 0 0 0 0 0 0 0 0 0 0
## [6,] 0 0 0 0 0 0 0 0 0 0
dim(yy)
## [1] 250 10
simUMF <- unmarkedMultFrame( y = yy,
yearlySiteCovs = list(year = year),
numPrimary=5)
summary(simUMF)
## unmarkedFrame Object
##
## 250 sites
## Maximum number of observations per site: 10
## Mean number of observations per site: 10
## Number of primary survey periods: 5
## Number of secondary survey periods: 2
## Sites with at least one detection: 136
##
## Tabulation of y observations:
## 0 1
## 2242 258
##
## Yearly-site-level covariates:
## year
## 01:250
## 02:250
## 03:250
## 04:250
## 05:250
### Fitting a model with constant parameters
The first model we fit assumes constant $$\Psi$$, $$\gamma$$, $$\epsilon$$ and $$p$$.
# Model with all constant parameters
m0 <- colext(psiformula= ~1,
gammaformula = ~ 1,
epsilonformula = ~ 1,
pformula = ~ 1,
data = simUMF, method="BFGS")
summary(m0)
##
## Call:
## colext(psiformula = ~1, gammaformula = ~1, epsilonformula = ~1,
## pformula = ~1, data = simUMF, method = "BFGS")
##
## Initial (logit-scale):
## Estimate SE z P(>|z|)
## -0.00337 0.236 -0.0143 0.989
##
## Colonization (logit-scale):
## Estimate SE z P(>|z|)
## -1.7 0.195 -8.71 3.01e-18
##
## Extinction (logit-scale):
## Estimate SE z P(>|z|)
## 0.214 0.318 0.672 0.502
##
## Detection (logit-scale):
## Estimate SE z P(>|z|)
## -0.654 0.17 -3.86 0.000115
##
## AIC: 1556.162
## Number of sites: 250
## optim convergence code: 0
## optim iterations: 46
## Bootstrap iterations: 0
We see that the estimate for $$\Psi_1$$, occupancy in the first year, is -0.00337. How is that related to the 0.6 we used in the simulation? This estimate is on the logit scale so we have to use the inverse logit function to transform it back to the probability scale. This can easily be done ‘by hand’ but unmarked has helperfunctions that become very convenient when backtransformation is a bit more complicated. Using this function, it is also easy to get confidence intervals.
# parameter estimates
exp(-0.00337)/(1+exp(-0.00337))
## [1] 0.4991575
names(m0)
## [1] "psi" "col" "ext" "det"
backTransform(m0, type="psi")
## Backtransformed linear combination(s) of Initial estimate(s)
##
## Estimate SE LinComb (Intercept)
## 0.499 0.0591 -0.00337 1
##
## Transformation: logistic
confint(backTransform(m0, type="psi"))
## 0.025 0.975
## 0.3854436 0.6129562
We estimated occupancy in the first year, $$\Psi_1$$. But what about occupancy probabilities in subsequent years? They are derived parameters, calculated as $$\Psi_t = \Psi_{t-1} \times (1-\epsilon_{t-1}) + (1-\Psi_{t-1}) \times \gamma_{t-1}$$.
We see that we need the estimates for $$\gamma$$ and $$\epsilon$$.
# derived parameters
backTransform(m0, type="ext")
## Backtransformed linear combination(s) of Extinction estimate(s)
##
## Estimate SE LinComb (Intercept)
## 0.553 0.0786 0.214 1
##
## Transformation: logistic
backTransform(m0, type="col")
## Backtransformed linear combination(s) of Colonization estimate(s)
##
## Estimate SE LinComb (Intercept)
## 0.154 0.0255 -1.7 1
##
## Transformation: logistic
When fitting a dynamic occupancy model with the colext function, the output already has the derived parameter estimates for the $$\Psi_t$$. They are stored in the slot projected.mean.
m0@projected.mean
## 1 2 3 4 5
## unoccupied 0.5008437 0.6997875 0.7580149 0.7750571 0.780045
## occupied 0.4991563 0.3002125 0.2419851 0.2249429 0.219955
What about confidence intervals? unmarked has a function that uses parametric bootstrap to work out confidence intervals. You need to choose the number of iterations (B=...), which should definitely be more than 10 but this can take a lot of time… For a publication, you probably want closer to 1000 iterations.
# confidence intervals for projected occupancy
m0 <- nonparboot(m0, B = 10)
cbind(projected=projected(m0)[2,], SE=m0@projected.mean.bsse[2,])
## projected SE
## 1 0.4991563 0.03243629
## 2 0.3002125 0.02345218
## 3 0.2419851 0.01802916
## 4 0.2249429 0.01764645
## 5 0.2199550 0.01811982
We saw that extinction probabilities are higher than colonisation probabilities. Does this mean that occupancy will keep declining and the species will eventually go extinct? Not necessarily. As fewer and fewer sites are occupied, there are fewer sites left from which the species can go extinct. On the other hand, there number of sites that are unoccupied and could be colonised increases. With constant extinction and colonisation probabilities, the expected number of sites from which the species goes extinct and the expected number of sites that it colonises will eventually be equal to each other. So occupancy will reach an equilibrium of sorts. It is given by $$\Psi_E = \frac{\gamma}{\gamma + \epsilon}$$.
# equilibrium occupancy
# --------------------
backTransform(m0, type="col")@estimate/(backTransform(m0, type="col")@estimate + backTransform(m0, type="ext")@estimate)
## [1] 0.217891
So estimated occupancy in year 5 is actually rather close to the equilibrium occupancy probability and if $$\epsilon$$ and $$\gamma$$ remain constant, we would not expect occupancy to change much further over time.
### A model with year-dependent extinction, colonisation and detection probabilities
When we simulated the data, we used time-varying extinction, colonisation and detection probabilities. Let’s fit the model that corresponds to the one we used to simulate the data.
# Model with gamma, epsilon and p year specific
# ------------------------------------
m1 <- colext(psiformula= ~1,
gammaformula = ~ year,
epsilonformula = ~ year,
pformula = ~ year,
data = simUMF, method="BFGS")
summary(m1)
##
## Call:
## colext(psiformula = ~1, gammaformula = ~year, epsilonformula = ~year,
## pformula = ~year, data = simUMF, method = "BFGS")
##
## Initial (logit-scale):
## Estimate SE z P(>|z|)
## 1.31 0.879 1.49 0.137
##
## Colonization (logit-scale):
## Estimate SE z P(>|z|)
## (Intercept) -5.50 11.9 -0.4616 0.644
## year 02 1.72 19.7 0.0871 0.931
## year 03 3.10 11.9 0.2597 0.795
## year 04 5.07 12.0 0.4233 0.672
##
## Extinction (logit-scale):
## Estimate SE z P(>|z|)
## (Intercept) -1.026 1.72 -0.5968 0.551
## year 02 0.876 2.21 0.3965 0.692
## year 03 -0.127 1.83 -0.0695 0.945
## year 04 -5.166 42.84 -0.1206 0.904
##
## Detection (logit-scale):
## Estimate SE z P(>|z|)
## (Intercept) -1.32 0.264 -4.99 5.98e-07
## year 02 -1.66 0.551 -3.02 2.54e-03
## year 03 1.19 0.375 3.17 1.51e-03
## year 04 1.32 0.387 3.40 6.64e-04
## year 05 -1.91 0.545 -3.50 4.74e-04
##
## AIC: 1444.535
## Number of sites: 250
## optim convergence code: 0
## optim iterations: 68
## Bootstrap iterations: 0
What are the estimated occupancy probabilities over the five years ($$\Psi_t$$) and corresponding confidence intervals now?
# parameter estimates
m1@projected.mean
## 1 2 3 4 5
## unoccupied 0.2129186 0.4198092 0.6789418 0.6996572 0.4244612
## occupied 0.7870814 0.5801908 0.3210582 0.3003428 0.5755388
m1 <- nonparboot(m1, B = 10)
cbind(projected=projected(m1)[2,], SE=m1@projected.mean.bsse[2,])
## projected SE
## 1 0.7870814 0.14062734
## 2 0.5801908 0.25232466
## 3 0.3210582 0.04982608
## 4 0.3003428 0.06936172
## 5 0.5755388 0.16971123
Which model is better? For some reason, we can’t use the familiar AIC function but unmarked has its own functions that calculate the model selection table.
AIC(m0,m1)
## Error in UseMethod("logLik"): no applicable method for 'logLik' applied to an object of class "c('unmarkedFitColExt', 'unmarkedFit')"
models <- fitList(' psi(.)gam(.)eps(.)p(.)' = m0,
' psi(.)gam(Y)eps(Y)p(Y)' = m1)
(ms <- modSel(models))
## nPars AIC delta AICwt cumltvWt
## psi(.)gam(Y)eps(Y)p(Y) 14 1444.53 0.00 1.0e+00 1.00
## psi(.)gam(.)eps(.)p(.) 4 1556.16 111.63 5.8e-25 1.00
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http://nasawavelength.org/resource-search?facetSort=1&topicsSubjects=Mathematics&resourceType=Instructional+materials%3AActivity&instructionalStrategies=Generating+and+testing+hypotheses
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## Narrow Search
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Filters: Your search found 10 results.
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# Design Challenge - Deploying the Satellites' Antennae
This is an activity about using models to solve a problem. Learners will use a previously constructed model of the MMS satellite to determine if the centrifugal force of the rotating MMS model is sufficient to push the satellite's antennae outward,... (View More)
Audience: Middle school
# Earth Exploration Toolbook: Exploring Air Quality in Aura NO2 Data
In this chapter, students will explore relationships between air quality and population density using the image visualization tool, Google Earth. You will learn how to download NO2 data and analyze them to develop a conceptual understanding of how... (View More)
Audience: Middle school, High school
Materials Cost: Free
# MY NASA DATA: Correlation of Variables by Graphing
Activities in this lesson promote a fundamental understanding of relationships between graphed data. Sample graphs allow students to become familiar with interpreting data and to recognize relationships between variables. Additional microsets of... (View More)
Keywords: Scatter plot; Time plot
Audience: Middle school, High school
Materials Cost: Free
# Determining the Nature, Size, and Age of the Universe
In this lesson, students measure the size of several galaxies to reproduce a plot of Hubble's Law. The goal of this lesson is to give students the chance to simulate the process that led to the notion that the universe is expanding, provide insight... (View More)
Keywords: Data graphing
# Using Mathematical Models to Investigate Planetary Habitability: Activity C The Role of Actual Data in Mathematical Models
Students explore how mathematical descriptions of the physical environment can be fine-tuned through testing using data. In this activity, student teams obtain satellite data measuring the Earth's albedo, and then input this data into a... (View More)
Audience: High school
Materials Cost: Free per student
# Design Challenge: How do You Keep Things from Getting Too Hot?
This is a lesson about designing and building an effective sunshade for a model MESSENGER craft. Learners will build a model of MESSENGER. They will use a scientific approach to solve problems and work as a cooperative team. They will discover their... (View More)
# Design Challenge: How to Keep Items Cool in Boiling Water?
This is a design challenge about heat transfer and insulation. Learners will apply the scientific method to design and build a container that will keep items cool when placed in boiling water. They will practice collaboration in team-building and in... (View More)
Audience: High school
Materials Cost: Over \$20 per group of students
# Snow Goggles and Limiting Sunlight
This is a lesson about radiation and the use of the scientific method to solve problems of too much radiation. Learners will build snow goggles similar to those used by the Inuit (designed to block unwanted light, while increasing the viewer's... (View More)
# Modeling your Water Balance
Students create a physical model illustrating soil water balance using drinking glasses to represent the soil column, and explain how the model can be used to interpret data and form predictions. Using data from the GLOBE Data Server, they calculate... (View More)
# Time Warp
In this inquiry investigation, students conclude that the motion of the Earth is linked to the changes we observe such as the length of the day. Students learn about the reason behind the Earth's time zones. An optional water clock and sand clock... (View More)
Keywords: Time; Orbit; Time zones
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https://app.seesaw.me/activities/m9m5by/wednesday-math-number-pairs-of-5-composing-and-decomposing-5
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Student Instructions
### Wednesday Math: Number Pairs of 5(composing and decomposing 5)
Tap to begin Page 1: Click to view the video on number pairs of 5. Then Go back to Page 2: Click on the black triangle to watch the directions. Page 3: How many different ways can you make the number 5? Use the move tool to drag the yellow and red counters to the squares to make 5 six different ways. Then on the red line, use the tool to write how many red counters you used and on the yellow line, write how many yellow counters you used. Now use the :mic
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https://www.convertit.com/Go/BusinessConductor/Measurement/Converter.ASP?From=CLP&To=currency
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New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```Chilean peso on 4-28-2017 = 1.5009147457007E-03 currency (currency)``` Related Measurements: Try converting from "CLP" to ADP (Andorran peseta on 4-28-2017), ANG (Neth. Antillian guilder on 4-28-2017), ATS (Austrian schilling on 4-28-2017), AUD (Australian dollar on 4-28-2017), BEF (Belgian franc on 4-28-2017), CNY (Chinese yuan renminbi on 4-28-2017), DJF (Djiboutian franc on 4-28-2017), EUR (European Union euro on 4-28-2017), FIM (Finnish markka on 4-28-2017), GBP (British pound sterling on 4-28-2017), HUF (Hungarian forint on 4-28-2017), INR (Indian rupee on 4-28-2017), LKR (Sri Lankan rupee on 4-28-2017), MXN (Mexican peso on 4-28-2017), NLG (Netherlands guilder on 4-28-2017), PTE (Portuguese escudo on 4-28-2017), SGD (Singapore dollar on 4-28-2017), SHP (Saint Helenian pound on 4-28-2017), tick, VND (Vietnamese dong on 4-28-2017), or any combination of units which equate to "currency" and represent currency. Sample Conversions: CLP = .00267162 ANG (Neth. Antillian guilder on 4-28-2017), .00200783 AUD (Australian dollar on 4-28-2017), .00267162 AWG (Aruban guilder on 4-28-2017), .05557677 BEF (Belgian franc on 4-28-2017), .00149261 CHF (Swiss franc on 4-28-2017), .01034813 CNY (Chinese yuan renminbi on 4-28-2017), 4.43 COP (Colombian peso on 4-28-2017), .26674257 DJF (Djiboutian franc on 4-28-2017), .01024988 DKK (Danish krone on 4-28-2017), .00137771 EUR (European Union euro on 4-28-2017), .0011602 FKP (Falkland pound on 4-28-2017), .0011602 GBP (British pound sterling on 4-28-2017), .0011602 GIP (Gibraltar pound on 4-28-2017), .0200293 NAD (Namibian dollar on 4-28-2017), .00303608 NLG (Netherlands guilder on 4-28-2017), .00486942 PEN (Peruvian nuevo sol on 4-28-2017), .00582505 PLN (Polish zloty on 4-28-2017), .0200293 SZL (Swazi lilangeni on 4-28-2017), .05196352 THB (Thai baht on 4-28-2017), .00150091 USD (United States dollar on 4-28-2017).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks!
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https://www.universetoday.com/66662/circumference-of-the-moon/amp/
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Categories: Guide to Space
# Circumference of the Moon
[/caption]
The equatorial circumference of the Moon is 10,916 km. And the circumference of the Moon in miles is 6,783 miles. So, if you wanted to drive your lunar rover around the Moon and return back where you started, you’d need to travel 10,916 kilometers.
Need some comparison? The equatorial circumference of the Earth is 40,075 km. That makes the size of the Moon’s circumference about 27.24% the size of the Earth.
The Moon isn’t the largest moon in the Solar System – it only has an equatorial radius of 1,737.4 km. The largest moon is Jupiter’s moon Ganymede, with an equatorial radius of 2,634 km. That means Ganymede’s circumference is 16,550 km; bigger than the Moon’s circumference by about 5,634 km.
Want some more measurements?
The circumference of the Moon in meters: 10,916,000 meters
The circumference of the Moon in centimeters: 1,091,600,000 centimeters
The circumference of the Moon in feet: 35,813,648 feet
The circumference of the Moon in inches: 429,763,780 inches
We’ve written many articles about the Moon for Universe Today. Here’s an article about the full Moon, and here’s an article about the atmosphere of the Moon.
If you’d like more info on the Moon, check out NASA’s Solar System Exploration Guide on the Moon, and here’s a link to NASA’s Lunar and Planetary Science page.
We’ve also recorded an entire episode of Astronomy Cast all about the Moon. Listen here, Episode 113: The Moon, Part 1.
Fraser Cain
Fraser Cain is the publisher of Universe Today. He's also the co-host of Astronomy Cast with Dr. Pamela Gay. Here's a link to my Mastodon account.
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https://physics.stackexchange.com/questions/156533/why-arent-transformations-caused-by-measurements-unitary?noredirect=1
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Why aren't transformations caused by measurements unitary?
It is said, that when measured, a quantum system undergoes "wave function collapse", which is a non-unitary transformation.
Why?
The wave function is
$\Psi = \alpha \left|0\right\rangle + \beta \left|1\right\rangle$
where
$\left|\alpha\right|^2 + \left|\beta\right|^2 = 1$
The probabilities sum after measurement is still 1, for example, if system collapsed to $\left|0\right\rangle$, then
$\left|1\right|^2 + \left|0\right|^2 = 1$
For example, if function was
$\Psi = \frac{1}{\sqrt{2}} \left|0\right\rangle + \frac{1}{\sqrt{2}} \left|1\right\rangle$
the transformation was
$\left[ \begin{array}{ c c } \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & 0 \end{array} \right]$
Isn't this transformation unitary?
• a unitary matrix has determinant $\pm1$ so that matrix can't be unitary. It is even degenerate – Phoenix87 Jan 4 '15 at 15:12
• But that part can easily be fixed by taking the unitary $\left(\begin{smallmatrix}1&1\\1&-1\end{smallmatrix}\right)/\sqrt{2}$. The point is that the unitary depends on the state $\Psi$. – Norbert Schuch Jan 4 '15 at 15:21
• As I said in one of the comments, I expect you could manually compute a unitary transformation that gives you the right answer. But you would need to already know the right answer, so it's a bit useless... – SuperCiocia Jan 4 '15 at 15:35
• A physical note: unitary => invertable. Physically we know measurements are not always invertable. – zzz Jan 4 '15 at 16:36
• @Dims you will find more relevant/useful points on this topic here. – Phonon Jan 4 '15 at 17:46
No.
As long as your state is $|\Psi \rangle = \alpha \left|0\right\rangle + \beta \left|1\right\rangle$, then as you said $\alpha$ and $\beta$ need to satisfy $\left|\alpha\right|^2 + \left|\beta\right|^2 = 1$, so say $\alpha = \beta = 1/\sqrt 2$.
If you perform a measurement and find that the system in the $\left|0\right\rangle$ state, then the new wavefunction will be $\Psi =\left|0\right\rangle$. You can write it as $\Psi = \alpha \left|0\right\rangle$ but because of normalisiation $|\alpha|^2$ needs to be 1, so $\alpha$ must be either 1 or a pure phase factor.
You had to change the normalisation by hand (changing $\alpha$ from $1/\sqrt 2$ to $1$). A unitary transformation on $|\Psi \rangle$ would affect only the kets and not the constants. The time evolution of any wavefunction is governed by the Schrodinger equation which, when solved, is effectively a unitary transformation -- unitary transformations leave the norm unchanged. The time evolution due to performing measurements, however, is something entirely different and does not follow the formalism of the Schrodinger equation.
A unitary transformation leaves the norm unchanged, since the norm of $U|\Psi \rangle$ is $\langle \Psi |U^{\dagger}U|\Psi \rangle = \langle \Psi | \Psi \rangle$ if $U$ is unitary. In your specific case this is true, but only because you had to manually change the normalisation of the post-measurement wavefunction. If QM included measurement, then there should be a deterministic way of computing how $\alpha$ or $\beta$ would change. But it doesn't, so you need to re-normalise it.
• See my edit please. So, am I correct thinking, that unitarity means that ANY given function will conserve length, while during collapse it is impposible to formulate one? – Dims Jan 4 '15 at 14:59
• With regards to your question in the comment: yes, measurement does not obey any unitary transformation. You have to put the measurement by hand, at least in the current framework of QM. I have to say I don't know whether it is impossible to formulate such a matrix, maybe you can formulate one that only works for a specific case, but that'd be a bit useless because, in order to construct it, you'd need to know what the final state is so you'd already know the answer. – SuperCiocia Jan 4 '15 at 15:05
• With regards to your edit: When you write out the matrix form of an operator you should always specify what vector representation you are using for your state vectors. I am assuming $|0\rangle = \left( \begin{array}{ c } 1 \\ 0 \end{array} \right)$ and $|1\rangle = \left( \begin{array}{ c } 0 \\ 1 \end{array} \right)$. Well this transformation is not unitary. Remember unitary means transpose and complex conjugate. The transpose of your matrix is not equal to the original one. – SuperCiocia Jan 4 '15 at 15:08
• @Dims : which length? Unitarity of a transformation on a wave-function preserves probabilities. The probability of obtaining the value $+\hbar/2$ from an electron polarized with spin up in the direction $z$, but whose spin projection is measured on the axis $x$, remains $1/2$, whatever unitary transformation you do that doesn't change the spin-up polarization of the electron. – Sofia Jan 4 '15 at 15:12
• Unitary transformations preserve the norm, i.e. the "length" of a vector in a Hilbert space $\langle \Psi | \Psi \rangle$. The norm can be interpreted as a probability if normalised to 1, in which case there would be a physical reason as to why it needs to be constant through any realistic time evolution. – SuperCiocia Jan 4 '15 at 15:18
I assume that you are referring to the outcome of any observable $O$ acting on a state $\psi$, since the act of measuring something is interpreted as "averaging" such operators on some state. General observables are self-adjoint operators which need not be unitary. Perhaps the simplest example of an observable is a projection, i.e. an operator $P$ with the property that $P^*P = P$ (idempotent and self-adjoint). Suppose that, in your case, $P = |0\rangle\langle0|$. The outcome of a measurement of $P$ on your state $\Psi$, when repeated $N$ times, is $|\alpha^2|N$ times YES (and hence $(1-|\alpha|^2)N$ times no. Moreover the result of $P\Psi$ is $\alpha|0\rangle$, which isn't a normalised vector, simply because $P$ is not a unitary.
• You can't say "an observable $O$ acting on a state $\psi$". An observable is a physical quantity, that you measure. But you can use the terminology "operator $\hat O$. Now, a measurement is not averaging. In each single measurement of the operator $\hat O$ you obtain one of its eigenvalues. Measuring $\hat O$ on many particles identically prepared, you obtain statistics, and from it you can calculate different things, one of them being the average value. Another thing: from operators describing observables we don't expect unitary, but self-adjointness. Unitarity we expect from transformations. – Sofia Jan 4 '15 at 15:05
• I don't distinguish between observables and self-adjoint operators. For me they are the very same thing. Also a measure is usually performed on a statistical ensemble, and this corresponds to the action of a state on the observable, i.e. $\omega(O)$. I don't see how the single outcome of an experiment can tell anything about the state of a system, unless you know a priori that you are dealing with a pure state. Finally I don't quite get your last remark. – Phoenix87 Jan 4 '15 at 15:10
• Idle question while reading this: Since self-adjoint and unitary operators are in one-to-one correspondence, has the exponential of a self-adjoint projection any physical relevant meaning? – ACuriousMind Jan 4 '15 at 15:22
• @ACuriousMind That correspondence holds between self-adjoint operators and one-parameter groups of unitaries. For more general unitaries you need more self-adjoint operators and you only hit the connected component of the identity: for every unitary $u$ homotopic to 1 there are self-adjoint operators $h_1,\ldots,h_n$ such that $u = e^{ih_1}\cdots e^{ih_n}$ – Phoenix87 Jan 4 '15 at 15:34
• @ACuriousMind Don't know why but I can't edit comments right now. As for the exponential of a projection you simply get $e^{iPt} = 1 + (e^{it}-1)P$. I think this can be interpreted physically if $P$ is some generator of some transformation (e.g. a continuous symmetry). Then its exponential is the generated flow. – Phoenix87 Jan 4 '15 at 15:46
Answering the question in the title: a measurement process is intrinsically non-unitary. One way to see this is to realise that the unitarity of a process is equivalent to its being reversible.
A measurement process is intrinsically non-reversible, as some information gets lost. For example, measuring $$(|0\rangle+|1\rangle)/\sqrt2$$ in the computational basis, you can get either $$|0\rangle$$ or $$|1\rangle$$. The same outputs can be obtained measuring a different state, e.g. $$(|0\rangle-|1\rangle)/\sqrt2$$. This means that, given a measurement result (say $$|0\rangle$$), there is no way to know from what state it came from. Some information is lost in the process. It follows that the process cannot be described by a unitary matrix.
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1 bassA = \relative cis, {
2 cis,4 gis' | cis gis |
3 cis gis | cis r8 cis, ~ |
4 cis4 gis' | cis gis |
5 cis gis |
7 dis'8 ais r d8 ~ d8 a4 d,8 |
8 dis'8 ais r d8 ~ d2 |
9 }
11 bassB = \relative cis, {
12 \time 2/4
13 cis4 gis cis gis cis gis |
14 \time 3/4
15 cis8 e8 gis, gis r gis |
16 \time 2/4
17 a4 e a e a e |
18 a c8 b r f f r |
19 }
21 bassSpooky = \relative fis,, {
22 \time 2/4
23 fis8 fis fis fis fis fis fis fis fis fis fis fis
24 \time 3/4
25 fis8 a8 cis, cis r cis |
26 \time 2/4
27 d8 d d d d d d d d d d d
28 d4 f8 e r ais, ais r
29 }
31 bassC = \relative cis, {
32 \time 2/4
33 b4. b8-. | b4. b8-. | b4. b8-. |
34 \time 3/4
35 b8 d8 fis, fis r fis |
36 \time 2/4
37 g4. d8-. | g4. d8-. | g4. d8-. |
38 g4 ais8 a r dis, dis r
39 }
41 bassWeird = \relative cis, {
42 \repeat volta 2 {
43 \time 5/4
44 g4 d g d g8 a |
45 \time 9/8
46 b4 fis cis'8 d cis r ais |
47 } \alternative {
48 { b4 fis cis'8 d cis r a | }
49 { b4 fis cis'8 d cis r a | }
50 }
51 }
53 bassDance = \relative cis, {
54 \time 5/4
55 e2. fis2 | b,2. fis'2 |
56 \time 3/4
57 e2. | fis8-. r fis, r r4_"D.S." \bar "|." |
58 }
60 bassBeauty = \relative cis, {
61 \repeat unfold 3 {
62 \time 4/4 { b,1 | }
63 %% TODO: natural harmonics
64 \time 3/4 { r8 r r r r r | }
65 }
66 d-. d-. d-. cis-. cis-. cis-. |
67 }
69 bass = {
70 \set Staff.midiInstrument = #"electric bass (finger)"
71 \clef "bass_8"
72 \key cis \minor
74 %% Intro
75 \time 2/4
76 \relative cis, { r4 r8 cis, ~ | }
77 \relative cis, {
78 \repeat volta 2 \bassA
79 \alternative {
80 { cis4-. gis-. | }
81 { cis4 gis | cis gis | cis gis | cis gis |}
82 }
83 }
85 %% Solo
86 \tempo 4 = 150
87 \bassB \bassB
89 %% Spooky
90 \key fis \minor
91 \repeat volta 2 \bassSpooky
93 %% Half time feel
94 \key b \minor
95 \repeat volta 2 {
96 \inStaffSegno % start repeat
97 \repeat volta 2 \bassC
98 \bassWeird
99 \bassDance
100 }
102 %% Beauty
103 \repeat volta 2 { \bassBeauty }
105 %% Arpeggio
106 %% TODO
107 }
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https://en.m.wikiversity.org/wiki/Metaphysics_(Alternate_View)/Creation/Fields
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# Metaphysics (Alternate View)/Creation/Fields
## space
After the Michelson-Morley experiment, people believed space was a void.
Q: If space is a void, then is our physical universe finite or infinite in size?
Q: If it is finite, then what is on the other side?
Q: If it is infinite, then is matter distributed equally throughout this infinite physical universe?
Q: If the answer is yes, then one must assume stars have been shinning for billions of years. If this is true, then why is the night sky so dark? This is Olbers’ Paradox. Heinrich Wilhelm Matthäus Olbers (1758-1840) was a German astronomer, physician and physicist.
Q: If the answer is no, then why is all the matter located in one place?
## gravity
Look at a sheet of polar graph paper. In the center is a mass of one. The accepted view has the circles representing the field of gravity that eminates from this mass.
Q: What happens to this gravity field when this mass of one converts to energy?
If the gravity field blinks out of existence, then its demise is faster than the speed of light. If the field still exists, then it is possible to have a gravity field without a mass.
## BASIC CYCLE
(An alternative view by Calgea 19:03, 16 June 2007 (UTC) 2007-6-17)
Creation (Alternate View) Before the Big Bang
### Space and Gravity
Look again at a sheet of polar graph paper. This alternative view says the circles represent expanding spheres of Space. These spheres expand from a mass center. The radial lines represent the field of Gravity. At the limit of Space, the points of Space convert to incoming lines of Gravity. At a mass center, opposing, incoming lines of Gravity convert to expanding spheres of Space. This relationship between Space and Gravity is the Basic Cycle. The cycle between one Big Bang and the next Big Bank is the Fundamental Cycle.
Rule 0: At a mass center in an expanding physical universe, opposing, incoming lines of Gravity convert to expanding spheres of Space. Unopposed, incoming lines of Gravity convert to Linear-motion.
When viewed from The Universe, the size of our physical universe is zero. From inside our physical universe, the size appears infinite with matter located near the origin of the Big Bang. The Big Bang generated vast amounts of expanding spheres of Space.
The most numerous items before the Big Bang were the centers trying to become Shells, that is, mass centers. These centers are low energy quanta of light. They are everywhere. However, as they move outward toward the surface of Space, their energies fade, and they become points of Space. In a similar manner, a starship heading for the surface of Space would likewise disintegrate and become part of the field of Space and return as Gravity. So in this cyclic sense, the field of Space is infinite.
Return to the polar graph paper and let the mass convert to Space and other fields at its center. In this case, the incoming lines of Gravity continue to remain perpendicular to the expanding spheres of Space and eventually end up at a center (mass or quanta) or at the original center of the Big Bang.
## ENERGY
The creation article proffered a set of fields including Space to define our physical universe. The other eight nonSpace fields are energy fields: Gravity, Ray, Kone, Electro, Magno, Spin, Shell, and Linear-motion.
At one time, people believed matter could not be created or destroyed. Later, people believed energy could not be created or destroyed. Now it seems, Space cannot be created or destroyed; it can only be temporarily stored in energy fields.
## GRAVITATION LAW
Newton's Universal Law of Gravitation states that any two objects exert a gravitational force of attraction on each other. The direction of the force is along the line joining the objects. The magnitude of the force is proportional to the product of the gravitational masses (m) of the objects, and inversely proportional to the square of the distance (d) between them.
• ${\displaystyle F=G(m_{1})(m_{2})/d^{2}}$
• where G is a constant.
Why is this equation correct?
Compressive force of Gravity
Consider the case where there are two mass centers a zillion light years apart. Apply rule zero. Opposing lines of Gravity enter each mass center and convert to expanding spheres of Space. Between these mass centers, there is a missing line of Gravity. The other mass center intercepted it. This unopposed incoming line of Gravity converts to Linear-motion, and the two mass centers move toward each other. In this case, the force of Gravity is a compressive force and not an attractive force. There is no ‘Action at a distance’.
Product of masses
Consider the case where there is a mass with two centers and a mass with three centers. Between these masses, there are six missing lines of incoming Gravity. Therefore, the force of Gravity between two objects is a product of the number of mass centers.
Field of Gravity
Look again at the sheet of polar graph paper. Place a mass of one at the center of the sheet and a new mass of one on a radial line just inside the outer circle. Each circle represents an expanding sphere of Space. The incoming lines of Gravity are perpendicular to these spheres. These spheres act as lenes and focus the lines of Gravity into the mass of one located at the center of the sheet. However, because of this focusing effect, some incoming lines of Gravity enter the center of the new mass. These are additional unopposed lines of Gravity, and they convert to Linear-motion. The same focusing effect happens to both masses. When the two masses touch, half the incoming Gravity goes into one mass center and half goes into the other mass center.
## RAY
Ray is energy in one dimension. Ray is everywhere. In space, Ray is a nutrino. In the sky, Ray is lightning. On Earth, Ray is electricity.
Tesla Coil's Ray
Let’s start with a Tesla coil. It’s named after Nikola Tesla (1856-1943). If one touches the top of an active coil, one can draw an arc. Directors use this idea in horror movies to have arcs of lightning hit actors and make their hair stand on end. Ray leaves the coil and hits the mass centers in the air. Ray converts to light at these centers, and one sees a harmless arc.
In the sky, Linear-motion converts to angular motion, that is, Spin, in the neutrons of Oxygen molecules. This buildup of energy continues until one or a few molecules convert their excess Spin to Ray. Ray usually fires at a mass center in the Earth. This release of Ray sets off a domino effect that travels with the speed of light.
Battery's Ray
During this process, Ray fires at other mass centers and converts to Linear-motion and light. Humans hear the air movement, thunder, and see a flash of lightning.
A similar event occurs with a battery. In this case, a chemical change occurs converting the orbital energy, Spin, to Ray. If the switch is open, then the change is slow and Ray leaks away. If the switch is closed, then Ray completes the loop and enhances the chemical reaction to produce more Ray. Volts are a measure of the orbital energy before and after a load like a lamp. Opening the switch produces an arc as in the Tesla coil as Ray continues to leave the battery.
The radiometer provides a different example. Shine an incandescent light on the vanes and the black vanes will move away. Shine a fluorescent light on the vanes and nothing happens. In the fluroescent tube, Ray moves from one end of the tube to the other end. In the incandescent light, the hot filiment emits Ray, and Ray hits the vanes. Shade the radiometer so the black vanes are on one side of the shade. Shine a light on the white vanes, and they will not move. In the white vanes, Ray converts to Spin and the orbiting neutrons increase their rotations. This excess Spin reconverts to Ray which leaves via the pedestal. In the black Carbon vanes, there are few if any orbiting neutrons. Ray converts to Linear-motion.
Testing Suggestion
Hold the north pole of a strong bar magnet near the black vanes of a nonmetalic radiometer.
## KONE
Let’s return to the Michelson-Morley experiment and think of a square box. In their experiment, they had a light source on the left that emitted light, which hit a semi-silvered mirror in the center of the box. Half of the light went straight ahead to hit a flat mirror on the right. Half of the light went up to the left to hit a flat mirror. The returning beams combined at the semi-silvered mirror and proceeded to a light detector at the bottom. Their device became an interferometer and was used to detect aether flow. They built their interferometer on top of a block of marble, which floated in a pool of mercury. They failed to detect aether flow thus proving space was a void.
A problem with this experiment was its pool of Mercury. Mercury has many mass centers, and these centers generated a vast number of expanding spheres of Space. In a sense, they conducted their experiment on a three dimensional artesian well of Space.
To understand Kone, start over and adjust one of the flat mirrors so as to produce total light interference. Do not reconduct their experiment. Just ask yourself: ‘Where is the energy’? Energy cannot be created or destroyed. It must be oscillating in fields other than the electro-magnetic field.
Testing suggestion
Duplicate the above total interference interferometer. Use the output of one device as the input to this duplicate device and recover the orginal light.
## ELECTRO
Electro-Magnetic wave
One associates the Electro field with the Electro-Magnetic (E-M) spectrum. This E-M spectrum starts with a center. Adding energy to this center produces a quantum of light. A stream of these moving quanta is the E-M spectrum. It is the stream that has wave properties. This spectrum includes radio and light waves. At the high energy end of this spectrum is Shell.
Dropping a letter from E and M provides the D-L spectrum or Dark-Light spectrum. Kone replaces Electro in this spectrum. Human eyes did not evolve to detect this D-L spectrum.
Sun spots are dark energy spots on our sun. A sun spot is energy in the D-L spectrum. It is not difficult to imagine a D-L sun and a solar system full of dark matter.
## MAGNO
Magno and Bar Magnet
Most people are familiar with a bar magnet and its magnetic field. In this article, Magno is this magnetic field. A bar magnet has a north and a south pole. Most people also know that like poles repel and unlike poles attract. Why?
Funnel effect of Magno
### Magnetic Funnel
To understand the why of this phenomenon, notice that this magnetic field produces a funnel at each end. Lines of Gravity enter these funnels and convert to expanding spheres of Space. Opposing lines of Gravity enter from all sides, but it is only the incoming lines at the poles that are of interest.
Compressive Force
Now place two bar magnets near each other with a north pole near a south pole. Notice that between the poles, the funnel all but disappears. The incoming lines of Gravity entering through the end funnels are unopposed and convert to Linear-motion. The two magnets move toward each other. People believe opposite poles attract, but that is not the case.
It is not surprising that the equation for the compressive magnetic force is similar to Newton’s force of Gravity equation.
## SHELL
Three sizes of Shell
A Shell is a Spin neutralized field. It is the upper limit of the E-M spectrum. It looks like a small ping-pong ball. Shell comes in three sizes: Normal, Small, and Smallest, AKA: Atom, Neutron, and Subneutron or Black Hole.
The combination of a normal Shell and a small Shell is known as a Hydrogen atom. Other combinations produce the elemets in the Periodic Table. The combination of a small Shell and a smallest Shell is a subHydrogen atom. Other combinations produce the subelements in the subPeriodic Table.
Many Hydrogen atoms produce a star like our sun. Gravity presses many Hydrogen atoms and molecules together so that they cannot move. They unSpin and convert to Light, Space and other energy fields.
Many subHydrogen atoms with Gravity produce a neutron star in a similar manner.
Many subNeutrons produce a Black Hole. Objects falling into a Black Hole unSpin and reconvert to Space.
### Hydrogen
Hydrogen & Deuterium
A Hydrogen atom comprises a neutron within a normal Shell. While most oppossing, incoming lines of Gravity convert to expanding spheres of Space, a few convert to Ray as they did during the compression period of our physical universe. Ray moves to the surface of Shell and converts to other fields. The surface disturbance is known as an electron. As Ray rotates, the electron appears to be in orbit. A viewer is on the other side of Ray and senses a negative electron. A viewer is on the same side of Ray for the central neutron and senses a positive charge on the central neutron and names it a proton.
The atomic weight of Hydrogen is about 1.008. Atomic weght is a time average count of the number of centers in an atom. Most of the time, the center of the neutron and the center of the normal Shell are co-located and count as one. When the centers are not co-located, they count as two.
Hydrogen molecule & Unpinned intersection
The atomic weight of Deuterium is about 2.014. Deuterium has an orbiting neutron.
The Hydrogen molecule comprises two Hydrogen atoms. While spinning at the speed of light, they can appear to each other to have poles. Gravity compresses their fields like pushing two magnets together. This molecule has an unpinned intersection.
### Carbon
Carbon atom
Carbon has an atomic weight of about 12.01. Carbon comprises six deuterium atoms in a circle. There is a neutron in each center of the normal Shells and a neutron in each intersection.
A diamond occurs when there are four horizontal Shells and a Shell on the top and a shell on the bottom. One needs pressure to force the many unpinned intersections.
Touch
Gas is the term one uses when there are few if any unpinned intersections.
Liquid is the term one uses when there are some unpinned intersections.
Solid is the term one uses when there are many unpinned intersections.
Touch is an awareness of Shells contacting other Shells.
### Helium
Helium atom
Helium has an atomic weight of about 4.0026. There are two Shells and four neutrons. Two neutrons are co-located at the center of the Shells, and two neurons are in the intersection of the Shells. This configuration inhibits interaction with other Shells.
Helium starts with four Hydrogen atoms. Two Shells unSpin during the fusion process.
### Oxygen
Oxygen atom
Oxygen has an atomic weight of just less than 16. As shown, there are two orbits. One contains eight neutrons, and one contains six neutrons. The fifteenth neutron acts as an anchor to prevent the orbits from flipping and remains in or near the center of the normal Shell.
When two Oxygen atoms combine with Carbon, there is an Oxygen atom on each side. Six neturons from each Oxygen atom pin the Oxygen atom to the Carbon atom. We sense the release of their orbital energies as heat.
Plants use Ray from the sun to resupply this energy and release the Oxygen into the air.
### Lithium
Lithium atom
Lithium has an atomic weight just less than 7. As shown, there are seven centers. So, an end neutron must occupy the center of the normal Shell often enough to reduce the count to six.
In this article, the interest in Lithium is it is a gyro in a normal Shell.
### Iron
Iron has an atomic weight of about 55.85. Iron comprises four Lithium atoms with four pinned intersections. Each intersection contains a gyro from a no longer existing Lithium atom.
Iron atom
Each of the four pinning gyros has seven neutrons to account for twenty-eight centers. Each of the other four gyros has six neutrons. There are four more centers, one in each of the four normal Shells. This brings the total to fifty-six. The count drops to fifty-five when one of the neutrons is co-loacated in its Shell’s center.
How can one Iron atom have a magnetic field and another Iron atom not have one?
Imagine all of these gyros pointing to the geometric center of the Iron atom. Lines of Gravity enter the many centers and convert to Rays. These Rays fire at the geometric center of the Iron atom. At this geometric center, the opposing Rays convert to Magno. The Magno lines curve upward and around to re-enter the bottom of the Iron atom. They then merge to form a perpendicular Ray, which leaves the atom.
Hitting a magnet reorients the gyros and reduces the strength of the magnetic field.
Note: The above speculations are not the accepted views.
Births Birth of our solar system, Life, Earth
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# Complete Riemannian manifold $M,$ $C$ closed subset of $M.$ Then for $p\in M$ there is $p^{'}\in C$ that is as close to $p$
I'm trying to prove the next:
If $C$ is a closed subset of a complete Riemannian manifold $M,$ then for each $p\in M,$ there is a point $p^{'}\in C$ that is as close to $p$ as possible.
So I'd like to prove that there is $p^{'}\in C$ such that $d(p,p^{'})=d(p,C):=\displaystyle\inf_{q\in C}d(p,q).$
I'm stuck proving this. If $C$ were bounded, the Hopf-Rinow theorem would imply the compactness of $C$ and we can get such point utilizing the sequence criteria of compactness.
However, $C$ only is closed. Then, since $(M,d)$ is metric space, then the distance from $p$ to $C$ is a continuous function. If we consider the boundary of $C,$ $d(p,C)$ attains a minimum in such set, but, what about the rest of $C?$
Is there another easier way to proof this?
Any kind of help is thanked in advanced
• Pick $q$ in $C$ and consider only the points $p'$ in $C$ with $d(p,p') \leq d(p,q)$. – Catalin Zara May 30 '18 at 2:06
• Thanks to answer @CatalinZara. Let $A:=\{p^{'}\in C:d(p,p^{'})\leq d(p,q)\}$ where $q\in C$ and $p\in M$ fixed point. Such set is closed because of continuity of Riemannian metric, and is bounded because $d(p,q)$ is fixed positive number with $q$ fixed, right? Then $A$ is compact for Hopf-Rinow and we have the case mentioned above. But why this is enogh to finish the proof? – Squird37 May 30 '18 at 2:19
• Or I don't understand your idea @CatalinZara? – Squird37 May 30 '18 at 2:26
As suggested in the comment, one can consider the
$$C_R = C\cap B_R(p).$$
For $R$ large enough, $C_R$ is nonempty. Since $M$ is complete, the Hopf-Rinow theorem implies that $B_R(p)$ (and thus $C_R$) is compact. Thus there is $p' \in C_R$ so that $$d(p,p') = \inf_{s\in C_R} d(p,s).$$
Since $d(p, p')\le R$, we also have
$$d(p,p') = \inf_{s\in C} d(p,s).$$
• Thanks @JohnMa. Now everything is clear. – Squird37 May 30 '18 at 2:57
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Tuesday, May 4, 2021 11:41:31 PM
Surface Area Of Composite Figures Worksheet Pdf
File Name: surface area of composite figures worksheet .zip
Size: 2374Kb
Published: 05.05.2021
How many faces are there? What shapes are the faces?
perimeter of composite figures worksheet pdf
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Maestro and other valuable worksheets we find area of composite shapes below. Them to use math is that the area shapes worksheet pdf worksheets available for 6th. Word problems from calculating area composite shapes pdf worksheet will be able to a page! Use khan academy does perimeter utilizing the area shapes worksheet has a rectangular components. Most important geometric shapes you need for 7th and label them without using these together to derive and circumference of composite worksheet pdf worksheet. Required for 6th grade children build good study habits and 8th grade enhance practice worksheets available for some complex figures and area composite shapes word.
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Area Of Composite Shapes Worksheet Pdf
In this worksheet, we will practice finding the surface area of a composite solid using the formulas for lateral or total surface areas of a single solid. The given figure is made by placing a cube of side length 13 cm on the top of another cube of side length 18 cm. Find its surface area. If a part of a cube, whose edge length is 7 cm , is cut to form a cuboid with side lengths of 3 cm , 4 cm , and 4 cm , find the surface area of the remaining part of the cube. The object shown is made from two cubes, one of side 3. If you must paint this object including the base of the combined solid , what area do you report?
Compose or decompose the compound shapes and acquaint with finding the surface area of composite figures with this collection of printable worksheets.
Area Of Composite Shapes Worksheet Pdf
Distribute copies of the Composite Figures activity sheet. State your answers correct to 2 decimal places where appropriate. We use formulae to calculate area. Some of the worksheets below are Area Of Composite Figures Worksheets, use your knowledge of the area of rectangles, parallelograms, and triangles to find the area of a composite figure. Composite Figures Area And Perimeter.
The shapes that we see around us are often a combination of two or more shapes. Count on our surface area of composite figures worksheets for an adequate practice in finding the surface area of non-overlapping rectangular prisms, compound shapes made of cubes, cones, cylinders, hemispheres, prisms, pyramids, and circumscribed figures with solids within solids. Follow the step-by-step process of decomposing, finding the SA of individual shapes, adding their surface areas, and subtracting the area of common parts and you will be good to go!
Pinyon Script While we talk about Area Compound Shapes Worksheet, scroll the page to see various related images to complete your ideas. It is easy to work out the area of composite or compound shapes - just break them up into basic shapes and go from there. What is included in this area of compound shapes worksheet?
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5 Comments
QuerubГn C. 05.05.2021 at 15:07
With many such interesting shapes for your 7th grade and 8th grade students to determine the surface area of compound solids, these pdf worksheets are a class.
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2) Identify what parts of each figure are on the surface of the solid. 3) Calculate the surface area of composite shapes. WHAT YOU'LL ikafisipundip.orgasmath.com/protected/content/ipe/grade%/07/g7_07_pdf. Solve the following.
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surface area of a composite solid? Work with a partner. You are manufacturing scale models of old houses. a. Name the four basic solids of this composite figure.
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Bluebulb Projects presents:
Enter a measurement to see comparisons
Equivalents in other units
How fast is 503.58 centimeters per minute?
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It's about one-and-three-tenths times as fast as a Sloth. In other words, the speed of a Sloth is 0.77 times 503.58 centimeters per minute. (for Brown-throated three-toed sloth, Bradypus variegatus)The three-toed sloth moves along the ground at an average speed of 400 centimeters per minute. Long thought to be lengthy sleepers, a 2008 study concluded that sloths sleep an average of only 9.6 hours per day. It's about one-fifth as fast as an Iceberg. In other words, the speed of an Iceberg is 5 times 503.58 centimeters per minute. (a.k.a. Berg) (Newfoundland iceberg average)Moved by ocean currents and wind, icebergs can drift at speeds of about 3,000 centimeters per minute. The largest iceberg ever recorded was a found near Baffin Island, Nunavut and was estimated to be nine billion metric tons. It's about one-fifteenth as fast as Walking Pedestrians (in Manhattan). In other words, the speed of Walking Pedestrians (in Manhattan) is 15 times 503.58 centimeters per minute. (Manhattan; average speed; 8,978 person-sample)A 2006 Study by the New York City Department of City Planning found that pedestrians in that city walk at an average rate of 7,800 centimeters per minute. Pedestrians wearing headphones, the study went on to find, walk at a slightly faster 8,500 centimeters per minute It's about one-twenty-fifth as fast as Michael Phelps. In other words, 503.58 centimeters per minute is 0.043207 times the speed of Michael Phelps, and the speed of Michael Phelps is 23.144 times that amount. (at the Beijing Olympics, 2008; 200 m freestyle) (a.k.a. Michael Fred Phelps) (swimmer; 1985-)Setting a world record, Michael Phelps swam the 200 m freestyle in 1:42.96 for an average speed of 11,655 centimeters per minute. Phelps would go on to win nine gold medals individually in the 2008 Olympics - more than all but eight of the competing nations. It's about one-thirtieth as fast as a Crocodile. In other words, the speed of a Crocodile is 32 times 503.58 centimeters per minute. (American Crocodile, Crocodylus acutus) (swimming speed)An American crocodile can reach speeds in the water of up to 17,000 centimeters per minute. On land, larger crocodiles can "gallop" when fleeing danger at speeds of up to 33,000 centimeters per minute. It's about one-seventy-fifth as fast as a Bull. In other words, the speed of a Bull is 77 times 503.58 centimeters per minute. (for animals involved in the Running of the Bulls, a.k.a. Encierro, San Fermin, Pamplona, Spain) (herd average speed)The herd of the annual Encierro in Pamplona, Spain runs at an average speed of 40,000 centimeters per minute. The Encierro is run annually from July 7th through July 14th and involves 42 bulls, 77 oxen, and an estimated 17,000 runners over the course of the event. It's about one-ninetieth as fast as Noah Ngeny. In other words, 503.58 centimeters per minute is 0.01108 times the speed of Noah Ngeny, and the speed of Noah Ngeny is 90.25 times that amount. (in Rieti, Italy; 1999) (sprinter; 1978-)Setting a world record at the Rieti Grand Prix in 1999, Noah Ngeny ran 1,000 m in 2:11.96 for an average speed of 45,470 centimeters per minute. According to some reports, Ngeny did not begin running competitively until just three years before setting the record.
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+ Text Only Site
+ Non-Flash Version
+ Contact Glenn
Pitot Tube Activity
If so instructed by your teacher, print out a worksheet page for these problems.
DEFINITIONS: Define the following terms.
Force (F) -
Mass (m) -
Pressure (p) -
Velocity (V) -
Density (r) -
UNIT ANALYSIS: Complete the table giving the correct unit of measurement in both the English and the Metric systems. You will use this information when solving the problems shown below.
QUANTITY ENGLISH UNIT METRIC UNIT LENGTH ______________ ______________ AREA ______________ ______________ VOLUME ______________ _______________ FORCE ______________ _______________ MASS ______________ _______________ PRESSURE ______________ _______________ VELOCITY ______________ _______________ DENSITY ______________ _______________
*NOTE: For practical purposes, velocity is measured in KNOTS. One knot = 1.69 feet/second.
PROBLEMS: Solve each problem. Make sure all units agree!!
1. Find the total air pressure detected by the pitot tube for a Cessna 182 airplane flying at a pressure altitude of 10,000 feet (static pressure of 1455.6 lb/ft2 ) and a speed of 120 knots if the air density is 0.00162 slugs/ft3.
2. Find the difference in pressure (pt-ps) for the same conditions described above if the airplane speeds up to 160 knots.
3. If the total air pressure for the same Cessna 182 is measured at 1478.7 lb/ft2, what is the new velocity in knots of the airplane?
Related Pages:
Standards
Worksheet
Lesson Index
Aerodynamics Index
+ Inspector General Hotline + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act + Budgets, Strategic Plans and Accountability Reports + Freedom of Information Act + The President's Management Agenda + NASA Privacy Statement, Disclaimer, and Accessibility Certification Editor: Tom Benson NASA Official: Tom Benson Last Updated: Thu, May 13 02:38:26 PM EDT 2021 + Contact Glenn
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## Search found 217 matches
Wed Jan 09, 2013 7:49 pm
Forum: Geometry
Topic: IMO 2007 Problem 4
Replies: 4
Views: 8931
### Re: IMO 2007 Problem 4
Sun Jan 06, 2013 6:15 pm
Forum: Introductions
Topic: Hello Everybody
Replies: 3
Views: 2365
### Re: Hello Everybody
Hello everyone.I am Zadid Hasan from the town of Mymensingh,Bangladesh.I joined this forum on Oct 27,2011.My present favourite color is BLUE.
Btw welcome charityst021.
Tiham,What do you mean your forum?
Having post number too high doesn't make it yours.
Fri Jan 04, 2013 10:14 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 58748
### Re: Secondary and Higher Secondary Marathon
Some moderator edit this.
Thu Jan 03, 2013 2:44 pm
Forum: Number Theory
Topic: IMO Shortlist 2007 N1
Replies: 2
Views: 1273
### Re: IMO Shortlist 2007 N1
My steps:
Thu Jan 03, 2013 2:41 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 58748
### Re: Secondary and Higher Secondary Marathon
Here goes another easy one.
Problem $23$:Find all continuous functions from the set of real numbers to itself satisfying $f(x + y) = f(x) + f(y) + f(x)f(y)$.
Thu Jan 03, 2013 3:08 am
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 58748
### Re: Secondary and Higher Secondary Marathon
It seems you folks lack stamina for a marathon.But resting is enough already i suppose.Let the marathon commence again with an easier problem. Problem $22$:Let $CH$ be the altitude of triangle $ABC$ with $∠ACB = 90°$. The bisector of $∠BAC$ intersects $CH$, $CB$ at $P$, $M$ respectively. The bisecto...
Fri Dec 21, 2012 3:52 am
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 58748
### Re: Secondary and Higher Secondary Marathon
Problem $21$: Determine all positive rationals $x,y,z$ such that $x+y+z$,$xyz$,$\frac{1} {x}+\frac{1} {y}+\frac{1} {z}$ are all integers.
Fri Dec 21, 2012 12:47 am
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 58748
### Re: Secondary and Higher Secondary Marathon
This sequence must be periodic with period at most $9$.Actually this is periodic with period $6$.So $A(30)=A(6)=77777770000000$
Wed Dec 19, 2012 11:05 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 58748
### Re: Secondary and Higher Secondary Marathon
Oops.Edited.
Wed Dec 19, 2012 3:40 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 64124
### Re: IMO Marathon
To revive the marathon here goes a problem.
Problem $14$:Let $a$ and $b$ be positive integers such that $a^n +n$ divides $b^n +n$ for
every positive integer $n$. Show that $a = b$.
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# Painting
To paint the pool with dimensions: 2 meters depth, 3m x 4m we bought paint to 50 meters square. How many "paint" will be a waste?
Result
x = 10 m2
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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## Next similar examples:
1. Painting a hut
It is necessary to paint the exterior walls of hut whose layout is a rectangle of 6.16 m x 8.78 m wall height is 2.85 meters. Cottage has five rectangular windows; three have dimensions of 1.15 m x 1.32 m and two 0,45 m x 0.96 m. How many m2 is necessary
2. Basen
How many square meters of tiles we need to tile the walls and floor of the pool 15 meters long, six meters wide and two meters?deep
3. Field on plan
Plan has a scale of 1: 2500.Determine dimensions in centimeters which will have on plan field with a length of 310 meters and a width of 182.5 meters.
4. Cardboard
How many m2 of cardboard are needed to make the cuboid with dimensions 40 cm 60 cm and 20 cm?
5. Glass aquarium
How many m2 of glass are needed to produce aquarium with bottom dimensions of 70 cm x 40 cm and 50 cm high?
6. The carpet
How many meters of carpet 90 cm wide need to cover floor room which has a rectangular shape with a lengths 4.8 m and 2.4 m if the number of pieces on the carpet is needed to be lowest?
7. Metal rails
Dad needs to improve the edges of the wooden boxes - to be reinforced with metal rails. How many cms of rails will he need if the box has the shape of a prism with the length of the edges 70cm 70cm and 120cm?
8. Surface area
Calculate the surface area of a four-sides 2-m high prism which base is a rectangle with sides 17 cm and 1.3 dm
9. Tiles path
A rectangular park measures 24mX18m. A path 2.4m wide is constructed all around it. Find the cost of paving the path with cement tiles of size 60cm X 40cm at the rate of \$5.60 per tile ?
10. Aquarium II
Calculate how much glass we need to build an aquarium with a rectangular shape with base 70 cm × 70 cm and a height of 70 cm, if the waste is 2%. Aquarium haven't top glass.
11. Pipe
How long is the pipe with an outside diameter of 1.44 m if his coloring consumed 44 kg of color. 1 kg of color coverage is 9 m2.
12. Greenhouse
Garden plastic greenhouse is shaped half cylinder with a diameter of 6 m and base length 20 m. At least how many m2 of plastic is need to its cover?
13. Bathroom 2
A bathroom is 2.4 meters long and 1.8 meters wide. How many square tiles 1 dm on each side are to be used to cover it?
14. Florist's
The florist got 72 white and 90 red roses. How many bouquets can bind from all these roses when each bouquets should have the same number of white and red roses?
15. Log
Worker cut his thick log to 6 pieces for 30 min. How long he cut log to 12 pieces?
16. Equation
?
17. One-third
A one-third of unknown number is equal to five times as great as the difference of the same unknown number and number 28. Determine the unknown number.
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Friday
May 6, 2016
# Posts by matt
Total # Posts: 2,126
Spanish
Esta mañana nos (1) _____ temprano en carro sin mapa. Al (2) _____ de la ciudad, Jorge (3) _____ al empleado de una gasolinera dónde (4) _____ el carril para acceder a la autopista. El empleado no (5) _____ contestarle porque no (6) _____ a dónde (7) _____...
September 25, 2015
SPAnish
Modifique el texto por cambiar los verbos italizados de lo familiar (informal) a lo formal Prepara (1) _____ para salir de viaje el lunes por la mañana. Para ello, haz (2) ______ las maletas, sal (3) _____ de la habitación y métete (4) _____ en el coche. ...
September 25, 2015
probability
The National Center for Educational Statistics reported that in 2009 the ACT college placement and admission examination were normally distributed and had a mean of 21.1 and standard deviation of 5.1. A college requires applicants to have an ACT score in the top 20% of all ...
September 21, 2015
AP Physics I
A placekicker is about to kick a field goal. The ball is 26.6 m from the goal post. The ball is kicked with an initial velocity of 19.0 m/s at an angle θ above the ground. Between what two angles, θ1 and θ2, will the ball clear the 2.8-m-high crossbar? (Hint: ...
September 20, 2015
Spanish
1 of 15 I tried to get there on time, but there was an accident on the freeway: Yo ____(1)____ llegar a tiempo, pero ____(2)____ un accidente en la autopista. Options 1-quise or queria 2-hubo or habia Answers 1- quise 2 -habia 2 of 15 They refused to admit that you already ...
September 19, 2015
probability
You and your friend just rented a car for an 8,000 mile cross-country road trip. Your rental car may be one of three different types: new (N), nearly one year old (O), and old (L). If the car you receive is brand new, it will break down with probability 0.08. If the car is ...
September 11, 2015
Gabriele enters an east-w straight bike path at the 3.0-km mark and rides w at a constant speed of 8.0 m/s. At the same time, Xena rides east from the1.0-km mark at a constant speed of 6.0 m/s. 1. Write function x(t) that describe the position of Gabriele as a function of time...
September 11, 2015
probability
You and your friend just rented a car for an 8,000 mile cross-country road trip. Your rental car may be one of three different types: new (N), nearly one year old (O), and old (L). If the car you receive is brand new, it will break down with probability 0.08. If the car is ...
September 10, 2015
SPANISH
EL PRETÉRITO. Modifique el párrafo por cambiar los verbos italizados al pretérito. Ejemplo: Tenemos que preguntarle al revisor. Tuvimos que preguntarle al revisor. Voy___(1)___ a coger el tren para irme de vacaciones a la sierra. Ya no me queda___(2)___ ...
September 10, 2015
probability
You and your friend just rented a car for an 8,000 mile cross-country road trip. Your rental car may be one of three different types: new (N), nearly one year old (O), and old (L). If the car you receive is brand new, it will break down with probability 0.08. If the car is ...
September 9, 2015
math
An executive drove from home at an average speed of 45 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at an average speed of 120 mph. The entire distance was 250 miles; the entire trip took three hours. ...
September 9, 2015
math
The weight of a bag of bricks plus 20% of its weight is 30 pounds. How much does the bag of bricks weigh, in pounds?
September 9, 2015
Stats
Explain what it means to have a deviation score of -18
September 8, 2015
Statistics
Assume that the unit of measure for each of the values in the sample is in dollars. What would be the units of measure for the sample: a) mean? b) range? C) Variance? d) Standard deviation?
September 8, 2015
probability
Alice and Bob each choose at random a number between zero and two. Consider the events events A: The magnitude of the difference of the two numbers is greater than 1/4. Event B: At least one of the numbers is greater than 1/4. Event C: Alice’s number is greater than 1/4. ...
September 7, 2015
probability
You and your friend just rented a car for an 8,000 mile cross-country road trip. Your rental car may be one of three different types: new (N), nearly one year old (O), and old (L). If the car you receive is brand new, it will break down with probability 0.08. If the car is ...
September 7, 2015
probability
Anne and Bob each have a deck of playing cards. Each flips over a randomly selected card. Assume that all pairs of cards are equally likely to be drawn. Determine the following probabilities: (1) The probability that at least one card is an ace, (2) the probability that the ...
September 7, 2015
World Geography
if 1/2 inch equals to ten miles how many inches are there in thirity miles ?
September 2, 2015
Algebra
Write 3 different algebraic expressions that uses x,y, and z and simplifies to the given value when x=-3, y=-2, and z=-1. The expression should involve only multiplication or division. Values are: 1, -16, and 12.
September 1, 2015
Algebra 2
1. A bag contains 12 red checkers and 6 black checkers. You will randomly select two checkers, one at a time, without replacement. (a) In how many ways can you select two checkers so that at least one of the checkers is red? List the ways. (b) What is the probability that you ...
August 26, 2015
math
A plane is flying at 168 miles per hour with a heading of 43.5° from due north. The wind is blowing a constant 33 miles per hour at 133.5° from due north. Find the ground speed and true course of the plane.
August 17, 2015
science
i think its A
August 17, 2015
science
Which of the following chemicals is insoluble in water? A. C6H6 {benzene) B. HCI (hydrochloric acid) C. NH3 (ammonia) D. NaHC03 (sodium bicarbonate)
August 17, 2015
Sophia
650ft/117.3ft/s=5.5 Is NOT the answer. Nor is 40.6 nor 7.4 nor 8.8 nor 16.2 nor 6.5. Nor is 0! Thinking they didn't stop at all before the crash!
August 12, 2015
Sophia
Algebra
August 12, 2015
Sophia
The following formula relates the distance S (in feet) traveled by a vehicle during an emergency stop to the time T (in seconds) it takes to arrive at a complete stop: s(t)=16t^{2} At the scene of an accident, Bill uses skid marks to determine the braking distance of one of ...
August 12, 2015
History
Ms Sue I am not understanding that is the Maccabean Revolt written for the gentiles to read in the future? Or is it for the Jews so they are aware of the past
July 19, 2015
History
What was the purpose of account of the Maccabean Revolt? Who was the intended audience?
July 19, 2015
experiment physics
Stress ke cari jawapan x jumpe
June 30, 2015
Chemistry
A solution of methanol and water has a vapour pressure of 213 Torr. What would you predict as the mole fractions of each component, assuming ideal behaviour? The vapour pressures of methanol and water are 256 Torr and 55.3 Torr, respectively
June 4, 2015
Chem.
Thank you
June 4, 2015
Chemistry
Do you mean 44.8L of Zn?
June 4, 2015
geometry
what is the surface area of a cube with side of 2 inches? what is the volume?
May 15, 2015
Math
Find the effective rate of interest (APR) for a loan with a loan amount of \$6,300, a time of 270 days, and interest of \$685.13. a. 14.5% b. 6.5% c. 9.3% d. 12.3%
April 21, 2015
math
Thank you
April 17, 2015
math
Simplify 6P2 A.30**** B.15 C.12 D.720 Simplify 8c5 A.6,720 B.336 C.56**** D.10
April 17, 2015
Science
What volume of aluminium has the same number of atoms as 8.0cm^3 of mercury?
April 16, 2015
math
NO ONE REPLIED.
April 13, 2015
math
How do i find the 12th term of the sequence 36,18,9.... using a formula
April 13, 2015
math
THANK YOU SO MUCH REINY.
April 13, 2015
math
How do i find the 12th term of the sequence 36,18,9.... using a formula
April 13, 2015
math
t12=36(1/3)^11 36(1/2048) =0.017578125 right or wrong?
April 13, 2015
HELPP
The volume of a rectangular prism can be computed using the formula V = lwh. What is the width of a prism that has a volume of 1664 cubic centimeters, a length of 16 centimeters, and a height of 8 centimeters? A. 11 cm B. 13 cm C. 26 cm D. 768 cm
April 7, 2015
math
write teh equation of the line: a line that goes through the point (-1,5) and has a slope of 2/3
April 6, 2015
physics
A 2.50m -long, 450g rope pulls a 10.0kg block of ice across a horizontal, frictionless surface. The block accelerates at 2.50m/s2 . How much force pulls forward on the rope?
March 27, 2015
Physics
You drop a rock from a 300-meter cliff. How long will it take to reach the ground? Assume it has no significant air resistance
March 26, 2015
Math Help
A right triangle has a leg B of length 7 and a hypotenuse of length 11. What is the length of the other leg A? Round to the nearest tenth, if necessary. A. 2.0 B. 8.5 C. 13.0 D. 9.6 Please help, I was gone for this lesson, just need someone to explain how they got the answer.
March 16, 2015
Dynamics
an 800 lb cannon fires a 5 lb cannonball in a horizontal direction. the cannon sits on a skid; the kinetic coefficient of friction between the skid and the horizontal surface is .25. if the cannonball has a speed of 650 ft/s when it leaves the cannon, determine a) the recoil ...
March 15, 2015
algebra
you research the cost of a gallon of gasoline over several years to look for trends the table shows the data you have collected what is the equation of the line of best fit how much would you expect to pay for gasoline in 2029 a)y=0.538x+1.36; \$33.10 b) y=0.289x+1.75:\$18.80 C ...
March 14, 2015
algebra
you research the cost of a gallon of gasoline over several years to look for trends the table shows the data you have collected what is the equation of the line of best fit how much would you expect to pay for gasoline in 2029 a)y=0.538x+1.36; \$33.10 b) y=0.289x+1.75:\$18.80 C ...
March 14, 2015
Calculus
A curve passes through the point (7,6) and has the property that the slope of the curve at every point P is 4 times the y-coordinate of P. What is the equation of the curve? Simplify the equation as much as possible.
March 9, 2015
physics
A painter is pressing a 125g paint pad against the ceiling with a force of 8.3 N at an angle of 30° from the vertical. The pad moves at a constant velocity. What is the coefficient of friction between the pad and ceiling?
March 5, 2015
Energy&Society
How do i find the fuel cost per energy unit (different energy sources). I don't have to calculate it. I just want to know from where I can get it.
March 5, 2015
Chemistry
Can we use moles of HCl instead of B to calculate limiting reactant?
February 27, 2015
Chemistry
When 0.500 moles of boron trichloride react with 1.20 moles hydrogen gas to produce elemental boron and hydrogen chloride gas, the actual yield of boron was 66.4 % of the theoretical yield. The mass of boron obtained was _____ g. Please help I'm stuck on it.
February 27, 2015
chemistry
sodium burns in chlorine to produce sodium chloride according to the reaction: 2Na(s) + Cl2(g) = 2NaCl(s). A student is asked to predict the mass of sodium chloride produced when 71 g of sodium is placed in a container of 200 g of chorine. The container is heated until the ...
February 23, 2015
S.S.
what was "king" and the main topic of conversation in the south? politics sugarcane cotton slavery I was thinking COTTON was KING so C
February 19, 2015
science
can someone explain the life cycle of the animalia kingdom please
February 11, 2015
physics
a tsunami originating in japan has a wavelength of 10^5 m and a period of 10 mins. how long does it take to travel 8000km?
February 11, 2015
Math
The company needs at least \$300,000 fire and theft insurance and at least \$4,200,000 and liability for the from these plans. How many units should be purchased from H plan to minimize the cost of the premiums? What is the minimum premium?
February 7, 2015
physics
Four identical point charges (+2.51 nC) are placed at the corners of a rectangle, which measures 4.00 m by 7.00 m. If the electric potential is taken to be zero at infinity, what is the potential at the geometric center of this rectangle?
February 6, 2015
Calculus 2
wouldn't it be the integral of sqrt(1+u^2)du instead of integral u(sqrt (1+u^2)du because the du takes care of the outside u?
February 4, 2015
Calculus 2
integral of e^x(sqrt(1+e^(2x)))dx
February 4, 2015
Calculus 2
use the method of equating coefficients. integral of (x^3+5x^2+12)/((x^2)(x^2+4)) dx
February 4, 2015
physics
Air becomes conducting when the electric field strength in a region exceeds 3x106 V/m. if a spark jumps between a charged object and a grounded object when the two objects are 10 cm apart, what is the potential difference between the two objects? If a lightning strike is 4x103...
February 2, 2015
Physics
A hollow metal sphere has inner and outer radii of 20.0 cm and 30.0 cm, respectively. As shown in the figure, a solid metal sphere of radius 10.0 cm is located at the center of the hollow sphere. The electric field at a point P, a distance of 15.0 cm from the center, is found ...
January 29, 2015
physics
A +1.4-nC point charge is placed at one corner of a square (1.8 m on a side), and a -2.5-nC charge is placed on the corner diagonally opposite. What is the magnitude of the electric field at either of the other two corners?
January 29, 2015
algebra
The mean on a Advanced Algebra test was 78 with a standard deviation of 8. If the test scores are normal distributed, find the interval about the mean that contains 99.7% of the scores. Use the empirical rule.
January 28, 2015
Math, Algebra
I've done. y-(-3)=(5/2)(x-(-3)) y-(-3)=(5/2(-15/2) y=(5/2x)-(-15/2)+3 y=(5/2x)+(9/2) Is that correct?
January 19, 2015
Math, Algebra
I normally use point slope. I'm used to having to sets of points not just one.
January 19, 2015
Math, Algebra
Can someone help me figure this out for my review? Find the equation of the line that passes through (2,-3) and has a slope of (5/2)
January 19, 2015
Chemistry (Error Propagation)
Calculate, with uncertainty, and express with a reasonable number of significant figures. a) 5.4 (±0.2) – 3.7 (±0.4) = b) 6.84 (±0.04) × 0.7 (±0.04) =
January 19, 2015
math
The hire purchase price for a T.V set is made up of \$4,000 deposit and 18 equal monthly installments of \$1,200 each. Amos was given a discount of 20% on the hire purchase price on paying the cash for the T.V set. How much did he pay?
January 17, 2015
physics
F=ma 47.6 N=25.7 x a a=47.6/25.7 a=1.85 m/s^2
January 15, 2015
Math
A rectangle is 26 meters long and 18 meters wide.What is the length of the diagonal of the rectangle to the nearest meter? equation: A2 + B2= C2 I don't know where to put the numbers, but I think the answer may be 32. Thank you for any help!
January 12, 2015
Chemistry (Acid-Base)
I'm a tad confused by the question, The protonation constant for the acid HA has the value K=2*10^6. (a) what is the principal species at pH 6.00? (b) what is the principal species at pH 9.00? (c) what is the quotient [A-]/[HA] at pH 6.3 and 7.5?
January 12, 2015
math
adi, john ,and hailee plant seedlings in their neighberhood park. adi plants 40% of the total number of seedlings, john plants 45% of the total number of seedlings and hailee plants the rest of the seedlings.if hailee plants 87 seedliings how maany seedlings do they plant ...
January 10, 2015
Math
A certain infinite geometric series has first term 7 and sum 4. What is the result when the third term is divided by the second term?
January 8, 2015
Geometry
In rectangle ABCD, AC= 8x-10 and BD=2x+20. Find the length of BD
January 3, 2015
Geometry
So the answer is y= 2/3x +11 ?
January 3, 2015
Geometry
Find an equation of the line passing through the point (6,5) and perpendicular to the line whose equation is 2y+3x=6
January 3, 2015
Spanish
I have a question regarding #6. Why would this not be "antónimos"? To conceal and to declare are the opposite, are they not?
December 27, 2014
Math (PreCalculus)
The second hand of a clock is 4 inches long. Describe the height of the tip of the second hand from the bottom of a clock. (The radius of the clock is 1 inch more than the length of the second hand.)
December 24, 2014
AP Calc
Find the value of c which satisfies Rolle’s Theorem for the function f(x)=sin(x²) on (0,√π).
December 20, 2014
math
Wanda bought a pair of jeans for 32.50. She can purchase a second pair doe 40% off the regular price. How much will she spend on two pair of jeans?
December 19, 2014
Math
Part (a): Find the sum a+(a+1)+(a+2)+.....+(a+n-1) in terms of a and n Part (b): Find all pairs of positive integers (a,n) such that n>= 2 and a+a+1)+(a+2)+....+(a+n-1)=100.
December 15, 2014
Math
Let a1,a2,......a10 be an arithmetic sequence. If a1+a3+a5+a7+a9=17 and a2+a4+a6+a8+a10=15 then find a1.
December 15, 2014
Math
Penn writes a 2013-term arithmetic sequence of positive integers, and Teller writes a different 2013-term arithmetic sequence of integers. Teller's first term is the negative of Penn's first term. Each then finds the sum of terms in his sequence. If their sums are ...
December 15, 2014
Math
Penn writes a 2013-term arithmetic sequence of positive integers, and Teller writes a different 2013-term arithmetic sequence of integers. Teller's first term is the negative of Penn's first term. Each then finds the sum of terms in his sequence. If their sums are ...
December 14, 2014
Algebra
Suppose that |a-b|+|b-c|+|c-d|+.......+|m-n|+|n-o|+............|x-y|+|y-z|+|z-a|=20. What is the maximum possible value of |a-n|?
December 13, 2014
Algebra
Suppose that |a-b|+|b-c|+|c-a|=20. What is the maximum possible value of |a-b|?
December 13, 2014
Algebra: Reiny Correction
It is the fiftieth term not the fifteenth so would it be t50=a+49d
December 13, 2014
Algebra
The second and ninth terms of an arithmetic sequence are 2 and 30 respectively. What is the fiftieth term.
December 13, 2014
Algebra
Evaluate [3.2]. It is a floor and ceiling function. Thanks.
December 12, 2014
Abnormal Psychology
I need help diagnosis this case for my abnormal psychology review. Mr. B is a 20-year-old male college student who was recently arrested for possession of marijuana, which was detected when he was stopped for un¬safe driving. He was charged with driving while intoxicated ...
December 9, 2014
Chem-Diprotic Acid
Ascorbic Acid is an organic diprotic acid, H2A, that is found in many natural materials. KA1=7.94x10^-5 and KA2=1.62x10^-12. If you start with a 0.75 M solution of H2A, what is the concentration of ALL species in the solition at ph=3.8? I have no idea how to even start this ...
December 7, 2014
Math
The quantity sqrt(45)- 2sqrt(5) +sqrt(360/2) can be expressed as sqrt(N), where N is an integer. Find N. * the whole fraction 360/2 is squarerooted.
December 5, 2014
Math
Solve for the positive value of x such that 3sqrt(x^2-4x+4)=16. The 3 is the small number next to the sqrt sign.
December 4, 2014
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fraction.hpp
Go to the documentation of this file.
1 //=======================================================================
3 //
5 // accompanying file LICENSE_1_0.txt or copy at
7 //=======================================================================
17 #ifndef PAAL_FRACTION_HPP
18 #define PAAL_FRACTION_HPP
19
20 #include "paal/utils/floating.hpp"
21
22 namespace paal {
23 namespace data_structures {
24
31 template <class A, class B> struct fraction {
33 using num_type = A;
35 using den_type = B;
37 A num;
39 B den;
41 fraction(A num, B den) : num(num), den(den) {}
42 };
43
56 template <class A, class B, class C, class D>
57 bool operator<(const fraction<A, B> &f1, const fraction<C, D> &f2)
58 {
59 return f1.num * f2.den < f2.num * f1.den;
60 }
61
74 template <class A, class B, class C, class D>
75 bool operator>(const fraction<A, B> &f1, const fraction<C, D> &f2)
76 {
77 return f2 < f1;
78 }
79
92 template <class A, class B, class C, class D>
93 bool operator<=(const fraction<A, B> &f1, const fraction<C, D> &f2)
94 {
95 return !(f2 < f1);
96 }
97
110 template <class A, class B, class C, class D>
111 bool operator>=(const fraction<A, B> &f1, const fraction<C, D> &f2)
112 {
113 return !(f1 < f2);
114 }
115
130 template<class A, class B, class C, class D, class EPS = A>
131 bool are_fractions_equal(const fraction<A, B>& f1, const fraction<C, D>& f2, EPS eps = A{})
132 {
133 auto x = f1.num * f2.den - f2.num * f1.den;
135 return cmp.e(x, 0);
136 }
137
148 template <class A, class B>
150 {
151 return fraction<A, B>(a, b);
152 }
153
165 template<class A, class B, class C>
166 auto operator*(C c, const fraction<A, B>& f) {
167
168 return make_fraction(c * f.num, f.den);
169 }
170
171 }
172 }
173
174 #endif // PAAL_FRACTION_HPP
auto operator*(C c, const fraction< A, B > &f)
operator*
Definition: fraction.hpp:166
bool are_fractions_equal(const fraction< A, B > &f1, const fraction< C, D > &f2, EPS eps=A{})
operator==
Definition: fraction.hpp:131
bool operator>=(const fraction< A, B > &f1, const fraction< C, D > &f2)
operator>=
Definition: fraction.hpp:111
bool operator>(const fraction< A, B > &f1, const fraction< C, D > &f2)
operator>
Definition: fraction.hpp:75
fraction< A, B > make_fraction(A a, B b)
make function for fraction
Definition: fraction.hpp:149
B den_type
denominator type
Definition: fraction.hpp:35
Class for comparing floating point.
Definition: floating.hpp:26
fraction(A num, B den)
constructor
Definition: fraction.hpp:41
simple class to represent fraction
Definition: fraction.hpp:31
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# C Program to find the frequency of characters in a String
devquora
Posted On: Jan 27, 2023
### Overview:
As we know that time is the most important concern in our life. To do any task easily, We have to use software to perform it in a simpler manner. Software is a set of programs with related documents. The Software Program or Program is a set of instructions to do a specific task. Here, our specific task is to write a C Program to Find the Frequency of Characters in a String. First, I want to explain What is String? Then, I would like to explain What is the Frequency of Characters in a String? Furthermore, I would like to demonstrate how to Find the Frequency of Characters in a String. After all, I will write the logic of a C Program to Find the Frequency of Characters in a String with output.
### What is a String?
String is a combination of characters that may be any digit from 0 to 9 and Alphabets from A to Z. The repetition of characters does not matter. For instance, a String Conax uses 5 characters i.e. C,o,n,e,and x.
### What is the Frequency of Characters in a String?
The frequency of a character in a given string is how many times a particular character is present in a given string. For example, if a user enters the string as Welcome User, and wants to check for the frequency of a character say e. Then it will be 3. Because e occurs 3 times in the given string Welcome User.
### Logic to Find the Frequency of Characters in a String
To concatenate two strings str1 and str2, you will copy all characters of str2 at the end of str1. Below is the step by step descriptive logic to concatenate two strings:
• Step 1: Initialize the variables.
• Step 2: Accept the input.
• Step 3: Initialize a for loop and terminate it at the end of string.
• Step 4: This for loop will be used to count the number of time each character is present.
• Step 5: Initialize another for loop to print the frequency if it is at least 1.
### C Program to Find the Frequency of Characters in a String
```#include <stdio.h>
int main()
{
//Initializing variables.
char str[100];
int i;
int freq[256] = {0};
//Accepting inputs.
printf("Enter the string: ");
gets(str);
//Calculating frequency of each character.
for(i = 0; str[i] != '\0'; i++)
{
freq[str[i]]++;
}
//Printing frequency of each character.
for(i = 0; i < 256; i++)
{
if(freq[i] != 0)
{
printf("The frequency of %c is %d\n", i, freq[i]);
}
}
return 0;
}
```
Save this program with the filename and extension .c like filename.c. Then compile it on turbo C or another compiler.
The output of the Program:
Enter the string: Welcome User
The frequency of is 1
The frequency of U is 1
The frequency of W is 1
The frequency of c is 1
The frequency of e is 3
The frequency of l is 1
The frequency of m is 1
The frequency of o is 1
The frequency of r is 1
The frequency of s is 1
### Conclusion:
To calculate the Frequency of Characters in a String we will use a for loop that will iterate each character of the string and then store the count of each of the characters in an array. Then we will use another for loop to print the frequency of each character that we have stored in an array.
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Damped systems - deriving equation
I am having some troubles deriving the formula for the roots for different types systems.. I am not quite sure if they are correct (pretty sure they aren't).
$y(s) = \frac{s+2\zeta\omega_n}{s^2 + 2\zeta\omega_ns +\omega_n^2}$
I don't see how i should derive them for different scenarios
In my book it is stated that for an underdamped system $\zeta < 1$ is the roots S1,2 = $-\zeta\omega_n \pm i\omega_n \sqrt{1-\zeta^2}$ which i don't get.
I've derived it to be $\frac{-2\zeta\omega_n \pm \sqrt{4\zeta^2\omega_n^2- 4\omega_n^2}}{2}$
I don't see how the hell they got the other one, and how they are for the other cases!!??
Well, just look at the root!
$$\frac{-2\zeta\omega_n \pm \sqrt{4\zeta^2\omega_n^2- 4\omega_n^2}}{2} =\frac{-2\zeta\omega_n \pm 2\omega_n\sqrt{\zeta^2- 1}}{2}$$ And that is: $$-\zeta\omega_n \pm i\omega_n\sqrt{1-\zeta^2}$$ Elaboration: $$\sqrt{4\zeta^2\omega_n^2- 4\omega_n^2}=\sqrt{4\omega_n^2(\zeta^2-1)}=$$ $$\sqrt{4\omega_n^2}\sqrt{\zeta^2-1}=2\omega_n\sqrt{\zeta^2-1}$$ Different scenarios:
• $\zeta=0$ no dampening, formula $S_{12}=i\omega_n$
• $\zeta<1$ dampening, solution is complex, oscillatory
• $\zeta=1$ critical dampening, the formula will be $S_{12}=-\zeta\omega_n$
• $\zeta>1$ overdamped, two different real roots, will take longer to converge than criticaly dampened, formula $S_{12}=-\zeta\omega_n \pm \omega_n\sqrt{\zeta^2-1}$
• Well.. that makes sense, but how come for the other cases, they do not exist in my text book.. Apr 18, 2014 at 13:09
• The other cases arise when you put $\zeta$=1 for crit dampened, and $\zeta$>1 for the overdamping. And well I forgot to put the imaginary unity in front of it. Apr 18, 2014 at 13:14
• They are just cases, $\zeta$ is the dampening factor. For different $\zeta$ you will get different behaviour. The easiest way to see this is by putting the values of $\zeta$ for the different scenarios in the formula, and looking how it changes. Apr 18, 2014 at 13:22
• Yes.. I know that the other cases arrives for different values of the damping ratio, but do they have a formula like the underdamped here? and $i$ is stil missing from the $4\omega_n^2$ Apr 18, 2014 at 13:22
• Ahh... ofCourse, $\sqrt{\zeta^2 - 1} =>\sqrt{-1(1 -\zeta^2)} => \sqrt{-1} \sqrt{(1 -\zeta^2)}$ Apr 18, 2014 at 13:40
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   Â
     Â
# How to Find the Volume of a Triangular Prism?
We study different Types of Prisms in mathematics like rectangular Prism, square prism etc. Here we will discuss triangular prism. A Solid figure that has two identical ends and also has flat sides is known as prism. Now we will see how to find the volume of a triangular prism. Formula to find volume of a triangular prism is given as:
Volume of Triangular Prism = al; Where, ‘a’ represents the area of one triangular end face and 'l' denotes the length of prism.
Let’s follow steps to find volume of triangular prism.
Step 1: To find the volume of prism first it is necessary to find area and length of prism. Suppose we have area and length of prism is 120 inch2, 15 inch respectively.
Step 2: Then we put these values in formula. As we know that volume of triangular prism is given by: Volume of Triangular Prism = AL.
So volume of triangular prism = 120 * 15, on solving further we get:
Volume of prism = 1800 inch2. This is how we find volume of a triangular prism. Now we will see some formulas based on prism. Formula to find surface area of triangular prism is given by: Surface area of Triangular Prism = 2B + PH, or we can also write it as: Surface area = BH + (s1 + s2 + s3) H;
Here, ‘B’ denotes area of base, ‘p’ is the perimeter of base, ‘h’ is the height of triangular prism, and s1 s2 s3 are sides of triangle.
## Further Read
Math Topics
Top Scorers in Worksheets
Want to know your friend’s score card! Login with Facebook.
Related Worksheet
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# West-southwest: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
(Redirected to Boxing the compass article)
A 16-point compass rose
A 32-point compass rose
A historical compass card
Boxing the compass is the action of naming all thirty-two principal points of the compass in clockwise order. Such names, formed by the initials of the cardinal directions and their intermediate ordinal directions, are accepted internationally, even though they have their origin in the English language, and are very handy to refer to a heading (or course or azimuth) in a general or colloquial fashion, without having to resort to computing or recalling degrees.
The set of 32 named points can be further divided into a set of 128 named directions using quarter-points,[1] although for communicating heading these fractional points have been superseded by degrees measured clockwise from North.
## Compass points
A simple algorithm can be used to convert a heading to an approximate compass point:
1. Divide the heading in degrees by 11.25 (360/32) to get to the case of 32 named points.
2. Add 1.5 to center the named points in their respective sectors on the circle, since north is 1 in the table instead of 0. If the result is 33 or more, subtract 32 to keep within the 32-point set.
3. Now look up the integer part of the result in the table below.
For example:
A heading of 75°, divided by 11.25 gives 6.67, added to 1.5 gives 8.17, truncated to give 8. 8 in the table below corresponds to east by north.
# Compass point Abbr. Traditional wind point Lowest Middle Highest
1 North N Tramontana 0.00° 5.62°
2 North by east NbE 5.63° 11.25° 16.87°
3 North-northeast NNE 16.88° 22.50° 28.12°
4 Northeast by north NEbN 28.13° 33.75° 39.37°
5 Northeast NE Greco or Bora 39.38° 45.00° 50.62°
6 Northeast by east NEbE 50.63° 56.25° 61.87°
7 East-northeast ENE 61.88° 67.50° 73.12°
8 East by north EbN 73.13° 78.75° 84.37°
9 East E Levante 84.38° 90.00° 95.62°
10 East by south EbS 95.63° 101.25° 106.87°
11 East-southeast ESE 106.88° 112.50° 118.12°
12 Southeast by east SEbE 118.13° 123.75° 129.37°
13 Southeast SE Sirocco 129.38° 135.00° 140.62°
14 Southeast by south SEbS 140.63° 146.25° 151.87°
15 South-southeast SSE 151.88° 157.50° 163.12°
16 South by east SbE 163.13° 168.75° 174.37°
17 South S Ostro 174.38° 180.00° 185.62°
18 South by west SbW 185.63° 191.25° 196.87°
19 South-southwest SSW 196.88° 202.50° 208.12°
20 Southwest by south SWbS 208.13° 213.75° 219.37°
21 Southwest SW Libeccio 219.38° 225.00° 230.62°
22 Southwest by west SWbW 230.63° 236.25° 241.87°
23 West-southwest WSW 241.88° 247.50° 253.12°
24 West by south WbS 253.13° 258.75° 264.37°
25 West W Poniente or Zephyrus 264.38° 270.00° 275.62°
26 West by north WbN 275.63° 281.25° 286.87°
27 West-northwest WNW 286.88° 292.50° 298.12°
28 Northwest by west NWbW 298.13° 303.75° 309.37°
29 Northwest NW Mistral 309.38° 315.00° 320.62°
30 Northwest by north NWbN 320.63° 326.25° 331.87°
31 North-northwest NNW 331.88° 337.50° 343.12°
32 North by west NbW 343.13° 348.75° 354.37°
1 North N Tramontana 354.38° 360.00°
## References
1. ^ Great War Primary Documents Archive, "Boxing the Compass" Accessed November 12, 2008
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21 April, 14:01
# A student measures the speed of yellow light in water to be 2.00x10^8
+3
1. 21 April, 15:06
0
NOTE: The given question is incomplete.
The complete question is given below.
A student measures the speed of yellow light in water to be 2.00 x 10⁸ m/s. Calculate the speed of light in air.
Solution:
Speed of yellow light in water (v) = 2.00 x 10⁸ m/s
Refractive Index of water with respect to air (μ) = 4/3
Refractive Index = Speed of yellow light in air / Speed of yellow light in water
Or, The speed of yellow light in air = Refractive Index * Speed of yellow light in water
or, = (4/3) * 2.00 x 10⁸ m/s
or, = 2.67 * 10⁸ m/s ≈ 3.0 * 10⁸ m/s
Hence, the required speed of yellow light in the air will be 3.0 * 10⁸ m/s.
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# Rounding to significant figures (Part 1)
In this lesson you will be introduced to significant figures and understand the concept of degrees of accuracy. You will also learn how to round whole numbers to a given significant figure.
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
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## Q6.Why is the approximation to this calculation greater than the actual answer?
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This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
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# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
Quiz:
# Rounding to Significant Figures
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
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## Q6.What does significant figure tell us?
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This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
Quiz:
# Rounding to Significant Figures
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
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# Lesson summary: Rounding to significant figures (Part 1)
### It looks like you have not completed one of the quizzes.
To share your results with your teacher please complete one of the quizzes.
### Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Walk
On the spot:
Chair yoga
### Take part in The Big Ask.
The Children's Commissioner for England wants to know what matters to young people.
Share your views and have your voice heard.
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# Special Graphs
## Contents
#### Graphing Rational Functions
To graph a rational function, we must determine three things:
• Zeros-- x values for which the numerator equals 0 (but not the denominator).
• Vertical asymptotes-- x values for which the denominator equals 0 (but not the numerator).
• Holes-- x values for which the numerator and the denominator equal 0.
Note: If a value of x makes a squared term in the denominator equal to 0, that value is called a "double asymptote." For example, f (x) = has a double asymptote of x = 4 .
Here are the steps to graphing a rational function:
1. Plot zeros.
2. Graph vertical asymptotes. These divide the graph into "sections."
3. Start from the right side of the graph. If the degree of the numerator is greater than the degree of the denominator, start from the upper right corner (or the lower right corner if the function is negative). If the degree of the numerator is less than the degree of the denominator, start just above the x -axis (or just below if the function is negative). If the degree of the numerator is equal to the degree of the denominator, start just above the line y = k , where k is the leading coefficient (or just below if negative).
4. Cross over any zeros, and approach the first asymptote.
5. If the asymptote is a single asymptote, approach on the opposite side of the asymptote from the opposite direction (up if the last asymptote led down, and vice versa). If the asymptote is a double asymptote, approach from the same direction.
6. Cross over any zeros, and approach the next asymptote.
7. Repeat steps 5 and 6 until the end of the graph is reached.
8. Remove all holes.
Example: Graph f (x) = .
• Zeros: x = - 1 , x = 0 (double), x = 5
• Asymptotes: Single: x = 4 . Double: x = - 2 .
• Holes: x = 3 .
• Degree of numerator = 5. Degree of denominator = 4.
Steps 1 and 2
Step 3
Step 4
Steps 5 and 6 (section 2)
Steps 5 and 6 (section 3)
Step 8
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# [R] replacing a for-loop with lapply
Daniel Berg daniel at nr.no
Mon May 9 18:31:50 CEST 2005
```Dear All,
I am trying to compute a goodness-of-fit statistic for a copula, based on an
empirical density estimate of this copula.
To do this I can use the following code:
> n <- dim(data)[1]
> d <- dim(data)[2]
> Chat <- rep(0,n)
> for(i in 1:n)
+ Chat[i] <- sum(apply(t(data)<=data[i,],2,prod))/(n+1)
However, I have a feeling this can be done more effectively than using a
for-loop. I have also tried the following:
> tmp1 <- lapply(1:n,function(i) t(data)<=data[i,])
> tmp2 <- lapply(1:n,function(i) apply(tmp1[[i]],2,prod))
> Chat <- as.numeric(lapply(1:n, function(i) sum(tmp2[[i]])))
but there is no improvement. I ran the following timing test:
> data <- matrix(runif(300),100,3)
> n = dim(data)[1]
> d = dim(data)[2]
> Chat = vector("numeric",n)
> M <- 30
> a <- rep(0,M)
> for(m in 1:M){
+ a[m] <- system.time({
+ tmp1 <- lapply(1:n,function(i) t(data)<=data[i,])
+ tmp2 <- lapply(1:n,function(i) apply(tmp1[[i]],2,prod))
+ Chat <- as.numeric(lapply(1:n, function(i) sum(tmp2[[i]])))})[3]}
> b <- rep(0,M)
> for(m in 1:30){
+ b[m] <- system.time(
+ for (i in 1:n)
+ Chat[i] = sum(apply(t(data)<=data[i,],2,prod))/(n+1))[3]}
> summary(a)
> summary(b)
and the output was:
> summary(a)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.8500 0.8700 0.8900 0.9013 0.9300 0.9800
> summary(b)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.8400 0.8600 0.8800 0.8883 0.9075 0.9900
Is there any way I can code this more efficiently in R or will I have to
turn to C? The data sets, on which I am actually going to run this code,
will be of sizes up to (5000x100) and I need hundreds of realizations...
Rgds,
Daniel
```
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Margin of Error, Confidence Interval & Sample Size
Need assistance with solving statistic problems
1. Find the margin of error for the given values of c, s and n.
c = 0.90 s = 2.7 n = 49
E = ? Round to three decimal places as needed.
2. Contruct the confidence interval for the population mean u.
C= 0.98, x=8.2, s=0.4 and n=53
A 98% confidence interval for u is?
Round to two decimal places as needed
3. Contruct the confidence interval for the population mean u.
C= 0.90, x=15.7, s=5.0 and n=75
A 90% confidence interval for u is?
Round to two decimal places as needed
4. Use the confidence interval to find the estimated margin of error. The find the sample mean. A Biologist reports a confidence interval of (1.6, 2.8) when estimation the mean height ( in centimeters) of a sample of seedlings. The estimated margin of error is?
5. Find the minimum sample size n needed to estimate u for the given values of c, s, and E. C=0.98, s=6.6 and E=1. Assume that a preliminary sample has at least 30 members. N=? Round up to the nearest whole number.
6. You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 40 home theater system has a mean price of \$114.00 and a standard deviation is \$18.80. Construct a 90% confidence interval for the population mean. The 90% confidence interval is? Round to two decimal places as needed.
7. You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Which interval is wider? If convenient, use technology to construct the confidence intervals. A random sample of 35 gas grills has a mean price of \$634.10 and a standard deviation of \$ 56.10. The 90% confidence interval is? Round to two decimal places as needed.
8. You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Which interval is wider? If convenient, use technology to construct the confidence intervals. A random sample of 40 eight-oz servings of different juice drinks has a mean price of 90.3 calories and a standard deviation of 43.9 calories. The 90% confidence interval is? Round to two decimal places as needed.
9. People were polled on how many books they read the previous year. How many subjects are needed to estimate the number of books read the previous year within one book with 90% confidence? Initial survey results indicate that o=12.3 books. How many subject s does a 90 % confidence level requires ? Round up to the nearest whole number as needed.
10. A doctor wants to estimate the HDL cholesterol of all 20 to 29 year old females. How many subjects are needed to estimate the HDL cholesterol within 2 points with 99% confidence assuming o= 17.3? Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size required? How many subject s does a 99 % confidence level requires ? Round up to the nearest whole number as needed.
11. Construct the indicated confidence interval for the population mean u using (a) a t-distrubution. (b) If you had incorrectly used a normal distribution, which interval would be wider?
C= 0.90, x=13.3, s=3.0, n=8
(a) The 90% confidence interval using a t-disribution is?
12. In the following situation, assume the random variable is normally distributed and use a norma distributionor a t-distribution to construct a 90% confidence interval for the population mean. If convenient, use technology to construct the confidence interval.
(a) In a random sample of 10 adults from a nearby county, the mean waste generated per person per day was 3.65 pounds and the standard deviation was 1.98 pounds. (b) Repeat part (a), assuming the same statistics came from a sample size of 400. Compare the results.
(a) For the sample of 10 adults, the 90% confidence interval is? Round to tow decimal places as needed.
13. Use the given confidence interval to find the margin of error and the sample proportion.
(0.768,0794) E = ?
14. In a survey of 649 males ages 18-64, 399 say they have gone to the dentist in the past year. Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. The 90% confidence interval for the population proportion p is? Round the final answers to the nearest thousandth as needed. Round all intermediate values to the nearest thousandth as needed.
15. In a survey of 9000 women, 6431 say they change their nail polish once a week. Construct a 95% confidence interval for the population proportion of women who change their nail polish once a week. A 95% confidence interval for the population proportion is? Round the final answers to the nearest thousandth as needed. Round all intermediate values to the nearest thousandth as needed.
16. A researcher wishes to estimate, with 95% confidence, the proportion of adults who have high-speed Internet access. Her estimate must be accurate within 2% of the true proportion.
a) Find the minimum sample size needed, using a prior study that found that 52% of the respondents said they have high-speed Internet access.
b) No preliminary estimate is available. Find the minimum sample size needed.
A) What is the minimum sample size needed using a prior study that found that 52% of the respondents said they have high-speed Internet access? N=? (Round up to the nearest whole number as needed)
17. The table below shows the results of a survey in which 2563 adults from Country A, 1107 adults from Country B, and 1050 adults from Country C were asked if human activity contributes to global warming. Complete parts (a), (b), and (c).
(a)Construct a 99% confidence interval for the proportion of adults from Country A who say human activity contributes to global warming.
Table
Adults who say that human activity contributes to global warming
Country A 69%
Country B 86%
Country C 91%
18. The table shows the results of a survey in which separate samples of 400 adults each from the East, South, Midwest, and West were asked if traffic congestion is a serious problem in their community. Complete parts (a) and (b).
Adults who say that traffic congestion is a serious problem.
East 36%
South 33%
Midwest 26%
West 56%
This question has the following supporting file(s):
File Viewer (Click To Zoom)
Solution Summary
The solution provides step by step method for the calculation of margin of error, confidence interval and sample size. Formula for the calculation and Interpretations of the results are also included. Interactive excel sheet is included. The user can edit the inputs and obtain the complete results for a new set of data.
\$2.19
• Plain text
• Cited sources when necessary
• Attached file(s)
• Confidence Interval & Sample Size.docx
• CI & SS.xlsx
Basil Sunny, MSc
Rating 4.9/5
Active since 2008
BSc, Mahatma Gandhi University Kottayam, India
MSc, Mahatma Gandhi University Kottayam, India
PhD (IP), Mahatma Gandhi University Kottayam, India
Responses 2662
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# linear programming Max 5A+5B
• Apr 13th 2008, 06:32 PM
wgr
linear programming Max 5A+5B
New to the forum, just wanted to see if someone can check my answer to the following...solve the following using graphical solution procedure:
Max. 5A+5B
s.t.
1A<=100
1B<=80
2A+4B<=400
A,B>=0
after using the graphical solution procedure to find the optimal solution for the equations....I arrived at point C with coordinates of 200,0 which, when plugged into the maximizing equation gives 1000.
Am I even close?
Gary
• Apr 13th 2008, 08:24 PM
CaptainBlack
Quote:
Originally Posted by wgr
New to the forum, just wanted to see if someone can check my answer to the following...solve the following using graphical solution procedure:
Max. 5A+5B
s.t.
1A<=100
1B<=80
2A+4B<=400
A,B>=0
after using the graphical solution procedure to find the optimal solution for the equations....I arrived at point C with coordinates of 200,0 which, when plugged into the maximizing equation gives 1000.
Am I even close?
Gary
A=200, B=0, is not feasible (that is it does not satisfy the constraints)
The vertices of the feasible region are (0,0), (100,0), (0,80) (40,80) and (100,50). The optimum occurs at one of these.
RonL
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numpy.ma.diff#
ma.diff(a, /, n=1, axis=-1, prepend=<no value>, append=<no value>)[source]#
Calculate the n-th discrete difference along the given axis. The first difference is given by `out[i] = a[i+1] - a[i]` along the given axis, higher differences are calculated by using `diff` recursively. Preserves the input mask.
Parameters:
aarray_like
Input array
nint, optional
The number of times values are differenced. If zero, the input is returned as-is.
axisint, optional
The axis along which the difference is taken, default is the last axis.
prepend, appendarray_like, optional
Values to prepend or append to a along axis prior to performing the difference. Scalar values are expanded to arrays with length 1 in the direction of axis and the shape of the input array in along all other axes. Otherwise the dimension and shape must match a except along axis.
Returns:
The n-th differences. The shape of the output is the same as a except along axis where the dimension is smaller by n. The type of the output is the same as the type of the difference between any two elements of a. This is the same as the type of a in most cases. A notable exception is `datetime64`, which results in a `timedelta64` output array.
`numpy.diff`
Equivalent function in the top-level NumPy module.
Notes
Type is preserved for boolean arrays, so the result will contain False when consecutive elements are the same and True when they differ.
For unsigned integer arrays, the results will also be unsigned. This should not be surprising, as the result is consistent with calculating the difference directly:
```>>> u8_arr = np.array([1, 0], dtype=np.uint8)
>>> np.ma.diff(u8_arr)
fill_value=999999,
dtype=uint8)
>>> u8_arr[1,...] - u8_arr[0,...]
255
```
If this is not desirable, then the array should be cast to a larger integer type first:
```>>> i16_arr = u8_arr.astype(np.int16)
>>> np.ma.diff(i16_arr)
fill_value=999999,
dtype=int16)
```
Examples
```>>> a = np.array([1, 2, 3, 4, 7, 0, 2, 3])
>>> x = np.ma.masked_where(a < 2, a)
>>> np.ma.diff(x)
masked_array(data=[--, 1, 1, 3, --, --, 1],
mask=[ True, False, False, False, True, True, False],
fill_value=999999)
```
```>>> np.ma.diff(x, n=2)
masked_array(data=[--, 0, 2, --, --, --],
mask=[ True, False, False, True, True, True],
fill_value=999999)
```
```>>> a = np.array([[1, 3, 1, 5, 10], [0, 1, 5, 6, 8]])
```>>> np.ma.diff(x, axis=0)
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# A_PointyThink
A_PointyThink is an action which is used as the thinker for the Pointy. The actor will first search for the nearest living player which it can see, which it will set as its target. The actor will then move at a speed defined by the Object's `Speed` value, mirroring its target player's movements – if the player is moving closer to the actor, the actor will move forwards; if the player is moving away however, the actor will move backwards instead.
This action also handles the position and movement of the Pointy's spikeballs, assuming that there are four of them and that they are all linked in a chain starting with the actor, each of them except the last tracer-ing the next spikeball. As the actor moves, each of the spikeball Objects in turn are rotated vertically accordingly. The actor's `ReactionTime` determines the multiple of the actor's radius to use as the distance the spikeball Objects are from the actor, i.e. spikeball distance = `ReactionTime`×`Radius`.
`Damage` determines the spikeballs' speed of rotation around the actor when the player moves – note that this is not directly given in degrees/tic, but in a scale where 8192 is 360°/tic, 4096 is 180°/tic, 2048 is 90°/tic, etc. For example, the Pointy has a `Damage` value of 128, which converts to 5.625 °/tic.
Object property Use
`Speed` Sets how fast the actor moves forwards/backwards when the player is moving
`Radius` Sets the distance the spikeballs are from the actor (`Radius`×`ReactionTime`)
`ReactionTime` Sets the distance the spikeballs are from the actor (`Radius`×`ReactionTime`)
`Damage` Sets how fast the spikeballs move around the actor when the player is moving
Actions – Enemy thinkers [view] A_BuzzFly • A_CrawlaCommanderThink • A_DetonChase • A_EggShield • A_FaceStabChase • A_GuardChase • A_HoodThink • A_JetbThink • A_JetChase • A_JetgShoot • A_JetgThink • A_JetJawChomp • A_JetJawRoam • A_MinusCheck • A_MinusDigging • A_MinusPopup • A_PointyThink • A_SharpChase • A_SharpSpin • A_SkimChase • A_SnailerThink • A_VultureCheck • A_VultureVtol
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http://territorios-luchas.tk/rrrl/math-homework-help-square-roots-vysi.php
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Skip Nav
# Square Root Worksheets
❶Solving Systems by Graphing Method 3- An exact answer.
Key to Algebra offers a unique, proven way to introduce algebra to your students. New concepts are explained in simple language, and examples are easy to follow. Word problems relate algebra to familiar situations, helping students to understand abstract concepts.
Students develop understanding by solving equations and inequalities intuitively before formal solutions are introduced. Students begin their study of algebra in Books using only integers. Books introduce rational numbers and expressions. Books extend coverage to the real number system. Square Root Worksheet Generator Basic worksheets: Only perfect squares Allow non-perfect squares Answer rounded to decimals.
Only simplify, no answers as decimals Take roots from decimals up to decimal digits. Besides taking a square root, include also: Basic instructions for the worksheets Each worksheet is randomly generated and thus unique. To get a different worksheet using the same options: Square Root Worksheet Generator. For more on that, check the links below. When simplifying a square root, I find that it is easiest to create a factor tree.
A factor tree breaks down a number into it's prime factors. Prime numbers are whole numbers that only have factors of 1 and itself. Examples include 2, 3, 5, 7, 11, 13, 17, 19, See the attached image to view the factor tree, which breaks down 27 into all of its prime factors.
For every pair of prime numbers that are part of the factor tree, you can take one outside of the radical. When evaluating a square root, you first determine between which 2 integers the square root falls using perfect squares. Then find the halfway point between the square root of 25 and 36 which is Thus, the square root of 27 is between 5 and 5. The square root of 26 would be approximately 5.
If you only need an estimate, find numbers near 27 that have whole-number square roots. For 27, the numbers 25 and 36 are around 27 and have whole-number square roots 5 and 6. And since 27 is closer to 25 than 26, you can also estimate that your decimal answer should be closer to 5 than 6, and will likely even be under 5.
Method 3- An exact answer. The answer above shows how to find an exact answer to the square root of If you are having to find the square root of a larger number and are not sure what square numbers go into it, another option is to find the prime factorization of the number breaking the number into all the smallest prime numbers that multiply to make it. For example, the prime factorization of 27 would be 3, 3, 3. The prime factorization of 30 would be 2, 3, 5.
## Main Topics
Free math lessons and math homework help from basic math to algebra, geometry and beyond. Students, teachers, parents, and everyone can find solutions to their math problems instantly.
### Privacy FAQs
Homework resources in Squares and Square Roots - Middle Grades - Math Military Families The official provider of online tutoring and homework help to .
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# 3259 (number)
3259 is an odd four-digits prime number following 3258 and preceding 3260. In scientific notation, it is written as 3.259 × 103. The sum of its digits is 19. It has a total of one prime factor and 2 positive divisors. There are 3,258 positive integers (up to 3259) that are relatively prime to 3259.
## Basic properties
• Is Prime? yes
• Number parity odd
• Number length 4
• Sum of Digits 19
• Digital Root 1
## Name
Name three thousand two hundred fifty-nine
## Notation
Scientific notation 3.259 × 103 3.259 × 103
## Prime Factorization of 3259
Prime Factorization 3259
Prime number
Distinct Factors Total Factors Radical ω 1 Total number of distinct prime factors Ω 1 Total number of prime factors rad 3259 Product of the distinct prime numbers λ -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ 8.08918 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 3259 is 3259. Since it has only one prime factor, 3259 is a prime number.
## Divisors of 3259
2 divisors
Even divisors 0 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ 2 Total number of the positive divisors of n σ 3260 Sum of all the positive divisors of n s 1 Sum of the proper positive divisors of n A 1630 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 57.0877 Returns the nth root of the product of n divisors H 1.99939 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 3259 can be divided by 2 positive divisors (out of which none is even, and 2 are odd). The sum of these divisors (counting 3259) is 3260, the average is 1630.
## Other Arithmetic Functions (n = 3259)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 3258 Total number of positive integers not greater than n that are coprime to n λ 3258 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 465 Total number of primes less than or equal to n r2 0 The number of ways n can be represented as the sum of 2 squares
There are 3,258 positive integers (less than 3259) that are coprime with 3259. And there are approximately 465 prime numbers less than or equal to 3259.
## Divisibility of 3259
m n mod m
2 1
3 1
4 3
5 4
6 1
7 4
8 3
9 1
3259 is not divisible by any number less than or equal to 9.
## Classification of 3259
• Arithmetic
• Prime
• Deficient
### Expressible via specific sums
• Polite
• Non hypotenuse
• Prime Power
• Square Free
## Base conversion 3259
Base System Value
2 Binary 110010111011
3 Ternary 11110201
4 Quaternary 302323
5 Quinary 101014
6 Senary 23031
8 Octal 6273
10 Decimal 3259
12 Duodecimal 1a77
20 Vigesimal 82j
36 Base36 2ij
## Basic calculations (n = 3259)
### Multiplication
n×y
n×2 6518 9777 13036 16295
### Division
n÷y
n÷2 1629.5 1086.33 814.75 651.8
### Exponentiation
ny
n2 10621081 34614102979 112807361608561 367639191482300299
### Nth Root
y√n
2√n 57.0877 14.8261 7.55564 5.04216
## 3259 as geometric shapes
### Circle
Diameter 6518 20476.9 3.33671e+07
### Sphere
Volume 1.44991e+11 1.33468e+08 20476.9
### Square
Length = n
Perimeter 13036 1.06211e+07 4608.92
### Cube
Length = n
Surface area 6.37265e+07 3.46141e+10 5644.75
### Equilateral Triangle
Length = n
Perimeter 9777 4.59906e+06 2822.38
### Triangular Pyramid
Length = n
Surface area 1.83963e+07 4.07931e+09 2660.96
## Cryptographic Hash Functions
md5 a87d27f712df362cd22c7a8ef823e987 ed86f246f734646d674dd346e5bb88f9d2efbabf b26ac45aca5d09d6a563f169df1cba8b19156c70f1bf2912ca980d6aaef55990 0f0eb07b8042ce815843e29fba501cb7311a69998f42cabd045a12cd3c9dc69ac93d4682932745eb6dbd252c95b545e7bf3916fabcb0f9c752bc797e468f9d09 f4c75ebebf2b2d4f7d4187b7563fecddb9061c43
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# Convert Pechus to Hour Angles
### Kyle's Converter > Angle > Pechus > Pechus to Hour Angles
Pechus * Hour Angles Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Hour Angles to Pechus
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 Pechus:
1 Pechus is equal to about 2° (or possibly 2½°). If a pechus is 2° then it is approximately equivalent to π/90 radians (SI).
1 Hour Angle:
1 Hour angle is 1 turn/24 or 15°. In terms of SI units an hour angle is π/12 radians.
Conversions Table
1 Pechus to Hour Angles = 0.133370 Pechus to Hour Angles = 9.3333
2 Pechus to Hour Angles = 0.266780 Pechus to Hour Angles = 10.6667
3 Pechus to Hour Angles = 0.490 Pechus to Hour Angles = 12
4 Pechus to Hour Angles = 0.5333100 Pechus to Hour Angles = 13.3333
5 Pechus to Hour Angles = 0.6667200 Pechus to Hour Angles = 26.6667
6 Pechus to Hour Angles = 0.8300 Pechus to Hour Angles = 40
7 Pechus to Hour Angles = 0.9333400 Pechus to Hour Angles = 53.3333
8 Pechus to Hour Angles = 1.0667500 Pechus to Hour Angles = 66.6667
9 Pechus to Hour Angles = 1.2600 Pechus to Hour Angles = 80
10 Pechus to Hour Angles = 1.3333800 Pechus to Hour Angles = 106.6667
20 Pechus to Hour Angles = 2.6667900 Pechus to Hour Angles = 120
30 Pechus to Hour Angles = 41,000 Pechus to Hour Angles = 133.3333
40 Pechus to Hour Angles = 5.333310,000 Pechus to Hour Angles = 1333.3333
50 Pechus to Hour Angles = 6.6667100,000 Pechus to Hour Angles = 13333.3333
60 Pechus to Hour Angles = 81,000,000 Pechus to Hour Angles = 133333.3333
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https://www.suss.edu.sg/courses/detail/mth310
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# Singapore University of Social Sciences
## Linear Optimisation Methods and Applications (MTH310)
Applications Open: To be confirmed
Applications Close: To be confirmed
Next Available Intake: To be confirmed
Language: English
Duration: 6 months
Fees: To be confirmed
Area of Interest: Science & Technology
Schemes: Lifelong Learning Credit (L2C)
Funding: To be confirmed
### Synopsis
The formulation of optimization models is considered, together with applications, and the various solution techniques widely used to solve the underlying linear objective function under a constrained systems of linear equations. The computer software accompanying this course is used as a powerful tool to solve various linear optmimization models.
Level: 3
Credit Units: 5
Presentation Pattern: Every January
E-Learning: BLENDED - Learning is done MAINLY online using interactive study materials in Canvas. Students receive guidance and support from online instructors via discussion forums and emails. This is supplemented with SOME face-to-face sessions. If the course has an exam component, This will be administered on-campus.
### Topics
• Systems of linear equations.
• Solution space.
• Formulation of linear programming models.
• Matrix formulation of simplex method.
• Principles of the simplex solution method.
• One-phase and two-phase simplex method.
• Integer programming models.
• Solution methods.
• Applications of linear programming.
• Integer programming.
• Optimization problem-solving.
• Case studies of industrial and applied problems.
### Learning Outcome
• Apply LU decomposition and examine conditions for convergence of solutions of linear systems.
• Formulate linear programming models. Apply the graphical method to find the optimum solution of two-variables linear programming models.
• Use the standard one-phase simplex method to solve linear optimisation models.
• Analyse linear optimisation problems with the two-phase simplex solution technique.
• Employ the branch and bound method to solve integer programming and 0-1 variable models.
• Compute the optimum solution of a large linear programming model with the modern interior point method.
• Construct a range of mathematical techniques to solve a variety of quantitative problems.
• Formulate solutions to problems individually and/or as part of a group.
• Analyze and solve a number of problem sets within strict deadlines.
• Verify solutions related to linear optimisation using Mathcad.
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https://math.stackexchange.com/questions/2652998/function-calculating-the-curvature-of-earth/2653027
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# Function calculating the curvature of earth
I am interested in calculating a box, based on a center point and its ∆ values. The box is an excerpt of the earth' surface. Note that I am not working with a projection of the earth (2D), but with values accounting for its original form - being a sphere (3D).
Given is the center (c) point - in geographical coordinates (latitude (x) / longitude (y)) -> c(x/y). Another information I am having is ∆ latitude (∆y) and ∆ longitude (∆x).
Here an abstraction of the earth and how the points are located.
_______90°______
/ \
∆(y) \
/ \
c(x/y) --- ∆(x)--------\
/ \ Longitude
/ \
|--------0° Equator --------|
\ /
\ /
\----- latitude ------/
\ /
\ /
------ -90°----
Given this information, I can calculate x/y from p1..p4, i.e. p2 = x + ∆x or p4 = x - ∆x. Once I have p1...p4, I have the coordinates for x1...x4 which are the corners.
x1------p1------x2
| | |
| ∆y |
| | |
p4--∆x--c(x/y)--p2
| | |
| | |
| | |
x4------p3------x3
When I draw the polygon created from x1...x4, it covers more than it should. This gets more extreme the closer I am to north - or south pole. I assume this is because of the curvature of the earth. The more I am moving north or south, the more distance is between two latitude lines
How would I correct the calculated polygon regarding the curvature of the earth? I tried a linear function f(x) = 1/90 * x but that only approximates it roughly.
• The "geographical coordinates" of course behave very badly near the poles. Why do you want to use them? From the mathematical standpoint, in order to get reasonable results you should use several different charts to cover the surface of the earth. If you like "latitudes and longitudes" consider a "multipolar world" where in addition to the north and south poles, you have few more (at least two more). – Moishe Kohan Feb 16 '18 at 13:49
• @MoisheCohen I do not have choice. The only information I am getting are geographical coordinates... I know its not good - but that is the only information I am having. Could you explain the chart approach? – Stophface Feb 16 '18 at 13:50
• You mean that for the center point $c$ you have the "standard" coordinates, but that's OK. My suggestion is, as I said, to use several different charts. You would have to compute the "transition maps" near your center point $c$. You thus, will have, say, two different "boxes" centered at $c$: One is standard and the other is obtained from the first one by a rotation. In a band which avoids the standard poles you will use one box (the standard one) while near the "standard" poles you will use another one. The more precision you need, the more boxes/charts you would have to use. – Moishe Kohan Feb 16 '18 at 13:56
$$dS=r^2\sin\phi\, dr\,d\phi\,d\theta$$ so that with a constant $\theta$ step, the $\phi$ step should grow like $\csc\phi$. The function that achieves such a behavior is the antiderivative,
$$\int\dfrac{d\phi}{\sin\phi}=-\log(\cot(\phi)+\csc(\phi)).$$
This function has vertical asymptotes at $\phi=0$ and $\phi=\pi$, which was to be expected as it is impossible to achieve constant area at the poles. A possible solution is to use plain disks around the poles, locally dropping the mesh topology.
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# Quick Answer: What Is An Example Of Unit Rate?
## What part of a graph tells you the unit rate?
On a linear graph, the unit rate is obtained by calculating the slope of the line.
On a curve, the unit rate is the slope of the line tangent to the….
## What is the formula of death rate?
CRUDE DEATH RATE is the total number of deaths to residents in a specified geographic area (country, state, county, etc.) divided by the total population for the same geographic area (for a specified time period, usually a calendar year) and multiplied by 100,000.
## How do you write a rate as a unit rate?
A unit rate is a rate with 1 in the denominator. If you have a rate, such as price per some number of items, and the quantity in the denominator is not 1, you can calculate unit rate or price per unit by completing the division operation: numerator divided by denominator.
## How do I calculate rates?
Use the formula r = d/t. Your rate is 24 miles divided by 2 hours, so: r = 24 miles ÷ 2 hours = 12 miles per hour. Now let’s say you rode your bike at a rate of 10 miles per hour for 4 hours.
## What does unit mean?
1 : a single thing, person, or group forming part of a whole There are 36 units in my apartment building. 2 : the least whole number : one. 3 : a fixed quantity (as of length, time, or value) used as a standard of measurement An inch is a unit of length.
## What is meant by rate?
a certain quantity or amount of one thing considered in relation to a unit of another thing and used as a standard or measure: at the rate of 60 miles an hour. a fixed charge per unit of quantity: a rate of 10 cents a pound.
## How do you simplify a unit rate?
With a unit rate, you are comparing a quantity to one. Some common unit rates are miles per gallon, price per pound, and pay rate per hour. To find a unit rate, simplify the ratio so that you have a 1 in the denominator. You can simply divide the first number in the ratio by the second.
## What are unit rates used for?
Unit Rates. A rate is a ratio that is used to compare different kinds of quantities. A unit rate describes how many units of the first type of quantity corresponds to one unit of the second type of quantity. Some common unit rates are miles (or kilometers) per hour, cost per item, earnings per week, etc.
## How do you find the rate and unit rate?
Lesson Summary A unit rate is a ratio between two different units with a denominator of 1. To calculate the unit rate, divide the numerator by the denominator. The resulting decimal number is the unit rate.
## What is the difference between a rate and a unit rate?
A rate is a special ratio in which the two terms are in different units. For example, if a 12-ounce can of corn costs 69¢, the rate is 69¢ for 12 ounces. … When rates are expressed as a quantity of 1, such as 2 feet per second or 5 miles per hour, they are called unit rates.
## How do you solve rate problems?
To solve a problem involving two travelers, follow these steps:Figure out which quantity related to the travelers is equal (a time, distance, or rate).Write two expressions for that quantity, one using each “traveler.”Set the two expressions equal to each other and solve the equation.
## What are some examples of unit rates?
For example, 60 miles in 2 hours is a rate. Students learn that a unit rate is a rate in which the second rate is 1 unit. For example, 30 miles in 1 hour, or 30 miles per hour, is a unit rate. In the problems in this lesson, students are given a rate, and are asked to find the corresponding unit rate.
## What is the definition of unit rate math is fun?
How much of something per 1 unit of something else. Examples: • 100 cars pass by in 2 hours. The unit rate is 50 cars per hour.
## What is a unit rate 7th grade math?
Ratios & Unit Rates Ratio is a comparison between two quantities by division. A rate is a ratio that compares quantities in different units. A unit rate is a rate with a denominator of 1.
## How do I calculate simple interest rate?
Simple interest is calculated by multiplying the daily interest rate by the principal, by the number of days that elapse between payments. Simple interest benefits consumers who pay their loans on time or early each month.
## What are unit rates?
Unit Rate Definitions Unit rate is used to compare quantities in which the second quantity is one. Common examples of unit rate would include, miles per hour earnings per hour cost per gallon. –
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# A formulation of continuity for bilinear forms
• quasar987
In summary, the conversation discusses the conditions for continuity in a bilinear form. The first two conditions, that the form is linear in both arguments and that there exists a constant M such that |a(x,y)|<M||x||||y||, imply continuity. However, the question arises as to whether continuity also implies the second condition. It is then mentioned that there is a theorem stating that a linear functional on a Hilbert space is continuous if and only if it is bounded. The conversation concludes that in this case, continuity does imply the second condition, as the norm of (x,y) must be less than 1 for the bilinear form to be bounded.
quasar987
Homework Helper
Gold Member
[SOLVED] A formulation of continuity for bilinear forms
## Homework Statement
My HW assignment read "Let H be a real Hilbert space and a: H x H-->R be a coninuous coersive bilinear form (i.e.
(i) a is linear in both arguments
(ii) There exists M>0 such that |a(x,y)|<M||x|| ||y||
(iii) there exists B such tthat a(x,x)>a||x||^2"
So apparently, condition (ii) is the statement about continuity. But I fail to see how this statement is equivalent to "a is continuous".
I see how (ii) here implies continuous, but not the opposite.
## The Attempt at a Solution
Let z_n = (x_n,y_n)-->0. Then x_n-->0 and y_n-->0. So |a(x,y)|<M||x|| ||y|| implies a(x,y)-->0. a is thus continuous at 0, so it is so everywhere, being linear.
Last edited:
I'm not quite sure what the question is. It seems to have been omitted. But just looking at your argument, z_n=(x_n,y_n)->0 doesn't imply x_n->0 or y_n->0. Does it?
Well, I'm making use of the fact that if (M,d) is metric space, then the product topology on M x M is generated by the metric
D((x1,y1),(x2,y2))=[d(x1,x2)² + d(y1,y2)²]^½
I conclude that the norm on H x H is
||(x,y)|| = [||x||² + ||y||²]^½
And now if (x_n,y_n)-->0, this means that
[||x_n||² + ||y_n||²]^½ --> 0,
which can only happen if x_n-->0 and y_n-->0.
------------
You're right, I have not actually typed the question entirely. It is because I was confused by the fact that they seem to imply that condition (ii) is equivalent to continuity. While (ii) implies continuity, does continuity implies (ii)?
Last edited:
Well, there is the theorem that a linear functional on a Hilbert space is continuous if and only if it's bounded...
You mean bounded in the unit ball?
In this case, you're right, it works. Because ||(x,y)|| = [||x||² + ||y||²]^½ <1 ==> ||x||, ||y||<1 and (ii) implies|a(x,y)|<M for all (x,y) in H x H such that ||(x,y)||<1.
## 1. What is a bilinear form?
A bilinear form is a mathematical function that takes two inputs and produces a single output. It is linear in each of its arguments, meaning that the function satisfies the properties of additivity and homogeneity.
## 2. How is continuity defined for bilinear forms?
Continuity for bilinear forms is defined in terms of boundedness and convergence. A bilinear form is continuous if it is bounded and the inputs converge to a limit point, the output also converges to a limit.
## 3. What is the importance of continuity in mathematics?
Continuity is important in mathematics because it allows us to make predictions and draw conclusions about functions and their behavior. It is a fundamental concept in calculus and analysis, and is used to study the properties of functions and their derivatives.
## 4. How does the formulation of continuity for bilinear forms differ from other forms of continuity?
The formulation of continuity for bilinear forms is unique because it takes into account the behavior of the function in two separate directions, rather than just one. This allows for a more comprehensive understanding of the function and its properties.
## 5. Are there any practical applications of continuity for bilinear forms?
Yes, continuity for bilinear forms has many practical applications in fields such as physics, engineering, and economics. It is used to model and analyze various physical systems and processes, and is essential for understanding the behavior of these systems over time.
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Saturday, December 31, 2011
Fast Calculation Of Prime Numbers
1. Take any number which is a multiple of 9 starting at zero ( 0 ) ( 0, 9, 18, 27, etc. ) .
2. Add 1, 2, 4, 5. 7, 8 to the number so that the first column of the number ends in ( 1, 3, 7, 9 ) . For example ( 11, 13, 17, 19 ) .
3. Test to see if it is a prime . A prime number is defined as being only evenly divisible by itself and 1.
Sunday, December 25, 2011
Time Is Gravity, Entropy, Chaos In Disguise
Gravity appears to us as a weak force, but it is really the effects of time in disguise. Einstein said that mass bends space around the mass and that mass and energy are the same thing in two forms. The more that the mass / energy flexes the space string, the greater the gravity which is seen by us as a force . Time follows the flexed space string at differing velocities depending on the amount of flex in the space string near the mass / energy . Time has a numerical value of nine ( 9 ) and can be broken up into 9 segments ( 1/9 ). The 9 segments can have different velocities and can effect what happens. We can see this phenomena in our own life. Each of us, have adventures in life which may appear spontaneous or random . The things that happen to you or I aren't always at the same time. This is because the energy around us is flexed arbitrarily by either outside , random , or self induced actions that flexes the energy string . If you and I experience the same thing, the length of time may be different because the time segments travel at different speeds. In addition, if something bad happens, it usually follows in units of three ( 3 ), because the universe runs on the number three ( 3 ) and tries to stabilize around the number three ( 3 ) when things get out of whack.
In summary:
1. Mass
2. Weight
3. Time
are all related to space strings. If the space strings flex, time follows the flexed string at different velocities and we see the effects of this flexing as gravity if a mass is involved or entropy / chaos if a mass isn't involved but the results are mild or severe.
Friday, December 09, 2011
Being Good
I got to thinking about being good. For most people, the connotation of being good is equivalent to having a dull and boring life. Most religions stress the need to be good in life and to help others. I also got to thinking about God or in other words the person who designed this universe so you and I would have a reasonable chance of survival. Looking at it from a science perspective it suddenly occurred to me, that being good is necessary to survive. Our universe is filled with energy and mass. Einstein said that energy and mass are equivalent which means that energy and mass can be converted from one to the other and back again. String theory says that all the energy is basically in the format of a string. A vibrating string in a loop gives the mass including us its' properties and behavior. A flexed string around a mass, such as the earth gives the earth its' gravity. The wind or other force is simply strings of energy or mass in motion. Velocity or acceleration is just a form of rolled up energy string ( think car here ) disturbing the energy fields as the mass ( car ) moves along. You and I are essentially a mass doing its' own thing by moving in a space occupied by strings. If we design something such as a car we have to round out its' rough edges so it won't cause any disturbance to the energy strings commonly known as turbulence. Similarly we want to round out our own rough edges in order to be acceptable to other people. Time in our universe is just energy that is still creating strings of space from dark energy and dark matter as it moves in a forward direction. Sometimes dark energy and dark matter break off from time before it gets converted into space. This broken off dark energy and dark matter sometimes brings disaster to ordinarily good people. This is the chance that keeps our world from getting too dull.
Wednesday, November 30, 2011
The Quantum World
The quantum world is made up of individual particles that don't have to travel through space to go somewhere. This world looks like foam continuously bubbling away. Travel in the quantum world is through the action of cloning, superposition and entanglement. Cloning is the method by which the individual particle reproduces itself at new locations. Superposition means that more than one particle can occupy the same space at the same time because space hasn't a physical reality in the quantum world. Quantum particles don't have to travel through space to get from A to B because space doesn't exist. Entanglement is the method by which a particle transfers information from one point to another point . A quantum particle can bubble many copies of itself at the same approximate location. This means that the bubbling quantum particle at the new location forms a cloud. This cloud may also appear as a wave which we can measure. We can also measure the position of the quanta, but we can't do both measurements at the same time. Different quanta may come together to form units of something else. If these quanta can't be separated out then we say a strong nuclear ( cloud ) force exists between them. This strong nuclear ( cloud ) force can also be considered as an unbreakable wave. If these quanta can be separated out in a reaction then we can say it has a weak nuclear force and a breakable wave that forms relationships with other quanta. Sometimes quanta form a rotational cloud with each other and these clouds carry plus and minus charges. These are your electromagnet quanta forces.
Tuesday, November 22, 2011
Time
Time is the basis of everything. When the universe was initially created, time separated out into dark matter, dark energy and string space. Dark matter acts like a thick honey which tends to keep everything in its' place like inertia. Dark Energy acts as an accelerator for the universe. String space when forming circles creates things and the string's circular vibration creates different characteristics which we see as that thing's properties. Unattached curved strings, when vibrating, appear to us a waves which when modulated transmit information. Radios use that technology. A non-vibrating curved string is gravity or mass. Inertia is related to the mass of the object. Spaceless time from another universe that existed before the creation of our universe is now our quantum world. In the quantum world there is time but not space. Einstein spoke of spooky action at a distance which is true in the abstract but not existing physically. If space doesn't exist in the quantum world, you don't have any delays. The absence of space means you can have superposition because more than one thing can exist literally on / in the same spot. You can also have entanglement which means that if something happens in one location the something immediately happens / clones in the other location. This happens because the something doesn't have to travel through physical space. The speed of change is infinite ( timeless ), because the limitation on the speed of light only applies when it is moving through space and not time. To take it one step further, suppose that something went wrong and spaceless time in another universe cloned space creating an expanding bubble resulting in our universe. Since our space includes classical time, and at the time of the creation of our universe everything wasn't immediately converted there would be more space than we first calculated from our perspective of logical events. Also not all the time would be converted which meant dark time, dark matter and dark energy still exists. In fact there is much more dark time, dark matter and dark energy in comparison to matter and anti-matter.
Friday, November 18, 2011
Holographic Projections, Black Holes & Event Horizons
The whole universe is organized and run by time which has a mathematical value of nine ( 9 ). Space is composed of a series of strings whose basic value is one to nine ( 1 – 9 ). The zero ( 0 ) string holds imaginary numbers. Everything that isn't time, is associated with a string that is equal to the one digit sum of its' digits . For instance ( 97 ) contains the digits 9 and 7 which total 16 ( 9 + 7 = 16 ) 16 contains the digits 1 and 6 which total 7 ( 1 + 6 = 7 ). Therefore something with the value of 97 rests on the 7th string. If you take the numbers from 1 to 99, and add the digits until one digit is left you will find that the sequence of the one digit numbers go from 1 to 9 and then repeat themselves. These strings can be thought of as strings of space. These space strings rest on a complex woven background string whose value is zero ( 0 ) and is equivalent to imaginary numbers since you can't show any values on string zero ( 0 ) except zero ( 0 ). The world around us ,including you and I, appears to be solid because our vision can't see finer detail on its' own. You and I including the quantum world is really like a swarm of flies. If you look at a swarm of flies, and someone asks you its' location, you tend to answer in terms of where the densest part of the fly swarm is located. In essence when someone asks you where someone or something is located, you answer in terms where the densest part of someone or something is located. Obviously in this instance the someone or something appears solid because that is where our sight begins and ends. If everything is basically similar to a fly swarm, you can also say it is a holograph of what exists on the average at that location. Dark matter in which the universe rests is really sticky time acting as inertia and gravity while dark energy is the force of time or entropy pulling / accelerating the expansion of the universe. Dark matter as disguised time also appears to us as time or inertia which tends to keep things from falling apart. Dark energy or expanding time appears to us as entropy which tends to pull things apart which we also see as aging, force or deterioration. If you think of the universe as having direction, west to east is the expansion of the strings ( 1 to 9 ) into larger numbers totaling the value of that particular string . For instance the east to west expansion of string seven could create the value of 250 ( 2 + 5 + 0 = 7 ). The value of 250 would be on string seven ( 7 ) since the digits of ( 250 ) would total 7 ( 2 + 5 + 0 = 7 ). This east to west expansion would form a holographic pattern of information or processes which time would then project north and south which we would see as an image, process or new facts depending on our location. Time only moves outward in our universe which we see as entropy, otherwise called deterioration, force and aging. If a holographic projection reaches a black hole, the first thing it sees is an event horizon. At this point the holographic projection is changed because only time and location enter the black hole while space information stays at the event horizon. If time and location leave the holographic projection, the only thing left is the space information string 7 since ( 2 + 5 + 0 = 7 ). Thus the information at the event horizon is proportional to the area of the event horizon rather than the volume of the black hole since the black hole doesn't contain space except from our perspective. The time and location which enters the black hole from the holographic projection is a non – physical distorted quantum representation of the holographic projection without the space string which represents physical space in our seen world. Thus in a spaceless quantum world represented by time and location you get superposition, entanglement and cloning which is instantaneous. If the black hole contracts in size, time and location reverses itself and rejoins its' string which appears to us as Hawking radiation due to the creation of , to us , a new holographic projection. Time and location for the old particle is destroyed in the black hole so conservation of energy is preserved.
Wednesday, November 16, 2011
Holographic Universe
Let's suppose the whole universe is organized and run by time which has a value of nine ( 9 ). Everything that isn't time, is associated with a string that is equal to the one digit sum of its' digits . For instance ( 97 ) contains the digits 9 and 7 which total 16 ( 9 + 7 = 16 ) 16 contains the digits 1 and 6 which total 7 ( 1 + 6 = 7 ). Therefore something with the value of 97 rests on the 7th string. If you take the numbers from 1 to 99, and add the digits until one digit is left you will find that the sequence of the one digit numbers go from 1 to 9 and then repeat themselves. These strings can be thought of as strings of space. These space strings rest on a complex woven background string whose value is zero ( 0 ) and is equivalent to imaginary numbers since you can't show any values on string zero ( 0 ) except zero ( 0 ). The world around us ,including you and I, appears to be solid because our vision can't see finer detail on its' own. You and I including the quantum world is really like a swarm of flies. If you look at a swarm of flies, and someone asks you its' location, you tend to answer in terms of where the densest part of the fly swarm is located. In essence when someone asks you where someone or something is located, you answer in terms where the densest part of someone or something is located. Obviously in this instance the someone or something appears solid because that is where our sight begins and ends. If everything is basically similar to a fly swarm, you can also say it is a holograph of what exists on the average at that location. Dark matter in which the universe rests is really sticky time acting as inertia and gravity while dark energy is the force of time or entropy pulling / accelerating the expansion of the universe. Dark matter as disguised time also appears to us as time or inertia which tends to keep things from falling apart. Dark energy or expanding time appears to us as entropy which tends to pull things apart which we also see as aging, force or deterioration. If you think of the universe as having direction, west to east is the expansion of the strings ( 1 to 9 ) into larger numbers totaling the value of that particular string . For instance the east to west expansion of string seven could create the value of 250 ( 2 + 5 + 0 = 7 ). The value of 250 would be on string seven ( 7 ) since the digits of ( 250 ) would total 7 ( 2 + 5 + 0 = 7 ). This east to west expansion would form a holographic pattern of information or processes which time would then project north and south which we would see as an image, process or new facts depending on our location. Time only moves outward in our universe which we see as entropy, otherwise called deterioration, force and aging.
Sunday, November 06, 2011
Riemann's Hypothesis – Calculating Zeros On ( y = ½ )
Sometimes in life you stumble upon something that is basically true and it's up to other people to prove it. Riemann, in his Riemann's Hypothesis, said that the zeros in his hypothesis all have the value of ½ and lie on the line ( y = ½ ) . Riemann also said that the zeros can be manipulated ( positions changed or zeros added ) in order to get the primes closer to their true location. Riemann provided a formula to prove his belief which was ( s = ½ + it ). The current method of calculating the position of the zeros is by plugging numbers into the formula and calculating. I don't have any quarrel with this method except that there isn't any guarantee that you haven't missed a zero which is supposed to be on the line ( y = ½ ) and is actually off the line ( y = ½ ) . The rub in the calculation is that if you miss a “t” or an “i”, it might be the number that proves Riemann's Hypothesis is incorrect. Personally, I don't think it matters because Riemann accidentally stumbled onto something that is correct so missed numbers in “t” or “i” are academic. I don't know what numbers they are using but the available literature says the first few zeroes were calculated around 14.1344725, 21.022040, 25.010858, 30.424876, 32.935062, and 37.586178. Even if you proved that all the zeros lay on the line ( y = ½ ), that proof still doesn't tell you how all these zeros relate to the position of the primes on the line ( y = ½ ) in useable terms.
Here's my proof for what it is worth:
Riemann said that all his non – trivial zeros lie on the line ( y = ½ ) . These zeros are at the location that the primes form a right angle line to ( y = ½ ) or in other words, are vertical to the line ( y = ½ ) . Prime numbers are defined as numbers that can only be divided by themselves and ( 1 ). An example of one digit primes are ( 1, 2, 3, 5, 7 ). You can see from these numbers that primes aren't linear ( 1, 2, 3, 4 ) but form harmonics. If all the primes were played musically, you would get a tune. The physical distance between the primes would be similar to the value of the rests in a musical piece. Calculations have been done on the location that some of Riemann's zeros lie on the line ( y = ½ ) using Riemann's formula ( ½ + it ). The first few zeroes were calculated around 14.1344725, 21.022040, 25.010858, 30.424876, 32.935062, and 37.586178. These numbers are not close to the position of the beginning one digit primes ( 1, 2, 3, 5, 7 ). You will see that the distance between ( 1, 2, 3, ) is ( 1 ) and the distance between 3 & 5, 5 & 7, is 2. Riemann, in his Riemann's Hypothesis, said that the zeros in his hypothesis all have the value of ½ and lie on the line ( y = ½ ) . Riemann also said that the zeros can be manipulated ( positions changed or zeros added ) in order to get the primes closer to their true location. This manipulation changes the physical distance between the primes which is equivalent to altering the value of the rests in a musical piece.
1. 1 X .99 = .99 ( 1 )
2. 2 X .99 = 1.98 ( 2 )
3. 3 X .99 = 2.97 ( 3 )
If we add the digit ( ½ ) or ( .5 ) in Riemann's line ( y = ½ ) and Riemann's zero ( 0 ) to .99, we form the number ( .509999 ).
1. 5 X ( .509999 ) = 2.549995 (4)
2. 7 X ( .509999 ) = 3.569993 ( 5 )
3. 11 X ( .509999 ) = 5.609989 ( 6 )
This basic principle calculates the position of the primes fairly accurately up to prime 31.
Using the same multipliers on the next prime ( 37 ) we get ( 37 X .509999 = 18.869963. Prime ( 37 ) is actually the 13th prime. Riemann said that the zeros can be manipulated ( positions changed or zeros added ). We still need the ( ½ ) in Riemann's line ( y = ½ ) and the ( 9's ). If we take ( .509999 ) and multiply it by itself ( ( .509999 X .509999 = .26009898 ).
1. ( 37 X .509999 X ,509999 = 9.62366226 ( 13 )
This calculation shows that the calculated position is short of the true position by about ( 13 – 9.62366226 = 3.37633774 ). The difference is approximately equal to Pi ( 3.141592654 ). In some calculations the difference is about the natural number ( e ) ( 2.71828`828 ).
In summary, Riemann's intuition told him that the zeros had a real value of ( ½ ) which was true since the digit ( ½ ) is used in the calculation. Riemann's intuition also told him that the zeros can be manipulated ( position changed or zeros added ) which is also true. What Riemann missed was that the calculation involved the number 9 and that Pi ( 3.141592654 ) and the natural number ( e ) ( 2.718281828 ) might have to be added or subtracted from the final answer. Riemann also missed that powers would also have to be used.
In general for calculating the location of any prime you:
1. Count the number of digits in a prime number. For instance 7919 has 4 digits. Subtract 1 from the number of digits ( 4 - 1 = 3 ) for 7919. Form another number equal to the number of digits in 7919 ( 4 ) by putting ( .5 ) in the far left column and 9 in the far right column. ( .5—9 ). Fill the middle with Riemann Hypothesis zeros ( 0 ) forming a four digit number ( .5009 ). Raise ( .5009 ) to the power of 3 ( which is the number of digits in 7919 ( 4 ) minus 1 ( 4 - 1 = 3 ). ( .5009 ) ^ 3 = .125676215. Multiply 7919 X .125676215 which equals 995.2299524. 7919 is the 1000th prime. The answer is out by approximately 5. Adjust the error by adding or subtracting Pi or (e).
The Clay Mathematics Institute is offering a \$1,000,000 prize for the solution to the Riemann Hypothesis. From my reading, it seems to involve proving whether or not all the zeros lie on the line ( y = ½ ) . Since I have shown how the Riemann Hypothesis relates to the location of the primes and by extension how many primes precede that prime ( counting 1, 2, 3, 4 ), it seems to me it is largely academic as to whether all the zeros lie on the line ( y = ½ ) . I have also found the missing pieces to Riemann's Hypothesis which was correct as far as it went.
Monday, October 31, 2011
The Riemann Hypothesis Resolved ( Hopefully finally)
Riemann said that all his non – trivial zeros lie on the line ( y = ½ ) . These zeros are at the location that the primes form a right angle line to ( y = ½ ) or in other words, are vertical to the line ( y = ½ ) . Prime numbers are defined as numbers that can only be divided by themselves and ( 1 ). An example of one digit primes are ( 1, 2, 3, 5, 7 ). You can see from these numbers that primes aren't linear ( 1, 2, 3, 4 ) but form harmonics. If all the primes were played musically, you would get a tune. The physical distance between the primes would be similar to the value of the rests in a musical piece. Calculations have been done on the location that some of Riemann's zeros lie on the line ( y = ½ ) using Riemann's formula ( ½ + bi ). The first few zeroes were calculated around 14.1344725, 21.022040, 25.010858, 30.424876, 32.935062, and 37.586178. These numbers are not close to the position of the beginning one digit primes ( 1, 2, 3, 5, 7 ). You will see that the distance between ( 1, 2, 3, ) is ( 1 ) and the distance between 3 & 5, 5 & 7, is 2. Riemann, in his Riemann's Hypothesis, said that the zeros in his hypothesis all have the value of ½ and lie on the line ( y = ½ ) . Riemann also said that the zeros can be manipulated ( positions changed or zeros added ) in order to get the primes closer to their true location. This manipulation changes the physical distance between the primes which is equivalent to altering the value of the rests in a musical piece.
1. 1 X .99 = .99 ( 1 )
2. 2 X .99 = 1.98 ( 2 )
3. 3 X .99 = 2.97 ( 3 )
If we add the digit ( ½ ) or ( .5 ) in Riemann's line ( y = ½ ) and Riemann's zero ( 0 ) to .99, we form the number ( .509999 ).
1. 5 X ( .509999 ) = 2.549995 (4)
2. 7 X ( .509999 ) = 3.569993 ( 5 )
3. 11 X ( .509999 ) = 5.609989 ( 6 )
This basic principle calculates the position of the primes fairly accurately up to prime 31.
Using the same multipliers on the next prime ( 37 ) we get ( 37 X .509999 = 18.869963. Prime ( 37 ) is actually the 13th prime. Riemann said that the zeros can be manipulated ( positions changed or zeros added ). We still need the ( ½ ) in Riemann's line ( y = ½ ) and the ( 9's ). If we take ( .509999 ) and multiply it by itself ( ( .509999 X .509999 = .26009898 ).
1. ( 37 X .509999 X ,509999 = 9.62366226 ( 13 )
This calculation shows that the calculated position is short of the true position by about ( 13 – 9.62366226 = 3.37633774 ). The difference is approximately equal to Pi ( 3.141592654 ). In some calculations the difference is about the natural number ( e ) ( 2.71828`828 ).
In summary, Riemann's intuition told him that the zeros had a real value of ( ½ ) which was true since the digit ( ½ ) is used in the calculation. Riemann's intuition also told him that the zeros can be manipulated ( position changed or zeros added ) which is also true. What Riemann missed was that the calculation involved the number 9 and that Pi ( 3.141592654 ) and the natural number ( e ) ( 2.718281828 ) might have to be added or subtracted from the final answer. Riemann also missed that powers would also have to be used.
In general for calculating the location of any prime you:
1. Count the number of digits in a prime number. For instance 7919 has 4 digits. Subtract 1 from the number of digits ( 4 - 1 = 3 ) for 7919. Form another number equal to the number of digits in 7919 ( 4 ) by putting ( .5 ) in the far left column and 9 in the far right column. ( .5—9 ). Fill the middle with Riemann Hypothesis zeros ( 0 ) forming a four digit number ( .5009 ). Raise ( .5009 ) to the power of 3 ( which is the number of digits in 7919 ( 4 ) minus 1 ( 4 - 1 = 3 ). ( .5009 ) ^ 3 = .125676215. Multiply 7919 X .125676215 which equals 995.2299524. 7919 is the 1000th prime. The answer is out by approximately 5. Adjust the error by adding or subtracting Pi or (e).
The Clay Mathematics Institute is offering a \$1,000,000 prize for the solution to the Riemann Hypothesis. From my reading, it seems to involve proving whether or not all the zeros lie on the line ( y = ½ ) . Since I have shown how the Riemann Hypothesis relates to the location of the primes and by extension how many primes precede that prime ( counting 1, 2, 3, 4 ), it seems to me it is largely academic as to whether all the zeros lie on the line ( y = ½ ) . I have also found the missing pieces to Riemann's Hypothesis which was correct as far as it went.
Sunday, October 30, 2011
The Universe Is Built On Strings
There is an old joke which said that all the things I needed to know were covered in kindergarten. Strings appear to be in that category. If you try to measure the definitive length of a string you will find that the task is impossible. First of all the string isn't rigid so you don't know where it begins and ends. Secondly, the length of the string depends on your measuring instrument. Different measuring instruments give different results. To add to the fun, the measuring instrument is not always accurate. Lastly the measurer may need glasses or alternately disturbs the measurement when he observes it ( quantum measurement ). If you tie the ends of a string so it forms a loop and vibrate it, you have illustrated that vibrating strings have different properties when they vibrate and by extension represent different things. For instance, metals, bricks, sand, etc. all look different to each of us because we can't see all their detail. If you hold a string straight it represents quantum time which doesn't have space. Therefore, without space to travel through, in the quantum world, you have cloning, entanglement and superposition .Everything is instantaneous and reproducable without space,having to be traveled and superposition because everything can be in the same location because there is no space to interfere with location. If you hold a string in an arc without moving it, you have gravity which is otherwise inertia. The closer the ends of the string are together in a loop the stronger the gravity ( inertia ). If a string is held in a loop and moved while in a loop, the movement of the string represents entropy, force, acceleration and velocity. The best demonstration of this phenomena is watching a car accelerate from a standing start. First of all the back end dips until the force of gravity or inertia is broken. Then entropy, force, acceleration and velocity happen. You can see this phenomena as soon as the front end of the car comes down and entropy, acceleration and velocity begin. Entropy is the deterioration of the car from position to position as it goes down the track. You and I can't see it because we can't see fine detail.
In summary:
1. A straight string represents spaceless quantum time leading to cloned copies, entanglement and superposition.
2. A tied vibrating looped string represents various properties which we see as things ( bricks, buildings, people ).
3. A static untied looped string represents gravity and inertia.
4. A dynamic, moving untied looped string represents entropy, force, velocity and acceleration.
Wednesday, October 26, 2011
Patterns, Harmonics, Disorder, Chaos
Our universe is based on fundamental patterns. The use of fundamental patterns in our universe is necessary because they help us to organize our surroundings into something that is understood. An example of patterns is language, mathematics and speech. We may have variations on the fundamental pattern such as language, accents and spelling. Mathematics is broken down into adding, subtracting, multiplying and dividing as well as algebra, geometry, calculus and trigonometry. Speech is broken down into sounds that convey a message. Wild animals use speech in the form of sound to communicate. These variations on the fundamental patterns could be called harmonics, similar to harmonics in music as a common example. We also have the opposite to patterns which could be called disorder or in the extreme, chaos. Disorder can be as innocuous as designing a different type of car, building, or inventing a new word. Chaos is basically extreme disorder. An example of this is death or other forms of extensive destruction. Common examples of disorder leading to eventual chaos is aging in ourselves, our possessions or the environment ( read climate change here ). The problem with chaos is that you don't always know it is present unless you can see it or are in the midst of it. Our universe is based on mathematics. Theoretically if you mapped all the incoming data, you could see when chaos is about to start or has started. That is a great theory but it breaks down in its' applicability when the incoming data is overwhelming. A simple example is counting from one ( 1 ) to infinity. If you graph all the numbers you will see it climbs up in one direction like climbing a hill. A simple method is to add all the digits in a number until you get a one column number ( for instance 476, ( 4 + 7 + 6 = 17, ( 1 + 7 =8 ) ). You will find that all the numbers from one to infinity total from ( 1 to 9 ) and then repeat themselves in a pattern. The resulting pattern looks like a saw tooth. You could call this a fractal pattern since it repeats itself indefinitely. Prime numbers are an example of a seemingly patternless chaos in our world, but they have an underlying complicated fractal pattern . Prime Numbers are defined as numbers that can be only divided by themselves and one ( 1 ). Prime Numbers follow a complicated fractal pattern. First of all, if you look at a list of prime numbers you will find that they always have the numbers 1, 3, 7, 9 in column zero or otherwise known as the far right column ( 11, 13, 17, 19 ). The second thing you will notice is that all numbers ending in ( 1, 3, 7, 9 ) aren't prime numbers ( 21, 33, 27, 39 ). This is our first pattern. The second thing you will notice is that if the sum of the digits of any number ending in 1, 3, 7, 9 total a multiple of 3 , except for prime number 3, ( for instance 6, 9, 12, etc. ) it isn't a prime number. If a number ending in 1, 3, 7, 9 in column zero ( far right column ) isn't a prime number it can usually be evenly divided by a number with 1, 3, 7, 9 in column zero ( far right column ). These are more patterns. Lastly, except for the one digit prime number 3 in the one digit prime number series ( 1, 2, 3, 5, 7 ), you will find if you continuously add the digits of a prime number you will find the column zero or far right column one digit totals for prime numbers are ( 1, 2, 4, 5, 7, 8 ). These numbers ( 1, 2, 4, 5, 7, 8 ) are the strange attractors of the prime numbers.
In summary:
1. Prime numbers, if they are prime numbers, have the pattern numbers 1, 3, 7, 9 in column 0 ( farthest right column ).
2. If the sum of the digits of any number ending in 1, 3, 7, 9, total a multiple of 3, except for prime number 3, ( for instance total 6, 9, 12, etc. ) it isn’t a prime number.
3. If a number ending in 1, 3, 7, 9 in column zero (0), isn’t a prime number it can usually be evenly divided by a number with 1, 3, 7, 9 in column (0).
4. If you add the digits and there is more than one total ( for instance 97 has a first digit total of 16 ( 9 + 7 = 16 )) then 16 is a harmonic of ( 7 ), since ( 1 + 6 = 7 ).
5. The column zero or far right column one digit totals for primes are ( 1, 2, 4, 5, 7, 8 ).
6. The one digit totals ( 1, 2, 4, 5, 7, 8 ) are all strange attractors of the complex prime number fractals.
The adding of these digits is a neat way of quantizing energy levels or Riemann numbers, which really is only a series of numbers in which we simplify the pattern in order to study it. Any chaotic system can be quantized in this manner similar to any harmonic oscillator or anything else for that matter.
Thursday, October 20, 2011
The Riemann Digits ½ , Zero ( 0 ) & My 9.
Prime numbers and the Riemann Hypothesis digits are all intertwined. A prime number is defined as a number that can only be divided by itself and one ( 1 ) ( 1, 2, 3, 5, 7 ).
Here's the skinny on prime numbers:
1. Prime numbers, if they are prime numbers, have the digits 1, 3, 7, 9 in column 0 ( farthest right column ) ( 11, 13, 17, 19 ).
2. If the sum of the digits of any number ending in 1, 3, 7, 9, total a multiple of 3, except for 3, ( for instance total 6, 9, 12, etc. ) it isn’t a prime number.
3. If a number ending in 1, 3, 7, 9 in column zero (0), isn’t a prime number it can usually be evenly divided by a number with 1, 3, 7, 9 in column (0). If not try other numbers.
The calculation of the location of the prime numbers on a line or calculating the number of previous primes on a line involve the digits ( ½, 0 & 9 ). Riemann in his Riemann's Hypothesis said that the zeros in his hypothesis all have the value of ½ and lie on the line ( y = ½ ) . Riemann also said that the zeros can be manipulated ( positions changed or zeros added ) in order to get the primes closer to their true location. I have separately discovered that the location of the prime numbers on the line ( y = ½ ) or any other line for that matter involves the digit 9.
The primary number for the location of a prime on any line from the Riemann Hypothesis is ( .509999 ). For the primes ( 1, 2, 3, ) you multiply them by ( .99 ) since their actual positions are ( 1, 2, 3 ). For the primes (5 to 23 ) you multiply them by ( .509999 ). For the primes 29 & 31 you multiply them by ( .509999 ) and subtract Pi ( 3.141592654 ) to bring them closer to their true position.
For the rest of the primes up to 97 you raise ( .509999 ) to the power of 2 and adjust using either Pi or the natural number e ( 2.718281828 ).
In summary, Riemann anticipated the number ( ½ ) & ( zero ( 0 ) and zero adjustment, but missed my discovery of the number 9 and the necessity of Pi and ( e ).
For calculating the location of any prime you:
1. Count the number of digits in a prime number. For instance 7919 has 4 digits. Subtract 1 from the number of digits ( 4 - 1 = 3 ) for 7919. Form another number equal to the number of digits in 7919 ( 4 ) by putting ( .5 ) in the far left column and 9 in the far right column. ( .5—9 ). Fill the middle with Riemann Hypothesis zeros ( 0 ) forming a four digit number ( .5009 ). Raise ( .5009 ) to the power of 3 ( which is the number of digits in 7919 ( 4 ) minus 1 ( 4 - 1 = 3 ). ( .5009 ) ^ 3 = .125676215. Multiply 7919 X .125676215 which equals 995.2299524. 7919 is the 1000th prime. The answer is out by approximately 5. Adjust the error by adding or subtracting Pi or (e).
It would seem from the above, that Riemann anticipated the zeros ( 0 ) and the value of those zeros being ( ½ ) since they were on the line ( y = ½ ). Riemann didn't anticipate the number 9 or the option of adjusting the calculation by adding Pi or ( e ).
The Clay Mathematics Institute is offering a \$1,000,000 prize for the solution to the Riemann Hypothesis. From my reading, it seems to involve proving whether or not all the zeros lie on the line ( y = ½ ) . Since I have shown how the Riemann Hypothesis relates to the location of the primes and by extension how many primes precede that prime ( counting 1, 2, 3, 4 ), it seems to me it is largely academic as to whether all the zeros lie on the line ( y = ½ ) .
Tuesday, October 18, 2011
The Riemann Hypothesis Resolved Using ½ , 0 and 9
Like a lot of people, I've been fooling around trying to prove Riemann's Hypothesis without much success. Let's suppose for the sake of argument that all the zeros lie on the line ( y = ½ ) . If they do, the solution for the Riemann Hypothesis brings us no closer to showing what the deep connection is between the zeta function and the distribution of the primes. My reading of the literature on the Riemann Hypothesis indicates that the zeros lie on the line ( y = ½ ) and that if the zeroes are manipulated in some fashion you can calculate the location of the primes to a greater and greater accuracy. If you know the location of one prime you know how many primes precede it because you are counting ( 1, 2, 3, 4 etc. ). Riemann's genius or intuition was that the distribution of the primes involve the number zero ( 0 ) and ½. The line is important because if you count something, that something is being placed on a line. If the something is a prime it lies on a line. Riemann's Hypothesis says that all the zeros lie on a line so for the sake of argument you have primes on a line and zeros ( 0 ) on a line. Riemann said that if you manipulate the zeros ( 0 ), you can bring the prime sitting on the line closer and closer to its' true position. What the line ( y = ½ ) means is a mystery, although the number ( ½ ) may prove useful. The Riemann Hypothesis equation brings the zeros ( 0 ) to the line ( y = ½ ) depending on what you are using for ( t ) or the other variables. Let's suppose the primes lie on a straight line which also has zeros ( 0 ) on it.
The first five primes are:
1. 1 ( Yeah, I know it shouldn't be included )
2. 2
3. 3
4. 5
5. 7
The closest you can get these primes to their true position is multiplying them by ( .99 ) if we ignore multiplying them by 1 which will tend to distort their position after prime 3 .
1. 1 X .99 = .99
2. 2 X .99 = 1.98
3. 3 X .99 =2.97
4. 5 X .99 = 4.95
5. 7 X .99 = 6.93
The primes from 6 to 10 are:
6. 11
7. 13
8. 17
9. 19
10. 23
Up to this point we have the number 9, and the Riemann Hypothesis numbers zero ( 0 ) and ( ½ ) . Riemann said we can also manipulate the zeros ( 0 ) to bring the calculated location of the primes closer to their true location.
For example prime number 11's true location is 6. If we multiply ( 11 X Riemann's Number ( ½ ) including Riemann's ( 0 ) and some of the missed number ( 9 ) we get ( 11 X .5099999 = 5.609989 ). Riemann said that we can also manipulate the zeros ( 0 ) which means we can add zeros or put them somewhere else in the multiplier. ( 11 X .5909999 = 6.5009989 ) or (11 X .50990099 = 5.608910891). You will see from these calculations that the best multiplier is .5099999 since the multiplication ( answer ) is the closest to 6.
The above example is ridiculous because the prime number is so small, but the principle is the same. The calculation of a prime's true position is more complicated, but this is the basic idea.
In summary you need:
1. ½
2. 0
3. 9
I calculate that the best usable number for the location of a prime is ( .5099999 ). Riemann said that the proportion of primes is about ( 1 / ln (x) ) where ( ln(x) is the natural logarithm of x to base ( e ) or ( 2.718281828 ) . Using Riemann's formula the proportion of primes below prime number 11 is ( 1 / ln(11) or ( .417032391 ). The number of primes below ( 11 ) using this formula is ( 11 X .4107032391 = 4.587356306) or in round numbers ( 5 ). My calculation using ( .5099999 ) is ( 6 ) in round numbers which is the exact location of prime number ( 11 ). Riemann worked out that if the zeros really do lie on the critical line, then the primes stray from the ( 1/ln(x) ) distribution exactly as much as a bunch of coin tosses stray from the 50:50 distribution law. I have proved that the non-trivial zeros have to lie on the line, because my calculation for the location of the primes uses Riemann's non-trivial zeros that have to be on the line ( y = ½ ) or a line since you are calculating the location of the primes in a sequence ( or the number of preceding primes ).
The Clay Mathematics Institute is offering a \$1,000,000 prize for the solution to the Riemann Hypothesis. So far no one has solved it. From my limited understanding of the Riemann Hypothesis the available calculations don't always show where the zeros cross the line ( y = ½ ) in terms of where each prime is situated. The Riemann Hypothesis only conjectures the zeros are on the line ( y = ½ ) . The primes aren't evenly spaced so it is doubtful that one equation is going to provide a decent formula for the location of all the primes. My solution is strictly arithmetic so it isn't as elegant as a function, but it works with some tweaking which is exactly what Riemann conjectured when he said the zeros can be manipulated. I believe that I have proved that the Riemann zeros are all on the line whether it be ( y = ½ ) or some other line, because those zeros have to be used in a calculation while being on a line to prove that the location of the primes are on a straight line and not somewhere else in space.
Sunday, October 16, 2011
The Position Of Primes Using The Riemann Hypothesis Numbers & The One That Was Missed
Like a lot of people, I've been fooling around trying to prove Riemann's Hypothesis without much success. Let's suppose for the sake of argument that all the zeros lie on the line ( y = ½ ) . If they do, the solution for the Riemann Hypothesis brings us no closer to showing what the deep connection is between the zeta function and the distribution of the primes. My reading of the literature on the Riemann Hypothesis indicates that the zeros lie on the line ( y = ½ ) and that if the zeroes are manipulated in some fashion you can calculate the location of the primes to a greater and greater accuracy. If you know the location of one prime you know how many primes precede it because you are counting ( 1, 2, 3, 4 etc. ). Riemann's genius or intuition was that the distribution of the primes involve the number zero ( 0 ) and ½. The line is important because if you count something, that something is being placed on a line. If the something is a prime it lies on a line. Riemann's Hypothesis says that all the zeros lie on a line so for the sake of argument you have primes on a line and zeros ( 0 ) on a line. Riemann said that if you manipulate the zeros ( 0 ), you can bring the prime sitting on the line closer and closer to its' true position. What the line ( y = ½ ) means is a mystery, although the number ( ½ ) may prove useful. The Riemann Hypothesis equation brings the zeros ( 0 ) to the line ( y = ½ ) depending on what you are using for ( t ) or the other variables. Let's suppose the primes lie on a straight line which also has zeros ( 0 ) on it.
The first five primes are:
1. 1 ( Yeah, I know it shouldn't be included )
2. 2
3. 3
4. 5
5. 7
The closest you can get these primes to their true position is multiplying them by ( .99 ) if we ignore multiplying them by 1 which will tend to distort their position after prime 3 .
1. 1 X .99 = .99
2. 2 X .99 = 1.98
3. 3 X .99 =2.97
4. 5 X .99 = 4.95
5. 7 X .99 = 6.93
The primes from 6 to 10 are:
6. 11
7. 13
8. 17
9. 19
10. 23
Up to this point we have the number 9, and the Riemann Hypothesis numbers zero ( 0 ) and ( ½ ) . Riemann said we can also manipulate the zeros ( 0 ) to bring the calculated location of the primes closer to their true location.
For example prime number 11's true location is 6. If we multiply ( 11 X Riemann's Number ( ½ ) including Riemann's ( 0 ) and some of the missed number ( 9 ) we get ( 11 X .5099999 = 5.609989 ). Riemann said that we can also manipulate the zeros ( 0 ) which means we can add zeros or put them somewhere else in the multiplier. ( 11 X .5909999 = 6.5009989 ) or (11 X .50990099 = 5.608910891).
The above example is ridiculous because the prime number is so small, but the principle is the same. The calculation of a prime's true position is more complicated, but this is the basic idea.
In summary you need:
1. ½
2. 0
3. 9
The Clay Mathematics Institute is offering a \$1,000,000 prize for the solution to the Riemann Hypothesis. So far no one has solved it. From my limited understanding of the Riemann Hypothesis the available calculations don't always show where the zeros cross the line ( y = ½ ) in terms of where each prime is situated. The Riemann Hypothesis only proves the zeros either cross the line or are on the line ( y = ½ ). The primes aren't evenly spaced so it is doubtful that one equation is going to provide a decent formula for the location of all the primes. My solution is strictly arithmetic so it isn't as elegant as a function, but it works with some tweaking which is exactly what Riemann conjectured when he said the zeros can be manipulated. The most fascinating part is that the tweaking involves Pi and the natural number ( e ) but that is another story.
Saturday, October 08, 2011
Riemann Hypothesis & Strings
Riemann's Hypothesis is that all the non-trivial zeros lie on the line ( y = ½ ) and these zeros have something to do with prime numbers. The calculations done up to the present time have shown that the zeros are on the line ( y = ½ ) and there are yet more calculations to go. Let us suppose that each of Riemann's zeros are part of separate strings of numbers and each time the Riemann equation crosses the line ( y = ½ ) it does so where a zero is located. To make it simple, suppose that the zeros are in the middle of the number and the number 1 is on both ends ( 101, 1001, 10001 etc. ). If we add up the digits in these numbers we find the total is always 2 ( 1 + 0 + 1 = 2 ), ( 1 + 0 + 0 + 1 = 2 ), ( 1 + 0 + 0 + 0 +1 = 2 ). To take it one step further we could say these numbers are on the line ( y = 2 ) since all the numbers total 2. Riemann's Hypothesis is that the non- trivial zeros lie on the line ( y = ½ ). If we flip our analogy, the numbers on the line ( y = ½ ) become ( 1 / 101, 1/1001, 1/ 10001, etc. ). Riemann's equation still crosses the line ( y = ½ ) where the zeros are located. Riemann's equation isn't exact in terms of the positions of the zeros. It has been calculated that the first position of the zeros is around 14 and the second calculated position around 21. Our analogy is exact because we can create these numbers to infinity and the principle is the same. Riemann said that the zeros have a real value of ½ or .5 . If we add ( ½ + 1/101 ) we obtain ( .50990099 ). Prime number ( 11 ) is the 6th prime ( 1, 2, 3, 5, 7, 11 ). ( 11 X .50990099 = 5.6089108911 ) or 6 rounded to one digit. The zeros in these numbers are Riemann Zeros and the 9's are Riemann 9's and can be adjusted to get a more accurate estimate. As a matter of interest prime number 97 is the 26th prime. If you multiply ( 97 X ( .50990099 ) X ( .50990099 )) you get ( 25.22969903 )which is very close to 26. If you adjust the Riemann zeros to form the number ( .509999 ) and do the multiplication ( 97 x (.509999 X .509999) ) you get ( 25.22960106 ) which is further away from 26. You can see from these calculations that if the Riemann zeros are manipulated as Riemann has suggested to obtain the location of the prime ( or calculate the number of primes up to that point ) you can vary the distance to the true position of the prime. As a matter of interest Riemann's equation first cuts the line ( y = ½ ) at a position just over 14. The 14th prime is 41. ( 41 X ( ½ + 1/ 101) X ( ½ + 1 / 101 ) is ( 10.664099959 ) which is just off 14 by the value of Pi ( 3.141592554 ). The 21st prime is 71. It's position is calculated at 18 using the same method ( 71 X ( ½ + 1/ 101 ) X ( ½ + 1 / 101 )) which is very close to 21. The calculation of the prime numbers and their positions are a little bit more complicated than this illustration but this is the basics.
Wednesday, October 05, 2011
Riemann Hypothesis Solved Using A Quantum Mechanical System ( Revised )
Riemann's Hypothesis is that all the non-trivial zeros lie on the line ( y = ½ ) and these zeros have something to do with prime numbers. Since Riemann's Hypothesis includes the line ( y = ½ ), let's chose a quantum energy system in which the energy levels are 2. Riemann's equation also included complex numbers. In a quantum energy system Riemann's complex numbers could be represented by zero's ( 0 ) since ( 2 + 0 = 2 ). ( 1 + 1= 2 ) also equals 2. Riemann also went on to say that the prime number locations were influenced by the position of the zeros. To extend our ( 1 + 1 = 2 ) analogy, we now have ( 1 + 0 + 1 = 2 ). We can convert these additions to numbers ( 11, 101, etc. ). We can now say that we have a real axis ( y = 2 ) with infinite numbers with zeros between their ones ( 11, 101, 1001, 10001, etc. ) on an infinite real axis line ( y = 2 ). We can also say that each of the zero's in ( 101, 1001, 10001, etc. ) are on the line ( y = 2) since the numbers ( 101, 1001, 10001, etc. are also on this line. Riemann also said that the values of the zeros ( 0's ) on the line are equal to ( ½ ) which is also true when the line ( y = 2 ) is flipped. We could also extend this argument to one which says that the value of the ones ( 1's ) are also ( ½ ) since the reciprocal of one ( 1 ) is ( 1 ) and the fraction ( 1 / 1 ) when added ( 1 / 101 ) has the value ½ on the line ( y = ½ ) . These numbers would also be evenly spaced on the real axis line ( y = 2 ) since each infinite number would total energy level 2 which means the distance between the numbers would be ( 2 + 2 = 4 ) or a spacing of 4. Flipped the distance between the numbers would be one ( 1 ) since ( ½ + ½ = 1 ). Riemann's Hypothesis says that all his formula's non-trivial zero's are on the line ( y = ½ ) and this is what we've proved up to now since the non-trivial zeros are in the numbers ( 101, 1001, 10001, etc. ). If we flip our infinite real axis line ( y = 2 ), we create a real axis line ( y = ½ ) with flipped infinite real numbers ( 1/11, 1/ 101, 1/1001, etc. ) all the way to infinity. If we add ( ½ + 1/101 ) we obtain ( .50990099 ). Prime number ( 11 ) is the 6th prime ( 1, 2, 3, 5, 7, 11 ). ( 11 X .50990099 = 5.6089108911 ) or 6 rounded to one digit. The zeros in these numbers are Riemann Zeros and the 9's are Riemann 9's and can be adjusted to get a more accurate estimate.. As a matter of interest prime number 97 is the 26th prime. If you multiply ( 97 X ( .50990099 ) X ( .50990099 )) you get ( 25.22969903 )which is very close to 26. If you adjust the Riemann zeros to form the number ( .509999 ) and do the multiplication ( 97 x (.509999 X .509999) you get ( 25.22960106 ) which is further away from 26. You can see from these calculations that if the Riemann zeros are manipulated as Riemann has suggested to obtain the location of the prime ( or calculate the number of primes up to that point ) you can vary the distance to the true position of the prime. The calculation of the prime numbers and their positions are a little bit more complicated than this illustration but this is the basics.
Sunday, October 02, 2011
Riemann Hypothesis Solved Using A Quantum Mechanical System
Riemann's Hypothesis is that all the non-trivial zeros lie on the line ( y = ½ ) and these zeros have something to do with prime numbers. Since Riemann's Hypothesis includes the line ( y = ½ ), let's chose a quantum energy system in which the energy levels are 2. Riemann's equation also included complex numbers. In a quantum energy system Riemann's complex numbers could be represented by zero's ( 0 ) since ( 2 + 0 = 2 ). ( 1 + 1= 2 ) also equals 2. Riemann also went on to say that the prime number locations were influenced by the position of the zeros. To extend our ( 1 + 1 = 2 ) analogy, we now have ( 1 + 0 + 1 = 2 ). We can convert these additions to numbers ( 11, 101, etc. ). We can now say that we have a real axis ( y = 2 ) with infinite numbers with zeros between their ones ( 11, 101, 1001, 10001, etc. ) on an infinite real axis line ( y = 2 ). These numbers would also be evenly spaced on the real axis line ( y = 2 ) since each infinite number would total energy level 2 which means the distance between the numbers would be ( 2 + 2 = 4 ) or a spacing of 4. Riemann's Hypothesis says that all his formula's non-trivial zero's are on the line ( y = ½ ). If we flip our infinite real axis line ( y = 2 ), we create a real axis line ( y = ½ ) with flipped infinite real numbers ( 1/11, 1/ 101, 1/1001, etc. ) all the way to infinity. If we add ( ½ + 1/11+ 1/ 101 + 1001 + etc. ) we eventually start to get numbers resembling ( .509999. .500999, etc. ) The zeros in these numbers are Riemann Zeros and the 9's are Riemann 9's. If you raise these numbers to the power of 2, you get somewhere over ¼ . As a matter of interest prime number 97 is the 26th prime. If you multiply ( 97 X ( .5099999 ) X ( .5099999 ) you get very close to 26. The calculation of the prime numbers are a little bit more complicated than this illustration but this is the basics.
Tuesday, September 27, 2011
The Complicated Fractal Nature Of Prime Numbers
Mandelbrot, the inventor of fractals, when he worked for IBM was presented with a problem involving interference in the transmission of information. Every so often parts of the transmitted information would seem to randomly drop off / scramble which, needless to say, caused problems with the transmitted informational message. The problem had to be fixed, but as the information drop off / scramble appeared to be random , everyone was flummoxed because of the lack of a recognizable pattern. Mandelbrot thought about it and started experimenting. One time he decided to take a straight line and divide it into 1/3rd . He discarded the middle 1/3rd and kept the other 2/3rd separated by a space ( ------ ------ ). He continued on and discovered that the pattern produced by this method matched the pattern of the informational message drop off / scramble. This discovery proved that the informational drop off / scramble wasn't random but followed a fractal pattern.
Prime Numbers also follow a fractal pattern. Prime Numbers are defined as numbers that can be only divided by themselves and one ( 1 ).
Prime Numbers follow a complicated fractal pattern. First of all, if you look at a list of prime numbers you will find that they always have the numbers 1, 3, 7, 9 in column zero or otherwise known as the far right column ( 11, 13, 17, 19 ). The second thing you will notice is that all numbers ending in ( 1, 3, 7, 9 ) aren't prime numbers ( 21, 33, 27, 39 ).
The second thing you will notice is that if the sum of the digits of any number ending ending in 1, 3, 7, 9 total a multiple of 3 ( divide by 1/3rd and discard the potential prime just like Mandelbrot discarded his string sections ), except for 3, ( for instance 6, 9, 12, etc. ) it isn't a prime number. If a number ending in 1, 3, 7, 9 in column zero ( far right column ) isn't a prime number it can usually be evenly divided by a number with 1, 3, 7, 9 in column zero ( far right column ).
Lastly, except for the one digit prime number 3 in the one digit prime number series ( 1, 2, 3, 5, 7 ) you will find if you continuously add the digits of a prime number ( for instance 97 = ( 9 + 7 = 16 ) ( 1 + 6 = 7 ) you will find the column zero or far right column one digit totals are ( 1, 2, 4, 5, 7, 8 ). All the rest of the columns are zero ( 01, 02, 04, 05, 07, 08 ).
In summary:
1. Prime numbers, if they are prime numbers, have the numbers 1, 3, 7, 9 in column 0 ( farthest right column ).
2. If the sum of the digits of any number ending in 1, 3, 7, 9, total a multiple of 3, except for 3, ( for instance total 6, 9, 12, etc. ) it isn’t a prime number. If a number ending in 1, 3, 7, 9 in column zero (0), isn’t a prime number it can usually be evenly divided by a number with 1, 3, 7, 9 in column (0).
3. Except for the one digit prime number 3 in the one digit prime number series ( 1, 2, 3, 5, 7 ) if you add the digits of a prime number ( for instance 97 = ( 9 + 7 = 16 ) ( 1 + 6 = 7 ) the column zero or far right column one digit totals are ( 1, 2, 4, 5, 7, 8 ).
Thursday, September 22, 2011
One Small Step For The Riemann Hypothesis
The Riemann Hypothesis says that all the non-trivial zeros ( 0 ) are on the line ( y = ½ ) and that this hypothesis has something to do with prime numbers. Another way of expressing it, is by saying that the magnitude of the oscillations of primes around their expected position is controlled by the real parts of the zeros of the zeta function. In particular, the error term in the prime number theorem is closely related to the position of the zeros.
In summary:
1. We have a line ( y = ½ ) .
2. We have some zeros that are real and we can use them since they touch / cross the line ( y = ½ )
3. The error term in the prime number theorem is related to the position of the zeros.
The Prime Number Theorem is concerned with the number of primes preceding a number. Due to the nature of mathematics when you multiply, it is probably better to have the number as a prime number for considerations of accuracy. In other words, if you can calculate the location of a prime number in the prime number series you automatically know how many prime numbers precede it.
We have the line ( y = ½ ) . Therefore let the first digit of the multiplier be ½ or .5. We have some zeros that are real and we can use them since they touch / cross the line ( y = ½ ) . Therefore let the middle digits be zeros ( 0 ). We now have the number .50. To finish our multiplier add the digit 9 to the end forming the number ( .5099999999 ).
The first prime numbers from 1 to 12 in order are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31. If we multiply these numbers by ( .509999999 ) our answer is very close to their actual position ( 23 X .509999999 = 11.72999999 ). The actual position of prime number 23 is 10. There are 9 prime numbers preceding 23 ( 1, 2, 3, 5,7,11,13, 17, 19 ).
The prime numbers from 13 to 26 in order are ( 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 ). If you multiply these primes by ( .509999999 ) you will find that the prime number positions are hugely incorrect. You will see in the “ In summary “ list that I indicated under ( 3. ) that the error term in the prime number theorem is related to the position of the zeros. You can increase the number of zeros by either adding them between ½ and 9 (for instance .5000999999) or by raising ( .509999999 ) to the power of 2 ( .509999999 X.509999999 = ( .2600999 )). You will see from ( .2600999 ) that we have adjusted the error term in the prime number theorem by adjusting the zeros ( 0 ) from one to two. Multiply the prime numbers by ( .2600999 ) to obtain the prime number location.
If you do the multiplication, you will find that some of the prime number locations are still out. For instance, ( 37 X .2600999 = 9.623699981). The actual location is 13. For some inexplicable reason if you add Pi ( 3.141592654 ) to this number you get ( 12.76529263 ). This trick of either adding or subtracting Pi works in the majority of cases. In some cases adding or subtracting the natural number ( 2.718281828 ) also works.
Here’s how the system works for numbers in general.
1. Count the number of digits in a prime number. For instance 7919 has 4 digits. Subtract 1 from the number of digits ( 4 - 1 = 3 ) for 7919. Form another number equal to the number of digits in 7919 ( 4 ) by putting ( .5 ) in the far left column and 9 in the far right column. ( .5—9 ). Fill the middle with Riemann Hypothesis zeros ( 0 ) forming a four digit number ( .5009 ). Raise ( .5009 ) to the power of 3 ( which is the number of digits in 7919 ( 4 ) minus 1 ( 4 - 1 = 3 ). ( .5009 ) ^ 3 = .125676215. Multiply 7919 X .125676215 = 995.2299524. 7919 is the 1000th prime number. The calculation is short by approximately the value of Pi ( 3.141592654 ). Pi + 995.2299524 is 998.3715451 which is very close to 1000.
It can be seen from these calculations that the magnitude of the oscillations of the primes around their expected position is controlled by the zeros ( 0’s) in the multiplier. The error term is closely related to the position of the zeros in the number ( .509999999 ). The error term can be controlled by either adding zeros ( .500999999, .50009999 ) or by raising these numbers to a power ( multiply the numbers by themselves ) thereby increasing the zeros. A further adjustment can be made by adding or subtracting Pi ( 3.141592654 ) or the natural number “e” ( 2.718281828 ).thereby creating a range. The present method of proving the Riemann Hypothesis consists of calculating whether or not zeros cross or touch the line ( y = ½ ) . While I appreciate the effort, it seems to me that I have proved the Riemann Hypothesis because my zeros can be infinitely added between the digits ½ and 9 ( .5----9 ). It can be seen that the calculation of the position of a prime and the number of primes preceding that prime all depend on the Riemann Hypothesis' real zeros. Thus the Riemann Hypothesis has been proved by illustrating its’ real relationship to the position of the primes and the number of primes preceding it.
Monday, September 19, 2011
Theory Of Everything
Here's The Theory Of Everything in a nutshell without the mathematics. The classical world that you and I see every day consists of:
1. Time
2. Energy / Weight / Mass
3. Space
Einstein in his equation E = MC^2 said that energy and mass were equivalent. Weight is the pull of gravity on an object. That is why Newton's apple fell from the tree since the pull of the earth ( gravity ) is greater than the pull of the apple ( gravity ) on the earth. Mass is the same thing as weight without gravity pulling on you and I. As far as space is concerned, Einstein said that space doesn't exist unless something extends into it. If you look at something you are extending your sight ( observation ) into space. If you drive a car you are extending it into space ( distance ). If we didn't have space there wouldn't be velocity, acceleration or force ( Force = Mass X Acceleration ( into space) ). Time is a marker for something happening ( The apple fell from the tree at 2:00 a.m.. ). Time is also a measure of distance ( I traveled 30 kilometers / miles in 30 minutes ).
In the quantum world there is time but not space. Einstein spoke of spooky action at a distance which is true in the abstract but not existing physically. If space doesn't exist in the quantum world, you don't have any delays. The absence of space means you can have superposition because more than one thing can exist literally on / in the same spot. You can also have entanglement which means that if something happens in one location the something immediately happens / clones in the other location. This happens because the something doesn't have to travel through physical space. The speed of change is infinite ( timeless ), because the limitation on the speed of light only applies when it is moving through space and not time. Information also travels using the same principle. If we attempt to measure the quantum / spaceless world from our classical / space world, by using particles ( light / x-ray etc. ) we disturb the something we are measuring by adding quanta / energy ( particles ) to it. This energy addition changes the original quantum something so that we aren't measuring the original.
In summary quantum time world without classical space is:
1. Cloning
2. Superposition
3. Entanglement
In summary our classical world is:
1. Time
2. Space
3. Energy / Weight / Mass
Wednesday, July 06, 2011
Quantum Gravity
One of the mysteries of life is gravity. Gravity is a peculiar form of energy but you can't do much with it because all it does is curve string space around a mass to create a force. Essentially, the denser the mass, the more the string space around the mass is curved and the stronger the force of gravity. The further you travel from the mass with its' gravity, the less the string space is curved and hence the weaker the force of gravity but it never goes away. Time in our reality is primarily used as a marker for the passing of, presence of, or future of, objects, particles, actions and events . Space in our universe is composed of flexible strings. Einstein says that space doesn't exist until something extends itself into it. That extension into space causes the string space to flex creating gravity. In the quantum world strings do not function as space. Strings function as quantum gravity which connects rolled up quantum Time that appears to us as various particles. This phenomena creates a graininess below the Planck length. Quantum gravity provides instantaneous entanglement and superposition. Entanglement means everything is instantaneous because there isn't any space which has to be traveled. Superposition means that more than one thing can be in the same location, because string space doesn't exist. This was the situation at the instance of the birth of our universe. Quantum Time connected by quantum gravity which we see as quantum particles. Once our universe started, quantum time converted into particles, objects, actions and events. Quantum gravity converted into string space, string forces and string energy. All of quantum time and quantum gravity wasn't converted, so we now see quantum gravity as nuclear force and quantum time as various nuclear particles.
Wednesday, June 22, 2011
The Helping Principle
One of the toughest principles is the helping principle. It could also be called the father's principle, the mother's principle, the ex-lover's principle, the friend's principle or any other principle with an appropriate adjective. Thinking about it, it seems to me it also could be called the God Principle. Whether we like it or not, each of us is programmed to help each other. Most of us are also Jack or Jill Average. This means that most of us will live in obscurity, be known only to our families and associates and when we die, unremembered except for our family on odd occasions. Thinking about it, the helping principle, to work effectively, requires you to have guts, fortitude, and the realization that there isn't any glory. The helping principle isn't about you. It also means that most of the time you will have to eventually let go, yet be available for any stumbles. One of the toughest components is the realization that you mustn't let them consciously realize that you are helping them. The helping principle demands that you, yourself, open up emotionally and empathetically. Another part, is that you can't screw up their life, for the self gratification of staying around. If you think that what I'm saying is all mostly airy – fairy just look at the social media. One of the social media's advantages is that you can make it private or eliminate friends. It is one thing in life to suspect that you don't quite measure up to someone's idea of perfection, but quite another to have to cope with the idea that someone's private doesn't include you and to rub it in you are no longer a friend or alas, moved further into the friendless column. One of the most debilitating aspects, is that you may be extremely sensitive to what is coming next, yet you have no chance of bailing. The easiest example, is the Christians that were thrown to the lions for their principles. I, like you, have principles, but if I were about to be thrown to the lions, I'd start to think that maybe I should have been slightly less principled. I'm a Christian. You may be of another faith, but the principle may be the same. One of the people in my faith is Jesus Christ. The older I get, the more I suspect he had more guts than most of us will ever have. I sense that God told him what was going to happen in excruciating detail and that furthermore, that he had no choice in the matter. I'm told that Jesus spent 40 days in the desert but the Bible doesn't say why. I sense that Jesus wanted to discuss what was coming and was it absolutely necessary that he had to do it. Perhaps, Jesus pointed out to God that he wasn't particularly qualified. I, sad to say, am following in Jesus' footsteps. I'm telling God that I'm too old for The Helping Principle and she's telling me, that I've got too much experience to retire. God is going to be the death of me, yet.
Sunday, June 19, 2011
Quantum Gravity
One of the mysteries of life is gravity. Gravity is a peculiar form of energy but you can't do much with it because it is a property of mass. Essentially, the bigger the mass, the more the space around the mass is curved and the stronger the gravity according to Einstein. The further you travel from the mass with its' gravity, the weaker the force of gravity but it never goes away. Time in our reality is primarily used as a marker for the passing of, presence of, or future of, objects, particles, actions and events . Space in our universe is also unique. Einstein says that space doesn't exist until something extends itself into it. In the quantum world space doesn't exist at all. This means that quantum time doesn't have any space yet we have entanglement and superposition. If space doesn't exist, yet we have entanglement and superposition, then we need a spaceless quantum force. The spaceless quantum force is quantum gravity because it's instantaneous and spaceless. This was the situation at the instance of the birth of our universe, but as soon as space was created, quantum gravity or at least the part we see, became a property of mass.
Monday, June 13, 2011
Quantum Time
The hardest thing to grasp about quantum time is that it hasn't any associated space. If quantum time doesn't have space, it means that anything existing in time, exists anywhere in all places and combinations at the same time. This phenomena is at the root of the multi-universe argument. This phenomena is also at the root of the concept of God in the christian religion. This concept is that God is everywhere at the same time and knows everything. I'm not familiar with all religions, but I suspect that most religions have the same expressed belief. Similarly, if you or I exist in quantum space, we exist everywhere and know everything at the same time. The knowing everything at the same time is due to quantum entanglement because knowledge hasn't any space to travel through from one point to another. Death doesn't exist in quantum time, because death requires space in which to operate and space doesn't exist in quantum time. That is why you and I would live forever. This concept is also the basis of heaven in the christian religion and probably in other religions too. Quantum Time also involves superposition, which is different things existing literally at the same location. The different things can be possibilities as well as things. This is possible because space doesn't exist in quantum time.
Quantum Time can be summarized as :
1. Space less.
2. Entanglement
3. Superposition
Sunday, March 20, 2011
The Quantum World In A Nutshell
In our everyday universe, we all have to travel through space to get from A to B. Our travel through space essentially involves our making a decision about:
1. Velocity
2. Acceleration
3. Trajectory ( direction ).
The quantum world doesn't require us to travel through space to get from A to B. To get around this problem, the quantum world uses a phenomena called entanglement. Entanglement is basically a reproductive mechanism which reproduces us / information from point A to point B's without travel through space being involved. Since travel through space isn't an option, entanglement reproduces us / information in all directions:
1. Past
2. Present
3. Future
In the quantum world, we can exist in all places at the same time, whereas in our everyday universe, we can only exist in all places at the same time through our imagination. This means that if we can exist in all places at the same time in a quantum world we can choose a preferred location from the locations generated by quantum entanglement, while our other selves choose their locations and proceed on their own paths and adventures.
Existing in all places at the same time, means that we are now subject to probability ( ½ ), possibility ( yes, no, maybe = 1/3rd ) and Likelihood ( 1/9 ( When does Time ( equal to 9 ) kick in ??? )).
Probability, possibility, and likelihood, in the quantum world relies on randomness for a decision. These random happenings in the quantum world appear in our familiar universe as bad or good luck. A probability, possibility and likelihood choice is also instigated as soon as we observe / try to measure the quantum world because we have to use energy in some form to do it.
In summary:
1. Entanglement
2. Probability, Possibility, Likelihood
3. Randomness / outside forces.
Wednesday, March 16, 2011
Quantum Slits, Quantum Information & Schrodinger's Cat
Suppose we have a quantum photon whose electron volt value is 632 electron volts. The quantum photon can be represented by a ball of string. It's surrounding electromagnetic wave can be represented by curved strings. This ball of string ( photon ) with its' surrounding waves ( curved strings ) represent a quantum electron particle and its' waves create an electromagnetic field. The surrounding electromagnetic field has a value of 11 ( add the digits in 632 ( 6 + 3 + 2 = 11 ). The electromagnetic field ( value = 11 )will radiate out forever in a circle from the photon unless something interferes with its' path. If you put a single vertical slit in a barrier, you will find that the electromagnetic field ( value 11 ) shows the most light behind the slit and dimmer light to the side since the electromagnetic wave starts to radiate outward after it leaves the back of the slit.
If you add more slits beside the first one, you will see bright bars and dark bars where no light seems to hit at all. The electromagnetic wave ( value 11 ) went through all the slits and it started to radiate outward after it left the slits. Unfortunately, the radiating wave from each of the slits either interfered with other slits or reinforced other slits. Thus you either got light ( reinforcement ) or dark bands ( interference ).
The fun starts after you decide to measure the electromagnetic field behind some slits as the electromagnetic field comes through all the slits. The problem arises over how you intend to do it. In our universe we have to use energy in some form to measure something. This energy usually comes in the form of a particle ( electrons or photons ). If the electromagnetic field is strong enough the use of electrons or photons or anything else, usually doesn't prevent ( or significantly corrupt ) our measurement. If we try to measure something in the quantum world, our added measurement energy changes the electromagnetic field we are trying to measure. In this case, our measurement destroys the electromagnetic field coming through some slits and we get a series of one slit results ( shows the most light behind the non-interfered slit and dimmer light through the interfered with slit since the electromagnetic wave starts to radiate outward after it leaves the back of the non-interfered slit ).
Most things in our universe is either yes or no. One choice out of 2 possibilities ( either yes or no ). One choice out of two possibilities is ( ½ ) . There is a third choice, which I like to think of as an unknown result ( one choice out of 3 possibilities ( 1/3rd ). So poor old Schrodinger’s cat goes into the old box. After awhile, the old cat is either alive or dead or maybe somewhere in between or not there at all. Anyway there is a third unknown ( can't guess ) possibility. In our universe, we don't know until we look. If Schrodinger's cat has shrunk into a quantum state, we'll never know for sure, because we can't measure without destroying the state of the electromagnetic field containing the information surrounding the cat.
The most interesting part is that if there are many possibilities contained in the third possibility ( sort of all things to all people ) then we can extract different possibilities ( information ) from the same electromagnetic field. Of course, what is right and what is wrong, is the question. It is sort of like a crowd of people coming to different conclusions after seeing the same thing in our world .
Monday, March 14, 2011
Particles & Waves
Both the real and the quantum world consists of:
1. Particles
2. Waves
3. Time
In the quantum world a particle generates electromagnet waves and electron waves. In the real world a particle ( us ) generates charisma which is the equivalent to an electromagnet wave. Sometimes a particle ( us ) generates a confidence and belief aura which is equivalent to an electron wave. What we call it is dependent on our language and culture but it is basically the same thing in two universes. The quantum world hasn't any space and in our world Einstein says that something extends into space, but space doesn't exist on its' own even though you and I instinctively see space. There isn't any space in the quantum world, so Time operates like a possibility or probability. That means that Time automatically comes into existence and at that point something statistically happens. The general concept of Time in our universe is that Time marks events when it is used as a clock ( meet me at 1:00 pm. at Frank's ) or as a measure of distance. ( 50 minutes in the oven at 350 degrees ). Quantum Particles or Atoms are also associated with their own personal waves. These quantum particles or mass can be represented by a ball of string. Surround the ball of string with curved strings which represent the ball of string's wave. This ball of string with its' surrounding waves ( curved strings ) represent a quantum particle and its' waves creating an electromagnetic field. Let's assume for the sake of argument that the quantum particle mass is 23. Since it has surrounding curved strings representing an electromagnetic field ( because quantum particle masses are measured in electron volts ) we can let this electromagnetic field be represented by 5 ( add the digits in 23 ( 2 + 3 = 5 ). Since we now have a quantum particle mass of 23 electron volts with a surrounding personal electromagnetic field of 5 we can assume it will interact with other quantum particle masses and their personal electromagnetic fields forming new particles and fields. Suppose we have another quantum particle whose mass is ( 47 ). Its' electromagnet string wave is ( 11 ) ( 4 + 7 = 11 ) or 2 ( 1 + 1 = 2 ). The electromagnet wave fields of the 23 mass electron ( 5 ) and the 47 mass electron ( 2 ) become intertwined. The combined electromagnetic fields have a value of ( 5 + 2 ) = 7. If Space – less quantum Time doesn't switch on, we will continue to have two separate quantum particles whose value is 23 and 47 with separate intertwined electromagnetic fields of 5 and 11(or 2). If Space-less quantum Time, with a quantum value of (9) and a possibility / probability of ( 1/9 ), switches on the two electron volt masses will combine into one mass of 70 electron volts ( 23 + 47 = 70 ). The new electron volt mass of 70 will have an electromagnetic wave of 7 ( 7 + 0 = 7 ).
Thursday, March 10, 2011
Quantum Atom Theory
Quantum Atom Theory is associated with both Quantum Chaos and the Nature Of Reality. Each is organized and run by Time moving in a forward direction. A minor offshoot of these processes is entropy, a process which allows us to disassemble things / processes and rearrange them into new forms of usefulness like communication. The opposite of entropy is gravity which tends to hold things together. The general concept of Time in our universe is that Time marks events when it is used as a clock ( meet me at 1:00 pm. at Frank's ) or as a measure of distance. ( 50 minutes in the oven at 350 degrees ). Quantum Particles or Atoms are also associated with their own personal waves. These quantum particles or mass can be represented by a ball of string. Surround the ball of string with curved strings which represent the ball of string's wave. This ball of string with its' surrounding waves ( curved strings ) represent a quantum particle and its' waves creating an electromagnetic field. Let's assume for the sake of argument that the quantum particle mass is 23. Since it has surrounding curved strings representing an electromagnetic field ( because quantum particle masses are measured in electron volts ) we can let this electromagnetic field be represented by 5 ( add the digits in 23 ( 2 + 3 = 5 ). Since we now have a quantum particle mass of 23 electron volts with a surrounding personal electromagnetic field of 5 we can assume it will interact with other quantum particle masses and their personal electromagnetic fields forming new particles and fields. The problem is we don't know when ( Time ). This is the so called arrow of Time because Time is unidirectional in an outward ( forward ) direction. We can see the same phenomena in our real world. In the quantum world it is called Heisenberg's Uncertainty Principle and in our real world fate or bad luck. In addition, everything in our universe is associated with Space ( particles, actions, events extended into space ) and Time ( when did it happen ??? ). Time itself is being extended outward in a unidirectional direction, but due to the electromagnetic nature of quantum particles, our space per time or Space / Time is being curved. It can be said that Time is a hidden variable of Quantum Mechanics. Each interaction creates an individual Space / Time or reference Frame relative to its' energy and mass. The greater the mass, the slower Time will run in that reference frame ( not speed ( distance / time ) because there is no space in the quantum world ). Going back to our example, the quantum particle mass is 23. Its' electromagnetic field is 5 ( 2 + 3 = 5 ). Subtract ( mass – string ( 23 – 5 = 18 ) to get the Space / Time Frame of 18. The Time string is rather unique. It doesn't have any space in the traditional sense, but the Time string is related to the number 9. The number 9 in the Time string can pop up at any time which means we neither know when it will happen nor the consequences ( Heisenberg again ). Mathematically ( Particle 23 – wave 5 = Space / Time Frame 18 ). ( Frame 18 / Time Value 9 = Space 2 ). The Space that must be covered in the Space / Time Frame is 2. So the dimensions of this Space / Time Frame is Constant Time Value ( 9 ) X Space ( 2 ) = Frame ( 18).
If we have a greater quantum mass particle ( 47 ), its' electromagnet string wave is ( 11 ) ( 4 + 7 = 11 ). Frame ( 47 – 11 = 36 ). The Space / Time Frame is 36 and the Space is ( 36 / 9 = 4 ) in the Space / Time Frame. Therefore it can be seen that the greater the mass ( 47 vs. 23 ), the slower Time ( not speed ( distance / time )) because there is no space in the quantum world. The slower part isn't speed ( distance / time ) but rather the length of Time the phenomena will take to appear in both the quantum and our real world before generating the division ( Frame / Time ) or in this case (36 / 9 = 4 ).
You can also look at it in terms of Space in our real world. If we have a greater quantum mass particle ( 47 ), its' electromagnet string wave is ( 11 ) ( 4 + 7 = 11 ). Frame is ( 47 – 11 = 36 ). The Space / Time Frame is 36 and the Space is ( 36 / 9 = 4 ) in the Space / Time Frame. Therefore it can be seen that the greater the mass ( 47 vs. 23 ), the slower Time ( not speed ( distance / time )) will appear to happen in terms of Space ( 2 vs. 4 ). The slower part isn't speed ( distance / time ) but rather the length of Time the phenomena will take to appear in both the quantum and our real world before generating the division ( Frame / Time ) or in this case (36 / 9 = 4 ).
This basic principle is why things that are better made last longer, because the thing's quantum world can hold together longer before Time generates the ( Frame / Time ) division creating a Space for things to happen.
This principle also generates uncertainty, probability, what will happen ( future ) in both our world and the quantum world because:
1. Don't know when ( Frame / Time = Space ) division will happen.
2. Probability – ½.
3. What will happen ( future particle / event / action ).
Here's another generator of uncertainty:
If we have a mass particle ( 47 ), it can have two electromagnetic waves of 11 and 2 ( 4 + 7 = 11 ), ( 1 + 1 = 2 ). Frame 1 is ( 47 – 11 = 36 ). Frame 2 is ( 47 – 2 = 45 ).The Space / Time Frame 1 is 36 and the Space is ( 36 / 9 = 4 ) in the Space / Time Frame 1 . The Space / Time Frame 2 is 45 and the Space is ( 45 / 9 = 5) in the Space / Time Frame 2 . Therefore we have two potential outcomes and the probability that either one will occur is 1 result out of two ( 2 ) choices or ½. Most decisions / processes / choices in our world are calculated on the probability of ½ or one (1 ) out of two ( 2 ) choices .
Lastly, future events in terms of what will happen in the quantum world ( Heisenberg ) and our real world ( bad luck / good luck ) is based on the interaction of electron volt quantum particles and their electromagnetic waves which can have different values and consequently different actions.
The funniest part is that in most religions we are taught to love ( help ) one another and to be generally good. The reason for it is the way God set up the universe's real and quantum worlds. The key to success and less aggravation is to prevent the electromagnetic fields of the quantum particles from being disturbed to our detriment. Yoga, anyone?????
Oh by the way, telling your ( hyperactive ) friend to “Chill” is based on Quantum Atom Theory and science in general, much to everyone's surprise, including my own!!!
Monday, March 07, 2011
Strings Solve The Riemann Hypothesis
Let's look at the Riemann Hypothesis from the aspect of a string. Suppose we have 9 strings numbered from one to nine ( 1 – 9 ). The only rule for any number on any particular string is that the total of the digits total the value of that string. For instance, string 2, would have digits totaling 2 ( 2, 11, 101, 1001, etc. ) since the digits in all the numbers total 2 ( 1 + 0 + 1 = 2 ) etc.. The question remaining is what do we do about string zero ( 0 ). String zero ( 0 ) in strings has the same function as √ -1 in particle math using numbers, formulas, and symbols. This is so because string zero ( 0 ) cannot be realistically graphed so it means anything as a real string. If string ( 0 ) is imaginary it is on the imaginary plane at right angles to string 2 or any other string for that matter. Each string number ( 2, 11, 101, 101 ) has a tail ( 2 ) on string 2 because the digits total 2 ( 101 = ( 1 + 0 + 1 = 2 )). The next question is concerning the distance between each number that is on string 2. If we add the number 11 ( 2 ) and 101 ( 2 ) number we get 4 ( 2 + 2 = 4 ) so we have a distance of 4 between each number similar to the distance in Calculus. At each location of 4 units we have an imaginary String 0 running vertically from String 2. The Riemann Hypothesis says that all the non-trivial zeros ( 0 ) are on the line ( y = ½ ) running off a line represented by the equation 1/2 + it. Our non-trivial zeros ( 0 ) become real when they cross string 2 at the “4” locations. If we can add zero to the addition of the digits without changing the digit total our zeros ( 0 ) are real. The Riemann Hypothesis, on the other hand, using particle math of numbers, formulas and symbols has to prove the zeros ( 0 ) have to cross ( y = ½ ) . Riemann's formula also uses fractions and imaginary numbers which he adds. Strings can also use fractions and imaginary String 0 which is equivalent to Riemann's method.
String 2 Whole Numbers:
2
101
1001
10001
100001
etc.
String 2 Fractions:
½
( 1/101)
(1 / 1001)
( 1 / 10001)
( 1/ 100001)
etc.
Take the fractions ( 1 / 101 ) etc. and raise them by the power of 2, since the numbers are on string 2. Fraction ( 1 / 101 ) ^ 2 = .2583624924.
Sum the powers which peak around .2583635005.
Prime number ( 97 ) which is the last of the two digit primes is the 26th prime. If we multiply ( 97 X .2583635005 ) we get ( 25.0612595501).
The Riemann Hypothesis says that the location of the prime numbers is dependent on the position of the zeros in the calculation.
If we take the square root of ( 25.0612595501) we get ( .5082946985 ).
Experimentation will show that the square root containing the most accurate zero (s) for the calculation is ( .509999999 ).
Prime number ( 97 ) which is the last of the two digit primes is the 26th prime. If we take the new square root ( .509999999 ) and raise it to the power of 2 ( .509999999 ^2 ) we get ( .260099999). If we multiply ( 97 X .260099999 ) we get ( 25.229699995) which is closer to 26 than using ( .5082946985) in the calculation.
Prime number 29 is the 11th prime. If we multiply ( 29 X (( .50999999 )^1) we get ( 14.78999997 ) which is approximately off 11 by the value of Pi ( 3.141592654 ). Therefore subtract the value of Pi.
Prime number 43 is the 15th prime. If we multiply ( 43 X (( .50999999 )^1) we get ( 11.18429998 ) which is approximately off 15 by the value of Pi ( 3.141592654 ). Therefore add the value of Pi.
Prime number 149 is the 36th prime. If we multiply ( 149 X (( .50999999 )^2) we get ( 38.75489992) which is approximately off 36 by the value of e^1 ( 2.718281828 ). Therefore subtract ( 2.718281828 ).
In summary:
1. The Riemann Hypothesis says that the location of the prime numbers is dependent on the position of the zeros in the calculation. The basic number which involves zero(s) in the calculation is the number ( .50999999 ). The prime number 191 is the 44th prime number The basic number which involves zero(s) in the calculation for the position ( 44th ) of the prime number 191 is the number ( .50099999 ).
2. You may have to add or subtract Pi ( 3.141592654 ) from the calculated position to obtain the most accurate position.
3. You may have to add or subtract the natural number ( e ) or ( 2.718281828 ) from the calculated position to obtain the most accurate position.
Friday, March 04, 2011
Energy / Time Theory
Space / Time generally means that you can't travel through space without traveling through time. Space in our universe is length, width and height. Time in our universe is a marked by a measure of time ( clock ) or a distance in time. Einstein extended this idea when he brought observation into the picture explaining what happened when two observers saw the same phenomena when either at rest or moving in the same frame or different frames. Later on, Einstein brought in his famous equation ( E = mc^2 ) which said that Mass and Energy were the same thing in different forms.
So in summary we now have:
1. Space / Time.
2. Mass / Energy.
But there is something missing and that is Energy / Time. Time and Energy don't contain any space. This may seem very hard to believe because if you and I look out a window we can see time and energy existing in various forms in space. In our world everything has the illusion of being solid simply because our eyes don't have the ability to see into the quantum world without assistance. Even if we could see into the quantum world, the information that we would gain would be very limited because our eyes run on photons for information and photons are so large that they can't always bounce off of what they hit and return to our eye. Sometimes the photons smother the quantum object which we want to see and therefore can't bounce back to our eye with information. It seems to me that in a quantum world all we have are strings which appear to us as Time and Energy. Strings do not contain space in the sense that strings have to travel through it in order to go from A to B. Rolled up Energy strings are a mass in our world and electron volts in a quantum world which we call atoms or particles because there isn't any space in a quantum world. Since the rolled up strings are flexed in a ball you get gravity without movement or inertia with movement. Also, the rolled up energy string ball we call Mass flexes the surrounding strings that aren't part of the ball thus creating gravity that distorts the path of passing photons causing things to appear to be elsewhere. Time is a unidirectional outward going string that doesn't react with the seen universe. Energy is also a string which when moving outward provides force, velocity and acceleration and when acting inward provides the strong and weak nuclear force depending on its' location. If the energy string flexes rapidly you get frequency or different property characteristics if associated with a mass ( energy string ball ). If you attempt to get information from the quantum world you are adding energy strings to it which affects what you are trying to measure or alternately changes its' characteristics ( properties ) so it now appears to be either a wave ( string ) or particle ( a ball of energy string we call a mass ). Since quantum space doesn't exist, how does Energy /Time work together ???? You can think of Energy, and Time as being parallel strings when they are quiescent in a quantum world. If you could see into a quantum world you would notice that the quantum world had areas of cloudiness which we call nuclear particles and atoms ( rolled up energy strings ). These nuclear particles and atoms are rolled up energy strings joined together by energy strings which we call the strong and weak nuclear force. There is no quantum space in the quantum world so everything is instantaneous because of the Time string which hasn't space. In addition because there isn't any space we measure energy in electron volts. An electron is a rolled up energy string with an electron charge on it. The Time string is rather unique. It doesn't have any space in the traditional sense, but the Time string is related to the number 9. The number 9 in the Time string can pop up at any time and so we can relate it to an infinite number of space measurements in our world.
Here's an example:
The common battery which you can readily buy at a store is 1.5 volts or 1.5 electron volts. If you put the 1.5 volt battery in something you will find that over time ( Time string ) it will cease to operate. If you measure the remaining voltage in the battery you will find it is approximately 1.34 volts. If you add the digits in 1.34 volts it will total ( 1 + 3 + 4 = 8 ). Adjust the number 8 so it can be subtracted from 1.34 volts ( 1.34 – 0.08 = 1.26 ). The value of the Time / Energy Frame is 1.26. Divide the Time / Energy Frame ( 1.26 ) by 9 which is the value of the Time String when it appeared ( how long did the battery last ??? ) ( 1.26 / 9 = 0.14 ). If you add ( 1.34 + 0.14 = 1.48 ) which is approximately 1.5.
You can see what happened, but it is an illusion because it actually happened in the quantum world where there isn't any ( quantum ) space. The most interesting thing of all is that probability is birthed in the quantum world when the Time String pops a nine and energy is released into the quantum world. This energy is seen by us as energy, action and events. There are many Time strings in the quantum world busy popping 9's at any time from our space / time perspective. Since the result of all this popping has to go somewhere in a quantum spaceless world we see the results as dark matter, dark mass and dark energy because we can't use our Space / Time to react with it. It is also interesting that our whole universe was created when one of the Time strings in the quantum world popped and didn't do what it was supposed to do.
So, in summary:
1. Space / Time.
2. Energy / Time.
3. Mass / Energy.
Tuesday, March 01, 2011
Theory Of Everything
The Theory Of Everything presupposes that there is one standard equation which when the right numbers are inserted, you have a mathematical explanation for everything. I think the one standard mathematical equation for the explanation for everything is the mathematical representation of a string. I'm not bright enough to write that equation, but I can tell you how the common string represents / can explain everything in our 4 dimensional universe as well as our quantum universe.
The basic construction of the universe in terms of a string is:
1. Energy
2. Mass
3. Time
If you hold out a piece of string in front of you, you will see that the string occupies space, but the string itself has no space because you can't stretch it. Einstein said that things occupy space, but space doesn't exist on its' own as a separate entity. If you try to stretch the string, you are applying an outward motion or force. Eventually, if you are strong enough, it will suddenly break and at the instance of the breakage, you will experience acceleration of the string. The string won't accelerate forever, but it will reach a constant speed which we call velocity ( distance / time ). Conversely, if you take a string and push it inward, you will see that you can't break it. This happens because the string on a quantum level is being held together by a strong nuclear force. Lastly if the string can be combined chemically to make something, then the new combination of whatever you are making is held together by the weak nuclear force on a quantum level. All this is energy in some form or another.
Mass is demonstrated by creating a ball of string. Each string in the ball of string is curved in some degree and not all of the strings are equally curved. A curved or flexed string represents gravity. This is why gravity is notoriously weaker than any other energy force associated with a ball of string. Also, since the curve in all the strings in the ball of string aren't exact you get a different gravity pull just like it happens on earth. Lastly, if you lay out a series of strings in any direction around the ball you will find that the strings surrounding the ball are generally curved representing the pull of gravity for anything traveling close to the ball of curved strings. Finally, if you could make the ball of curved string vibrate you could read its' characteristics. If you could make a string extended into space vibrate in a pattern you could transmit information at a designated frequency. This is the basis of electronics which uses a charged particle ( electron ) to vibrate its' string in a pattern which we call communication.
Time seems to me to be dark matter. It moves outward in one direction. We can't interact with it in terms of changing its' direction or stopping it. We also can't travel backward or forward along a string of time. You can think of Energy, and Time as being parallel strings when they are quiescent in a quantum world. If you could see into a quantum world you would notice that the quantum world had areas of cloudiness which we call nuclear particles and atoms. These nuclear particles and atoms are rolled up mass strings joined together by energy strings which we call the strong and weak nuclear force. There is no quantum space in the quantum world so everything is instantaneous because of the Time string which hasn't space. In addition because there isn't any space we measure energy in electron volts. An electron is a rolled up mass string with an electron charge on it. The Time string is rather unique. It doesn't have any space in the traditional sense, but the Time string is related to the number 9. The number 9 in the Time string can pop up at any time and so we can relate it to an infinite number of space measurements in our world.
Here's an example:
The common battery which you can readily buy at a store is 1.5 volts or 1.5 electron volts. If you put the 1.5 volt battery in something you will find that over time ( Time string ) it will cease to operate. If you measure the remaining voltage in the battery you will find it is approximately 1.34 volts. If you add the digits in 1.34 volts it will total ( 1 + 3 + 4 = 8 ). Adjust the number 8 so it can be subtracted from 1.34 volts ( 1.34 – 0.08 = 1.26 ). The value of the Time / Energy Frame is 1.26. Divide the Time / Energy Frame ( 1.26 ) by 9 which is the value of the Time String when it appeared ( how long did the battery last ??? ) ( 1.26 / 9 = 0.14 ). If you add ( 1.34 + 0.14 = 1.48 ) which is approximately 1.5.
You can see what happened, but it is an illusion because it actually happened in the quantum world where there isn't any ( quantum ) space. The most interesting thing of all is that probability is birthed in the quantum world when the Time String pops a nine and energy is released into the quantum world. This energy is seen by us as space, energy, action and events. There are many Time strings in the quantum world busy popping 9's at any time from our space / time perspective. Since the result of all this popping has to go somewhere in a quantum spaceless world we see the results as dark matter, dark mass and dark energy because we can't use our Space / Time to react with it. It is also interesting that our whole universe was created when one of the Time strings in the quantum world popped and didn't do what it was supposed to do.
So, in summary:
1. Space / Time.
2. Energy / Time.
3. Mass / Energy.
It's interesting that Einstein proved that Mass / Energy were the same thing and that Space / Time are related. Since Time has the common value of 9 in the quantum world it seems that Space / Time and Energy / Time are the same thing since both can be divided by Time which has the same value ( 9 ).
So, once again:
1. Space
2. Energy
3. Mass
are three peas in the same pod called Time whose value is 9.
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Catalog: 2015–2016 Catalog Year
MATH 127: Precalculus II
General Information
• Name: Precalculus II
• Discipline: Mathematics (MATH)
• Units (Credits): 3
• Transfer Information: Courses with numbers 100 to 299: This course is designed to apply toward a WNC degree and/or transfer to other schools within the Nevada System of Higher Education, depending on the degree chosen and other courses completed. It may transfer to colleges and universities outside Nevada. For information about how this course can transfer and apply to your program of study, please contact a counselor.
Prerequisites and Recommended Courses
• Prerequisites: MATH 126 or three units of high school mathematics at the level of algebra and above, or consent of instructor
Course Outline
I: Catalog Course Description
Studies circular functions, trigonometric identities and equations, conic sections, complex numbers, and discrete algebra.
II: Course Objectives
Upon completion of this course successful students will be able to:
• Use geometry to work with symmetrical triangles.
• Solve problems using the distance formula and the pythagorean theorem.
• Compute trigonometric functions of special angles and use them to solve for the unknown part(s) of right triangles.
• Use the Laws of Sines and Cosines to solve for the unknown parts of triangles.
• Compute vector sums and differences.
• Solve trigonometric identities and equations.
• Graph trigonometric functions.
• Compute the values of inverse trigonometric functions.
• Graph equations and functions in polar coordinates.
• Use DeMoivre's theorem to find the powers and roots of complex numbers.
Linkage of course to educational program mission and at least one educational program outcome.
General Education Mission: This course addresses the fourth bullet under goal one of the college's mission to, "Provide instruction that contributes to a student's abilities to think critically and solve problems; to reason mathematically and apply computational skills."
Math 127 satisfies the General Education Requirement for any degree or certificate program and addresses the following learning objectives of the General Education Requirement by ensuring that successful students:
• Are able to apply appropriate college-level mathematical skills to real life applications
• Have developed adequate problem solving, creative reasoning, and critical thinking skills.
Educational Objectives for A.A. and A.S. Degrees: Math 127 addresses the following learning objectives of the A.A. and A.S. degree programs by ensuring that successful students:
• Will know the subject matter to a level that is appropriate to the emphasis of the degree program.
• Can succeed at their transfer institutions.
Educational Objectives for A.G.S. Degree: Math 127 addresses the following learning objectives of the A.G.S. degree by ensuring that successful students:
• Will know the subject matter to a level that is appropriate for their desired fields of study.
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# Pendulum Period
## Homework Statement
Prove the equation: T = 2 pi * sqrt(L/g), then determine the period of a "seconds pendulum" (period = 2 sec on Earth) on the surface of the moon.
T = 2pi * r/v
T = 2pi/ω
ar = rω2
## The Attempt at a Solution
Do you assume ar is ag?
T = 2pi/ω
= 2 * pi / sqrt(a/r)
= 2 * pi * sqrt(r/a)
...Apparently, L is supposed to be the length of the pendulum, so somewhere along the line, I completely screwed up.
I can solve the second part of the question just fine, but since I've never actually seen T = 2 pi * sqrt(L/g) used or proved, I'd rather not work with it until I understand where it comes from and how it works.
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# The Role of Writing in the Teaching of Mathematics (Part 1)
Last week while visiting Elon University, I had the opportunity to participate in a number of interesting conversations about the role of writing in the teaching of mathematics. This is a topic I’ve been thinking about since my first-year writing seminar on cryptography was approved. I learned a few lessons about the teaching of writing that semester, and I drew on those lessons, as well as my colleague Patrick Bahls’ fantastic new book, Student Writing in the Quantitative Disciplines, during my conversations at Elon. I’ll share a few thoughts on the teaching of writing in math courses in a future blog post. For now, I would like to focus on the nature of mathematical writing.
When considering the role of writing in mathematics, a natural first inclination is to think about the writing of proofs. This is the kind of writing in which research mathematicians engage: carefully constructed chains of logical reasoning in which new theorems are shown to be true based on existing axioms and theorems. A proof is a kind of argument, but proof writing is not like the kind of argumentative writing used in other disciplines. Mathematicians, unlike those in the humanities and social sciences and even natural sciences, don’t support their arguments with evidence or data; they use if-then statements and other forms of logical reasoning. There are no other points of view to consider, nor is there need for statistical significance. A sound proof leaves no room for doubt about the veracity of its theorem.
As quick as mathematicians are to go to the writing of proofs when thinking about mathematical writing, our statistician colleagues move just as quickly to valuing expository writing, at least those at Elon did. It seems undergraduate statistics majors often go on to write reports for non-statisticians. Such expository writing requires statisticians to communicate very technical ideas to a relatively non-technical audience, to justify their results in ways that those with much less statistical training can understand and believe. Mathematicians engage in this kind of writing, too, sometimes. See the works of Keith Devlin for examples of fine expository writing about mathematics.
As I thought about these two types of mathematical writing, I realized that they have something in common with other (non-mathematical) kinds of writing. To write mathematics well, one must know one’s audience. This is perhaps more obvious with expository writing, in which the goal is to help a “lay” audience understand fairly arcane mathematical results. To do this well, one has to know the mathematical background of one’s reader and to provide a sort of scaffolding built on that mathematical background that assists the reader in following the arguments one wants to make. (This is not unlike the kind of scaffolding that effective teaching provides to novices in a domain.)
Audience is important in the writing of proofs, too, although it’s harder to see that importance. Research mathematicians write to their peers, who are assumed to be equally versed in the mathematical content of a paper as the writer is. Imagine the mathematical ideas required to understand a particular proof as a ladder, one idea on top of another. A freshman calculus student might be on the lower rungs of this ladder, a math major a little higher up, a math grad student higher still, and colleagues in one’s sub-field of mathematics just a rung or two from the top. When research mathematicians write papers, they usually don’t try to provide any scaffolding to help the reader along since their audience is assumed to be only one or two rungs away. This means, as any grad student can attest, a research paper can be very hard to understand for those not at the top of the ladder.
In expository writing, there is an intentional choice to write for an audience well down the ladder of mathematical ideas. In the writing of proofs, there’s also a choice as to which audience to write for. Mathematicians could write papers that, say, grad students find easier to read and understand, but they don’t, and I don’t think they’re generally very intentional about this choice. Mathematicians place a high value on writing that is precise and concise, and writing that is aimed at those further down the latter is seen to lack those qualities. To be fair, mathematicians also value what they call elegance in the writing of proofs, which is a somewhat nebulous combination of cleverness and clarity. Truly elegant proofs often make sense to readers at many different rungs on the latter.
What implications does all of this have for the teaching of writing in mathematics? Teaching math students to write well involves teaching them to explain their mathematical ideas in ways their audience can understand, and that requires students to make intentional choices about their target audience. It’s important to put these two elements, choice of audience and the use of appropriate explanations (scaffolding) for that audience, front and center in the teaching of writing in math. Perhaps the ladder metaphor is one that would help students understand these key “moves” in mathematical writing.
More on this topic in a future post!
Image: “my view for the next week,” Atle Brunvoll, Flickr (CC)
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This presentation is the property of its rightful owner.
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# Science Assignment PowerPoint PPT Presentation
Science Assignment. Ashlee Marshall and Ashley Lenger. Bedroom. Radio Computer/laptop T.V Phone Charges Lava Lamps Lamps Lights Electric Blanket Air Conditioner Fan Heater . AC vs. DS. Lava lamps are an AC electricity items.
Science Assignment
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## Science Assignment
Ashlee Marshall and
Ashley Lenger
### Bedroom
• Computer/laptop
• T.V
• Phone
• Charges
• Lava Lamps
• Lamps
• Lights
• Electric Blanket
• Air Conditioner
• Fan
• Heater
### AC vs. DS
• Lava lamps are an AC electricity items.
• It is AC because the Lava Lamps needs than 240v to power it, and DC can’t supply that amount of voltage.
### Electrical Specifications
• The Voltage for a Lava Lamp is 240 volts because it is AC electricity.
• The currents is 3 Amp.
• To work out the Power ratings to have to divide the Voltage by the Currents. That is 240v divided by 3 Amp.
• The power rating is 80 Watts of Power.
### The lava lamp circuit
Thermostat
Switches
240v A.C
Light Bulb
Heating Component
### How the Circuit Works
• The circuit works by the electrons travelling around the circuit.
• First the electrons get to the switch, if the switch is open the electrons will not be able to pass. It is the same if the Thermostat, it is there to stop the Lava Lamp over heating.
• Next the electrons come to the Heating Component and globe the electrons will separate evenly because the paths are the same size.
• The electrons will then met up again and the circuit will start again.
Electrical
Theory
### What is an electrical Current??
• An electric current of Electrons, flowing through wires and electronic components. As the electrons are pushed through pipes by a pump, electric current is pushed through wires by a battery.
• An Electrical Current can only flow through a closed circuit
### Good metals Conductors and why?
• Good metal conductors are metals like Copper, aluminium and silver.
• In a conductor, electric current can flow freely. Unlike insulators.
### Insulators
• Insulators are materials that don’t allow the electrons to pass though the material.
• Also insulators have no changes.
• Insulators to keep heat or electricity either in or out of a container.
• Good Insulators are materials like glass, wool and rubber.
• Most non-metallic solids are good insulators.
### Where and how is electricity generated?
• A power station contains large machines called turbines, which are turn very quickly.
• Power stations need large amounts of energy to turn the turbines. The spinning turbine causes large magnets to turn within wire coils( these are the generators).
• The moving magnets within the coil of wire causes the electrons to move within the coil of wire.
• Electricity can be generated most anywhere on the land, underwater, in the skyand in outer space.
### 3 other ways
Three other ways to generate electricity can be.
• Solar energy.
• Wind energy
• Magnetic energy. Uses magnetic force to induce perpetual motion by propelling itself generating electricity.
### Bibliography
• http://en.wikipedia.org/wiki/Lava_lamp
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http://andrewgelman.com/2008/10/27/what-is-the-probability-your-vote-will-make-a-difference/
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## What is the probability your vote will make a difference?
Nate Silver, Aaron Edlin, and I estimated the probability that a single vote in any state will be decisive in the presidential election. It was Aaron’s idea, Nate supplied the simulations, and I calculated the probabilities and made the graphs.
Here’s our article describing what we did, here’s the abstract:
One of the motivations for voting is that one vote can make a difference. In a presidential election, the probability that your vote is decisive is equal to the probability that your state is necessary for an electoral college win, times the probability the vote in your state is tied in that event. We compute these probabilities for each state in the 2008 presidential election, using state-by-state election forecasts based on the latest polls. The states where a single vote is most likely to matter are New Mexico, Virginia, New Hampshire, and Colorado, where your vote has an approximate 1 in 10 million chance of determining the national election outcome. On average, a voter in America has a 1 in 60 million chance of being decisive in the presidential election.
and here are some graphs:
There are more graphs if you follow the link to the article.
As Aaron, Noah, and I have discussed, it can be rational to vote even when the probability of decisive vote is 1 in 10 million.
P.S. Typo in Figure 1 caption above fixed.
P.P.S. See here for more discussion of why we can compute the probability of a decisive vote, even though the election might be decided by a recount
1. Robert Kern says:
Shouldn’t the map’s caption read “States with lighter colors…”
2. […] certainly not, of course. But Andrew Gelman has posted a chart showing some interesting calculations about how likely your vote is to be decisive in any […]
3. […] Yglesias makes a case for the National Popular Vote with an accompanying visual chart (courtesy of Andrew Gellman) illustrating how much your vote is likely to matter depending on where you […]
4. […] Fathers weren’t perfect This post at Red State Blue State Rich State Poor State makes a strong argument for why the electoral […]
5. […] Nate Silver (the brilliant electoral scientist behind FiveThirtyEight) and Aaron Edlin put together a chart that “estimated the probability that a single vote in any state will be decisive in the […]
6. […] determined that a single vote in New Mexico, New Hampshire, Virginia, and Colorado has about a 1 in 10 million […]
7. Abe says:
Actually, the chance that your vote would be decisive is zero. Even if the “true” vote were exactly 50/50, limitations in verifying such the count would never allow for a single vote to be decisive. Recount after recount would ensue, and none would be definitive to that degree of accuracy–not to mention the legal battles that would ensue.
8. LnGrrrR says:
Young, military NH voter here, putting my ballot in for Obama…
9. Nick says:
This is kind of an obnoxious question since I haven’t really taken the time to understand your methodology but…
If I use a simple indpendent binomial model for voting in California, starting with a uniform distribution on the percentage of Obama voters and then conditioning on the last four poll results (giving McCain everyone who didn’t respond for Obama), I seem to be getting the probability of a tied vote in California as around 10^-19. This is a bit lower than the roughly 10^-8 you have in your paper. Obviously there might be some problems with the independent binomial model but I think I was fairly generous in my calculation in other ways (e.g. the uniform prior and giving McCain all undecided/third-party votes). Is there a simple explanation for why I should believe that the probability of a tied election in California is closer to 10^-8/10^-9 than to 10^-19?
10. James H says:
This gives me a lot to think about between now and voting day. Thanks for putting such a good effort into it.
James H.
http://serviceafol.blogspot.com/
11. tom0063 says:
Again no breakdown of Nebraska by Congressional districts.
Omaha – the only city in the USA with its own electoral vote, and a very even split between Obama and McCain, should certainly be up near the top of this chart.
We’ll split NE’s vote and bring Electoral College reform to the rest of America.
12. Andrew says:
Abe: See recent blog entry for an explanation.
Nick: Read the linked article. Our calculations come directly from the uncertainty distributions derived from Nate’s simulations. The binomial distribution is irrelevant to the question.
Tom0063: Well, we do mention Nebraska and Maine in a footnote . . .
13. […] computed the probability that one single vote will swing the election in each state. In Pennsylvania, the odds that your vote will be the single, deciding vote is somewhere around 0.4 […]
14. […] United States of NM, NH, and VA Filed under: Politics — Tags: electoral college — Jon @ 12:37 pm There are lots of great reasons to live in a place like California or DC — having a direct say in the government isn’t exactly one of them. […]
15. […] and I have bonded over coffee): protest/third-party votes aren’t without their purpose. And since your vote counts but doesn’t really, think about it. (Or just go read the article and call it even.) Posted in politics | Tags: […]
16. […] Noah Using the latest polls, the folks at Red State, Blue State, Rich State, Poor State have calculated the probability that a single vote would decide the outcome of the presidential election, i.e.: […]
17. […] Here are some graphs from Andrew Gelman at Columbia University that illustrate that point: […]
18. […] likely is it your vote will count? Depends where you live. In New Mexico 1 in 10 million. In Utah 1 in 100 million. i don’t care, I still felt […]
19. […] Andrew Gelman, Nate Silver, and Aaron Edlin estimate the the probability of any given vote being decisive in Tuesday’s US election is about one in 60 million. Residents of a few states, like New Mexico, Virginia, New Hampshire, and Colorado, are especially important: their chance is one in 10 million. Oh, did I mention that your vote doesn’t matter? […]
20. Nick says:
Andrew: I had read the paper, though not as thoroughly as I perhaps should have. Clearly you didn’t actually use the independent binomial model but I didn’t understand why it should give such wildly different results. I’m guessing that the answer to that, though, is wrapped up inside the simulations which aren’t described in this paper, hence the lack of enlightenment on my part. (Reading around a bit, it looks to me like perhaps a lot of the difference comes down to putting too much confidence in the polls.)
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http://www.cut-the-knot.org/Curriculum/Fallacies/HooperParadox.shtml
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## Hooper's Paradox:How Is It Possible?
This page is dedicated to the valiant effort by Douglas Rogers to establish the true date of the first appearance of what Martin Gardner called Hooper's paradox. In his small book, M. Gardner indicates that the paradox can be found in the 4th volume (p. 286) of the 1794 edition of William Hooper's Rational Recreations. The point of the inquiry is that the reference is to the 4th edition of the book, the first having appeared 20 years earlier, in 1774. Was the paradox included (or even known) at the time of the original writing, or was it conceived some time in the intervening twenty years before the 4th edition? I trust we'll know the truth soon.
The truth, according to Douglas Rogers, is that the puzzle has been included in the first edition of Hooper's book with a conspicuous error in the diagram. Instead of a 2×6 rectangle Hooper drew a 3×6 shape. The error has been corrected in later editions. But a question arose, How does one make such a pronounced mistake? Driven by curiosity, Professor Rogers embarked on a search that led him to a discovery that the puzzle in this form has appeared in a 1769-1770 collection Nouvelles récréations physiques et mathématiques by the French author Edmé Gilles Guyot. Guyot has corrected his mistake in the second edition of his work even before the appearance of Hooper's first edition. Comparison of the two works by expert librarians at the University of London and University of Aberdeen proved beyond a reasonable doubt that Hooper cribbed the puzzle from Guyot's text probably along with much else.
So what is the puzzle?
A 3×10 rectangle is cut into two equal triangles that form a 2×6 rectangle and two equal trapezoids that combine into a 4×5 rectangle.
Thus an area of 30 square units is being transformed into 32 by mere cutting and rearrangement of the pieces. Not for nothing Hooper called his paradox "geometric money." As M. Gardner observed, Hooper's paradox can be given an infinite number of forms by varying the proportions of the figures and the degree of slope of the diagonal. It can be constructed so that the loss and gain is 1 square unit, 2, 3, 4, 5, and so on up to infinity. Obviously, the smaller the difference the more difficult is to detect the deceiving nature of the construction.
[Frederickson, pp. 271-273] credits Sebastiano Serlio (1475-1554), an Italian architect and mathematician, with one of the variants that transformed the same 3×10 rectangle into a 4×7 and 1×3 rectangles. Curiously, as Prof. David Singmaster has observed, Serlio has been solving a practical problem of converting one rectangle into another and has not noticed that the two had different areas.
The applet below presents a slight modification of Hooper's construction. The four shapes comprising the original rectangle can be dragged to the right into their designated positions on the right.
What if applet does not run?
Three parameters are under the user control: the sides of the rectangle and the length of the horizontal leg of the triangles.
### References
1. G. N. Frederickson, Dissections: Plane & Fancy, Cambridge University Press, 1997
2. M. Gardner, Mathematics Magic and Mystery, Dover, 1956, pp. 131-132
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http://mathoverflow.net/questions/104371/validity-of-generalized-reidemeister-moves-for-a-virtual-knot/104375
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# Validity of generalized Reidemeister moves for a virtual knot
I am studying virtual knot theory. A virtual knot is a knot diagram with real or virtual crossing information. The equivalence relation includes generalized Reidemeister moves. There are premitted virtual Reidemeister moves, and forbbiden moves.
Question
Do the forbidden moves have some validity? How are generalized Reidemeister moves made? For example, If we admit the moves the virtual knot is trivial. Therefore, we make some of the moves forbidden. The moves have two real crossings and one virtual crossing. However, instead of these moves, why the other moves are not selected? For instance, may we exclude classical Reidemeister I move instead of the forbidden moves? I am confused!
-
I've posted an answer, but this strikes me as more of a math.SE question- also, note that googling "virtual Reidemeister move" would have answered the question. – Daniel Moskovich Aug 9 '12 at 20:06
If you remove Reidemeister I you get the theory of framed knots. Deciding which moves to include determines which objects you are studying. – Daniel Moskovich Aug 9 '12 at 20:11
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https://codegolf.stackexchange.com/questions/tagged/array-manipulation?sort=newest&page=2
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# Questions tagged [array-manipulation]
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480 questions
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Home Metamath Proof ExplorerTheorem List (p. 303 of 424) < Previous Next > Bad symbols? Try the GIF version. Mirrors > Metamath Home Page > MPE Home Page > Theorem List Contents > Recent Proofs This page: Page List
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Theorem List for Metamath Proof Explorer - 30201-30300 *Has distinct variable group(s)
TypeLabelDescription
Statement
Theorembnj151 30201* Technical lemma for bnj153 30204. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & (𝜃 ↔ ((𝑅 FrSe 𝐴𝑥𝐴) → ∃!𝑓(𝑓 Fn 𝑛𝜑𝜓))) & (𝜏 ↔ ∀𝑚𝐷 (𝑚 E 𝑛[𝑚 / 𝑛]𝜃)) & (𝜁 ↔ ((𝑅 FrSe 𝐴𝑥𝐴) → (𝑓 Fn 𝑛𝜑𝜓))) & (𝜑′[1𝑜 / 𝑛]𝜑) & (𝜓′[1𝑜 / 𝑛]𝜓) & (𝜃′[1𝑜 / 𝑛]𝜃) & (𝜃0 ↔ ((𝑅 FrSe 𝐴𝑥𝐴) → ∃𝑓(𝑓 Fn 1𝑜𝜑′𝜓′))) & (𝜃1 ↔ ((𝑅 FrSe 𝐴𝑥𝐴) → ∃*𝑓(𝑓 Fn 1𝑜𝜑′𝜓′))) & (𝜁′[1𝑜 / 𝑛]𝜁) & 𝐹 = {⟨∅, pred(𝑥, 𝐴, 𝑅)⟩} & (𝜑″[𝐹 / 𝑓]𝜑′) & (𝜓″[𝐹 / 𝑓]𝜓′) & (𝜁″[𝐹 / 𝑓]𝜁′) & (𝜁0 ↔ (𝑓 Fn 1𝑜𝜑′𝜓′)) & (𝜁1[𝑔 / 𝑓]𝜁0) & (𝜑1[𝑔 / 𝑓]𝜑′) & (𝜓1[𝑔 / 𝑓]𝜓′) (𝑛 = 1𝑜 → ((𝑛𝐷𝜏) → 𝜃))
Theorembnj154 30202* Technical lemma for bnj153 30204. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑1[𝑔 / 𝑓]𝜑′) & (𝜑′ ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) (𝜑1 ↔ (𝑔‘∅) = pred(𝑥, 𝐴, 𝑅))
Theorembnj155 30203* Technical lemma for bnj153 30204. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓1[𝑔 / 𝑓]𝜓′) & (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖 ∈ 1𝑜 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) (𝜓1 ↔ ∀𝑖 ∈ ω (suc 𝑖 ∈ 1𝑜 → (𝑔‘suc 𝑖) = 𝑦 ∈ (𝑔𝑖) pred(𝑦, 𝐴, 𝑅)))
Theorembnj153 30204* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & (𝜃 ↔ ((𝑅 FrSe 𝐴𝑥𝐴) → ∃!𝑓(𝑓 Fn 𝑛𝜑𝜓))) & (𝜏 ↔ ∀𝑚𝐷 (𝑚 E 𝑛[𝑚 / 𝑛]𝜃)) (𝑛 = 1𝑜 → ((𝑛𝐷𝜏) → 𝜃))
Theorembnj207 30205* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜒 ↔ ((𝑅 FrSe 𝐴𝑥𝐴) → ∃!𝑓(𝑓 Fn 𝑛𝜑𝜓))) & (𝜑′[𝑀 / 𝑛]𝜑) & (𝜓′[𝑀 / 𝑛]𝜓) & (𝜒′[𝑀 / 𝑛]𝜒) & 𝑀 ∈ V (𝜒′ ↔ ((𝑅 FrSe 𝐴𝑥𝐴) → ∃!𝑓(𝑓 Fn 𝑀𝜑′𝜓′)))
Theorembnj213 30206 First-order logic and set theory. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
pred(𝑋, 𝐴, 𝑅) ⊆ 𝐴
Theorembnj222 30207* Technical lemma for bnj229 30208. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑁 → (𝐹‘suc 𝑖) = 𝑦 ∈ (𝐹𝑖) pred(𝑦, 𝐴, 𝑅))) (𝜓 ↔ ∀𝑚 ∈ ω (suc 𝑚𝑁 → (𝐹‘suc 𝑚) = 𝑦 ∈ (𝐹𝑚) pred(𝑦, 𝐴, 𝑅)))
Theorembnj229 30208* Technical lemma for bnj517 30209. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑁 → (𝐹‘suc 𝑖) = 𝑦 ∈ (𝐹𝑖) pred(𝑦, 𝐴, 𝑅))) ((𝑛𝑁 ∧ (suc 𝑚 = 𝑛𝑚 ∈ ω ∧ 𝜓)) → (𝐹𝑛) ⊆ 𝐴)
Theorembnj517 30209* Technical lemma for bnj518 30210. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (Proof shortened by Mario Carneiro, 22-Dec-2016.) (New usage is discouraged.)
(𝜑 ↔ (𝐹‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑁 → (𝐹‘suc 𝑖) = 𝑦 ∈ (𝐹𝑖) pred(𝑦, 𝐴, 𝑅))) ((𝑁 ∈ ω ∧ 𝜑𝜓) → ∀𝑛𝑁 (𝐹𝑛) ⊆ 𝐴)
Theorembnj518 30210* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜏 ↔ (𝜑𝜓𝑛 ∈ ω ∧ 𝑝𝑛)) ((𝑅 FrSe 𝐴𝜏) → ∀𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅) ∈ V)
Theorembnj523 30211* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝐹‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜑′[𝑀 / 𝑛]𝜑) & 𝑀 ∈ V (𝜑′ ↔ (𝐹‘∅) = pred(𝑋, 𝐴, 𝑅))
Theorembnj526 30212* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜑″[𝐺 / 𝑓]𝜑) & 𝐺 ∈ V (𝜑″ ↔ (𝐺‘∅) = pred(𝑋, 𝐴, 𝑅))
Theorembnj528 30213 Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
𝐺 = (𝑓 ∪ {⟨𝑚, 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅)⟩}) 𝐺 ∈ V
Theorembnj535 30214* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑′ ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑚 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅)⟩}) & (𝜏 ↔ (𝜑′𝜓′𝑚 ∈ ω ∧ 𝑝𝑚)) ((𝑅 FrSe 𝐴𝜏𝑛 = (𝑚 ∪ {𝑚}) ∧ 𝑓 Fn 𝑚) → 𝐺 Fn 𝑛)
Theorembnj539 30215* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝐹‘suc 𝑖) = 𝑦 ∈ (𝐹𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜓′[𝑀 / 𝑛]𝜓) & 𝑀 ∈ V (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑀 → (𝐹‘suc 𝑖) = 𝑦 ∈ (𝐹𝑖) pred(𝑦, 𝐴, 𝑅)))
Theorembnj540 30216* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑁 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜓″[𝐺 / 𝑓]𝜓) & 𝐺 ∈ V (𝜓″ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑁 → (𝐺‘suc 𝑖) = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅)))
Theorembnj543 30217* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑′ ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑚 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅)⟩}) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜎 ↔ (𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝𝑚)) ((𝑅 FrSe 𝐴𝜏𝜎) → 𝐺 Fn 𝑛)
Theorembnj544 30218* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑′ ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑚 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅)⟩}) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜎 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝𝑚)) ((𝑅 FrSe 𝐴𝜏𝜎) → 𝐺 Fn 𝑛)
Theorembnj545 30219 Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑′ ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & 𝐷 = (ω ∖ {∅}) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅)⟩}) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜎 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝𝑚)) & ((𝑅 FrSe 𝐴𝜏𝜎) → 𝐺 Fn 𝑛) & (𝜑″ ↔ (𝐺‘∅) = pred(𝑥, 𝐴, 𝑅)) ((𝑅 FrSe 𝐴𝜏𝜎) → 𝜑″)
Theorembnj546 30220* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
𝐷 = (ω ∖ {∅}) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜎 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝𝑚)) & (𝜑′ ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑚 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) ((𝑅 FrSe 𝐴𝜏𝜎) → 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅) ∈ V)
Theorembnj548 30221* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & 𝐵 = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐾 = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝐶⟩}) & ((𝑅 FrSe 𝐴𝜏𝜎) → 𝐺 Fn 𝑛) (((𝑅 FrSe 𝐴𝜏𝜎) ∧ 𝑖𝑚) → 𝐵 = 𝐾)
Theorembnj553 30222* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑′ ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑚 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅)⟩}) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜎 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝𝑚)) & 𝐶 = 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝐶⟩}) & 𝐵 = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐾 = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐿 = 𝑦 ∈ (𝐺𝑝) pred(𝑦, 𝐴, 𝑅) & ((𝑅 FrSe 𝐴𝜏𝜎) → 𝐺 Fn 𝑛) (((𝑅 FrSe 𝐴𝜏𝜎) ∧ 𝑖𝑚𝑝 = 𝑖) → (𝐺𝑚) = 𝐿)
Theorembnj554 30223* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜂 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝 ∈ ω ∧ 𝑚 = suc 𝑝)) & (𝜁 ↔ (𝑖 ∈ ω ∧ suc 𝑖𝑛𝑚 = suc 𝑖)) & 𝐾 = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐿 = 𝑦 ∈ (𝐺𝑝) pred(𝑦, 𝐴, 𝑅) & 𝐾 = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐿 = 𝑦 ∈ (𝐺𝑝) pred(𝑦, 𝐴, 𝑅) ((𝜂𝜁) → ((𝐺𝑚) = 𝐿 ↔ (𝐺‘suc 𝑖) = 𝐾))
Theorembnj556 30224 Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜎 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝𝑚)) & (𝜂 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝 ∈ ω ∧ 𝑚 = suc 𝑝)) (𝜂𝜎)
Theorembnj557 30225* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
𝐷 = (ω ∖ {∅}) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅)⟩}) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜎 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝𝑚)) & (𝜂 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝 ∈ ω ∧ 𝑚 = suc 𝑝)) & (𝜁 ↔ (𝑖 ∈ ω ∧ suc 𝑖𝑛𝑚 = suc 𝑖)) & 𝐵 = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐶 = 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅) & 𝐾 = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐿 = 𝑦 ∈ (𝐺𝑝) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝐶⟩}) & (𝜑′ ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑚 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & ((𝑅 FrSe 𝐴𝜏𝜎) → 𝐺 Fn 𝑛) ((𝑅 FrSe 𝐴𝜏𝜂𝜁) → (𝐺𝑚) = 𝐿)
Theorembnj558 30226* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
𝐷 = (ω ∖ {∅}) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅)⟩}) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜎 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝𝑚)) & (𝜂 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝 ∈ ω ∧ 𝑚 = suc 𝑝)) & (𝜁 ↔ (𝑖 ∈ ω ∧ suc 𝑖𝑛𝑚 = suc 𝑖)) & 𝐵 = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐶 = 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅) & 𝐾 = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐿 = 𝑦 ∈ (𝐺𝑝) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝐶⟩}) & (𝜑′ ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑚 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & ((𝑅 FrSe 𝐴𝜏𝜎) → 𝐺 Fn 𝑛) ((𝑅 FrSe 𝐴𝜏𝜂𝜁) → (𝐺‘suc 𝑖) = 𝐾)
Theorembnj561 30227 Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜎 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝𝑚)) & (𝜂 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝 ∈ ω ∧ 𝑚 = suc 𝑝)) & ((𝑅 FrSe 𝐴𝜏𝜎) → 𝐺 Fn 𝑛) ((𝑅 FrSe 𝐴𝜏𝜂) → 𝐺 Fn 𝑛)
Theorembnj562 30228 Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜎 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝𝑚)) & (𝜂 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝 ∈ ω ∧ 𝑚 = suc 𝑝)) & ((𝑅 FrSe 𝐴𝜏𝜎) → 𝜑″) ((𝑅 FrSe 𝐴𝜏𝜂) → 𝜑″)
Theorembnj570 30229* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
𝐷 = (ω ∖ {∅}) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜂 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝 ∈ ω ∧ 𝑚 = suc 𝑝)) & (𝜌 ↔ (𝑖 ∈ ω ∧ suc 𝑖𝑛𝑚 ≠ suc 𝑖)) & 𝐾 = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝐶⟩}) & ((𝑅 FrSe 𝐴𝜏𝜂) → 𝐺 Fn 𝑛) & (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑚 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) ((𝑅 FrSe 𝐴𝜏𝜂𝜌) → (𝐺‘suc 𝑖) = 𝐾)
Theorembnj571 30230* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
𝐷 = (ω ∖ {∅}) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅)⟩}) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜎 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝𝑚)) & (𝜂 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝 ∈ ω ∧ 𝑚 = suc 𝑝)) & (𝜁 ↔ (𝑖 ∈ ω ∧ suc 𝑖𝑛𝑚 = suc 𝑖)) & 𝐵 = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐶 = 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅) & 𝐾 = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐿 = 𝑦 ∈ (𝐺𝑝) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝐶⟩}) & (𝜑′ ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑚 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & ((𝑅 FrSe 𝐴𝜏𝜎) → 𝐺 Fn 𝑛) & (𝜌 ↔ (𝑖 ∈ ω ∧ suc 𝑖𝑛𝑚 ≠ suc 𝑖)) & ((𝑅 FrSe 𝐴𝜏𝜂) → 𝐺 Fn 𝑛) & (𝜓″ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝐺‘suc 𝑖) = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅))) ((𝑅 FrSe 𝐴𝜏𝜂) → 𝜓″)
Theorembnj605 30231* Technical lemma. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜃 ↔ ∀𝑚𝐷 (𝑚 E 𝑛[𝑚 / 𝑛]𝜒)) & (𝜑″[𝑓 / 𝑓]𝜑) & (𝜓″[𝑓 / 𝑓]𝜓) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜂 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝 ∈ ω ∧ 𝑚 = suc 𝑝)) & 𝑓 ∈ V & (𝜒′ ↔ ((𝑅 FrSe 𝐴𝑥𝐴) → ∃!𝑓(𝑓 Fn 𝑚𝜑′𝜓′))) & (𝜑″ ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓″ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & ((𝑛 ≠ 1𝑜𝑛𝐷) → ∃𝑚𝑝𝜂) & ((𝜃𝑚𝐷𝑚 E 𝑛) → 𝜒′) & ((𝑅 FrSe 𝐴𝜏𝜂) → 𝑓 Fn 𝑛) & ((𝑅 FrSe 𝐴𝜏𝜂) → 𝜑″) & ((𝑅 FrSe 𝐴𝜏𝜂) → 𝜓″) ((𝑛 ≠ 1𝑜𝑛𝐷𝜃) → ((𝑅 FrSe 𝐴𝑥𝐴) → ∃𝑓(𝑓 Fn 𝑛𝜑𝜓)))
Theorembnj581 30232* Technical lemma for bnj580 30237. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Unnecessary distinct variable restrictions were removed by Andrew Salmon, 9-Jul-2011.) (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜒 ↔ (𝑓 Fn 𝑛𝜑𝜓)) & (𝜑′[𝑔 / 𝑓]𝜑) & (𝜓′[𝑔 / 𝑓]𝜓) & (𝜒′[𝑔 / 𝑓]𝜒) (𝜒′ ↔ (𝑔 Fn 𝑛𝜑′𝜓′))
Theorembnj589 30233* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) (𝜓 ↔ ∀𝑘 ∈ ω (suc 𝑘𝑛 → (𝑓‘suc 𝑘) = 𝑦 ∈ (𝑓𝑘) pred(𝑦, 𝐴, 𝑅)))
Theorembnj590 30234 Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) ((𝐵 = suc 𝑖𝜓) → (𝑖 ∈ ω → (𝐵𝑛 → (𝑓𝐵) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))))
Theorembnj591 30235* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜃 ↔ ((𝑛𝐷𝜒𝜒′) → (𝑓𝑗) = (𝑔𝑗))) ([𝑘 / 𝑗]𝜃 ↔ ((𝑛𝐷𝜒𝜒′) → (𝑓𝑘) = (𝑔𝑘)))
Theorembnj594 30236* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑓 Fn 𝑛𝜑𝜓)) & 𝐷 = (ω ∖ {∅}) & (𝜑′ ↔ (𝑔‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑔‘suc 𝑖) = 𝑦 ∈ (𝑔𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒′ ↔ (𝑔 Fn 𝑛𝜑′𝜓′)) & (𝜃 ↔ ((𝑛𝐷𝜒𝜒′) → (𝑓𝑗) = (𝑔𝑗))) & ([𝑘 / 𝑗]𝜃 ↔ ((𝑛𝐷𝜒𝜒′) → (𝑓𝑘) = (𝑔𝑘))) & (𝜏 ↔ ∀𝑘𝑛 (𝑘 E 𝑗[𝑘 / 𝑗]𝜃)) ((𝑗𝑛𝜏) → 𝜃)
Theorembnj580 30237* Technical lemma for bnj579 30238. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑓 Fn 𝑛𝜑𝜓)) & (𝜑′[𝑔 / 𝑓]𝜑) & (𝜓′[𝑔 / 𝑓]𝜓) & (𝜒′[𝑔 / 𝑓]𝜒) & 𝐷 = (ω ∖ {∅}) & (𝜃 ↔ ((𝑛𝐷𝜒𝜒′) → (𝑓𝑗) = (𝑔𝑗))) & (𝜏 ↔ ∀𝑘𝑛 (𝑘 E 𝑗[𝑘 / 𝑗]𝜃)) (𝑛𝐷 → ∃*𝑓𝜒)
Theorembnj579 30238* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) (𝑛𝐷 → ∃*𝑓(𝑓 Fn 𝑛𝜑𝜓))
Theorembnj602 30239 Equality theorem for the pred function constant. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝑋 = 𝑌 → pred(𝑋, 𝐴, 𝑅) = pred(𝑌, 𝐴, 𝑅))
Theorembnj607 30240* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜃 ↔ ∀𝑚𝐷 (𝑚 E 𝑛[𝑚 / 𝑛]𝜒)) & (𝜑″[𝐺 / 𝑓]𝜑) & (𝜓″[𝐺 / 𝑓]𝜓) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜂 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝 ∈ ω ∧ 𝑚 = suc 𝑝)) & 𝐺 ∈ V & (𝜒′ ↔ ((𝑅 FrSe 𝐴𝑥𝐴) → ∃!𝑓(𝑓 Fn 𝑚𝜑′𝜓′))) & (𝜑″ ↔ (𝐺‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓″ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝐺‘suc 𝑖) = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅))) & ((𝑛 ≠ 1𝑜𝑛𝐷) → ∃𝑚𝑝𝜂) & ((𝜃𝑚𝐷𝑚 E 𝑛) → 𝜒′) & ((𝑅 FrSe 𝐴𝜏𝜂) → 𝐺 Fn 𝑛) & ((𝑅 FrSe 𝐴𝜏𝜂) → 𝜑″) & ((𝑅 FrSe 𝐴𝜏𝜂) → 𝜓″) & (𝜑 ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜑0[ / 𝑓]𝜑) & (𝜓0[ / 𝑓]𝜓) & (𝜑1[𝐺 / ]𝜑0) & (𝜓1[𝐺 / ]𝜓0) ((𝑛 ≠ 1𝑜𝑛𝐷𝜃) → ((𝑅 FrSe 𝐴𝑥𝐴) → ∃𝑓(𝑓 Fn 𝑛𝜑𝜓)))
Theorembnj609 30241* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜑″[𝐺 / 𝑓]𝜑) & 𝐺 ∈ V (𝜑″ ↔ (𝐺‘∅) = pred(𝑋, 𝐴, 𝑅))
Theorembnj611 30242* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑁 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜓″[𝐺 / 𝑓]𝜓) & 𝐺 ∈ V (𝜓″ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑁 → (𝐺‘suc 𝑖) = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅)))
Theorembnj600 30243* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & (𝜒 ↔ ((𝑅 FrSe 𝐴𝑥𝐴) → ∃!𝑓(𝑓 Fn 𝑛𝜑𝜓))) & (𝜃 ↔ ∀𝑚𝐷 (𝑚 E 𝑛[𝑚 / 𝑛]𝜒)) & (𝜑′[𝑚 / 𝑛]𝜑) & (𝜓′[𝑚 / 𝑛]𝜓) & (𝜒′[𝑚 / 𝑛]𝜒) & (𝜑″[𝐺 / 𝑓]𝜑) & (𝜓″[𝐺 / 𝑓]𝜓) & (𝜒″[𝐺 / 𝑓]𝜒) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅)⟩}) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜎 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝𝑚)) & (𝜂 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝 ∈ ω ∧ 𝑚 = suc 𝑝)) & (𝜁 ↔ (𝑖 ∈ ω ∧ suc 𝑖𝑛𝑚 = suc 𝑖)) & (𝜌 ↔ (𝑖 ∈ ω ∧ suc 𝑖𝑛𝑚 ≠ suc 𝑖)) & 𝐵 = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐶 = 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅) & 𝐾 = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐿 = 𝑦 ∈ (𝐺𝑝) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝐶⟩}) (𝑛 ≠ 1𝑜 → ((𝑛𝐷𝜃) → 𝜒))
Theorembnj601 30244* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & (𝜒 ↔ ((𝑅 FrSe 𝐴𝑥𝐴) → ∃!𝑓(𝑓 Fn 𝑛𝜑𝜓))) & (𝜃 ↔ ∀𝑚𝐷 (𝑚 E 𝑛[𝑚 / 𝑛]𝜒)) (𝑛 ≠ 1𝑜 → ((𝑛𝐷𝜃) → 𝜒))
Theorembnj852 30245* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) ((𝑅 FrSe 𝐴𝑋𝐴) → ∀𝑛𝐷 ∃!𝑓(𝑓 Fn 𝑛𝜑𝜓))
Theorembnj864 30246* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & (𝜒 ↔ (𝑅 FrSe 𝐴𝑋𝐴𝑛𝐷)) & (𝜃 ↔ (𝑓 Fn 𝑛𝜑𝜓)) (𝜒 → ∃!𝑓𝜃)
Theorembnj865 30247* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & (𝜒 ↔ (𝑅 FrSe 𝐴𝑋𝐴𝑛𝐷)) & (𝜃 ↔ (𝑓 Fn 𝑛𝜑𝜓)) 𝑤𝑛(𝜒 → ∃𝑓𝑤 𝜃)
Theorembnj873 30248* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & (𝜑′[𝑔 / 𝑓]𝜑) & (𝜓′[𝑔 / 𝑓]𝜓) 𝐵 = {𝑔 ∣ ∃𝑛𝐷 (𝑔 Fn 𝑛𝜑′𝜓′)}
Theorembnj849 30249* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (Proof shortened by Mario Carneiro, 22-Dec-2016.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & (𝜒 ↔ (𝑅 FrSe 𝐴𝑋𝐴𝑛𝐷)) & (𝜃 ↔ (𝑓 Fn 𝑛𝜑𝜓)) & (𝜑′[𝑔 / 𝑓]𝜑) & (𝜓′[𝑔 / 𝑓]𝜓) & (𝜃′[𝑔 / 𝑓]𝜃) & (𝜏 ↔ (𝑅 FrSe 𝐴𝑋𝐴)) ((𝑅 FrSe 𝐴𝑋𝐴) → 𝐵 ∈ V)
Theorembnj882 30250* Definition (using hypotheses for readability) of the function giving the transitive closure of 𝑋 in 𝐴 by 𝑅. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} trCl(𝑋, 𝐴, 𝑅) = 𝑓𝐵 𝑖 ∈ dom 𝑓(𝑓𝑖)
Theorembnj18eq1 30251 Equality theorem for transitive closure. (Contributed by Mario Carneiro, 22-Dec-2016.) (New usage is discouraged.)
(𝑋 = 𝑌 → trCl(𝑋, 𝐴, 𝑅) = trCl(𝑌, 𝐴, 𝑅))
Theorembnj893 30252 Property of trCl. Under certain conditions, the transitive closure of 𝑋 in 𝐴 by 𝑅 is a set. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
((𝑅 FrSe 𝐴𝑋𝐴) → trCl(𝑋, 𝐴, 𝑅) ∈ V)
Theorembnj900 30253* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} (𝑓𝐵 → ∅ ∈ dom 𝑓)
Theorembnj906 30254 Property of trCl. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
((𝑅 FrSe 𝐴𝑋𝐴) → pred(𝑋, 𝐴, 𝑅) ⊆ trCl(𝑋, 𝐴, 𝑅))
Theorembnj908 30255* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑥, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & (𝜒 ↔ ((𝑅 FrSe 𝐴𝑥𝐴) → ∃!𝑓(𝑓 Fn 𝑛𝜑𝜓))) & (𝜃 ↔ ∀𝑚𝐷 (𝑚 E 𝑛[𝑚 / 𝑛]𝜒)) & (𝜑′[𝑚 / 𝑛]𝜑) & (𝜓′[𝑚 / 𝑛]𝜓) & (𝜒′[𝑚 / 𝑛]𝜒) & (𝜑″[𝐺 / 𝑓]𝜑) & (𝜓″[𝐺 / 𝑓]𝜓) & (𝜒″[𝐺 / 𝑓]𝜒) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅)⟩}) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜎 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝𝑚)) & (𝜂 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝 ∈ ω ∧ 𝑚 = suc 𝑝)) & (𝜁 ↔ (𝑖 ∈ ω ∧ suc 𝑖𝑛𝑚 = suc 𝑖)) & (𝜌 ↔ (𝑖 ∈ ω ∧ suc 𝑖𝑛𝑚 ≠ suc 𝑖)) & 𝐵 = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐶 = 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅) & 𝐾 = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅) & 𝐿 = 𝑦 ∈ (𝐺𝑝) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑚, 𝐶⟩}) ((𝑅 FrSe 𝐴𝑥𝐴𝜒′𝜂) → ∃𝑓(𝐺 Fn 𝑛𝜑″𝜓″))
Theorembnj911 30256* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) ((𝑓 Fn 𝑛𝜑𝜓) → ∀𝑖(𝑓 Fn 𝑛𝜑𝜓))
Theorembnj916 30257* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & (𝜒 ↔ (𝑓 Fn 𝑛𝜑𝜓)) (𝑦 ∈ trCl(𝑋, 𝐴, 𝑅) → ∃𝑓𝑛𝑖(𝑛𝐷𝜒𝑖𝑛𝑦 ∈ (𝑓𝑖)))
Theorembnj917 30258* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) (𝑦 ∈ trCl(𝑋, 𝐴, 𝑅) → ∃𝑓𝑛𝑖(𝜒𝑖𝑛𝑦 ∈ (𝑓𝑖)))
Theorembnj934 30259* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜑′[𝑝 / 𝑛]𝜑) & (𝜑″[𝐺 / 𝑓]𝜑′) & 𝐺 ∈ V ((𝜑 ∧ (𝐺‘∅) = (𝑓‘∅)) → 𝜑″)
Theorembnj929 30260* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜑′[𝑝 / 𝑛]𝜑) & (𝜑″[𝐺 / 𝑓]𝜑′) & 𝐷 = (ω ∖ {∅}) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) & 𝐶 ∈ V ((𝑛𝐷𝑝 = suc 𝑛𝑓 Fn 𝑛𝜑) → 𝜑″)
Theorembnj938 30261* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
𝐷 = (ω ∖ {∅}) & (𝜏 ↔ (𝑓 Fn 𝑚𝜑′𝜓′)) & (𝜎 ↔ (𝑚𝐷𝑛 = suc 𝑚𝑝𝑚)) & (𝜑′ ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑚 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) ((𝑅 FrSe 𝐴𝑋𝐴𝜏𝜎) → 𝑦 ∈ (𝑓𝑝) pred(𝑦, 𝐴, 𝑅) ∈ V)
Theorembnj944 30262* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜑′[𝑝 / 𝑛]𝜑) & (𝜑″[𝐺 / 𝑓]𝜑′) & 𝐷 = (ω ∖ {∅}) & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) & (𝜏 ↔ (𝑓 Fn 𝑛𝜑𝜓)) & (𝜎 ↔ (𝑛𝐷𝑝 = suc 𝑛𝑚𝑛)) (((𝑅 FrSe 𝐴𝑋𝐴) ∧ (𝜒𝑛 = suc 𝑚𝑝 = suc 𝑛)) → 𝜑″)
Theorembnj953 30263 Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & ((𝐺𝑖) = (𝑓𝑖) → ∀𝑦(𝐺𝑖) = (𝑓𝑖)) (((𝐺𝑖) = (𝑓𝑖) ∧ (𝐺‘suc 𝑖) = (𝑓‘suc 𝑖) ∧ (𝑖 ∈ ω ∧ suc 𝑖𝑛) ∧ 𝜓) → (𝐺‘suc 𝑖) = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅))
Theorembnj958 30264* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) ((𝐺𝑖) = (𝑓𝑖) → ∀𝑦(𝐺𝑖) = (𝑓𝑖))
Theorembnj1000 30265* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑁 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜓″[𝐺 / 𝑓]𝜓) & 𝐺 ∈ V & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) (𝜓″ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑁 → (𝐺‘suc 𝑖) = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅)))
Theorembnj965 30266* Technical lemma for bnj852 30245. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑁 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜓″[𝐺 / 𝑓]𝜓) & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) (𝜓″ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑁 → (𝐺‘suc 𝑖) = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅)))
Theorembnj964 30267* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜓′[𝑝 / 𝑛]𝜓) & (𝜓″[𝐺 / 𝑓]𝜓′) & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) & (((𝑅 FrSe 𝐴𝑋𝐴) ∧ (𝜒𝑛 = suc 𝑚𝑝 = suc 𝑛) ∧ (𝑖 ∈ ω ∧ suc 𝑖𝑝 ∧ suc 𝑖𝑛)) → (𝐺‘suc 𝑖) = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅)) & (((𝑅 FrSe 𝐴𝑋𝐴) ∧ (𝜒𝑛 = suc 𝑚𝑝 = suc 𝑛) ∧ (𝑖 ∈ ω ∧ suc 𝑖𝑝𝑛 = suc 𝑖)) → (𝐺‘suc 𝑖) = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅)) (((𝑅 FrSe 𝐴𝑋𝐴) ∧ (𝜒𝑛 = suc 𝑚𝑝 = suc 𝑛)) → 𝜓″)
Theorembnj966 30268* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & 𝐷 = (ω ∖ {∅}) & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) & (((𝑅 FrSe 𝐴𝑋𝐴) ∧ (𝜒𝑛 = suc 𝑚𝑝 = suc 𝑛)) → 𝐶 ∈ V) & (((𝑅 FrSe 𝐴𝑋𝐴) ∧ (𝜒𝑛 = suc 𝑚𝑝 = suc 𝑛)) → 𝐺 Fn 𝑝) (((𝑅 FrSe 𝐴𝑋𝐴) ∧ (𝜒𝑛 = suc 𝑚𝑝 = suc 𝑛) ∧ (𝑖 ∈ ω ∧ suc 𝑖𝑝𝑛 = suc 𝑖)) → (𝐺‘suc 𝑖) = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅))
Theorembnj967 30269* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & 𝐷 = (ω ∖ {∅}) & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) & (((𝑅 FrSe 𝐴𝑋𝐴) ∧ (𝜒𝑛 = suc 𝑚𝑝 = suc 𝑛)) → 𝐶 ∈ V) (((𝑅 FrSe 𝐴𝑋𝐴) ∧ (𝜒𝑛 = suc 𝑚𝑝 = suc 𝑛) ∧ (𝑖 ∈ ω ∧ suc 𝑖𝑝 ∧ suc 𝑖𝑛)) → (𝐺‘suc 𝑖) = 𝑦 ∈ (𝐺𝑖) pred(𝑦, 𝐴, 𝑅))
Theorembnj969 30270* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & 𝐷 = (ω ∖ {∅}) & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & (𝜏 ↔ (𝑓 Fn 𝑛𝜑𝜓)) & (𝜎 ↔ (𝑛𝐷𝑝 = suc 𝑛𝑚𝑛)) (((𝑅 FrSe 𝐴𝑋𝐴) ∧ (𝜒𝑛 = suc 𝑚𝑝 = suc 𝑛)) → 𝐶 ∈ V)
Theorembnj970 30271 Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & 𝐷 = (ω ∖ {∅}) (((𝑅 FrSe 𝐴𝑋𝐴) ∧ (𝜒𝑛 = suc 𝑚𝑝 = suc 𝑛)) → 𝑝𝐷)
Theorembnj910 30272* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜑′[𝑝 / 𝑛]𝜑) & (𝜓′[𝑝 / 𝑛]𝜓) & (𝜒′[𝑝 / 𝑛]𝜒) & (𝜑″[𝐺 / 𝑓]𝜑′) & (𝜓″[𝐺 / 𝑓]𝜓′) & (𝜒″[𝐺 / 𝑓]𝜒′) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) & (𝜏 ↔ (𝑓 Fn 𝑛𝜑𝜓)) & (𝜎 ↔ (𝑛𝐷𝑝 = suc 𝑛𝑚𝑛)) (((𝑅 FrSe 𝐴𝑋𝐴) ∧ (𝜒𝑛 = suc 𝑚𝑝 = suc 𝑛)) → 𝜒″)
Theorembnj978 30273* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜃 ↔ (𝑅 FrSe 𝐴𝑋𝐴𝑦 ∈ trCl(𝑋, 𝐴, 𝑅) ∧ 𝑧 ∈ pred(𝑦, 𝐴, 𝑅))) & (𝜃𝑧 ∈ trCl(𝑋, 𝐴, 𝑅)) ((𝑅 FrSe 𝐴𝑋𝐴) → TrFo( trCl(𝑋, 𝐴, 𝑅), 𝐴, 𝑅))
Theorembnj981 30274* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) (𝑍 ∈ trCl(𝑋, 𝐴, 𝑅) → ∃𝑓𝑛𝑖(𝜒𝑖𝑛𝑍 ∈ (𝑓𝑖)))
Theorembnj983 30275* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) (𝑍 ∈ trCl(𝑋, 𝐴, 𝑅) ↔ ∃𝑓𝑛𝑖(𝜒𝑖𝑛𝑍 ∈ (𝑓𝑖)))
Theorembnj984 30276 Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} (𝐺𝐴 → (𝐺𝐵[𝐺 / 𝑓]𝑛𝜒))
Theorembnj985 30277* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜒′[𝑝 / 𝑛]𝜒) & (𝜒″[𝐺 / 𝑓]𝜒′) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) (𝐺𝐵 ↔ ∃𝑝𝜒″)
Theorembnj986 30278* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & 𝐷 = (ω ∖ {∅}) & (𝜏 ↔ (𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = suc 𝑛)) (𝜒 → ∃𝑚𝑝𝜏)
Theorembnj996 30279* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜃 ↔ (𝑅 FrSe 𝐴𝑋𝐴𝑦 ∈ trCl(𝑋, 𝐴, 𝑅) ∧ 𝑧 ∈ pred(𝑦, 𝐴, 𝑅))) & (𝜏 ↔ (𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = suc 𝑛)) & (𝜂 ↔ (𝑖𝑛𝑦 ∈ (𝑓𝑖))) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} 𝑓𝑛𝑖𝑚𝑝(𝜃 → (𝜒𝜏𝜂))
Theorembnj998 30280* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜃 ↔ (𝑅 FrSe 𝐴𝑋𝐴𝑦 ∈ trCl(𝑋, 𝐴, 𝑅) ∧ 𝑧 ∈ pred(𝑦, 𝐴, 𝑅))) & (𝜏 ↔ (𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = suc 𝑛)) & (𝜑′[𝑝 / 𝑛]𝜑) & (𝜓′[𝑝 / 𝑛]𝜓) & (𝜒′[𝑝 / 𝑛]𝜒) & (𝜑″[𝐺 / 𝑓]𝜑′) & (𝜓″[𝐺 / 𝑓]𝜓′) & (𝜒″[𝐺 / 𝑓]𝜒′) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) ((𝜃𝜒𝜏𝜂) → 𝜒″)
Theorembnj999 30281* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜑′[𝑝 / 𝑛]𝜑) & (𝜓′[𝑝 / 𝑛]𝜓) & (𝜒′[𝑝 / 𝑛]𝜒) & (𝜑″[𝐺 / 𝑓]𝜑′) & (𝜓″[𝐺 / 𝑓]𝜓′) & (𝜒″[𝐺 / 𝑓]𝜒′) & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) ((𝜒″𝑖 ∈ ω ∧ suc 𝑖𝑝𝑦 ∈ (𝐺𝑖)) → pred(𝑦, 𝐴, 𝑅) ⊆ (𝐺‘suc 𝑖))
Theorembnj1001 30282 Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜏 ↔ (𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = suc 𝑛)) & (𝜂 ↔ (𝑖𝑛𝑦 ∈ (𝑓𝑖))) & 𝐷 = (ω ∖ {∅}) & ((𝜃𝜒𝜏𝜂) → 𝜒″) ((𝜃𝜒𝜏𝜂) → (𝜒″𝑖 ∈ ω ∧ suc 𝑖𝑝))
Theorembnj1006 30283* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜃 ↔ (𝑅 FrSe 𝐴𝑋𝐴𝑦 ∈ trCl(𝑋, 𝐴, 𝑅) ∧ 𝑧 ∈ pred(𝑦, 𝐴, 𝑅))) & (𝜏 ↔ (𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = suc 𝑛)) & (𝜂 ↔ (𝑖𝑛𝑦 ∈ (𝑓𝑖))) & (𝜑′[𝑝 / 𝑛]𝜑) & (𝜓′[𝑝 / 𝑛]𝜓) & (𝜒′[𝑝 / 𝑛]𝜒) & (𝜑″[𝐺 / 𝑓]𝜑′) & (𝜓″[𝐺 / 𝑓]𝜓′) & (𝜒″[𝐺 / 𝑓]𝜒′) & 𝐷 = (ω ∖ {∅}) & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) & ((𝜃𝜒𝜏𝜂) → (𝜒″𝑖 ∈ ω ∧ suc 𝑖𝑝)) ((𝜃𝜒𝜏𝜂) → pred(𝑦, 𝐴, 𝑅) ⊆ (𝐺‘suc 𝑖))
Theorembnj1014 30284* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} ((𝑔𝐵𝑗 ∈ dom 𝑔) → (𝑔𝑗) ⊆ trCl(𝑋, 𝐴, 𝑅))
Theorembnj1015 30285* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & 𝐺𝑉 & 𝐽𝑉 ((𝐺𝐵𝐽 ∈ dom 𝐺) → (𝐺𝐽) ⊆ trCl(𝑋, 𝐴, 𝑅))
Theorembnj1018 30286* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜃 ↔ (𝑅 FrSe 𝐴𝑋𝐴𝑦 ∈ trCl(𝑋, 𝐴, 𝑅) ∧ 𝑧 ∈ pred(𝑦, 𝐴, 𝑅))) & (𝜏 ↔ (𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = suc 𝑛)) & (𝜑′[𝑝 / 𝑛]𝜑) & (𝜓′[𝑝 / 𝑛]𝜓) & (𝜒′[𝑝 / 𝑛]𝜒) & (𝜑″[𝐺 / 𝑓]𝜑′) & (𝜓″[𝐺 / 𝑓]𝜓′) & (𝜒″[𝐺 / 𝑓]𝜒′) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) & (𝜒″ ↔ (𝑝𝐷𝐺 Fn 𝑝𝜑″𝜓″)) & ((𝜃𝜒𝜏𝜂) → 𝜒″) & ((𝜃𝜒𝜏𝜂) → (𝜒″𝑖 ∈ ω ∧ suc 𝑖𝑝)) ((𝜃𝜒𝜂 ∧ ∃𝑝𝜏) → (𝐺‘suc 𝑖) ⊆ trCl(𝑋, 𝐴, 𝑅))
Theorembnj1020 30287* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜃 ↔ (𝑅 FrSe 𝐴𝑋𝐴𝑦 ∈ trCl(𝑋, 𝐴, 𝑅) ∧ 𝑧 ∈ pred(𝑦, 𝐴, 𝑅))) & (𝜏 ↔ (𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = suc 𝑛)) & (𝜂 ↔ (𝑖𝑛𝑦 ∈ (𝑓𝑖))) & (𝜑′[𝑝 / 𝑛]𝜑) & (𝜓′[𝑝 / 𝑛]𝜓) & (𝜒′[𝑝 / 𝑛]𝜒) & (𝜑″[𝐺 / 𝑓]𝜑′) & (𝜓″[𝐺 / 𝑓]𝜓′) & (𝜒″[𝐺 / 𝑓]𝜒′) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) & (𝜒″ ↔ (𝑝𝐷𝐺 Fn 𝑝𝜑″𝜓″)) ((𝜃𝜒𝜂 ∧ ∃𝑝𝜏) → pred(𝑦, 𝐴, 𝑅) ⊆ trCl(𝑋, 𝐴, 𝑅))
Theorembnj1021 30288* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜃 ↔ (𝑅 FrSe 𝐴𝑋𝐴𝑦 ∈ trCl(𝑋, 𝐴, 𝑅) ∧ 𝑧 ∈ pred(𝑦, 𝐴, 𝑅))) & (𝜏 ↔ (𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = suc 𝑛)) & (𝜂 ↔ (𝑖𝑛𝑦 ∈ (𝑓𝑖))) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} 𝑓𝑛𝑖𝑚(𝜃 → (𝜃𝜒𝜂 ∧ ∃𝑝𝜏))
Theorembnj907 30289* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜃 ↔ (𝑅 FrSe 𝐴𝑋𝐴𝑦 ∈ trCl(𝑋, 𝐴, 𝑅) ∧ 𝑧 ∈ pred(𝑦, 𝐴, 𝑅))) & (𝜏 ↔ (𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = suc 𝑛)) & (𝜂 ↔ (𝑖𝑛𝑦 ∈ (𝑓𝑖))) & (𝜑′[𝑝 / 𝑛]𝜑) & (𝜓′[𝑝 / 𝑛]𝜓) & (𝜒′[𝑝 / 𝑛]𝜒) & (𝜑″[𝐺 / 𝑓]𝜑′) & (𝜓″[𝐺 / 𝑓]𝜓′) & (𝜒″[𝐺 / 𝑓]𝜒′) & 𝐷 = (ω ∖ {∅}) & 𝐵 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & 𝐶 = 𝑦 ∈ (𝑓𝑚) pred(𝑦, 𝐴, 𝑅) & 𝐺 = (𝑓 ∪ {⟨𝑛, 𝐶⟩}) ((𝑅 FrSe 𝐴𝑋𝐴) → TrFo( trCl(𝑋, 𝐴, 𝑅), 𝐴, 𝑅))
Theorembnj1029 30290 Property of trCl. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
((𝑅 FrSe 𝐴𝑋𝐴) → TrFo( trCl(𝑋, 𝐴, 𝑅), 𝐴, 𝑅))
Theorembnj1033 30291* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜃 ↔ (𝑅 FrSe 𝐴𝑋𝐴)) & (𝜏 ↔ (𝐵 ∈ V ∧ TrFo(𝐵, 𝐴, 𝑅) ∧ pred(𝑋, 𝐴, 𝑅) ⊆ 𝐵)) & (𝜂𝑧 ∈ trCl(𝑋, 𝐴, 𝑅)) & (𝜁 ↔ (𝑖𝑛𝑧 ∈ (𝑓𝑖))) & 𝐷 = (ω ∖ {∅}) & 𝐾 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & (∃𝑓𝑛𝑖(𝜃𝜏𝜒𝜁) → 𝑧𝐵) ((𝜃𝜏) → trCl(𝑋, 𝐴, 𝑅) ⊆ 𝐵)
Theorembnj1034 30292* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜃 ↔ (𝑅 FrSe 𝐴𝑋𝐴)) & (𝜏 ↔ (𝐵 ∈ V ∧ TrFo(𝐵, 𝐴, 𝑅) ∧ pred(𝑋, 𝐴, 𝑅) ⊆ 𝐵)) & (𝜁 ↔ (𝑖𝑛𝑧 ∈ (𝑓𝑖))) & 𝐷 = (ω ∖ {∅}) & 𝐾 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & (∃𝑓𝑛𝑖(𝜃𝜏𝜒𝜁) → 𝑧𝐵) ((𝜃𝜏) → trCl(𝑋, 𝐴, 𝑅) ⊆ 𝐵)
Theorembnj1039 30293 Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜓′[𝑗 / 𝑖]𝜓) (𝜓′ ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅)))
Theorembnj1040 30294* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑′[𝑗 / 𝑖]𝜑) & (𝜓′[𝑗 / 𝑖]𝜓) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜒′[𝑗 / 𝑖]𝜒) (𝜒′ ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑′𝜓′))
Theorembnj1047 30295 Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜌 ↔ ∀𝑗𝑛 (𝑗 E 𝑖[𝑗 / 𝑖]𝜂)) & (𝜂′[𝑗 / 𝑖]𝜂) (𝜌 ↔ ∀𝑗𝑛 (𝑗 E 𝑖𝜂′))
Theorembnj1049 30296 Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜁 ↔ (𝑖𝑛𝑧 ∈ (𝑓𝑖))) & (𝜂 ↔ ((𝜃𝜏𝜒𝜁) → 𝑧𝐵)) (∀𝑖𝑛 𝜂 ↔ ∀𝑖𝜂)
Theorembnj1052 30297* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜃 ↔ (𝑅 FrSe 𝐴𝑋𝐴)) & (𝜏 ↔ (𝐵 ∈ V ∧ TrFo(𝐵, 𝐴, 𝑅) ∧ pred(𝑋, 𝐴, 𝑅) ⊆ 𝐵)) & (𝜁 ↔ (𝑖𝑛𝑧 ∈ (𝑓𝑖))) & 𝐷 = (ω ∖ {∅}) & 𝐾 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & (𝜂 ↔ ((𝜃𝜏𝜒𝜁) → 𝑧𝐵)) & (𝜌 ↔ ∀𝑗𝑛 (𝑗 E 𝑖[𝑗 / 𝑖]𝜂)) & ((𝜃𝜏𝜒𝜁) → ( E Fr 𝑛 ∧ ∀𝑖𝑛 (𝜌𝜂))) ((𝜃𝜏) → trCl(𝑋, 𝐴, 𝑅) ⊆ 𝐵)
Theorembnj1053 30298* Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜑 ↔ (𝑓‘∅) = pred(𝑋, 𝐴, 𝑅)) & (𝜓 ↔ ∀𝑖 ∈ ω (suc 𝑖𝑛 → (𝑓‘suc 𝑖) = 𝑦 ∈ (𝑓𝑖) pred(𝑦, 𝐴, 𝑅))) & (𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & (𝜃 ↔ (𝑅 FrSe 𝐴𝑋𝐴)) & (𝜏 ↔ (𝐵 ∈ V ∧ TrFo(𝐵, 𝐴, 𝑅) ∧ pred(𝑋, 𝐴, 𝑅) ⊆ 𝐵)) & (𝜁 ↔ (𝑖𝑛𝑧 ∈ (𝑓𝑖))) & 𝐷 = (ω ∖ {∅}) & 𝐾 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} & (𝜂 ↔ ((𝜃𝜏𝜒𝜁) → 𝑧𝐵)) & (𝜌 ↔ ∀𝑗𝑛 (𝑗 E 𝑖[𝑗 / 𝑖]𝜂)) & ((𝜃𝜏𝜒𝜁) → ∀𝑖𝑛 (𝜌𝜂)) ((𝜃𝜏) → trCl(𝑋, 𝐴, 𝑅) ⊆ 𝐵)
Theorembnj1071 30299 Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
𝐷 = (ω ∖ {∅}) (𝑛𝐷 → E Fr 𝑛)
Theorembnj1083 30300 Technical lemma for bnj69 30332. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)
(𝜒 ↔ (𝑛𝐷𝑓 Fn 𝑛𝜑𝜓)) & 𝐾 = {𝑓 ∣ ∃𝑛𝐷 (𝑓 Fn 𝑛𝜑𝜓)} (𝑓𝐾 ↔ ∃𝑛𝜒)
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# Covariance is Symmetric
## Theorem
Let $X$ and $Y$ be random variables.
Suppose $\cov {X, Y}$ and $\cov {Y, X}$ exist.
Then $\cov {X, Y} = \cov {Y, X}$.
## Proof
$\ds \cov {X, Y}$ $=$ $\ds \expect {\paren {X - \expect X} \paren {Y - \expect Y} }$ Definition of Covariance $\ds$ $=$ $\ds \expect {\paren {Y - \expect Y} \paren {X - \expect X} }$ Real Multiplication is Commutative $\ds$ $=$ $\ds \cov {Y, X}$ Definition of Covariance
$\blacksquare$
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# What would the winning jackpot system consist of?
Topic closed. 9 replies. Last post 8 years ago by rcbbuckeye.
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United States
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October 22, 2005
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Posted: July 15, 2009, 8:28 pm - IP Logged
I assume we are all looking that "holy grail" of systems that would get you out of debt or win you the jackpot.
Assuming you found the winning system, how would you theorize it would work?
What characteristic do you think it would have?
(I am refering to jackpot games, PB/MM to be specific)
Northern Ohio
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Posted: July 16, 2009, 1:23 pm - IP Logged
I'll start: It would have to have a higher "hit" frequency. That is, I'd say half the chances to win any given prize. For example, for the MM, the overall chance to win a prize would be 1 in 40; with a good system, it would be at least 1 in 20.
mid-Ohio
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Posted: July 16, 2009, 3:20 pm - IP Logged
I'll start: It would have to have a higher "hit" frequency. That is, I'd say half the chances to win any given prize. For example, for the MM, the overall chance to win a prize would be 1 in 40; with a good system, it would be at least 1 in 20.
Most states recognize the chances are good that a player could reduce his odds of winning by a half by observing the history of past drawings so the payout amounts are adjusted accordingly. For example, MegaMillions overall odds of winning something are 1:39.8 and the most likely prize is a 0+1 which pays \$2 so if a player reduced his overall odds of winning to 1:20 he would still be spending \$20 to win \$2. To break even, a player would have to reduce his overall odds of winning close to 1:2 which would be the equivalent of reducing each number pool by 500% and playing a 5/11 + 1/9 game instead of a 5/56 + 1/46 game.
* you don't need to buy more tickets, just buy a winning ticket *
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October 22, 2005
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Posted: July 16, 2009, 9:16 pm - IP Logged
I'll start: It would have to have a higher "hit" frequency. That is, I'd say half the chances to win any given prize. For example, for the MM, the overall chance to win a prize would be 1 in 40; with a good system, it would be at least 1 in 20.
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October 22, 2005
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Posted: July 16, 2009, 9:45 pm - IP Logged
I think the winning system have to show some kind of format based on historical facts.
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Amarillo, Texas
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Posted: July 16, 2009, 10:43 pm - IP Logged
I have had a lot of success with this method.
I consider each ball a separate game. For instance, Position One is a different game than Position Two, etc. I make charts for each position and have had some considerable success this way. I play one dollar per day and usually hit one or two numbers. I frequently also miss a number by one digit. Try this and see how it rocks your world.
Llanoman
g
New Mexico
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Posted: July 17, 2009, 10:37 am - IP Logged
Can you give us an example of what you might play for tonight?
mid-Ohio
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19901 Posts
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Posted: July 17, 2009, 2:36 pm - IP Logged
I have had a lot of success with this method.
I consider each ball a separate game. For instance, Position One is a different game than Position Two, etc. I make charts for each position and have had some considerable success this way. I play one dollar per day and usually hit one or two numbers. I frequently also miss a number by one digit. Try this and see how it rocks your world.
Llanoman
g
Are you talking about a pick3/4 game rather than a jackpot type game like PB and MM?
* you don't need to buy more tickets, just buy a winning ticket *
Australia
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Posted: July 18, 2009, 5:26 am - IP Logged
5 winning numbers and the powerball number
2014 = -1016; 2015= -1409; 2016 = -1171; 2017 = ? TOT = -3596
keno historic = -2291 ; 2015= -603; 2016= -424; 2017 = ? TOT = - 3318
Texas
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Posted: July 18, 2009, 8:33 am - IP Logged
I have had a lot of success with this method.
I consider each ball a separate game. For instance, Position One is a different game than Position Two, etc. I make charts for each position and have had some considerable success this way. I play one dollar per day and usually hit one or two numbers. I frequently also miss a number by one digit. Try this and see how it rocks your world.
Llanoman
g
I use this method also, (if I'm understanding you right). I'll hit 1 number with it occasionally. I track each number by position and frequency of hits in each position. The problem is that while the number 2 has hit 38 times in the first position and the number 13 had hit 23 times in the second, and so on, it doesn't mean they are going to all play on the same line, or even in the same drawing. But, I play this way, along with a line of random numbers because I haven't figured out a better method.
I sure ain't going to spend 20 bucks every draw just to win 2. I play 2 lines and megaply them. I'm really just trying to hit 5 of 5 for a million.
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Functions are used throughout mathematics and have applications in many disciplines. However, we are often interested in finding the extreme values, such as function maximums.
So, what do you need to know about function maximums? A function can have multiple local maximum values, but it can have only one absolute (or global) maximum value. However, the maximum value (a y-value) can occur at more than one x-value. Every absolute maximum is a local maximum, but not every local maximum is an absolute maximum.
Of course, certain functions may only have one local maximum, and others may have no absolute maximum.
Let’s get started.
## 7 Common Questions About Function Maximums
A function can have multiple local maximum values, but it can have only one absolute (global) maximum value. However, the maximum value (a y-value) can occur at more than one x-value.
This still raises the question of when a local maximum is an absolute maximum, so we’ll start there.
### Can A Local Maximum Be An Absolute Maximum?
A local maximum can be an absolute maximum, and every absolute maximum is also a local maximum. However, not every local maximum is an absolute maximum.
The Venn diagram below illustrates this idea: there are always at least as many absolute maximums as there are local maximums for a function.
Let’s look at an example to make this idea more clear.
As you can see, there are two local maximums on this graph. However, the one on the right has the higher y-value, so it is also an absolute maximum for the function.
### Does A Linear Function Have A Maximum?
A linear function does have a maximum in some cases (when we restrict its domain). However, a linear function may not have a maximum if the domain is unbounded.
For example, the function f(x) = x is unbounded on the set of real numbers. The reason is that we can always plug in a larger value of x to get a larger output (y-value).
However, this same function has a maximum of y = 5 at x = 5 on the domain [0, 5].
Remember that for a linear function with a maximum, the maximum will occur:
• At the right endpoint if the slope is positive.
• At the left endpoint if the slope is negative.
• At all points if the slope is zero.
For example, the linear function g(x) = -5x on the domain [-4, 4] has a maximum value of 20 at its left endpoint, x = -4.
### Does Every Function Have An Absolute Maximum?
Not every function has an absolute maximum. As mentioned above, some linear functions do not have a maximum value if the domain is unrestricted.
There are other functions with no absolute maximum if their domains are unrestricted.
For example, the quadratic function y = x2 + 1 (whose graph is a parabola, pictured below) has no absolute maximum on the set of real numbers. The reason is that we can always plug in larger x values to get larger outputs (y-values).
There are even functions that have no absolute maximum on a bounded domain. For example, the function f(x) = 1/x on the domain [0, 1] has no global maximum.
The reason is that 1/x is unbounded as x approaches zero.
### Can A Maximum Be Negative?
A maximum can be negative in some cases. For example, the quadratic function f(x) = -x2 – 3 has an absolute maximum value of y = -3 at x = 0 (the vertex of the parabola).
This maximum (a y-value) is negative. We can also have maximums where the x values are negative.
For example, the function g(x) = -(x + 1)2 – 4 has an absolute maximum value of y = -4 at x = -1 (the vertex of the parabola).
### Can A Maximum Be An Inflection Point?
A maximum can also be an inflection point. We can come up with a piecewise function with a point that satisfies both of these conditions.
For example, consider the piecewise function:
This piecewise function is pictured below.
Note that:
• This function is continuous at x = 0 (both functions have a limit of y = 0 as x approaches 0, and the value of y is 0 at x = 0). (You can learn more about zero limits in my article here).
• The first derivative is zero at x = 0, and it changes signs from positive to negative at x = 0, so we have a function maximum there.
• Finally, the second derivative is zero at x = 0, and it changes signs from negative to positive at x = 0, so we have an inflection point there as well.
### Can A Quadratic Function Have A Maximum & Minimum?
A quadratic function on the set of real numbers has either a maximum or a minimum, but not both. However, a quadratic function can have a maximum and a minimum on a restricted domain.
For example, consider the quadratic function
• g(x) = x2
defined on the interval [1, 2].
The minimum value of the quadratic function on this interval is y = 1, when x = 1. The maximum value of the quadratic function on this interval is y = 4, when x = 2.
The graph below helps to illustrate this.
This still leaves the question of whether a quadratic function has a maximum or a minimum when defined on the entire set of real numbers.
#### How To Tell If A Quadratic Function Has A Maximum Or Minimum
Remember that a quadratic function has the form
• f(x) = ax2 + bx + c
where a is not zero.
The value of a determines the nature of the quadratic function’s maximum or minimum over the real numbers:
• If a is positive, then the parabola is concave up (opens up), and the quadratic function has a minimum at the vertex of the parabola. However, it has no maximum value.
• If a is negative, then the parabola is concave down (opens down), and the quadratic function has a maximum at the vertex of the parabola. However, it has no minimum value.
For example, the graph of the parabola f(x) = x2 – 1 (shown below) has a minimum of y = -1 at x = 0. It has a minimum since the leading coefficient has a value of 1, which is positive (a > 0).
On the other hand, the graph of the parabola f(x) = -x2 – 1 (shown below) has a maximum of y = -1 at x = 0. It has a maximum since the leading coefficient has a value of -1, which is negative (a < 0).
The table below summarizes the relationship between the sign of a in a quadratic function and the nature of the extreme values (maximum or minimum) of a parabola.
ax2 + bx + c
## Conclusion
Now you know more about function maximums, along with when and how they can occur. You also know the answers to some common questions about function maximums.
You can learn about how to use derivatives and graphs to find function maximums here.
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# Change from baseline mixed effects models
I am trying to analyze a study that has three treatment groups and the measurements are conducted on the same subjects over time. The first time point is a baseline measurement and then there are 7 additional time points. I have taken the primary data and have fit a mixed effects model: resp ~ baseline + treat * time + (time|subject) using the syntax for lmer. Once I have the model fit I used the package lsmeans to obtain model estimates with confidence limits at the different time points.
The question here is whether with this model I am able to obtain estimates for changes from baseline. Alternatively, I could subtract the baseline values for each subject and refit the model on the change from baseline data, rather than the original data. What is more appropriate?
Related to the previous question, if the data fitted are the primary data rather than the change from baseline, is there a way to get from lsmeans the estimates for change from baseline values with the corresponding 95% confidence limits?
• I hope that now is clearer. Thank you for the suggestion. Apr 26, 2016 at 4:20
If you fit the model with offset(baseline) instead of just baseline, then the lsmeans will indeed be in terms of changes from baseline. This is due to the fact that offset in essence puts in a covariate with its regression coefficient constrained to be 1 -- hence it is equivalent to subtracting it from the response.
But with baseline as an ordinary covariate, I am having trouble understanding what you would even mean by "change from baseline" in such a model. You could test the hypothesis $H_0: \beta_1=1$ versus $H_1: \beta_1\ne1$, where $\beta_1$ is the coefficient for baseline; and if rejected, it would suggest that change from baseline is not all that meaningful when explaining the factor effects.
I suppose another approach would be to add the argument
lsmeans( ..., cov.reduce = baseline ~ treat)
which would put in separate baseline predictions for each treatment; the results would then be predictions using predicted baseline values as a moderating variable. The LS means will be (means of) predictions from the model at those baseline values. However if you subtract 1 from $\hat\beta_1$, you obtain changes from those baseline values, because
$$\beta_0 + (\beta_1 - 1)x + \mbox{factor effects} = (\beta_0 + \beta_1 x + \mbox{factor effects}) - x$$
where $x$ is the baseline value. You can implement this by devilish hacking, something like this:
lsm = lsmeans(...) # with or without that cov.reduce thing
lsm@bhat[2] = lsm@bhat[2] - 1
summary(lsm)
This assumes that baseline is the first term in the model (second to the intercept). The statistical output is still valid because the variance of $\hat\beta_1-1$ is the same as the variance of $\hat\beta_1$, as is also true of the covariances of this with other regression coefficients.
But, as I said, this may or may not be what you mean by change from baseline when baseline is used as a covariate.
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https://www.physicsforums.com/threads/golf-club-swing-weight-and-club-head-rotation.942825/
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# Golf club "Swing Weight" and club head rotation
• I
for those Who Dont know golf. Swing weight is a measurment how Heavy the clubhead Feels. If i put on a heavier grip the swingweight falls. Heavier head swing weight goes up etc.
If i Want to swing the
Club with as little clubhead rotation as possible (i know forarm creats the rotation) Would i benefit from a lighter or heavier club? Or lighter or heavier swingweight? (Move weight to grip or head)
To me If i swing a 7 iron at 90mph using a lighter club must be easier to hold against the rotation? Am i thinking wrong?
## Answers and Replies
Related Classical Physics News on Phys.org
PT1
The standard answer is to see your local pro. I am not a pro so I can't offer advice. If, however, you are asking to explore the physics, again I am not a pro but I have given some thought to the problem. Hopefully I can help you gain some insight .
First consider what happens when you swing the club. To do this, I suggest you hold the club out in front of you, eye high and horizontal. Look down the shaft and you should see that the club-head sits above the shaft. Consider that if you swing the shaft thru a perfect plane and rotate, the club-head will move down to join the shaft in the plane.
If however you swing the club-head thru a perfect plane and rotate, the shaft will rise up slightly to join the club-head in the plane. Try doing these slowly until you can feel how different they are. Manipulating the club thru these two exercises feels very different to me. Swinging the whole club without rotation will feel different again.
The following is based on the assumption of trying to swing the shaft on plane. Now if you can imagine swinging the shaft thru a perfect plane, with 90% rotation of the club-head then any momentum built up in the club-head would tend to keep the shaft on plane as you complete the back-swing. If you could manage to swing the shaft on plane without rotation of the club-head then any momentum in the club-head would tend to pull the shaft slightly off the plane. (the bulk of the club-head being outside the intended plane) .It would follow that the effect of an increase in club-head weight would depend on the mechanics of your swing, but it could adversely affect your swing-plane if you do not rotate.
It doesn't immediately occur to me that increasing the weight of the grip might cause some similar problem, but again, I am not a pro so ask your pro. (Unless you are that rare breed that has a perfect swing-plane). Even if you can manage to learn to swing with little rotation, squaring up the club-face will still be required. The more I say the more I realize that I should have quit at 'see your local pro'. Over-analysis does lead to paralysis.
Last edited by a moderator:
It doesn't immediately occur to me that increasing the weight of the grip might cause some similar problem, but again, I am not a pro so ask your pro. (Unless you are that rare breed that has a perfect swing-plane). Even if you can manage to learn to swing with little rotation, squaring up the club-face will still be required. The more I say the more I realize that I should have quit at 'see your local pro'. Over-analysis does lead to paralysis.
to sum up, would you think a lighter total weight club or heavier total club, or lighter or heavier swingweight would help? ^^
PT1
I don't really feel qualified to make recommendations As I said, I am not a pro. What would work for one person may not for another. A pro should be able to identify any areas of your swing that would help, before deciding that you need to try changes to your clubs. Pros also know how to identify the club specs that should work for you. If you do not have access to a golf professional, it may be economical to go to a thrift store and try to find a club with the specs you are considering. Prices can be as low as 3 to 5 dollars for a single club.Then try it out for a while to see if it helps. What I posted about the swing was not intended to suggest anything more than a way that you could try to learn for yourself what feels right. Did any of what I said make any sense?
A.T.
If i Want to swing the Club with as little clubhead rotation as possible
Rotation around which axis, at what time point (swing or impact)?
Rotation around which axis, at what time point (swing or impact)?
at impact i mean.
A.T.
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http://www.chegg.com/homework-help/questions-and-answers/stephens-electronics-is-considering-a-change-in-its-target-capital-structure-which-current-q3468062
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## Firm's cost of equity
Stephens Electronics is considering a change in its target capital structure, which currently consists of 25% debt and 75% equity. The CFO believes the firm should use more debt, but the CEO is reluctant to increase the debt ratio. The risk-free rate, rRF, is 5.0%, the market risk premium, RPM, is 6.0%, and the firm's tax rate is 40%. Currently, the cost of equity, rs, is 11.5% as determined by the CAPM. What would be the estimated cost of equity if the firm used 60% debt? (hint: You must first find the current beta and then the unlevered beta to solve the problem.)
• rf=5%
RPM=6%
rs=rf+RPM*βe
11.5=5+6*βe
on solving βe(levered)=1.08
this is beta levered when leverage ratio is 25/75 = 1/3=0.33
β(unlevered) = (βe(levered)*E)/(D(1-t)+E) + β(debt)*D(1-t)/(D*(1-t) + E)
since β(debt)=0, that is debt is risk free
hence β(unlevered) = (βe(levered)*E)/(D(1-t)+E)
β(unlevered)=1.08/(D/E*(1-t) + E)
=1.08/(1/3*(1-.4) + 1)
= 1.08/1.2 = 0.9
now when firm uses 60% debt,40% equity then D/E ratio becomes = 60/40=1.5
hence β(unlevered) = (βe(levered)*E)/(D(1-t)+E)..now use new D/E ratio
0.9 = βe(levered)/(D/E*(1-t) + E)
0.9 = βe(levered)/(1.5*(1-.4)+1)
on solving βe(levered)=1.71
hence when debt is 60% βe(levered) increases from 1.08 to 1.71
• As given in the question,
rRf=5%
RPM=6%
Since we know that,rs=rRf+RPM*ße
11.5=5+6*ße
ße=1.08
ße = > leverage ratio is 25/75 = 1/3=0.33
We know that,
ß = (ße*E)/(D(1-t)+E) + ßd*D(1-t)/(D*(1-t) + E)
As given in question ßd=0, since debt is risk free
Therefore,
ß = (ße*E)/(D(1-t)+E)
ß = 1.08/(D/E*(1-t) + E)
= 1.08/(1/3*(1-.4) + 1)
= 1.08/1.2 = 0.9
now,debt = 60%,equity = 40% => D/E = 60/40=1.5
hence ß = (ße*E)/(D(1-t)+E)..now use new D/E ratio
0.9 = ße/(D/E*(1-t) + E)
0.9 = ße/(1.5*(1-.4)+1)
on solving ße=1.71
hence when debt is 60% ße increases from 1.08 to 1.71
Get homework help
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http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=EXMT
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Back to list of Stocks See Also: Seasonal Analysis of EXMTGenetic Algorithms Stock Portfolio Generator, and Best Months to Buy/Sell Stocks
# Fourier Analysis of EXMT (Anything Technologies Media)
EXMT (Anything Technologies Media) appears to have interesting cyclic behaviour every 87 weeks (60.6522*sine), 79 weeks (58.6388*sine), and 87 weeks (40.8674*cosine).
EXMT (Anything Technologies Media) has an average price of 100.67 (topmost row, frequency = 0).
Click on the checkboxes shown on the right to see how the various frequencies contribute to the graph. Look for large magnitude coefficients (sine or cosine), as these are associated with frequencies which contribute most to the associated stock plot. If you find a large magnitude coefficient which dramatically changes the graph, look at the associated "Period" in weeks, as you may have found a significant recurring cycle for the stock of interest.
## Fourier Analysis
Using data from 1/3/2000 to 4/23/2018 for EXMT (Anything Technologies Media), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
0100.6664 0
1170.1918 71.25862 (1*2Ï€)/872872 weeks
2112.1477 94.27399 (2*2Ï€)/872436 weeks
371.43814 82.75186 (3*2Ï€)/872291 weeks
451.47632 62.49289 (4*2Ï€)/872218 weeks
549.29627 41.94843 (5*2Ï€)/872174 weeks
661.38754 34.66663 (6*2Ï€)/872145 weeks
768.45997 45.07693 (7*2Ï€)/872125 weeks
861.22815 56.50122 (8*2Ï€)/872109 weeks
950.35148 59.91081 (9*2Ï€)/87297 weeks
1040.8674 60.65221 (10*2Ï€)/87287 weeks
1130.1957 58.63876 (11*2Ï€)/87279 weeks
1222.23733 51.79918 (12*2Ï€)/87273 weeks
1319.26501 45.33963 (13*2Ï€)/87267 weeks
1416.62135 40.97242 (14*2Ï€)/87262 weeks
1514.27413 34.60106 (15*2Ï€)/87258 weeks
1616.7498 28.12023 (16*2Ï€)/87255 weeks
1721.30696 26.66362 (17*2Ï€)/87251 weeks
1823.41297 28.09294 (18*2Ï€)/87248 weeks
1924.2576 30.2122 (19*2Ï€)/87246 weeks
2022.77713 34.1314 (20*2Ï€)/87244 weeks
2116.88668 36.21386 (21*2Ï€)/87242 weeks
2211.02654 33.16364 (22*2Ï€)/87240 weeks
238.87481 28.61179 (23*2Ï€)/87238 weeks
248.20338 25.342 (24*2Ï€)/87236 weeks
257.82442 22.05415 (25*2Ï€)/87235 weeks
269.10593 18.77283 (26*2Ï€)/87234 weeks
2711.87564 17.60669 (27*2Ï€)/87232 weeks
2813.50424 19.17715 (28*2Ï€)/87231 weeks
2912.34651 21.09844 (29*2Ï€)/87230 weeks
3010.09125 21.40616 (30*2Ï€)/87229 weeks
317.91246 21.38446 (31*2Ï€)/87228 weeks
324.06797 20.88249 (32*2Ï€)/87227 weeks
33-.46365 16.93051 (33*2Ï€)/87226 weeks
34-1.58799 10.27742 (34*2Ï€)/87226 weeks
35.95102 5.07183 (35*2Ï€)/87225 weeks
364.1626 2.47282 (36*2Ï€)/87224 weeks
376.82164 1.15896 (37*2Ï€)/87224 weeks
388.9006 .59444 (38*2Ï€)/87223 weeks
3910.13081 .19121 (39*2Ï€)/87222 weeks
4010.96248 -.89952 (40*2Ï€)/87222 weeks
4112.56458 -2.66616 (41*2Ï€)/87221 weeks
4215.52973 -3.96557 (42*2Ï€)/87221 weeks
4319.21799 -3.46111 (43*2Ï€)/87220 weeks
4421.73127 -.74392 (44*2Ï€)/87220 weeks
4521.24376 2.28029 (45*2Ï€)/87219 weeks
4619.12874 2.54394 (46*2Ï€)/87219 weeks
4718.9368 .73789 (47*2Ï€)/87219 weeks
4820.35228 -.01737 (48*2Ï€)/87218 weeks
4921.23957 -.26405 (49*2Ï€)/87218 weeks
5022.97083 -1.09831 (50*2Ï€)/87217 weeks
5125.91685 -.06397 (51*2Ï€)/87217 weeks
5226.77319 2.94038 (52*2Ï€)/87217 weeks
5325.40032 4.36542 (53*2Ï€)/87216 weeks
5424.87644 4.05937 (54*2Ï€)/87216 weeks
5525.07734 4.09199 (55*2Ï€)/87216 weeks
5625.11771 3.75024 (56*2Ï€)/87216 weeks
5726.38235 3.1545 (57*2Ï€)/87215 weeks
5828.12185 4.12151 (58*2Ï€)/87215 weeks
5928.52638 5.62873 (59*2Ï€)/87215 weeks
6028.76722 6.26752 (60*2Ï€)/87215 weeks
6129.6095 7.47797 (61*2Ï€)/87214 weeks
6229.26691 9.32649 (62*2Ï€)/87214 weeks
6328.18832 9.83102 (63*2Ï€)/87214 weeks
6428.30415 9.65538 (64*2Ï€)/87214 weeks
6528.92711 10.45808 (65*2Ï€)/87213 weeks
6628.85374 11.608 (66*2Ï€)/87213 weeks
6728.57163 12.39788 (67*2Ï€)/87213 weeks
6828.50519 13.12025 (68*2Ï€)/87213 weeks
6928.51724 14.24269 (69*2Ï€)/87213 weeks
7027.86642 15.78621 (70*2Ï€)/87212 weeks
7126.31904 16.66437 (71*2Ï€)/87212 weeks
7225.17576 16.48774 (72*2Ï€)/87212 weeks
7324.78957 16.51856 (73*2Ï€)/87212 weeks
7424.10029 16.8113 (74*2Ï€)/87212 weeks
7523.57088 16.54227 (75*2Ï€)/87212 weeks
7623.85005 16.71883 (76*2Ï€)/87211 weeks
7723.53638 17.95117 (77*2Ï€)/87211 weeks
7821.94072 18.70123 (78*2Ï€)/87211 weeks
7920.42174 18.19045 (79*2Ï€)/87211 weeks
8019.72666 17.36333 (80*2Ï€)/87211 weeks
8119.28416 16.74566 (81*2Ï€)/87211 weeks
8218.90587 15.80837 (82*2Ï€)/87211 weeks
8319.42015 14.56513 (83*2Ï€)/87211 weeks
8421.01259 14.22222 (84*2Ï€)/87210 weeks
8522.35005 15.49352 (85*2Ï€)/87210 weeks
8622.24789 17.41578 (86*2Ï€)/87210 weeks
8720.859 18.69636 (87*2Ï€)/87210 weeks
8819.15063 18.6257 (88*2Ï€)/87210 weeks
8918.37945 17.55231 (89*2Ï€)/87210 weeks
9018.79119 16.81827 (90*2Ï€)/87210 weeks
9119.37666 16.9394 (91*2Ï€)/87210 weeks
9219.78636 17.46742 (92*2Ï€)/8729 weeks
9319.95399 18.55124 (93*2Ï€)/8729 weeks
9419.13699 19.87792 (94*2Ï€)/8729 weeks
9517.5912 20.27784 (95*2Ï€)/8729 weeks
9616.61042 19.75315 (96*2Ï€)/8729 weeks
9716.4896 19.51757 (97*2Ï€)/8729 weeks
9816.14461 20.09439 (98*2Ï€)/8729 weeks
9914.87136 20.54416 (99*2Ï€)/8729 weeks
10013.43904 20.03164 (100*2Ï€)/8729 weeks
10112.59207 19.15311 (101*2Ï€)/8729 weeks
10211.79437 18.20952 (102*2Ï€)/8729 weeks
10311.39093 16.4047 (103*2Ï€)/8728 weeks
10412.83173 14.57143 (104*2Ï€)/8728 weeks
10515.28668 14.94977 (105*2Ï€)/8728 weeks
10616.01426 17.24017 (106*2Ï€)/8728 weeks
10714.58518 18.77454 (107*2Ï€)/8728 weeks
10813.06213 18.62909 (108*2Ï€)/8728 weeks
10912.41956 18.09012 (109*2Ï€)/8728 weeks
11011.89459 17.80941 (110*2Ï€)/8728 weeks
11111.27592 17.13339 (111*2Ï€)/8728 weeks
11211.38182 16.13631 (112*2Ï€)/8728 weeks
11312.09865 15.80352 (113*2Ï€)/8728 weeks
11412.48089 16.09876 (114*2Ï€)/8728 weeks
11512.5734 16.33397 (115*2Ï€)/8728 weeks
11612.77265 16.745 (116*2Ï€)/8728 weeks
11712.55579 17.6317 (117*2Ï€)/8727 weeks
11811.45 18.2844 (118*2Ï€)/8727 weeks
11910.08995 17.91751 (119*2Ï€)/8727 weeks
1209.48703 16.87418 (120*2Ï€)/8727 weeks
1219.57789 16.12455 (121*2Ï€)/8727 weeks
1229.66775 15.64574 (122*2Ï€)/8727 weeks
12310.07339 14.99312 (123*2Ï€)/8727 weeks
12411.28026 15.01422 (124*2Ï€)/8727 weeks
12511.97319 16.597 (125*2Ï€)/8727 weeks
12610.65076 18.3624 (126*2Ï€)/8727 weeks
1278.20728 18.27102 (127*2Ï€)/8727 weeks
1286.85045 16.35845 (128*2Ï€)/8727 weeks
1297.40256 14.48829 (129*2Ï€)/8727 weeks
1308.72025 13.97503 (130*2Ï€)/8727 weeks
1319.51793 14.43856 (131*2Ï€)/8727 weeks
1329.69749 14.97889 (132*2Ï€)/8727 weeks
1339.75964 15.45622 (133*2Ï€)/8727 weeks
1349.57334 16.20329 (134*2Ï€)/8727 weeks
1358.70767 16.77048 (135*2Ï€)/8726 weeks
1367.85201 16.50014 (136*2Ï€)/8726 weeks
1377.85574 16.31335 (137*2Ï€)/8726 weeks
1387.49575 17.19061 (138*2Ï€)/8726 weeks
1395.61781 17.60231 (139*2Ï€)/8726 weeks
1403.89482 16.14756 (140*2Ï€)/8726 weeks
1413.911 14.37715 (141*2Ï€)/8726 weeks
1424.41836 13.81758 (142*2Ï€)/8726 weeks
1434.2183 13.52091 (143*2Ï€)/8726 weeks
1444.06233 12.72369 (144*2Ï€)/8726 weeks
1454.37336 12.06179 (145*2Ï€)/8726 weeks
1464.60991 11.7443 (146*2Ï€)/8726 weeks
1474.76754 11.33654 (147*2Ï€)/8726 weeks
1485.19836 11.10502 (148*2Ï€)/8726 weeks
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1504.56021 11.67553 (150*2Ï€)/8726 weeks
1513.49071 10.66987 (151*2Ï€)/8726 weeks
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1534.88339 7.37178 (153*2Ï€)/8726 weeks
1546.59165 7.20735 (154*2Ï€)/8726 weeks
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1596.87616 10.06147 (159*2Ï€)/8725 weeks
1606.57518 8.82423 (160*2Ï€)/8725 weeks
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1628.65715 8.11354 (162*2Ï€)/8725 weeks
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1688.88608 10.51072 (168*2Ï€)/8725 weeks
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1728.05086 9.86395 (172*2Ï€)/8725 weeks
1738.9128 9.34402 (173*2Ï€)/8725 weeks
1749.66522 9.85556 (174*2Ï€)/8725 weeks
1759.64123 10.51453 (175*2Ï€)/8725 weeks
1769.34421 10.73689 (176*2Ï€)/8725 weeks
1779.13909 10.70712 (177*2Ï€)/8725 weeks
1789.04179 10.44251 (178*2Ï€)/8725 weeks
1799.39377 10.00952 (179*2Ï€)/8725 weeks
18010.26816 9.95305 (180*2Ï€)/8725 weeks
18111.17528 10.62293 (181*2Ï€)/8725 weeks
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18311.1721 12.95363 (183*2Ï€)/8725 weeks
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1918.36538 13.91243 (191*2Ï€)/8725 weeks
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1938.52268 14.48425 (193*2Ï€)/8725 weeks
1948.65329 15.39286 (194*2Ï€)/8724 weeks
1957.85858 16.60147 (195*2Ï€)/8724 weeks
1966.32627 16.94413 (196*2Ï€)/8724 weeks
1975.2127 16.39641 (197*2Ï€)/8724 weeks
1984.66552 15.76601 (198*2Ï€)/8724 weeks
1994.4075 15.06097 (199*2Ï€)/8724 weeks
2004.8396 14.47702 (200*2Ï€)/8724 weeks
2015.52459 14.90267 (201*2Ï€)/8724 weeks
2025.26274 15.92932 (202*2Ï€)/8724 weeks
2034.26539 16.35873 (203*2Ï€)/8724 weeks
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2101.07733 15.78332 (210*2Ï€)/8724 weeks
211.6679 15.18851 (211*2Ï€)/8724 weeks
212.58086 15.16532 (212*2Ï€)/8724 weeks
213-.06026 15.27872 (213*2Ï€)/8724 weeks
214-.71341 14.85248 (214*2Ï€)/8724 weeks
215-.89711 14.51945 (215*2Ï€)/8724 weeks
216-1.28811 14.59112 (216*2Ï€)/8724 weeks
217-2.07125 14.42752 (217*2Ï€)/8724 weeks
218-2.74528 13.93516 (218*2Ï€)/8724 weeks
219-3.32495 13.42188 (219*2Ï€)/8724 weeks
220-3.97281 12.69689 (220*2Ï€)/8724 weeks
221-4.29636 11.61392 (221*2Ï€)/8724 weeks
222-4.04737 10.67417 (222*2Ï€)/8724 weeks
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225-3.42581 10.03995 (225*2Ï€)/8724 weeks
226-3.78887 10.1104 (226*2Ï€)/8724 weeks
227-4.55705 9.74552 (227*2Ï€)/8724 weeks
228-5.09907 8.75522 (228*2Ï€)/8724 weeks
229-5.10274 7.61567 (229*2Ï€)/8724 weeks
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# Portfolio
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### Portfolio
1. 1. NAME: CHEW UNG HENG ID: 0315397 GROUP SESSION: MR.WAN DESIGN COMMUNICATION ARC1713
2. 2. SKETCHES 4 HATCHING Through this assignment, I had learnt the proper way to sketch the chosen perspective site. Besides, I had experienced how to sketch on site which make me feel very excited and interesting. Lecturer also taught me the suitable technique to render those sketches and make my sketches look more nicer. It is no doubt that I had learnt lot of hatching skills through this project which taught by our lecturer. I keep exercising on the 4 main hatching skills by choosing an object. By doing the hatches, I get to know that we need to consider about the place where the sunlight come from and how the shadow produced.
3. 3. ORTHOGRAPHIC Along this assignment, we learnt to draw the plan , elevations and sections of different sides of the house. We have to follow all the measurement of the heights and widths of the spaces and furnitures of the house to be drawn . Then, we need to using suitable line weights while tracing on it .
4. 4. AXONOMETRIC PERSPECTIVE Through this assignment, I found it was the hardest project as it need our full attention while doing it We need to draw the exterior and interior of the house in the same time with a specific angle. I also learnt to use the proper line weight on it. I also added the shadow to make it more details. With this project, lecturer taught us how to draw the perspective in different way which is by plan and elevations. Both one-point and two-point perspective had drawn by following the plan. We need to focus on the proportion of the perspective and also the way to render the perspective drawn .
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# A cone of h
1. Radius of the cone = r = 5 cm
Height of the cone = h = 20 cm
Let R be the radius of the sphere.
Since Volume of a solid remains unchanged if transformed from one shape to other.
Volume of cone = Volume of sphere
πr2h = πR3
r2h = 4R3
(25 × 20)/4 = R3
R3 = 125
R = 5
R = 5 cm.
Thus, diameter of the sphere = 5 × 2 = 10 cm.
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UNIT18
# UNIT18 - Unit 18 Determinants Associated with each square...
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Unit 18 Determinants Associated with each square matrix is a number called its determinant. We defne the determinant and look at some oF its properties in this section. Defnition 19.1. G ivenan( n × n ) matrix A we defne the submatrix A ij oF A as the matrix obtained by deleting the i th row and j th column oF A . Example 1. IF A = 1213 2131 2534 3789 then A 11 = 131 534 789 and A 23 = 123 254 379 Defnition 19.2. The Determinant Defnition-Part 1 Let A =( a ij )bean( n × n ) matrix then the determinant oF A is defned as : (i) det ( A )= a 11 iF n = 1. (i.e. iF A =[ a ]then det ([ a ]) = a ) (ii) det ( A a 11 a 22 - a 12 a 21 iF n = 2. (i.e. det ± ab cd ² = ad - bc ) (iii) det ( A a 11 ( detA 11 ) - a 12 det ( A 12 ) ... +( - 1) 1+ j a 1 j det ( A 1 j )+ ...+( - 1) 1+ n a 1 n det ( A 1 n ) 1
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= n j =1 ( - 1) j +1 a 1 j ( det ( A 1 j )) if n> 2 This process is called expanding the determinant along the frst row of A . Example 2. Suppose A = ± 12 25 ² Then by (ii) above we have: det( A ) = (1)(5) - (2)(2) = 1 Example 3. Suppose A = 123 213 321 Find det ( A ) Solution: Looking at the de±nition of the determinant we apply (iii) to get: det ( A )= ( - 1) 1+1 (1) det ( A 11 )+( - 1) 1+2 (2) det ( A 12 - 1) 1+3 (3) det ( A 13 ) =(1) det ( A 11 ) - (2) det ( A 12 )+(3) det ( A 13 ) det ± 13 21 ² - (2) det ± 23 31 ² +(3) det ± 32 ² =(1)( - 5) - (2)( - 4) + (3)(1) = 6 Suppose we are asked to ±nd the determinant of a matrix such as A = 1234 2131 2000 3789 Then det ( A )=( - 1) 2 (1) det ( A 11 - 1) 3 (2) det ( A 12 - 1) 4 (3) det ( A 13 - 1) 5 (4) det ( A
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UNIT18 - Unit 18 Determinants Associated with each square...
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https://lovelace.augustana.edu/q2a/index.php/1072/can-we-change-a-double-variable-to-an-int-variable?show=1131
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Can we change a double variable to an int variable?
Yes, we can.
This is an example of how you convert:
int i = 3; // i is 3
double d = (double) i; // d = 3.0
+2
Is it the same for the other way to int?
+2
yeah, just switch their places
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# Examine This Report on stats project help
Visually depict two exclusive fractions which might be comparable to a offered fraction. The fractional worth is revealed on the range line when you shade within the fraction. Equivalent Fraction Pointer is amongst the Interactivate evaluation explorers.
Stage in the era of Sierpinski's Triangle -- a fractal produced from subdividing a triangle into four smaller sized triangles and cutting the center a single out. Take a look at range styles in sequences and geometric Homes of fractals.
Make your own polygon and remodel it from the Cartesian coordinate method. Experiment with reflections throughout any line, revolving around any line (which yields a three-D impression), rotations about any place, and translations in any path.
Reactions are an easy way for you personally to precise your response to a put up. We use these alerts to help badge, categorize, and feature the top content material. You merely get three for every publish so rely on them wisely!
Enter a established of data factors, then derive a purpose to suit those points. Manipulate the purpose on a coordinate aircraft utilizing slider bars. Find out how Each and every consistent and coefficient impacts the ensuing graph.
Although this is a sizeable variety, it is actually most likely an important less than-representation of the actual quantity of GHB/analog-connected deaths. This is due to: one) plan toxicology exams used by hospitals and ME’s never detect GHB/analogs; 2) quite a few coroners and ME’s are unfamiliar with GHB/analogs and don't know to ask for specific exams to detect them; 3) here are the findings restricted money can be obtained for Loss of life investigations, so toxicology screening normally stops when other medicine are detected; four) there aren't any centralized databases to point when and the place GHB-involved deaths are detected; 5) Dying documents are often un-searchable, on account of restricted technological know-how and/or cash; and 6) entry to Loss of life data is, in certain states, confined by privacy problems.
Study estimation by way of modeling of the forest fire. This action will allow the user to burn up a virtual forest, then estimate the variety, the %, or perhaps the portion of trees burned. Hearth Assessment is amongst the Interactivate assessment explorers.
Learners Participate in a generalized Variation of link 4, gaining the chance to place a chunk just after simplifying fractions, converting fractions to decimals and percentages, and answering algebra thoughts involving fractions.
See all of that neat stuff under? Guess what? It really is free of charge. If you believe what we're doing is worthwhile, and such as point we offer a plethora of marketing components to you at no cost, you should fall a suggestion within the jar.
Visualize components via making rectangular spots with a grid. As you attract Each individual component set on the grid, the elements might be outlined. Factorize two has become the Interactivate evaluation explorers.
Test your algebra capabilities by answering inquiries. This quiz asks you to unravel algebraic linear and quadratic equations of one variable. Opt for difficulty degree, query varieties, and cut-off date. Algebra Quiz has become the Interactivate evaluation quizzes.
Mixtures permits exploration of percents through two piles of coloured and uncolored chips. The user have to decide the amount of chips to color to create the specified share of colored chips in comparison with the entire pile. Mixtures is amongst the Interactivate assessment explorers.
Establish your personal additional reading polygon and completely transform it while in the Cartesian coordinate technique. Experiment with reflections throughout any line, revolving close to any line (which yields a three-D image), rotations about any level, and translations in almost any direction.
Generate graphs of features and sets of requested pairs on precisely the same coordinate aircraft. This is like a graphing calculator with advanced viewing possibilities.
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# Automatic Differentiation
Deep.Net performs automatic reverse accumulation differentiation on symbolic expressions to calculate the derivatives of the user-specified model.
In most cases, the differentiation functions are invoked by the optimizer. However, sometimes it is desired to obtain an expression for the derivative.
### A sample expression
We define an expression
$\mathbf{f}(\mathbf{x}, \mathbf{y}) = \frac{1}{(\sin x)^2 + y} \,.$
Here we do not use the model builder, because our intent is not to build a full model with parameter and optimizer support. Instead we define the symbolic sizes directly using the SizeSpec.symbol function and declare variables using Expr.var.
1: 2: 3: 4: 5: 6: 7: 8: open ArrayNDNS open SymTensor let n = SizeSpec.symbol "n" let x = Expr.var "x" [n] let y = Expr.var "y" [n] let f = 1.0 / ((sin x) ** 2.0 + y)
## Computing derivatives
We can now compute the derivatives of $$\mathbf{f}(\mathbf{x}, \mathbf{y})$$. To do so, we call the Deriv.compute function with the expression we want to differentiate. The function value and the input variables can be of any dimensionality and shape.
1: let df = Deriv.compute f
The derivative object df now contains the derivative of f w.r.t. all variables that occur in that expression.
To access a specific derivative use the Deriv.ofVar function on df and pass the requested variable.
1: 2: let dfdx = df |> Deriv.ofVar x let dfdy = df |> Deriv.ofVar y
We now have expressions for $$\partial f / \partial x$$ and $$\partial f / \partial y$$.
## Evaluating the expressions
We evaluate the expressions using the (slow) host interpreter. Because we do not use the model builder, we have to invoke Func.make directly to create a callable function from an expression.
1: 2: 3: 4: 5: let cmplr = DevHost.Compiler, CompileEnv.empty let fnF = Func.make cmplr f |> arg2 x y let fnDfdx = Func.make cmplr dfdx |> arg2 x y let fnDfdy = Func.make cmplr dfdy |> arg2 x y
Func.make expects two arguments: the first is the compiler (or interpreter) to use to transform the expression into a function. We use the host interpreter (DevHost.Compiler) without any optional options (CompileEnv.empty). The second argument is the expression to compile.
Func.make returns a function taking a variable environment VarEnvT (essentially a map from variable names to values) and returning a tensor value. To avoid having to build the variable environment explicitly, we use the arg2 function that modifies the resulting function to take two tensor arguments instead.
We can now generate some test values for the variables $$\mathbf{x}$$ and $$\mathbf{y}$$.
1: 2: let valX = seq { 0.1 .. 0.2 .. 1.0 } |> ArrayNDHost.ofSeq let valY = seq { 1.1 .. 0.2 .. 2.0 } |> ArrayNDHost.ofSeq
And compute the function values as well as derivatives.
1: 2: 3: 4: 5: 6: printfn "Using x = %A" valX printfn "Using y = %A" valY printfn "f(x, y) = %A" (fnF valX valY) printfn "" printfn "J_x f = \n%A" (fnDfdx valX valY) printfn "J_y f = \n%A" (fnDfdy valX valY)
This prints
1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: Using x = [ 0.1000 0.3000 0.5000 0.7000 0.9000] Using y = [ 1.1000 1.3000 1.5000 1.7000 1.9000] f(x, y) = [ 0.9009 0.7208 0.5781 0.4728 0.3978] J_x f = [[ -0.1613 0.0000 0.0000 0.0000 0.0000] [ 0.0000 -0.2934 0.0000 0.0000 0.0000] [ 0.0000 0.0000 -0.2812 0.0000 0.0000] [ 0.0000 0.0000 0.0000 -0.2203 0.0000] [ 0.0000 0.0000 0.0000 0.0000 -0.1541]] J_y f = [[ -0.8117 0.0000 0.0000 0.0000 0.0000] [ 0.0000 -0.5196 0.0000 0.0000 0.0000] [ 0.0000 0.0000 -0.3342 0.0000 0.0000] [ 0.0000 0.0000 0.0000 -0.2235 0.0000] [ 0.0000 0.0000 0.0000 0.0000 -0.1583]]
As expected, the Jacobians are diagonal because we computed the derivatives of an element-wise function.
## Meaning of the derivative matrix
The derivative is always returned in the shape of a Jacobian, i.e. the derivative is always a matrix. If $$\mathbf{f}$$ and $$\mathbf{x}$$ are vectors, this means
$(J_\mathbf{x} \mathbf{f})_{ij} = \frac{\partial f_i}{\partial x_j}$
and $$J_\mathbf{x} \mathbf{f}$$ will be an $$n \times m$$ matrix where $$n$$ is the length of $$\mathbf{f}$$ and $$m$$ is the length of $$\mathbf{y}$$.
If the function or an argument has the value of a matrix, the Jacobian will still be a matrix. Consider, for instance, that $$X$$ is a $$k \times l$$ matrix and $$G(X)$$ is an $$n \times m$$ matrix-valued function. Then the Jacobian $$J_G X$$ computed by Deep.Net will be a matrix of shape $$k l \times n m$$. The derivative is computed as if $$G$$ and $$X$$ were flattened into vectors (using row-major order). Thus the derivatives of the individual elements are given by
$\frac{\partial G_{i,j}}{\partial X_{v,w}} = (J_\mathbf{X} \mathbf{G})_{im + j, vl + w}$
This is also true for higher-order tensors, i.e. the derivative will be computed as if any higher order tensor were flattened into a vector using row-major order. Likewise, a scalar-valued function will produce a Jacobian matrix with one row.
### Chain rule
Using matrices to store the derivatives has the advantage that the chain rule is always valid.
Consider a vector-valued function $$\mathbf{f} (G (\mathbf{x}))$$. Given the derivatives $$J_\mathbf{x} G$$ and $$J_G \mathbf{f}$$ we can compute the derivative $$J_\mathbf{x} \mathbf{f}$$ by
$J_\mathbf{x} \mathbf{f} = J_G \mathbf{f} \cdot J_\mathbf{x} G$
where $$\cdot$$ represent the matrix dot product.
val n : obj
Full name: Diff.n
val x : float
Full name: Diff.x
val y : float
Full name: Diff.y
val f : float
Full name: Diff.f
val sin : value:'T -> 'T (requires member Sin)
Full name: Microsoft.FSharp.Core.Operators.sin
val df : obj
Full name: Diff.df
val dfdx : obj
Full name: Diff.dfdx
val dfdy : obj
Full name: Diff.dfdy
val cmplr : obj * obj
Full name: Diff.cmplr
val fnF : (obj -> obj -> obj)
Full name: Diff.fnF
val fnDfdx : (obj -> obj -> obj)
Full name: Diff.fnDfdx
val fnDfdy : (obj -> obj -> obj)
Full name: Diff.fnDfdy
val valX : obj
Full name: Diff.valX
Multiple items
val seq : sequence:seq<'T> -> seq<'T>
Full name: Microsoft.FSharp.Core.Operators.seq
--------------------
type seq<'T> = System.Collections.Generic.IEnumerable<'T>
Full name: Microsoft.FSharp.Collections.seq<_>
val valY : obj
Full name: Diff.valY
val printfn : format:Printf.TextWriterFormat<'T> -> 'T
Full name: Microsoft.FSharp.Core.ExtraTopLevelOperators.printfn
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https://rdrr.io/github/BenBarnard/slidR/src/inst/mvNormData.R
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# inst/mvNormData.R In BenBarnard/slidR: Heteroscedastic Linear Dimension Reduction
# Data Simulation File This file contains the function to generate 1 sample of
# multivariate normal data. In: a list of means and a list of covariances, both
# from the Parameter_Configurations file. Out: a list of two data frames - the
# testing and training data frames. Each data frame will have p columns plus an
# additional column containing the true population classification for training
# and comparison.
mvNormData <- function(params, N = 5000, testPCT = 0.995){
# We expect m populations, so the length of means_ls and covs_ls should be m
# means_ls is the list of p-dimensional mean vectors
# covs_ls is the list of p x p covariance matrices
# N = is the simulation sample size ~ from each population ~ . This should be
# very large to overcome simulation standard error.
# testPCT is the percentage of N to be held aside for testing purposes.
library(MASS)
library(plyr)
library(dplyr)
means_ls <- params\$Mu
covs_ls <- params\$Sigma
# Create a wrapper for mvrnorm
mvrnorm_wrap <- function(mean, cov){
data <- MASS::mvrnorm(n = N, mu = mean, Sigma = cov)
as.data.frame(data)
}
# This is a list of m data frames, each of N observations over p columns. We
# then combine each data frame into one data frame with a label column. Thus,
# we have a data frame of p + 1 columns and m * N rows
data_df <- Map(mvrnorm_wrap, means_ls, covs_ls) %>% ldply
names(data_df)[1] <- "category"
# Now we split the data frame into testing and training data.
test_df <- data_df %>%
dplyr::group_by(category) %>%
dplyr::sample_frac(testPCT) %>%
as.data.frame
# This line requires the latest version of dplyr. It errors otherwise.
train_df <- dplyr::setdiff(data_df, test_df)
# Return statement
list(Test = test_df, Train = train_df)
}
##### Testing the Function ####################################################
# Test this function (using "params" object defined at the end of file 1)
# data_ls <- mvNormData(means_ls = params\$Mu, covs_ls = params\$Sigma)
# test_df <- data_ls\$Test
# train_df <- data_ls\$Train
# # Remove the original list to save space
# rm(data_ls)
BenBarnard/slidR documentation built on Jan. 2, 2018, 4:32 p.m.
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# Nuclear Reactions
### After reading this section you will be able to do the following:
• Describe an atom using its chemical notation.
• Begin to understand nuclear reaction equations.
What is the difference between chemical reactions and nuclear reactions?
Nuclear reactions can be described mathematically in much the same way as chemical reactions. We commonly express these reactions by equations, although there is a unique difference in the nature of the reactions. The principle difference between them lies in how the reaction occurs, specifically how the atom is affected. Chemical reactions involve an atom’s electrons while nuclear reactions involve the atom’s nucleus.
Writing a nuclear reaction equation
In order to write an equation for a nuclear reaction, we must first establish some basic rules. Each of the elements involved in the reaction is identified by the chemical symbol. Two numbers are attached to the symbol. The number at the upper right is the mass number, also known as the ‘A’ number. The 'A' number describes the atomic weight of the atom and identifies the number of protons and neutrons in the nucleus. The number at the lower left is the atomic number, or ‘Z’ number. The 'Z' number describes the number of protons in the nucleus and determines the type of atom.
The symbol for Uranium-238 is $^{238}_{92}U$
This shows you that Uranium has a mass number of 238 and an atomic number of 92. Symbols are also utilized to represent alpha and beta particles. The symbol for an alpha particle is $^{4}_{2}He$. The symbol for a beta particle is $^{0}_{-1}e$. The chemical symbol for a neutron is $^{1}_{0}n$. Can you determine the mass number and atomic number of the neutron?
Now that we know what these symbols represent, let's see how they can be applied to a nuclear equation. Uranium-238 is an isotope, which undergoes alpha decay to produce Thorium and gamma rays. This is expressed mathematically by the following equation:
$^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He + Gamma$ $Rays$
Note that when the mass numbers on each side of the equation are added together that they are equal. The same principle is true for the atomic numbers, and it shows that none of the atomic particles have been lost. One way to check to see if you have written the proper nuclear equation is to make sure both sides of the equation have the same number or atomic particles represented.
### Review:
1. A nuclear reaction can be described by an equation, which must be balanced.
2. The symbol for an atom or atomic particle includes the symbol of the element, the mass number, and the atomic number.
3. The mass number, which describes the number of protons and neutrons, is attached at the upper left of the symbol.
4. The atomic number, which describes the number of protons in the nucleus, is attached at the lower left of the symbol.
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Terrapin Resources
# What do the different buttons do?
The “See & Say” Bee-Bot has three switches underneath, for power, sound, and sensor.
# Moving the Bee-Bot
Bee-Bot moves forward and back, one step at a time. One step is equal to 6 inches, 15 centimeters, or 1 length of Bee-Bot.
TRY IT: Press the Forward button one time, then press the GO button. Bee-Bot will move forward one unit.
TRY IT: Press the X button to clear out those commands, and to put in a new set of commands. Press the Forward button two times, then press Go. Bee-Bot will move forward two units, one at a time.
TRY IT: Clear the previous commands. Press the Back button two times, then press Go. Bee-Bot will move backwards two units.
Press the X button to clear all previously entered commands. Otherwise each new command entered gets added to the end of what has already been entered.
Bee-Bot moves right or left 90 degrees.
TRY IT: Press the Right button one time, then select Go. Bee-Bot will turn right 90 degrees.
TRY IT: Combine at least three commands, for example, Forward 1, Left 1 , Back 1. Don’t forget to clear out your old program by selecting the X (Clear) button. Where did Bee-Bot go?
# Lesson 1: What is the distance of one unit or step of Bee-Bot?
#### Overview
Students will determine the distance of 1 unit of Bee-Bot. This will be helpful when using Bee-Bot and solving future challenges.
• Estimation
• Distance
• Measurement
#### Materials Needed
• Bee-Bot
• String or other resource for measuring such as yarn or ribbon
• Marker
• Scissors
#### Activity
Place Bee-Bot on the floor.
1. Have a student place one end of string next to the front of Bee-Bot.
2. Another student presses the Forward arrow once and then presses Go. Bee-Bot will move forward one unit.
3. Now mark the string where the front of Bee-Bot stopped.
4. Cut the string where marked. This represents the distance of one unit of Bee-Bot.
#### Extension
• Have students create a name for the unit of measurement.
• Compare the string to a ruler and see how many inches/centimeters one unit of Bee-Bot represents.
• Estimate how many units it will take to move Bee-Bot to a designated location in the classroom. Try the program with the robot and see if you were correct, or if you underestimated or overestimated the distance.
# Lesson 2: Getting to Know You
#### Overview
Students will apply units of measurement to move Bee-Bot to another location.
• Estimation
• Distance
• Measurement
• Direction
• Bee-Bot
#### Activity
1. Have students sit in a circle.
2. One student picks another student sitting in the circle and estimates how many units it will take Bee-Bot to reach the selected student. The first student will estimate, test, and adjust if necessary.
3. Once Bee-Bot reaches the other student, have that student answer a question before selecting yet another student to program Bee-Bot to reach. The teacher can ask a question that is general, such as what their favorite animal, movie or book is. Or you can incorporate questions from current units of study.
#### Extension
Addition problems: Write each student’s initial estimate on a chalkboard, dry erase board, or somewhere else that all can see. For example if a student estimates “forward 5”, write 5. If the estimate is too short, put a plus sign next to the 5, creating an addition problem where you can add the additional units needed to reach the desired location. Place Bee-Bot back where it began, enter the additional units and select Go. (Bee-Bot remembers the previous entry.) OR clear Bee-Bot’s memory by selecting the Clear (X) key, and then enter the new estimated total.
Subtraction problems: Apply the same concept for subtraction. If Bee-Bot goes too far students can use the back button to take away steps. Make sure you place Bee-Bot back where it started.
Don’t forget that Bee-Bot holds all previous commands entered, and adds new ones onto the end. You can delete previous commands by selecting the X (clear) button.
# What is the distance of one unit of Bee-Bot?
One unit is equal to 6 inches, 15 centimeters, of 1 length of Bee-Bot.
# Best way to care for your Bee-Bot
Use this link and click on “How Do I Get Started with my ‘Bot?”
# How many commands can Bee-Bot hold?
The original Bee-Bot can hold up to 40 commands at one time. The newer “See & Say” Bee-Bot can hold up to 200 commands at a time.
# How do I clear the Bee-Bot’s program?
Press the X button to clear all previous commands given to the Bee-Bot.
# Where is Bee-Bot’s power switch?
The power switch is located on the bottom side of Bee-Bot.
# Where is the battery located?
The battery is located on the bottom of the Bee-Bot.
# How do I turn the sound off?
The switch to turn the sound on and off is located on the bottom side of Bee-Bot.
The “See & Say” Bee-Bot has three switches underneath, for power, sound, and sensor.
# How do I charge Bee-Bot?
Use this link and click on “How Do I Charge my ‘Bot?”
# My Bee-Bot won’t charge
Use this link and click on “My Bee-Bot/Blue-Bot Won’t Charge!”
# What is the best way to maximize battery life?
Use this link and click on “How Do I Maximize Battery Life?”
# Tips to troubleshoot robot problems
Use this link to get to the Troubleshooting Tips page, and click on the applicable headings.
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Muthukumaran
Posted on
# Reactive Programming Basics - adding of two numbers and beyond
Disclaimer : I am not going to explain observables, Observer,Subscriptions,Subscriber,preventing memory leaks… Just simple reactive programming.
``````a=74
b=18
c= a +b
``````
The value of c => 92
If our program is completed here , we don’t have to go into reactive programming.
Now consider
``````a=a + 1
``````
Stop..
It appears lame to use variables like a,b,c usage and incrementing one of it.
We will consider practical situation
Scenario
Donald Trump born on 14 June 1946 & Cheguara on same day (14 June) of 1928 , the age difference between these two is 18 years
Now we can write these as
``````let ageOfTrump = 74
let ageOfCheGuara = ageOfTrump + 18
``````
so ageOfCheGuara would be 92
And the long kept secret about Age is , it always kept incrementing on birthday 😊
So if there is a method happyBirthdayTrump which increase age of trump
``````function happyBirthDayToTrump(){
ageOfTrump=ageOfTrump + 1
}
``````
What should be the value of ageOfCheGuara after this method has been called?
``````Still 92,
``````
The contract (ageOfCheGuara = ageOfTrump + 18) denotes
ageOfCheGuara should always be sum of ageOfTrump & 18 , But unfortunately it was not always.
How to fix it
So to fix it , Whenever Trump got a birthday and his age gets incremented
We should also be updating ageOfCheGuara ,something like this
``````function happyBirthDayToTrump(){
ageOfTrump=ageOfTrump + 1
ageOfCheGuara = ageOfTrump + 18
}
``````
Now things are fine, Because i keep on insisting that relationship again.
Is there a way , I could specify only once the relationship and use it across the application
if I could update ageOfTrump, ageOfCheGuara should automatically be updated.
This is a common problem, if you are from C or C++ background you could have used pointer to look into address than its value, which will help us to solve these kindda issues.
But This is Javascript, we don’t do that. How can we do ?
By making every relationship as sacred and final and no unwanted updates for sake of programming, you have already been making your system reactive.
Making Reactive with RxJS
There are many ways to achieve it , We use Rx Libraries to achieve it here
So the
ageOfTrump = 74
ageOfCheGuara = ageOfTrump + 18
instead of using as variables we will have something like
``````const ageOfTrump = new BehaviorSubject(74)
``````
BehaviourSubject helps us to set a default value 74
We can retrieve its current value by using getValue method
Don’t worry about these syntax, There are more than 100’s of operators. Just focus on the concept.
``````const ageOfCheGuara = ageOfTrump.pipe(
map(val=>val + 18)
)
``````
by using pipe operator with ageOfTrump , we are specifying ageOfCheGuara depends on it.map operator accepts a function , which will receive the value from trump and transform the value
The happyBirthDayToTrump method will be
``````function happyBirthDayToTrump(){
const incrementedAge =ageOfTrump.getValue() + 1
ageOfTrump.next(incrementedAge)
}
``````
The currentValue to be retrieved with getValue method
The value should be set with next method
Now we have made these variables reactive.
The entire source code is available on Link
The same has been demonstrated in video and available on
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Lunar Distance (astronomy)
Get Lunar Distance Astronomy essential facts below. View Videos or join the Lunar Distance Astronomy discussion. Add Lunar Distance Astronomy to your Like2do.com topic list for future reference or share this resource on social media.
Lunar Distance Astronomy
Lunar distance
A lunar distance, , is the Moon's average distance to Earth. The actual distance varies over the course of its orbit. The image compares the Moon's apparent size when it is closest and farthest from Earth.
Unit information
Unit system astronomy
Unit of distance
Symbol LD or ${\textstyle \Delta _{\oplus L}}$
Unit conversions
SI base unit
Metric system
English units
Lunar distance (LD or ${\textstyle \Delta _{\oplus L}}$), also called Earth–Moon distance, Earth-Moon characteristic distance, or distance to the Moon, is a unit of measure in astronomy. It is the average distance from the center of Earth to the center of the Moon. More technically, it is the mean semi-major axis of the geocentric lunar orbit. It may also refer to the time-averaged distance between the centers of the Earth and the Moon, or less commonly, the instantaneous Earth–Moon distance. The lunar distance is approximately a quarter of a million miles .[1]
The mean semi-major axis has a value of 384,402 km (238,856 mi).[2] The time-averaged distance between Earth and Moon centers is 385,000.6 km (239,228.3 mi). The actual distance varies over the course of the orbit of the Moon, from 356,500 km (221,500 mi) at the perigee to 406,700 km (252,700 mi) at apogee, resulting in a differential range of 50,200 km (31,200 mi).[3]
Lunar distance is commonly used to express the distance to near-Earth object encounters.[4] Lunar distance is also an important astronomical datum; the precision of this measurement to a few parts in a trillion has useful implications for testing gravitational theories such as general relativity,[5] and for refining other astronomical values such as Earth mass,[6]Earth radius,[7] and Earth's rotation.[8] The measurement is also useful in characterizing the lunar radius, the mass of the Sun and the distance to the Sun.
Millimeter-precision measurements of the lunar distance are made by measuring the time taken for light to travel between LIDAR stations on the Earth and retroreflectors placed on the Moon. The Moon is spiraling away from the Earth at an average rate of 3.8 cm (1.5 in) per year, as detected by the Lunar Laser Ranging Experiment.[9][10][11] By coincidence, the diameter of corner cubes in retroreflectors on the Moon is also .[12][13]
Distance between the Earth and Moon - sizes and distance to scale.
## Value
Lunar distance expressed in selected units
Unit Mean value Uncertainty Ref
meter [2]
kilometer [2]
mile 238,856 0.043 in [2]
AU = [15][16]
light-second [2]
## Variation
The instantaneous lunar distance is constantly changing. In fact the true distance between the Moon and Earth can change as quickly as ,[3] or more than in just 6 hours, due to its non-circular orbit.[17] There are other effects that also influence the lunar distance. Some factors are described in this section.
### Perturbations and eccentricity
The distance to the Moon can be measured to an accuracy of over a 1-hour sampling period,[18] which results in an overall uncertainty of for the average distance. However, due to its elliptical orbit with varying eccentricity, the instantaneous distance varies with monthly periodicity. Furthermore, the distance is perturbed by the gravitational effects of various astronomical bodies - most significantly the Sun and less so Jupiter. Other forces responsible for minute perturbations are: gravitational attraction to other planets in the solar system and to asteroids; tidal forces; and relativistic effects.[19] The effect of radiation pressure from the Sun contributes an amount of ± to the lunar distance.[18]
Although the instantaneous uncertainty is sub-millimeter, the measured lunar distance can change by more than from the mean value throughout a typical month. These perturbations are well understood[20] and the lunar distance can be accurately modeled over thousands of years.[19]
The Moon's distance from the Earth and moon phases in 2014.
Moon phases: 0 (1) - new moon, 0.25 - first quarter, 0.5 - full moon, 0.75 - last quarter.
Variation of the distance between the centers of the Moon and the Earth over 700 days.
### Tidal dissipation
Through the action of tidal forces, angular momentum is slowly being transferred from the Earth's rotation to the Moon's orbit. The result is that Earth's rate of spin is imperceptibly decreasing (at a rate of ),[21] and the lunar orbit is gradually expanding. The current rate of recession is .[20] However, it is believed that this rate has recently increased, as a rate of would imply that the Moon is only 1.5 billion years old, whereas scientific consensus assumes an age of about 4 billion years.[22] It is also believed that this anomalously high rate of recession may continue to accelerate.[23]
It is predicted that the lunar distance will continue to increase until (in theory) the Earth and Moon become tidally locked. This would occur when the duration of the lunar orbital period equals the rotational period of Earth. The two bodies would then be at equilibrium, and no further rotational energy would be exchanged.However models predict that 50 billion years would be required to achieve this configuration,[24] which is significantly longer than the expected lifetime of the solar system.
### Orbital history
The average lunar distance is increasing, which implies that the Moon was closer in the past. There is geological evidence that the average lunar distance was about 52 R? (i.e. 52 x the radius of the Earth, or 205,000 miles) during the Precambrian Era; 2,500 million years BP.[22]
The giant impact hypothesis, a widely accepted theory, states that the Moon was created as a result of a catastrophic impact between Earth and another planet, resulting in a re-accumulation of fragments at an initial distance of 3.8 R? (15,000 miles).[25] In this theory, the initial impact is assumed to have occurred 4.5 billion years ago.[26]
## History of measurement
Until the late 1950s all measurements of lunar distance were based on optical angular measurements: the earliest accurate measurement was by Hipparchus in the 2nd century BC. The space age marked a turning point when the accuracy of our knowledge of this value was much improved. During the 1950s and 1960s, there were experiments using radar, lasers, spacecraft, and computer modeling.[27]
This section is intended to illustrate some of the historically significant or otherwise interesting methods of determining the lunar distance, and is not intended to be an exhaustive or all-encompassing list.
### Parallax
The oldest method of determining the lunar distance involved measuring the angle between the Moon and a chosen reference point from multiple locations, simultaneously. The synchronization can be coordinated by making measurements at a pre-determined time, or during an event which is observable to all parties. Before accurate mechanical chronometers, the synchronization event was typically a lunar eclipse, or the moment when the Moon crossed the meridian (if the observers shared the same longitude). This measurement technique is known as lunar parallax.
For increased accuracy, certain adjustments must be made, such as adjusting the measured angle to account for refraction and distortion of light passing through the atmosphere.
#### Lunar eclipse
Early attempts to measure the distance to the Moon exploited observations of a lunar eclipse combined with knowledge of Earth's radius and an understanding that the Sun is much further than the Moon. By observing the geometry of a lunar eclipse, the lunar distance can be calculated using trigonometry.
The earliest accounts of attempts to measure the lunar distance using this technique were by Greek astronomer and mathematician Aristarchus of Samos in the 4th century BC[28] and later by Hipparchus, whose calculations produced a result of 59-67 R? (233,000-265,000 miles).[29] This method later found its way into the work of Ptolemy,[30] who produced a result of R? (253,000 miles) at its farthest point.[31]
#### Meridian crossing
An expedition by French astronomer A.C.D. Crommelin observed lunar transits through the meridian (the moment when the Moon crosses an imaginary great circle that passes directly overhead and through the celestial poles) on the same night) from two different locations. Careful measurements from 1905-1910 measured the angle of elevation at the moment when a specific lunar crater (Mösting A) crossed the meridian, from stations at Greenwich and at Cape of Good Hope, which share nearly the same longitude.[32] A distance was calculated with an uncertainty of , and this remained the definitive lunar distance value for the next half century.
#### Occultations
By recording the instant when the Moon occults a background star, (or similarly, measuring the angle between the moon and a background star at a predetermined moment) the lunar distance can be determined, as long as the measurements are taken from multiple locations of known separation.
Astronomers O'Keefe and Anderson calculated the lunar distance by observing four occultations from nine locations in 1952.[33] They calculated a mean distance of ; however this value was refined in 1962 by Irene Fischer, who incorporated updated geodetic data to produce a value of .[7]
An experiment was conducted in 1957 at the U.S. Naval Research Laboratory that used the echo from radar signals to determine the Earth-Moon distance. Radar pulses lasting were broadcast from a diameter radio dish. After the radio waves echoed off the surface of the Moon, the return signal was detected and the delay time measured. From that measurement, the distance could be calculated. In practice, however, the signal-to-noise ratio was so low that an accurate measurement could not be reliably produced.[34]
The experiment was repeated in 1958 at the Royal Radar Establishment, in England. Radar pulses lasting were transmitted with a peak power of 2 megawatts, at a repetition rate of 260 pulses per second. After the radio waves echoed off the surface of the Moon, the return signal was detected and the delay time measured. Multiple signals were added together to obtain a reliable signal by superimposing oscilloscope traces onto photographic film. From the measurements, the distance was calculated with an uncertainty of .[35]
These initial experiments were intended to be proof-of-concept experiments and only lasted one day. Follow-on experiments lasting one month produced a mean value of ,[36] which was the most accurate measurement of the lunar distance at the time.
### Laser ranging
An experiment which measured the round-trip time of flight of laser pulses reflected directly off the surface of the Moon was performed in 1962, by a team from Massachusetts Institute of Technology, and a Soviet team at the Crimean Astrophysical Observatory.[37]
During the Apollo missions in 1969, astronauts placed retroreflectors on the surface of the Moon for the purpose of refining the accuracy of this technique. The measurements are ongoing and involve multiple laser facilities. The instantaneous accuracy of the Lunar Laser Ranging experiments can exceed sub-millimeter resolution, and is the most reliable method of determining the lunar distance, to date.
### Amateur astronomers and citizen scientists
Due to the modern accessibility of accurate timing devices, high resolution digital cameras, GPS receivers, powerful computers and near-instantaneous communication, it has become possible for amateur astronomers to make high accuracy measurements of the lunar distance.
On May 23, 2007 digital photographs of the Moon during a near-occultation of Regulus were taken from two locations, in Greece and England. By measuring the parallax between the Moon and the chosen background star, the lunar distance was calculated.[38]
A more ambitious project called the "Aristarchus Campaign" was conducted during the lunar eclipse of 15 April 2014.[17] During this event, participants were invited to record a series of five digital photographs from moonrise until culmination (the point of greatest altitude).
The method took advantage of the fact that the Moon is actually closest to an observer when it is at its highest point in the sky, compared to when it is on the horizon. Although it appears that the Moon is biggest when it is near the horizon, the opposite is true. This phenomenon is known as the moon illusion. The reason for the difference in distance is that the distance from the center of the Moon to the center of the Earth is nearly constant throughout the night, but an observer on the surface of Earth is actually 1 Earth radius from the center of Earth. This offset brings them closest to the Moon when it is overhead.
Modern cameras have now reached a resolution level capable of capturing the Moon with enough precision to perceive and more importantly to measure this tiny variation in apparent size. The results of this experiment were calculated as LD = R?. The accepted value for that night was 60.61, which implied a 3% accuracy. The benefit of this method is that the only measuring equipment needed is a modern digital camera (equipped with an accurate clock, and a GPS receiver).
Other experimental methods of measuring the lunar distance that can be performed by amateur astronomers involve:
• Taking pictures of the Moon before it enters the penumbra and after it is completely eclipsed.
• Measuring, as precisely as possible, the time of the eclipse contacts.
• Taking good pictures of the partial eclipse when the shape and size of the Earth shadow are clearly visible.
• Taking a picture of the Moon including, in the same field of view, Spica and Mars - from various locations.
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## Pi-day
March the 14th. has officially been named the international π-day to honour this magical number which equals the ratio of the circmference to the diameter for all circles.
http://www.wikihow.com/Celebrate-Pi-Day
In 1882 the german mathematician Ferdinand Lindemann showed that pi is a transcendental number http://en.wikipedia.org/wiki/Transcendental_number
Such numbers have an infinitesimal decimal expansion with no periodicity in the numbers.
It has been said that the decimalexpansion of pi is the perfect generator of random numbers.
The record is ten million decimals generated with a computer using the Taylor-series expansion of arcustangens π/4 =1 and then solving for pi.
Perhaps the strangest quality of pi is that it surfaces in many areas of science other than geometry.
Probability calculus, imaginary numbers, infinitesimal series, calculus e.g..
So in the infinite number of decimals of pi are hidden the answers to many scientific enigmas. http://en.wikipedia.org/wiki/Pi
Annons
## Om mattelararen
Licentiate of Philosophy in atomic Physics Master of Science in Physics
Detta inlägg publicerades i Geometri, Gymnasiematematik(high school math). Bokmärk permalänken.
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## Assignment on electric potential and potential energy
### Question 1
A charge of 2C is taken from a point where the potential is -20V to point where the potential is V. If 400 J of work is done in the process, find V.
Ans: 180V
### Question 2
Toe point charges of -10μC and +40μC are kept 15 cm apart. At which point on the line joining them will the electric potential be zero.
Ans: 3 cm from -10μC charge
### Question 3
Three point charges Q1=25μC, Q2=50μC Q3=100μC are kept at the corners A, B and C respectively of an equilateral triangle ABC having each side of 7.5 m. find the total electrostatic potential energy of the system.
Ans: 10.5 J
### Question 4
How much kinetic energy is gained by an electron, initially at rest when it moves through a potential difference of 1000 V?
Ans: 103eV or 1.6 x 10-16 J
### Question 5
Two positive charges of +12μC and +8μC are placed 10 cm apart. Find the work done in bringing them 4 cm closer.
Ans: 5.7 J
### Question 6
A charge of 5μC is placed at a point. What is the work required to carry a charge of 2C at a distance of 6 cm from it in a circular path around it to the diametrically opposite end?
Ans: Zero
### Question 7
An electron is liberated from the lower of the two large parallel metal plates separated by a distance of 20 mm. The upper plate has a potential of +2400 V relative to the lower plate. How long does the electron take to reach the upper plate? (e/m = 1.8 x 1011 C/kg)
Ans: 1.36 x 10-9 s
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# PROFILE
Check out all the problems this user has already solved.
Problem Problem Name Ranking Submission Language Runtime Submission Date
1052 Month 23867º 19081250 C 0.008 8/6/20, 2:13:45 PM
1051 Taxes 17858º 19081170 C 0.000 8/6/20, 2:05:58 PM
1050 DDD 24409º 19080918 C 0.000 8/6/20, 1:45:34 PM
1049 Animal 13906º 19080835 C 0.000 8/6/20, 1:35:27 PM
1048 Salary Increase 21688º 19080499 C 0.000 8/6/20, 12:47:26 PM
1047 Game Time with Minutes 15893º 19080416 C 0.000 8/6/20, 12:31:34 PM
1046 Game Time 22600º 19080274 C 0.000 8/6/20, 12:05:06 PM
1045 Triangle Types 20475º 19080032 C 0.000 8/6/20, 11:10:07 AM
1044 Multiples 29039º 19079782 C 0.000 8/6/20, 10:19:49 AM
1043 Triangle 25063º 19079739 C 0.000 8/6/20, 10:16:25 AM
1042 Simple Sort 26845º 19079657 C 0.000 8/6/20, 10:06:08 AM
1080 Highest and Position 17996º 19078063 C 0.000 8/6/20, 3:26:28 AM
1070 Six Odd Numbers 20104º 19077988 C 0.000 8/6/20, 3:13:51 AM
1041 Coordinates of a Point 28181º 19052282 C 0.000 8/3/20, 2:52:23 PM
1040 Average 3 26291º 19052175 C 0.000 8/3/20, 2:32:24 PM
1038 Snack 35433º 19051809 C 0.000 8/3/20, 1:57:02 PM
1037 Interval 32905º 19051699 C 0.000 8/3/20, 1:35:12 PM
1036 Bhaskara's Formula 33016º 19051554 C 0.000 8/3/20, 1:25:33 PM
1035 Selection Test 1 37192º 19051423 C 0.000 8/3/20, 1:02:33 PM
1021 Banknotes and Coins 22126º 18994147 C 0.000 7/28/20, 2:16:32 PM
1020 Age in Days 37437º 18993084 C 0.000 7/28/20, 12:35:41 PM
1019 Time Conversion 38203º 18993018 C 0.000 7/28/20, 12:26:32 PM
1018 Banknotes 35733º 18992962 C 0.000 7/28/20, 12:16:52 PM
1017 Fuel Spent 39556º 18992754 C 0.000 7/28/20, 11:34:03 AM
1016 Distance 38680º 18992688 C 0.000 7/28/20, 11:21:41 AM
1015 Distance Between Two Points 43152º 18992573 C 0.000 7/28/20, 10:53:53 AM
1014 Consumption 43579º 18992471 C 0.000 7/28/20, 10:28:35 AM
1013 The Greatest 41833º 18992433 C 0.000 7/28/20, 10:19:37 AM
1012 Area 44024º 18992361 C 0.000 7/28/20, 10:05:27 AM
1011 Sphere 46517º 18991377 C 0.000 7/28/20, 6:46:41 AM
1 of 2
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# 14932 (number)
14932 is an even five-digits composite number following 14931 and preceding 14933. In scientific notation, it is written as 1.4932 × 104. The sum of its digits is 19. It has a total of 3 prime factors and 6 positive divisors. There are 7,464 positive integers (up to 14932) that are relatively prime to 14932.
## Basic properties
• Is Prime? no
• Number parity even
• Number length 5
• Sum of Digits 19
• Digital Root 1
## Name
Name fourteen thousand nine hundred thirty-two
## Notation
Scientific notation 1.4932 × 104 14.932 × 103
## Prime Factorization of 14932
Prime Factorization 22 × 3733
Composite number
Distinct Factors Total Factors Radical ω 2 Total number of distinct prime factors Ω 3 Total number of prime factors rad 7466 Product of the distinct prime numbers λ -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 14932 is 22 × 3733. Since it has a total of 3 prime factors, 14932 is a composite number.
## Divisors of 14932
1, 2, 4, 3733, 7466, 14932
6 divisors
Even divisors 4 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ 6 Total number of the positive divisors of n σ 26138 Sum of all the positive divisors of n s 11206 Sum of the proper positive divisors of n A 4356.33 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 122.197 Returns the nth root of the product of n divisors H 3.42765 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 14932 can be divided by 6 positive divisors (out of which 4 are even, and 2 are odd). The sum of these divisors (counting 14932) is 26138, the average is 4356.333.
## Other Arithmetic Functions (n = 14932)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 7464 Total number of positive integers not greater than n that are coprime to n λ 3732 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 1750 Total number of primes less than or equal to n r2 8 The number of ways n can be represented as the sum of 2 squares
There are 7,464 positive integers (less than 14932) that are coprime with 14932. And there are approximately 1,750 prime numbers less than or equal to 14932.
## Divisibility of 14932
m n mod m
2 0
3 1
4 0
5 2
6 4
7 1
8 4
9 1
The number 14932 is divisible by 2 and 4.
• Deficient
• Polite
## Base conversion 14932
Base System Value
2 Binary 11101001010100
3 Ternary 202111001
4 Quaternary 3221110
5 Quinary 434212
6 Senary 153044
8 Octal 35124
10 Decimal 14932
12 Duodecimal 8784
20 Vigesimal 1h6c
36 Base36 bis
## Basic calculations (n = 14932)
### Multiplication
n×y
n×2 29864 44796 59728 74660
### Division
n÷y
n÷2 7466 4977.33 3733 2986.4
### Exponentiation
ny
n2 222964624 3329307765568 49713223555461376 742317854130149266432
### Nth Root
y√n
2√n 122.197 24.6248 11.0543 6.83634
## 14932 as geometric shapes
### Circle
Diameter 29864 93820.5 7.00464e+08
### Sphere
Volume 1.39458e+13 2.80186e+09 93820.5
### Square
Length = n
Perimeter 59728 2.22965e+08 21117
### Cube
Length = n
Surface area 1.33779e+09 3.32931e+12 25863
### Equilateral Triangle
Length = n
Perimeter 44796 9.65465e+07 12931.5
### Triangular Pyramid
Length = n
Surface area 3.86186e+08 3.92363e+11 12191.9
## Cryptographic Hash Functions
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# The true reason why computer was invented
in life • 6 months ago
(image courtesy by William Gartner, Documentary Films)
Computers are not really invented for playing games, surfing the internet, chatting, socializing in social media such as FaceBook, Twitter, Instagram etc...
The word itself "COMPUTER". Let's remove the letter 'R' so the word now "COMPUTE"
Compute, compute, compute, compute and compute...
So in short computer was designed and invented for one purpose only
to solve and compute "Mathematical" complex problems.
Way back in World War 2.
If two battleship cruisers aim to shoot each other they often missed.
So to hit one battleship on the ocean is just mere luck.
Usually Battleship Cruiser has giant cannon guns and the bullets
are also giant. So meaning these giant bullets are expensive.
So every time a Battleship Cruiser fire its guns and missed a target
then it's a waste of money.
So to solve this problem they needed to write the mathematics (Calculus)
equation on paper. But the downside it takes someone 2 minutes or
1 minute before someone can solve the problem on paper.
Before they can fire the giant guns the enemy is already on the move.
So meaning the solved equation on paper and its answer is wrong because
the enemy ship is already on a different location.
The solution to write the mathematical equation on paper
is not really the solution. So they needed to come up ideas
to solve this problem.
They needed a machine to compute this mathematical problem in one second only.
Take note ONE SECOND only, and that is the "Computer".
Currently, an ordinary computer can compute thousands of computation in just one second. Now, how about "SUPER COMPUTER" it can compute millions of computations in just ONE SECOND.
How about human can it compute ONE SECOND only? LOL, of course not.
Of course, as time passed by the "Computer" evolves and now everyone can use and enjoy it. Especially the internet.
Thank you very much, please upvote, resteem, comment and
P.S. I hope my Filipino steemians support this blog, because I'm a
pinoy.
Sort Order:
Seriously enjoyed that, I'm going to be telling that story to people in the future for sure. Well not "story", but fact lol.
Hi there! Welcome back :)
I've been away for past 3 weeks (short holiday) and finally im back :) Loads of catching up is awaiting me now.
I checked your profile and im glad to see that you're still very active on steemit.
How have you been doing?
(I upvoted number of posts today so my voting power is runnig very low. For that reason I cant upvote your post today but i will be following you closely)
Cheers,
Piotr
ps.
I loved your article. never knew that this is real computers history
@iridion9 you were flagged by a worthless gang of trolls, so, I gave you an upvote to counteract it! Enjoy!!
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MATLAB Examples
# Splitting the vertex set of a graph
We demonstrate how to split the vertex set of a graph into two parts using the split method. The goal is to separate the vertices of the graph into two natural clusters. The splitting is based on the Fiedler vector of the Laplacian matrix.
## Create a random tree
```g = graph; random_tree(g,25); ```
## Find an embedding for the tree
Note: This requires the Optimization Toolbox
```distxy(g); ```
```Optimization terminated: relative function value changing by less than OPTIONS.TolFun. Embedding score = 28.62 Elapsed time is 3.134483 seconds. ```
## Partition the vertices
```p = split(g) ```
```{ {1,2,3,4,8,9,13,15,16,17,20,22,24} {5,6,7,10,11,12,14,18,19,21,23,25} } ```
## Draw the result
```cdraw(g,p); ```
## Partition a grid graph
```grid(g,3,11); p = split(g); clf; % erase the previous drawing cdraw(g,p) ```
## Release storage
```free(g); ```
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# Equation maker online
This Equation maker online provides step-by-step instructions for solving all math problems. We will also look at some example problems and how to approach them.
## The Best Equation maker online
We'll provide some tips to help you choose the best Equation maker online for your needs. Solving exponential functions is essential for future engineers, scientists and mathematicians. Understanding how to solve exponential equations is the foundation for all other math problems, so it’s important to get a solid grasp of the basics. One of the simplest ways to solve an exponential equation is by breaking it down into fractions. This can be done by either rearranging the equation or recognizing that each term is divisible by the unknown number in the bottom-left corner. This may seem like a minor detail, but it can be surprisingly useful when you start seeing exponential expressions everywhere. For example, the time between two events can be expressed as an exponential function if you know that each event is equal to 1 divided by t (t is the time increment). Or if you understand that 2c = Km where c is velocity and m is mass, then you can use this formula to solve any number of non-linear math problems.
The best math problem scanner is software that can take a picture of your child’s hand and compare it to a database of pictures. It can then tell you whether your child has the right number of fingers, or if they have too many or too few. The downside is that these programs are expensive and often only work with certain phones. But they are worth the investment if you want peace of mind when it comes to your child’s hand development.
In order to solve inequality equations, you have to first make sure that every variable is listed. This will ensure that you are accounting for all of the relevant information. Once you have accounted for all variables, you can start to solve the equation. When solving inequality equations, keep in mind that multiplication and division are not commutative operations. For example, if you want to find the value of x in an inequality equation, you should not just divide both sides by x. Instead, you should multiply both sides by the reciprocal of x: To solve inequality equations, it is best to use graphing calculators because they can handle more complex mathematics than simple hand-held calculators can. Graphing calculators can also be used to graph inequalities and other functions such as t and ln(x).
Depending on the application, solver types can be categorized by how they solve the problem at hand (e.g., deterministic or stochastic), how they compute solutions (e.g., matrix or vectorial), and how computationally efficient they are (e.g., linear or nonlinear). One of the most common types of solver is a heuristic algorithm. Heuristic algorithms are designed to solve problems by using a combination of past experience and intuition to make an educated guess as to what approach will work best. For example, if you've seen how certain ingredients combine before without ending up with something bad, you can assume that they're unlikely to combine in a way that would cause an undesirable result - which is why heuristic algorithms will often use these past experiences as starting points in their calculations when solving new problems. While heuristic algorithms may not be perfect, they are often fast and easy to use since there isn't any need for complex calculations behind them. Another type of solver is
Perfect for people with problems with math such as mine, but I think that you should be able to see why in some problems by watching a video, good job!! Dis app has helped me to solve more complex and complicate math question and has helped me improve in my grade
Thalia Bryant
This app is so good for students like me who has bad teacher that can't make me understand anything. I have the plus version in the other phone and I'd just say the explanation is excellent for making me understand more. Great app! Totally recommended
Karen Stewart
Solving for a variable Fraction calculator algebra help Inequalities solver How to solve system of inequalities Direct variation equation solver Math help with steps
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1. Math
2. Statistics And Probability
3. module 8 question 14 the following data are the monthly...
# Question: module 8 question 14 the following data are the monthly...
###### Question details
Module 8 Question 14: The following data are the monthly salaries y and the grade point averages x for students who obtained a bachelor's degree in business administration. Please include work.
GPA Monthly Salary ($) 2.7 3,500 3.4 3,900 3.6 4,300 3.2 3,800 3.4 4,100 2.9 2,300 The estimated regression equation for these data is ŷ = -1,205.2 + 1,517.2x and MSE =304,957. a. Develop a point estimate of the starting salary for a student with a GPA of 3.0 (to 1 decimal).$ ___
b. Develop a 95% confidence interval for the mean starting salary for all students with a 3.0 GPA (to 2 decimals).
$( ___ , ___ ) c. Develop a 95% prediction interval for Ryan Dailey, a student with a GPA of 3.0 (to 2 decimals).$ ( ___ , ___ )
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+0
t = 14 is a solution to the equation (t/7)= 2.
0
181
1
t = 14 is a solution to the equation (t/7)= 2.
Guest Mar 14, 2012
Sort:
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