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# Variable Acceleration - Clueless
by ssjcory
Tags: integration, variable accel'n
P: 11 1. The problem statement, all variables and given/known data The problem is labeled Variable Acceleration. The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared and is given (in SI units) by the expression Ax = -3.00 Vx^2. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to one-half of it's initial value. 2. Relevant equations I'm not sure (I suspect that the equations of motion are used?) 3. The attempt at a solution The professor has given us the answers so the answer is .222 seconds. I am not able to attempt a solution I can only restate the variables in more readable terms. Final speed (Vf) = .75 Initial speed (V0) = 1.50 Initial time (T0) = 0 Initial position(X0) = 0 Equation for var accel (Ax) = -3.00 Vx^2 Final time (Tf) = ? this is the goal I know the problem has to do with integration but I'm not sure where to start. Can someone help me out? Thanks, Cory
HW Helper
P: 2,950
Quote by ssjcory 1. The problem statement, all variables and given/known data The problem is labeled Variable Acceleration. The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared and is given (in SI units) by the expression Ax = -3.00 Vx^2. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to one-half of it's initial value. 2. Relevant equations I'm not sure (I suspect that the equations of motion are used?) 3. The attempt at a solution The professor has given us the answers so the answer is .222 seconds. I am not able to attempt a solution I can only restate the variables in more readable terms. Final speed (Vf) = .75 Initial speed (V0) = 1.50 Initial time (T0) = 0 Initial position(X0) = 0 Equation for var accel (Ax) = -3.00 Vx^2 Final time (Tf) = ? this is the goal I know the problem has to do with integration but I'm not sure where to start. Can someone help me out? Thanks, Cory
Have you learnt differential equations? You'll definitely need to set one up to solve this.
P: 11
Quote by Curious3141 Have you learnt differential equations? You'll definitely need to set one up to solve this.
I did differentiation and integration in calc 1 but I'm really not sure how to apply it here.
Emeritus
HW Helper
PF Gold
P: 7,808
Variable Acceleration - Clueless
Quote by ssjcory 1. The problem statement, all variables and given/known data The problem is labeled Variable Acceleration. The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared and is given (in SI units) by the expression Ax = -3.00 Vx^2. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to one-half of it's initial value. ... Can someone help me out? Thanks, Cory
Hopefully you know that $\displaystyle a=\frac{dv}{dt}\,.$
Substituting (-3.00)v2 in for a gives:
$\displaystyle (-3.00)v^2=\frac{dv}{dt}\,.$
Divide both sides by v2 and integrate with respect to time.
P: 11
Quote by SammyS Hopefully you know that $\displaystyle a=\frac{dv}{dt}\,.$ Substituting (-3.00)v2 in for a gives:$\displaystyle (-3.00)v^2=\frac{dv}{dt}\,.$Divide both sides by v2 and integrate with respect to time.
I am definitely missing something big here.
I see that $\displaystyle a=\frac{dv}{dt}\,.$
and I know $\displaystyle (-3.00)v^2=\frac{dv}{dt}\,.$
because we are given $\displaystyle a=(-3.00)v^2$
As you said divide each side by v^2 (I guess so that we have all the variables on the right.
So we get $\displaystyle -3.00= \frac{dv}{(dt)(v^2)}\,$
It can be rewritten to have dt * v^-2 on top and dt on the bottom.
So now -3 = (dv*v^-2) / dt ... I am given initial and final velocity which I know have to go into this equation somehow. Initial = 1.5, final=.75
I am completely lost again :(
By virtue of knowing the answer I know that 1.5^-2 + .75^-2 ... yields the answer ... but I have no true idea why because I don't understand where that dv/dt went and I didn't really integrate (aka add one to the exponent and divide by that exponent)
I know I am not seeing something clearly. It has been a few years since calculus but I feel like its not the physics or the calculus that are messing me up, but where they intersect.
Cory
Emeritus
HW Helper
PF Gold
P: 7,808
Quote by ssjcory I am definitely missing something big here. I see that $\displaystyle a=\frac{dv}{dt}\,.$ and I know $\displaystyle (-3.00)v^2=\frac{dv}{dt}\,.$ because we are given $\displaystyle a=(-3.00)v^2$ As you said divide each side by v^2 (I guess so that we have all the variables on the right. So we get $\displaystyle -3.00= \frac{dv}{(dt)(v^2)}\,$ It can be rewritten to have dt * v^-2 on top and dt on the bottom. ... Thanks for your time, Cory
OK! You have $\displaystyle -3.00= v^{-2}\frac{dv}{dt}\,.$
Now integrate:
$\displaystyle \int-3.00\,dt= \int v^{-2}\frac{dv}{dt}dt$
$\displaystyle =\int v^{-2} dv$
Proceed on.
P: 11
Quote by SammyS OK! You have $\displaystyle -3.00= v^{-2}\frac{dv}{dt}\,.$ Now integrate:$\displaystyle \int-3.00\,dt= \int v^{-2}\frac{dv}{dt}dt$ $\displaystyle =\int v^{-2} dv$Proceed on.
Wouldn't the integration of -3 be -3X(or -3T in this circumstance?)
And on the right side v^-2 should become v^(n+1)/n+1 ... Yielding (v^-1)/1 ... or v^-1 ?
So -3T = v^-1? And plug in the final and then initial velocity?
-3T = 1.5^-1 === -.222
-3T = .75^-1 === -.444
This is very confusing to me would it be -.444(equation of time with final velocity) minus -.222(equation of time with initial velocity) ... = -.222 ... Do I just disregard the negative because it is the time? Or have I completely screwed this up?
Thanks!
Cory
HW Helper
P: 2,950
Quote by ssjcory Wouldn't the integration of -3 be -3X(or -3T in this circumstance?) And on the right side v^-2 should become v^(n+1)/n+1 ... Yielding (v^-1)/1 ... or v^-1 ? So -3T = v^-1? And plug in the final and then initial velocity? -3T = 1.5^-1 === -.222 -3T = .75^-1 === -.444 This is very confusing to me would it be -.444(equation of time with final velocity) minus -.222(equation of time with initial velocity) ... = -.222 ... Do I just disregard the negative because it is the time? Or have I completely screwed this up? Thanks! Cory
The indefinite integral of $v^{-2}$ wrt v is $-v^{-1} + C= -\frac{1}{v} + C$. You missed out the negative sign in yours (remember that the "n+1" here is MINUS one) and the constant of integration.
I've always found it better to use definite integration on both sides and incorporate the initial conditions that way. This way you don't have to bother to solve for the constant of integration.
So rewrite as:
$$\int_0^t (-3)dt = \int_{1.5}^v v^{-2}dv$$
and here, you're basically saying that at t=0, v = 1.5 m/s (lower bound). At an unknown time t, the velocity is just v (upper bound). Although this is not the best way to write the integral from a purist mathematical perspective, it's very common in physics, and hence is acceptable.
Now integrate and impose the bounds:
$$-3t = -\frac{1}{v} + \frac{1}{1.5}$$
See how the bounds were incorporated there. On the left hand side, the term for t=0 vanishes. On the right hand side, you need to subtract off the -1/v expression at v = 1.5, but subtracting a negative is like adding it (minus*minus = plus).
All you're required to do now is to find t when v is half the initial velocity, or 0.75 m/s.
P: 11
Quote by Curious3141 The indefinite integral of $v^{-2}$ wrt v is $-v^{-1} + C= -\frac{1}{v} + C$. You missed out the negative sign in yours (remember that the "n+1" here is MINUS one) and the constant of integration. I've always found it better to use definite integration on both sides and incorporate the initial conditions that way. This way you don't have to bother to solve for the constant of integration. So rewrite as: $$\int_0^t (-3)dt = \int_{1.5}^v v^{-2}dv$$ and here, you're basically saying that at t=0, v = 1.5 m/s (lower bound). At an unknown time t, the velocity is just v (upper bound). Although this is not the best way to write the integral from a purist mathematical perspective, it's very common in physics, and hence is acceptable. Now integrate and impose the bounds: $$-3t = -\frac{1}{v} + \frac{1}{1.5}$$ See how the bounds were incorporated there. On the left hand side, the term for t=0 vanishes. On the right hand side, you need to subtract off the -1/v expression at v = 1.5, but subtracting a negative is like adding it (minus*minus = plus). All you're required to do now is to find t when v is half the initial velocity, or 0.75 m/s.
So I was just a tad bit off for the integration (meant to put the negative in front of the v^-1)
So once we have the basic form
-3T = -v^-1
Which I think I was able to get to.
You have to add the initial velocity since that is where the equation starts?
I'm guessing that's something you just have to think about and know to do.
After that I plugged in .75 into v and ended up with t = .222 which is the right answer :)
I double checked by plugging in the initial values (T=0) and (V=1.5) to make sure that both sides of the equation balanced. They did so I knew it was correct :D
I have one last question for ya.
The dv or dt don't really matter do they? Are they just an indicator of what we are integrating in respect to?
Thanks so much you are a life saver. I know one of these questions is going to be on the exam (Only 8 questions on it!)
Thanks,
Cory
HW Helper
P: 2,950
Quote by ssjcory So I was just a tad bit off for the integration (meant to put the negative in front of the v^-1) So once we have the basic form -3T = -v^-1 Which I think I was able to get to. You have to add the initial velocity since that is where the equation starts? I'm guessing that's something you just have to think about and know to do. After that I plugged in .75 into v and ended up with t = .222 which is the right answer :) I double checked by plugging in the initial values (T=0) and (V=1.5) to make sure that both sides of the equation balanced. They did so I knew it was correct :D
It's unclear what you're doing, so I can't tell if you're really doing the right thing. I don't understand what you mean by "basic form" here. The velocity at time t is *not* given by "-3T = -v^-1" because you haven't taken into account the fact that the initial velocity is 1.5 m/s. You either have to consider the constant of integration (and then solve for it using the initial conditions), or you have to set up the definite integral as I did. Either way, what I posted is the actual relation between v and t.
I have one last question for ya. The dv or dt don't really matter do they? Are they just an indicator of what we are integrating in respect to? Thanks so much you are a life saver. I know one of these questions is going to be on the exam (Only 8 questions on it!) Thanks, Cory
Of course they matter! They're the variables of integration. Manipulating them algebraically to bring them on either side of the equation (called separating the variables) is just a way to solve this very simple differential equation.
Related Discussions General Physics 1 Classical Physics 5 Classical Physics 8 Introductory Physics Homework 1 Introductory Physics Homework 11
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# How Hedge Funds Get Rich (Hint: It’s Not Their Returns)
I used to think that those that ran hedge funds got rich because of their incredible returns. Then I heard about the 2 and 20 fee structure that most hedge funds charged. The typical hedge fund fee structure (historically) is 2% of assets under management and 20% of all positive returns.
Therefore, if you gave a hedge fund \$1 million and they got a 10% return on it, their total take in fees would be:
[\$1 million * 0.02] + [(\$1 million * 0.1) * 0.2] = \$20,000 + \$20,000 = \$40,000.
This represents \$20,000 for managing your money and \$20,000 for their positive performance (i.e. the \$100,000 they earned with your money). Despite their \$40,000 fee, you would end up \$60,000 richer than before you met them. Good deal, right?
However, what if I told you that in less than 20 years the hedge fund would have more money than you (regardless of the size of your initial investment). Shocking, but true.
I created a simulation model in R that illustrates this (program is here). It assumes that the hedge fund will receive a market return each year (~10% average with a 20% standard deviation) and reinvest all fees it gets along with its clients’ money.
After running 10,000 simulations of this model, the average amount of time before the hedge fund’s capital exceeds the client capital is about 17 years. You can see the hedge fund’s capital as a percentage of the client capital for the first 100 simulations here:
The horizontal black line represents the point at which the hedge fund has as much money as its clients. The dashed gray vertical line is the point where the hedge fund, on average, would have more capital than its clients.
You can see that most of the 100 simulations cross the 100% line before the 20 year mark. The luckiest hedge fund is richer than its clients within 13 years and the unluckiest within 23 years.
Now, compare this to a lower cost ETF/index fund option (see plot below). If the client were to purchase an S&P 500 ETF/index fund from Vanguard or Charles Schwab with an annual expense ratio of 5 basis points (0.05%), the time it would take for the fund to have more capital than its client is almost 1,500 years!
Remember, this is the exact same simulation as above, but with lower fees.
The stark contrast between the hedge fund fee structure and a low cost ETF should illustrate how extractive the 2 and 20 hedge fund fee structure is for investors. This is true even when the hedge fund has a watermark (a performance benchmark it must hit before it gets a performance fee) or whether the hedge fund can outperform the market.
I ran sensitivities on both of these factors and they made little difference. No matter what initial capital you give the hedge fund to start with, the hedge fund will become richer than you since its real talent is transferring your wealth into its coffers.
The good news is it seems that investors have caught on to this fact as hedge funds saw increased outflows in 2016 even as they dropped their fees. However, the current fee averages (1.4% management and 17% performance) don’t change the math enough to make much of a difference.
With these fees, a hedge fund would have more capital than its client within 22 years instead of 17. These are both a far cry from the ~1,500 years that would be required by a low cost index fund/ETF.
How You Can Get Rich (Hint: Pay Lower Fees)
The moral of the story is that fees matter, a lot. When it comes to investing, a little more in fees can go a long way (for those charging them). So though 2% might sound small now, within a short time frame it can add up very quickly.
If you liked this post, consider signing up for my newsletter.
This is post 01. Any code I have related to this post can be found here with the same numbering: https://github.com/nmaggiulli/of-dollars-and-data
This content, which contains security-related opinions and/or information, is provided for informational purposes only and should not be relied upon in any manner as professional advice, or an endorsement of any practices, products or services. There can be no guarantees or assurances that the views expressed here will be applicable for any particular facts or circumstances, and should not be relied upon in any manner. You should consult your own advisers as to legal, business, tax, and other related matters concerning any investment.
The commentary in this “post” (including any related blog, podcasts, videos, and social media) reflects the personal opinions, viewpoints, and analyses of the Ritholtz Wealth Management employees providing such comments, and should not be regarded the views of Ritholtz Wealth Management LLC. or its respective affiliates or as a description of advisory services provided by Ritholtz Wealth Management or performance returns of any Ritholtz Wealth Management Investments client.
References to any securities or digital assets, or performance data, are for illustrative purposes only and do not constitute an investment recommendation or offer to provide investment advisory services. Charts and graphs provided within are for informational purposes solely and should not be relied upon when making any investment decision. Past performance is not indicative of future results. The content speaks only as of the date indicated. Any projections, estimates, forecasts, targets, prospects, and/or opinions expressed in these materials are subject to change without notice and may differ or be contrary to opinions expressed by others.
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# Interest Coverage Ratio (ICR)
Written by True Tamplin, BSc, CEPF®
Updated on June 21, 2021
## What is Interest Coverage Ratio (ICR)? – Definition
The interest coverage ratio is aimed at indicating how well can a company service its long term loans. ICR is calculated by dividing net profit (before deducting the interest) by the total interest expenses and is expressed in times. The interest coverage ratio is also known as the debt service ratio or debt service coverage ratio. Times interest earned or ICR is a measure of a company’s ability to honor its debt payments. It may be calculated as either EBIT or EBITDA divided by the total interest expense of the company.
## Explanation
This is an important and much-studied ratio, especially when borrowing is high relative to shareholders’ fund. This situation, known as being highly geared, is explained here. It is also particularly significant when the interest charge is high relative to profits. Obviously a company that cannot pay its interest charge has severe problems and might not be able to carry on, at least not without a fresh injection of funds.
Interest cover is profit before interest and tax divided by the interest charge. the higher the resulting number the more easily the business is managing to pay the interest charge.
## Interpretation of interest coverage ratio:
If the Interest coverage ratio is high, it shows that interest payments are not a major part of the company’s total expenses and the company is therefore likely to be able to service them comfortably. However, if the ICR is low, it means even a small drop in the company’s operation levels can make payment of interest difficult for the company. Lenders are therefore very concerned about this ratio.
A highly geared company, i.e. a company with a high level of borrowings, will generally have a low-interest coverage ratio. Conversely, a low geared company will generally have a high ratio.
## Formula to calculate interest coverage ratio
The formula of interest coverage ratio is written as follows:
Where:
(i) Earnings before interest and tax are the operating profit of the company.
(ii) Fixed interest expenses indicate all the payable interest on borrowings like bonds, loans, etc.
## Interest coverage ratio example:
The following information has been extracted from the Income statement of John Trading Company.
• Interest expenses: \$25,000
• Earnings before interest and tax: \$300,000
Required: Calculate the interest coverage ratio of John trading company.
### Solution:
Interest coverage ratio = Earnings before interest and tax / Fixed interest expenses
= \$300,000 / \$25,000
= 12 times
The earnings are 12 times more than the interest expenses of John trading company which shows, the company is comfortable enough for the payment of interest expenses on its borrowings.
## Example 2:
Suppose, A company X has the following data:
Total Revenue = \$20,000,000
Cost of goods sold: \$10,00,000
Operating Expenses:
Salaries = \$2,20,000
Rent = \$5,00,000
Utilities = \$3,00,000
depreciation = \$1,50,000
Interest expenses = \$3,000,000
Required:
Prepare the Income Statement of the Company and calculate Interest Coverage Ratio (ICR).
## Solution
The income statement of Company X is as below:
* the annual income tax expense of the company is 20%.
## Interest Coverage Ratio Calculator
### 1 thought on “Interest Coverage Ratio (ICR)”
1. Good way to find/calculate Interest coverage ratio
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# Calculating trajectory with PhysX drag
Between posts at
http://forum.unity3d.com/threads/drag-factor-what-is-it.85504/
and more from stackoverflow which I cannot manage to locate again, I’ve been working on creating a toolset for calculating trajectory to match PhysX’s systems. To that end, I have the calculations completed without drag factored in.
In short, I calculate the velocity to launch from point A to point B in X seconds with
``````HorizontalVelocity = HorizontalDistance * TimeInFlight;
VerticalVelocity = (VerticalDistance + ((0.5f * GravityMagnitude) * (TimeInFlight * TimeInFlight))) / TimeInFlight;
``````
and then calculate resultant positions along that curve by multiplying that initial velocity vector (per direction) by a chosen timestep and any applicable gravitational influence.
``````HorizontalPoint = HorizontalVelocity * TimeStep;
VerticalPoint = (VerticalVelocity * TimeStep) - ((0.5f * GravityMagnitude) * (TimeStep * TimeStep));
``````
From the aforementioned post, Physx is very definitely defining motion which can be recreated with:
``````void FixedUpdate()
{
velocity = velocity + (Physics.gravity * Time.fixedDeltaTime);
velocity = velocity * (1 - (GetComponent<Rigidbody>().drag * Time.fixedDeltaTime));
transform.position += (velocity * Time.fixedDeltaTime);
}
``````
What I have been unable to manage to do, however, is implement that drag into the trajectory calculations. I would presume that drag should be able to be simplified into something like
``````Mathf.Clamp01(Mathf.Pow(1.0f - (drag * Time.fixedDeltaTime), targetTime / Time.fixedDeltaTime))
``````
in order to simulate the effects of drag over time, but I have had no success in joining it together with either of the initial velocity calculation or the stepped calculations (to create preview lines, for example) in a way which matches up with the actual physics calculations.
In general, I’ve been aiming for a single formula for these calculations (represented by essentially only using starting point and velocity vector, then getting points at specific times) rather than working incrementally (for example, calculate each physics step one at a time to catch up to the destination point). Am I overlooking something in how I’m attempting to factor in drag ahead of time, or is it even more complicated than I’m realizing?
Your mentioned Math.Pow would work for your HorizontalPoint but not for VerticalPoint. The problem is that the calculations involved are addition and multiplication each step. An addition is a “fix-amount-change” while the multiplication is a “percentage-change”. Since those things happen each step they are alternating.
So drag isn’t a factor of the whole progression but of each step since each step is based on the last value plus the gravity vector the whole thing scaled down by drag. You can write it as endless progression like that:
``````((((initialVel + g)*d+g)*d+g)*d+g)*d...
// btw d = 1 - drag * Time.fixedDeltaTime
``````
Just look at the result after 1, 2, 3 iterations:
``````// i = initialVel
0 iterations i
1 iterations i*d + g*d
2 iterations i*d*d + g*d*d + g*d
3 iterations i*d*d*d + g*d*d*d + g*d*d + g*d
4 iterations i*d^4 + g*d^4 + g*d^3 + g*d^2 + g*d
5 iterations i*d^5 + g*d^5 + g*d^4 + g*d^3 + g*d^2 + g*d
...
``````
This progression can’t be liniarized. So i see no way beside iterating it.
Floating point math nowadays is extremely fast. You can do thousands or millions of those calculations without much problems.
Also keep in mind that this doesn’t even consider friction or collisions (which of course can’t be predicted without iterating).
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# How does stress affect a building?
## How does stress affect a building?
Compressive stress can change the shape of structural members and in struts, if it reaches a level of stress that is too much for the member, usually called buckling.
## What is stress on a building?
A material under stress is in a state that has resulted from the application of a force or forces. These forces can also be called stresses. Stress patterns in structural elements can be complex but they usually comprise just three basic types of stress: Tensile. Compressive.
What are stress in structural members?
When a structural member is loaded, deformation of the member takes place and resistance is set up against deformation. This resistance to deformation is known as stress. The stress is defined as force per unit cross sectional area.
What is stress and strain in building structures?
When a material is loaded with a force, it produces a stress, which then causes a material to deform. Engineering strain is defined as the amount of deformation in the direction of the applied force divided by the initial length of the material.
### Is stress a strength?
Stress is a measure of how much force an object experiences per unit area, and strength is a material’s ability to withstand stress.
### Is the stress that produces twisting?
Torsion is the stress that produces twisting. The torsion strength of a material is its resistance to twisting or torque. Shear is the stress that resists the force tending to cause one layer of a material to slide over an adjacent layer. [Figure 1D] Two riveted plates in tension subject the rivets to a shearing force.
What is stress in statics?
Stress is the measure of an external force acting over the cross sectional area of an object. Stress has units of force per area: N/m2 (SI) or lb/in2 (US).
What causes structural stress?
stress causes strain and strain results in structures, different physical conditions create different structures, inferring stress from faults, and. the relationship between analogs and reality.
#### What is stress structural?
The structural stress can also be regarded as the sum of the axial and bending stress in a plate or shell, i.e. the stress at the surface after a stress linearization in the thickness direction.
#### What is a stress in SOM?
In physics, stress is the force acting on the unit area of a material. The effect of stress on a body is named as strain. Stress can deform the body. Stress can be categorized into three categories depending upon the direction of the deforming forces acting on the body.
How does stresses affect structural members in commercial buildings?
If ties don’t have these properties, the roof will sag as a result. Tensile stress affects structural members such as ties in a way that it can cause a deformity such as a bend. This, in simpler terms, would be the member being stretched, resulting in a dip.
How are stress and deformation related in structural engineering?
Previous articles introduced the concept of stress as a fundamental engineering calculation. Another useful analysis determines the deformation, or strain, experienced in structural members or assemblies. While the term “strain” can loosely refer to processes involving pasta and muscles, structural strain is defined much more rigorously.
## How are stress and strain related in engineering?
1 We’ll Cross That Bridge…. Previous articles introduced the concept of stress as a fundamental engineering calculation. 2 Engineering Strain. Engineering strain can be defined as the deformation of a material as the result of an applied force or load. 3 Strain Analysis.
## How is structural stress used in structural analysis?
The structural stress may especially be used to optimise the shape parameters of the structural member. Qualitative assessments, relative considerations and trend statements are the prevailing results of the common finite element analysis which does not take the notch effect into account.
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# Minor loss in pipe formula
1. Apr 25, 2016
### foo9008
1. The problem statement, all variables and given/known data
what is the meaning of number of diameter N ? why the author make L = ND ? and then divide it by D ?
2. Relevant equations
3. The attempt at a solution
#### Attached Files:
• ###### Capture.PNG
File size:
27.6 KB
Views:
85
2. Apr 25, 2016
### BvU
This is a practical way to do it: you get an equivalent length for fittings, elbows etc. that you can add to the sum of lengths of straight sections and use in friction factor formulas (e.g. Darcy).
Head loss is a function of $L\over D$ .
Didactically the sheet you show is indeed rather ready for improvement. I find it confusing.
3. Apr 25, 2016
### foo9008
can you explain about what is ND / D ? i am confused
4. Apr 25, 2016
### BvU
ND is L so ND / D is L/D. That is the factor that appears in the friction factor equations such as Darcy and Fanning (*). The approach exploits the observed similarity in flow properties between a flow in a pipe of 100 m and 1 m diameter and a flow in a pipe of 10 m with a diameter of 10 cm.
(*)
And I would almost wish one of the two never existed . Now you have to be really careful if you divide 16 or 64 by Re for laminar flow.....
5. Apr 25, 2016
### foo9008
why L = ND ? i dont understand it
6. Apr 25, 2016
### BvU
Length of the pipe expressed in number of diameters. Nicely dimensionless. What can I say ?
7. Apr 25, 2016
### foo9008
what does it mean by number of diameter ?
8. Apr 25, 2016
### BvU
The two pipes in #4 have the same ${L\over D} = 10$ so they will show the same pressure drop for a given fluid with widely different volume flows (factor 100) but the same flow velocity.
9. Apr 25, 2016
### foo9008
ys , they have L/ D of factor 100 ,why they will have the same pressure drop ?
10. Apr 25, 2016
### BvU
Allright, L/D = 100 .
That's what has been observed to be the case .
'Apparently' $\Delta p$ is a function of L/D, something that probably also comes out of similarity considerations.
Something with ${\rm Re} = {\rho v D\over \mu}$
11. Apr 25, 2016
### foo9008
ok , how does the case that you mentioned relate yo number of diameter ?
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Test: Theory Of Production- 1
# Test: Theory Of Production- 1
Test Description
## 30 Questions MCQ Test Business Economics for CA Foundation | Test: Theory Of Production- 1
Test: Theory Of Production- 1 for CA Foundation 2022 is part of Business Economics for CA Foundation preparation. The Test: Theory Of Production- 1 questions and answers have been prepared according to the CA Foundation exam syllabus.The Test: Theory Of Production- 1 MCQs are made for CA Foundation 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Theory Of Production- 1 below.
Solutions of Test: Theory Of Production- 1 questions in English are available as part of our Business Economics for CA Foundation for CA Foundation & Test: Theory Of Production- 1 solutions in Hindi for Business Economics for CA Foundation course. Download more important topics, notes, lectures and mock test series for CA Foundation Exam by signing up for free. Attempt Test: Theory Of Production- 1 | 30 questions in 30 minutes | Mock test for CA Foundation preparation | Free important questions MCQ to study Business Economics for CA Foundation for CA Foundation Exam | Download free PDF with solutions
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Test: Theory Of Production- 1 - Question 1
### When output decreases by 20% due to increase in inputs by 20%, this stage called the law of _______
Detailed Solution for Test: Theory Of Production- 1 - Question 1
By increasing inputs, output cannot decrease, so it is hypothetical statement and hence no law is present for this situation
Test: Theory Of Production- 1 - Question 2
### Consider the following table: What is the total output, when 2 labour are employed?
Detailed Solution for Test: Theory Of Production- 1 - Question 2
Marginal product is the result of total output of 2nd - total output of 1st. So let the total output of 2nd be x , then x-100=80, value of x is 180
Test: Theory Of Production- 1 - Question 3
### Law of diminishing returns is applicable in _________
Test: Theory Of Production- 1 - Question 4
If a firm’s output is zero, then
Detailed Solution for Test: Theory Of Production- 1 - Question 4
If the firm is operating at a level of output where the market price is at a level higher than the zero-profit point, then price will be greater than average cost and the firm is earning profits. If the price is exactly at the zero-profit point, then the firm is making zero profits.
Test: Theory Of Production- 1 - Question 5
Production activity in the short run is analysed by
Detailed Solution for Test: Theory Of Production- 1 - Question 5
The law of variable proportions which states marginal physical product of a variable factor eventually diminishes, even if it increases in the beginning.
Test: Theory Of Production- 1 - Question 6
Look carefully at the table which represents a firm's short-run total cost schedule. When output goes up from four to five shirts the marginal cost is:
Test: Theory Of Production- 1 - Question 7
_________ shows the overall output generated at a given level of input :
Test: Theory Of Production- 1 - Question 8
At the point of inflexion, the marginal product is:
Detailed Solution for Test: Theory Of Production- 1 - Question 8
At point of inflexion marginal product is maximum and here after it starts decreasing.
Test: Theory Of Production- 1 - Question 9
Change in total revenue due to incremental change in quantity supplied is called:
Test: Theory Of Production- 1 - Question 10
If the marginal product of labour is below the average product of labour. It must be true that:
Test: Theory Of Production- 1 - Question 11
Increasing returns to scale can be explained in terms of:
Test: Theory Of Production- 1 - Question 12
Isoquants are equal to :
Detailed Solution for Test: Theory Of Production- 1 - Question 12
An isoquant is a firm’s counterpart of the consumer’s indifference curve. An isoquant is a curve that shows all the combinations of inputs that yield the same level of output. ‘Iso’ means equal and ‘quant’ means quantity. Therefore, an isoquant represents a constant quantity of output. The isoquant curve is also known as an “Equal Product Curve” or “Production Indifference Curve” or Iso-Product Curve.”
Test: Theory Of Production- 1 - Question 13
Given production is 1,00,000 units, fixed costs is Rs 2,00,000 Selling price is Rs 10 per unit and variable cost is Rs 6 per unit. Determine profit using technique of marginal costing.
Test: Theory Of Production- 1 - Question 14
Law of diminishing returns is applicable in :
Test: Theory Of Production- 1 - Question 15
What is Production in Economics:-
Test: Theory Of Production- 1 - Question 16
An Isoquant is ________ to an iso cost line at equilibrium point:
Detailed Solution for Test: Theory Of Production- 1 - Question 16
Least Cost Factor Combination or Producer's Equilibrium or Optimal Combination of Inputs.The point of tangency between the isocost and an isoquant is an important but not a necessary condition for producer's equilibrium. The essential condition is that the slope of the isocost line must equal the slope of the isoquant .
Test: Theory Of Production- 1 - Question 17
Innovation theory of entrepreneurship is propounded by
Detailed Solution for Test: Theory Of Production- 1 - Question 17
Pronounced as one of the greatest economists of the 20th century, Joseph Alois Schumpeter breathed life into the concepts of innovation and entrepreneurship. According to his theory, innovation can be leveraged in: Launch of a new product or an upgraded version of an existing product.
Test: Theory Of Production- 1 - Question 18
The concept of returns to scale is related with _________.
Test: Theory Of Production- 1 - Question 19
External economies can be achieved through:
Detailed Solution for Test: Theory Of Production- 1 - Question 19
External economies of scale occur outside of a firm but within an industry.
For example investment in a better transport network servicing an industry will resulting in a decrease in costs for a company working within that industry
Investment in industry-related infrastructure including telecommunications can cut costs for all Another example is the development of research and development facilities in local universities that several businesses in an area can benefit from
Likewise, the relocation of component suppliers and other support businesses close to the centre of manufacturing are also an external cost saving.
Agglomeration economies may also result from the clustering of businesses in a distinct geographical location e.g. software in Silicon Valley or investment banks in the City of London
Test: Theory Of Production- 1 - Question 20
Economies and diseconomies of scale explain why the:
Detailed Solution for Test: Theory Of Production- 1 - Question 20
Test: Theory Of Production- 1 - Question 21
The other names for capital formation is
Detailed Solution for Test: Theory Of Production- 1 - Question 21
The other names for capital formation is investments
Test: Theory Of Production- 1 - Question 22
Production activity in the short period is analysed with the help of
Test: Theory Of Production- 1 - Question 23
The marginal product curve is above the average product curve when the average product is:
Test: Theory Of Production- 1 - Question 24
Which one of the following is not a characteristic of land?
Detailed Solution for Test: Theory Of Production- 1 - Question 24
Characteristics of land
Fixed Quantity:
Land is Permanent:
Land is a Primary Factor of Production
Land is a Passive Factor of Production
Land is Immovable
Land Differs in Fertility
Land has Many Uses
Hence, an active production is not a charterstic of land.
Test: Theory Of Production- 1 - Question 25
Which of the following is not a characteristics of Land?
Detailed Solution for Test: Theory Of Production- 1 - Question 25
Characteristics of land
Fixed Quantity:
Land is Permanent:
Land is a Primary Factor of Production
Land is a Passive Factor of Production
Land is Immovable
Land Differs in Fertility
Land has Many Uses
Test: Theory Of Production- 1 - Question 26
With a view to increase his production, Hariharan a manufacturer of shoes, increases all the factors of production in his unit by 100%. But at the end of the year he finds that instead of an increase of 100%, his production has increased by only 80%. Which law of returns to scale is operating in this case?
Test: Theory Of Production- 1 - Question 27
External economies are enjoyed:
Detailed Solution for Test: Theory Of Production- 1 - Question 27
The correct option is Option B.
Industry refers to a number of markets grouped together. So when an industry expands, there are a lot of sectors that are benefitted. All the firms in the economy get external economies i.e., they get the benefit of large scale business and they tend to grow further.
Test: Theory Of Production- 1 - Question 28
Diminishing marginal returns implies :
Detailed Solution for Test: Theory Of Production- 1 - Question 28
Diminishing returns to labour occurs when marginal product of labour starts to fall. This means that total output will be increasing at a decreasing rate. The law of diminishing returns implies that marginal cost will rise as output increases. Eventually, rising marginal cost will lead to a rise in average total cost.
Test: Theory Of Production- 1 - Question 29
Law of variable proportion is valid when:
Detailed Solution for Test: Theory Of Production- 1 - Question 29
The law of variable proportions come into being when there is a fixed factor and a variable factor. The law of variable proportions states that as the quantity of one factor is increased, keeping the other factors fixed, the marginal product of that factor will eventually decline.
Test: Theory Of Production- 1 - Question 30
If LAC curve falls as output expands, this is due to ______:
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# Milind Amga
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Times 2 - START HERE
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## Friday, June 11, 2021
### Exploring the Ternary Operator
I’ve never had a lot of use for the ternary operator. Until recently, I think I under appreciated it. I Don’t recall ever using it in professional development though that may be largely related to using languages that didn’t support it. I never taught it to beginners either. Between time available and topics to be covered it never rose to a high enough priority for me. I’m starting to rethink that. But first, what am I talking about?
In C-style languages the syntax is:
`(expression-1) ? expression-2 : expression-3`
There is a Boolean expression inside the parenthesis, followed by a ? and a value for a true case, a : and a value for a false case. For example:
`(Player == WHITE) ? "White" : "Black";`
Other languages have different formats. For example, in Python the true option comes before the Boolean expression.
`[on_true] if [expression] else [on_false]`
```>>> x, y = 5, 6
>>> print("x" if x> y else "y")```
We actually see this sort of thing in spreadsheets though I never really thought of it as a type of ternary operator. This example from Excel is really a ternary operation
`=IF(G10 < G11 ,"big","small")`
Back in the day we had something not all that different in FORTRAN. There was (probably still is) an IF statement in FORTRAN that branches to one of three lines depending on if an arithmetic value is less than, equal to, or greater than zero.
`IF ( N ) 10, 20, 30 `
But enough history. What really changed my mind about this operator? As is typical for me I revalued it when I found a good use for it. I have been writing a version of the Reversi program in .NET and C#. I borrowed a bunch of Java code from my friend Tom Indelicato (read about that program here) Steal from the best is my motto and Tom is an outstanding programmer. Along with the syntactic changes moving from Java to C# I wrote a lot of code to make the game work in a graphical user interface. I made some different design decisions as well which changed how some things are handled. I created a user control class for the game squares for starters. As part of this I used three different integer values to indicate whether the square held a black disk,a white disk, or was empty.
In the past I have often used a Boolean value to indicate which of two players was the current player. Switching players is pretty simple with a Boolean value.
`currentPlayer = !curentPlayer;`
That wasn’t going to work if my player indicators were integers. I decided to use the ternary operator because I didn’t want to write multiple lines of code for something that simple. So I wrote:
`Player = (Player == WHITE) ? BLACK : WHITE;`
WHITE and BLACK are defined constants. The next thing I knew, I was finding uses for this operator in all sorts of places. The really cool thing (ok for a geek like me) was that I could use this statement inside other statements.
`message = “No legal moves for " + ((Player == BLACK) ? "Black" : "White");`
I guess this old dog can learn new (to him) tricks. Do you use/teach this operator? Are there other language features you have ignored until one day you realized they would be just right for something you were doing?
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# What were the dimensions of the universe during Planck epoch?
Wikipedia says:
It is believed that, due to the extraordinarily small scale of the universe at the time, quantum effects of gravity dominated physical interactions.
But I wonder whether there is any indication that the dimensions of the universe were small at the time rather than being infinite?
Undoubtedly it was very dense but very dense does not necessary mean "small".
Is Wikipedia wrong on this point?
• Comment to the question (v1): It seems that OP is essentially pondering if the volume of 3-dimensional space is infinite or finite. Related: physics.stackexchange.com/q/9419/2451 and links therein. Dec 20 '13 at 13:58
• The (visible) universe is currently approximately $10^{27}$ m in diameter. From supernova surveys, we know that the universe is accelerating its expansion. Extrapolate backwards for about 13.8 billion years and you get a diameter that is infinitesimal. Dec 20 '13 at 14:01
• Saying that the scale was small doesn't necessarily imply that the spatial dimensions themselves of the whole universe were small. IMHO, that sentence from Wikipedia would be better if it used the term "scale factor". On a related note, see physics.stackexchange.com/a/136861/123208 Oct 19 '20 at 23:48
If the universe is infinite now, it must also have been infinite during it's earliest phase greater than t=0. If the universe is finite now, it it must also have been finite during it's earliest phase greater than t=0. What happens at t=0 is undefined since the GR equations for the universe have a singularity at t=0.
One of the comments mentioned the finite "visual" universe, also more commonly called the observable universe. This is always finite independently of whether the (whole) universe is finite or infinite.
Sometimes you may see in an article the term "universe" (carelessly) used when the intended meaning is "observable universe".
• Well, given the universe if approaching de Sitter space, there is no question about its finitness (it depends on the choice of the coordinates). Oct 19 '20 at 17:29
• @Anixx I am confused by "approaching de Sitter space". A 3D boundary to a 4D sphere is a de Sitter space. It is a solution of the Friedman equation with $\Omega_\Lambda=0$ and $\Omega_m>1$. It never approaches a de Sittrer space because it is always a de Sitter space.
– Buzz
Oct 30 '20 at 22:22
• @Anixx Is your question about wanting to know the volume of the observable universe at the Planck time? This value depends on whether or not you want to take into account the concept of an early inflation period. If you are willing to omit inflation, I can calculate a roughly approximate volume at Planck time.
– Buzz
Nov 3 '20 at 18:55
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my subscriptions
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13ass13ass 2 points
Just spitballing here.
One naive approach would be to filter the data using three different thresholds, each threshold capturing the three different states.
Determine the threshold by eye, just like you’ve been doing. Eg with respect to your sample data, filter 1 is anything above 0.6, filter 2 is between 0.6 and 0.25, filter 3 is below 0.25. Then store each filtered dataset separately. Now use your standard methods to analyze the trends for each dataset.
Now you’ve got your trends for all three cases. What’s wrong with this approach?
tubbstosterone 2 points
I tried this earlier by subsetting by standard deviation (the most naive of the naive) but there was data bleed from other curves. For instance data from the slow grouping would bleed into medium grouping. The tough part here is that the values vary per project configuration, so the values will all have their own thresholds, which is harder to automate.
I will say, though, if I can't find a way to separate the data via curves, hard coded thresholds may be the only way to go
13ass13ass 2 points
Just so we’re on the same page, I’m imagining three horizontal lines on your sample data: a low one about 0.1* on the y axis, a medium one in the middle, and a high one at about 0.8.
Would you be interested in a model that could automatically give you those horizontal lines? Or are you talking about something else?
If those lines are okay, you could probably find them procedurally with 3-means clustering.
• edit - changed from 0.25 to 0.1
tubbstosterone 2 points
That's exactly what I had in mind! Thanks.
tubbstosterone commented on a post in r/funny
95.3k
Oh Christ that drove me insane. Had a math class at my local University (UNM) and the professor was a Chinese foreign national. Being from the southwest, I can work with Latino accents despite being one of the most gringo gringo of all the gringos in town, but a Chinese accent? Shit, on top of that she went a mile a minute. I suffered through and finally withdrew after 7 weeks of her class, only to get outright frustrated with college and pull out my first semester of freshman year.
Word of advice to all high schoolers going into college this fall: get reviews on RateMyProfessor. You'll thank yourselves later.
tubbstosterone 1 point
I had a professor in college who spoke great English, but said everything in extreme Chinese tonality. Not even the students from China understood what the hell he was talking about.
tubbstosterone commented on a post in r/AskReddit
13.0k
tubbstosterone 1 point
Being in any Greek organization on campus. Shit is an additional \$4k a semester and that doesn't include food, board, all the swaps you are required to go to, all the t-shirts you are required to buy, etc. Plus, these kids all seem to have recent cars. Sure, only about half of them had mustangs, top of the line wranglers, gigantic, lifted, brand new trucks, and land rovers, but I'm a fully functioning adult and I still have yet to own a vehicle less than 8 years old.
tubbstosterone commented on a post in r/AskReddit
tubbstosterone 1 point
It went on and off for close to 5 years. Despite not being interested in the beginning and breaking things off a couple times, I didn't like being alone and tried my hardest to get it to work.
The final break was a little fucked up, but I laughed off the broken engagement and, despite the hit to my self esteem, have done fantastically ever since. Such a waste of time, money, and effort.
tubbstosterone commented on a post in r/funny
78.6k
tubbstosterone 129 points
I have three coworkers who all managed to emigrate to the US shortly before Thanksgiving. Despite all coming in from different corners of the world, at different times (80's, 90's, 00's), they all had the same first reaction: "WHAT THE FUCK IS WRONG WITH THAT CHICKEN?!"
Turns out not everyone is used to the idea of a Turkey.
tig999 10 points
None of them were European were thay? Cuz usually most Europeans still have it for Christmas
tubbstosterone 2 points
Colombia, Bangladesh, and Pakistan. My Colombian coworker said that, despite the fact that he has had/seen Turkey before, he'd never seen one of that scale; his were always a little bigger than Chickens. He said he felt like a king walking around Disneyworld with a turkey the size of his forearm.
tubbstosterone commented on a post in r/AdviceAnimals
9.9k
Captain_Aizen 53 points
This is great advice and is sure to help maintain strong family & friend ties. All I have to do... is just not participate in things that matter greatly to others! Why didn't I think of that. Thanks you fucking duck.
tubbstosterone 5 points
I announced to my parents this year that my wife and I aren't having Thanksgiving with them. It's been an absolute clusterfuck ever since and I'm not sure if it actually hurt us more than just going and forcing a smile. The random text messages telling us how hurtful we are, telling me that I'm a bad son, and asking why I hate God have been absolutely wonderful.
tubbstosterone commented on a post in r/videos
26.1k
tubbstosterone 1 point
Is this an inappropriate place for my version of a #metoo? Even though I used to joke about it, it kinda warped me for a little bit. I used to feel safe around her, but I just didn't know she preyed on nerds.
I've already spent at least a solid 5 minutes debating on whether or not to press that save button.
tubbstosterone commented on a post in r/FFXV
40
tubbstosterone 4 points
I'm really glad they did that (I needed it), but they really shouldn't have needed to. Character development of side characters really suffered during development. I'd rather have that lack rather than deficiencies in other departments, but it's a real shame. I loved the game, but (having completed it months ago) I couldn't name anyone other than the central characters.
tubbstosterone commented on a post in r/AskReddit
7.5k
umkhunto 305 points
This always makes me laugh. The idea that a bunch of civvies with AR-15's think they will be able to resist, if their government decides to go full Fidel Castro on them.
You can have all the guns in the world, but if the best trained, best funded, most efficient and terrible war machine, we have ever seen in human history, gets turned on its public - you're fucked.
tubbstosterone 1 point
If there were to be a conflict where there WAS an organized force rebelling against the government in reasonable numbers, the fight would be unwinnable for the US government unless the government decides to glass rebelling territory. Guerrilla warfare is notoriously hard to combat. No amount of tanks and drones can win that battle. Hell, the Union had to literally set the Confederacy on fire to win the American Civil War. We even have monuments down here in Alabama for buildings that survived Sherman.
To understand this argument, I believe it's important to remember that the intention for the civvies to have access to the equipment for a rebellion is for cases like leaders like Al-Assad or Gaddafi. I highly doubt we'll see anything like that on American soil any time soon (if at all), but the idea was that we could be prepared. Whether you agree with the idea or not is a totally different conversation.
tubbstosterone commented on a post in r/AskReddit
39.8k
tubbstosterone 1 point
If Sandman has taught me anything, it's that you'd be begging for death after a couple centuries. After centuries of switching between poverty, power, being enslaved, and back again, all while watching all you come to love crumble into dust, I think I'd be happy with my short time here.
tubbstosterone commented on a post in r/AskReddit
19.9k
tubbstosterone 1 point
Throw chicken tenders (not the breaded frozen kind; raw chicken cut into strips from the grocery store) into the oven on a cookie sheet lined with foil (seasoned with cajun seasoning of course) at 375/400 for 20 or so minutes. I then microwave a frozen steamer bag of whatever veggies (usually broccoli or one of those mixed veggie bags). Simple, easy, and helped me lose 50lbs.
Super lazy? I just order in using waitr.
tubbstosterone commented on a post in r/pcgaming
299
tubbstosterone 76 points
I think everyone's missed the fact that Wolfenstein II being leaked from a German address is amazing.
iffoh 3 points
why? not like they will leave the nazis or any of the symbols in the german version anyways.
tubbstosterone 8 points
That's exactly why it's interesting; the country that requires that the villains and imagery to be stripped of context or meaning is also the one responsible for leaking a game that features said villains and imagery right at the forefront.
It's just an interesting twist.
tubbstosterone commented on a post in r/KingOfTheHill
28
tubbstosterone 4 points
KOTH is an odd show because it has a fairly even distribution of good episodes, in my opinion. There's several episodes I loathe towards the beginning ("Hilloween" for instance) and several episodes I absolutely love at the very end ("I loved To Sirloin with Love"). The show changing hands was never a big deal. Yeah, they changed up some origin stories, but I kind of appreciated some of the changes. My only real complaint is that I just did not get enough of later series Cotton. He just got better as the show went on.
mandatheangrypanda 8 points
If you have an opinion to share, why don't you go in the kitchen and put it in a bunt cake.
LOL Cotton was king!
tubbstosterone 7 points
I can't choose my favorite, so here's two:
1. "We'll see who can't drive their grandson at night without glasses or a license, using a mop to press the pedals!"
2. "In my day, the principal was the meanest sum-bitch God ever put on one leg. He'd lean on a desk with both hands, and swing his leg at ya! Then, when you were standing there shocked that a one-legged man had kicked ya ... he'd bite ya!"
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# u/tubbstosterone
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Following this user will show all the posts they make to their profile on your front page.
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# Calculating Fields in Tables
Article Number:040510
Fields in a table can be calculated using operators or the SUM function. This page explains the following two calculation methods.
## Calculate fields in a table by row
You can automatically calculate fields in a table by row.
### Example of calculating numeric values in a table by row
The following example shows a formula that automatically calculates a total price by multiplying the "Quantity" and "Unit Price" fields in a table.
Place a "Calculated" field or a "Text" field inside the table and specify the following formula.
The content has been copied.
``Quantity\*Unit_Price``
The "Total" field in which this formula is set displays the total price, which is calculated by multiplying the values of the "Quantity" and "Unit Price" fields.
### Example of calculating time values in a table by row
The following example shows a formula that uses the "Start" and "End" fields in a table to automatically display the time taken to respond to an inquiry.
Place a "Calculated" field inside the table, specify the following formula in it, and set the calculation result to display a time.
The content has been copied.
``End-Start``
The "Service Time" field in which this formula is set displays the time that results from subtracting the "Start" field value from the "End" field value.
## Sum the numeric or time values in a table
You can sum the values of the fields in a table.
### Example of summing the numeric values in a table
The following example shows a formula that automatically sums the "Subtotal" fields in a table.
Place a "Calculated" field or a "Text" field outside the table and specify the following formula.
The content has been copied.
``SUM(Subtotal)``
The "Total" field in which this formula is set displays the sum of the values in the "Subtotal" fields.
### Example of summing the time values in a table
The following example shows a formula that automatically sums the values in the "Service Time" fields in a table.
Place a "Calculated" field outside the table, specify the following formula in it, and set the calculation result to display a time.
The content has been copied.
``SUM(Service_Time)``
The "Total Service Time" field in which this formula is set displays the sum of the "Service Time" field values.
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https://mathoverflow.net/questions/303452/a-infinity-modules/303468
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# A-infinity modules
Using: https://arxiv.org/pdf/math/9910179.pdf as a reference...
My question involves spelling out explicitly the comment in 4.2 -
"Equivalently, the datum of an $A_\infty$-structure on a graded space $M$ is the datum of a differential $m_1$ on $M$ and of a morphism of $A_\infty$-algebras from $A$ to the opposite of the dg-algebra $Hom_k(M,M)$ (cf. 3.3)."
$Q1$: What is the $A_\infty$ structure on $Hom_k(M,M)^\text{op}$? (Is it just $m_1$ is the usual differential and $m_2$ is composition in opposite order with all higher $m_i$ vanishing for $i>2$..?) Could they possibly mean any other $A_\infty$-structure?
$Q2$: If my answer for Q1 is correct how does this imply that an $A_\infty$-module structure on $M$ is the same as an $A_\infty$ map $\alpha:A \to Hom_k(M,M)^\text{op}$ given that:
a map of $A_\infty$-algebras is the same as having
$$\alpha_m:A^{\otimes m}\to Hom_k(M,M)^\text{op}$$
for all $m>0$ s.t.
$$\sum(-1)^{r+st}\alpha_u(1^{\otimes r}\otimes m^A_s \otimes 1^{\otimes t})=\sum(-1)^{s}m^M_r(\alpha_{i_1} \otimes \cdots \otimes \alpha_{i_r} )$$
where the left hand side is a sum over all decompositions $m=r+s+t$ where $u=r+1+t$, $s\geq1$, and $r,t\geq 0$. The right hand side is a sum over all $1\leq r \leq m$ and over all $i_1+\cdots +i_r=m$ and the sign on the right is given in the paper in 3.4.
$m^M_r$ are the higher multiplications on $Hom_k(M,M)$ that are referenced in Q1.
Ps. Obviously, there will have to be some use of tensor-hom adjunction, I just can't quite spell it out.
Let $(A,(m_i)_i)$ be an $A_\infty$-algebra and $(M,d_M)$, a (co)chain complex.
$Q1$: We endow $L:=Hom_k^*(M,M)$ with an $A_\infty$-structure where
$$m_1^L(f):=d_M\circ f - (-1)^{\text{deg} f} f\circ d_M$$
$$m_2^L(f_1\otimes f_2):=f_2\circ f_1$$
$$m_i^L=0 \text{ for all }i\geq3$$
Now we show that Q2 puts an $A$-$A_\infty$-module structure on $M$.
$Q2$: Assume we have an $A_\infty$ map $\alpha:A\to L$.
This is equivalent to having maps for all $n\geq 1$:
$$\alpha_n:A^{\otimes n}\to L$$ of (cohomological) degree $1-n$ such that (*):
$$\sum(-1)^{r+st}\alpha_u(1^{\otimes r} \otimes m_s \otimes 1^{\otimes t})= m_1^L(\alpha_n) + \sum_{i+j=n}(-1)^{i-1} m_2^L(\alpha_i \otimes \alpha _j)$$ $$=d_M\circ \alpha_n - (-1)^{1-n} \alpha_n \circ d_M +\sum_{i+j=n}(-1)^{i-1}\alpha_j \circ \alpha_i$$ as maps from $A^{\otimes n}$ to $L$.
Now if we redefine our maps so that $\beta_1=d_M$ and for all $n\geq 2$ $$\beta_n:M\otimes A^{\otimes n-1}\to M$$
$$m\otimes a_1\otimes \cdots \otimes a_{n-1} \mapsto [\alpha_{n-1}(a_1\otimes \cdots \otimes a_{n-1})](m)$$
then from the equation (*) above:
$$\sum(-1)^{r+st}\beta_{u+1}(m\otimes a_1 \otimes \cdots \otimes a_r \otimes m_s(a_{r+1}\otimes \cdots \otimes a_{r+s})\otimes a_{r+s+1} \otimes \cdots \otimes a_{n-1})$$
$$=\beta_1(\beta_n(m\otimes a_1 \otimes \cdots \otimes a_{n-1})) - (-1)^{1-n} \beta_n(\beta_1(m)\otimes a_1 \otimes \cdots \otimes a_{n-1}) + \sum_{i+j=n-1} (-1)^{i-1} \beta_{j+1} ( \beta_{i+1}(m\otimes a_1 \otimes \cdots \otimes a_{i})\otimes a_{i+1} \otimes \cdots \otimes a_{n-1})$$ for all $m\in M$ and all $a_1,\ldots,a_{n-1}\in A$.
This shows that $M$ has the structure of a right $A$-$A_\infty$-module.
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Grand Canonical Ensemble and Criteria for Equilibrium
Presentation on theme: "Grand Canonical Ensemble and Criteria for Equilibrium"— Presentation transcript:
Grand Canonical Ensemble and Criteria for Equilibrium
Lecture 7 Grand Canonical Ensemble and Criteria for Equilibrium Problem 7.2 Grand Canonical Ensemble Entropy and equilibrium
Problem 7.2 From definitions of canonical ensemble averages calculate
and and show that
Probabilities in Grand Canonical Ensemble
Number of particle can vary Where Ξ is the grand canonical partition function Which can be also written as Where Q(N) is canonical partition function for system with N particles
Formula for Number of Particles
Number of particle by definition Since
Entropy - 1 Consider Quantity S’ in terms of state probabilities
Where k is a constant. What is S’ when p1=1 and rest of p=0? In general pis are many and very small and thus S’ is large and positive
Entropy - 2 Consider differential of S’ in terms of state probabilities But since therefore Consider changing two states, j and k probabilities a bit - to conserve total probability dpj=-dpk
Entropy - 3 In microcanonical ensemble all pis are the same
By differentiating the above equation second time with respect to pj Thus S’ has a maximum in equilibrium for isolated system. Also, since pi =1/Ω, for microcanonical ensemble
Other thermodynamic functions and equilibrium
In microcanonical we showed that reaches maximum in equilibrium. Following similar procedures one can show that has minimum in equilibrium for canonical ensemble, and has maximum for grand canonical ensemble
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A114647 Expansion of (-3+4*x+3*x^2)/((1-x)*(x+1)*(x^2+2*x-1)); a Pellian-related sequence. 5
3, 2, 7, 12, 31, 70, 171, 408, 987, 2378, 5743, 13860, 33463, 80782, 195027, 470832, 1136691, 2744210, 6625111, 15994428, 38613967, 93222358, 225058683, 543339720, 1311738123, 3166815962, 7645370047, 18457556052, 44560482151, 107578520350, 259717522851 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Generating floretion: - 1.5'i + 'j + 'k - .5i' + j' + k' + .5'ii' - .5'jj' - .5'kk' - 'ij' + 'ik' - 'ji' + .5'jk' + 2'ki' - .5'kj' + .5e LINKS Colin Barker, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (2,2,-2,-1). FORMULA a(n) = A000129(n+1)+2*A059841(n). - R. J. Mathar, Nov 10 2009 From Colin Barker, May 26 2016: (Start) a(n) = (1+(-1)^n+(-(1-sqrt(2))^(1+n)+(1+sqrt(2))^(1+n))/(2*sqrt(2))). a(n) = 2*a(n-1)+2*a(n-2)-2*a(n-3)-a(n-4) for n>3. (End) PROG (PARI) Vec((-3+4*x+3*x^2)/((1-x)*(x+1)*(x^2+2*x-1)) + O(x^50)) \\ Colin Barker, May 26 2016 CROSSREFS Cf. A100828, A114688, A114689, A114695, A114696, A114697, A000129. Sequence in context: A143332 A255919 A212189 * A234750 A052546 A260016 Adjacent sequences: A114644 A114645 A114646 * A114648 A114649 A114650 KEYWORD easy,nonn AUTHOR Creighton Dement, Feb 18 2006 STATUS approved
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Last modified September 20 04:04 EDT 2020. Contains 337264 sequences. (Running on oeis4.)
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https://plainmath.net/algebra-ii/58788-what-is-the-standard-form-of-y-equal-x-5-8x-2
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Julianne Martinez
2022-02-03
What is the standard form of y=(-x-5)(8x-2)?
supletiv2i4
Expert
Explanation:
Standard form for an expression is listing the terms , beginning with the term with the highest exponent of the variable followed by decreasing exponents until the last term , usually a constant.
begin by distributing the brackets.
Each term in the 2nd bracket must be multiplied by each term in the 1st.This can be done as follows.
thus: -x(8x-2)-5(8x-2)
hence $-8{x}^{2}+2x-40x+10=-8{x}^{2}-38x+10$
This expression is in standard form.
$-8{x}^{2}-38x+10$
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https://www.numerade.com/questions/the-light-beam-shown-in-figure-p2219-makes-an-angle-of-200circ-with-the-normal-line-n-nprime-in-the-/
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### Discussion
You must be signed in to discuss.
### Video Transcript
for this problem. On the topic of reflection and refraction of light, we have shone a light beam in the figure and this makes an angle of 20 degrees with the normal line in the linseed oil. We want to find data, the angle of incidence and theta prime the angle of refraction if we are given the refractive index index for linseed oil, so we can first use Snell's law to find the angle theta and using Snell's law, we get data Is the ox sign of the effective index for the linseed oil in oil times sign data in the oil divided by the effective index in air. And so if we substitute our values in, this becomes the oxen of 1.48 times sine of 20 degrees over the refractive index for air, which is one and so calculating. We get the angle of incidence at the oil interface data to be 30 0.4 degrees, and now similarly, we can calculate the angle of refraction as the light enters the water and this is the prime. So theater prime is the oxen of the reflective index for oil times. Sign data for oil over the reflective index of water. And so this is the ox sign of one 0.48 times again, the sign of 20 degrees divided by the effective index of water, which is one 0.333 So calculating we get the angle of refraction of the light enters the water. Peter Prime to be 22 0.3 degrees.
University of Kwazulu-Natal
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https://planetmath.org/CyclicGroup
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# cyclic group
A group is said to be cyclic if it is generated by a single element.
Suppose $G$ is a cyclic group generated by $x\in G$. Then every element of $G$ is equal to $x^{k}$ for some $k\in\mathbb{Z}$. If $G$ is infinite, then these $x^{k}$ are all distinct, and $G$ is isomorphic to the group $\mathbb{Z}$. If $G$ has finite order (http://planetmath.org/OrderGroup) $n$, then every element of $G$ can be expressed as $x^{k}$ with $k\in\{0,\dots,n-1\}$, and $G$ is isomorphic to the quotient group $\mathbb{Z}/n\mathbb{Z}$.
Note that the isomorphisms mentioned in the previous paragraph imply that all cyclic groups of the same order are isomorphic to one another. The infinite cyclic group is sometimes written $C_{\infty}$, and the finite cyclic group of order $n$ is sometimes written $C_{n}$. However, when the cyclic groups are written additively, they are commonly represented by $\mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$.
While a cyclic group can, by definition, be generated by a single element, there are often a number of different elements that can be used as the generator: an infinite cyclic group has $2$ generators, and a finite cyclic group of order $n$ has $\phi(n)$ generators, where $\phi$ is the Euler totient function.
Some basic facts about cyclic groups:
Title cyclic group Canonical name CyclicGroup Date of creation 2013-03-22 12:23:27 Last modified on 2013-03-22 12:23:27 Owner yark (2760) Last modified by yark (2760) Numerical id 21 Author yark (2760) Entry type Definition Classification msc 20A05 Related topic GeneralizedCyclicGroup Related topic PolycyclicGroup Related topic VirtuallyCyclicGroup Related topic CyclicRing3 Defines cyclic Defines cyclic subgroup Defines infinite cyclic Defines infinite cyclic group Defines infinite cyclic subgroup
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https://stats.stackexchange.com/questions/88010/categorical-variable-with-a-very-large-number-of-categories-as-a-predictor/101090
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# Categorical variable with a very large number of categories as a predictor
I am trying to use a categorical variable as a predictor in a supervised learning setting, but there are too many categories for the classification algorithm to handle, something like over a 1000 categories.
What are some ways to get a manageable number of categories, is there a standard way of binning these categories?
I suppose this binning should be performed on a training set disparate from a test set, to get a truer measure of out-of-sample error? If cross-validation is employed, I suppose the procedure should be run on on each fold.
## 2 Answers
Binning is really painful - many just say it's not a right thing to do, many others offer grouping by looking at the response, anyway you will feel a little uncomfortable:)
From the comments, I see that it's geographical data that you want to use as predictor (ZIP codes). Then consider kriging - I have used it for a similar problem - predicting the price of apartment from the address solely and was very satisfied, particularly since it solved one more important problem - predicting the outcome in case of a new predictor level (since the predictor becomes continuous rather than categorical, you will be able to predict the price of delivery even there was no delivery for a given ZIP). Beautiful heatmaps is another bonus.
Here is a nice lecture (with a nice Italian accent) by Fabio Veronesi with examples of kriging with R. http://www.fabioveronesi.net/r-course/lesson4.html
One problem that you will face is geocoding the GPS coordinates, and you can use the function geocode from ggmap package in R (using free service from Google up to some limit of queries per day, I queried for a week to get all done). http://cran.r-project.org/web/packages/ggmap/ggmap.pdf
Hopefully, this was useful, happy predicting.
Neural networks have been used with well over 1000 categories (see the Google paper on image recognition). But yes, it is not easy, and you may need much much more data to learn.
Have you considered aggregating categories into larger groups instead?
• This is really the heart of my question, what techinques can be used to accomplish aggregating categories into larger groups in a supervised environment? How will this technique affect the accuracy of the classifier?
– PT83
Feb 27, 2014 at 2:55
• Ideally, your application provides a hierarchy of classes, not a flat class scheme. Otherwise, what good is finding a super category, when your application needs a more detailed category instead? Feb 27, 2014 at 8:40
• And if there is no apriori hierarchy of classes? I dont know how detailed or not I need to get, this is what I need to determine. Let me give you an example, suppose I have zipcodes by state or region (like the Northeast), these are my categories. I want to predict some cost. What technique can I use to group these zipcodes so I can effectively use the groups as predictors for cost?
– PT83
Feb 27, 2014 at 14:48
• In this case, you are treating your zip codes an categorial input variable, not as output class variable. However, even with categorial variables one should usually try to first preprocess them into something more meaningful than a string of digits. Feb 27, 2014 at 16:10
• Right, into groups, and how should one do that to get a effective predictor for cost? Hierarchical clustering perhaps?
– PT83
Feb 27, 2014 at 17:52
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https://pressbooks.bccampus.ca/chem1114langaracollege/chapter/oxidation-reduction-reactions/
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Chapter 6. Chemical Reactions and Equations
# 6.4 Oxidation-Reduction Reactions
### Learning Objectives
By the end of this section, you will be able to:
• Define oxidation and reduction.
• Assign oxidation numbers to atoms in simple compounds.
• Recognize a reaction as an oxidation-reduction reaction.
• Recognize composition, decomposition, combustion and single replacement reactions.
• Predict the products of a combustion reaction.
## Redox Reactions
Earth’s atmosphere contains about 20% molecular oxygen, O2, a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemical reactions involving O2, but its meaning has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions. A few examples of such reactions will be used to develop a clear picture of this classification.
Some redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride:
$2\text{Na}(s) + \text{Cl}_2(g) \longrightarrow 2\text{NaCl}(s)$
It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a half-reaction:
$2\text{Na}(s) \longrightarrow 2\text{Na}^{+}(s) + 2\text{e}^{-}$
$\text{Cl}_2(g) + 2\text{e}^{-} \longrightarrow 2\text{Cl}^{-}(s)$
These equations show that Na atoms lose electrons while Cl atoms (in the Cl2 molecule) gain electrons, the “s” subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:
$\begin{array}{r @ {{}={}} l} \pmb{\text{oxidation}} & \text{loss of electrons} \\[1em] \pmb{\text{reduction}} & \text{gain of electrons} \end{array}$
In this reaction, then, sodium is oxidized and chlorine undergoes reduction. Viewed from a more active perspective, sodium functions as a reducing agent (reductant), since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant), as it effectively removes electrons from (oxidizes) sodium.
$\begin{array}{r @ {{}={}} l} \pmb{\text{reducing agent}} & \text{species that is oxidized} \\[1em] \pmb{\text{oxidizing agent}} & \text{species that is reduced} \end{array}$
Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding NaCl:
$\text{H}_2(g) + \text{Cl}_2(g) \longrightarrow 2 \text{HCl}(g)$
The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound was ionic.
The following guidelines are used to assign oxidation numbers to each element in a molecule or ion:
1. The oxidation number of an atom in an elemental substance is zero.
2. The oxidation number of a monatomic ion is equal to the ion’s charge.
3. Oxidation numbers for common non-metals are usually assigned as follows:
• Hydrogen: +1 when combined with nonmetals, −1 when combined with metals
• Oxygen: −2 in most compounds, sometimes −1 (so-called peroxides, O22−), very rarely $-\frac{1}{2}$ (so-called superoxides, O2), positive values when combined with F (values vary)
• Halogens: −1 for F always, −1 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values)
4. The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.
Note: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., 2+), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2). This convention aims to emphasize the distinction between these two related properties.
### Example 1
Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:
a) H2S
b) SO32−
c) Na2SO4
Solution
a) According to guideline 1, the oxidation number for H is +1.
Using this oxidation number and the compound’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:
$\text{charge on H}_2 \text{S} = 0 = (2 \times +1) + (1 \times x)$
$x = 0 = - (2 \times +1) = -2$
b) Guideline 3 suggests the oxidation number for oxygen is −2.
Using this oxidation number and the ion’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:
${\text{charge on SO}_3}^{2-} = -2 = (3 \times -2) + (1 \times x)$
$x = -2 - (3 \times -2) = +4$
c) For ionic compounds, it’s convenient to assign oxidation numbers for the cation and anion separately.
According to guideline 2, the oxidation number for sodium is +1.
Assuming the usual oxidation number for oxygen (-2 per guideline 3), the oxidation number for sulfur is calculated as directed by guideline 4:
${\text{charge on SO}_4}^{2-} = -2 = (4 \times -2) + (1 \times x)$
$x = -2 -(4 \times -2) = +6$
Test Yourself
Assign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions:
a) KNO3 b) AlH3 c) NH4+ d) H2PO4
a) N, +5 b) Al, +3 c) N, −3 d) P, +5
### Example 2
Assign oxidation numbers to the atoms in each substance.
a) Br2 b) SiO2 c) Ba(NO3)2
Solution
a) Br2 is the elemental form of bromine. Therefore, by rule 1, each atom has an oxidation number of 0.
b) By rule 3, oxygen is normally assigned an oxidation number of −2. For the sum of the oxidation numbers to equal the charge on the species (which is zero), the silicon atom is assigned an oxidation number of +4.
c) The compound barium nitrate can be separated into two parts: the Ba2+ ion and the nitrate ion. Considering these separately, the Ba2+ ion has an oxidation number of +2 by rule 2. Now consider the NO3 ion. Oxygen is assigned an oxidation number of −2, and there are three oxygens. According to rule 4, the sum of the oxidation number on all atoms must equal the charge on the species, so we have the simple algebraic equation
x + 3(−2) = −1
where x is the oxidation number of the nitrogen atom and −1 represents the charge on the species. Evaluating,
x + (−6) = −1
x = +5
Thus, the oxidation number on the N atom in the nitrate ion is +5.
Test Yourself
Assign oxidation numbers to the atoms in H3PO4.
H = +1, O = −2, P = +5
Using the oxidation number concept, an all-inclusive definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which one or more elements involved undergo a change in oxidation number. While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist Example 5c). Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:
$\pmb{\text{oxidation}} = \text{increase in oxidation number}$
$\pmb{\text{reduction}} = \text{decrease in oxidation number}$
Returning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl2 to −1 in NaCl). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in H2 to +1 in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl2 to −1 in HCl).
## Classification of Redox Reactions
Four classifications of chemical reactions will be reviewed in this section. Predicting the products in some of them may be difficult, but the reactions are still easy to recognize.
1 – A composition reaction (sometimes also called a combination reaction or a synthesis reaction) produces a single substance from multiple reactants. A single substance as a product is the key characteristic of the composition reaction. There may be a coefficient other than one for the substance, but if the reaction has only a single substance as a product, it can be called a composition reaction. In the reaction
2 H2(g) + O2(g) $\longrightarrow$ 2 H2O(ℓ)
water is produced from hydrogen and oxygen. Although there are two molecules of water being produced, there is only one substance—water—as a product. So this is a composition reaction.
2 – A decomposition reaction starts from a single substance and produces more than one substance; that is, it decomposes. One substance as a reactant and more than one substance as the products is the key characteristic of a decomposition reaction. For example, in the decomposition of sodium hydrogen carbonate (also known as sodium bicarbonate),
2 NaHCO3(s) $\longrightarrow$ Na2CO3(s) + CO2(g) + H2O(ℓ)
sodium carbonate, carbon dioxide, and water are produced from the single substance sodium hydrogen carbonate.
Composition and decomposition reactions are difficult to predict; however, they should be easy to recognize.
### Example 3
Identify each equation as a composition reaction, a decomposition reaction, or neither.
a) Fe2O3 + 3 SO3 $\longrightarrow$ Fe2(SO4)3
b) NaCl + AgNO3 $\longrightarrow$ AgCl + NaNO3
c) (NH4)2Cr2O7 $\longrightarrow$ Cr2O3 + 4 H2O + N2
Solution
a) In this equation, two substances combine to make a single substance. This is a composition reaction.
b) Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction.
c) A single substance reacts to make multiple substances. This is a decomposition reaction.
Test Yourself
Identify the equation as a composition reaction, a decomposition reaction, or neither.
C3H8 $\longrightarrow$ C3H4 + 2 H2
decomposition
3 – Combustion reactions in which the reductant, also called a fuel, and oxidant, molecular oxygen, react vigorously and produce significant amounts of heat, and often light, in the form of a flame. Combustion reactions produce oxides of all other elements as products; any nitrogen in the reactant is converted to elemental nitrogen, N2. Many reactants, called fuels, contain mostly carbon and hydrogen atoms, reacting with oxygen to produce CO2 and H2O. For example, the balanced chemical equation for the combustion of methane, CH4, is as follows:
CH4 + 2 O2 $\longrightarrow$ CO2 + 2 H2O
Kerosene can be approximated with the formula C12H26, and its combustion equation is
2 C12H26 + 37 O2 $\longrightarrow$ 24 CO2 + 26 H2O
Sometimes fuels contain oxygen atoms, which must be counted when balancing the chemical equation. One common fuel is ethanol, C2H5OH, whose combustion equation is
C2H5OH + 3 O2 $\longrightarrow$ 2 CO2 + 3 H2O
If nitrogen is present in the original fuel, it is converted to N2, not to a nitrogen-oxygen compound. Thus, for the combustion of the fuel dinitroethylene, whose formula is C2H2N2O4, we have
2 C2H2N2O4 + O2 $\longrightarrow$ 4 CO2 + 2 H2O + 2 N2
### Example 4
Complete and balance each combustion equation.
a) the combustion of propane, C3H8
b) the combustion of ammonia, NH3
Solution
a) The products of the reaction are CO2 and H2O, so our unbalanced equation is
C3H8 + O2 $\longrightarrow$ CO2 + H2O
Balancing (and you may have to go back and forth a few times to balance this), we get
C3H8 + 5 O2 $\longrightarrow$ 3 CO2 + 4 H2O
b) The nitrogen atoms in ammonia will react to make N2, while the hydrogen atoms will react with O2 to make H2O:
NH3 + O2 $\longrightarrow$ N2 + H2O
To balance this equation without fractions (which is the convention), we get
4 NH3 + 3 O2 $\longrightarrow$ 2 N2 + 6 H2O
Test Yourself
Complete and balance the combustion equation for cyclopropanol, C3H6O.
C3H6O + 4 O2 $\longrightarrow$ 3 CO2 + 3 H2O
Watch a brief video showing the test firing of a small-scale, prototype, hybrid rocket engine planned for use in the new Space Launch System being developed by NASA. The first engines firing at 3 s (green flame) use a liquid fuel/oxidant mixture, and the second, more powerful engines firing at 4 s (yellow flame) use a solid mixture.
4 – Single-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the acid oxidation of certain metals:
$\text{Zn}(s) + 2\text{HCl}(aq) \longrightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$
Metallic elements may also be oxidized by solutions of other metal salts; for example:
$\text{Cu}(s) + 2 \text{AgNO}_3(aq) \longrightarrow \text{Cu(NO}_3)_2(aq) + 2 \text{Ag}(s)$
This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting Cu2+ ions dissolve in the solution to yield a characteristic blue color (Figure 2).
### Example 5
Identify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant.
a) $\text{ZnCO}_3(s) \longrightarrow \text{ZnO}(s) + \text{CO}_2(g)$
b) $2\text{Ga}(l) + 3\text{Br}_2(l) \longrightarrow 2\text{GaBr}_3(s)$
c) $2\text{H}_2 \text{O}_2(aq) \longrightarrow 2\text{H}_2 \text{O}(l) + \text{O}_2(g)$
d) $\text{BaCl}_2(aq) + \text{K}_2 \text{SO}_4(aq) \longrightarrow \text{BaSO}_4(s) + 2\text{KCl}(aq)$
e) $\text{C}_2 \text{H}_4(g) + 3\text{O}_2(g) \longrightarrow 2\text{CO}_2(g) + 2\text{H}_2 \text{O}(l)$
Solution
Redox reactions are identified per definition if one or more elements undergo a change in oxidation number.
a) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.
b) This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in Ga(l) to +3 in GaBr3(s). The reducing agent is Ga(l). Bromine is reduced, its oxidation number decreasing from 0 in Br2(l) to −1 in GaBr3(s). The oxidizing agent is Br2(l).
c) This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called disproportionation reaction). Oxygen is oxidized, its oxidation number increasing from −1 in H2O2(aq) to 0 in O2(g). Oxygen is also reduced, its oxidation number decreasing from −1 in H2O2(aq) to −2 in H2O(l). For disproportionation reactions, the same substance functions as an oxidant and a reductant.
d) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.
e) This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from −2 in C2H4(g) to +4 in CO2(g). The reducing agent (fuel) is C2H4(g). Oxygen is reduced, its oxidation number decreasing from 0 in O2(g) to −2 in H2O(l). The oxidizing agent is O2(g).
Test Yourself
This equation describes the production of tin(II) chloride:
$\text{Sn}(s) + 2\text{HCl}(g) \longrightarrow \text{SnCl}_2(s) + \text{H}_2(g)$
Is this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidant and reductant.
Yes, a single-replacement reaction. Sn(s) is the reductant, HCl(g) is the oxidant.
## Key Concepts and Summary
Chemical reactions are classified according to similar patterns of behavior. Redox reactions involve a change in oxidation number for one or more reactant elements. There are four classifications of chemical reactions: composition, decomposition, combustion and single displacement.
### Exercises
1. Is the reaction
2 K(s) + Br2(ℓ) $\longrightarrow$ 2 KBr(s)
2. In the reaction
2 Ca(s) + O2(g) $\longrightarrow$ 2 CaO
indicate what has lost electrons and what has gained electrons.
3. In the reaction
2 Li(s) + O2(g) $\longrightarrow$ Li2O2(s)
indicate what has been oxidized and what has been reduced.
4. Assign oxidation numbers to each atom in each substance.
a) P4 b) SO2
c) SO22− d) Ca(NO3)2
5. . Assign oxidation numbers to each atom in each substance.
a) CO b) CO2
c) NiCl2 d) NiCl3
6. Assign oxidation numbers to each atom in each substance.
a) CH2O b) NH3
c) Rb2SO4 d) Zn(C2H3O2)2
7. Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.
2 NO + Cl2 $\longrightarrow$ 2 NOCl
8. Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.
2 KrF2 + 2 H2O $\longrightarrow$ 2 Kr + 4 HF + O2
9. Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.
2 K + MgCl2 $\longrightarrow$ 2 KCl + Mg
10. Indicate what type, or types, of reaction each of the following represents:
a) $\text{Ca}(s) + \text{Br}_2(l) \longrightarrow \text{CaBr}_2(s)$
b) $\text{Ca(OH)}_2 (aq) + 2\text{HBr}(aq) \longrightarrow \text{CaBr}_2(aq) + 2\text{H}_2 \text{O}(l)$
c) $\text{C}_6 \text{H}_{12}(l) + 9\text{O}_2(g) \longrightarrow 6\text{CO}_2(g) + 6\text{H}_2 \text{O}(g)$
11. Indicate what type, or types, of reaction each of the following represents:
a) $\text{H}_2 \text{O}(g) + \text{C}(s) \longrightarrow \text{CO}(g) + \text{H}_2(g)$
b) $2\text{KClO}_3(s) \longrightarrow 2\text{KCl}(s) + 3\text{O}_2(g)$
c) $\text{Al(OH)}_3(aq) + 3\text{HCl}(aq) \longrightarrow \text{AlCl}_3(aq) + 3\text{H}_2 \text{O}(l)$
d) $\text{Pb(NO}_3)_2(aq) + \text{H}_2 \text{SO}_4(sq) \longrightarrow \text{PbSO}_4(s) + 2\text{HNO}_3(aq)$
12. Silver can be separated from gold because silver dissolves in nitric acid while gold does not. Is the dissolution of silver in nitric acid an acid-base reaction or an oxidation-reduction reaction? Explain your answer.
13. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.
a) H3PO4 b) Al(OH)3 c) SeO2
d) KNO2 e) In2S3 f) P4O6
14. Classify the following as acid-base reactions or oxidation-reduction reactions:
a) $\text{Na}_2 \text{S}(aq) + 2 \text{HCl}(aq) \longrightarrow 2 \text{NaCl}(aq) + \text{H}_2 \text{S}(g)$
b) $2\text{Na}(s) + 2\text{HCl}(aq) \longrightarrow 2\text{NaCl}(aq) + \text{H}_2(g)$
c) $\text{Mg}(s) + \text{Cl}_2(g) \longrightarrow \text{MgCl}_2(aq)$
d) $\text{MgO}(s) + 2\text{HCl}(aq) \longrightarrow \text{MgCl}_2(s) + \text{H}_2 \text{O}(l)$
e) $\text{K}_3 \text{P}(s) + 2\text{O}_2(g) \longrightarrow \text{K}_3 \text{PO}_4(s)$
f) $3\text{KOH}(aq) + \text{H}_3 \text{PO}_4(aq) \longrightarrow \text{K}_3\text{PO}_4(aq) + 3 \text{H}_2 \text{O}(l)$
15. Complete and balance the following acid-base equations:
a) HCl gas reacts with solid Ca(OH)2(s).
b) A solution of Sr(OH)2 is added to a solution of HNO3.
16. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.
a) $\text{Al}(s) + \text{F}_2(g) \longrightarrow$
b) $\text{Al}(s) + \text{CuBr}_2(aq) \longrightarrow \;\text{(single displacement)}$
c) $\text{P}_4(s) + \text{O}_2(g) \longrightarrow$
d) $\text{Ca}(s) + \text{H}_2 \text{O}(l) \longrightarrow \;\text{(products are a strong base and a diatomic gas)}$
17. The military has experimented with lasers that produce very intense light when fluorine combines explosively with hydrogen. What is the balanced equation for this reaction?
18. Great Lakes Chemical Company produces bromine, Br2, from bromide salts such as NaBr, in Arkansas brine by treating the brine with chlorine gas. Write a balanced equation for the reaction of NaBr with Cl2.
19. Lithium hydroxide may be used to absorb carbon dioxide in enclosed environments, such as manned spacecraft and submarines. Write an equation for the reaction that involves 2 mol of LiOH per 1 mol of CO2. (Hint: Water is one of the products.)
20. Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:
a) $\text{Ca(OH)}_2(s) + \text{H}_2 \text{S}(g) \longrightarrow$
b) $\text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{S}(g) \longrightarrow$
21. Write balanced chemical equations for the reactions used to prepare each of the following compounds from the given starting material(s). In some cases, additional reactants may be required.
a) solid ammonium nitrate from gaseous molecular nitrogen via a two-step process (first reduce the nitrogen to ammonia, then neutralize the ammonia with an appropriate acid)
b) gaseous hydrogen bromide from liquid molecular bromine via a one-step redox reaction
c) gaseous H2S from solid Zn and S via a two-step process (first a redox reaction between the starting materials, then reaction of the product with a strong acid)
22. Which is a composition reaction and which is not?
a) NaCl + AgNO3 $\longrightarrow$ AgCl + NaNO3
b) CaO + CO2 $\longrightarrow$ CaCO3
23. Which is a composition reaction and which is not?
a) 2 SO2 + O2 $\longrightarrow$ 2 SO3
b) 6 C + 3 H2 $\longrightarrow$ C6H6
24. Which is a decomposition reaction and which is not?
a) HCl + NaOH $\longrightarrow$ NaCl + H2O
b) CaCO3 $\longrightarrow$ CaO + CO2
25. Which is a decomposition reaction and which is not?
a) Na2O + CO2 $\longrightarrow$ Na2CO3
b) H2SO3 $\longrightarrow$ H2O + SO2
26. Which is a combustion reaction and which is not?
a) C6H12O6 + 6 O2 $\longrightarrow$ 6 CO2 + 6 H2O
b) 2 Fe2S3 + 9 O2 $\longrightarrow$ 2 Fe2O3 + 6 SO2
27. Which is a combustion reaction and which is not?
a) P4 + 5 O2 $\longrightarrow$ 2 P2O5
b) 2 Al2S3 + 9 O2 $\longrightarrow$ 2 Al2O3 + 6 SO2
28. Is it possible for a composition reaction to also be a combustion reaction? Give an example to support your case.
29. Complete and balance each combustion equation.
a) C4H9OH + O2 $\longrightarrow$ ?
b) CH3NO2 + O2 $\longrightarrow$ ?
1. Yes; both K and Br are changing oxidation numbers.
2. Ca has lost electrons, and O has gained electrons.
3. Li has been oxidized, and O has been reduced.
4. a) P: 0
b) S: +4; O: −2
c) S: +2; O: −2
d) Ca: 2+; N: +5; O: −2
5. a) C: +2; O: −2
b) C: +4; O: −2
c) Ni: +2; Cl: −1
d) Ni: +3; Cl: −1
6. a) C: 0; H: +1; O: −2
b) N: −3; H: +1
c) Rb: +1; S: +6; O: −2
d) Zn: +2; C: 0; H: +1; O: −2
7. N is being oxidized, and Cl is being reduced.
8. O is being oxidized, and Kr is being reduced.
9. K is being oxidized, and Mg is being reduced.
10. a) oxidation-reduction (addition); b) acid-base (neutralization); c) oxidation-reduction (combustion)
11. a) single replacement; b) decomposition; c) acid-base; d) precipitation
12. It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction.
13. a) H +1, P +5, O −2; b) Al +3, H +1, O −2; c) Se +4, O −2;
d) K +1, N +3, O −2; e) In +3, S −2; f) P +3, O −2
14. a) acid-base; b) oxidation-reduction: Na is oxidized, H+ is reduced;
c) oxidation-reduction: Mg is oxidized, Cl2 is reduced; d) acid-base;
e) oxidation-reduction: P3− is oxidized, O2 is reduced; f) acid-base
15. a) $2\text{HCl}(g) + \text{Ca(OH)}_2(s) \longrightarrow \text{CaCl}_2(s) + 2\text{H}_2 \text{O}(l)$;
b) $\text{Sr(OH)}_2(aq) + 2\text{HNO}_3(aq) \longrightarrow \text{Sr(NO}_3)_2(aq) + 2\text{H}_2 \text{O}(l)$;
16. a) $2\text{Al}(s) + 3\text{F}_2 \longrightarrow 2\text{AlF}_3(s)$;
b) $2\text{Al}(s) + 3\text{CuBr}_2(aq) \longrightarrow 3\text{Cu}(s) + 2\text{AlBr}_3(aq)$;
c) $\text{P}_4(s) + 5\text{O}_2(g) \longrightarrow \text{P}_4 \text{O}_{10}(s)$;
d) $\text{Ca}(s) + 2\text{H}_2 \text{O}(l) \longrightarrow \text{Ca(OH)}_2(aq) + \text{H}_2(g)$;
17. $\text{H}_2(g) + \text{F}_2(g) \longrightarrow 2\text{HF}(g)$
18. $2\text{NaBr}(aq) + \text{Cl}_2(g) \longrightarrow 2\text{NaCl}(aq) + \text{Br}_2(l)$
19. $2\text{LiOH}(aq) + \text{CO}_2(g) \longrightarrow \text{Li}_2 \text{CO}_3(aq) + \text{H}_2 \text{O}(l)$
20. a) $\text{Ca(OH)}_2(s) + \text{H}_2 \text{S}(g) \longrightarrow \text{CaS}(s) + 2\text{H}_2\text{O}(l);$
b) $\text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{S}(g) \longrightarrow \text{Na}_2 \text{S}(aq) + \text{CO}_2(g) + \text{H}_2 \text{O}(l)$
21. a) step 1: $\text{N}_2(g) + 3\text{H}_2(g) \longrightarrow 2\text{NH}_3(g)$,
step 2: $\text{NH}_3(g) + \text{HNO}_3(aq) \longrightarrow \text{NH}_4 \text{NO}_3(aq) \longrightarrow \text{NH}_4 \text{NO}_3\;\text{(s) (after drying)}$
b) $\text{H}_2(g) + \text{Br}_2(l) \longrightarrow 2\text{HBr}(g)$
c) $\text{Zn}(s) + \text{S}(s) \longrightarrow \text{ZnS}(s) \;\text{and} \;\text{ZnS}(s) + 2\text{HCl}(aq) \longrightarrow \text{ZnCl}_2(aq) + \text{H}_2 \text{S}(g)$
22. a) not composition b) composition
23. a) composition b) composition
24. a) not decomposition b) decomposition
25. a) not decomposition b) decomposition
26. a) combustion b) combustion
27. a) combustion b) combustion
28. Yes; 2 H2 + O2 $\longrightarrow$ 2 H2O (answers will vary)
29. a) C4H9OH + 6 O2 $\longrightarrow$ 4 CO2 + 5 H2O
b) 4 CH3NO2 + 3 O2 $\longrightarrow$ 4 CO2 + 6 H2O + 2 N2
## Glossary
combustion reaction: vigorous redox reaction producing significant amounts of energy in the form of heat and, sometimes, light
half-reaction: an equation that shows whether each reactant loses or gains electrons in a reaction.
oxidation: process in which an element’s oxidation number is increased by loss of electrons
oxidation-reduction reaction: (also, redox reaction) reaction involving a change in oxidation number for one or more reactant elements
oxidation number: (also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic
oxidizing agent: (also, oxidant) substance that brings about the oxidation of another substance, and in the process becomes reduced
reduction: process in which an element’s oxidation number is decreased by gain of electrons
reducing agent: (also, reductant) substance that brings about the reduction of another substance, and in the process becomes oxidized
single-displacement reaction: (also, replacement) redox reaction involving the oxidation of an elemental substance by an ionic species
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# Of the 50 students in a class, 35 students took a course in speech and
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Of the 50 students in a class, 35 students took a course in speech and [#permalink]
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02 Feb 2019, 01:17
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Of the 50 students in a class, 35 students took a course in speech and 26 students took a course in writing. At least how many students took both a course in speech and a course in writing?
(A) 9
(B) 11
(C) 15
(D) 24
(E) 26
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Re: Of the 50 students in a class, 35 students took a course in speech and [#permalink]
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02 Feb 2019, 01:40
rohan2345 wrote:
Of the 50 students in a class, 35 students took a course in speech and 26 students took a course in writing. At least how many students took both a course in speech and a course in writing?
(A) 9
(B) 11
(C) 15
(D) 24
(E) 26
making a grid would be much desirable however here I am taking another approach
There are 4 categories in terms of students choosing the courses.
Only Speech (a) + Only Writing (b) + both Course (c) + none course (d) = 50
Given,
a+c = 35 i.e. b+d = 15
b+c = 26 i.e. a+d = 24
To minimize c we need to maximize a and b and minimize d which may be zero
now, 35+26-c = 50
i..e. c = 11
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Re: Of the 50 students in a class, 35 students took a course in speech and [#permalink]
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02 Feb 2019, 01:41
35+26-50=11
Ans:b
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Re: Of the 50 students in a class, 35 students took a course in speech and [#permalink] 02 Feb 2019, 01:41
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Originally Posted by JimmyP
That company showed up and did some measurements and came up with 38000 BTU's/ hr heat loss. So a 3 or 3.5 ton system needed.
My electrical utility bill average for Dec to Feb was 1500kw /month this is 170500 btu's day and 12 hours / day ( darkness without solar gain) of usage would be 14208 btu's/hr. There is no way I would need 38000btu's hr.
Maybe he likes to size for -6 'F with an assumed HIGH infiltration rate.
.........................................
One needs to review electric usage by the hour if you wish to establish heat capacity required
based on power usage
and equipment heat output
or one must compensate for Heating Degree Days (HDD) versus Design Temperature.
PEAK LOAD might be 1.5 to 2.5 X the Average usage depending of the day, week or month.
For example, Design Temperature = 10'F and Average Temperature = 30'F ( HDD = 65 - 30 = 35)
dT Design = 70 -10 = 60
dT HDD = 70 - 35 = 35
60 / 35 = 1.7
dT HDD = 70 - 45 = 25 for a warmer month, week or day
60/ 25 = 2.4
If your average temperature for Dec - Febr was 30'F ( HDD = 35 * 90 days = 3150),
one would use 1.7 * portion of electric bill used for heating IF electric strip was used.
X C.O.P. if heat pump is used.
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Dan
For here the average daily low for that period 16F and the daily average is 23F. That would make dt HDD 65-16= 49. 49 x 90Days = 4410 HDD
A design of 10 would be Design dT of 60. So 60/49 =1.2 that would be 1500 x 1.2 average or 60 kw /day.
Last night I used 16 kw in 9 hours 10 pm to 7 am with an outside temp of 28 and an inside temperature of 63.
Jimmy
Last edited by JimmyP; 04-10-2013 at 06:38 AM. Reason: add information.
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Originally Posted by hvacvegas
we don't base a load on average.
you may keep your thermostat at a different temperature than the manual J suggested temp.
What is the design temperature based on? Average or what ?
4. Originally Posted by JimmyP
What is the design temperature based on? Average or what ?
It's based on an expectation of how cold/hot it normally gets. Not average temperatures.
Also, what he designed your indoor temp makes a big difference also.
Both these numbers have standards based on location, that can be deviated.
All this crazy math stuff is why we make the little bucks, and you should listen to the professionals that come out to your home.
I personally think you already have decided what you want in your house.
Your just looking for someone whos going to agree with you.
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Originally Posted by JimmyP
Dan
For here the average daily low for that period 16F and the daily average is 23F. That would make dt HDD 65-16= 49. 49 x 90Days = 4410 HDD
A design of 10 would be Design dT of 60. So 60/49 =1.2 that would be 1500 x 1.2 average or 60 kw /day.
Last night I used 16 kw in 9 hours 10 pm to 7 am with an outside temp of 28 and an inside temperature of 63.
Jimmy
So what's the equipment?
Strip ?
or heat pump?
Mfg :
Model:
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Originally Posted by JimmyP
What is the design temperature based on? Average or what ?
Temperature that is only exceeded for 175 hours [2% of 8,760] / year _ 30 year average.
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Originally Posted by JimmyP
Dan
For here the average daily low for that period 16F and the daily average is 23F. That would make dt HDD 65-16= 49. 49 x 90Days = 4410 HDD
A design of 10 would be Design dT of 60. So 60/49 =1.2 that would be 1500 x 1.2 average or 60 kw /day.
Last night I used 16 kw in 9 hours 10 pm to 7 am with an outside temp of 28 and an inside temperature of 63.
Jimmy
WHY is your inside temp dropping to 63?
Where does 1500 come from?
If you wish to Use average all meaningful data is destroyed.
Don't you have your electric usage for the last 50 months or so?
Each month can be correlated to HDD.
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Originally Posted by dan sw fl
WHY is your inside temp dropping to 63?
Where does 1500 come from?
If you wish to Use average all meaningful data is destroyed.
Don't you have your electric usage for the last 50 months or so?
Each month can be correlated to HDD.
I set it to 63 for night time from 11pm to 6 AM
Actual usge in KW from electrical bills for 1500/month for 3 month period used 4500 in 90 days.
Yes for 24 months. it uses electric baseboard heat.
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Originally Posted by hvacvegas
It's based on an expectation of how cold/hot it normally gets. Not average temperatures.
Also, what he designed your indoor temp makes a big difference also.
Both these numbers have standards based on location, that can be deviated.
All this crazy math stuff is why we make the little bucks, and you should listen to the professionals that come out to your home.
I personally think you already have decided what you want in your house.
Your just looking for someone whos going to agree with you.
I have done a load calculation so I should have an idea what is required. He didn't put in insulation values. He did put in running wall length and window sizes. I have a vaulted ceiling that is triangular on top. He put in total height for that wall of 21 ft and he should have used 15.5 ft He didn't know what manual J was, He said he puts the numbers in the program and it gives a recomendation . It may be a manual J program. I don't know. The guy designing the Duckwork did not know what Manual D was. He had a slide card giving him pressure loss for pipe size and length.
The equipment sizing was based on heating load required, from what I gather is wrong according to the experts.
Jimmy
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Dan
Some numbers from my electrical bills. Electric baseboard heat only. Total usage 6824kwh Usage for electrical only is 4784kwh ( 500 /month for other usage) HDD for that peroid 1834.
http://stjohns.weatherstats.ca/metrics/hdd.html.
Last month used 1295kwh for heat and 552 HDD's.
Jimmy
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Originally Posted by JimmyP
Actual usge in KW from electrical bills ...
Yes for 24 months. it uses electric baseboard heat.
With EACH of the 24 months data being Kept a Secret, NO ONE CAN DO A REAL LOAD CALC based on Actual Usage.
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Originally Posted by JimmyP
Dan
Some numbers from my electrical bills.
Electric baseboard heat only.
Total usage 6824kwh Usage for electrical only is 4784kwh ( 500 /month for other usage) HDD for that peroid 1834.
http://stjohns.weatherstats.ca/metrics/hdd.html.
Last month used 1295kwh for heat and 552 HDD's. Jimmy
Last months data of 1,284 KWH FOR HEAT with 552 HDD equates to 21,000 BTU/HR
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## Sorry no secret
Originally Posted by dan sw fl
With EACH of the 24 months data being Kept a Secret, NO ONE CAN DO A REAL LOAD CALC based on Actual Usage.
Here are the numbers:
The estimated values are from when they can't read the meter if the road or drivway is not yet plowed after a heavy snowfall and this happens a few times during the winter as we are located in a remote area and sometimes the road is not plowed right away.
Date (Y/M/D) Energy Reading Read / Estimate kwh Used Days Usage Per Day HDD
2013-03-21 46657 Read 1795 29 62 562
2013-02-20 44862 Read 2906 30 97 592
2013-01-21 41956 Estimated 2003 33 61 674
2012-12-19 39953 Estimated 1312 26 50 567
2012-11-23 38641 Read 1068 31 34 391
2012-10-23 37573 Read 438 32 14 269
2012-09-21 37135 Estimated 760 30 25 99
2012-08-22 36375 Read 633 29 22 30
2012-07-24 35742 Read 956 34 28 32
2012-06-20 34786 Read 733 27 27 246
2012-05-24 34053 Read 1214 31 39 313
2012-04-23 32839 Read 1314 32 41 381
2012-03-22 31525 Read 1633 29 56 1865 (Jan-March)
2012-02-22 29892 Read 2646 30 88
2012-01-23 27246 Read 2694 34 79
2011-12-20 24552 Read 1774 28 63 1317 (Oct -Dec)
2011-11-22 22778 Estimated 866 28 31
2011-10-25 21912 Read 532 34 16
2011-09-21 21380 Estimated 541 29 19
2011-08-23 20839 Read 598 30 20 321 (July - Sep)
2011-07-24 20241 Read 561 33 17
2011-06-21 19680 Read 829 32 26
2011-05-20 18851 Read 1198 30 40
2011-04-20 17653 Read 3242 29 112 1077 ( April-June)
2011-03-22 14411 Estimated 1336 29 46
2011-02-21 13075 Estimated 1568 32 49
2011-01-20 11507 Read 1519 31 49 1825 (Jan-Mar)
Page 4 of 5 First 12345 Last
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1. Gamma and factorial
Is this expression true?
$\displaystyle (n+s)! \Gamma (s+1)= s! \Gamma (n+s+1)$
i was looking at the proof of $\displaystyle J_{-n} (x)= (-1)^n J_n (x)$ and i deduced that expression must be true for the proof provided to be sound, the proof was not from a rigourous source so it may have cut corners & have botched up somewhere.
2. Originally Posted by phycdude
Is this expression true?
$\displaystyle (n+s)! \Gamma (s+1)= s! \Gamma (n+s+1)$.
For this to make sense requires that $\displaystyle s$ and $\displaystyle n$ be natural numbers. In which case the gammas may be replaced by factorials and both sides are equal to $\displaystyle s!(n+s)!$
CB
3. Originally Posted by phycdude
Is this expression true?
$\displaystyle (n+s)! \Gamma (s+1)= s! \Gamma (n+s+1)$
What makes sense if s is non-integer and n is positive integer is the following: $\displaystyle \Gamma(n+s+1)=(n+s)\Gamma(n+s)=(n+s)(n+s-1)\Gamma(n+s-1)$ $\displaystyle =\cdots=(n+s)\cdots (s+1)\Gamma(s+1)$. On the other hand, it is not uncommon to define $\displaystyle s!=\Gamma(s+1)$ for non-integer $\displaystyle s$, in which case your expression is trivially true.
4. Originally Posted by Laurent
What makes sense if s is non-integer and n is positive integer is the following: $\displaystyle \Gamma(n+s+1)=(n+s)\Gamma(n+s)=(n+s)(n+s-1)\Gamma(n+s-1)$ $\displaystyle =\cdots=(n+s)\cdots (s+1)\Gamma(s+1)$. On the other hand, it is not uncommon to define $\displaystyle s!=\Gamma(s+1)$ for non-integer $\displaystyle s$, in which case your expression is trivially true.
thanks , makes lots of sense now
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# Afamily of 2 adults and 3 children goes to a play. admission cost \$8 per adult and \$5 per child. what expression would show the total admission cost for the family? (smart people:
masonorourke · 08.07.2019 09:00
29.06.2019 22:10
just watch memes
step-by-step explanation:
it me < 3
28.06.2019 03:10
Hello!
27.06.2019 10:00
the population of other specie of birds will decline or other specie of birds will find some alternate source of food to survive.
explanation:
according to the competitive exclusion principle, which is also known as gause's law, two specie living in the same niche and feeding on the same resource cannot co-exist stably.
as we have well studied the theory of natural selection which states that only those organisms survive which have better tendency to compete for food and adapt to the environmental conditions.
coming towards the question, what wi ll happen when one species outcompetes the other for the small fish , the other specie will not get enough food and will start to die or decline in population. eventually either other specie needs to find another alternative food source for itself or migrate at some other location where there is less competition otherwise it will just not exist anymore at that niche.
hope it !
24.06.2019 01:40
Idk idk idk idk idk idk idk idk
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# An Introduction To Linear Algebra by Kenneth Kuttler
By Kenneth Kuttler
Similar algebra & trigonometry books
Mathematics for Machine Technology
The mathematical strategies that has to be understood and utilized within the desktop trades and production are awarded in transparent, actual global phrases within the new version of this most sensible promoting booklet. the knowledge of these ideas is under pressure in either the presentation and alertness in all issues, from easy to complicated.
Dynamical Systems: Stability, Symbolic Dynamics, and Chaos
This new text/reference treats dynamical structures from a mathematical point of view, centering on multidimensional structures of actual variables. history fabric is thoroughly reviewed because it is used during the booklet, and concepts are brought via examples. a number of workouts aid the reader comprehend offered theorems and grasp the options of the proofs and subject into consideration.
Lately, the invention of the relationships among formulation in Łukasiewicz good judgment and rational polyhedra, Chang MV-algebras and lattice-ordered abelian roups, MV-algebraic states and coherent de Finetti’s checks of constant occasions, has replaced the examine and perform of many-valued common sense.
Additional info for An Introduction To Linear Algebra
Example text
Also if x is a scalar, xA = (cij ) where cij = xaij . The number Aij will typically refer to the ij th entry of the matrix, A. The zero matrix, denoted by 0 will be the matrix consisting of all zeros. Do not be upset by the use of the subscripts, ij. The expression cij = aij + bij is just saying that you add corresponding entries to get the result of summing two matrices as discussed above. Note there are 2 × 3 zero matrices, 3 × 4 zero matrices, etc. In fact for every size there is a zero matrix.
When this has been done, B = A−1 . 20. 25 Let A = 1 −1 1 . Find A−1 . 1 1 −1 Form the augmented matrix, 1 1 1 0 −1 1 1 1 0 0 1 0 1 0 . −1 0 0 1 Now do row operations untill the n × n matrix on the left becomes the identity matrix. This yields after some computations, 1 1 1 0 0 0 2 2 0 1 0 1 −1 0 0 0 1 1 − 21 − 12 and so the inverse of A is the matrix on the right, 1 1 0 2 2 1 −1 0 . 1 1 − 2 − 21 Checking the answer is easy. Just multiply the matrices and see if 1 1 1 0 1 0 1 0 2 2 1 −1 1 1 −1 0 = 0 1 1 1 −1 0 0 1 − 12 − 12 Always check your answer because if mistake.
Amn Bnj Bnj . The second entry of this m × 1 matrix is m A2k Bkj . k=1 which is a m × 1 matrix or column vector which equals A11 A12 A1n A21 A22 A2n .. B1j + .. B2j + · · · + .. . . . 1. MATRICES 39 Similarly, the ith entry of this m × 1 matrix is m Ai1 B1j + Ai2 B2j + · · · + Ain Bnj = Aik Bkj . 3. This motivates the definition for matrix multiplication which identifies the ij th entries of the product. 7 Let A = (Aij ) be an m × n matrix and let B = (Bij ) be an n × p matrix.
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## Axiomatic system...
Hello guys!!I am from Greece so sorry for my bad english...
I have a problem at propositional logic...specifically at the proofing system...
the excercise which made me crazy is the following...
proove : $\vdash ((\phi \rightarrow{\psi})\rightarrow{\neg{\chi}})\rightar row{(\chi} \rightarrow{\neg{(\phi \rightarrow{\psi})}})
$
we can use the following axioms and theorems...
$\vdash ((\phi \rightarrow{(\psi \rightarrow{\sigma})})\rightarrow{((\phi \rightarrow{\psi})\rightarrow{(\phi \rightarrow \sigma})})$
$\vdash \phi \rightarrow{(\psi \rightarrow{\phi})}$
$\vdash(\neg{\phi} \rightarrow{\neg \psi} )\rightarrow{((\neg \phi \rightarrow{\psi})\rightarrow{\phi})}$
$\phi \rightarrow \psi \vdash \neg{\psi} \rightarrow \neg{\phi}$
$\vdash \phi \rightarrow \psi \equiv \phi \vdash \psi$
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Yes and no. Clearly, stock prices (or prices of any asset) are not observed continuously. This applies to both, the value (price) dimension and the time dimension.
This however does not mean that we can't model stock prices as a time and space continuous process. Frequently, time and space continuous approaches are more elegant and yield nicer results. Furthermore, continuous models are often limits of the discrete models. Anyway, the implementation of continuous models requires you to discretise the model (because you only have discrete data sets and computers can only work with discrete sets).
Note that we can, by the way, record prices with higher precision than cents or pence. Look at currencies which can be traded with a finer grid and we could (technically) use arbitrary fine partitions of the positive real axis. But yes, you can never observe a stock trading at \\pi\$. But models are simply easier and nicer if you allow for a continuous range.
Whether you see stock prices as a continuous process which we merely record discretely or whether you believe stock prices are discrete objects which we simply model continuously is almost a philosophical question.
Here some examples:
• Discrete time, discrete state space
Simple Random Walk
• Discrete time, continuous state space
Gaussian Random Walk
• Continuous time, discrete state space
Poisson Conting Process
• Continuous time, continuous state space
Brownian motion
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# Time step for wind turbine analysis
Register Blogs Members List Search Today's Posts Mark Forums Read
March 23, 2011, 17:30 Time step for wind turbine analysis #1 New Member Amool A Raina Join Date: May 2010 Posts: 14 Rep Power: 8 Hi all, I am using a unsteady density based implicit solver for simulating the flow around a 50m radius wind turbine blade. The turbine rotates at 14 rpm. I would like to know if there is a specific way by which I could determine the time step size for my simulation. I started off with a low time step size 10^-5 and based on the convergence behavior, I kept increasing the size of the time step to 0.01. how do i know of the most appropriate time step size? Any help would be greatly appreciated! Thanks, aar
March 24, 2011, 01:22
#2
Senior Member
Raashid Baig
Join Date: Mar 2010
Location: Bangalore, India
Posts: 136
Rep Power: 8
Quote:
Originally Posted by amoolraina Hi all, I am using a unsteady density based implicit solver for simulating the flow around a 50m radius wind turbine blade. The turbine rotates at 14 rpm. I would like to know if there is a specific way by which I could determine the time step size for my simulation. I started off with a low time step size 10^-5 and based on the convergence behavior, I kept increasing the size of the time step to 0.01. how do i know of the most appropriate time step size? Any help would be greatly appreciated! Thanks, aar
Hi Everyone,
I have this doubt too about the automatic timestep (how is it calculated)
and physical time step (how it is calculated) and what is the difference between the physical and auto time scale.
Raashid
March 24, 2011, 10:29
#3
Member
Join Date: Dec 2009
Posts: 41
Rep Power: 8
Quote:
Originally Posted by amoolraina Hi all, I am using a unsteady density based implicit solver for simulating the flow around a 50m radius wind turbine blade. The turbine rotates at 14 rpm. I would like to know if there is a specific way by which I could determine the time step size for my simulation. I started off with a low time step size 10^-5 and based on the convergence behavior, I kept increasing the size of the time step to 0.01. how do i know of the most appropriate time step size? Any help would be greatly appreciated! Thanks, aar
You have to give a look at some paper about experimental studies in wind gallery.From these papers you could get info about the Strouhal and from the Strouhal obtain the eddies relase frequency and form that obtain the period (inverse).Then you have to consider a time step about 100 times inferior to that period.
March 24, 2011, 14:28 #4 New Member Amool A Raina Join Date: May 2010 Posts: 14 Rep Power: 8 do u have any references for wind turbines?
March 25, 2011, 03:58 #5 Member Join Date: Dec 2009 Posts: 41 Rep Power: 8
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# The perimeter of an isosceles triangle is 20cm. if each equal side id double the base find the sides
2
by khushalhinduja
what is "id" in question
what is id
Its a typo, I think it means "is" .
2016-04-10T16:40:10+05:30
equal sides= 8cm
base= 4cm
perimeter = 8 + 8 + 4 = 20cm
2016-04-10T16:44:29+05:30
Let the isosceles triangle be ΔABC and AB = AC.
Let the lengths of each of two equal sides be 2x.
∴AB = AC = 2x
Also, each equal side is double than the base.
So, BC = x
Now, AB + BC + AC = 20 cm
∴2x + x + 2x = 20
∴5x = 20
∴x = 4cm
Thus,
AB = AC = 2x = 2*4 = 8 cm
BC = x = 4cm
Thus, the lengths of two equal sides is 8 cm and the length of base is 4 cm.
Please mark it as the brainliest
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# Is a forecast a prediction of the future often based on?
## Is a forecast a prediction of the future often based on?
Forecasting refers to the practice of predicting what will happen in the future by taking into consideration events in the past and present. Basically, it is a decision-making tool that helps businesses cope with the impact of the future’s uncertainty by examining historical data and trends.
## What is the general process of changing values in cells to see the effect on formulas in other cells?
What-If Analysis is the process of changing the values in cells to see how those changes will affect the outcome of formulas on the worksheet.
See also What is the opposite of the Scared?
## What is a function that is contained inside another function?
A nested function is a function that is completely contained within a parent function.
## Is a set of values that Excel saves and can substitute automatically in your worksheet?
A Scenario is a set of values that Excel saves and can substitute automatically on your worksheet. You can create and save different groups of values as scenarios and then switch between these scenarios to view the different results.
## What are the 4 common types of forecasting?
Technique Use
1. Straight line Constant growth rate
2. Moving average Repeated forecasts
3. Simple linear regression Compare one independent with one dependent variable
4. Multiple linear regression Compare more than one independent variable with one dependent variable
## What is used to predict the future?
Methods including water divining, astrology, numerology, fortune telling, interpretation of dreams, and many other forms of divination, have been used for millennia to attempt to predict the future.
## When you copy or move a formula to other cells the cell changes automatically?
The correct answer is Relative Reference. With relative cell referencing, when we copy a formula from one area of the worksheet to another, it records the position of the cell relative to the cell that originally contained the formula. This is the default mode of referencing in a spreadsheet.
## What is a cell with a formula that will be solved for specific results called?
A cell with a formula that will be solved for specific results is called a(n): Objective cell.
## How to enter a formula to display the value of a cell from another sheet in Excel?
Create a cell reference to another worksheet Click the cell in which you want to enter the formula. , type = (equal sign) and the formula you want to use. Click the tab for the worksheet to be referenced. Select the cell or range of cells to be referenced.
See also What is a flexibility contract?
## Which of the following methods will not enter data in a cell?
The correct answer is Pressing the Esc key.
## Which one is not a function of Excel?
ALT is not a function in Microsoft Excel spreadsheets. ALT is actually an abbreviation for the key Alternate, which is used in keyboard shortcuts to access certain commands and functions. It is not a standalone function that can be used in formulas or calculations in Excel.
## Which function can be used to get the total of series of numbers?
You can use a simple formula to sum numbers in a range (a group of cells), but the SUM function is easier to use when you’re working with more than a few numbers. For example =SUM(A2:A6) is less likely to have typing errors than =A2+A3+A4+A5+A6.
## Which forecast is prediction about future is based on the assumption that firm does not change the course of its action?
Passive forecasting – forecast prediction about future is based on the assumption that the firm does not change the course of its action.
## Is forecasting a prediction?
Forecasting involves estimating future events or trends based on historical and statistical data. Predictions make educated guesses or projections without relying on historical data or statistical methods. Forecasting predicts outcomes over a longer time frame, often over months, years, or even decades.
## What is forecast accuracy based on?
Forecast accuracy is the measure of how accurately a given forecast matches actual sales. Forecast bias describes how much the forecast is consistently over or under the actual sales. Common metrics used to evaluate forecast accuracy include Mean Absolute Percentage Error (MAPE) and Mean Absolute Deviation (MAD).
See also How do I track my DHL pickup?
## What are forecasts and projections based on?
Projections outline financial outcomes based on what might possibly happen (in theory), whereas forecasts describe financial outcomes based on what you expect actually will happen, given current conditions, plans, and intentions.
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Research website of Vyacheslav Gorchilin
2016-09-13
All articles
The magnetic component of energy of standing waves
More Amps noticed that when you make the conductor perpendicular to the magnetic field lines in the conductor begins to flow current. Moreover, the velocity vector must be perpendicular as the magnetic field lines and the direction of the current. Ie in fact we have three perpendicularly between the forces passing one another depending on the conditions of the experiment. What if one of the forces, for example magnetic, will form a standing wave, the energy cost of maintaining a minimum? In this work we will try to answer this question.
1. The square of the displacement amplitude of the standing wave
Let's look at two examples example 1 and example 2 In both cases we observe a standing wave, but first wave is visually, and the second moves. In more detail the process of conversion of transverse waves into longitudinal look here.
Since the animation is built with maximum approximation to real lines on the page examples you need to wait for some time until a stationary regime.
For some tasks it is necessary to determine the magnitude of such movement, and as the wave moves across the area, it is obvious that you will find. We agree that further square displacement of a wave we will call PSV, and looking ahead to say that under certain conditions this value — is the numerical expression of so-called free energy. But first we introduce the definition of a wave, and then find her PSV.
The equation of the standing wave
To represent momentum in a long line (DL) will use the standard expression for the falling of $$A_1$$ and reflected $$A_2$$ waves [1]. $A_1 = \frac{a}{2} \sin(\omega t - k x), \qquad A_2 = \frac{a}{2} \sin(\omega t + k x) \qquad (1.1)$ in Parallel we will consider a variant of cosine: $A_1 = \frac{a}{2} \cos(\omega t - k x), \qquad A_2 = \frac{a}{2} \cos(\omega t + k x) \qquad (1.2)$ where: $$a$$ is the wave amplitude, $$\omega$$ is angular frequency, $$t$$ is time, $$k$$ — dimensional multiplier, $$x$$ — distance. As in the further calculations we will use the relative units, the $$\omega = k = 2\pi$$. Also, in these calculations we will assume FOR the ideal, without attenuation. Then the resulting equation of the standing wave will be expressed as $A = A_1 + A_2 = \frac{a}{2} (\sin(\omega t - k x) + \sin(\omega t + k x)) = a\cos(k x)\sin(\omega t) \qquad (1.3)$ $A = A_1 + A_2 = \frac{a}{2} (\cos(\omega t - k x) + \cos(\omega t + k x)) = a\cos(k x)\cos(\omega t) \qquad (1.4)$ the Example described in equation (1.4) waves, which is half-wave DL, is here.
For our purposes we need the complex form of the pulse, therefore, for greater generalization we will present it in the form of the Fourier series. While we consider it a special case — the expansion of the sines and cosines. So, sine and cosine equation of the standing wave in a more General case would be: $A(x,t) = \sum_{i=1}^N (a_i \cos(ik x) \sin(i\omega t)) \qquad (1.5)$ $A(x,t) = \sum_{i=1}^N (a_i \cos(ik x) \cos(i\omega t)) \qquad (1.6)$ where $$N$$ is the number of harmonics $$a_i$$ is the amplitude of the ith harmonic. Now we can represent more complex pulses such as these.
PSV
Next, we need to find the area of the displacement wave which goes on the DL, while moving in time. We denote it by the symbol $$S$$. Its value should be proportional to the displacement of the entire area of the pulse in space and in time. If the momentum in space is fixed, and varies only in time, then the value of $$S$$ must be zero.
The following figure shows the same pulse, but for two different values of time (1) and (2), Pulse forms FOR a standing wave, so the length FOR $$\lambda$$ can take values $$\frac14, \frac12, \frac34, 1$$, etc.
We seek the change in the square wave space-time in this form: $\Delta S' = \sum_i {\Delta A_x \over \Delta x } \Delta A_t \qquad (1.7)$ where: $$\Delta A_x$$ is the change in the amplitude of the pulse at the point $$x$$, $$\Delta x$$ is the change in the spatial coordinates in the point $$\Delta A_t$$ is the change in the amplitude of the pulse at offset $$\Delta t$$, $$i$$ is the index that passes all the values of the spatial axis, from zero to $${\lambda \over \Delta x }$$. Then the opposite is true, the expression: $\Delta S' = \sum_i {\Delta A_t \over \Delta t } \Delta A_x$ Turning to the differentials and the integrals, we obtain the following expression: $d S' = \int_0^\lambda {\partial A_x \over \partial x } \partial A_t = \int_0^\lambda {\partial A_x \over \partial x} {\partial A_t \over \partial t} dt \qquad (1.8)$ where: $${\partial A_x \over \partial x} {\partial A_t \over \partial t}$$ the partial derivatives in the spatial and temporal coordinate. Then the total PSV will be true [2]: $S = \iint_0^\lambda {\partial A_x \over \partial x} {\partial A_t \over \partial t} dx\,dt \qquad (1.9)$
Nimirum PSV
To understand what the pulse shape gives a higher PSV, we need to normalize. For this we use a theorem Parseval [3] for Fourier series and find the quantity surveyor: $\Psi = \iint_0^1 A(x,t)^2 \, dx\,dt \qquad (1.10)$ so now our desired formula will look like this: $\bar S = \frac {S} {\Psi} = \frac {\iint_0^\lambda {\partial A_x \over \partial x} {\partial A_t \over \partial t} dx\,dt} {\iint_0^1 A(x,t)^2 \, dx\,dt} \qquad (1.11)$ Need to say that $$\bar S$$ convenient relative value for comparison. For example, for the classical Tesla transformer [4], where $$\lambda = \frac14$$ and only one of the first harmonic, $$\bar S = 1$$ (example). So all the other PSV we can compare it with this case. For $$\lambda = \frac12$$ and the first two equal-amplitude harmonics of $$\bar S = 7$$ here it is clearly seen the wave motion in DL (example). If we add harmonics, and this further strengthen the wave motion, it is possible to obtain even greater value $$\bar S$$. So for example this option would give $$\bar S = 64$$.
$$S$$ and $$\bar S$$ will always be taken modulo, as it is important for us the relative and not the absolute value of this quantity. In the formulas, the module is not shown, but implied.
The calculation according to the formula (1.11) easier to carry out the math editor, for example in MathCAD. For this editor and this formula program is here. For $$N=10$$ the editor on a computer with a processor 2.2 GHz, calculates the equation (1.11) for 12 seconds, for $$N=15$$ for 25 seconds. Below we show the simplified formula for a particular case, but which is rendered by the computer in milliseconds.
Half-wave DL
Using the formula (1.11) it is possible to find the relative area of offset for any signal, but since we will only be concerned with pulses in the form (1.5) or (1.6), we can narrow down the problem and to find a more simple equation for a single raspostranenija particular case.
When $$\lambda = \frac12$$ is set FOR half-wave mode, for which the formula (1.11) obtained a simpler form. For the wave (1.5) the simplified formula would be: $S \approx 8 \sum_{i=1}^{N} a_{i}a_{i-1}{ i^3 (i-1) \over 4i^2-1} \quad \qquad (1.12)$ For the wave (1.6) — like this: $S \approx 8 \sum_{i=1}^{N} a_{i}a_{i-1}{ i^2 (i-1)^2 \over (2i-1)^2} \quad \qquad (1.13)$ They work with an accuracy of 10% in the range $$N \le 12$$. From the formulas, in particular, it follows that $$S$$ is zero if the spectrum of the pulse contains only one harmonic (example, example) or in the spectrum contains only odd harmonics (example). As seen in these examples, wave in FOR not moving. Further summaries will be clear that the odd harmonics are responsible for the steepness of the decline-the rise of the pulse, and even for wave motion in LONG. Also give an example with even and odd harmonics, where it is clearly seen the wave motion.
Nimirum approximate formulas
From formula (1.10) we take the rule of normalization and output it for impulse (1.5) and (1.6): $\Psi = \frac14 \sum_{i=1}^N a_i^2 \qquad (1.14)$ Thus, the total approximate formula for the case of $$\lambda = \frac12$$ and wave (1.5) is this: $\bar S \approx 32 {\sum_{i=1}^{N} a_{i}a_{i-1}{ i^3 (i-1) \over 4i^2-1} \over \sum_{i=1}^N a_i^2} \qquad (1.15)$ and for the wave (1.6) is this: $\bar S \approx 32 {\sum_{i=1}^{N} a_{i}a_{i-1}{ i^2 (i-1)^2 \over (2i-1)^2} \over \sum_{i=1}^N a_i^2} \qquad (1.16)$ If $$a_{i}$$ are the same, then the formula is quite simplified: $\bar S \approx {32 \over N} \sum_{i=1}^{N} { i^3 (i-1) \over 4i^2-1 } \qquad (1.17)$ $\bar S \approx {32 \over N} \sum_{i=1}^{N} { i^2 (i-1)^2 \over (2i-1)^2} \qquad (1.18)$ From the formulas (1.17, 1.18) implies the interesting fact that the PSV for the sine and cosine waveforms are slightly different.
If the number is equal to the amplitude of the harmonics is large enough $$N \ge 12$$, then the PSV for sine and cosine waveform be approximately equal, and the formula is simplified further [5]: $\bar S \approx \frac43 (2N + 1)(N + 1) \qquad (1.19)$ we must not forget that the approximate formulas (1.12-1.19) derived for the particular case of $$\lambda = \frac12$$, which we will consider in future. For chipremoving pulse PSV calculation is given here.
A game of the mind, or energy pyramid?
This particular case brings us to an interesting observation. In the formula (1.19) was found relative (normalized) value of PSV, but the absolute value of the square displacement will be: $S \approx \frac{a^2}{3} N(2N + 1)(N + 1) = 2 a^2 \left({N^3 \over 3} + {N^2 \over 2} + {N \over 6}\right) \qquad (1.20)$ The expression before the parentheses is the square of the wave amplitude or the kind of unit energy, for example, a single block, and in brackets the formula is for building the whole pyramid [6]. In this interpretation of $$N$$ is the number of its tiers, and $$S$$ is the total energy of the pyramid.
But our task is to learn to use a PSV to obtain the free energy, which will be discussed in the second part of this article.
The materials used
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# Solving indefinite series
• Jan 25th 2010, 09:23 AM
PJani
Solving indefinite series
I dont know how to solve this series with limit of partial sum.
$\sum_{n=1}^{\infty}\frac{1}{(3n-2)(3n+1)}$
Thanx for any help
• Jan 25th 2010, 09:37 AM
Jhevon
Quote:
Originally Posted by PJani
I dont know how to solve this series with limit of partial sum.
$\sum_{n=1}^{\infty}\frac{1}{(3n-2)(3n+1)}$
Thanx for any help
Hint: this is a telescoping sum
Note that $\frac 1{(3n - 2)(3n + 1)} = \frac 13 \cdot \frac {3n + 1 - (3n - 2)}{(3n - 2)(3n + 1)} = \frac 13 \left( \frac 1{3n - 2} - \frac 1 {3n + 1} \right)$
(instead of using algebraic manipulation as i did above, you can use partial fractions)
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https://www.teacherspayteachers.com/Product/4th-Grade-Task-Cards-Perimeter-and-Area-4MD3-3880709
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These task cards help students practice perimeter and area. With these cards, students practice mathematical thinking with HIGHER ORDER THINKING SKILLS!
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Thread: Need some help with few trig equations...
1. Need some help with few trig equations...
Find all solutions of the given equation:
1.) 2 sin x + 1 = 0
2.) sin 2x - cos x = 0
3.) sin² x - sin x = 0
**Not looking for the entire answer, just how to start.
Prove that the given trig identity is true:
cos (a - b) = (cos a) (cos b) + (sin a) (sin b)
**I know that it's true because it states the formula in the book but I just don't know how to prove it, lol.
2. 1) $\displaystyle \sin x = - \frac{1} {2}.$
2) $\displaystyle \sin 2x = 2\sin x\cos x$ & factorise.
3) Factorise and proceed as 2).
---
About last formula, it requires a geometric reasoning to prove it, or you can also use complex numbers.
3. Could you explain a little further on...all of them?
For #1, -1/2 is what I got on my own, but I don't know how to put the answer in terms of (ex. from another problem: x= pi/3 + 2n(pi) for any int n).
As for #2 and #3, can you give me the first step to factorizing? I'm clueless on where to start.
And #4...I still have no clue.
4. Originally Posted by toop
Could you explain a little further on...all of them?
For #1, -1/2 is what I got on my own, but I don't know how to put the answer in terms of (ex. from another problem: x= pi/3 + 2n(pi) for any int n).
As for #2 and #3, can you give me the first step to factorizing? I'm clueless on where to start.
And #4...I still have no clue.
Krizalid seems to be away for the moment so I'll jump in...
1) $\displaystyle sin x = - \frac{1}{2}$
$\displaystyle x = sin^{-1} ( \frac{1}{2} )$ (The value of x must be in quadrants 3 and 4.
2) $\displaystyle 2 \ sin x \ cos x - cos x= 0$
$\displaystyle cos x( 2 sin x - 1) = 0$
$\displaystyle Cos x = 0$
OR
$\displaystyle 2 Sin x - 1 = 0$
3) $\displaystyle sin^{2} x - sin x = 0$
$\displaystyle sin x(sin x - 1) = 0$
$\displaystyle Sin x = 0$
OR
$\displaystyle sin x - 1 = 0$
4) This is a long process of proving. Look it up on the net, or it is most probably in your textbook.
5. Originally Posted by toop
cos (a - b) = (cos a) (cos b) + (sin a) (sin b)
First of all, we're gonna find a formula for $\displaystyle \cos (\alpha + \beta ).$
Consider the following sketch:
Construction:
• Consider a rectangle $\displaystyle NOPR.$
• Extend $\displaystyle \overline{ON}.$ We get point $\displaystyle M.$
• Join $\displaystyle M$ & $\displaystyle P.$
• Extend $\displaystyle \overline{NR}.$ We get point $\displaystyle Q.$
• Draw $\displaystyle \overline {QP} \perp\overline {MP} ,$ now join $\displaystyle M$ & $\displaystyle Q.$
• Let $\displaystyle \measuredangle \,OMP = \alpha$ & $\displaystyle \measuredangle \,QMP = \beta .$
Start by showing that $\displaystyle \measuredangle \,OMP = \measuredangle \,PQR = \alpha .$
We have
$\displaystyle \cos (\alpha + \beta ) = \frac{{\overline {MN} }} {{\overline {QM} }} = \frac{{\overline {OM} - \overline {ON} }} {{\overline {QM} }} = \frac{{\overline {OM} }} {{\overline {QM} }} \cdot \frac{{\overline {MP} }} {{\overline {MP} }} - \frac{{\overline {RP} }} {{\overline {QM} }} \cdot \frac{{\overline {QP} }} {{\overline {QP} }}.$
Rearrange
$\displaystyle \cos (\alpha + \beta ) = \frac{{\overline {MP} }} {{\overline {QM} }} \cdot \frac{{\overline {OM} }} {{\overline {MP} }} - \frac{{\overline {QP} }} {{\overline {QM} }} \cdot \frac{{\overline {RP} }} {{\overline {QP} }} = \cos \beta \cos \alpha - \sin \beta \sin \alpha \quad\blacksquare$
Finally, write $\displaystyle \cos (\alpha - \beta ) = \cos \left[ {\alpha + ( - \beta )} \right]$ and you'll get the desired formula.
---
Now, a proof with complex numbers.
$\displaystyle \cos (\alpha - \beta ) = \text{Re} \Big[ {e^{i(\alpha - \beta )} } \Big] = \text{Re} \Big[ {(\cos \alpha + i\sin \alpha )(\cos \beta - i\sin \beta )} \Big].$
Finally
$\displaystyle \cos (\alpha - \beta ) = \text{Re}\, (\cos \alpha \cos \beta - i\cos \alpha \sin \beta + i\sin \alpha \cos \beta + \sin \alpha \sin \beta ).$
And we happily get $\displaystyle \cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\quad\blacksquare$
As desired.
6. Originally Posted by toop
Prove that the given trig identity is true:
cos (a - b) = (cos a) (cos b) + (sin a) (sin b)
The best way is to use $\displaystyle \cos z = = \frac{1}{2}(e^{iz} + e^{-iz})$ --- but that is unfair.
Here is another geometric proof that I remember from 10th grade.
1)Construct the circle $\displaystyle x^2+y^2 = 1$.
2)Construct those red lines touching at $\displaystyle P_1,P_2$.
3)$\displaystyle a$ is the bigger angle from the x-axis.
4)$\displaystyle b$ is the smaller angle from the [tex]x-axis.
5)Coordinates of $\displaystyle P_1$ is $\displaystyle (\cos b,\sin b)$.
6)Coordinates of $\displaystyle P_2$ is $\displaystyle (\cos a,\sin a)$
7)The distance from $\displaystyle P_1 \to P_2$ is $\displaystyle \sqrt{(\cos b-\cos a)^2 + (\sin b - \sin a)^2}$.
8)The angle between the two red lines is $\displaystyle a-b$.
9)The length of $\displaystyle P_1P_2$ using law of cosines is $\displaystyle \sqrt{1^2+1^2 - 2 \cos (a-b)}$.
10)Equality of #7 and #9 leads to ...
$\displaystyle \sqrt{(\cos b-\cos a)^2 + (\sin b - \sin a)^2} = \sqrt{2 - 2\cos (a-b)}$
$\displaystyle (\cos b - \cos a)^2 + (\sin b - \sin a)^2 = 2 - 2 \cos (a-b)$
$\displaystyle \cos^2 b - 2\cos b\cos a + \cos^2 a + \sin^2 b - 2\sin a \sin b + \sin^2 a = 2 - 2 \cos (a-b)$
$\displaystyle \cos (a-b) = \cos a \cos b + \sin a \sin b$
7. Wow, thanks for the help on #4.
For #1 I got:
$\displaystyle \frac{{7\pi }} {6} + 2n\pi$
Does that look correct to you guys?
For #2 I got:
$\displaystyle x = \frac{\pi } {2} + 2n\pi$
For #3:
$\displaystyle x = 2n\pi ;\frac{\pi } {2} + 2n\pi$
Thanks a lot guys!
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## CCO '20 P2 - Exercise Deadlines
View as PDF
Points: 15 (partial)
Time limit: 1.0s
Memory limit: 1G
Author:
Problem types
Bob has programming exercises that he needs to complete before their deadlines. Exercise only takes one time unit to complete, but has a deadline time units from now.
Bob will solve the exercises in an order described by a sequence , such that is the first exercise he solves, is the second exercise he solves, and so on. Bob's original plan is described by the sequence . With one swap operation, Bob can exchange two adjacent numbers in this sequence. What is the minimum number of swaps required to change this sequence into one that completes all exercises on time?
#### Input Specification
The first line consists of a single integer . The next line contains space-separated integers .
For 17 of the 25 available marks, .
#### Output Specification
Output a single integer, the minimum number of swaps required for Bob to solve all exercises on time, or -1 if this is impossible.
#### Sample Input 1
4
4 4 3 2
#### Sample Output 1
3
#### Explanation for Output for Sample Input 1
One valid sequence is , which can be obtained from by three swaps.
#### Sample Input 2
3
1 1 3
#### Sample Output 2
-1
#### Explanation of Output for Sample Input 2
There are two exercises that are due at time , but only one exercise can be solved by this time.
• commented on July 23, 2020, 12:16 p.m.
can an editorial be opened for this problem?
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# reddit is a platform for internet communities
[–] -1 points0 points (0 children)
There is a strategy consisting on betting double the last bet in the same color until you win. This strategy works because the decision is taken before actually seeing the results. With every spin of the wheel the conditional probabilities are unchanged, this is, the spins are independent from each other; however, before the spins start the probability of getting n reds in a row is .486n, in other words, the probability of getting the same color decreases exponentially with the number of spins. The fallacy is thinking the probabilities are changing with every spin.
[–] 0 points1 point (0 children)
I'm new to R so I don't know strong data manipulation there. STATA has the wide and long commands to help you there. I managed to do this in excel though, this is the file If you need more tables tell me is a very simple method
[–] 0 points1 point (0 children)
Introducing the revolutionary 1080p
[–] 0 points1 point (0 children)
Difference of means with t-test or anova.. though you need controls or you may detect mere tendency. Take a look at this thread: http://www.reddit.com/r/statistics/comments/23g2sx/spss_question_about_variables_in_linear/
[–][S] 0 points1 point (0 children)
He is online
[–][S] 0 points1 point (0 children)
Thank you
[–][S] 0 points1 point (0 children)
I am using NA time zone
[–][S] 0 points1 point (0 children)
We can update probabilities conditioned to not being released the day of the spotlight so there is a 85% prob. he is released today (Monday).
[–] 1 point2 points (0 children)
[–] 1 point2 points (0 children)
Which is supposed to be a simple way of doing multivariate analysis? What does this even mean??
[–] 0 points1 point (0 children)
Go to the statistics bureau of your country and download macroeconomic series or whatever series you like. For example for USA: http://www.bea.gov/
[–] 0 points1 point (0 children)
I would say, read the book carefully I don't think you understand what they are trying to solve. Anyway, an ARDL model can be generalized to have as many covariates as you want. With 4 lags however, I wouldn't even bother to run a regression, your estimates will have huge variance even when your variables come from stationary distributions. That's why your dickey fuller test rejects the hypothesis of not unit root.
[–] 0 points1 point (0 children)
It depends on the specifics of your database. It would be interesting to see which of these factors explains more variability in the probability of finding a match or which factor "pays" more in the market lets say with a probit model.
[–] 0 points1 point (0 children)
Actually you can use a model to correct for selection bias (or at least reduce the bias) because you know the estimates in the dating sample and you know the population value. See for example http://en.wikipedia.org/wiki/Heckman_correction
[–] 1 point2 points (0 children)
Run the other types of crimes and see which ones are the most affected by the treatment.
[–] 1 point2 points (0 children)
Be sure to understand well the diff and diff approach. Your control group helps to keep a lot of effects controlled.
[–] 1 point2 points (0 children)
They are in the sex crime regression I sent you. Constant is the average sex crime of treated schools in the first 12 years. You have to think of each coefficient as the effect of each variable holding every other variable fixed.
[–] 1 point2 points (0 children)
One remark. The interpretation I gave is causal but only you know exactly what other factors are playing here, the fact that everything that affects crime is being held fixed is a strong assumption. You need to support your hypothesis with some theory about why you think calendar choice is affecting crime.
[–] 1 point2 points (0 children)
I sent you the documents to your email. Basically I changed the data base to run the Diff in Diff in a regression in order to not only see the estimator of the treatment effect but also the estimated variance.
You have to run each model and see the results. I ran the sex crime case in SPSS and this is what I found: The model specification is Crime(it) = B0 + B1 * Period2 + B2 * Control + B3*Treatment + e
Period 2 is a dummy for the second period, where some schools changed the treatment. Control is a dummy for control schools. Treatment is a dummy for cases where the treatment was applied, this is when non control schools changed the calendar.
The estimates of each one are: Constant = 142 Period 2 = -22 Control = 123 Treatment = 77.8
Only period2 is not statistically significant. We are interested in B3 which is the ATE. It is shown that the change in the calendar (your treatment) increases the average sex related crime rate per 100,000 residents in almost 78 crimes with p-value = .002
If you have questions about the interpretation or methodology feel free to ask.
[–] 0 points1 point (0 children)
Var (u2) = Var(x1)/16 + Var(x2)/9 + Var(x3)/16
`````` =7/16+13/9+20/16 which is greater than 1
``````
Assuming all covariances are zero
[–] 0 points1 point (0 children)
yeah, cov(S&P, Stock) / Var(S&P)
[–] 0 points1 point (0 children)
You are talking precisely about the modern portfolio theory and the CAPM model. http://en.wikipedia.org/wiki/Capital_asset_pricing_model
[–] 0 points1 point (0 children)
I think your sample is too small to make good statistical inference, remember that the estimated variance of the estimators is inversely proportional to the sample size. Plus, it doesn't help too much that you try to assign 1 of 6 possible values to the outcome. A t-test can work since you are interested in each treatment separated from the rest, an F test which tells you if all of them are different from the control as a group is not particularly interesting.
With the correct amount of data, you would be able to run an ordered logit model if you assume the conditional probability of the outcome follows a logistic function or an ordered probit if you assume it follows a normal.
http://en.wikipedia.org/wiki/Ordered_logit
[–] 0 points1 point (0 children)
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# Homework Help: DA/dt of expanding circle
1. Oct 24, 2009
### synergix
1. The problem statement, all variables and given/known data
a) if A is the area of circle with radius r and the circle expands as time passes, find dV/dt in terms of dr/dt
b)Suppose oil spills from a ruptured ranker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 1 m/s, how fast is the area of the spill increasing when the radius is 30 m?
2. Relevant equations
A=(pi)r2
3. The attempt at a solution
dA/dt= ((pi)r2)'=(r2+2(pi)r)
dA/dt= ((pi)2(30) + 302)
is this correct? it seems like an awfully big number what have I done wrong?
2. Oct 24, 2009
### synergix
oh wait I am treating pi like a variable not a constant. durh
3. Oct 24, 2009
### synergix
so dA/dt = 2*30*pi
4. Oct 24, 2009
### Dick
You've got it dA/dt=pi*2*r*dr/dt.
5. Oct 24, 2009
### synergix
OK so I would need to have multiplied 2*pi*r*1m/s but since its one it doesn't matter but that is what is happening correct?
I am multiplying the derivative of the expression by the derivative of r?
6. Oct 24, 2009
### Dick
It matters if you are keeping track of units. That's where the m/s came from.
7. Oct 24, 2009
### Dick
You are using the chain rule. d/dt(f(r))=d/dr(f(r))*dr/dt. You knew that, right?
8. Oct 24, 2009
### synergix
I know that now. I missed a good couple classes (long story short) because I had no other choice. Now I am trying to catchup. I guess I better do some reading..thanks
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Free Version
Moderate
# Time to Charge Capacitor
SATPHY-R141QC
The circuit below shows a resistor, $R$, an initially uncharged capacitor, $C$, and a battery with a voltage $V$.
The following tables give the resistance, capacitance, and voltage values for the resistor, capacitor, and battery respectively.
Which of the following tables describes the set of values for the circuit in which the capacitor will take the longest time to charge?
A
Capacitance Voltage Resistance
100 mF 3.0 V 25, 000 $\Omega$
B
Capacitance Voltage Resistance
100 mF 1.0 V 50, 000 $\Omega$
C
Capacitance Voltage Resistance
10 mF 2.0 V 250, 000 $\Omega$
D
Capacitance Voltage Resistance
100 mF 10.0 V 10, 000 $\Omega$
E
Capacitance Voltage Resistance
100 mF 4.0 V 25, 000 $\Omega$
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UCSB Science Line
How many moles would 6.022•1022 molecules of water represent? Question Date: 2020-03-13 Answer 1: One mole of a substance is defined as having exactly 6.02214076*1023 (*see note below) constitutive particles of that substance. Those particles can be atoms, ions, molecules, etc. To determine the number of moles for some given number of molecules, one can use a simple but powerful technique known as dimensional analysis. Dimensional analysis is essentially a method of converting units, involving multiplying given units by conversion factors and "canceling" like units that appear in both the numerator and denominator to reach the desired measure. Since this method works for any units, it is applicable for a wide range of problems. [Assuming the starting and ending units are of a consistent "type". For example, distances can only be measured in length units; trying to convert to 3 miles to pounds is nonsensical and impossible.] For the present case, the given units are molecules of water and the desired units aremoles of water. The conversion factor is the definition of the mole, i.e., 1 mol H2O / 6.022...*1023 molecules H2O. Thus, 6.022*1022 molecules H2O * (1 mol H2O / 6.022...*1023 molecules H2O) ≈ 0.1 mol H2O. *The mole used to be defined as the amount of a substance containing as many constitutive particles as the number of atoms of carbon-12 in 12 grams of carbon 12. The definition was changed in 2018/2019 to be 1 mole = 6.02214076*1023 (exact) particles. The US standards institute (NIST) has a fascinating (lengthy) article on this redefinition. Because the change is so recent, many articles/sites/textbooks will likely have the earlier definition still. The responses to this question on ScienceLine and this chemistry page contain more information on using the mole/Avogadro's Number. Answer 2:Since 1 mole of particles is 6.022 x 1023, then the answer is: Number of moles = (6.022 x 1022) / (6 x 1023) = 1 x 10-1 = 0.1 moles of water molecules. Answer 3: A mole of anything is 6.022•1023 particles. So, divide the number of molecules by Avogadro's number, and that's the number of moles. Answer 4: That's 0.1 mole of water, because 6 x 1023 is the number of molecules or atoms in a mole. Answer 5:Since one mole of anything (molecules, cars, bananas) is defined as 6.022*1023 of that thing, we can use this number to calculate how many moles of that thing we have. In this case, we can just divide 6.022*102 water molecules by 6.022*1023 molecules per mole. You should get an answer that is something like 10-x moles of water molecules.Click Here to return to the search form.
Copyright © 2020 The Regents of the University of California, All Rights Reserved. UCSB Terms of Use
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## hurewicz theorem proof
So the above diagram commutes for any f, i.e., his natural.
Then the hurewicz homomorphism h W n.X/ ! Note. Need an account? Let X G= T n<!
The proof is long and intricate, but worth studying even if . This is proved in, for example, Whitehead (1978) by induction, proving in turn the absolute version and the Homotopy Addition Lemma. Repeat until you reach the rst nonzero homology group. Then the Hurewicz theorem [HW41, p. 91, Theorem VI 7] states that Ask Question Asked 6 months ago. Lemma. A special thanks goes out to Kristine Bauer and Marie-Andr ee B. Langlois, for proof-reading my thesis. The following topologicalresults are used in the proof of Theorem 2.1.For any Hurewicz bration f : X Y , if Y is path-connected, then bydenition the bers are homotopy equivalent to each other. Want to take part in these discussions? . Let X be any .n1/-connected based space. We can remove CHfrom Theorem 2 by strengthening the hypothesis. Proof of Blakers-Massey, Eilenberg-Mac Lane spaces. We give an elementary proof of the rational Hurewicz theorem and compute the rational cohomology groups of Eilenberg-MacLane spaces and the rational homotopy groups of spheres. e-mail:klaus@mfo.de and MATTHIASKRECK Mathematisches Institut, Universitat Heidelberg, Im Neuenheimer Feld188, Week 10. Log In Sign . a Cech-hurewicz isomorphism theorem . Theorem 5.25 (Hurewicz). For n E J, the pointed n-movability of (X, x) is defined in a similar way to one given by Borsuk [1], that only the category of pointed compacta is considered. Then h induces an isomorphism H1(X) = 1(X,x0)ab. Minducing an isomorphism on homology groups (this step needs M simply-connected, not just H 1M= 0). To do so, we de ne a map K: C i(X) ! There exists proofs of the Hurewicz theorem in which one constructs a concrete isomorphism between the spaces, but in this thesis we avoid the . 2 The Fundamental Group and First Homology Group If a is irrational, then there are infinitely many rational numbers p/q satisfying p 1 a-- < (1) q Aq2' Let us start with the notation and some preliminary definitions. An application is a Connectivity Theorem: Localize away from (p1)! Let A be a constant satisfying 0 < A < /35. Modified 6 months ago.
The most common version of its proof consists of showing that the composition of the homotopy group functors with the infinite symmetric product defines a reduced homology theory. gluel : Q (y:Y ) sm(x 0;y . For any CW-complex X, He n(X) =He n(Xn+1). Then the Hurewicz homomorphism k(X) He k(X) is an isomorphism if k = m and is an epimorphism if k = m+1. or reset password. the switch in dimension theory from subsets of Cartesian to general separable metric spaces is due to Hurewicz. It seems that it is induced by the map . Several questions about the Hurewicz theorem have been asked on this site earlier, but this is not a duplicate . Let i n+1. The theorem Dold-Thom theorem. The proofs in [17] and [16] have the advantage of reducing the shape theoretic result to the classical one.
Proof: This follows directly from Proposition 7.10 and the description of the Hurewicz homomorphism at the prime 2 provided by Corollary 9.22 and Remark 9.23 (see [21, Th eor` eme 5.15] for more details). For a compact metrizable space X we denote its topological dimension (Lebesgue covering dimension) by dimX. The goal of the remaining lectures is to sketch the proof of Smale's theorem. Theorem. In algebraic topology, the Dold-Thom theorem states that the homotopy groups of the infinite symmetric product of a connected CW complex are the same as its reduced homology groups. If n>2, this is 0, and then we apply the Hurewicz theorem at level 3. The proofs are very easy to follow; virtually every step and its justification is spelled out, even elementary and obvious ones. Log In with Facebook Log In with Google. Let f : X Y be a continuous map between compact metrizable spaces. Theorem 3.6 If f: X!Y is an open, perfect mapping from a space X onto a Hurewicz space Y, then Xis mildly Hurewicz. 3 (without of course any referenceto the Hurewicz Theorem)andthe relative homotopy-addition theorem. Theorem 0.1 (Hurewicz theorem). Let (U n) n2N be a sequence of clopen covers of X. The proof uses the same initial simplications as in Scheepers's proof of Hurewicz'sTheorem [9, Theorem 13]. Hurewicz theorem (Theorem 17) in the form that the fibre of a map is like the co-fibre. Our results (and proofs) apply to all topological spaces Xin which each open set is a union of countably many clopen sets, and the spaces considered are assumed to have this property.1 Fix a topological space X. (Hurewicz) Let X be a path-connected topological space. Close Log In. The Hurewicz theorem is one of the funda-mental results in the classical topological dimension theory. Since Sn is simply connected, Theorem 5.28 gives that 2(Sn) = H2(Sn). In order to de ne the isomorphism appearing in TheoremS, we must give a bilinear map n(X) ! Let i n+1. A quick proof of the rational Hurewicz theorem and a computation of the rational homotopy groups of spheres - Volume 136 Issue 3. The first type of theorem considers local fibrations where local is in terms of closed covers of the . Proof. that CT C U . The following lemma is obvious from [7]. Short description: Gives a homomorphism from homotopy groups to homology groups In mathematics, the Hurewicz theorem is a basic result of algebraic topology, connecting homotopy theory with homology theory via a map known as the Hurewicz homomorphism. 3. The proof of our main theorem makes much use of composition properties of this ltration and its interaction with Topological Andre-Quillen homology. Proof. If X is a Lusin set, then M(X) is undetermined. The key idea that permits this refinement is a comparison of unstable and stable computations of $${\mathbb {A}}^1$$-homotopy sheaves.The proof of Theorem 4 relies on the beautiful computation of the first stable $${\mathbb {A}}^1$$-homotopy sheaf of the motivic sphere spectrum by Rndigs-Spitzweck-stvr [].After the discussion of Sect. In this paper we prove any regular almost Lipschitz submersion constructed by Yamaguchi on a collapsed Alexandrov space with curvature bounded below is a Hurewicz fibration (Theorem A), and any two such fibrations on one collapsed Alexandrov space are homotopy equivalent to each other (Theorem B). 1. ], or [Ha2, Theorem 5.8]), at least for the classes of nitely generated abelian groups and nite abelian groups (the proof is a generalization of the one we gave for the classical version of the theorem). First, by the CW Approximation Theorem we may assume that X is a CW complex with a single 0-cell, based attaching maps, and no q-cells for 1 q<n. HUREWICZ, WITOLD(b, Lodx, Russian Poland, 29 June 1904; d. Uxmal, Mexico, 6 September 1956)topology. Note. 4. C -* and -* are m-, -connected. I can't see where the map came from. 5.5. A quick proof of the rational Hurewicz theorem and a computation of the rational homotopy groups of spheres By STEPHANKLAUS Mathematisches Forschungsinstitut, Oberwolfach, Lorenzenhof, 77709Oberwolfach-Walke, Germany. The classical Hurewicz theorem has been transposed into shape theory by several authors and at various levels of generality [9], [22], [5], [17], [16]. Corollary 3. First, recall that given two pointed types Xand Y, the smash product X^Y is de ned to be the higher inductive type with constructors: sm : X Y !X^Y. Viewed 126 times 2 $\begingroup$ For . The proof explicitly uses the Hawaiian earring and the authors mention in Remark 2.7 that . If X is a presheaf 1-topos, this follows from the classical Hurewicz theorem. Thethird proof uses homotopicalhomologyonly andis there fore best adapted to a purelyhomotopical approachto algebraic topo logy. Lemma. Proof; Step 1: Pick an orthonormal basis e 1;e 2;:::;e n for Rn, and consider the map v!e i vfrom Rn to Rn. The details of this procedure will follow. 2, Suslin's conjecture can be viewed as a . rational Hurewicz theorem 18.906 Problem Set 8 Due Wednesday, April 11 in class In this problem set, we'll cover a lot of steps in constructing (most) Pontriagin classes. Show that $\ker h =[\pi_1(X,x_0), \pi_1(X,x_0)]$ in the proof of Hurewicz Theorem. The proof is based on the main ideas of V.~Kapovitch, A.~Petrunin, and W.~Tuschmann, and the following results: (1) We prove that any regular almost Lipschitz submersion constructed by Yamaguchi on a collapsed Alexandrov space with curvature bounded below is a Hurewicz fibration. Proof of the main theorem 8 6. It is also very useful that there exists an isomorphism : n SP(X) H n (X) which is compatible with the Hurewicz homomorphism h: n (X) H n (X), meaning that one has a commutative diagram Hurwitz's Theorem Richard Koch February 19, 2015 Theorem 1 (Hurwitz; 1898) Suppose there is a bilinear product on Rnwith the property that jjv wjj= jjvjjjjwjj Then n= 1;2;4;or 8. Theorem (Hurewicz). For any CW-complex X, He n(X) =He n(Xn+1). Hn+1 (Xn+1;Xn) is an iso-morphism, and together with a lemma from Homological . An Easy Proof of Hurwitz's Theorem Manuel Benito and J. Javier Escribano We provide an easy proof, based on the Brocot series, of a well-known theorem of Hurwitz. arrangement and twisted Hurewicz maps Masahiko Yoshinaga Kobe University. A classical theorem of Hurewicz characterizes spaces with the Hurewicz covering prop- . Stable homotopy groups, Hurewicz theorem, homology Whitehead theorem. The most general result (Theorem 5) is the one in [16]. S2yields the isomorphism 2(S2) 1(S1). auxr : X^Y. Waner uses rather sophisticated geometric techniques, based on his notion of a F-CW complex for a representation V , to prove his results. Postnikov . The idea is that a generator of C k(X) is a cube whose boundary may map anywhere in X, and we have to modify it, via a chain homotopy, to obtain a cube whose boundary maps to x 0. This approach has the advantages of (1) portability - in the sense that one could equally . THE HUREWICZ THEOREM IN HOMOTOPY TYPE THEORY 2 Now we explain the homology groups that appear on the right-hand-side of the Hurewicz isomorphism. C.T.C.Wall has shown me a proof of this Corollary for n>2, using covering spaces and the relative Hurewicz theorem. To do so, we de ne and study a more a) For all positive integers k, there is a natural Hurewicz map h k: . Hurewicz's theorem. HUREWICZ FIBRATIONS BY JAMES E. ARNOLD, JR. . native proof of the celebrated Hurewicz theorem in the case that the topo-logical space is a CW-complex. Your contributions have been a huge help! . Onlythe generalproperties of homologytheory as obtained fromhomotopytheory ( [2], [10]) are required; they amount essen tially to the Eilenberg-Steenrod axioms. Proposition 0.6.
Let X2X be an n-connective object for some n 1. 93 Theorem 2. We rst show that h is a group homomorphism. A 0-dimensional space . The proof is given by a standard technique (cf. (This may be seen by considering to be obtained from A by adding cells of dimension ^ m +1, Finally, we state the full form of the Hurewicz theorem (without proof). or. Lemma 2 (Hurewicz theorem). Sign Up with Apple. For a connected CW complex X one has n SP(X) H n (X), where H n denotes reduced homology and SP stands for the infite symmetric product.. In [6] we de ned a space to be indestructibly productively Lindel of if it remained productively Lindel of in any countably closed forcing extension.
Proof. for all but nitely many n, i.e. Corollary 2. Such a proof is not available for n= 2, essentially because of the non-abelian nature of crossed modules. While in higher degrees the Hurewicz homomorphism is in general far from being an isomorphism, the thrust of the Hurewicz theorem is to show that high connectivity is a sufficient condition to ensure that it is. If (X,x0) is a connected movable pointed metric compactum {satisfying the condition mentioned above) then, if there is > 1 such that {{,0) = 0 for l^i<n, the Hurewicz homomorphism h : {,0) -> Hn{X\ ) is an isomorphism and HfX ; ) = 0 for 1 ^ i < n. Proof of Theorem 1. The so-called Sperner proof of the invariance of dimension was independently and simultaneously found and published by . The proof of the following theorem will be given next time. Formally speaking, we rst show that if X0 X1 Xn1 Xn = X is the CW decomposition of X, then the Hurewicz homomorphism n+1 (Xn+1;Xn) ! Let A ;B be families of covers of . Proof. Recall that SO(n) is the group of n n orthogonal matrices with determinant +1, and SO(n) is connected. Twisted Hurewicz maps. First, it preserves identity elements since if we consider the constant loop at x0as a 1-cycle, we can express it as the boundary of the constant 2-simplex at x0. 1 Randell's result Thm. The wholly original proof of the main theorem by Hurewicz rests upon the utilization of the space E;+m of mappings of X ~En +m , as defined by Frechet and the proof that the mappings of the desired type are dense in E:+m. The suspension isomorphism is also natural, so h= h . If f: X !Y is a This approach has the advantages of (1) portability - in the sense that one could equally . Theorem 9. We discuss some applications throughout the paper. The unoriented and complex cobordism rings are also computed in a similar fashion. Corollary 1.3 (Relative Hurewicz Theorem as in [BH6] ) Let (V;W) be an (n 1)-conn-ected pair of connected spaces. - Nonresonance theorem for local system homology groups. In mathematics, the Hurewicz theorem is a basic result of algebraic topology, connecting homotopy theory with homology theory via a map known as the Hurewicz homomorphism.The theorem is named after Witold Hurewicz, and generalizes earlier results of Henri Poincar. Homology of classifying spaces 4 2. Week 8. Homotopy pullbacks, Homotopy Excision, Freudenthal suspension theorem. G n, where G The Relative Hurewicz Theorem states that if both and are connected and the pair is -connected then for and is obtained from by factoring out the action of .
football trends and facts
#### hurewicz theorem proof
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# super(RationalMap) -- get the rational map whose target is a projective space
## Synopsis
• Function: super
• Usage:
super phi
rationalMap(phi,Dominant=>null)
rationalMap phi
• Inputs:
• phi, , whose target is a subvariety $Y\subset\mathbb{P}^n$
• Outputs:
• , the composition of phi with the inclusion of $Y$ into $\mathbb{P}^n$
## Description
So that, for instance, if phi is a dominant map, then the code rationalMap(super phi,Dominant=>true) yields a map isomorphic to phi.
i1 : phi = specialQuadraticTransformation 7; o1 : RationalMap (quadratic birational map from PP^8 to 8-dimensional subvariety of PP^10) i2 : phi' = super phi; o2 : RationalMap (quadratic rational map from PP^8 to PP^10) i3 : describe phi o3 = rational map defined by forms of degree 2 source variety: PP^8 target variety: complete intersection of type (2,2) in PP^10 dominance: true birationality: true projective degrees: {1, 2, 4, 8, 16, 22, 20, 12, 4} number of minimal representatives: 1 dimension base locus: 3 degree base locus: 10 coefficient ring: QQ i4 : describe phi' o4 = rational map defined by forms of degree 2 source variety: PP^8 target variety: PP^10 image: complete intersection of type (2,2) in PP^10 dominance: false birationality: false degree of map: 1 projective degrees: {1, 2, 4, 8, 16, 22, 20, 12, 4} number of minimal representatives: 1 dimension base locus: 3 degree base locus: 10 coefficient ring: QQ i5 : describe rationalMap(phi',Dominant=>true) o5 = rational map defined by forms of degree 2 source variety: PP^8 target variety: complete intersection of type (2,2) in PP^10 dominance: true birationality: true projective degrees: {1, 2, 4, 8, 16, 22, 20, 12, 4} number of minimal representatives: 1 dimension base locus: 3 degree base locus: 10 coefficient ring: QQ
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# Time and Distance- Aptitude Questions and Answers - in Hindi
How many minutes does a car take to cover a distance of 200 m, if it runs at a speed of 60 kmph? 60 किमी प्रति घंटे की चाल से चल रही एक कार 200 मीटर की दूरी तय करने में कितना समय लेती है? A) 1/3 minutes B) 1/5 minutes C) 1/6 minutes D) 1/8 minutes Correct Answer : 1/5 minutes Explanation :Time taken to cover 60 km = 1 hour that is 60000 m = 60 minutes (1 km=1000 m,1 hour=60 minutes) Time taken to cover 1 m = 60/60000 minutes = 1/1000 minutes Time taken to cover 200 m = 200 * 1/1000 minutes = 1/5 minutes. Post/View Answer Post comment Cancel Thanks for your comment.! Write a comment(Click here) ...
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Tags: cryptography
Rating:
## Tyrannosaurus Rex
The flag was encoded with the following proecess:
base64encode -> enc function -> hexlify
So we can decode flag with the following process:
unhexlify -> dec function -> base64decode
The dec function can be found by studying the enc function.
enc function:
Lets say we have an array e = [a,b,c,d] , a new array *z* is formed
by xoring two consecutive elements. The last element is xored with
the first. So we get z = [a^b, b^c, c^d, d^a]. Then z is hexlified.
dec function:
Observation 1: The beginning of base64 of a string is
always the same. It is just the end that differs.
Observation 2: a ^ a = 0. Therefore, a ^ a ^ b = b.
With these observations in mind, we can now construct the dec function.
We can base64 encode "flag" and grab the ascii value of the first 6bit value. A simple python code to do that is base64.b64encode(b'flag')[0]. We then unhexlify the encoded flag and perform observation 2 on it.
We then finally base64 decode the result.
#### Code Snippet
def dec():
cur = 90 #base64.b64encode(b'flag')[0]'
u = binascii.unhexlify(c)
s=""
for i in u:
v = cur ^ i #Perform a ^ a ^ b
s += chr(v)
cur = v
s = s[len(s)-1]+s[0:len(s)-1]
print(base64.b64decode(s))
[Original writeup](https://github.com/AmosAidoo/ctf-writeups/tree/master/H%40ctivityCon_CTF)
Original writeup (https://ctftime.org/team/128582).
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# Gödel's Second Incompleteness Theorem Explained in Words of One Syllable
First of all, when I say "proved", what I will mean is "proved with the aid of the whole of math". Now then: two plus two is four, as you well know. And, of course, it can be proved that two plus two is four (proved, that is, with the aid of the whole of math, as I said, though in the case of two plus two, of course we do not need the whole of math to prove that it is four). And, as may not be quite so clear, it can be proved that it can be proved that two plus two is four, as well. And it can be proved that it can be proved that it can be proved that two plus two is four. And so on. In fact, if a claim can be proved, then it can be proved that the claim can be proved. And that too can be proved. Now, two plus two is not five. And it can be proved that two plus two is not five. And it can be proved that it can be proved that two plus two is not five, and so on. Thus: it can be proved that two plus two is not five. Can it be proved as well that two plus two is five? It would be a real blow to math, to say the least, if it could. If it could be proved that two plus two is five, then it could be proved that five is not five, and then there would be no claim that could not be proved, and math would be a lot of bunk. So, we now want to ask, can it be proved that it can't be proved that two plus two is five? Here's the shock: no, it can't. Or, to hedge a bit: if it can be proved that it can't be proved that two plus two is five, then it can be proved as well that two plus two is five, and math is a lot of bunk. In fact, if math is not a lot of bunk, then no claim of the form "claim X can't be proved" can be proved. So, if math is not a lot of bunk, then, though it can't be proved that two plus two is five, it can't be proved that it can't be proved that two plus two is five. By the way, in case you'd like to know: yes, it can be proved that if it can be proved that it can't be proved that two plus two is five, then it can be proved that two plus two is five.
George Boolos, Mind, Vol. 103, January 1994, pp. 1 - 3.
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https://www.quesba.com/questions/use-present-value-of-1-table-determine-present-value-of-1-received-one-year-1841031
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# Use the Present Value of? \$1 table to determine the present value of? \$1 received one year from...
Use the Present Value of? \$1 table to determine the present value of? \$1 received one year from now. Assumea12?% interest rate. Use the same table to find the present value of? \$1 received two years from now. Continue this process for a total of five years. Round to three decimal places.
Requirement 1. NOTE: Table is in attachment.
Requirement 1. What is the total present value of the cash flows received over the? five-year period?
Calculate the total present value of? \$1 received each year. ?(Round to three decimal? places, X.XXX.)
Present Value
One Year From Now =
Two Years From Now =
Three Year From Now =
Four Years From Now =
Five Years From Now =
Total Present Value =
Requirement 2. Could you characterize this stream of cash flows as an? annuity? Why or why? not?
The stream of cash flows ?
is
is not
an annuity because it is a stream of ?
equal
unequal
different
equal
of time intervals.
Oct 22 2022| 07:43 PM |
Earl Stokes Verified Expert
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# Number Zero Riddle
Author: Anonymous
6 years ago
Riddle: What number can you subtract half from to obtain a result of zero?
Answer: The number 8. It's made up of two zeros, one on top of the other.
VOTE
Source: https://www.riddles.com/5379
COMMENT
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# Symmetric Encryption - PowerPoint PPT Presentation
IT 352 : Lecture 2- part2. Symmetric Encryption. Najwa AlGhamdi , MSc – 2012 /1433. Outline. What is cryptography Symmetric-Key Encryption Symmetric-Key Encryption Algorithm. Cryptography. “Confidentiality service”
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IT 352 : Lecture 2- part2
Symmetric Encryption
NajwaAlGhamdi , MSc – 2012 /1433
Outline
• What is cryptography
• Symmetric-Key Encryption
• Symmetric-Key Encryption Algorithm
Cryptography
• “Confidentiality service”
• Cryptography: comes from Greek means “secret writing”. It is the art and science of secret writing.
• Crypto-analysis: breaking of the code.
• Cryptology (Crypto): studying both aspects..
• Encryption: the original goal of cryptography تشفير
• Decryption فك التشفير
Cryptography
• can characterize cryptographic system by:
• type of encryption operations used
• substitution
• transposition
• product
• number of keys used
• single-key or private
• two-key or public
• way in which plaintext is processed
• block
• stream
Cryptography
• Three kinds of cryptographic algorithms
• Symmetric (Secret Key) Cryptography (DES, RCx, AES)
• Asymmetric (Public Key) Cryptography (RSA, Diffie-Hellman, DSS)
• Message Digests (MD4, MD5, SHA-1)
plaintext
plaintext
Ciphertext
encryption
decryption
Symmetric Encryption
• sender and recipient share a common key.
• all classical encryption algorithms are private-key.
• was only type prior to invention of public-key in 1970’s
• and by far most widely used.
Symmetric Encryption- Requirement
• two requirements for secure use of symmetric encryption:
• a strong encryption algorithm
• a secret key known only to sender / receiver
• mathematically have:
Y = E(K, X)
X = D(K, Y)
• assume encryption algorithm is known
• implies a secure channel to distribute key
Cryptanalysis
• objective to recover key not just message
• general approaches:
• cryptanalytic attack
• brute-force attack
• if either succeed all key use compromised
• Cryptanalysis Attack
• ciphertext only
only know algorithm & ciphertext, is statistical, know or can identify plaintext
• known plaintext
know/suspect plaintext & ciphertext
• chosen plaintext
select plaintext and obtain ciphertext
• chosen ciphertext
select ciphertext and obtain plaintext
• chosen text
select plaintext or ciphertext to en/decrypt
Cryptanalysis – Brute-Force Attack
• always possible to simply try every key
• most basic attack, proportional to key size
• assume either know / recognise plaintext
Feistel Cipher Structure
• Horst Feistel devised the feistel cipher
• based on concept of invertible product cipher
• partitions input block into two halves
• process through multiple rounds which
• perform a substitution on left data half
• based on round function of right half & subkey
• then have permutation swapping halves
• implements Shannon’s S-P net concept
Feistel Cipher Design Elements
• block size
• key size
• number of rounds
• subkey generation algorithm
• round function
• fast software en/decryption
• ease of analysis
1.Data Encryption Standard (DES)
• most widely used block cipher in world
• adopted in 1977 by NBS (now NIST)
• as FIPS PUB 46
• encrypts 64-bit data using 56-bit key
• has widespread use
• has been considerable controversy over its security
1.Data Encryption Standard (DES)
• although DES standard is public
• was considerable controversy over design
• in choice of 56-bit key (vs Lucifer 128-bit)
• and because design criteria were classified
• subsequent events and public analysis show in fact design was appropriate
• use of DES has flourished
• especially in financial applications
• still standardised for legacy application use
DES Design Controversy
• clear a replacement for DES was needed
• theoretical attacks that can break it
• demonstrated exhaustive key search attacks
• AES is a new cipher alternative
• prior to this alternative was to use multiple encryption with DES implementations
• Triple-DES is the chosen form
Multiple Encryption & DES
• clear a replacement for DES was needed
• theoretical attacks that can break it
• demonstrated exhaustive key search attacks
• AES is a new cipher alternative
• prior to this alternative was to use multiple encryption with DES implementations
• Triple-DES is the chosen form
Double-DES
• could use 2 DES encrypts on each block
• C = EK2(EK1(P))
• issue of reduction to single stage
• and have “meet-in-the-middle” attack
• works whenever use a cipher twice
• since X = EK1(P) = DK2(C)
• attack by encrypting P with all keys and store
• then decrypt C with keys and match X value
• can show takes O(256) steps
Triple-DES with Three-Keys
• although are no practical attacks on two-key Triple-DES have some indications
• can use Triple-DES with Three-Keys to avoid even these
• C = EK3(DK2(EK1(P)))
• has been adopted by some Internet applications, eg PGP, S/MIME
2. The Advanced Encryption Standard (AES)
• clear a replacement for DES was needed
• have theoretical attacks that can break it
• have demonstrated exhaustive key search attacks
• can use Triple-DES – but slow, has small blocks
• US NIST issued call for ciphers in 1997
• 15 candidates accepted in Jun 98
• 5 were shortlisted in Aug-99
• Rijndael was selected as the AES in Oct-2000
• issued as FIPS PUB 197 standard in Nov-2001
2.The AES Cipher - Rijndael
• designed by Rijmen-Daemen in Belgium
• has 128/192/256 bit keys, 128 bit data
• an iterative rather than feistel cipher
• processes data as block of 4 columns of 4 bytes
• operates on entire data block in every round
• designed to be:
• resistant against known attacks
• speed and code compactness on many CPUs
• design simplicity
AES Structure
• data block of 4 columns of 4 bytes is state
• key is expanded to array of words
• has 9/11/13 rounds in which state undergoes:
• byte substitution (1 S-box used on every byte)
• shift rows (permute bytes between groups/columns)
• mix columns (subs using matrix multiply of groups)
• add round key (XOR state with key material)
• view as alternating XOR key & scramble data bytes
• initial XOR key material & incomplete last round
• with fast XOR & table lookup implementation
Random Numbers
• many uses of random numbers in cryptography
• nonces in authentication protocols to prevent replay
• session keys
• public key generation
• keystream for a one-time pad
• in all cases its critical that these values be
• statistically random, uniform distribution, independent
• unpredictability of future values from previous values
• true random numbers provide this
• care needed with generated random numbers
Pseudorandom Number Generators (PRNGs)
• often use deterministic algorithmic techniques to create “random numbers”
• although are not truly random
• can pass many tests of “randomness”
• known as “pseudorandom numbers”
• created by “Pseudorandom Number Generators (PRNGs)”
Stream Cipher Properties
• some design considerations are:
• long period with no repetitions
• statistically random
• depends on large enough key
• large linear complexity
• properly designed, can be as secure as a block cipher with same size key
• but usually simpler & faster
3. RC4
• a proprietary cipher owned by RSA DSI
• another Ron Rivest design, simple but effective
• variable key size, byte-oriented stream cipher
• widely used (web SSL/TLS, wireless WEP/WPA)
• key forms random permutation of all 8-bit values
• uses that permutation to scramble input info processed a byte at a time
RC4 Key Schedule
• starts with an array S of numbers: 0..255
• use key to well and truly shuffle
• S forms internal state of the cipher
for i = 0 to 255 do
S[i] = i
T[i] = K[i mod keylen])
j = 0a
for i = 0 to 255 do
j = (j + S[i] + T[i]) (mod 256)
swap (S[i], S[j])
RC4 Encryption
• encryption continues shuffling array values
• sum of shuffled pair selects "stream key" value from permutation
• XOR S[t] with next byte of message to en/decrypt
i = j = 0
for each message byte Mi
i = (i + 1) (mod 256)
j = (j + S[i]) (mod 256)
swap(S[i], S[j])
t = (S[i] + S[j]) (mod 256)
Ci = Mi XOR S[t]
Resources
• Network Security Essential , chapter 2 .
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### Home > ACC7 > Chapter cc310 > Lesson cc310.1.1 > Problem10-12
10-12.
1. Find the cube root of the following numbers. Use your calculator as needed and round any decimal answers to the nearest hundredth. Identify each answer as a rational or irrational number. Homework Help ✎
1. 1728
2. 54
3. 0.125
If this was the volume of a cube, how long would its side be?
Consider the same situation from part (a).
Irrational numbers are cube roots of non-perfect cubes.
3.78, irrational
See parts (a) and (b).
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# How does an atom's electric field overcome an electron's inertia?
An electron has mass, and therefore has inertia. How does an atom's electric field perpetually overcome an electron's inertia, necessary to hold it in its shell? Does this require continual work to be performed by an atom's electric field, and represent a conservation of energy violation? Do electron elemental and molecular bonds represent an energy violation in the same respect? Is the atom's field not performing work while it influences the motions of electrons?
I understand General Relativity proposes curved space to reconcile energy conservation laws in respect of overcoming an orbiting body's inertia by stating that gravity is not a force, but just objects traveling in straight lines through curved space. However there is no such proposition for the atom?
• Without invoking neither GR, nor QM, there is no violation of conservation laws, when a massive object (a star) keeps a planet in orbit, as the forces acting on the planet are roughly perpendicular to its velocity. Similarly, if you think of an atom as a classical object, an electron is roughly in a circular orbit with the Coulomb attraction acting perpendicular to its motion, thus performing no work. Of course, the electron is experiencing acceleration, which would suggest, that it ought to radiate EM waves, however quantization of allowed energy levels prevents that. Commented Feb 16, 2016 at 8:42
All spinning and orbiting objects overcome the inertia of their outer elements without continually doing work. Wheels, balls, planets, etc. They have to otherwise they would stop.
For example, if you have a ball and a piece of string and spin it around your head, the tension in the string (and the ball) provides a force that accelerates the ball inwards and overcomes (or rather redirects) its forward inertia.
Work is required to spin the ball up and create a tension in the string, but once moving it is not necessary to add additional energy to the system. The system is in balance. The ball is continually trying to move away from the string (maintaining tension) and the tension in the string is continually pulling it back.
Although the ball-and-string model is not a great analogy of electrons in orbit around a nucleus, the principle is the same in terms of how energy is conserved.
The "tension" between the electron and the nucleus is provided by the electromagnetic force (which indeed is also behind the tension in the piece of string).
The planetary model for an atom is a classical concept, and has many flaws, first noted by Rutherford. Bohr offered the first quantum fix, which was much improved by de Broglie.
Jumping ahead we have electron shells. The electrons exist as quantum probability waves, spread out throughout their shell. Calculation tells you the likelihood of a pointlike interaction with a specific electron, though wave interference effects must be included for accurate analysis.
• Thank you Peter. I am generally aware of modern proposed structure of atoms, electron probability waves/shells etc. The maths works, and so it must be that it represents something of reality. However I am not aware of the conversations which must have taken place, concerning implications for electron inertia. As electrons have mass, and how can a theory implicated with mass be disassociated with inertia? If this is so, then I would be very interested in the explanation, if it exists? Commented Feb 16, 2016 at 12:17
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# Solving Geometry Problems Involving System of Equations In Two Variables
Solving Geometry Problems Involving System of Equations In Two Variables
Algebra has also found wide applications in Geometry. Among its common problems are problems involving perimeters and angles. In this hub I included five sample problems involving perimeter and angles.
Sample Problem Number One :
A rectangular garden has a perimeter of 100 meters. The length is four times its width. Find the dimensions of the garden.
Solution :
Let X = width
Let Y = length
P = 2L + 2w
Equation one : 2X + 2Y = 100
Equation Two : Y = 4X
Substitute equation two in equation one :
2X + 2 (4X) = 100
2X + 8X = 100
10X = 100
(1/10) 10X = 100 (1/10)
X = 10 ==è width
4X = 4 (10) = 40 ==è lenth
Sample Problem Two :
The perimeter of a rectangular picture frame is 80 centimeter. Two times the width is equal to the length increased by five. Find its dimensions.
Solution :
Let X = width
Let Y = length
P = 2L + 2W
Equation one : 2X + 2Y = 80
Equation two: 2X = Y + 5
We use elimination method by subtraction to solve for the unknown variables.
2X + 2Y = 80
(-)2X (+) -Y = (-) 5
3Y = 75
(1/3) 3Y = 75 (1/3)
Y = 25 ==è length
To solve for width substitute in equation two
2X = 25 + 5
2X = 30
X = 15 ====è width
Sample Problem Three:
The perimeter of a rectangular flower garden is 120 meters. The length is ten meters greater than its width. Find its dimensions.
Solution :
Let X = width
Let Y = length
P = 2L +2W
Equation one : 2X + 2Y = 120
Equation two : Y = X + 10
Substitute equation two in equation one :
2X + 2 (X + 10 ) = 120
2X + 2X + 20 = 120
4X = 120 – 20
4X = 100
(1/4) 4X = 100 (1/4)
X = 25 ==è width
Y = 25 + 10 = 35 ===è lenth
Sample Problem Number Four :
The sum of the two nonright angles in a right triangle is of course 90 degrees. If twice the first is 40 degrees more than three times the second. Find the measurement of the angles of the right triangle.
Solution :
Let X = first angle
Let Y = second angle
Equation one : X + Y = 90
Equation two : 2X = 3Y + 40
From equation one we derive Y = 90 – X the substitute this in equation two.
2X = 3(90-X) + 40
2X = 270 – 3X + 40
2X + 3X = 270 + 40
5X = 310
(1/5) 5X = 310 (1/5)
X = 62 degrees ===è first angle
Y = 90 – 62 = 28 ===è second angle
Sample Problem Number Five :
Two angles are supplementary. The bigger angle is thrice as large as the smaller angle. Find the measurement of the angles.
Solution :
Let X = smaller angle
Let Y = bigger angle
The two angles are supplementary it means that their sum is equal to 180 degrees.
Equation one : X + Y = 180
Equation two : Y = 3X
Substitute equation two in equation one :
X +3X = 180
4X = 180
(1/4) 4X = 180 (1/4)
X = 45 ==èsmaller angle
Y = 3(45) = 135 ====. Bigger angle
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DeBorrah K. Ogans 5 years ago
Cristina, Nice Geometry lesson... You are so smart intellectually and gifted spiritually! What a double blessing that you also Love the Lord!
Thank You for sharing, In HIS Love, Grace, Joy, Peace & Blessings!
cristina327 5 years ago from Manila Author
Hi DeBorrah I am glad to hear from you again. Thank you for appreciating this hub. Your visit and comments are much appreciated. You have been really a great encouragement to me. Blessings to you and your family. Best regards.
John Peterson 4 years ago
Trolololololololol
catherine 2 years ago
it is very great i really like iit
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# Course 1, Week 4, Assignment 1, Exercise 9 - where is dZ?
I’m confused about Exercise 9 - L_model_backward. I think I might understand how to get current_cache (though I’m not sure because none of my code is working at all), but the next line I don’t understand:
`dA_prev_temp, dW_temp, db_temp = ...`
I’m pretty sure I’m supposed to use the linear_backward function here, but I need dZ as input for that, which I don’t have. Am I supposed to nest a sigmoid_backward function inside this? That’s the only place I remember calculating dZ before, since in Exercise 7 where we used linear_backward, we were given dZ. Or am I way off base here? In which case, can someone please point me in the right direction?
Hi @parrotox for the backprop step of the network you need to differentiate back the through the non linear step and the linear step of each layer. In the noteboook there is a function, `linear_activation_backward` (that uses `linear_backward`) and `Implement the backward propagation for the LINEAR->ACTIVATION layer.` Note that the `linear_backward` function on excercise 7` Implement the linear portion of backward propagation for a single layer (layer l)`
1 Like
YES! Thank you, I got it now
Hey, i’m having trouble with current_cache, i don´t know how to called, someone can help me please, i got this error.
TypeError Traceback (most recent call last)
in
1 t_AL, t_Y_assess, t_caches = L_model_backward_test_case()
----> 2 grads = L_model_backward(t_AL, t_Y_assess, t_caches)
3
4 print("dA0 = " + str(grads[‘dA0’]))
5 print("dA1 = " + str(grads[‘dA1’]))
in L_model_backward(AL, Y, caches)
41 current_cache = caches
42
—> 43 dA_prev_temp, dW_temp, db_temp =linear_activation_backward(dAL, current_cache, activation = “sigmoid”)
44 grads[“dA” + str(L-1)] = dA_prev_temp
45 grads[“dW” + str(L)] = dW_temp
in linear_activation_backward(dA, cache, activation)
33 # dA_prev, dW, db = …
34 # YOUR CODE STARTS HERE
—> 35 dZ = sigmoid_backward(dA, activation_cache)
36 dA_prev, dW, db = linear_backward(dZ, linear_cache)
37
~/work/release/W4A1/dnn_utils.py in sigmoid_backward(dA, cache)
74 Z = cache
75
—> 76 s = 1/(1+np.exp(-Z))
77 dZ = dA * s * (1-s)
78
TypeError: bad operand type for unary -: ‘tuple’
Hi @paolaruedad in the `linear_activation_backward` where you are getting the error you are using the sigmoid function. Think about in which layers you are using the sigmoid so you can decide what’s the right content for `current_cache` at that step.
If you have a look at the `for` loop below where ReLU is used it may help you out too.
Though, now I have a problem with what I believe is the cache of the loop. Here’s my error message.
``````IndexError Traceback (most recent call last)
``````
in
1 t_AL, t_Y_assess, t_caches = L_model_backward_test_case()
----> 2 grads = L_model_backward(t_AL, t_Y_assess, t_caches)
3
4 print("dA0 = " + str(grads[‘dA0’]))
5 print("dA1 = " + str(grads[‘dA1’]))
in L_model_backward(AL, Y, caches)
66 # YOUR CODE STARTS HERE
67 current_cache = caches[l]
—> 68 dA_prev_temp, dW_temp, db_temp = linear_activation_backward(grads[“dA” + str(l + 1)], current_cache, “relu”)
69 grads[“dA” + str(l)] = dA_prev_temp + str(l)
70 grads[“dW” + str(l + 1)] = dW_temp + str(l + 1)
in linear_activation_backward(dA, cache, activation)
21 # dA_prev, dW, db = …
22 # YOUR CODE STARTS HERE
—> 23 dZ = relu_backward(dA, activation_cache)
24 dA_prev, dW, db = linear_backward(dZ, linear_cache)
25
~/work/release/W4A1/dnn_utils.py in relu_backward(dA, cache)
54
55 # When z <= 0, you should set dz to 0 as well.
—> 56 dZ[Z <= 0] = 0
57
58 assert (dZ.shape == Z.shape)
IndexError: too many indices for array
I have tried every combination reasonable to try of cache and cache[y] but can’t find an answer. Now, I am out of idea to solve this problem. Can you help me?
Hi, as the error is happening when calling the `relu_backward` function I would suggest that you temporarily edit that function (which is defined in the file `dnn_utils.py`) to print `dZ`, `dZ.shape` and `Z.shape` in that context.
And I say temporarily because you are not expected to modify this file, it is correct as it is, so you only may want to do it in order to debug this issue.
i’m sorry, but i still not getting it, i just changed a lot of parameters, it is backward propagation, i don’t know what to put in the cache o how to called it, please some idea. it has been a day already, i’m just stuck.
The above is assigning to `current_cache` the full list of `caches` but how many layers do you have with sigmoid activation? Just one, so you have to assign to `current_cache` the right index from `caches`. Does it help?
1 Like
Ok, yes i understand, but i think the problem it is how to called that in python, i’m doing this nameofdelist[+ str(L)], for the sigmoide because it is the 3 layer, and for the relu are two, nameofdelist[+ str(l)], i assumed it is l, because the for loop is doing the iteration, but i still getting the error, how do i supposed to take just the respective layers in python?
I think an example could help.
Let’s assume there are 3 layers, the `caches` list would have the indexes: 0, 1, 2. So for the last layer, you would should assign `caches[2]`. Obviously you don’t need to hardcode the numbers you have to use `L`. Note that the indexes of the list are integers.
@albertovilla I have no idea how to do this
Why should I change the source code?
@lachainone your error is taking place in this line:
This is using a function defined in the file `W4A1/dnn_utils.py`, in particular the error you are getting is in the statement `dZ[Z <= 0] = 0` but you don’t know how `dZ` relates to your inputs to the function because your parameters are `dA` and `activation_cache`.
In order to understand why the error is happening I would suggest you open the `dnn_utils.py` file and add some print statements so you can backtrace where is the error and then correct the problem in your code.
You can open that file by clicking on the Jupyter logo, you will see a folder `release` and from there you can navigate to the file and open it.
Alternatively, you could skip editing this file and try to replicate the problem in your Jupyter notebook by noticing how is `dZ` calculated in that function:
``````dZ = np.array(dA, copy=True)
``````
The full function is:
def relu_backward(dA, cache):
“”"
Implement the backward propagation for a single RELU unit.
``````Arguments:
dA -- post-activation gradient, of any shape
cache -- 'Z' where we store for computing backward propagation efficiently
Returns:
dZ -- Gradient of the cost with respect to Z
"""
Z = cache
dZ = np.array(dA, copy=True) # just converting dz to a correct object.
# When z <= 0, you should set dz to 0 as well.
dZ[Z <= 0] = 0
assert (dZ.shape == Z.shape)
return dZ
``````
@albertovilla
The issue had nothing to do with the cache.
It’s solved now
@lachainone, how did You solve this problem. I’ve been stuck with it for 2 days now and have no idea how to solve it
Did anynone have error like above?
1 Like
I’m having the same error on my side. I think its something related to the value of dA inside the loop. When I I put it as dAL its gives me the same error but I want to put it as a variable to change each time I loop inside a layer but I dont know how to put it as a variable containing l. did you solve it?
Hello team,
I’m facing the similar sort of problem too while running the codes for exercise 9 (Week 4/Assignment 1). The traceback is hitting the most recent call and error on (dAL). I am not able to figure out what is wrong with the codes in this case? Kindly help. Thanks and regards.
Hi @Rashmi , have a look at your input parameters of the linear_activation_backward function. It requires the gradient which you initialized, not grads[“dAL”] which is the full dictionary of gradients.
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Top
# Probability
The probability is the quantitative representation of the likelihood of an event to occur. In terms of probability, an event is referred to a possible set of outcomes when an experiment is performed and the probability is measured for the events. The probability of an event falls between $0$ and $1$. Here, $0$ represents an impossible event and $1$ implies a certain event. The chances of occurring of an event are more if the probability is higher.
Related Calculators Calculation of Probability Binomial Distribution Probability Calculator Binomial Probability Calculator Coin Toss Probability Calculator
## How do you find the Probability?
Numerically, the probabilities can be calculated by dividing the number of favorable or desired outcomes with the total number of possible outcomes. This is applicable in the cases when the experiments performed are well defined and random.
Let us take the simplest example. When an unbiased or fair coin is tossed in air, there are only two possible outcomes - head and tail. So, the probability to getting a head is equal to the $\frac{1}{2}$, where 1 is the favorable outcome (head) and $2$ represents total number of outcomes (head and tail). The coin being fair, the probability of both outcomes (head and tail) is same.
## Probability Formula
Probability of an Event:
$P(E)$ = $\frac{Number\ of\ desired\ outcomes}{Number\ of\ total\ outcomes}$
Complementary Events:
The events that have only two possible outcomes are called complementary events.
Example: Tossing a coin $(\frac{head}{tail})$, getting $2$ and not $2$ on rolling a die.
$P(A) + P(A^C)$ = $1$
The probability of happening of either event A or event B is:
$P(A$ or $B)$ = $P(A) + P(B) - P(A$ and $B)$
We may even write this formula as:
$P(A \cup B)$ = $P(A) + P(B) - P(A \cap B)$
Mutually Exclusive or Disjoint Events:
The events that cannot happen together at the same time are mutually exclusive. Two events $A$ and $B$ are mutually exclusive if
$P(A \cap B)$ = $0$
So, law of addition in that case is:
$P(A \cup B)$ = $P(A) + P(B)$
Independent Events:
Two events are independent if happening of one doesn’t affect that of another. So, events $A$ and $B$ are independent if
$P(A \cap B)$ = $P(A) \cdot P(B)$
Conditional Probability:
Conditional probability refers to the probability of an event when another event has already occurred. Probability of event B, provided that A has already occurred is:
$P(A|B)$ = $\frac{P(A\ \cap\ B)}{P(B)}$
Bayes Law:
$P(A|B)$ = $\frac{[P(B|A)\ \cdot\ P(A)]}{P(B)}$
## Examples
Example 1:
On a roll of a fair $6$ sided die, what is the probability of getting of rolling a $1$ or a $4$?
Solution:
Total number of outcomes = $6$
Probability of getting a $1$
$P(1)$ = $\frac{1}{6}$
Probability of getting a $4$
$P(4)$ = $\frac{1}{6}$
Since both events are mutually-exclusive,
$P(1$ or $4)$ = $P(1) + P(4)$ = $\frac{1}{6}$ + $\frac{1}{6}$ = $\frac{1}{3}$
Example 2:
Determine the probability of drawing a jack and an ace consecutively without replacement from a well-shuffled deck of cards.
Solution:
Total outcomes = $52$
There are $4$ jacks. So
$P(Jack)$ = $\frac{4}{52}$ = $\frac{1}{13}$
Since another card is drawn without replacement, hence now total outcomes are $51$.
$P(Ace)$ = $\frac{4}{51}$
Required probability = $P(Jack) \times P(Ace)$ = $\frac{1}{13}$ $\times$ $\frac{4}{51}$ = $\frac{4}{663}$
More topics in Probability Probability Problems Probability Formulas Probability Terms Probability Rules Types of Probability Binomial Distribution Random Variable Uniform Distribution Permutations Combinations Bayes Theorem Probability
*AP and SAT are registered trademarks of the College Board.
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# Physics
posted by .
Hi,
I honestly don't know how to approach this question and what equation to use.
A 50.0 g Super Ball traveling at 28.0 m/s bounces off a brick wall and rebounds at 18.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.50 ms, what is the magnitude of the average acceleration of the ball during this time interval? (Note: 1 ms = 10-3 s.)
Can anyone help me? What equation would I use and would I use -g or a +g?
• Physics -
accelerlation is change of velocity/time
= (vf-vi)/time
Now for the details. Let the direction of vi be positive, so vf is negative.
= (-18-28)/time watch the units of time.
• Physics -
Force = rate of change of momentum
= (momentum out - momentum in)/time
momentum out = -50*10^-3*18 kg m/s
momentum in = +50*10^-3*28
momentum out - momentum in = -50*10^-3*46
Force = -50*10^-3*46/3.5*10^-3 s
acceleration = Force/mass = -46/3.5*10^-3
= -13.1*10^3 = 1.31*10^4 m/s^2
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Wednesday
May 4, 2016
# Homework Help: business
Posted by Anonymous on Saturday, June 22, 2013 at 11:37pm.
Catherine borrowed \$19,000 on June 20, at 10% interest. If the loan was due on September 17, what was the amount of interest on the loan using the exact interest method? (Round to the nearest cent) (Points : 2)
\$500.01
\$273.06
\$463.00
\$463.29
## Answer This Question
First Name: School Subject: Answer:
## Related Questions
More Related Questions
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# Excel Hint 6: The FV function and 401(k) fees
Hint 6: The Excel FV function is used to calculate the future value of a dormant 401(k) or IRA when the only difference between the two accounts is the level of fees.
The situation: Finance tip #6 considers potential benefits realized by rolling over funds from a high-cost 401(k) to a low-cost IRA. The retirement account is not accepting new contributions. The person at age 50 will either leave \$500,000 in a 401(k) or move \$500,000 to a Roth for 15 years. The pre-fee annual rate of return on both the 401(k) and the IRA is 6.0 percent. The 401(k) has an annual fee of 1.3 percent. The IRA has an annual fee of 0.03 percent.
Question on use of Excel: How does one calculate the value of the retirement account after 15 years?
Analysis:
This calculation can be completed with the FV function. The FV function has arguments – Rate, Nper, Pmt, Pv, and Type.
• The Rate is .047 for the 401(k) and 0.057 for the IRA.
• The Nper or holding period is 15 for both accounts.
• The PMT is 0 for both accounts. (The worker is no longer making contributions.)
• The PV is the initial account balance, \$500,000.
• The type is 0 because the \$500,000 exists in the account at the beginning of the period.
Results
• FV(0.047,15,0,500,000,1) is \$995,796.
• FV(0.057,15,0,500,000,1) is \$1,148,404.
Remember to go to finance tip 6 for a more complete discussion of factors impacting the choice between a high-fee 401(k) and a low-fee IRA.
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Cody
# Problem 449. Grandpa's telescope
Solution 255871
Submitted on 6 Jun 2013 by Jacek Ho
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% d=200; b=100; p=0.5; mm = telescope(d,b,p) mm_correct = 21 assert(abs(round(abs(mm))- mm_correct)<1000*eps)
mm_correct = 21 mm = 21 mm_correct = 21
2 Pass
%% d=100; b=200; p=0.5; mm = telescope(d,b,p) mm_correct = 21 assert(abs(round(abs(mm))- mm_correct)<1000*eps)
mm_correct = 21 mm = 21 mm_correct = 21
3 Pass
%% d=100; b=100; p=0.5; mm = telescope(d,b,p) mm_correct = 0 assert(round(abs(mm))== mm_correct)
mm_correct = 0 mm = 0 mm_correct = 0
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Select Page
Studio 33 – Party Compilation 01 – 26
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. with the international music hall sensation, The Voice Kids on Thursday. Check out the album from Audio.The present invention relates to an improvement in a control circuit for controlling a variable current source.
FIGS. 7 and 8 show a conventional control circuit for controlling a variable current source.
A current source 2 is connected to a transistor 15 via a resistor 10. The base of the transistor 15 is supplied with a control voltage from a resistor 11. The base of the transistor 15 is connected to the junction of the current source 2 and the resistor 10 via a diode 12. A parallel circuit of a resistor 13 and a diode 14 is connected to the emitter of the transistor 15.
The collector of the transistor 15 is connected to a resistance unit 15a which in turn is connected to a transistor 17 via a diode 16. The emitter of the transistor 17 is connected to the positive terminal of a DC power source 18. The base of the transistor 17 is supplied with a predetermined voltage from a resistance unit 19.
The resistance unit 19 is connected to a digital-to-analog converter (hereinafter referred to as D/A converter) 20 which in turn is connected to the negative terminal of the DC power source 18. The emitter of the transistor 17 is also connected to the D/A converter 20 via a resistance 21.
The resistance unit 19 comprises a plurality of resistors 21a-21h connected in parallel, and a switch 25 formed by combining a transistor 25a and a diode 25b. The transistor 25a is connected to the emitter of the transistor 17 and the D/A converter 20. The switch 25 is connected to the base of the transistor 17. The resistance unit 19 provides the resistance required of the transistor 17. The emitter of the transistor 17 is connected to the D/A converter 20 via the resistance 21. The base of the transistor 17 is supplied with a voltage obtained by dividing the output voltage of the D/A converter 20 by n, where n is a predetermined constant, via the resistance 21. The emitter of the transistor 17 is also connected to the D/A converter 20 via a resistor 22 having a resistance of 1/n. The base of the transistor 17 is supplied with a predetermined reference voltage Vref via a resistor 23 having a resistance of 1/n.
The base of the transistor 17 is supplied with a voltage obtained by
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New Delhi, May 12 (ANI): Court sources have revealed that a three-member special TADA court, set up to try Sohrabuddin and his wife Kauser Bi in 2008 and then in 2012, widened its scope by also trying former Gujarat Home Minister Amit Shah in the case.
The expanded special TADA court, which is hearing the cases against Kauser Bi and Shah, widened its scope by also trying former Gujarat Home Minister Amit Shah in the case.
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It is also being said that Special
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# Use the definition of a derivative to show that if $f(x) = 1/x,$ then $f'(x) = -1/x^2.$ (This proves the Power Rule for the case $n = -1.$ )
## $$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\lim _{h \rightarrow 0} \frac{x-(x+h)}{h x(x+h)}=\lim _{h \rightarrow 0} \frac{-h}{h x(x+h)}=\lim _{h \rightarrow 0} \frac{-1}{x(x+h)}=-\frac{1}{x^{2}}$$
Derivatives
Differentiation
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##### Kristen K.
University of Michigan - Ann Arbor
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University of Nottingham
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Idaho State University
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### Video Transcript
Hey, it's clear, So enumerated here. So we have up of access equal Thio one over X. Yeah, if the door If it is, it's equal to limit. As each approaches Ciro one over X Plus H minus one over X over each becomes equal to the limit knows each approaches Ciro for a negative H over X X plus H over each. This is equal to limit as each approaches. So for negative one over X terms X plus H, which is equal to negative one over X square.
#### Topics
Derivatives
Differentiation
##### Kristen K.
University of Michigan - Ann Arbor
##### Samuel H.
University of Nottingham
##### Michael J.
Idaho State University
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# How to use the Spatial Containers
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Revision as of 11:03, 5 May 2012 (view source)Nelson (Talk | contribs)← Older edit Latest revision as of 15:23, 17 July 2015 (view source)m (→Calling the Constructor) (9 intermediate revisions by 2 users not shown) Line 7: Line 7: comprising the problem. comprising the problem. − + This would be very CPU intensive, and would limit the overall performance of the application. To reduce CPU requirements of processing contact interaction, it is necessary to eliminate couples of discrete elements that are far from each other and are not in contact. A set procedures designed to detect bodies that are close to each other is usually called a contact detection − This would be very CPU intensive, and would limit the application that use and need to evaluate + − the contact forces ,comprising a very small number (a few thousand) of separates bodies. + − To reduce CPU requirements of processing contact interaction, it is necessary to eliminate couples of discrete elements that are far from each other and are not in contact. A set procedures designed to detect + − bodies that are close to each other is usually called a contact detection + algorithm, or sometimes a contact search algorithm. algorithm, or sometimes a contact search algorithm. − + With this purpose a set contact search algorithms based on spatial decomposition are provided in Kratos, so they can be used for any application. − With this purpose was created in Kratos Program a contact search algorithm based on spatial decomposition, + − so it can be used for any application and is unique in that it was implemented in a generic way. + Therefore it requires a contact search algorithm having the following characteristics: Therefore it requires a contact search algorithm having the following characteristics: Line 23: Line 17: *RAM efficient *RAM efficient − The space decomposition implemented in Kratos are: + The space decomposition algorithms implemented in Kratos are: − * Trees : Kd-Trees, Octrees + * '''Trees''' : Kd-Trees, Octrees − * Bins + * '''Bins''' : Dynamic Bins, Static Bins There are two kind of searches algorithms in Kratos: There are two kind of searches algorithms in Kratos: − * Points Searches: Used commonly for searching near points in a radius and can be used for + * '''Points Searches''': Used commonly for searching near points in a radius and can be used for searches contacts for spheric and circle geometries. Use the Kd_tree,Octree and bins for spatial decomposition. − searches contacts for spheric and circle geometries. Use the Kd_tree,Octree and bins for spatial + − decomposition. + − + − * Object Searches: The searches is based in the concept of objects. Is a generic class, thats mean, we can use either objects type even points. For now, use the bins for spatial decomposition. It could be interesting proof hybrid methods would be a + − and future investigation lines. + − + − == Spatial decomposition based in bins. Main methods == + − + − Exist implemented in Kratos two versions of domain decomposition based in bins: + − dynamic and static bins. The difference between them is the dynamic allocation of memory. + − However both define the methods. The most common methods used are presented and described below: + − + − * Bins based in points : + − *SearchNearestPoint : Calculates the nearest point of a point. + − *SearchInRadius : Get the list of point in a certain radius. + − + − * Bins based in Objects : + − * SearchObjects : Get the list of contacts of a certain objects even itself. + − * SearchObjectsInner : The same as above, just without return itself. + − * SearchConcts : Get the list of all pair contacts. + − + − + − == Using the search algorithm based in objects == + + * '''Object Searches''': The searches is based in the concept of objects. Is a generic class, thats mean, we can use either objects type even points. For now, use the bins for spatial decomposition. == The User Configure File == == The User Configure File == − It say that is the main file created by the user, + The configuration file is an user-made file where the specific characteristics of the problem. − which defines the types, and operations containers + In simple terms, it can be considered as a small template class that the user needs to fill with − necessary to make the spatial decomposition and the search for contacts. + the types, operations and containers necessary for the strategy to search the contacts. − Is then the parameter that defines Bins. + − + This file is divided of four parts: − This consists of four parts: + *Definition of the types and their containers. *Definition of the types and their containers. Line 73: Line 44: Suppose that we have a set of discrete elements like a spheres. Suppose that we have a set of discrete elements like a spheres. − In our file called "discrete_configure_file.h" need to create the class, + The first step is to create a configuration file. The easiest way to do that is − types and methods before mentioned. Is useful to create this file using + copying the template found in "KRATOS_ROOT/kratos/templates/header_template.h". − the template of Kratos found in kratos/templates directory. Only we need to do is + In this example we are going to name the file *"example_configure_file.h"*. − changes the default names and directives by the name of our file. Normally this file is stored + − in the custom_utilities directory of the application. + + === Creating the Skeleton === + The first thing we have to do is changing the default names and directives of the file. + This normal implies changing two tokens that appear repeatedly across the file: − * Change the name of directives + KRATOS_FILENAME_H_INCLUDED − #if !defined(KRATOS_DISCRETE_PARTICLE_CONFIGURE_INCLUDED) + − #define KRATOS_DISCRETE_PARTICLE_CONFIGURE_INCLUDED''' + − + − *Change the name of class + − ///@} + − ///@name Kratos Classes + − ///@{ + − template + − class DiscreteParticleConfigure{''' + + ClassName − * Typing the typenames, objects and container of the particle + that in our case would become: − ///@} + KRATOS_EXAMPLE_CONFIGURE_INCLUDED − ///@name Type Definitions + − ///@{ + ExampleConfigure − typedef Point PointType; + − typedef std::vector::iterator DistanceIteratorType; + − typedef ModelPart::ElementsContainerType::ContainerType ContainerType; + − typedef ContainerType::value_type PointerType; + − typedef ContainerType::iterator IteratorType; + − typedef ModelPart::ElementsContainerType::ContainerType ResultContainerType; + − typedef ResultContainerType::value_type ResultPointerType; + − typedef ResultContainerType::iterator ResultIteratorType; + − typedef ContactPair ContactPairType; + − typedef std::vector ContainerContactType; + − typedef ContainerContactType::iterator IteratorContactType; + − typedef ContainerContactType::value_type PointerContactType; + − typedef std::vector::iterator PointerTypeIterator; + + === Definition of the types and containers === − {{Note| + Once our file *"example_configure_file.h"* is prepared we need to fill it with the − Another types and container different to the model part can be used. + types and operations. − }} + + The proper place to define the files is under the "Type Definitions" section and the list of + types that need to be defined is the following: − * Computing the bounding box + *'''PointType''': Type of the points that form our element. + *'''DistanceIteratorType''': Type of the iterator for the distances between contacts. + *'''ContainerType''': Type of the elements container. + *'''ResultContainerType''': Type of the elements container for the results. This is normally the same as '''ContainerType''' but it can vary in case our results are different elements. For example if we want to search all the tetrahedrons in a X distance from a set of triangles. + *'''PointerType''': Type of the pointer to our elements + *'''IteratorType''': Type of the iterator to our elements + *'''ResultPointerType''': Type of the pointer to our result elements + *'''ResultIteratorType''': Type of the iterator to our result elements + *'''ContactPairType''': Type of the contact pair + *'''ContainerContactType''': + *'''IteratorContactType''': + *'''PointerContactType''': + *'''PointerTypeIterator''': Type of the pointer iterator − '''static inline void CalculateBoundingBox'''(const PointerType& rObject, PointType& rLowPoint, PointType& rHighPoint) + In our example, a possible definition of these types can be: − { + − rHighPoint = rLowPoint = rObject->GetGeometry().GetPoint(0); + + ///@} + ///@name Type Definitions + ///@{ + + typedef Point<Dimension, double> PointType; + typedef std::vector<double>::iterator DistanceIteratorType; + typedef ModelPart::ElementsContainerType::ContainerType ContainerType; + typedef ModelPart::ElementsContainerType::ContainerType ResultContainerType; + typedef ContainerType::value_type PointerType; + typedef ContainerType::iterator IteratorType; + typedef ResultContainerType::value_type ResultPointerType; + typedef ResultContainerType::iterator ResultIteratorType; + typedef ContactPair<PointerType> ContactPairType; + typedef std::vector<ContactPairType> ContainerContactType; + typedef ContainerContactType::iterator IteratorContactType; + typedef ContainerContactType::value_type PointerContactType; + typedef std::vector<PointerType>::iterator PointerTypeIterator; {{Note| {{Note| − It is necessary the name ''' static inline void CalculateBoundingBox ''' is well type, beacause + Please, note that these are only suggestions and any other types and containers different to ModelPart can be used. − the function is called by the bins. + }} }} + === Definition of the operations === + Along with the definition of the types of the previous section one also needs to define a set of basic operations that + will be used by the strategy. This operations involve mainly interaction between objects which have to be defined apart + as they are bound to the characteristics of the elements. + The list of functions to be defined are: + + * Computing the bounding box. * The Intersection function between objects. * The Intersection function between objects. + * Intersection cell and objects. + * Distance function between two objects. − static inline bool Intersection(const PointerType& rObj_1, const PointerType& rObj_2){ + Please note that the signature of all functions is '''mandatory''' as the strategy will only be able to use them if they appear exactly as − array_1d rObj_2_to_rObj_1 = rObj_1->GetGeometry().GetPoint(0) - rObj_2->GetGeometry().GetPoint(0); + described. − double distance_2 = inner_prod(rObj_2_to_rObj_1, rObj_2_to_rObj_1); + − const double& radius_1 = rObj_1->GetGeometry()(0)->GetSolutionStepValue(RADIUS); + − const double& radius_2 = rObj_2->GetGeometry()(0)->GetSolutionStepValue(RADIUS); + − double radius_sum = radius_1 + radius_2; + − bool intersect = (distance_2 - radius_sum * radius_sum) <= 0; + − return intersect; + − } + + ==== BoundingBox ==== − * Intersection cell and objects + *'''static inline void CalculateBoundingBox'''(const PointerType& rObject, PointType& rLowPoint, PointType& rHighPoint): + This method is used to calculate the boundingbox of the element (rObject). Once calculated the Low and High points of the + boundingbox need to be stored in '''rLowPoint''' and '''rHighPoint''' − static inline bool IntersectionBox(const PointerType& rObject, const PointType& rLowPoint, const PointType& rHighPoint){ + *'''static inline void CalculateBoundingBox'''(const PointerType& rObject, PointType& rLowPoint, PointType& rHighPoint, const double& Radius): − array_1d center_of_particle = rObject->GetGeometry().GetPoint(0); + This method is used to calculate the boundingbox of the element (rObject) given an imposed radius. Once calculated the Low and High points of the − const double& radius = rObject->GetGeometry()(0)->GetSolutionStepValue(RADIUS); + boundingbox need to be stored in '''rLowPoint''' and '''rHighPoint''' − bool intersect = (rLowPoint[0] - radius <= center_of_particle[0] && rLowPoint[1] - radius <= center_of_particle[1] && rLowPoint[2] - radius <= center_of_particle[2] && + − rHighPoint[0] + radius >= center_of_particle[0] && rHighPoint[1] + radius >= center_of_particle[1] && rHighPoint[2] + radius >= center_of_particle[2]); + − return intersect; + − } + + As an example, for our spheric elements: + − {{Note| + static inline void CalculateBoundingBox(const PointerType& rObject, PointType& rLowPoint, PointType& rHighPoint) { − Intersection cell and objects: If we do not have this function a simple return true is enougth. + − }} + rHighPoint = rLowPoint = rObject->GetGeometry().GetPoint(0); + double radius = rObject->GetGeometry()(0)->GetSolutionStepValue(RADIUS); + + for(std::size_t i = 0; i < 3; i++){ + rLowPoint[i] += -radius; + rHighPoint[i] += radius; + } + } − == Calling the Constructor == + static inline void CalculateBoundingBox(const PointerType& rObject, PointType& rLowPoint, PointType& rHighPoint, const double& Radius) { + + rHighPoint = rLowPoint = rObject->GetGeometry().GetPoint(0); + + for(std::size_t i = 0; i < 3; i++){ + rLowPoint[i] += -Radius; + rHighPoint[i] += Radius; + } + } − Once the configure file is done, we are able to use Contact Search Algorithm, + ==== Object-Object Intersection ==== − based in a domain decomposition. In the fail where it will be called , we need to add + − in the directives the corresponding files, those are the configure file and the + − bins. Maybe is required some redefinition of the typedef defined above in the configure file. + − static const std::size_t space_dim = 2; + *'''static inline bool Intersection'''(const PointerType& rObj_1, const PointerType& rObj_2): − typedef DiscreteConfigure Configure; + This method is used to test if two given elements (rObj_1, rObj_2) intersect with each other. If the intersection − typedef Configure::PointType PointType; + is not defined the default return value must be '''True'''. − typedef PointType::CoordinatesArrayType CoordinatesArrayType; + − typedef Configure::ContainerType ContainerType; + − typedef Configure::PointerType PointerType; + − typedef Configure::IteratorType IteratorType; + − typedef Configure::ResultContainerType ResultContainerType; + − typedef Configure::ResultPointerType ResultPointerType; + − typedef Configure::ResultIteratorType ResultIteratorType; + − typedef Configure::ContactPairType ContactPairType; + − typedef Configure::ContainerContactType ContainerContactType; + − typedef Configure::IteratorContactType IteratorContactType; + − typedef Configure::PointerContactType PointerContactType; + − typedef Configure::PointerTypeIterator PointerTypeIterator; + − typedef ContainerContactType ContainerContactPair; + − typedef IteratorContactType IteratorContainerContactPair; + − typedef PointerContactType PointerContainerContactPair; + + *'''static inline bool Intersection'''(const PointerType& rObj_1, const PointerType& rObj_2, const double& Radius): + This method is used to test if elements rObj_2 intersect with rObj_1 in a radius (Radius) . If the intersection + is not defined the default return value must be '''True'''. − After that we call the constructor as follow: − * Using Dynamic Bins + As an example: − + − BinsObjectStatic Bins(it_begin, it_end); + static inline bool Intersection(const PointerType& rObj_1, const PointerType& rObj_2){ − const std::size_t MaxNumberOfResults = 1000; + − std::size_t NumberOfResults = 0; + array_1d<double, 3> rObj_2_to_rObj_1 = rObj_1->GetGeometry().GetPoint(0) - rObj_2->GetGeometry().GetPoint(0); − ResultIteratorType begin; + double distance_2 = inner_prod(rObj_2_to_rObj_1, rObj_2_to_rObj_1); + + const double& radius_1 = rObj_1->GetGeometry()(0)->GetSolutionStepValue(RADIUS); + const double& radius_2 = rObj_2->GetGeometry()(0)->GetSolutionStepValue(RADIUS); + + double radius_sum = radius_1 + radius_2; + bool intersect = (distance_2 - radius_sum * radius_sum) <= 0; + return intersect; + } + static inline bool Intersection(const PointerType& rObj_1, const PointerType& rObj_2, const double& Radius){ + + array_1d<double, 3> rObj_2_to_rObj_1 = rObj_1->GetGeometry().GetPoint(0) - rObj_2->GetGeometry().GetPoint(0); + double distance_2 = inner_prod(rObj_2_to_rObj_1, rObj_2_to_rObj_1); + + const double& radius_2 = rObj_2->GetGeometry()(0)->GetSolutionStepValue(RADIUS); + + double radius_sum = Radius + radius_2; + bool intersect = (distance_2 - radius_sum * radius_sum) <= 0; + return intersect; + } + ==== Object-Cell Intersection ==== − {{Note| + *'''static inline bool IntersectionBox(const PointerType& rObject, const PointType& rLowPoint, const PointType& rHighPoint): − Note that we can use the dynamic or static bins. The difference between them + This method is used to test if a given element (rObject) intersect with a cell defined by rLowPoint and rHighPoint. − is the dynamic allocation of memory. Naturally, the static bins is faster + If the intersection is not defined the default return value must be '''True'''. − because the we know size of containers. + − }} + − *Lastly, we can get the list of pairs of contacts + *'''static inline bool IntersectionBox(const PointerType& rObject, const PointType& rLowPoint, const PointType& rHighPoint, const double& Radius): + This method is used to test if a given element (rObject) intersect with a cell defined by rLowPoint and rHighPoint in a radius (Radius). + If the intersection is not defined the default return value must be '''True'''. + + + As an example: + + + static inline bool IntersectionBox(const PointerType& rObject, const PointType& rLowPoint, const PointType& rHighPoint){ + + array_1d<double, 3> center_of_particle = rObject->GetGeometry().GetPoint(0); + const double& radius = rObject->GetGeometry()(0)->GetSolutionStepValue(RADIUS); + + bool intersect = ( rLowPoint[0] - radius <= center_of_particle[0] && + rLowPoint[1] - radius <= center_of_particle[1] && + rLowPoint[2] - radius <= center_of_particle[2] && + rHighPoint[0] + radius >= center_of_particle[0] && + rHighPoint[1] + radius >= center_of_particle[1] && + rHighPoint[2] + radius >= center_of_particle[2] + ); + + return intersect; + } + + static inline bool IntersectionBox(const PointerType& rObject, const PointType& rLowPoint, const PointType& rHighPoint, const double& Radius){ + + array_1d<double, 3> center_of_particle = rObject->GetGeometry().GetPoint(0); + + bool intersect = ( rLowPoint[0] - Radius <= center_of_particle[0] && + rLowPoint[1] - Radius <= center_of_particle[1] && + rLowPoint[2] - Radius <= center_of_particle[2] && + rHighPoint[0] + Radius >= center_of_particle[0] && + rHighPoint[1] + Radius >= center_of_particle[1] && + rHighPoint[2] + Radius >= center_of_particle[2] + ); + + return intersect; + } + + Please notice that if the object-cell function returns '''True''' the strategy will behave as a brute froce search. + + ==== Distance function ==== + + *'''static inline void Distance'''(const PointerType& rObj_1, const PointerType& rObj_2, double& distance): + This function returns the distance between to given elements (rObj_1, rObj_2). Result must be stored in "distance" + + For instance: + + + static inline void Distance(const PointerType& rObj_1, const PointerType& rObj_2, double& distance) { + + array_1d<double, 3> center_of_particle1 = rObj_1->GetGeometry().GetPoint(0); + array_1d<double, 3> center_of_particle2 = rObj_2->GetGeometry().GetPoint(0); + + distance = sqrt((center_of_particle1[0] - center_of_particle2[0]) * (center_of_particle1[0] - center_of_particle2[0]) + + (center_of_particle1[1] - center_of_particle2[1]) * (center_of_particle1[1] - center_of_particle2[1]) + + (center_of_particle1[2] - center_of_particle2[2]) * (center_of_particle1[2] - center_of_particle2[2]) ); + } + + == Calling the Constructor == + + Once the configure file is finished we are able to use Contact Search Algorithm. + + In the file where our search is going to be called we will need to include our configuration file along + with the desired strategy. The available strategies can be found in "KRATOS_ROOT/kratos/spatial_containers" and + basically are: + + * Bins_static + * Bins_static_objects + * Bins_dynamic + * Bins_dynamic_objects + + Typically dynamic strategies are slower but are not bound to a specific amount of memory. Static strategies are faster + but we need to anticipate the memory requirements. + + For our example we are going to select "bins_dinamyc_objects.h". Hence our file will have to include: + − *However, is faster to do a loop and get the bodies that are in contact with a particular element. + static const std::size_t space_dim = 2; − + − For(IteratorType it =it_begin; it!=it_end; it++) + typedef ExampleConfigure<space_dim> Configure; − rBinsObjectDynamic.SearchObjectsInner(*it, Result); + + typedef Configure::PointType PointType; + typedef PointType::CoordinatesArrayType CoordinatesArrayType; + typedef Configure::ContainerType ContainerType; + typedef Configure::PointerType PointerType; + typedef Configure::IteratorType IteratorType; + typedef Configure::ResultContainerType ResultContainerType; + typedef Configure::ResultPointerType ResultPointerType; + typedef Configure::ResultIteratorType ResultIteratorType; + typedef Configure::ContactPairType ContactPairType; + typedef Configure::ContainerContactType ContainerContactType; + typedef Configure::IteratorContactType IteratorContactType; + typedef Configure::PointerContactType PointerContactType; + typedef Configure::PointerTypeIterator PointerTypeIterator; + typedef Configure::ContainerContactType ContainerContactPair; + typedef Configure::IteratorContactType IteratorContainerContactPair; + typedef Configure::PointerContactType PointerContainerContactPair; − where ''Result'' is the list of contact for a particular element. + After that we can create the search strategy itself. + In order to create our search strategy we will need to pass all the elements in our domain: + + IteratorType it_begin = mElements.begin(); + IteratorType it_end = mElements.end(); + + BinsObjectDynamic<Configure> rBinsObjectDynamic(it_begin, it_end); − ==References== + Finally we call the search. For example if we want to search the neighbours of our elements in a Radius N: − * Munjiza. Ante, ''The Combined Finite-Discrete Element Method'', John Wiley & Sons, Ltd. 2004 + + + ContainerType SearchElements; + double * Radius; + ResultContainerType Results; + double * ResultDistance; + + int MaxNumberOfElements = MAX_RES; + + ... + + rBins.SearchObjectsInRadiusExclusive(SearElementPointerToGeometricalObjecPointerTemporalVector,Radius[i],Results.begin(),ResultsDistances.begin(),MaxNumberOfElements);
## Using the Spatial Search Algorithm
Large-scale in Finite Element Methods, Discrete Element Method and Combined Finite-Discrete Element Method simulations involve contact of a large number of separate bodies. Processing contact interaction for all possible contacts would involve a total number of operations proportional to N^2, where N is the total number of separates bodies comprising the problem.
This would be very CPU intensive, and would limit the overall performance of the application. To reduce CPU requirements of processing contact interaction, it is necessary to eliminate couples of discrete elements that are far from each other and are not in contact. A set procedures designed to detect bodies that are close to each other is usually called a contact detection algorithm, or sometimes a contact search algorithm.
With this purpose a set contact search algorithms based on spatial decomposition are provided in Kratos, so they can be used for any application. Therefore it requires a contact search algorithm having the following characteristics:
• Robust,
• CPU efficient,
• RAM efficient
The space decomposition algorithms implemented in Kratos are:
• Trees : Kd-Trees, Octrees
• Bins : Dynamic Bins, Static Bins
There are two kind of searches algorithms in Kratos:
• Points Searches: Used commonly for searching near points in a radius and can be used for searches contacts for spheric and circle geometries. Use the Kd_tree,Octree and bins for spatial decomposition.
• Object Searches: The searches is based in the concept of objects. Is a generic class, thats mean, we can use either objects type even points. For now, use the bins for spatial decomposition.
## The User Configure File
The configuration file is an user-made file where the specific characteristics of the problem. In simple terms, it can be considered as a small template class that the user needs to fill with the types, operations and containers necessary for the strategy to search the contacts.
This file is divided of four parts:
• Definition of the types and their containers.
• Function that calculates bounding box
• Function of intersection between objects
• Function of intercession between the object and cells.
## An Example Configure File
Suppose that we have a set of discrete elements like a spheres. The first step is to create a configuration file. The easiest way to do that is copying the template found in "KRATOS_ROOT/kratos/templates/header_template.h". In this example we are going to name the file *"example_configure_file.h"*.
### Creating the Skeleton
The first thing we have to do is changing the default names and directives of the file. This normal implies changing two tokens that appear repeatedly across the file:
``` KRATOS_FILENAME_H_INCLUDED
ClassName
```
that in our case would become:
``` KRATOS_EXAMPLE_CONFIGURE_INCLUDED
ExampleConfigure
```
### Definition of the types and containers
Once our file *"example_configure_file.h"* is prepared we need to fill it with the types and operations.
The proper place to define the files is under the "Type Definitions" section and the list of types that need to be defined is the following:
• PointType: Type of the points that form our element.
• DistanceIteratorType: Type of the iterator for the distances between contacts.
• ContainerType: Type of the elements container.
• ResultContainerType: Type of the elements container for the results. This is normally the same as ContainerType but it can vary in case our results are different elements. For example if we want to search all the tetrahedrons in a X distance from a set of triangles.
• PointerType: Type of the pointer to our elements
• IteratorType: Type of the iterator to our elements
• ResultPointerType: Type of the pointer to our result elements
• ResultIteratorType: Type of the iterator to our result elements
• ContactPairType: Type of the contact pair
• ContainerContactType:
• IteratorContactType:
• PointerContactType:
• PointerTypeIterator: Type of the pointer iterator
In our example, a possible definition of these types can be:
``` ///@}
///@name Type Definitions
///@{
typedef Point<Dimension, double> PointType;
typedef std::vector<double>::iterator DistanceIteratorType;
typedef ModelPart::ElementsContainerType::ContainerType ContainerType;
typedef ModelPart::ElementsContainerType::ContainerType ResultContainerType;
typedef ContainerType::value_type PointerType;
typedef ContainerType::iterator IteratorType;
typedef ResultContainerType::value_type ResultPointerType;
typedef ResultContainerType::iterator ResultIteratorType;
typedef ContactPair<PointerType> ContactPairType;
typedef std::vector<ContactPairType> ContainerContactType;
typedef ContainerContactType::iterator IteratorContactType;
typedef ContainerContactType::value_type PointerContactType;
typedef std::vector<PointerType>::iterator PointerTypeIterator;
```
Note: Please, note that these are only suggestions and any other types and containers different to ModelPart can be used.
### Definition of the operations
Along with the definition of the types of the previous section one also needs to define a set of basic operations that will be used by the strategy. This operations involve mainly interaction between objects which have to be defined apart as they are bound to the characteristics of the elements.
The list of functions to be defined are:
• Computing the bounding box.
• The Intersection function between objects.
• Intersection cell and objects.
• Distance function between two objects.
Please note that the signature of all functions is mandatory as the strategy will only be able to use them if they appear exactly as described.
#### BoundingBox
• static inline void CalculateBoundingBox(const PointerType& rObject, PointType& rLowPoint, PointType& rHighPoint):
This method is used to calculate the boundingbox of the element (rObject). Once calculated the Low and High points of the boundingbox need to be stored in rLowPoint and rHighPoint
• static inline void CalculateBoundingBox(const PointerType& rObject, PointType& rLowPoint, PointType& rHighPoint, const double& Radius):
This method is used to calculate the boundingbox of the element (rObject) given an imposed radius. Once calculated the Low and High points of the boundingbox need to be stored in rLowPoint and rHighPoint
As an example, for our spheric elements:
```static inline void CalculateBoundingBox(const PointerType& rObject, PointType& rLowPoint, PointType& rHighPoint) {
rHighPoint = rLowPoint = rObject->GetGeometry().GetPoint(0);
double radius = rObject->GetGeometry()(0)->GetSolutionStepValue(RADIUS);
for(std::size_t i = 0; i < 3; i++){
rLowPoint[i] += -radius;
rHighPoint[i] += radius;
}
}
```
```static inline void CalculateBoundingBox(const PointerType& rObject, PointType& rLowPoint, PointType& rHighPoint, const double& Radius) {
rHighPoint = rLowPoint = rObject->GetGeometry().GetPoint(0);
for(std::size_t i = 0; i < 3; i++){
rLowPoint[i] += -Radius;
rHighPoint[i] += Radius;
}
}
```
#### Object-Object Intersection
• static inline bool Intersection(const PointerType& rObj_1, const PointerType& rObj_2):
This method is used to test if two given elements (rObj_1, rObj_2) intersect with each other. If the intersection is not defined the default return value must be True.
• static inline bool Intersection(const PointerType& rObj_1, const PointerType& rObj_2, const double& Radius):
This method is used to test if elements rObj_2 intersect with rObj_1 in a radius (Radius) . If the intersection is not defined the default return value must be True.
As an example:
``` static inline bool Intersection(const PointerType& rObj_1, const PointerType& rObj_2){
array_1d<double, 3> rObj_2_to_rObj_1 = rObj_1->GetGeometry().GetPoint(0) - rObj_2->GetGeometry().GetPoint(0);
double distance_2 = inner_prod(rObj_2_to_rObj_1, rObj_2_to_rObj_1);
const double& radius_1 = rObj_1->GetGeometry()(0)->GetSolutionStepValue(RADIUS);
const double& radius_2 = rObj_2->GetGeometry()(0)->GetSolutionStepValue(RADIUS);
double radius_sum = radius_1 + radius_2;
bool intersect = (distance_2 - radius_sum * radius_sum) <= 0;
return intersect;
}
```
``` static inline bool Intersection(const PointerType& rObj_1, const PointerType& rObj_2, const double& Radius){
array_1d<double, 3> rObj_2_to_rObj_1 = rObj_1->GetGeometry().GetPoint(0) - rObj_2->GetGeometry().GetPoint(0);
double distance_2 = inner_prod(rObj_2_to_rObj_1, rObj_2_to_rObj_1);
const double& radius_2 = rObj_2->GetGeometry()(0)->GetSolutionStepValue(RADIUS);
double radius_sum = Radius + radius_2;
bool intersect = (distance_2 - radius_sum * radius_sum) <= 0;
return intersect;
}
```
#### Object-Cell Intersection
• static inline bool IntersectionBox(const PointerType& rObject, const PointType& rLowPoint, const PointType& rHighPoint):
This method is used to test if a given element (rObject) intersect with a cell defined by rLowPoint and rHighPoint. If the intersection is not defined the default return value must be True.
• static inline bool IntersectionBox(const PointerType& rObject, const PointType& rLowPoint, const PointType& rHighPoint, const double& Radius):
This method is used to test if a given element (rObject) intersect with a cell defined by rLowPoint and rHighPoint in a radius (Radius). If the intersection is not defined the default return value must be True.
As an example:
``` static inline bool IntersectionBox(const PointerType& rObject, const PointType& rLowPoint, const PointType& rHighPoint){
array_1d<double, 3> center_of_particle = rObject->GetGeometry().GetPoint(0);
const double& radius = rObject->GetGeometry()(0)->GetSolutionStepValue(RADIUS);
bool intersect = ( rLowPoint[0] - radius <= center_of_particle[0] &&
rLowPoint[1] - radius <= center_of_particle[1] &&
rLowPoint[2] - radius <= center_of_particle[2] &&
rHighPoint[0] + radius >= center_of_particle[0] &&
rHighPoint[1] + radius >= center_of_particle[1] &&
rHighPoint[2] + radius >= center_of_particle[2]
);
return intersect;
}
```
``` static inline bool IntersectionBox(const PointerType& rObject, const PointType& rLowPoint, const PointType& rHighPoint, const double& Radius){
array_1d<double, 3> center_of_particle = rObject->GetGeometry().GetPoint(0);
bool intersect = ( rLowPoint[0] - Radius <= center_of_particle[0] &&
rLowPoint[1] - Radius <= center_of_particle[1] &&
rLowPoint[2] - Radius <= center_of_particle[2] &&
rHighPoint[0] + Radius >= center_of_particle[0] &&
rHighPoint[1] + Radius >= center_of_particle[1] &&
rHighPoint[2] + Radius >= center_of_particle[2]
);
return intersect;
}
```
Please notice that if the object-cell function returns True the strategy will behave as a brute froce search.
#### Distance function
• static inline void Distance(const PointerType& rObj_1, const PointerType& rObj_2, double& distance):
This function returns the distance between to given elements (rObj_1, rObj_2). Result must be stored in "distance"
For instance:
``` static inline void Distance(const PointerType& rObj_1, const PointerType& rObj_2, double& distance) {
array_1d<double, 3> center_of_particle1 = rObj_1->GetGeometry().GetPoint(0);
array_1d<double, 3> center_of_particle2 = rObj_2->GetGeometry().GetPoint(0);
distance = sqrt((center_of_particle1[0] - center_of_particle2[0]) * (center_of_particle1[0] - center_of_particle2[0]) +
(center_of_particle1[1] - center_of_particle2[1]) * (center_of_particle1[1] - center_of_particle2[1]) +
(center_of_particle1[2] - center_of_particle2[2]) * (center_of_particle1[2] - center_of_particle2[2]) );
}
```
## Calling the Constructor
Once the configure file is finished we are able to use Contact Search Algorithm.
In the file where our search is going to be called we will need to include our configuration file along with the desired strategy. The available strategies can be found in "KRATOS_ROOT/kratos/spatial_containers" and basically are:
• Bins_static
• Bins_static_objects
• Bins_dynamic
• Bins_dynamic_objects
Typically dynamic strategies are slower but are not bound to a specific amount of memory. Static strategies are faster but we need to anticipate the memory requirements.
For our example we are going to select "bins_dinamyc_objects.h". Hence our file will have to include:
``` static const std::size_t space_dim = 2;
typedef ExampleConfigure<space_dim> Configure;
typedef Configure::PointType PointType;
typedef PointType::CoordinatesArrayType CoordinatesArrayType;
typedef Configure::ContainerType ContainerType;
typedef Configure::PointerType PointerType;
typedef Configure::IteratorType IteratorType;
typedef Configure::ResultContainerType ResultContainerType;
typedef Configure::ResultPointerType ResultPointerType;
typedef Configure::ResultIteratorType ResultIteratorType;
typedef Configure::ContactPairType ContactPairType;
typedef Configure::ContainerContactType ContainerContactType;
typedef Configure::IteratorContactType IteratorContactType;
typedef Configure::PointerContactType PointerContactType;
typedef Configure::PointerTypeIterator PointerTypeIterator;
typedef Configure::ContainerContactType ContainerContactPair;
typedef Configure::IteratorContactType IteratorContainerContactPair;
typedef Configure::PointerContactType PointerContainerContactPair;
```
After that we can create the search strategy itself. In order to create our search strategy we will need to pass all the elements in our domain:
``` IteratorType it_begin = mElements.begin();
IteratorType it_end = mElements.end();
BinsObjectDynamic<Configure> rBinsObjectDynamic(it_begin, it_end);
```
Finally we call the search. For example if we want to search the neighbours of our elements in a Radius N:
``` ContainerType SearchElements;
double * Radius;
ResultContainerType Results;
double * ResultDistance;
int MaxNumberOfElements = MAX_RES;
...
rBins.SearchObjectsInRadiusExclusive(SearElementPointerToGeometricalObjecPointerTemporalVector,Radius[i],Results.begin(),ResultsDistances.begin(),MaxNumberOfElements);
```
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# 6 cm to inches
## Understanding the Conversion from Centimeters to Inches
Centimeters and inches are both units of length measurement, but they belong to different systems. While centimeters are part of the metric system, inches are used in the imperial and US customary systems. Understanding the conversion from centimeters to inches is essential for anyone working with measurements in both systems or dealing with international units.
The centimeter-to-inch conversion can be a bit confusing at first, but it is a relatively straightforward process once you grasp the concept. One inch is equal to 2.54 centimeters, which means that to convert centimeters to inches, you need to divide the length in centimeters by 2.54. For example, if you have a measurement of 10 centimeters, you would divide it by 2.54 to get the equivalent length in inches, which is approximately 3.94 inches. This conversion factor stays constant, making it easier to apply across different measurements.
## The Importance of Knowing the Centimeter to Inch Conversion
One might question the importance of knowing the centimeter to inch conversion in today’s modern world where technology and globalization have seemingly made distance measurements less relevant. However, a basic understanding of this conversion can prove to be quite useful in many practical scenarios.
For one, knowing the centimeter to inch conversion allows for seamless communication and comprehension when dealing with international measurements. While many countries have transitioned to the metric system, the United States and few others still rely predominantly on the imperial system, which includes inches. Therefore, if you are involved in fields such as engineering, construction, or even international trade, understanding how to convert between centimeters and inches becomes crucial for accurate measurements and effective collaboration.
## Historical Context of the Centimeter and Inch Measurements
The history of measurement systems is a complex and fascinating subject, and the origins of the centimeter and inch measurements are no exception. The centimeter, or centimetre, is part of the metric system, a decimal-based system of measurement that was first introduced in France during the late 18th century. It was designed to provide a universal and standardized system of measurement for all scientific, industrial, and commercial purposes. The centimeter, which is equal to one hundredth of a meter, was chosen as the unit of length in this system due to its practicality and ease of use.
On the other hand, the inch is an imperial unit of length that can be traced back to ancient times. Its origins can be found in the early measurements made by ancient civilizations such as the Egyptians, Babylonians, and Romans. Over time, the inch became more standardized, particularly in England during the reign of Queen Elizabeth I. It was later officially defined in the British Weights and Measures Act of 1824, which established the inch as exactly 2.54 centimeters.
Understanding the historical context of the centimeter and inch measurements provides valuable insights into the evolution of measurement systems and their impact on various fields. By delving into their origins, we gain a deeper appreciation for the precision and convenience that these measurements bring to our daily lives.
## Exploring the Mathematical Relationship between Centimeters and Inches
Centimeters and inches are both measurements of length, albeit on different scales. To explore the mathematical relationship between centimeters and inches, we need to understand the conversion factor between the two units. The conversion factor is derived from the definition of one inch, which is equal to 2.54 centimeters. This means that one centimeter is approximately 0.3937 inches.
With this conversion factor in mind, we can establish a mathematical equation to convert centimeters to inches. To convert a given length in centimeters to inches, we simply multiply the length by the conversion factor. For example, if we have a length of 10 centimeters, we can calculate the equivalent length in inches by multiplying it by 0.3937. The result is approximately 3.937 inches. Conversely, to convert inches to centimeters, we would divide the length in inches by the conversion factor of 0.3937.
## Common Applications of the Centimeter to Inch Conversion
One common application of the centimeter to inch conversion is in the field of international trade and commerce. With the global economy becoming increasingly interconnected, businesses often need to convert measurements between different systems to ensure accuracy and consistency. The centimeter to inch conversion is particularly relevant in industries such as manufacturing and fashion, where products are often designed and produced using different measurement systems. By understanding and applying the centimeter to inch conversion, businesses can seamlessly communicate and meet the needs of customers and suppliers from different regions of the world.
Another important application of the centimeter to inch conversion is in the field of construction and engineering. Many architectural and engineering plans and blueprints are based on inch measurements, while some international standards and regulations may require the use of centimeters. By having a solid understanding of the centimeter to inch conversion, professionals in these industries can ensure that their designs and measurements align with the necessary standards and specifications. Whether it’s calculating the dimensions of a building or determining the spacing between objects, a precise conversion between centimeters and inches is essential for accurate results in construction and engineering projects.
## Converting Centimeters to Inches: Step-by-Step Guide
To convert centimeters to inches, follow these step-by-step instructions. First, we need to understand the conversion factor between the two units. One inch is equal to 2.54 centimeters. This ratio, derived from the International System of Units (SI), allows us to convert measurements accurately.
Begin by identifying the given measurement in centimeters that you want to convert to inches. Once you have the value, multiply it by the conversion factor of 2.54. The result will be the equivalent length in inches. Remember to carry out the multiplication accurately, as even a small error can lead to significant discrepancies in the final conversion. By following these steps, converting centimeters to inches becomes a straightforward process, enabling you to work with both metric and imperial measurements seamlessly.
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SCEV expression for ICmpInst
Hi,
i am playing the ScalarEvolution these days. i found the the ScalarEvolution will simply return a SCEVUnknow for a ICmpInst, so i think maybe great to add a new kind of SCEV to the ScalarEvolution framework.
for example, if i run ScalarEvolution on the bc file generate from the following C source file:
int f(int a, int b, int c, int d) {
return (2 * a + 5 * c + 2) > (4 * d - 3*b +3);
}
i will get a SCEVUnknow for the compare instruction, but it’s great if i could get something like 2 * a + 5 * c - 4 * d - 3*b - 1 > 0 for the compare instruction
In my opinion, we need only 3 kind of SCEV expression to express the ICmpInst: SCEVEqCond for equal condition, SCEVNeCond for not equal condition and SCEVGtCond for others. Because we can always transform A < B to B > A, and transform A >= B to A > B - 1 (or A + 1> B), and A <= B to A < B + 1 (or A - 1 < B). Furthermore, we can transform A > B to A - B > 0 and A != B to A - B != 0, so the SCEV for conditions will be very simple.
As there are already some functions such as “isKnownNonZero” in ScalarEvolution, so we can compute these condition easily.
With the SCEV for conditions, we may write more meaningful code:
SCEVEQCond *S = SE.getCondition(some_icmp_instruction);
if (some_cond.isAlwaysTrue(SE))
… do some thing …
else
… do some others thing …
Dose this make sense? or i just make things unnecessarily complex?
any comment is appreciated.
–best regards
ether
Hi,
i am playing the ScalarEvolution these days. i found the the ScalarEvolution will simply return a SCEVUnknow for a ICmpInst, so i think maybe great to add a new kind of SCEV to the ScalarEvolution framework.
for example, if i run ScalarEvolution on the bc file generate from the following C source file:
int f(int a, int b, int c, int d) {
return (2 * a + 5 * c + 2) > (4 * d - 3*b +3);
}
i will get a SCEVUnknow for the compare instruction, but it’s great if i could get something like 2 * a + 5 * c - 4 * d - 3*b - 1 > 0 for the compare instruction
oh, by the way, it seems that llvm neither optimize
(2 * a + 5 * c + 2) > (4 * d - 3*b +3 + 2 * a)
to
(5 * c + 2) > (4 * d - 3*b + 3)
nor
(5 * c) > (4 * d - 3*b + 1)
This kind of optimization maybe preform by SCEV for integer conditions
here is the original c source code:
int f(int a, int b, int c, int d) {
return (2 * a + 5 * c + 2) > (4 * d - 3*b +3 + 2 * a);
}
and here is the .ll file generate from the online demo (http://llvm.org/demo/index.cgi):
``````define i32 @f(i32 %a, i32 %b, i32 %c, i32 %d) nounwind readnone {
entry:
%0 = shl i32 %a, 1 ; <i32> [#uses=2] <- this is 2 * a, used twice
%1 = mul i32 %c, 5 ; <i32> [#uses=1]
%2 = add nsw i32 %0, 2 ; <i32> [#uses=1]
%3 = add nsw i32 %2, %1 ; <i32> [#uses=1] <- 2 * a used here for the left hand side expression
%4 = shl i32 %d, 2 ; <i32> [#uses=1]
%5 = mul i32 %b, 3 ; <i32> [#uses=1]
%6 = add i32 %0, 3 ; <i32> [#uses=1] <- 2 * a used here for the right hand side expression
%7 = sub i32 %6, %5 ; <i32> [#uses=1]
%8 = add nsw i32 %7, %4 ; <i32> [#uses=1]
%9 = icmp sgt i32 %3, %8 ; <i1> [#uses=1]
%10 = zext i1 %9 to i32 ; <i32> [#uses=1]
ret i32 %10
}
``````
Hi,
i am playing the ScalarEvolution these days. i found the the ScalarEvolution will simply return a SCEVUnknow for a ICmpInst, so i think maybe great to add a new kind of SCEV to the ScalarEvolution framework.
for example, if i run ScalarEvolution on the bc file generate from the following C source file:
int f(int a, int b, int c, int d) {
return (2 * a + 5 * c + 2) > (4 * d - 3*b +3);
}
i will get a SCEVUnknow for the compare instruction, but it's great if i could get something like 2 * a + 5 * c - 4 * d - 3*b - 1 > 0 for the compare instruction
In my opinion, we need only 3 kind of SCEV expression to express the ICmpInst: SCEVEqCond for equal condition, SCEVNeCond for not equal condition and SCEVGtCond for others. Because we can always transform A < B to B > A, and transform A >= B to A > B - 1 (or A + 1> B), and A <= B to A < B + 1 (or A - 1 < B). Furthermore, we can transform A > B to A - B > 0 and A != B to A - B != 0, so the SCEV for conditions will be very simple.
As was already pointed out, several of these replacement are not valid due to
overflow conditions. LLVM now has flags to track when addition overflow can
be considered undefined, and this makes more things possible, though not
everything, and it still requires careful overflow-aware reasoning.
As there are already some functions such as "isKnownNonZero" in ScalarEvolution, so we can compute these condition easily.
With the SCEV for conditions, we may write more meaningful code:
SCEVEQCond *S = SE.getCondition(some_icmp_instruction);
if (some_cond.isAlwaysTrue(SE))
... do some thing ...
else
... do some others thing ...
Dose this make sense? or i just make things unnecessarily complex?
I think it's unnecessarily complex :-).
If you want to simplify ICmp instructions, you can just call getSCEV on both of the
ICmp's operands and go from there. If you're writing a lot of code to do this, factoring
it out into utility routines would be natural. I don't think an ICmp expression would be
significantly more useful, because I expect all you'd do with it is just grab the
operands.
Dan
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https://www.sololearn.com/en/Discuss/3280640/precedence-of-comparison-and-logical-operators
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Precedence of comparison and logical operators | Sololearn: Learn to code for FREE!
0
# Precedence of comparison and logical operators
There is that one example in Introduction to Python, where it states that "You can put parentheses around the operations that should be done first. It makes the code easier to read." a = (3 > 2) or False Would it be correct, and is it possible to get into a situation, where evaluating a logical operation needs to precede the comparison operation, possibly in the same line? Are parentheses necessary in that case?
13th Jun 2024, 9:29 AM
Igor MatiÄ
+ 6
Comparison operators have higher precedence than Boolean operators so comparisons would be evaluated first. You may add parentheses to modify the evaluation. 2 == (3 and False) is a valid expression. It would first do (3 and False), which reduces to (False). Then it would compare 2 == False, which is False. In Booleans, any non-zero value is considered True. Zero is considered False. In the expression above, 3 is seen as True. And when using a Boolean in a mathematical expression, True gets promoted to integer 1, and False gets promoted to integer 0. print(5 + True) #outputs 6. --------- For a chart of operator precedence, see https://docs.python.org/3/reference/expressions.html#operator-precedence
13th Jun 2024, 5:36 PM
Brian
+ 5
Python would interpret numbers other than 0 as True (0 is False). so (3 and False) would become (True and False) which evaluates to False. Then it is 2 == False. False evaluates to 0 (True to 1), so that is interpreted as 2 == 0 which is False. so 2 == (3 and False) is False
14th Jun 2024, 2:07 AM
Bob_Li
+ 3
Yes. Examine this logical expression without parentheses and with them: print(False and False or True) #True print(False and (False or True)) #False
13th Jun 2024, 2:36 PM
Brian
+ 2
Thanks, Brian and Bob_Li , that helped a lot.
14th Jun 2024, 4:51 AM
Igor MatiÄ
0
You i know that, i don't know how it transfers into domain of logical operations.
13th Jun 2024, 11:34 AM
Igor MatiÄ
0
That would mean that comparison operators need to be solved first in any case? 2 == 3 and False 2 == (3 and False) would not be correct and can't happen in a program?
13th Jun 2024, 3:54 PM
Igor MatiÄ
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https://www.r-bloggers.com/mapping-anthony-bourdains-travels/
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# Mapping Anthony Bourdain’s Travels
June 22, 2019
By
[This article was first published on R programming – Journey of Analytics, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
Anthony Bourdain was an amazing personality – chef, author, world traveler, TV showhost. I loved his shows as much for the exotic locations as for the yummilicious local cuisine. So I was delighted to find a dataset that included all travel location data, from all episodes of his 3 hit TV shows. Dataset attributed to Christine Zhang for publishing the dataset on Github.
In today’s tutorial, we are going to plot this extraordinary person’s world travels in R. So our code will cover geospatial data mapping using 2 methods:
• Leaflets package to create zoomable maps with markers
• Airplane route style maps to see the paths traveled.
## Step 1 – Prepare the Workspace
Here we will load all the required library packages, and import the dataset.
places <- data.frame(fread(‘bourdain_travel_places.csv’), stringsAsFactors = F)
## Step 2 – Basic Exploration
Our dataset has data for 3 of Bourdain’s shows:
• No Reservations
• Parts Unknown – which I personally loved.
• The Layover
Let us take a sneak peak into the data:
How many countries did Bourdain visit? We can calculate this for the whole dataset or by show:
numshow <- sqldf(“select show , count(distinct(country)) as num_ctry from places group by show”) # Num countries by show.
numctry <- nrow(table(places\$country)) # Total countries visited
numstates <- nrow(table(places\$state[places\$country == ‘United States’])) ## Total states visited in the US.
Wow! Bourdain visited 93 countries overall, and 68 countries for his show “No Reservations”. Talk about world travel.
I did notice some records have state names as countries, for example California, Washington and Massachussets. But these are exceptions, and overall the dataset is extremely clean. Even disregarding those records, 80+ countries is nothing to be scoffed at, and I had never even heard of some of these exotic locations.
P.S.: You know who else gets to travel a lot? Data scientists earning \$100k+ per year. Here’s my new book which will help you how to land such a dream job.
## Step 3 – Create a Leaflet to View Sites on World Map
Thankfully, the data already has geographical coordinates, so we don’t need to add any processing steps. However, if you have cities which are missing coordinates then use the “worldcities” file from the Projects page under “Rent Analysis”.
We do have some duplicates, where Bourdain visited the same location in 2 or more shows. So we will de-duplicate before plotting.
Next we will add an info column to list the city and state name that we can use on the marker icons.
places4\$info <- paste0(places4\$city_or_area, “, “, places4\$country) # marker icons
mapcity <- leaflet(places4) %>%
setView(2.35, 48.85, zoom = 3) %>%
options = popupOptions(closeButton = T),
clusterOptions = markerClusterOptions())
mapcity # Show the leaflet
## Step 4 – Flight Route View
Can we plot the cities in flight view style? Yes, we can as long as we transform the dataframe where each record has a departure and arrival city. We do have the show and episode number so this is quite easy.
Once we do that we will use a custom function which basically plots a circle marker at the two cities and a curved line between the two.
plot_my_connection=function( dep_lon, dep_lat, arr_lon, arr_lat, …){
inter <- gcIntermediate(c(dep_lon, dep_lat), c(arr_lon, arr_lat), n=50, addStartEnd=TRUE, breakAtDateLine=F) inter=data.frame(inter) diff_of_lon=abs(dep_lon) + abs(arr_lon) if(diff_of_lon > 180){
lines(subset(inter, lon>=0), …)
lines(subset(inter, lon<0), …)
}else{
lines(inter, …)
}
} # custom function
For the actual map view, we will create a background world map image, then use the custom function in a loop to plot each step of Bourdain’s travels. Depending on how we create the transformed dataframe, we can plot Bourdain’s travels for a single show, single season or all travels.
Here are two maps separately for the show “Parts Unknown” and “The Layover” respectively. Since the former had more seasons, the map is a lot more congested.
par(mar=c(0,0,0,0)) # background map
map(‘world’,col=”#262626″, fill=TRUE, bg=”white”, lwd=0.05,mar=rep(0,4),border=0, ylim=c(-80,80) ) # other cols = #262626; #f2f2f2; #727272
for(i in 1:nrow(citydf3)){
plot_my_connection(citydf3\$Deplong[i], citydf3\$Deplat[i], citydf3\$Arrlong[i], citydf3\$Arrlat[i],
col=”gold”, lwd=1)
points(x=citydf\$long, y=citydf\$lat, col=”blue”, cex=1, pch=20) # add points and names of cities
text(citydf\$city_or_area, x=citydf\$long, y=citydf\$lat, col=”blue”, cex=0.7, pos=2) # plot city names
As always, the code files are available on the Projects Page. Happy Coding!
Call to Action:
If you read this far and also want a job or promotion in the DataScience field, then please do take a look at my new book “Data Science Jobs“. It will teach you how to optimize your profile to land great jobs with high salary; 100+ interview Qs and niche job sites which everybody else overlooks.
R-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
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| 2.671875
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https://calculatorsonline.org/rounding-numbers/what-is-382-rounded-to-the-nearest-hundredth
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# 382 rounded to the nearest hundredth
Here you will see step by step solution to round of the 382 to the nearest hundredth or round 382 to 2 decimal place . What is 382 rounded to the nearest hundredth? 382 rounded to the nearest hundredth is 382.00, check the explanation that how to rounding the 382 to 2 decimal place.
= 382.00
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https://stat.ethz.ch/pipermail/r-help/2013-April/352346.html
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[R] Selecting and then joining data blocks
arun smartpink111 at yahoo.com
Thu Apr 25 14:14:14 CEST 2013
```HI,
set.seed(24)
#creating the four matrix in a list
lst1<-lapply(1:4,function(x) matrix(sample(1:40,20,replace=TRUE),ncol=5))
names(lst1)<- paste0("B",1:4)
vec<- c(1,2,4,3,2,3,1)
res<-do.call(rbind,lapply(vec,function(i) lst1[[i]]))
dim(res)
#[1] 28 5
#or
B1<- lst1[[1]]
B2<- lst1[[2]]
B3<- lst1[[3]]
B4<- lst1[[4]]
res2<-do.call(rbind,lapply(vec,function(i) get(paste0("B",i))))
identical(res,res2)
#[1] TRUE
A.K.
----- Original Message -----
From: Preetam Pal <lordpreetam at gmail.com>
To: r-help at r-project.org
Cc:
Sent: Thursday, April 25, 2013 7:51 AM
Subject: [R] Selecting and then joining data blocks
Hi all,
I have 4 matrices, each having 5 columns and 4 rows .....denoted by
B1,B2,B3,B4.
I have generated a vector of 7 indices, say (1,2,4,3,2,3,1} which refers to
the index of the matrices to be chosen and then appended one on the top of
the next: like, in this case, I wish to have the following mega matrix:
B1over B2 over B4 over B3 over B2 over B3 over B1.
1> How can I achieve this?
2> I don't want to manually identify and arrange the matrices for each
vector of index values generated (for which the code I used is :
index=sample( 4,7,replace=T)). How can I automate the process?
Basically, I am doing bootstrapping , but the observations are actually 4X5
matrices.
Thanks,
Preetam
---
Preetam Pal
(+91)-9432212774
M-Stat 2nd Year, Room No. N-114
Statistics Division, C.V.Raman
Hall
Indian Statistical Institute, B.H.O.S.
Kolkata.
[[alternative HTML version deleted]]
______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
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# Our solutions
Leaning Tower of Pisa
This week's math challenge involved using playing cards. The aim of the challenge was to end up with all the cards in sequence from 10-2.
However the rules stated that cards had to start in three piles, only be moved to the bottom of each row if the value was lower and finally to only move one card at time.
Here's how Year 6 did.
## How we used the pattern
Up and Down Staircases
This week's challenge was to spot patterns in numbers when creating staircases with blocks. The children explored the staircases they made, counting the blocks used, then added the next step up. They had to investigate whether the number of blocks created a pattern and if so could they predict the next quantity needed without having to make the next staircase.
## part 2
Collaboration and visualising were the key to this week's math challenge.
Each group had to create a quadrilateral using string then find lines of symmetry within the shape.
## Tables without tens
Fractions in a box
This week's challenge required the children to use what they knew about fractions and square numbers to try and solve how many discs (of a certain colour) could be fitted into the box. Information was given but amounts were not. Here are our solutions.
## Fractions in a box
Make 37
This week's challenge had the children really puzzled, with questions such as 'How can I make an odd number answer out of 10 odd numbers?'
The children had a trial and error system of using exactly 10 numbers from the list 1,3,5 or 7 to make the total 37.
After many false claims of 'I've done it!' to be checked as 'your total is 38 or different , to you've used too many digits' we had a look online to see the solution.
Here are our explanations of our workings out.
## Making 37
A Puzzling Cube
For this week's challenge the children had to work out the net of a cube with only certain views of its faces.
Some children chose a hands on approach of sticking the faces on a cube and testing the positions. While other children were visualising the cube being folded and unfolded, as well as turning it to check the patterns matched the criteria.
## How did I solve it
An Egg problem
This week's task required the children to work systematically but also work backwards to find missing amounts.
## An Egg Problem
Money Bags
This week's challenge was aimed at using logical reasoning to find the amount of coins put in each bag. The children had to prove that they could pay for any amount between 1p and 15p by just using a bag with a set amount of pennies in.
## Money bags
This week's challenge was to try and spot patterns in numbers after following a set rule of adding and multiplying.
The children were then challenged to create their own set of rules and see if they could make a pattern emerge.
## Spotting patterns
Trebling
This week's challenge required using times table facts and the knowledge of how to exchange when calculating to solve the coded question.
Each letter represented a number and would only make the answer correct if all the above skills were used.
## Let me explain
• Spilsby Primary School
• Woodlands Avenue,
• Spilsby,
• Lincolnshire,
• PE23 5EP
• T:
Top
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https://learn.careers360.com/engineering/question-the-degree-of-dissociation-of-pcl5-is-60then-find-out-the-observed-molar-mass-of-the-mixtureoption-1-1303div-clas/
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The degree of dissociation of PCl5 is 60%,then find out the observed molar mass of the mixture.Option: 1 130.3Option: 2 135Option: 3 229.5Option: 4 206.5
we know this formula
Theoretical moles X Theoretical molar mass = Observed moles X Observed molar mass
for solving we have to find other values.
Below I have given the process.
We have been given that the degree of dissociation is 60% means 60/100 =0.6
$\mathrm{PCl}_{5(\mathrm{g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{g})}+\mathrm{Cl}_{2(\mathrm{g})}$?
Initial moles 1 0 0
At equilibrium 1-$\alpha$? $\alpha$? $\alpha$?
Where, $\alpha$? = degree of dissociation = 0.6
Total number of moles at equilibrium(observed moles) = 1 - $\alpha$? + $\alpha$? + $\alpha$
= 1+ $\alpha$ = 1.6
Total number of moles at equilibrium(theoretical moles) = initial moles
= 1 + 0 + 0 = 1
Theoretical molar mass = molar mass of PCl5 = 208.5
Now. from mass conservation we have
Theoretical moles X Theoretical molar mass = Observed moles X Observed molar mass
$\Rightarrow\mathrm{1 \times 208.5 = 1.6 \times M_{obs}}$?
$\therefore \mathrm{M_{obs} = \frac{208.5}{1.6}=130.3}$?
Therefore, option(1) is correct
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https://community.qlik.com/t5/QlikView-App-Dev/New-Buyers/m-p/994358
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Announcements
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Anonymous
Not applicable
Hi all,
I'm trying to show in a table the number of new buyers we have to each branch. New Buyers are buyers that have never bought before.
count(
{< UNIT.STOCKSTATUS = {'Sold'}, BUYER.B_NO = p({=\$(=\$(vRptYrMthSeqTY)-0)"}>}) * e({=\$(=\$(vRptYrMthSeqTY)-11)"}>}) * e({=\$(=\$(vRptYrMthSeqTY)-3)"}>}), BUYER.B_CATEGORY = {'*'}-{'PR'}
Could someone check to see if my set analysis is correct please?
Thanks
1 Solution
Accepted Solutions
Anonymous
Not applicable
Author
Hi all,
Thanks for everyone's responses I have managed to solve this now.
count(
{< UNIT.STOCKSTATUS = {'Sold'},BUYER.B_CATEGORY = {'*'}-{'PR'},VENDOR.C_GROUP = {'TCG'},SOLD.YrMthSeq = {"<=\$(=\$(vRptYrMthSeqTY)-0) >=\$(=\$(vRptYrMthSeqTY)-0)"}, BUYER.B_NO = e({=\$(=\$(vRptYrMthSeqTY)-11)"},BUYER.B_CATEGORY = {'*'}-{'PR'},UNIT.STOCKSTATUS = {'Sold'},VENDOR.C_GROUP = {'TCG'}>})
Thanks
9 Replies
Champion
Try like below one,
count({<
UNIT.STOCKSTATUS = {'Sold'},
SOLD.YrMthSeq = {'<=\$(vRptYrMthSeqTY))'},
What is the value of the variable vRptYrMthSeqTY.? Also can you explain below statement!!
p({<SOLD.YrMthSeq = {"<=\$(=\$(vRptYrMthSeqTY)-0) >=\$(=\$(vRptYrMthSeqTY)-0)"}>}) *
e({<SOLD.YrMthSeq = {"<=\$(=\$(vRptYrMthSeqTY)-4) >=\$(=\$(vRptYrMthSeqTY)-11)"}>}) *
e({<SOLD.YrMthSeq = {"<=\$(=\$(vRptYrMthSeqTY)-1) >=\$(=\$(vRptYrMthSeqTY)-3)"}>}),
It would be helpful to guide you.
Anonymous
Not applicable
Author
SOLD.YrMthSeq = {"<=\$(=\$(vRptYrMthSeqTY)-0) >=\$(=\$(vRptYrMthSeqTY)-0)"}>})
The above is a calendar we use -0 is September so -1 would be August -2 July and so on.
p({<SOLD.YrMthSeq = {"<=\$(=\$(vRptYrMthSeqTY)-0) >=\$(=\$(vRptYrMthSeqTY)-0)"}>}) *
e({<SOLD.YrMthSeq = {"<=\$(=\$(vRptYrMthSeqTY)-4) >=\$(=\$(vRptYrMthSeqTY)-11)"}>}) *
e({<SOLD.YrMthSeq = {"<=\$(=\$(vRptYrMthSeqTY)-1) >=\$(=\$(vRptYrMthSeqTY)-3)"}>}),
With the above set analysis I am trying to say the buyer who bought in September but exclude the ones that bought 11 months before September.
I could write this as e({<SOLD.YrMthSeq = {"<=\$(=\$(vRptYrMthSeqTY)-1) >=\$(=\$(vRptYrMthSeqTY)-11)"}>})
I am very new to e's and p's by the way so I could be completely wrong!!
Champion
Hi Gareth,
Ok. I just want to know the value of variable \$(vRptYrMthSeqTY).? I guess we don't need P() and E(). Simply we can mention the month numbers.
Creator
For a better performance, solve this in your script! See New/Lost/Returning/Loyal Customers
Creator III
Hi,
You may try the below expression:
count(
{<UNIT.STOCKSTATUS = {'Sold'},
BUYER.B_CATEGORY = {'*'}-{'PR'} , SOLD.YrMthSeq>} Sales_Amount)<=0"}
Idea is to find out the buyers having zero sales before September (or selected month) along with other sets. Considered that Sales_Amount is the field containing brought value by Buyer.
Regards,
Som
Anonymous
Not applicable
Author
Hi Tamil,
The value is a number so \$(vRptYrMthSeqTY). would equal 57 its the number of the month we are in.
Are you thinking about something like this,
count({
SOLD.YrMthSeq = {"<=\$(=\$(vRptYrMthSeqTY)-0) >=\$(=\$(vRptYrMthSeqTY)-0)"}>
-
<UNIT.STOCKSTATUS = {'Sold'}, BUYER.B_CATEGORY -= {'PR'},
SOLD.YrMthSeq = {"<=\$(=\$(vRptYrMthSeqTY)-1) >=\$(=\$(vRptYrMthSeqTY)-11)"}
>}
Champion
Hi Gareth,
Sorry for the late reply. Yes, i thought the same. Any issues in the above set analysis which you mentioned.
Edit: Did you alter your first expression like this and try.
count(
{<
UNIT.STOCKSTATUS = {'Sold'},
>}
Anonymous
Not applicable
Author
Hi all,
Thanks for everyone's responses I have managed to solve this now.
count(
{< UNIT.STOCKSTATUS = {'Sold'},BUYER.B_CATEGORY = {'*'}-{'PR'},VENDOR.C_GROUP = {'TCG'},SOLD.YrMthSeq = {"<=\$(=\$(vRptYrMthSeqTY)-0) >=\$(=\$(vRptYrMthSeqTY)-0)"}, BUYER.B_NO = e({=\$(=\$(vRptYrMthSeqTY)-11)"},BUYER.B_CATEGORY = {'*'}-{'PR'},UNIT.STOCKSTATUS = {'Sold'},VENDOR.C_GROUP = {'TCG'}>})
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https://networkpages.nl/cycling-to-57912-national-monuments-in-the-netherlands/
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Bezoek de website voor leraren en scholieren →
In March 2021 the largest roadmap instance of the traveling salesman problem ever was solved. Frans de Ruiter, consultant at the Dutch company CQM, together with Bill Cook, professor of applied mathematics at the John Hopkins University, and Keld Helsgaun, associate professor emeritus in computer science at Roskilde University in Denmark, managed to determine the shortest cycling route for a route past all 57,912 locations with National Monuments in the Netherlands.
This is a new record for the Travelling Salesman Problem. Although it is not very likely that you will take your bike and start cycling through the Netherlands this results shows how large instances of problems having a very high complexity, like the Travelling Salesman Problem, can be solved when optimization techniques are used. On Thursday 9 September we interviewed Frans who explained to us how this astonishing result was made possible.
Figure 1: The optimal cycling route, taken from the Travelling Salesman Problem website.
#### The mathematical map
"The inspiration to work on this problem came from an online talk I watched by Bill Cook. In the talk Bill was explaining how they managed to find the optimal walking route between 40.603 historic sites in the US and 49,687 pubs in the UK. During the talk he mentioned that a major challenge in such problems is to get a matrix with all the distances of the edges connecting the points. We are talking about a matrix with almost 1.6 billion entries, which is a huge amount of entries to just download from Google Maps. But creating such large distance matrices was exactly what our algorithms at CQM can do. In our projects we create matrices of similar size on a daily basis. The key in speeding up the computations is by efficiently preprocessing the map. In our case, we use natural boundaries that show up in road maps. For instance, rivers and lakes divide the Netherlands in several components connected by a limited number of bridges and ferries as shown on the map. When I realized that providing such a large distance matrix is possible I decided to contact Bill. And we started working on finding the optimal cycling tour to 57,912 Dutch monuments."
#### Almost in one shot
After creating the matrix of the cycling distances between the 57,912 Dutch monuments the quest for the optimal cycling tour started. This quest lasted almost seven months from which the three were just waiting for the computations to be complete!
"The first step of the analysis was to use heuristic methods to find an initial tour which would be close to optimal. Keld Helsgaun took over and used the LKH code to find an initial tour. The LKH code is an effective implementation of the Lin-Kernighan heuristic for solving the traveling salesman problem. The initial tour the heuristic gave had length equal to 20,253,564 meters. That is an astonishingly good result given that the optimal route, as it turned out in the end, is just 502 meters shorter than this one. A couple of weeks later we managed to improve the tour length by 186 meters, using a parallel version of LKH. What I found really astonishing in the LKH heuristic is that for the case of the UK pub tour it gave the optimal tour in one shot!"
Figure 2: The shortest possible tour to nearly every pub in the United Kingdom was computed in one shot using the LKH heuristic. Taken from the Travelling Salesman Problem website.
"We found no further improvements over the following weeks, hence Bill started working on the computations to show whether this route was the optimal one. This computation was carried out between September 2020 and March 2021 at the University of Waterloo. The full computation for this problem used a total of 96.9 years of computer time (if run on a single processor core of one of our Linux servers). During the process Bill and Keld managed to improve the initial route a couple of times and in the end obtain a cycling tour they couldn’t improve any more. Then Bill showed that this tour was indeed optimal!”
Figure 3: Frans de Ruiter set to bike the Dutch tour!
#### CQM
CQM is a Dutch consultancy company with a leading role in data analysis and quantitative methods. They use advanced algorithms to tackle highly complex problems from logistics, route and production planning, and data analysis. Frans explains in his talk “Analytics to improve mobility for elderly and disabled citizens” in the Analtyics for a Better World seminar series how these techniques are used in practice by CQM:
"A problem that I work on at CQM concerns the daily planning of a Dutch taxi company. CQM offers a route planning service for the fast calculation of the distances between large numbers of addresses, thus enabling automatic route planning. In the Netherlands almost 75% of the taxi market consists of contracted rides, which are meant for specific groups of people. Think of health insurers that organize the transport of people that need to go or leave from the hospital but also of elderly who cannot arrange regular transport on their own. From all contracted taxi rides we focus on the so-called Valys contracts. These are interregional relatively long-distance rides subsidized by the Dutch government for recreational and social purposes. Almost 200.000 people per year make use of such contracts and almost 90% of the contracts are made at least one day in advance. We compute every day the most efficient way to combine the taxi rides that all the customers need. These are about 2,500 rides on quiet days, up to 15,000 rides on Christmas day. Using our optimization techniques we can make this daily planning once we receive the bookings sparing up to thousands of kilometers for the taxi drivers!"
Have a look at the website of Professor Bill Cook for more optimal routes and information on the Travelling Salesman Problem. You can also watch this talk of Bill Cook where he explains the methods used to solve so large instances of the Travelling Salesman Problem. For a major theoretical breakthrough in the Euclidean Travelling Salesman Problem you can have a look at this article published a couple of weeks ago. The featured image is also taken from the website of Bill Cook on the Travelling Salesman Problem.
• Article
### A big breakthrough in the Euclidean Travelling Salesman Problem••
Recently there has been a breakthrough in the field of algorithms for geometric network problems, concerning the complexity of the Euclidean Travelling Salesman Problem.
• Article
### Contest: How fast can you travel around the Netherlands?•
. Suppose you are coming to the Netherlands for the first time, you want to enjoy your time in the country in the best possible way and probably visit and see as much as possible. So you rend a car and you decide to travel around. But now the challenge begins, planning such a trip!
• Article
### Finding the shortest route to your holiday destination: Dijkstra's algorithm••
Nowadays we have route planners such as TomTom and Google Maps to make driving to a holiday destination a lot simpler. In this article we explain the science behind these route planners.
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2018 NABTEB GCE Physics Practical Questions and Answers
NABTEB GCE Physics Practical Questions and Answers, NABTEB GCE 2018 Physics Practical Questions and Answers, NABTEB GCE 2018 Physics Practical Expo, NABTEB GCE 2018 Physics Practical Runs, NABTEB GCE 2018 Physics Practical Choke, NABTEB GCE 2018 Physics Practical Dubs, NABTEB GCE 2018 Physics Practical Link, NABTEB GCE Physics Practical Questions and Answers, NABTEB GCE Physics Practical Expo, NABTEB GCE Physics Runs, NABTEB GCE Physics Practical Choke, NABTEB GCE Physics Practical Dubs, NABTEB GCE Physics Practical Link
👑 2018 NABTEB GCE PHYSICS PRACTICAL ANSWERS👑
(1a)
COMPLETE THE TABLE WITH THIS:
Under S/N: 1, 2, 3, 4, 5, 6
Under p: 4.2, 5.6, 6.9, 8.1, 9.5, 10.6
Under q: 4.8, 6.9, 8.3, 9.9, 11.1, 12.6
Under P: 16.80, 22.40, 27.6, 32.4, 38.0, 42.4
Under Q: 19.20, 27.60, 33.20, 39.60, 44.40, 50.40
(1av)
(1avi)
Slope S = DQ/DP = 46.5 – 20.0/40.0 – 15.5
= 26.5/24.5
= 1.08
(1avii)
-I would avoid error due to parallax when using meter rule
-I would ensure that zero error was noted and corrected on metre rule
(1bi)
Density is the mass of a substance per its unit volume while relative density is the ratio of the density of that substance to a certain temperature and the density of water at the same temperature or some other temperature used as a reference.
(1bii)
Given V = 1.5×10^5m^3
m = 3.0×10^-2kg
.:. Density, P = m/v = 3.0 x 10^-2/1.5 x 10^-3
= 2.0 x10^3kgm^-3
(1biii)
PLhL = Pwhw
Where PL is density of liquid = ?
Pw is density of water = 1.0g/cm3
hL is the height of liquid = 15cm
hw is height of water = 10.ocm
PL x 15 = 1 x 10.o
PL = 1 x 10/15
PL = 10/15 = 0.67g/cm3
=======================================
(3a)
COMPLETE THE TABLE WITH THIS and add extra one column:
Under S/N: 1,2,3,4,5,6
Under I: 1.00, 1.50, 2.00, 2.50, 3.00, 3.50
Under V: 3.90, 3.10, 2.40, 1.70, 1.00, 0.50
Under R = V/I: 3.900, 2.067. 1.200, 0.680, 0.333, 0.143
(3av)
(3avi)
Slope, S = DI/DV = 3.75-1.00/3.80-0.00
= 2.75/3.80
= 0.724
Intercept C = 3.75
(3avii)
-I would avoid error due to parallax when using the ammeter
-I would ensure tight terminals
(3bi)
Ohm’s law states that the current through a metallic conductor between two points is directly proportional to the potential difference across the two points provided that temperature and other physical condition remains constant.
(3bii)
As the temperature of a metallic conductor increases, the kinetic energy of the electrons of the conductor also increases ,due to which more obstruction is offered to the flowing electrons and hence the current.Therefore,as more obstruction is offered to the flowing electrons, the resistance is increased.
(3biii)
Given E = 3.0V
r = 0.5 ohms
R = 4 ohms
E = I(R+r)
I = E/R+r
= 3.0/4+0.5
= 0.67A
Potential difference = E – IR
= 3.0 – 0.67 x 4
= 3.0 – 2.68
= 0.32V
=======COMPLETED======
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# STRENGTHS and WEAKNESSES of the THEORY OF RELATIVITY
Discussion in 'Alternative Theories' started by Sibilia, Aug 3, 2012.
1. ### SibiliaRegistered Senior Member
Messages:
723
MENTAL MAP OF TIME
The branch of the clocks corresponds to the field of the Physics. The branch of the calendar corresponds to the Astronomy. The branch of the concepts corresponds to the Philosophy. And the branch of theories corresponds to the Philochrony.
3. ### SibiliaRegistered Senior Member
Messages:
723
LAW OF THE SENSES (--><--)
According to the mathematical law of the senses time flows from the future to the present or becoming of moments (<----), but events are presented to us in a direction from the past to the present or to the future (---->). When we think about an event that happened we project to the past and when we think about an event will happen after now we project to the future.
......... B-A
b----------------->p<----------------
........ past .... becoming ... future
b is beginning, p is present, B is before and A is after. Present is a flowing and dimensionless point that the subject projects to the past or to the future. The becoming is the present. The whole scheme integrates the becoming-time.
5. ### SibiliaRegistered Senior Member
Messages:
723
THE SIGNS OF SPACE AND TIME
Peircean Philochrony (Charles S. Peirce)
Types of images or signs: icons, indexes and symbols
- The icon is similar to the object.
- The index is directly related to the object.
- The symbol is given by a law or convention.
- All words are symbols.
SPACE
The space (icon)
Coordinate axis (index)
Dimensions (Index)
----------------------------> Direction and sense.
TIME
The Clock (icon and index)
Calendar (Index)
The law of the senses (index)
........ B-A
b----------------->p<----------------
........ past .... becoming ... future
Note that an arrow in the space indicates direction and sense, but in the time indicates only sense.
7. ### SibiliaRegistered Senior Member
Messages:
723
THE INTERVAL IN THE PHILOCHRONY
An interval is the distance between two points in space or time. The definition given by dictionaries about interval is incorrect, because in the time there can be no space or distance, but continuous activity (CA) with orderly pace (OP). This is the CAOP. In a time interval there is activity, no space. The space alone is motionless. The CAOP is manifested in all phenomena: physical, chemical, biological, meteorological, etc. The measure of time (duration) does not measure distance, measures the CAOP.
Examples of CAOP are astronomical cycles:
- The succession of day and night.
- The succession of the seasons in the year.
- The phases of the Moon.
Conventionally, we have the succession of days and months (names).
In all processes we find the CAOP. This is an universal principle. This explanation is to establish the difference between the intervals of space and time. In the space bodies move from one place to another and in the time beings pass from one moment to another.
8. ### SibiliaRegistered Senior Member
Messages:
723
THE TIME SYMBOL
When we talk about activity in Philochrony we refer to the CAOP (continuous activity with orderly pace).
The CAOP is inherent to time.
When we see this symbol we must remember that in a time interval there is no distance or space, but activity.
NOTES:
- The apex of the A points to 12:00.
- The opposite ends to the apex point at 5:00 and 7:00 respectively.
- The center line marks 9:16.
9. ### SibiliaRegistered Senior Member
Messages:
723
TIME: ABSTRACT DISTANCE
Time is the perception of becoming, as well as the color is the perception of the wavelength of visible spectrum (light). Both are objective, but differ in that the color is particular and the time is abstract. A certain amount of time "t" is the product of an objective abstraction. The becoming-time duality is analogous to the wavelength-color duality (light). We already know that the first one is called duration.
The interpretation of the colors and their relation to emotions are subjective. Emotions can also affect time. Time seems to pass faster when we are happy or distracted than when we are sad or waiting for something.
When we think about an hour, we do not remember everything that it happened in that period, but we abstract the distance or interval between the beginning and the end of the hour. Time is an abstract but objective distance; the space is concrete. With age it makes easier to us to abstract longer time periods each time.
10. ### riverValued Senior Member
Messages:
9,506
The biggest weakness I have found with GR is the missing part of the Maxwell equation
Namely the part of the equation , the quaternion equation , which was originally included by Maxwell but was dismissed by Oliver Heaviside
That missing part of the equation is scalar potentials
11. ### James RJust this guy, you know?Staff Member
Messages:
30,511
What do the scalar potential/quaternion parts explain that the rest of the equations do not, river?
12. ### riverValued Senior Member
Messages:
9,506
The nucleus of the atom
The internal stress of the atom
Even though the equation =0
ask me more , this is new to me too
Last edited: Nov 7, 2014
13. ### riverValued Senior Member
Messages:
9,506
on pg. 171 of joseph P. Farrells' book , The Giza Death Star Deployed
the title of this segment is called , " Maxwell's Quaternion Electromagnetic Theory "
" there two types of effects that electromagnetic fields can have on charged particles : (1) translation and (2) stress "
14. ### James RJust this guy, you know?Staff Member
Messages:
30,511
What do the missing scalar/quaternion terms in the Maxwell equations have to do with the nucleus, exactly?
I don't know what you mean by that. Please elaborate.
Which equation?
15. ### riverValued Senior Member
Messages:
9,506
its complicated to me
anyway again to quote ( there is an equation given , I just can't duplicate it because of my computers limitations )
" Note that this standard zero vector equation says nothing at all about the internal stress on a particle as a result of the three interacting V, X, Y , summing to zero ".
" Two interpretations are possible. One is that nothing , translational or otherwise, is happening. If one is trained in physics to replace the resultant with a zero vector , then one implicitly is taught to assume no significant EM effects are occurring at all, if the only significant effect in view is translation, which is the only significant thing the vector analysis can model ! "
16. ### James RJust this guy, you know?Staff Member
Messages:
30,511
Where did that quote come from?
17. ### riverValued Senior Member
Messages:
9,506
to quote again
" However , the other interpretation is that translation effects are only one subset of possible effects, and that non-translational effects , non-linear effects---electromagnetic or otherwise -- maybe occurring "
18. ### riverValued Senior Member
Messages:
9,506
Joseph P. Farrells' book , pg 171 , The Giza Death Star Deployed
19. ### originTrump is the best argument against a democracy.Valued Senior Member
Messages:
9,921
This is apparently from a lunatic named Joseph P. Farrell. The quote is from his book The Giza Death Star Deployed, a book that argues that the pyramids were actually weapons for shooting down flying saucers or something.
20. ### latecurtisRegistered Member
Messages:
35
I have heard over and over how an objects mass increases when approaching the speed of light. Please explain in detail what physically happens to the object if it were a spaceship and was being observed close up. What would you see happening.
Messages:
21,475
The object's mass appears to increase from the point of view of a remote distant FoR.
It would appear shorter in the direction of travel....
and clocks on the spaceship would appear to be ticking slower....
Messages:
21,475
Joseph Patrick Farrell is an American theologian, scholar on the East–West Schism and the author of a number of books on alternative history, history, historical revisionism, archaeology, andscience/physics.
WIKI:
I'll raise your
, with a
and a
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0
The sum of a complex number and its conjugate?
Wiki User
2010-11-23 21:15:38
Given a complex number z = a + bi, the conjugate z* = a - bi, so z + z*= a + bi + a - bi = 2*a. Note that a and b are both real numbers, and i is the imaginary unit: +sqrt(-1).
Wiki User
2010-11-23 21:15:38
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Re: How to construct symmetric matrix from just a one
• To: mathgroup at smc.vnet.net
• Subject: [mg110375] Re: How to construct symmetric matrix from just a one
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Wed, 16 Jun 2010 05:36:23 -0400 (EDT)
```dragec wrote:
> symmetric matrix from given just a half matrix with diagonal.
> Eg:
>
> From:
> 1 0 0 0
> 2 3 0 0
> 4 9 5 0
> 2 2 3 4
>
> To give:
>
> 1 2 4 2
> 2 3 9 2
> 4 9 5 3
> 2 2 3 4
Any of these should do the job.
symmetrize1[mat_] := mat + Transpose[mat] -
DiagonalMatrix[Diagonal[mat]]
symmetrize2[mat_] := With[{n=Length[mat]},
Table[If[i<j,mat[[j,i]],mat[[i,j]]], {i,n}, {j,n}]]
symmetrize3[mat_] := With[{n=Length[mat]},
Table[Join[mat[[i,1;;i]],mat[[i+1;;n,i]]], {i,n}]]
Daniel Lichtblau
Wolfram Research
```
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# Ticket #7508(closed defect: fixed)
Opened 4 years ago
## hash collisions for derivatives of symbolic functions - act 3
Reported by: Owned by: burcin burcin major sage-4.3 symbolics pynac N/A Karl-Dieter Crisman Burcin Erocal sage-4.3.rc0
### Description
Reported by Alex Raichev on sage-support:
```----------------------------------------------------------------------
| Sage Version 4.2.1, Release Date: 2009-11-14 |
| Type notebook() for the GUI, and license() for information. |
----------------------------------------------------------------------
sage: X= var('x,y,z')
sage: f= function('f',*X); f
f(x, y, z)
sage: d= {}
sage: for l in [1..2]:
....: for s in UnorderedTuples(X,l):
....: print diff(f,s)
....: d[diff(f,s)]= 69
....:
D[0](f)(x, y, z)
D[1](f)(x, y, z)
D[2](f)(x, y, z)
D[0, 0](f)(x, y, z)
---------------------------------------------------------------------------
NotImplementedError Traceback (most recent call
last)
...
/Applications/sage/local/lib/python2.6/site-packages/sage/symbolic/
expression_conversions.py in derivative(self, ex, operator)
344 NotImplementedError: derivative
345 """
--> 346 raise NotImplementedError, "derivative"
347
348 def arithmetic(self, ex, operator):
NotImplementedError: derivative
```
This is another form of the problem I couldn't fix in #6243 and #6851.
## Change History
### comment:1 Changed 3 years ago by burcin
• Status changed from new to needs_review
• Report Upstream set to N/A
• Authors set to Burcin Erocal
This is fixed (hopefully, for good) in the new pynac package here:
attachment:trac_7508-fderivative_hash_collision_doctest.patch adds doctests for the fix.
Note that the new pynac version also contains fixes for #7264 and #7406. Tests should be run with the patches from those tickets also applied in this order:
This ticket now depends on #7490.
### comment:2 Changed 3 years ago by kcrisman
• Status changed from needs_review to positive_review
Positive review.
### comment:3 Changed 3 years ago by kcrisman
I should point out that #7264 has a problem, so the spkg should not be merged until that is resolved.
### comment:4 Changed 3 years ago by mhansen
• Status changed from positive_review to closed
• Reviewers set to Karl-Dieter Crisman
• Resolution set to fixed
• Merged in set to sage-4.3.rc0
Note: See TracTickets for help on using tickets.
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# Important MCQs Basic Statistics 1
The post is about the MCQs Basic Statistics. There are 20 multiple-choice questions with answers covering the topics of coefficient of variation, ungrouped and grouped data, distribution of data, primary and secondary data, data collection methods, variables, reports, sample and population, surveys, and data collection methods. Let us start with MCQs Basic Statistics quiz.
1. The first-hand and unorganized form of data is called
2. Given $X_1=12,X_2=19,X_3=10,X_4=7$, then $\sum_{i=1}^4 X_i$ equals?
3. Census reports used as a source of data is
4. The grouped data is also called
5. The listing of the data in order of numerical magnitude is called
6. A parameter is a measure which is computed from
7. A constant variable can take values
8. The data which have already been collected by someone are called
9. The mean of a distribution is 23, the median is 24, and the mode is 25.5. It is most likely that this distribution is:
10. The sum of dots, when two dice are rolled, is
11. Data collected by NADRA to issue computerized identity cards (CICs) are
12. If a distribution is abnormally tall and peaked, then it can be said that the distribution is:
13. A specific characteristic of a population is called
14. A chance variation in an observational process is
15. Primary data and __________ data are the same
16. The questionnaire survey method is used to collect
17. The mean of a distribution is 14 and the standard deviation is 5. What is the value of the coefficient of variation?
18. The number of accidents in a city during 2010 is
19. A variable that assumes any value within a range is called
20. According to the empirical rule, approximately what percent of the data should lie within $\mu \pm 2\sigma$?
### Online MCQs Basic Statistics with Answers
• The mean of a distribution is 14 and the standard deviation is 5. What is the value of the coefficient of variation?
• The mean of a distribution is 23, the median is 24, and the mode is 25.5. It is most likely that this distribution is:
• According to the empirical rule, approximately what percent of the data should lie within $\mu \pm 2\sigma$?
• If a distribution is abnormally tall and peaked, then it can be said that the distribution is:
• The sum of dots, when two dice are rolled, is
• The number of accidents in a city during 2010 is
• The first-hand and unorganized form of data is called
• The data which have already been collected by someone are called
• Census reports used as a source of data is
• The grouped data is also called
• Primary data and ——— data are the same
• The questionnaire survey method is used to collect
• Data collected by NADRA to issue computerized identity cards (CICs) are
• A parameter is a measure which is computed from
• Given $X_1=12,X_2=19,X_3=10,X_4=7$, then $\sum_{i=1}^4 X_i$ equals?
• A chance variation in an observational process is
• A constant variable can take values
• A specific characteristic of a population is called
• The listing of the data in order of numerical magnitude is called
• A variable that assumes any value within a range is called
Online MCQs Test Preparation Website
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1. ## Unique random numbers
ok i read the FAQ on random numbers but couldnt make much sense of it, i am writing code for an assigments, which is to create a lottery program, here is the code i have so far:
Code:
```
#include "stdafx.h"
#include "time.h"
int lottery[6];
void main()
{
srand((unsigned)time(NULL));
lottery[0] = rand()%49;
lottery[1] = rand()%49;
lottery[2] = rand()%49;
lottery[3] = rand()%49;
lottery[4] = rand()%49;
lottery[5] = rand()%49;
printf("%d\t%d\t%d\t%d\t%d\t%d\n\n", lottery[0], lottery[1], lottery[2], lottery[3], lottery[4], lottery[5]);
system("pause");
}```
can anyone help so i can create UNIQUE random numbers?
2. 1. Create the first random number, store it in array.
2. Create next random number.
3. Check if number generated exists in array, if not, add to array.
4. If array full, move on, otherwise go back to step 2.
3. how do you check if the number is the same in the array
4. You're probably going to need to use nested loops. An outer loop which keeps generating numbers until we have inserted X number of unique values, and an inner loop which checks the array from index 0 up to (not including) whatever the current count is at.
5. > void main()
int main
Perhaps you should consider say
int array[49];
store 1 to 49 in each respective slot
shuffle the array - randomly exchange pairs of array elements
exchanging disturbs the order, but preserves the uniqueness
take the first six elements of the array as your lottery draw
6. Originally Posted by aydin
how do you check if the number is the same in the array
i think that we are ending up confusing the guy,he probably is just starting out with C.i think that what hk_mp5kpdw said is nice enough.try to implement it on your own
7. Code:
```#include <stdlib.h>
#include <stdio.h>
int is_number_in_array(int lottery[],int size, int number){
//cicle the array and compare the number.
// return 1 when match is found (1 means true)
return 0;
}
int main(){
int lottery[6];
srand((unsigned)time(NULL));
/*************
LOOP1: loop till 6 number have been generated
LOOP2:
generate random number
check if it is in array//use a function
if so continue loop2
place number in array
******************/
printf("%d\t%d\t%d\t%d\t%d\t%d\n\n", lottery[0], lottery[1], lottery[2], lottery[3], lottery[4], lottery[5]);
system("pause");
return 0;
}```
Now the part you wanted (to check if a number is part of the array) you write it.
8. Salem's answer is probably the easist and most effective for this problem.
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Write the equation that satisfies the given conditions. The line passes through the two points (3, -1) and (-2, 5).
Question
Updated 4/16/2014 6:51:38 PM
This conversation has been flagged as incorrect.
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Original conversation
User: Write the equation that satisfies the given conditions. The line passes through the two points (3, -1) and (-2, 5).
Weegy: The equation of the line that passes through (4, -1) and (-2, 3) is y = 5/3 - (2 x)/3
shifa saleheen|Points 9966|
User: Write the equation that satisfies the given conditions. The line passes through the two points (3, -1) and (-2, 5).
Weegy: The equation of the line that passes through (4, -1) and (-2, 3) is y = 5/3 - (2 x)/3
shifa saleheen|Points 9966|
User: Given line L = ax + by + c = 0, b 0 What is the slope of this function?
Weegy: ax+b=0 taking the b on the other side ax= -b x= -b/a
Sting|Points 6697|
User: If A = {( x , y ): (2, -2), (4, -1), (6, 0), . . .}, complete the rule for set A . y= 1/2x - 3 -1/2x + 3 -1/2x - 1
Question
Updated 4/16/2014 6:51:38 PM
This conversation has been flagged as incorrect.
Flagged by yeswey [4/16/2014 6:46:23 PM]
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The equation of the line passing through the two points (3, -1) and (-2, 5) is y = (-6/5)x + 13/5
Confirmed by yumdrea [4/16/2014 6:57:14 PM]
3
ax + by + c = 0
by = -ax - c
y = (-a/b)x - c/b
The slope of the line L = ax + by + c = 0 is -a/b.
Confirmed by yumdrea [4/16/2014 6:57:15 PM]
3
If A = {( x , y ): (2, -2), (4, -1), (6, 0), . . .}, the rule for set A is: y= 1/2x - 3
Confirmed by yumdrea [4/16/2014 6:57:17 PM]
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# Bubble Sort Merge Sort and Quick Sort in Python
Sorting data is probably the most common tasks you’ll need to do in your programs. As a user of many different online applications, you are sorting data every day. Every single shopping website allows you to sort the data. For example, you can sort by lowest price, highest price, number of customer reviews, average review score, and so on. It doesn’t matter if its Amazon, Target, Barnes n Noble, or the Lego store. Its all pretty easy to do, all we have to do as the user is click a link, and our data is sorted. The sorting problem has been solved, and all modern programming languages have sorting logic already built-in, and they’re very efficient, so you will not need to implement your own sorting algorithm. It does help to understand the theory and application of some common sorting techniques, and that is what we will examine now.
## Bubble Sort
The bubble sort is the first sorting algorithm to learn about. You won’t likely use it in the wild, as it is not very efficient. It is easy to understand, however, and it’s a good first sorting routine to learn. A bubble sort starts by comparing the first two elements to each other to see which is larger. If the first element is bigger than the second, then the two elements get swapped. The bubble sort then advances and does the same operation on the next two elements. This continues until all items in the array have been inspected and the largest value has moved all the way to the right (top) of the array. The algorithm then repeats the entire process working on all the elements except the one before the last item and then the one before that and so on until the array is fully sorted.
This visualization is the first pass of the code example below.
• Easier to understand and implement
• Not very good performance: O(n2)
• Other sorting algorithms are higher performance
• A good first sorting algorithm to learn
#### Bubble sort Python code
```Original List: [90, 50, 10, 20, 70, 60, 40, 30, 80]
Iteration: 1 [50, 10, 20, 70, 60, 40, 30, 80, 90]
Iteration: 2 [10, 20, 50, 60, 40, 30, 70, 80, 90]
Iteration: 3 [10, 20, 50, 40, 30, 60, 70, 80, 90]
Iteration: 4 [10, 20, 40, 30, 50, 60, 70, 80, 90]
Iteration: 5 [10, 20, 30, 40, 50, 60, 70, 80, 90]
Iteration: 6 [10, 20, 30, 40, 50, 60, 70, 80, 90]
Iteration: 7 [10, 20, 30, 40, 50, 60, 70, 80, 90]
Iteration: 8 [10, 20, 30, 40, 50, 60, 70, 80, 90]
Sorted List: [10, 20, 30, 40, 50, 60, 70, 80, 90]
```
## Merge Sort
Merge sort is a divide and conquer algorithm that breaks an array into smaller pieces to operate on. Merge sort has better performance than the bubble sort. How it works is to successively break an array down until there are only individual arrays of one element each. At that point, the algorithm begins merging these arrays back up into each other until the original array is rebuilt fully sorted.
• Divide and conquer
• Breaks array into individual pieces
• Uses recursion to operate on the data
• Merges the pieces back into the array in sorted form
• Has good performance on large amounts of data
#### Merge sort Python code
```Original List: [90, 50, 10, 20, 70, 60, 40, 30, 80]
Sorted List: [10, 20, 30, 40, 50, 60, 70, 80, 90]
```
## Quick Sort
Quicksort is also a divide and conquer algorithm that uses recursion to perform its job, and often has better performance than Merge Sort. Quicksort completes the sorting of data in place in the existing array. The main feature of Quicksort is the selection of a Pivot Point. The pivot point is used to begin partitioning the array. The purpose of the partitioning process is to move items that are on the wrong side of the pivot value and figure out the point at which to split the array. In the quick sort, there is a lower index and an upper index. It begins by incrementing the lower index, as long as it is less than the upper index, and until it finds a value that’s larger than the pivot value. Then the upper index is decremented until it finds a value that is less than the pivot value as long as the upper index is greater than the lower index. When those two indexes cross each other, the array is split. The pivot value is then swapped with the upper index so the left side contains values below the pivot, and the right side contains values above the pivot. This process continues until the arrays can no longer be split. All of the sorting logic gets done in the partition step of the quick sort, and the data is sorted in place.
#### Quick sort Python code
```Original List: [90, 50, 10, 20, 70, 60, 40, 30, 80]
Sorted List: [10, 20, 30, 40, 50, 60, 70, 80, 90]
```
### Bubble Sort Merge Sort and Quick Sort in Python Summary
There are many algorithms to sort data. We had a look at three of the more common ones implemented in Python. Those are the Bubble sort, the merge sort, and the quick sort. Each has an algorithmic complexity associated with them, and varying degrees of performance and ease of use.
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# {Revised} Nineveh Constant {1}
Chatelain, Maurice, Our Cosmic Ancestors, Sedona, Arizona: Temple Golden Publications, 1988. Introduction, p. 5.:“A clay tablet covered with cuneiform script and discovered in the ruins of Nineveh shows a huge number, 195,955,200,000,000, which represents expressed in Seconds, an enormous period of time of 2268 million days or 6.3 million years. This period is an exact multiple of any astronomical Cycle known so far and must, therefore, have been an astronomical constant of the universe.”
The exactitude of the Nineveh constant is concise in one key quote from Chatelain.: “Every period of revolution or conjunction of all the solar system bodies calculated by the Constant of Nineveh corresponded exactly down to several decimal points with the values given in the modern tables of United States astronomers… I have not been able to find even a single period of revolution or conjunction of a solar system planet or satellite that would not be an exact fraction down to the fourth decimal point of the Great Constant of the Solar System”
As the late, great, John Michell, maintained within the inspirational moreover challenging “The Dimensions of Paradise”: “In the operations of simple arithmetic and throughout all the numerical manifestations of nature, such as the periods and intervals of the solar system, certain “nodal” numbers occur, providing a link between processes and phenomena which otherwise appear quite unconnected with each other”
Personally, I prefer the term Numerical Alphabet, given that via arranging the letters {numbers} additional interconnections emerge as with, for instance a crossword puzzle, otherwise anagrams etc. However via deference to the respected John Michell who refers to a Numerical Canon, I will employ his term denoted as: NC During the following article I will, certainly, employ the proper terms, such as Feet, Inches, Seconds, Area, Volume etc, frequently the numbers will allude to their own myriad interconnections, as it were, it is, in basic terms, a matter of “looking” at the numbers that will emerge to identify the appropriate path to follow. Furthermore of significance is the cyclic patterns that will appear within the digits, Moreover please make a note of the amount of times a number will end with the number 5, in view of the fact that whilst a number ends with a 5, {or zero}, it is divisible by 5, as I trust will be evidenced as we progress The Great Pyramid of Giza denoted as GP, whichever Gematria values, denoted as Gem I state the GP dimensions within numerous essays of this website, consequently I will employ a few GP units of measure, afterwards examine different cultures, in addition to different ages, in the hope that it will be observed, no matter the age, no matter the culture, the entire employed the identical “numbers” albeit in varying ratios Without a doubt to the displeasure of the “they never met each other” brigade, but there we are, life is not always as we expect it to be, nonetheless the numbers cannot lie. Certainly, numbers can be manipulated to fit the desired outcome of the author, however nowhere within this work will there be any “Nero” fiddling; the numbers are as they are. The numerical interconnections will begin virtually instantly, since all are utterly, faultlessly interlinked.
{A} The Earth year I always employ obtained via 1881 x 6080 x 30 = 343,094,400 ÷ 299,008 ÷ Pi = 365.24219891874… {Please observe 1881 and 6080 below}
{B} the Platonic: 25,920 x 365. 24219891874….x 20698594.035356453…= 195,955,200,000,000 the Nineveh Constant
{C} the Maya “Super Number”: 1,366,560 ÷ 20,698,594.035356453…x Pi x 7 = 1.451896875
{D} Then 18,144,000 ÷ 20,698,594.035356453…x Pi x 1881 ÷ 6080 x 299,008 ÷ 46,656 = 5.460125 this being the {C} 1.451896875 x 3.760683760683…that multiplied by the 399 days for Jupiter then by the Maya much-employed Calendar number 26 = 39,013.333…one-third of exactly 117,040 which is the 6080 x 19.25 this being Gem Simon Peter 1925 reduced and 1925 x 2.7428571…= 5280 Feet the Mile. The 2.7428571…x 396,423.8151041666…= 1,087,333.89285714…and the 396,423.815104166…multiplied by the PWS {Plato’s World-Soul} member 768 obtains exactly 304,453,490 which is the 3.760683760683….multiplied by exactly 80,956,950.75
The {E} 5,480,162,820, is this 80,956,950.75 x 67.692307… that multiplied by the 2.7428571…x 18,200 = 3,379,200 this is the Acres in one Square Mile 640 multiplied by 5280 the Mile, the 18,200 is 50 x 364 the Maya Temple steps, explained shortly, The 2.7428571…x 7 = 19.2 Feet the Wall height of the GP so-called “King’s Chamber” the first PWS member 384 is 19.2 x 20 and the {A} British Admiralty Nautical Mile of 6080 Feet is Gem Mary {Mary, Mare, Ocean?} 192 x 31.666…that multiplied by 0.6 = 19 the Metonic Cycle that multiplied by 21 = 399 days for Jupiter. The time interval between one opposition and the next is just about 780 days for Mars, 399 days for Jupiter, and 378 days for Saturn
{E} The Plato: Magnesia favoured number: 5040 x 1087333.89285714… = 5,480,162,820, the ancient fraction: the Rhind Papyrus via the Egyptian scribe: Ahmes is: 256:81 that produces 3.160493827…,the Royal Egyptian Cubit: of 1.714285… The Musical Scale: Pythagorean Limma {256:243} this I denote as FP {Fraction of Plato}, the Schooldays Pi is: of 3.142857… The Greek Foot is: 441 ÷ 440 = 1.0022727…,the longer value Greek Foot is 1.01376,
All Gematria Values denoted as: Gem
The 1,087,333.89285714… ÷ 3.142857… x FP x 7776 = 2,834,185,216 this, for example, is 101,376 x 27,957.1616… that multiplied by Gem: Lord Jesus Christ: 3168 = 88,568,288 this for example is Gem: Simon Peter: 1925 x 1.714285… x 3.160493827… x 1.0022727 …x 8472.73163265306…this multiplied by Gem: Founder of the City: 1225 = 10,379,096.25 this is the {E} 5,480,162,820 x 0.00189393… this multiplied by Gem: Mary: 192 = 0.3636 …which, to be brief as feasible is one eleventh of 4
Within the Great Pyramid of Giza {denoted as GP} the following “numbers” will emerge The GP Slant Angle: 51.7795679451499…The GP Edge Angle being: 41. 92109147699….the GP Edge Length: is the 516168 = 718. 448327995827.. The prior obtained {E} 5,480,162,820 divided by Pi the 51.7795679451499…then by the 41.92109147699…then the 718. 448327995827.. ÷ 3.0625 = 365.242199549…a tolerable Earth Year that multiplied by 360,000 Degrees then divided by the Mile of 5280 Feet obtains 24902.8772420181…Miles, a tolerable Earth Equatorial Circumference The {D} 5.460125 ÷ 3.0625 x 1225 = 2184.05 this divided by the {C} 1.451896875 = 1504.273504 this multiplied by 1881 x 26 = 73,568,000 this is 6080 x 12,100 that multiplied by 3.6 = 43,560 the Square Feet in one Acre
World -Soul
Via prudent analysis with reference to the dialogues of Plato {Timaeus 33-37… 43-45}, there will be established, a series of numbers with regard to the World Soul, as fashioned via the Creators hand {pertaining to the judgment of Plato at least} This series of numbers in my opinion ought to contain 36 terms; equally I denote the series as PWS {Plato’s World Soul} On the subject of PWS, subsequent to the formation of the World-Soul, a fraction being surplus in the ratio 256:243, this is the musical scale: the Pythagorean Limma, the celebrated author Robert Temple, within his stimulating book, “The Sirius Mystery” refers to this fraction as a Sacred fraction, whereas indicated I dub it FP {Fraction of Plato} An additional, oft times overlooked, ancient fraction being the Comma of Pythagoras {531,441:524,288} that I denote as Comma, and FP x FP x Comma = 1.125 which is, for example is 9 ÷ 8 The 9th member of PWS is 1024 that multiplied by the 4th PWS member: 512 = 524,288, obviously a Comma component, tother Comma component: 531,441 is 243 x 2187, which is the 17th PWS member However, the Square root of 531,441 is 729 which is 3 x 243 a component of the FP {256:243} in that case, upon the Rhind Papyrus circa 1650 BCE the Egyptian scribe Ahmes indicates a “Pi” version: 256:81, of 3.160493827,,, seeing as the 81 x 3 = 243, it follows that the “Rhind” is three times the FP {256:243} Consequently, I imagine the message is clear, every one of the foregoing and forthcoming, “numbers” are totally interconnected, of course we are required to employ Modern-day expressions such as Seconds, Feet, Inches, Area, Volume etc, nevertheless of more import being that they are no more than numbers, stitched within the rich and mystifying tapestry of number much akin to the letters alphabet, there exists a Numerical Alphabet
Nineveh Constant: 195,955,200,000,000
The predicament, although a pleasing predicament, is, that given that all number is undeniably interconnected, where to end? Since the Square root of the Comma component 531,441 is 729 moreover at commencement we observe: “195,955,200,000,000, which represents expressed in Seconds, an enormous period of time of 2268 million days…” However, the 2268 is the 729 x 3.111,,, which is 1.75 x 1.777,,, which is the Square root of: 3.160493827…the Rhind Papyrus fraction {256:81} The 1.75 pertains to an ancient ratio of measure: 175 ÷ 176 = 0.99431818,,, this multiplied by the 729 = 724.85795454,,, which is 632.8125 Egyptian Feet of 1.14545,,, or 724.85795454,, is 421.875 Egyptian Cubits of 1.71818,,,
The FP {256:243} component, namely 256 x 632.8125 = 162,000 The FP {256:243} component, namely 256 x 421.875 = 108,000 As a result, the 3.111… ÷ 0.99431818… = 3.12888… this is the 729 divided by 232.9900568181…and the 2268 ÷ 232.9900568181… = 9.73432098765… that multiplied by 132,860,250 = 1,293,304,320 which divided by the Modern-day Furlong of 660 = 1,959,552 the Nineveh Constant, albeit decreased The 132,860,250 divided by the Platonic Precession Cycle: 25,920 = 5125.78125 that multiplied by the FP {256:243} obtains 5400 And the Rhind: 3.160493827… x 62,001,450,000,000 = 195,955,200,000,000: Nineveh the 62,001,450,000,000, for example, divided by the: 25,920 = 2,392,031,250
The number 5040 is the most favoured number to be employed within the Utopian City-State as envisioned by Plato namely: Magnesia The 2,392,031,250 is 5040 x 474,609.375 that multiplied by the FP {256:243} obtains exactly 500,000 Then again, the Nineveh: 195,955,200,000,000 ÷ 5040 = 38,880,000,000 which is the Platonic: 25,920 x 1,500,000 Intriguing is that the Plato given area of Atlantis {minus the Central Shrine} is 4,561,920,000 however this multiplied by 42954.5454… = 195,955,200,000,000: Nineveh The 42954.5454 …is 37,500 Egyptian Feet of 1.14545… otherwise 42954.5454… is 25,000 Egyptian Cubits of 1.71818… etc Conversely: Nineveh: 195,955,200,000,000 ÷ 5544 = 35345454545.4545… that divided by the Egyptian Foot of 1.14545… = 30857142857.142857… that divided by the Schooldays Pi {22 ÷ 7} of 3.142857… = 9818181818.1818… that divided by the Egyptian Cubit of 1.71818 …= 5714285714.285714 …which is one seventh of 40,000,000,000
Or 35345454545.4545… x 33 = 1,166,400,000,000 this divided by the end PWS member: 10368 = 112,500,000 obviously 100,000,000 x 1.125 which is FP x FP x Comma Otherwise the 1,166,400,000,000 ÷ 1092266666666.666… = 1.06787109375 which is the FP x Comma Moreover the 1092266666666.666… x 3 = 800,000,000 which is: 10,368 x 316049382.716049382…. which is the Rhind {256:81} multiplied by 100,000,000 the numbers: 11664 and 32768 ought to be familiar to the musical scales’ aficionado
The employed 5544 minus 5040 = 504, the 5544 minus Gem: Divine Circle: 1224 = 4320 or 5544 minus the GP base perimeter of 3024 = 2520 that multiplied by 2 = 5040 The 30857142857.142857… x 7 = 216,000,000,000 that is 2,500,000 x 86,400 Seconds in 24 hours, otherwise 2160,00,000,000 x 12 = 2,592,000,000,000 yet this is 2,400,000,000 x 1080, which is the Lunar, Yin-Yang number The 9818181818.1818… x 11 = 108,000,000,000 since the GP base perimeter of 3024 is 2.8 x 1080 It follows that, as we view prior, the 9818181818.1818… divided by the Egyptian Cubit of 1.71818… = 5714285714.285714… a seventh of 40,000,000,000 also 30857142857.142857… ÷ 5714285714.285714… = 5.4 Otherwise 9818181818.1818… multiplied by the Yards in one Mile: 1760 = 17,280,000,000,000, there being 1728 Cubic Inches in one Cubic Foot.
To tie all together, in that case: Nineveh: 195,955,200,000,000 ÷ 108,000,000,000 = 1814.4 not only is 1814.4 the Platonic: 25,920 x 0.07, the 1814.4 is the GP base perimeter of 3024 x 0.6 Furthermore Nineveh: 195,955,200,000,000 ÷ 17,280,000,000,000 = 11.34 that multiplied by 266.666… = 3024 in addition to 266.666… x 4.05 = 1080, the Lunar, Yin-Yang number We will divide a reduced Nineveh Constant of 195,955,200 by various units of measure; it creates no divergence whether the units we employ are of time, distance, area, volume etc, since all are superbly interrelated Likewise, a quantity of the products of the divisions are units of measure, otherwise via multiplication, division etc will prove to be additional units of measure. For example, the below highlighted: 37112.7272… is the Roman Remen of: 1.202727… x 30857.142857… {one seventh of: 216,000} or 30857.142857… is the Schooldays Pi of 3.142857… x 9818.1818… which for example, is the Greek Cubit of: 1.50340909… x 6530.612244897959… this multiplied by Gem: Founder of the City: 1225 = 8,000,000
The below highlighted 322159.0909 is for example, the Roman Stade {600ft.} of: 577.30909… x 558.03571428… this multiplied by 7168 = 4,000,000 furthermore the 7168 is 6804 times the FP {Pythagorean Limma: 256:243} the GP base perimeter of 3024 x 2.25 = 6804 additional divisions of Nineveh will be observed within Nineveh Constant {Two}
The Nineveh Constant reduced to 195,955,200 will be divided by all the following,
The GP base side-length of 756 Feet = 259,200
The 6th and final PWS member 10,368 = 18,900
Cubic Inches in one cubic Yard 46,656 = 4200
Mile of 5280 Feet = 37112.7272
Plato/Magnesia number 5040 = 38,880
Greek Furlong 633.6 = 309272.7272…
Longer Roman Foot 0.9732096 = 201349431.8181…
Shorter Roman Foot 0.96768 = 202,500,000
Roman Furlong 608.256 = 322159.0909…
Greek Mile 5068.8 = 38659.0909…
Arabian Hashimi Foot 1.064448 = 184090909.0909…
Roman Mile 4866.048 = 40269.886363…
Roman Mile 4838.4 = 40,500
NC Earth Radius 20,901,888 Feet = 9.375
Moon Diameter 11,404,800 Feet = 17.1818…= 10 Egyptian Cubits
Greek Cubit 1.52064 = 128863636.3636…
Roman Cubit 1.4598144 = 134232954.5454…
Short Roman Cubit 1.45152 = 135,000,000
Short Greek Cubit 1.512 = 129,600,000
Long Egyptian Foot 1.152 = 170,100,000
Greek Foot 10.08 = 19,440,000
Lunar, Yin-Yang number 1080 = 181,440
Egyptian Foot 1.14545…= 171,072,000
Egyptian Cubit 1.71818…= 114,048,000
Greek Foot 1.0022727…= 195510857.142857…
Gematria Lord Jesus Christ 3168 = 61854.5454…
Common Egyptian Mile 4909.0909…= 39,916.8
NC Earth to Moon in Miles 237,600 = 824.7272…
NC Earth to Sun in Miles 93,312,000 = 2.1
Atlantis Area 4,561,920,000 = 0.04295454…
479,001,600 {=12!} = 0.4090…
GP Volume 91,445,760 Cubic Feet = 2.142857…
GP base Area Square Feet 571,536 = 342.857142…
Any product with the six numbers 124578 in any order, will be ratios of additional Units of Measure containing a cyclic 124578 in any order. Equally, the products containing a cyclic 72…or 81…or 09…or 36…and so on will be ratios of Units of Measure containing these cyclic numbers or other cyclic numbers for that matter. For example, the Greek Foot 1.0022727…produced 195510857.142857…this is exactly 49,500 x 3949.714285…the Earth Polar Radius in Miles, otherwise 195510857.142857…. is exactly 1,875,000 x 104.2724571428…the Stonehenge Sarsen Circle Outer Diameter in Feet, and so on. since they contain the numerical sequence 124578 albeit rearranged as it were
Earth Year and Circumference
Regarding the interconnections of the Nineveh Constant, we will employ Modern-day units of measure in the company of Gem: {Gematria Values} in addition to ancient fractions, in order that we listen to the music of the spheres Initially we require determining the Earth Year being considered, with reference to the Nineveh Constant The year I normally employ I unearthed via 1881 x 6080 x 30 = 343,094,400 this divided by 299,008 afterwards divided by Pi = 365.2421989187…a tolerable Earth Year that multiplied by 68.1818… {360,000 ÷ 5280} obtains an Earth Equatorial Circumference of 24902.8771990053… {Miles}
The 343,094,400 divided by the 365.2421989187…x 655351027.39726027… ÷ Pi = 195,955,200,000,000 the Nineveh Constant The 655351027.39726027… x 2559.9999744 = 1,677,698,613,360 this is 195,955,200,000,000 divided by 116.80000116 this multiplied by the 2559.9999744 = 299,008 the 2559.9999744 plus 0.0000256 = 2560 The 343,094,400 is 56,430 x 6080, the British Admiralty {or as I dub it} “Sea” Mile the 56430 is 2970 x 19 the Metonic Cycle, a prime number likewise Gematria: Eve is 19 accordingly 19 x 320 = 6080 The 2970 x 80 = 237,600, the NC {Numerical Canon} distance Earth to Moon, otherwise 2970 x 31418.1818… = 93,312,000 the NC distance Earth to Sun, the 31418.1818… is a resemblance of Pi otherwise 31418.1818… x 6,237,000,000 = 195,955,200,000,000 The 6,237,000,000 is the 56,430 x 110526.315789473…this multiplied by the Greek Cubit of 1.52064 x 19 = 3,193,344 this is the 237,600 x 13.44 etc
One example regarding the 343,094,400 the following ought to suffice, the 343,094,400 ÷ 25,920 ÷ Pi x 81 ÷ 365 ÷ 2.56 = 365.2421989187…The 299,008, for example is:…the Sothic Cycle: 1460 x 204.8 {The 16th PWS member is 2048}……The Maya Synodic period of Venus: 584 x 512… {The 4th PWS member is 512}……Gem: Abraxas, Mithras: 365 x 819.2 {The 33rd PWS member is 8192}……Gem: Son of Man: 2190 x 136.5333…etc
The Nineveh: 195,955,200,000,000 divided by the 136.5333… will obtain 1,435,218,750,000 which divided by the Platonic: 25,920 = 55,371,093.75, that multiplied by 6.196272761904… will regain the year employed number: 343,094,400 The 6.196272761904… divided by the Rhind: {256:81} of 3.160493827… = 1.960539428571… this is exactly 0.623808 x 3.142857 …which is the Schooldays Pi {22 ÷ 7 The 0.623808 multiplied by the 136.5333… = 85.1705856 this divided by Gem Mary: 192 = 0.4435968 that divided by the 0.623808 = 0.7111… one ninth of 6.4, likewise 192 x 2 = 384, the 1st PWS member, consequently the very last member 10,368 is 192 x 54 At this time we will employ: Gem: Church: 294
The Schooldays Pi of 3.142857…, the cubic Inches in one cubic Yard 46,656, the Greek Foot of 1.0022727… the NC {Numerical Canon} Earth to Sun distance: 93,312,000, Gem: Founder of the City: 1225 the Platonic precession: 25,920 and the GP base side length of 756 reduced to 0.756 The Nineveh: 195,955,200,000,000 reduced to 1,959,552 ÷ 294 = 6665.142857…. {one seventh of 46656} The 6665.142857… ÷ 3.142857… = 2120.7272… {933,120 ÷ 440} the 2120.7272… ÷ 1.0022727… = 2115.91836734693…and this multiplied by 1225 = 2,592,000 that multiplied by 0.756 = 1,959,552
Hence the GP base area {756 x 756} of 571,536 x 342857142.857142… = 195,955,200,000,000: The 342857142.857142… x 7 = 2,400,000,000 The GP Volume: 756 x 756 x 480 ÷ 3 = 91445760 Cubic Feet multiplied by 2142857.142857 …= 195,955,200,000,000. The 2142857.142857… x 7 = 15000000 or the Modern-day Mile of 5280 x 37112727272.7272… = 195,955,200,000,000. The 37112727272.7272… x 11 = 408,240,000,000 which is the GP base area of 571,536 x 714285.714285… one seventh of 5,000,000
Another Earth year and Circumference
The 195,955,200,000,000 ÷ 6209572.8210606…÷ 86,400 = 365.2425159276…that multiplied by 68.1818… {360,000 ÷ 5280} obtains an Earth Equatorial Circumference of 24902.8988132… {Miles} This Year of 365.2425159276…compares favourably to the year we calculated prior: 365.2421989187…yet without employing Pi The employed 6209572.8210606… multiplied by the Modern-day Mile of: 5280 Feet obtains 32,786,544,495.2 this divided by the GP base perimeter of 3024 then by 85159.85,583,168 = 127.3148148…this divided by the FP {256:243} obtains 120.849609375 that multiplied by 26.2144 = 3168, Gem: Lord Jesus Christ The 26.2144 x 20,000 = 524,288 this is a component of the Comma: 531,441:524,288 Then 6,300,000 x 365.2425159276…x 85159.85583168 = 195,955,200,000,000 Nineveh, the 85159.85583168 x 24902.8988132… = 2120727272.7272…one eleventh of 23,328,000,000 which is 900,000 x 25,920 Or 23,328,000,000 ÷ 2268 = 10285714.285714… one seventh of exactly 72,000,000 In addition to 195,955,200,000,000 ÷ 2268 = 86,400,000,000
The NC {Numerical Canon} Earth mean Circumference of 24883.2 divided by the GP Tan: {480 ÷ 378} of 1.269841…= 19595.52 obviously the Nineveh Constant reduced Otherwise 479,001,600 {=12!} divided by 1.269841…afterwards divided by Gem: Simon Peter 1925 = 195955.2 Nineveh Constant reduced Besides, as we recognize: the GP base side length of 756 x 259,200,000,000 = 195,955,200,000,000: Likewise, the Maya: Long Count days: 1,360,800 x 144,000,000 = 195,955,200,000,000 this indicates the Maya: 1,360,800 ÷ 756 = 1800
A Philosophy connection
The Àryabhatiya of Àryabhata: An Ancient Indian Work on Mathematics and Astronomy,
“In a Yuga, the revolutions of the Sun are 4,320,000 of the Moon: 57,753,336 of the Earth eastward: 1,582,237,500 of Saturn: 146,564 of Jupiter: 364,224 of Mars: 2,296,824…”
Likewise on page 14, Professor Clark writes: “…for the Circumference of the sky…it works out as 12,474,720,576,000
On page 8 Professor Clark writes… “Revolutions of the Moon in a Yuga {I, 1} …that is to say, 57,753,336”
I will denote whichever data of the Àryabhatiya of Àryabhata as AA, Since the sky Circumference is mentioned, it follows that the Nineveh Constant: 195,955,200,000,000 divided by the AA sky Circumference of 12,474,720,576,000 = 15.7081835064… The GP base perimeter of 3024 ÷ 15.7081835064…= 192.51112 exactly. Hence 360 Degrees divided by 15.7081835064… = 22.9179904761… this divided by the Rhind {256:81} of 3.160493827… = 7.2513954241… this divided by the Schooldays Pi {22 ÷ 7} of 3.142857… = 2.3072621803977272… this is 1.3428510044… Egyptian Cubits of 1.71818… The 1.3428510044… x 7 = 9.39995703125 that multiplied by 20.48 will regain the above 192.51112; the 16th PWS member is 2048
The Much-Maligned Metre
The Feet metres ratio: 1 ÷ 0.3048 = 3.28083989501…Feet in one Metre, the Mile of 5280 x 23.7333… = 125,312 this divided by a reduced Nineveh of: 1,959,552 then multiplied by 9,266,400 = 59257.989417… that divided by the GP Tan {480 ÷ 378} of 1.269841… = 46665.666 …this minus the cubic Inches in one cubic Yard: 46,656 = 9.666… which is one third of the prime number 29, otherwise the 46665.666… x 3 = 1,399,970 that divided by the 125,312 = 11.171875 that multiplied by 32 = 357.5 The 9,266,400 divided by the Platonic 25,920 = 357.5 Gem: Lord Jesus Christ: 3168 ÷ 357.5 = 8.8615384… that multiplied by Gem: YHWH: 26 = 230.4, the 18th PWS member is 2304 The GP Tan is 480 ÷ 378 = 1.269841… the 23.7333… x 15 = 356 that plus 9 = 365 Then 125,312 ÷ 9 ÷ 1.00137515625 ÷ 3168 ÷ FP ÷ 1.269841… will obtain 3. 28083989501.. In other words: 125,312 ÷ 38195.0976 = 3.28083989501…
The 38195.0976 divided by the longer value Greek Foot of 1.01376 = 22229.950984126 that divided by the GP Tan of 1.269841… = 17506.0864 which is Gem: 3168 x 5.5259111 that divided by the FP {256:243} obtains: 5.2452984375 that divided by the 1.00137515625 = 5.238095… which multiplied by 1008 = 5280 The Greek Foot is 10.08 and 1008 x 3 = 3024, the GP base perimeter, likewise the reduced Nineveh: 1,959,552 ÷ 1008 = 1944 which is the 15th PWS member This 5.238095… divided by the Rhind Papyrus {256:81} of 3.160493827… = 1.6573660714285… which is one seventh of 11.6015625 that multiplied by 20,480 = 237,600, the NC {Numerical Canon} distance Earth to Moon
The Nineveh Constant 195,955,200,000,000 divided by the AA: Moon: 57,753,336 = 3,392,967.63740193…that divided by the Platonic Precession Cycle of 25,920 obtains 130.901529220753…We identify that: the Maya: 1,360,800 x 144,000,000 = 195,955,200,000,000 the Maya Long Count days: 1,360,800 divided by the 130.901529220753…= 10,395.60048 that multiplied by 1,200,000,000 = 12,474,720,576,000 this is the AA: sky Circumference the 3,392,967.637401932…divided by Maya Tzolkins 5256 = 645.5417879379…this multiplied by 19324419904.47693714285… = 12,474,720,576,000
The 19,324,419,904.47693714285 …x 7 = 135,270,939,331.33856 that divided by 479,001,600 {=12!} then multiplied by the 130.901529220753..then by the Cubic Inches in one cubic Yard: 46,656 = 1,724,724,625.0909… this divided by the Egyptian Foot of 1.14545… = 1,505,711,974.285714… that divided by the Schooldays Pi {22 ÷ 7} of 3.142857… = 479,090,173.6363… that divided by the Egyptian Cubit of 1.71818 …= 278,835,550.793650… that divided by the AA: Moon: 57,753,336 = 4.8280423… this multiplied by a reduced GP base perimeter {3024} of 302.4 = 1460, the Sothic Cycle
Hence 4.8280423…. will obtain {a few examples} Gem: Abraxas, Mithras: 365 ÷ 4.8280423 …= 75.6 the GP base side length of 756 reduced. Gem: Son of Man: 2190 ÷ 4.8280423 = 453.6 Maya Synodic period of Venus: 584 ÷ 4.8280423… = 120.96 As indicated prior, the Earth Year and Equatorial Circumference I usually employ derived via 1881 x 6080 x 30 = 343,094,400 this divided by 299,008 ÷ Pi = 365.2421989187…this Year multiplied by 360,000 Degrees afterwards divided by the Mile of 5280 = 24902.8771990053…the employed 299,008 divided by the 4.8280423 …= 61931.52 which is the GP base side length of 756 x 81.92, the 33rd PWS member is 8192
Equally the 4.8280423… multiplied by Gem: Mary: 192 = 926.984126… this is 730 x 1.269841… the GP Tan 480 ÷ 378 The Maya “Super Number”: 1,366,560 is 730 x 1872 which is obviously an additional Maya calendar number albeit reduced, given that 260 x 72,000 = 18,720,000 Moreover Gem: Mary: 192 x 2 = 384 which is the first member of PWS and 384 x 27 = 10,368 this is the end PWS member, the 4th member: 512 multiplied by the Maya Synodic period of Venus: 584 will obtain the 29,9008 The GP Tan: 1.269841 …is 1.25 x 1.015873… which is the Eye of Horus fraction {64:63}
The 1.269841… is 1.2053571428… x 1.053497942386831…which is the FP {Pythagorean Limma 256:243} the 1.2053571428… x 7 = 8.4375 that multiplied by 3072 = 25,920 The 21st PWS member is 3072 The GP Tan: 1.269841… x 0.79824407100677490234375 = 1.0136432647705078125 which is the Comma of Pythagoras {531,441:524,288} Then 8.4375 ÷ 0.79824407100677490234375 = 10.5700753772…that multiplied by the Platonic: 25,920 = 273976.35377882…which is the “Comma” component: 524,288 divided by exactly 1.913625 that multiplied by the Pythagorean Limma {256:243} = 2.016 The GP Tan: 1.269841 x 2.4888… = 3.160493827… this is the Rhind Papyrus {256:81} the GP base perimeter of 3024 ÷ 2.4888… = 1215 that divided by the Lunar, Yin-Yang number 1080 = 1.125 The “Comma” multiplied twice by the “Limma” obtains exactly 1.125 which is 9 ÷ 8 Consequently the “Rhind” is three times the “Limma”
The interconnections never cease since the Nineveh reduced to 1.959552 divided by the GP Tan: 1.269841… = 1.5431472 which is 0.2592 x 5.9535 this multiplied by the 4.8280423… obtains 28.74375 that multiplied by 94,720 = 2,722,608, this divided by the Sea Mile of 6080 then multiplied by the prime number, also Gem: Eve: namely 19 = 8508.15 this divided by Gem: Abraxas, Mithras: 365 = 23.31 that multiplied by 13,000 = 303,030 obviously 30 x 10,101 which is 1001 x 10.0909… that multiplied by 66 = 666 otherwise the NC Earth mean Circumference {24883.2} reduced to 2.48832 divided by the GP Tan: 1.269841… obtains exactly 1.959552 Nineveh reduced
Sky Circumference
The AA sky Circumference 12,474,720,576,000 divided by the Platonic Precession Cycle of 25,920 {Years} obtains 481,277,800 this divided by 8.333… = 57,753,336 the AA: Moon The 8.333… x 43.2 = 360 Degrees, hence the AA: Earth eastward 1,582,237,500 ÷ 8.333… = 189,868,500 which is the Maya Calendar Long Count days: 1,360,800 x 139.527116402… and this multiplied by 567 then by 5256 Maya Tzolkins will obtain 415,812,015 this divided by Gem: Abraxas, Mithras: 365 = 1,139,211… that multiplied by 1.720095750479…= 1,959,552 the Nineveh reduced The 1.720095750479…x 139.527116402 = 240 that multiplied by the AA: revolutions of the Sun 4,320,000 = 1,036,800,000 which is the end PWS member 10,368 increased in addition to 10,368 is 0.4 x 25,920 otherwise the Maya 1,360,800 ÷ 10,368 = 131.25 that multiplied by 23.04 = 3024 the GP base perimeter, the 18th PWS member is 2304 The 567 multiplied by the Eye of Horus fraction {64:63} of 1.015873…; obtains 576, Gem: Eagle moreover the GP height is 5760 {480 x 12} Hence Nineveh 195,955,200,000,000 is 1,036,800,000 x 189,000 that multiplied by 0.016 = 3024
One method is: via the AA: Sun: 4,320,000
“In a Yuga, the revolutions of the Sun are 4,320,000,
Since 1 ÷ 4,320,000 then multiplied by 249,494,411,520,000 obtains 57,753,336 the AA: Moon the 249,494,411,520,000 is the Platonic “years” 2,5920 x 9,625,556,000 which is 57,753,336 x 166.666… one third of 500 Equally: 249,494,411,520,000 is 20 x 12,474,720,576,000 the AA Circumference of the sky Equally Gem: Eagle, Pneuma {Breath} is 576, the GP height is 5760 {480 x 12} furthermore we observe 576 within the AA: sky Circumference Hence 12,474,720,576,000 ÷ 576 = 21,657,501,000 which is 835551.736111… Platonic “Years” of 2,5920 The 835551.736111… divided by the AA: Moon: 57,753,336 = 0.014467592592…., which, due to the cyclic 592, multiplied by 25,920 = 375 which, for example, is 3 ÷ 0.008
The 835551.736111… multiplied by 506,316, afterwards divided by the below: 7325.1736111 …= 57,753,336 the 506,316 is 19.5337962962… Platonic Years of 25,920 and this 19.5337962962… divided by the above 0.014467592592… obtains 1350.176 this, divided by the longer value Greek Foot of 1.01376 = 1331.8497474… which divided by the 19.5337962962.. = 68.1818… that multiplied by 5.5 will regain the above 375 Accordingly, the AA: 57,753,336 is 2228.137962962 …Platonic “years” of 25,920 the 2228.137962962…. divided by the 19.5337962962…. subsequently multiplied by the 506,316 = 57,753,336 Therefore, employing the AA: Moon: 57,753,336 is an easier system Hence 1 ÷ 57,753,336 x 91,379,493,969,300,000 = 1,582,237,500 the AA: Earth eastward
The 91,379,493,969,300,000 is the AA: 12,474,720,576,000 x 7325.1736111… Then 7325.1736111… x 25,920 = 189,868,500 which is the 1,582,237,500 x 0.12 via this approach the product will forever be 0.12 Hence 1 ÷ 57,753,336 x 8,464,559,937,504 = 146,564 the AA: Saturn the 8,464,559,937,504 is the AA: 12,474,720,576,000 x 0.6785370370… this multiplied by 25,920 = 17587.68 which is the 146,564 x 0.12
As an aside: AA: Saturn 146,564 is 25,920 x 5.654475308… that divided by the Rhind Papyrus {256:81} of 3.160493827… = 1.789111328125 that multiplied by 9830.4 will regain the 17587.68 The 9830.4 x 1,268,994,199.21875 = 12,474,720,576,000 The 1268994199.21875 ÷ 146,564 ÷ 23.7047445679957…= 365.255738463…a tolerable Earth Year that multiplied by 360,000 Degrees then divided by the Mile of 5280 Feet obtains an Earth Equatorial Circumference of 24903.800349804… Miles. The 23.7047445679957…is Pi x 7.5454 which divided by the Egyptian Foot of 1.14545… = 6.587301… that divided by the Eye of Horus fraction {64:63} of 1.015873… will obtain exactly 6.484375, the longer value Egyptian Foot of 1.152 increased to 115.2 x 6.484375 = 747
This 747 pertains to Plato’s Law No: 747 {Laws V1}
“numbers have a use in respect of all the variations of which they are susceptible, both in themselves and as measures of height and depth, and in all sounds and in motions, as well as those which proceed in a straight direction, upwards or downwards, as in those which go round and round”
And 747 minus 343 {7 x 7 x 7} obtains 404 that plus 100 = 504, and as mentioned prior: “The number 5040 is the most favoured number to be employed within the Utopian City-State as envisioned by Plato namely: Magnesia
Anyway, to resume with: 1 ÷ 57,753,336 I may as well indicate the remainder Hence 1 ÷ 57,753,336 x 21,035,151,051,264 = 364,224 AA: Jupiter The 21,035,151,051,264 is the AA: sky 12,474,720,576,000 x 1.686222… subsequently 1.686222… x 25,920 = 43706.88 which is the 364,224 x 0.12 Hence 1 ÷ 57,753,336 x 132,649,248,204,864 = 2,296,824 AA: Mars The 132,649,248,204,864 is the AA: 12,474,720,576,000 x 10.633444… subsequently 10.633444… x 25,920 = 275,618.88 which is the 2,296,824 x 0.12
Various Gem {Gematria Value} interconnections
Gem: {Gematria Value} Only begotten son is 1176 The GP base perimeter of 3024 minus 1176 = 1848 that minus Gem: Divine Circle: 1224 = 624 this is 2 x 312, the Maya Third Sun the Maya Long Count {days} 1,360,800 x 144,000,000 = 195,955,200,000,000 Nineveh Actual Pi minus 0.1176 x 1000 = 3023.99265358…tolerable as 3024, the GP base perimeter Yet 10,000,000 divided by the AA: “Yuga the revolutions of the Sun are 4,320,000” = 2.3148148… this divided by the FP {Pythagorean Limma: {256:243} obtains exactly 2.197265625, the GP base perimeter of 3024 ÷ 2.197265625 = 1376.256 The 1376.256 x 142,382,812,500 = 195,955,200,000,000: the 142,382,812,500 multiplied by the FP {256 ÷ 243} obtains exactly 150,000,000,000
It follows that the AA: “Revolutions of the Moon in a Yuga {I, 1}…that is to say, 57,753,336 ” divided by the 2.3148148…. = 24949441.152, that multiplied by Pi then divided by the Platonic 25,920 = 3023.9576016…an improved version with reference to the GP base perimeter of 3024 in addition to 3023.9576016…divided by the 2.3148148… divided by Pi = 415.8240192, which is 7.2 x 57.753336 otherwise the 24949441.152 x 60,000 = 1,496,966,469,120 which is 25,920 x 57,753,336 the AA: Moon Hence 2.3148148… is the Platonic Year: 25,920 divided by exactly 11197.44 Otherwise 60,000 ÷ 25,920 = 2.3148148… this multiplied by 846526.464 = 1,959,552 Nineveh reduced, in addition to 846526.464 is the Platonic: 25,920 x 32.6592, that multiplied by the 2.3148148… = 75.6, the GP base side length decreased
The Maya 5256 divided by the 2.3148148… = 2270.592, which divided by the above 11197.44 = 0.202777… this multiplied by 1800 = 365 which is Gem: Abraxas, Mithras Hence 0.202777… will obtain: 584, 1460, 2190, 1,366,560, for example the Maya Synodic period of Venus: 584 ÷ 0.202777… = 2880 which divided by the 2.3148148… = 1244.16 which is the NC Earth mean Circumference of 24883.2 ÷ 20
In that case we employ: 2.3148148…
{1} 2.3148148… x 1,866,240 = 4,320,000 the AA: Sun
{2} 2.3148148… x 24,949,441.152 = 57,753,336 : the AA: Moon
{3} 2.3148148… x 683,526,600 = 1,582,237,500 the AA: Earth eastward:
{4} 2.3148148… x 63,315.648 = 146,564 the AA: Saturn
{5} 2.3148148… x 157,344.768 = 364,224 the AA: Jupiter
{6} 2.3148148… x 992,227.968 =: 2,296,824: the AA: Mars
{7} 2.3148148… x 5,389,079,288,832 = 12,474,720,576,000 the AA: sky Circumference
{Please see the numbers: ,1866,240, the 24,949,441.152, the 683,526,600, etc below}
As we observe prior a reduced Nineveh Constant of 1,959,552 ÷ 2.3148148… = 846,526.464 which divided by the Platonic: 25,920 = 32.6592 which divided by the GP base perimeter of 3024 = 0.0108, which is a reduced Lunar, Yin-Yang number: 1080 As we observe prior: the Platonic 25,920 is 2.3148148.. multiplied by 11197.44 and this divided by the GP base perimeter of 3024 = 3.70285714 ….which multiplied by 7,000 = 25,920
The 3.70285714 ..minus 0.56 = 3.142857… afterward 3.70285714 …minus 1.98857142… = 1.714285 {12 ÷ 7} …The 1.98857142… plus 1.15428571… = 3.142857…, The 1.15428571… x 0.875 = 1.01… the 1.1542857…1 x 2.72277… = 3.142857…, then the 2.72277… x 0.367272… = 1 The 0.367272… x 1.1 = 0.404 We observe 0.404 prior albeit as: “And 747 minus 343 {7 x 7 x 7} obtains 404 that plus 100 = 504, and as mentioned prior: “The number 5040 is the most favoured number to be employed within the Utopian City-State as envisioned by Plato namely: Magnesia”
It follows that the above figures equate as follows: the Nineveh Constant reduced to 19,595,520 x 2.3148148 = 45,360,000
{1A} the prior 1,866,240 ÷ 25,920 = 72 the 45,360,000 x 72 ÷ 25,920 = 126,000 that divided by the AA: Sun: 4,320,000 = 0.0291666…
{2A} the prior 24,949,441.152 ÷ 25,920 = 962.5556 the 45,360,000 x 962.5556 ÷ 25,920 = 1,684,472.3 that divided by the AA: Moon 57,753,336 = 0.0291666…
{3A} the prior 683,526,600 ÷ 25,920 = 26370.625 the 45,360,000 x 26,370.625 ÷ 2,5920 = 4,6148,593.75 that divided by the AA: Earth: eastward 1,582,237,500 = 0.0291666…
{4A} the prior 63315.648 ÷ 25,920 = 2.4427333 the 45,360,000 x 2.4427333 ÷ 25,920 = 4274.78333 that divided by the AA: Saturn: 146,564 = 0.0291666…
{5A} the prior 157344.768 ÷ 25,920 = 6.0704 the 45,360,000 x 6.0704 ÷ 25,920 = 10623.2 that divided by the AA: Jupiter: 364,224 = 0.0291666…
{6A} the prior 992227.968 ÷ 25,920 = 38.2804 the 45,360,000 x 38.2804 ÷ 25,920 = 66990.7 that divided by the AA: Mars: 2,296,824 = 0.0291666…
{7A} the prior 538,9079,288,832 ÷ 25,920 = 20,7912,009.6 the 45,360,000 x 20,7912,009.6 ÷ 25,920 = 36,3846,016,800 that divided by the AA: sky Circumference: 12,474,720,576,000 = 0.0291666…
As we observe prior: the Nineveh Constant 195,955,200,000,000 ÷ 25,920 = 7,560,000,000 Subsequently 1 ÷ 0.0291666… = 34.285714… one seventh of 240 Otherwise the previous continuous product of 0.12 ÷ 0.0291666… = 4.1142857… one seventh of 28.8 Hence the 34.285714… x 168 = 5760 the GP height {Inches} likewise 34.285714… x 264.6 = 9072 the GP base perimeter {Inches} the GP base area: 571536 is 2160 x 264.6
Music of the Spheres
We will merge numbers about AA {Àryabhatiya of Àryabhata} amid the Nineveh Constant in addition to various Egyptian Greek, Roman, etc units of measure; nonetheless it is unattainable to indicate all interconnections since there are myriad
The AA sky: 12,474,720,576,000 divided by a reduced Nineveh Constant of: 1,959,552 = 6,366,108.465608 and this divided by 10,742,808.03571428… = 0.592592… the cyclic 592 impels the employ of the Platonic Precession Cycle: 25,920 The 10742808.03571428… divided by the AA: Moon: 57,753,336 = 0.18601190476… this multiplied by 16.990814814 …obtains 3.160493827… this is the Rhind {256:81} Yet the employed 16.990814814… divided by 16.128 = 1.05349794238683…this is the FP {musical scale: Pythagorean Limma 256:243} which is one third of the “Rhind”
Therefore the 0.592592… x 25,920 = 15,360 the 13th PWS member is 1536 and this 15,360 divided by the GP base perimeter of 3024 = 5.079365 that multiplied by exactly 269041.5 = 1,366,560 which is the Maya “Super Number”: 260 x 5256 Tzolkins, otherwise the Maya days: 1,360,800 plus 5760 = 1,366,560 The GP height is 5760 Inches: {480 x 12}, likewise Gem:{Gematria Value} Eagle, Pneuma {Breath} is 576 that divided by the above 0.592592… = 972 which is 1.285714… x 756, the GP base side length, the 1.285714… x 7 = 9 Likewise the Maya: 1,360,800 x 144,000,000 = 195,955,200,000,000: Nineveh As a result we currently observe interconnections regarding the AA, the Maya, the Nineveh Constant, the Platonic Precession, furthermore the GP and much more besides [please see 5.079365 and 0.592592… below] The 8th PWS member is 972
Musical Chairs
Likewise, the FP {Pythagorean Limma {256:243} entered the arena given that the above 15,360 multiplied by the “Limma” obtains 16181.728395061… please make a note of the highlighted 1.728 since the cubic Inches in one cubic Yard is 46,656 {1728 x 27} Accordingly 46,656 x 16181.728395061… = 754,974,720 this is 25,920 x 26214.4 x 1.111… which is one ninth of 10 Mentioned prior is the ancient Comma of Pythagoras {531,441:524,288} the Comma component, namely 524,288 is 20 x 26214.4 Likewise, via examining the Timaeus of Plato, I discovered the “Whole” via which the World-Soul is formed to be 46,656 on the other hand, leaving aside appellations such as Feet, Inches, Area, Seconds etc, the 46,656 is 1.8 = 25,920 otherwise 46,656 x 1.851851… = 86,400 Seconds in 24 hours The 1.851851… x 27 = 50, thus 1.851851… x 0.54 = 1
There is, at this time, a requirement to validate the “numbers” employed Prior we observe: “The 10742808.03571428… divided by the AA: Moon: 57,753,336 = 0.18601190476…” The 10742808.03571428… is one seventh of 75,199,656.25 that divided by the prior 5.079365 = 14,804,932.32421875 this multiplied by 842605.714285… obtains the AA; sky: 12,474,720,576,000 the 842,605.714285… x 7 = 5,898,240 this divided by the prior 0.592592… = 9,953,280 which is the Seconds in 24 hours: 86400 x 115.2
The longer value Egyptian Foot is 1.152 however the 115.2 x 17,010 = 1,959,552 Nineveh reduced The 17,010 divided by the Mile of 5280 = 3.22159090… this divided by the Egyptian Foot of 1.14545 then by 0.9920634… = 2.835 that multiplied by the Pythagorean Limma {256:243} obtains 2.9866 that multiplied by 1012.5 = 3024, the GP base perimeter The 0.9920634… x 74,844 = 74,250 that divided by the Square Feet in one Acre: 43,560 = 1.704545… that multiplied by the Nineveh reduced 1,959,552 = 3340145.4545this divided by the 3.22159090… = 1,036,800 the end PWS member is 10,368 and 10,368 x 189 = 1,959,552 Nineveh reduced The employed 74,844 divided by the GP base perimeter of 3024 = 24.75 that multiplied by 245.6565… = 6080, the Sea Mile The 245.6565… divided by the above 0.592592… = 414.5454… the “Land” Mile of 5280 divided by this 414.5454… then multiplied by the prime number, and Gem: Eve, namely 19 = 242, this obviously obtains: 242 x 20 = 4840 Square Yards in one Acre, 242 x 180 = 43,560, Square Feet in one Acre
GP base area
We observe prior: “1,4804,932.32421875 this multiplied by 842,605.714285 …will obtain the AA; sky: 12,474,720,576,000” The Nineveh Constant: 195,955,200,000,000 x 242 = 47,421,158,400,000,000 that divided by the GP base area {756 x 756} of 571,536 = 82971428571.428571…. that divided by the 842605.714285… = 98470.05208333… this divided by the Greek Mile of 5068.8 = 19.4266990379050925925 that divided by the likewise above: 0.592592 obtains exactly 32.78255462646484375 The likewise above 1012.5 ÷ 32.78255462646484375 = 30.8853294545… that multiplied by 13.422082971643518518… = 414.5454…
The 13.4220829716435185185… x 1.769472 = 23.75 that multiplied by 256 = 6080, the Sea Mile The Nineveh Constant: 195,955,200,000,000 divided by this 414.5454… afterwards multiplied by the prime number, and Gem: Eve, namely 19 = 898,1280,000,000 which divided by the Square Feet in one Acre: 43,560 = 206181818.1818… which is 120,000,000 Egyptian Cubits of 1.71818… Hence 206181818.1818… x 950,400 = 195,955,200,000,000
The 950,400 is 25,920 x 36.666… one third of 110 {obviously 5280 ÷ 48} The 950,400 is 86,400 x 11 and since we espy the Plato-Magnesia favourite number 5040, then 950,400 is 5040 x 188.571428… one seventh of 1320, {obviously 5280 ÷ 4} The employed 1012.5 divided by Gem: Tree of Life: 1625 = 0.6230769… The prior 5.079365 divided by this 0.6230769 …= 8.15206741132667…please note the highlighted 1520674 that suggests the Greek Cubit of 1.52064 Accordingly, 8.15206741132667…x 1.52064 = 12.396359788… that divided by the Rhind Papyrus {256:81} of 3.160493827… = 3.922285714… that divided by the Schooldays Pi {22 ÷ 7} of 3.142857… = 1.248 which, to be brief is 13 x 0.096
Maya Connection
In addition to 13 will obtain: 26: YHWH……312 Maya Third Sun…..1625: Tree of Life…… furthermore 13 divided by a reduced: Nineveh of 1.959552 then multiplied by 152.6189538461… will regain the 1012.5 as we observe above Then 152.6189538461… multiplied by the Maya 5256 then by Gem: Tree of Life: 1625 = 1,30,3518,484.8 this divided by Gem: Founder of the City: 1225 = 1,064,096.722285714… which is the Schooldays Pi of: 3.142857… x 338576.2298181…, which is the Egyptian Cubit of 1.71818… x 197054.94857142… which divided by the Egyptian “Royal” Cubit of 1.714285… = 114,948.72 this divided by the Eye of Horus fraction {64:63} of 1.015873… obtains 113,152.64625 which divided by exactly 35006.59993359375 = 3.2323… this is the Sea Mile of 6080 divided by 1881: which is the sloping floor line length of the magnificent Grand Gallery, literally “entombed” within the GP {Great Pyramid} Otherwise, whilst this floor line length of 1881 is challenged, which is fine by me, in that case 1881 is the prime number and Gem: Eve: 19 x 99 Otherwise 1881 is Gem: Weapons of god: 665 x 2.8285714… = 1881 The 2.8285714… is one seventh of exactly 19.8, hence 2.8285714… will obtain: 237,600…3168…3960…43,560…633.6…7920…etc
The NC: Earth to Moon: 237,600 ÷ 2.8285714… = 84,000, likewise the NC Earth to Sun: 93,312,000 ÷ 2.8285714… = 32989090.9090… one eleventh of 362,880,000 although this divided by the GP base perimeter of 3024 = 120,000 the 362,880,000 divided by the NC: Moon diameter in Feet of: 11,404,800 = 31.8181… this is 7 ÷ 0.22 also 36,2880,000 x 540,000 = 195,955,200,000,000: Nineveh Subsequently the above 35006.59993359375 multiplied by the GP Tan {480 ÷ 378} of 1.269841… = 44452.8253125 that multiplied by 2.5858… regains the commencing 114948.72 the employed 2.5858… x 202,752 = 524,288 pertaining to the Comma {53,1441:524,288} the 202,752 x 9.66477272… = 1,959,552 a reduced Nineveh the 9.66477272… x 2681.904761… = 25,920 the 2681.904761… divided by the Rhind {256:81} of 3.160493827… = 848.571428
As indicated prior the Maya Long Count days: 1,360,800 x 144,000,000 = 195,955,200,000,000 the Nineveh Constant The Maya “Super Number” 1,366,560 x 1,360,800 = 1,859,614,848,000 this is the 195,955,200,000,000 x 0.00949 that multiplied by 1369.86301369… = 13 The Maya Third Sun: 312 is 24 x 13 The 1369.86301369… x 4.4384 = 6080, the “Sea” Mile A reduced Roman Mile {4838.4} of 4.8384 minus 4.4384 = 0.4
The Numerical Canon
The NC distance re Earth to Sun: 9,3312,000 Miles divided by the 848.571428 = 109963.6363… which is 64000 Egyptian Cubits of 1.71818… The reduced Nineveh: 19,59,552 is 1.71818 …x 1,140,480 this is one tenth of 11,404,800 which is the NC Moon diameter, since 11,404,800 is 5280 x 2160 The NC distance re Earth to Moon in Miles: 237,600 ÷ 848.571428 …= 280 The NC Earth Circumference: in Feet 131,383,296 ÷ 848.571428… = 154,828.8 this resembles the “Comma” component {531,441:524,288} namely 524,288 As a result 154,828.8 ÷ 524,288 = 0.2953125 that multiplied by 10,240 = 3024 which is the GP base perimeter, the 10240 is 40 x 256 which is a component of the “Limma” {256:243} in addition to the “Rhind” {256:81} likewise the 9th PWS member is 1024
Atlantis
The Plato given Atlantis area, minus the Central Shrine: 4,561,920,000 ÷ 848.571428 ..= 5,376,000 that divided by the GP base perimeter of 3024 = 1777.777…, the Square root of the Rhind “Pi” {256:81} of 3.160493827… is 1.777… The Nineveh Constant: 195,955,200,000,000 ÷ 4,561,920,000 = 42954.5454… which is 37,500 Egyptian Feet of 1.14545 …Then again: 4,561,920 minus 1,959,552 = 2,602,368, which is 2268 x 1147.428571… that divided by Pi then multiplied by 68.1818… {360,000 ÷ 5280} obtains an Earth Equatorial Circumference of 24902.5812255994…Miles and this multiplied by Gem: Founder of the City: 1225 afterwards multiplied by Pi = 95836363.6363… one eleventh of 1054,200,000 this divided by the Plato-Magnesia number: 5040 = 209166.666… that multiplied by 3.142857… x 200 = 131,476,190.476190… that divided by 5280 = 24900.793650… a tolerable Earth Equatorial Circumference Nonetheless the 131,476,190.476190 …divided by the Rhind fraction {256:81} of 3.160493827… = 41,599,888.39285714… that divided by the 1147.428571… obtains 36254.8828125 that multiplied by 5404932654.5454… = 195,955,200,000,000: Nineveh
The 5404932654.5454… divided by the commencing: 42954.5454… = 125,829.12, this divided by the NC Earth Circumference of 24883.2 = 5.0567901234… the first PWS member: 384 divided by the 5.0567901234… = 75.9375 that multiplied by the FP {256:243} component: 256 = 19,440, the 15th PWS member is 1944 and 1944 x 1.333 = 2592, the 19th member The prior 1147.428571… minus the likewise prior 848.571428… = 298.857142… one seventh of exactly 2092, this resembles the NC Earth: Feet of 20,901,888 which is 24883.2 x 840, or the NC {Numerical Canon} Earth mean Circumference {Feet} of 131,383,296 ÷ 20,901,888 = 6.285714… obviously 2 x 3.142857As a result the 2092 x 10,000 x 6.285714… = 131,497,142.857142… which is: 5280 x 24904.761904… that divided by the Rhind {256:81} of 3.160493827… = 7880.02232142857… that divided by the 298.857142 …= 26.3671875 that multiplied by the 1st PWS member: 384 = 10,125, which we observe prior albeit as 1012.5 The foregoing, although providing, hopefully, an insight with reference to the myriad interconnections, is not the least bit the sole sequence to employ, in view of the fact that each and every one of the “numbers” are totally interconnected
Blending the Planets
Professor Clark: “In a Yuga the revolutions of the Sun are 4.320,000 of the Moon: 57,753,336 of the Earth eastward: 1,582,237,500 of Saturn: 146,564 of Jupiter: 364,224 of Mars: 2,296,824…”
Professor Clark: “for the Circumference of the sky…it works out as 12,474,720,576,000
{A} Nineveh reduced: 1,959,552 is the AA: Sun: 4,320,000 x 0.4536 this is: 0.2592 x 1.75 The Saturn: 146,564 ÷ 1.75 = 83750.857142… this divided by the Schooldays Pi {22 ÷ 7} of 3.142857… = 26,648, this divided by the GP base perimeter of 3024 = 8.8121693… that multiplied by the: Nineveh reduced 1,959,552 = 17,267,904 this is the Platonic Precession: 25,920 x 666.2 that multiplied by 220 = 146,564 {Saturn}
{B} Nineveh: 195,955,200,000,000 divided by the AA: Earth eastward: 1,582,237,500 ÷ 25,920 then multiplied by 30674.47843253968.. = 146,564 {Saturn} The 30674.47843253968… divided by the GP base perimeter of 3024 afterwards divided by the {A} 8.8121693 = 1.15109871031746The shorter value Roman Cubit of 1.45152 increased to 145,152 x 1.15109871031746 = 167,084.28 and this divided by the Saturn: 146,564 = 1.14000900630441…that multiplied by the 30674.47843253968 = 34969.1816767857142… one seventh of 244784.2717375 The AA: Earth eastward: 1,582,237,500 divided by the 244784.2717375 then multiplied by 56.348245184… = 364,224 {of Jupiter}
The 56.348245184… divided by the Roman Cubit of 1.4598144 then multiplied by the AA: Mars: 2,296,824 = 88,656,477.0812615740740… this multiplied by 243 = 2,1543,523,930.7465625 this multiplied by 0.868735227515119……= 18,715,618,163.4545 …that divided by the Egyptian Foot of 1.14545… = 16,339,031,730 The 0.868735227515119…multiplied by the above 1.15109871031746… = The 16,339,031,730 divided by the AA: Circumference of the sky: 12,474,720,576,000 then multiplied by 16,448,308,914,474 will regain the above: 21,543,523,930.7465625 The 16,448,308,914,474 is 25,920 x 86,400 x 5280 x 1.391036710927 …
The reduced Nineveh ,,552 x 1.391036710927…x 5280 = 14,392,270,300.16475 this divided by the 21,543,523,930.7465625 then multiplied 86,449,903.333… = 57,753,336 the AA: Moon The 86,449,903.333 …divided by the AA: Mars: 2,296,824 = 37.6388 that multiplied by 360 = 13,550, and then the 364,224 {of Jupiter} divided by this 13,550 = 26.88 this multiplied by 112.5 = 3024, the GP base perimeter The reduced Nineveh 1,959,552 divided by the AA: Moon: 57,753,336 = 0.033929676374019…The GP base perimeter of 3024 ÷ 0.033929676374019…= 89125.518518… this divided by the musical scale: Pythagorean Minor Third {32:27} of 1.185185… = 75199.65625 this multiplied by the Grand Gallery 1881 then divided by the Sea Mile of 6080 = 23264.89365234375 The AA: Moon: 57,753,336 ÷ 23264.89365234375 = 2482.4242… this multiplied by the Modern-day Furlong of 660 Feet = 1,638,400 this divided by the end PWS member: 10,368 = 158.024691358 …this divided by the Eye of Horus fraction {64:63} of 1.015873… = 155.555… which is 7 ÷ 0.045 or 155.555… x 9 = 1400
Maya Calendar
The Maya indicates a period of 23,040,000,000 days {alautun} acknowledged via suitable scholars, this is 1,152,000,000 Kinchiltun Likewise the Maya “Long Count” is 1,872,000 days {260 x 7,200}, however of more import: the Maya Long Count days: 1,360,800 added to the GP 5760 = 1,366,560 the Maya “Super Number” the 18th PWS member is 2304 The GP height is 480 Feet or 5760 Inches, Gem: Eagle, Pneuma {Breath} is 576 The GP height: 480 x 480 = 230,400 in line with the Maya: 2,3040,000,000, the Square root of this is 151789.327688082207…that multiplied by exactly 0.1640625 = 24902.9365738259…a tolerable Earth Equatorial Circumference {Miles} that multiplied by the Modern-day Mile of 5280 then divided by 360,000 Degrees obtains an Earth Year of 365.243069749…
The 24902.9365738259…divided by 6.095238… = 4085.63803164332…which is exactly 0.02691650390625 x 151789.327688082…Subsequently half the GP base side length {756} of 378 ÷ 151789.327688082…= 0.00249029365738259…that obviously multiplied by 10000000 will regain the “Earth” of 24902.9365738259…Exactly 62.015625 ÷ 0.00249029365738259…= 24902.9365738259…The above 0.02691650390625 x 6.095238… = 0.1640625 this multiplied by 2304 = 378 Hence 0.1640625 x 378 = 62.015625 Hence 378 ÷ 62.015625 = 6.095238… this divided by the Rhind Papyrus {256:81} of 3.160493827… = 1.9285714… which is one seventh of 13.5
Then 4581822.675 ÷ 1.9285714… = 2375759.90555 …this divided by 1440.32777… = 1649.457812 …this multiplied by 1,892,158,107,840 = 3,121,034,973,382,080 The employed 1440.32777… divided by 666 = 2.162654320987…, which divided by the FP {Pythagorean Limma 256:243} obtains 2.05283203125 Otherwise the “Limma” component: 243 x 2.162654320987… = 525.525 exactly, which is 10.01 x 52.5 that multiplied by 14.4 = 756 the GP base side length However, as we observe prior, the GP base side length of 756 multiplied by the Platonic Precession Cycle of 25,920 = 19,595,520 this divided by the aforementioned Maya Long Count days: 1,360,800 = 14.4 The 19,595,520 is, noticeably the Nineveh Constant abridged The 19,595,520 divided by the 26th PWS member: 4608 = 4252.5 that divided by the above 0.1640625 = 25,920
Plato’s World Series
As indicated, the 26th PWS member is 4608 the 13th member is 1536 the PWS: {Plato’s World Series discussed anon previously, the 18th member is 2304 this we observe prior as 23,040,000,000 days {alautun} The 1,360,800 days: Maya Long Count {9.9.0.0.0} divided by 46,080 = 29.53125 that multiplied by 12.8 = 378 The 1,360,800 days indicating 400 Saturn and 378 Jupiter Synodic orbits it is also interesting that: 1,360,800 ÷ 400 = 3402 which is 9 x 378 hence the GP base side length of 9072 {Inches} is 3402 x 2.666… which is one third of 8
Professor Clark: “In a Yuga the revolutions of the Sun are 4.320,000 of the Moon: 57,753,336 of the Earth eastward: 1,582,237,500 of Saturn: 146,564 of Jupiter: 364,224 of Mars: 2,296,824…”
The AA: Saturn: 146,564
The 146,564 divided by the 3402 then divided by the Pythagorean Limma {256:243} obtains 40.8939732142857… one seventh of 286.2578125 that divided by the above 29.53125 = 9.69338624… the cyclic 338624 tail ends with 624 which is 2 x 312, the Maya Third Sun Hence the Platonic year of 25,920 ÷ 312 x 9.69338624… = 805.296703… that is the AA: Saturn: 146,564 divided by 182 which is one half of the Maya: 364, furthermore the Maya temples four sides of 91 steps each, plus the top platform itself counting as 1 obtains: 91 x 4 = 364 that added to the 1 = 365 this is Gem Abraxas, Mithras both Solar “gods” 365 also obtains the Maya numbers, 584, 1460, 5256, 18,980, 1,366,560 and so on
The AA: Jupiter: 364,224
The 1,360,800 days: Maya Long Count {9.9.0.0.0} divided by 378 = 3,600 Then: 364,224 ÷ 3,600 = 101.17333… that divided by the Pythagorean Limma {256:243} obtains exactly 96.035625 that divided by the above 29.53125 = 3.252 The Platonic year: 25,920 x 3.252 = 84291.84 which divided by the AA: Jupiter: 364,224 = 0.23142857… one seventh of 1.62, hence 0.23142857… will obtain: 324, 648, 972, 1134, 1296, 1944, 2592 etc
Likewise, 0.23142857… x 8,467,200 = 1,959,552 Nineveh reduced, the 8,467,200 is 25,920 x 326.666… The 1,360,800 days divided by this 326.666 …= 4165.714285… one seventh of 29,160 ample scope, in addition to the 20th PWS member is 2916 Then again, the GP height of 5760 divided by the above 3402 = 1.693121.. that multiplied by 907.2 = 1536, the 13th PWS member is 1536 as motioned above regarding PWS The 907.2 is 2.4 x 378 Likewise, the GP base area will be: 756 x 756 = 571,536 this plus 789,264 = 1,360,800 the Maya Long Count days The 789,264 minus 646,380 = 142,884 the Square root of this is 378 The 646,380 minus 571,536 = 74,844 we observe this prior as: “The employed 74,844 divided by the GP base perimeter of 3024 = 24.75 that multiplied by 245.6565… = 6080, the Sea Mile”
The opening two words of the Hebrew Torah: Bereshit Bara, obtained the Gem {Gematria Value} 1116, Bereshit Bara translates roughly as “In the beginning {god} created” The prior 74,844 minus 1116 = 73,728 which is 32 x 2304 the 18th PWS member The 789,264 also equates as: 789 plus 264 = 1053 the abridged Nineveh Constant 1,959,552 ÷ 1053 = 1860.923076… that multiplied by Gem: YHWH: 26 = 48,384 which is 10 Roman Miles of 4838.4, hence the 789,264 minus 3,600 = 785,664 which is 2304 x 341 The Tetractys number total: 8436 x 341 = 2,876,676 this divided by the FP {Pythagorean Limma: 256:243} obtains 2,730,594.796875 which is the 1053 x 2593.157451923076 The Nineveh 1,959,552 ÷ 2593.157451923076 = 755.662560538621…this commences with 755 hinting at the GP base side length of 756 Accordingly 755.662560538621…x 4.00178619123…= 3024 this is the GP base perimeter {756 x 4} The 4.00178619123… multiplied by 435,456 afterwards by the Maya Third Sun: 312 = 543,691,764
The GP base area {756 x 756} of 571,536 minus the 435,456 = 136,080 the Maya Long Count reduced. Likewise, the GP height of 5760 x 756 = 4,354,560 which is the Platonic Precession Cycle of 25,920 divided by exactly 168 that multiplied by 4.5 = 756 As we observe prior: “1649.457812… this multiplied by 1,892,158,107,840 will obtain exactly 3,121,034,973,382,080”
The 1649.457812… x 3024 = 4,987,960.424872 Then exactly 2,850,260,249,664 ÷ 4987960.424872 = 571,428 which is 666 x 858 which is the Mile of 5280 x 0.1625 this is a reduced Gem: Tree of Life: 1625
Astronomical Unit
At this time, we will employ the Greek Mile of 5068.8, the musical scale: Pythagorean Limma {256:243} The Rhind Papyrus {256:81} and much more One AU {Astronomical Unit} being the average distance between the Sun and the Earth, An AU is in the region of the length of the semi-major axis of the Earth’s elliptical orbit, dependent on which source is examined; it is usually given as 149,597,870.691 otherwise 149,597,870.6913241 Kilometre’s
Accordingly, 1 AU = 1 ÷ 0.3048 = 3.28083989501…x 149,597,870.691 ÷ 5.28 = 92,955,807.2674331901…Miles
Otherwise, 1 AU = 1 ÷ 0.3048 = 3.28083989501…x 149,597,870.6913241 ÷ 5.28 = 92,955,807.26763457657…Miles
The Nineveh Constant 195,955,200,000,000 ÷ 5.28 ÷ 4297.505270270… ÷ 57.7272… = 149,597,870.7094974042…then 0.2 ÷ 11 = 0.01818… the 149,597,870.7094974042…minus 0.01818…. = 14,9597,870.69131558601…more than adequate I would imagine The 4297.505270270… multiplied by the 57.7272… = 248,083.258783783… which is: 5.28 x 46985.4656787469… We will presently observe exactly 27.8784, the Nineveh: Constant 195,955,200,000,000 divided by the 46985.4656787469…= 4,170,549,278.78765243326…this is 27.8784 x 149,597,870.7094974042…as above. The 27.8784 x 0.1818 …= 5.0688 this is a reduced Greek Mile of 5068.8 and the 0.1818… is the above 0.01818… x 10 The employed 57.72727… x 11 = 635 that minus 270 = 365 Gem: Abraxas, Mithras The Egyptian Stade {500 ft.} is 572.7272…
The 4297.505270270 …x 219.78 = 944,505.7083 this for example, divided by the Greek Mile of 5068.8 = 186.337142578125 {the 219.78 is 666 x 0.33} The FP x 186.337142578125 = 196.3057962962… the cyclic 962 hints at the employ of 1296 As a result 1296 x 196.3057962962… = 254,412.312 within the digits of this exact 254,412.312 we espy 312, which is the Maya Third Sun, also Gem: YHWH: 26 x 12 = 312 Otherwise, Gem: Tree of Life: 1625 is Gem: YHWH: 26 x 62.5; accordingly, we will employ 5256 Maya Tzolkins which multiplied by 260 = 1,366,560, the Maya “Super Number”
As a result, 254412.312 ÷ 5256 = 48.40416894977… at this time within the cyclic we espy Gem: 1689: Chrysolite, moreover 1689 is Jasper: 501 + 2190 Gem: Son of Man, furthermore 2190 is 6 x 365 Gem: Abraxas, Mithras, in addition to 365 x 4 = 1460, the Sothic Cycle hence 1689 x 48.40416894977… = 81,754.64135616438… this divided by the commencing 4297.505270270 …= 19.023744292… that multiplied by 365 = 6943.666… one third of exactly 20831 this divided by the Sothic Cycle: 1460 = 14.2678082191… that multiplied by the cubic Inches in one cubic Yard: 46,656 = 665,678.860273972 …
The 81,754.64135616438… divided by the 665,678.860273972… then multiplied by the Greek Mile of 5068.8 = 622.5192819486152… which divided by the 48.40416894977 …= 12.860860… this multiplied by 5494.5 = 70,664 this divided by the prior 27.8784 = 2534.7222… this divided by Gem: Simon Peter: 1925 = 1.3167388… that multiplied by exactly 443.52 = 584, the Maya Synodic period of Venus The 5494.5 divided by this 443.52 = 12.38839285714… that multiplied by 53.76 = 666 The Greek Mile of 5068.8 divided by this employed 53.76 = 94.285714… one seventh of 660 the Modern-day Furlong Although the 53.76 divided by the longer value Greek Foot of 1.01376 = 53.0303… the Modern-day Mile of 5280 ÷ 53.0303… = 99.565714285… that is the 443.52 ÷ 4.4545… one eleventh of 49 and 49 x 25 = 1225 Gem: Founder of the City
One version of “Pi” is 864 ÷ 275 = 3.14181… this divided by the 4.454…5 then multiplied by Gem: Founder of the City: 1225 = 864 The Schooldays Pi {22 ÷ 7} of 3.142857… x 31.68 = 99.565714285… as above The 31.68 a reduced 3168: Gem Lord Jesus Christ, however 3168 x 1.666… = 5280 and the 1.666… x 3 = 5
Various Units of Measure
A sample re the properties of the 27.8784 given that it is impossible to indicate the entire The Roman Mile of 4866.048 ÷ 27.8784 = 174.5454 this divided by the Egyptian Foot of 1.14545… = 152.380952 that divided by the Rhind {256:81} of 3.160493827… = 48.2142857… the GP height of 5760 is 33 x 174.5454… that divided by 100 then by the Egyptian Cubit of 1.71818… = 1.015873… which is the Eye of Horus fraction {64:63} The 48.2142857… divided by the Schooldays Pi {22 ÷ 7} of 3.142857… = 15.340909… this divided by the Greek Foot {441:440} of 1.0022727… then multiplied by Gem: Founder of the City: 1225 = 18,750 which multiplied by a reduced Gem: Mary {192} of 0.0192 = 360 Then again Gem: Eagle, Pneuma {Breath} 576 divided by the 174.5454 = 3.3 Then 18750 divided by 3.3 then multiplied by the Yards in one Mile: 1760 = 10,000,000
The Roman Mile of 4838.4 ÷ 27.8784 = 173.55371900826… this multiplied by 3.9826285714… = 691.2 which is the Egyptian Stade, likewise the 31st PWS member is 6912 The 3.9826285714… x 7 = 27.8784 The 691.2 ÷ 27.8784 = 24.793388429…this multiplied by 0.40656 = 10.08 which is the Greek Foot The 27.8784 divided by the 0.40656 = 68.571428… which is one seventh of 480 The Greek Foot of 10.08 ÷ 27.8784 = 0. 3615702479338.. this multiplied by 1752.356571428… = 633.6 which is the Greek furlong The 1752.356571428… x 7 = 12266.496 which is 22.7272… x 539.725824…, this divided by the 27.8784 = 19.36 that multiplied by 250 = 4840 the Square Yards in one Acre
The Greek furlong of 633.6 ÷ 27.8784 = 22.7272… one eleventh of 250 The 27.8784 divided by the longer value Roman Foot of 0.9732096 = 28.6458333… that multiplied by 184.32 = 5280 the Modern-day Mile The 184.32 possesses myriad routes to follow since it is 259.2 x 0.7111 one ninth of 6.4 The 27.8784 divided by the shorter value Roman Foot of 0.96768 = 28.809523 this multiplied by 7 then by the above 184.32 = 37171.2 which divided by the NC Earth of 24883.2 then multiplied by 29,160 = 43,560, the Square Feet in one Acre
The 27.8784 divided by the Egyptian Foot of 1.14545… = 24.338285714 that multiplied by the 29,160 x 7 = 4,967,930.88 that multiplied by 2.2956841138659…= 11,404,800 the NC Moon diameter The Grand Gallery 1881 x 2.2956841138659…= 4318.1818… this multiplied by the Yards in one Mile: 1760 = 7,600,000 which is: 1250 x 6080 the Sea Mile The 27.8784 divided by the Egyptian Cubit of 1.71818 …= 16.225,523,809… that multiplied by the 7,600,000 x 567 = 69,919,027,200 which divided by the Sea Mile of 6080 then by the NC: Moon of 11,404,800 = 1.008333… that multiplied by 4800 = 4840 the Square Yards in one Acre The 27.8784 divided by the Greek Cubit of 1.52064 = 18.333 …that multiplied by 96 = 1760 Yards in one Mile
Numerical Canon
About the NC {Numerical Canon} the distance from Earth to Sun is 93,312,000 Miles The Nineveh Constant: 195,955,200,000,000 ÷ 93,312,000 = 2,100,000 The NC distance Earth to Moon is 237,600 Miles, the Nineveh Constant: 195,955,200,000,000 ÷ 237,600 = 824727272.7272… this is one eleventh of 9,072,000,000, the GP base side length is 9072 Inches Or 824727272.7272… is 4,800,000,00 x 1.71818… this is the Egyptian Cubit, the GP height is 480 Feet, or the shorter value Roman Cubit of 1.45152 increased to 14,515,200 x 56.8181 = 824727272.7272… The 56.8181… multiplied by the Yards in one Mile: 1760 = 100,000
The NC Moon diameter is 11,404,800 Feet The Nineveh Constant: 195,955,200,000,000 ÷ 11,404,800 = 17181818.1818… obviously 10,000,000 x 1.71818… the Egyptian Cubit However 1.71818… divided by the Greek Foot {441:440} of 1.0022727… = 1.714285… which is a Royal Egyptian Cubit Hence 195,955,200,000,000 ÷ 1.0022727… = 195,510,857,142,857.142857… that divided by the 1.714285… = 114,048,000,000,000 obviously the NC Moon diameter 11,404,800 increased The 195,510,857,142,857.142857 x 7 = 1,368,576,000,000,000 which is the Mile of 5280 x 259,200,000,000 an increased Platonic Precession
The 195,510,857,142,857.142857… is 62,208,000,000,000 x 3.142857… which is the Schooldays Pi The Nineveh: 195,955,200,000,000 divided by the Seconds in 24 hours: 86,400 = 2,2680,00,000, the 62,208,000,000,000 ÷ 2,268,000,000 = 27,428.571428…. which is one seventh of 192,000, the Gem Mary is 192 in addition to the British Admiralty, otherwise as I dub it Sea Mile of 6080 ÷ 192 = 31.666… this multiplied by 166.736842105… = 5280 obviously the “Land” Mile Gem: Eve: 19 x 166.736842105…= 3168, this is Gem: Lord Jesus Christ Hence the Nineveh: 195,955,200,000,000 divided by the 3168 = 61854545454.5454… that is 54,000,000,000 Egyptian Feet of 1.14545… or exactly 36,000,000,000Egyptian Cubits of 1.71818… Or 61854545454.5454… is 61,714,285,714.285714… Greek Cubits {441:440} of 1.0022727… The 61,714,285,714.285714 …x 7 = 432,000,000,000
The 195,955,200,000,000 divided by the British Admiralty Nautical Mile of 6080 Feet obtains 32229473684. 210526315….that divided by the prior 166.736842105…= 193295454.5454… that divided by the Egyptian Foot of 1.14545 = 168,750,000 that divided by the “Land” Mile of 5280 = 31960.22727… this divided by the Greek Foot of 1.0022727… = 31887.75510204081…this multiplied by Gem: Founder of the City 1225 = 39,062,500 Hence: the Nineveh: 195,955,200,000,000 divided by the 39,062,500 obtains exactly 5016453.12 that divided by the 166.736842105… = 30086.050909… which divided by the Egyptian Stade of 691.2 = 43.5272…7 that multiplied by the 39,062,500 = 1,700,284,090.9090…, this divided by the above 168,750,000 = 10.07575… that multiplied by 66 = 665 Gem: Weapons of god and 665 x 9.142857… = 6080, the Sea Mile the 9.142857… x 7 = 64
We observe prior: the “Royal” Egyptian Cubit 12 ÷ 7 = 1.714285… Hence: the Nineveh: 195,955,200,000,000 ÷ 1.714285… = 114,307,200,000,000 which is 2,59,200,000,000 x 441 that multiplied by 2 = 882, Gem: City of god and 882 x 10.285714… = 9072 the GP base side length the 10.285714… x 7 = 72 Within Nineveh Constant {Two} I will indicate the Nineveh: 195,955,200,000,000 sometimes abridged to 1,959,552 divided by numerous ancient and modern units of measure Gem {Gematria Values} etc
Until the next essay
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### gogateiiit's blog
By gogateiiit, history, 4 months ago, ,
Can anyone help me solve below problem?
Given an array with N numbers and given list of pair of integers u and v where each entry is two indices which cannot be together in chosen subset.Tell maximum possible size of such subset.
Don't worry about constraints tell your best solution.
» 4 months ago, # | 0 Maximum independent set. Polynomial (even linear) in some graphs.
• » » 4 months ago, # ^ | 0 Thanks AlexandruValeanu for the reference,Original problem which I simplified to above problem was given array with N numbers find subset such that product of any two numbers in subset is not cube. 1<=N<=100 and number in 1<=array<=1000000 and 1 is cube. Any thoughts on how to solve it.
• » » » 4 months ago, # ^ | ← Rev. 2 → +3 I think it's a bipartite graph, so Max independent set problem can be solved by creating the flow network and getting the max flow, so the answer to the problem is: Ans = TotalNodesNumber - MaxFlow
• » » » » 4 months ago, # ^ | 0 Can subset contain a cube? If it is so final graph can contain one clique of cubes in addition to bipartite part.
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support the project
# Calculation of Expansion Vessel
The height from the connection point of the tank to the highest water collection point
m
The maximum water pressure at the connection point of the expansion vessel (which is the maximum strength limit of the water supply system)
bar
Number of pumps
pcs
Maximum hourly water consumption
l/hour
Electric power of one pump
kW
Water reserve required:
Fill in this field only if having a reserve is more important than the frequency of the pump. For more information, please see below.
liters
## Calculation of Expansion Vessel
In water supply systems, an expansion vessel is used to solve various problems, and different methods are used to calculate it for each case. The above algorithm for calculating the expansion vessel allows you to choose tanks to solve the two most popular tasks and determines the initial pressure of the gas space, as well as the turn-on and turn-off pressure of the pump.
The first version of the expansion vessel calculation prioritizes the pump switching frequency. In systems with booster stations, well and booster pumps, an expansion vessel tank is necessary to reduce the frequency of switching on the pump.
Water cannot be compressed like air due to its physical properties. Therefore, even a short-term opening of one water tap in an apartment building can cause the pump to turn on. Frequent activation of the pump leads to its rapid wear and failure.
The switching frequency is related to the electric power of the pump. For example, pumps with a power of more than 8 kW are recommended to be turned on no more than 10 times per hour, pumps with a power of less than 5 kW - no more than 20 times per hour, and pumps whose power falls into the range of 5 to 10 kW a little more than 15 times per hour. This dependence is the basis of the above calculation algorithm.
In addition to the power of the pump, the permissible switching frequency is affected by many other factors. For example, the greater the mass of moving parts, the lower the permissible switching frequency. Therefore, pay attention to the correspondence between the values of the switching frequency obtained as a result of the selection of the expansion vessel and the optimal frequency for the selected pump, and if necessary, repeat the calculation of the tank.
The second option for calculating the expansion vessel prioritizes the stored water volume. It is recommended for systems in which the interruption of water supply is not allowed, but there are interruptions in the supply of electricity or water from centralized water supply systems.
## Selection of Expansion Vessel
The volume of the selected expansion vessel must be greater than or equal to the volume obtained as a result of the calculation. There are no negative consequences from overestimating the volume of the hydraulic accumulator beyond the calculated value, regardless of how much it is exceeded.
When choosing an expansion vessel, its temperature and strength characteristics should be taken into account. The maximum pressure of the tank must be greater than or equal to the maximum pressure at the point of its connection. If the installation of the expansion vessel is planned indoors, it should be taken into account that tanks with a diameter of more than 750mm and a height of more than 1.5m may not pass through the door, and mechanization means will be needed to move them. In this case, it is better to prefer not one but several expansion vessels of smaller capacity.
When choosing an expansion vessel, remember that the volume of water stored in it is on average 40-50% of the volume of the tank.
## Pressure Setting in Expansion Vessel
Before connecting to the water supply system, it is necessary to set the pressure of the gas space in the expansion vessel (Pg) according to the value given in the calculation results. For this, it may be necessary to depressurize it through a nipple or pump up the tank with a compressor, depending on the factory-set pressure of the gas space. After creating the pressure of the gas space, connect the tank to the water supply system.
The automatic control of the pump should be adjusted so that the start-up pressure and cut-off pressure of the pump correspond to the calculated values.
The maximum water pressure Pmax is taken to be no more than 4.5 bar, in some cases 6 bar. The minimum Pmin is the pressure at which the excess pressure at the upper water distribution valve is the minimum permissible value, usually (0.2-1.0 bar).
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# Polygon Analysis
63 %
38 %
Education
Published on October 14, 2008
Author: mrfollett
Source: authorstream.com
Polygon Analysis : Polygon Analysis Coordinate Geometry Parallel Lines : Parallel Lines Parallel lines have the same slope. 1 set Trapezoid 2 sets Parallelogram Example : Example Quadrilateral ABCD has the following slopes Is it a parallelogram, trapezoid, or neither? Parallelogram Example : Example Quadrilateral ABCD has the following slopes Is it a parallelogram, trapezoid, or neither? Trapezoid Perpendicular Lines : Perpendicular Lines Perpendicular lines have opposite inverse slopes. Example: All sides perpendicular Rectangle Example : Example Quadrilateral ABCD has the following slopes Is it a rectangle? Yes Example : Example Quadrilateral ABCD has the following slopes Is it a rectangle? No Length of Sides : Length of Sides Use the distance formula All same length Rhombus Example : Example Quadrilateral ABCD has the following side lengths Is it a rhombus? No Example : Example Triangle PQR has the following side lengths What type of triangle is it? Scalene Putting it all together : Putting it all together Putting it all together : Putting it all together Quadrilateral ABCD has the following info What is it? Trapezoid Putting it all together : Putting it all together Quadrilateral ABCD has the following info What is it? Square Triangles too : Triangles too Triangle ABC has the following info What is it? Right Scalene Triangle
User name: Comment:
March 28, 2020
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March 27, 2020
March 27, 2020
March 27, 2020
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PURPLEKITTY57
SparkPoints
# BMR Correction to Spark
### Monday, December 12, 2011
My personal trainer told me I am not eating enough to keep my body burning calories for the amount of exercise I am doing. I explained that spark calculated it; then I found out that Spark 'assumes everyone is sedentary' so only uses the 1.2 multiplier! So I found another site that explains how you should calculate your BMR based on your activity:
• If you rarely exercise, multiply your BMR by 1.2. THIS IS SPARKS CALC.
• If you exercise on 1 to 3 days per week, doing light activity, multiply your BMR by 1.375.
• If you exercise on 3 to 5 days per week, doing moderate activity, multiply your BMR by 1.55.
• If you exercise 6 to 7 days per week, doing vigorous activity, multiply your BMR by 1.725.
• If you exercise every day and have a physical job or if you often exercise twice a day, multiply your BMR by 1.9.
Added Note: so to be sure you are eating enough to keep up your strength while exercising, but NOT over eating, this is the guideline to use.
• MOBYCARP
That's interesting. I have no idea what formula SP uses to calculate calories needed, but I know that my current calorie range is calculated by telling SP that I plan to burn 6870 calories through exercise. What I report on SP isn't close to that amount (3504 last week), but I might have to increase it a bit to maintain my weight.
You've provided me with a conceptual framework for understanding what's going on with the SP fitness/nutritional tracker interaction. Thank you!
2189 days ago
• FROGGGY13
Well, I go by Spark guidelines, but I know they are not entirely accurate. I am losing faster than Spark predictions while eating by the guidelines and exercising less than I had set it up for. I think you really need to tweakby your body's reactions.
2190 days ago
• BREWMASTERBILL
Well, not quite. SP does use a multiplier, however, if you give SP a goal weight and a goal date, you can add expected calories burned per week from fitness and it will calculate a daily recommendation based on the BMR, burn rate and date. I think the multiplier is arbitrary. I actually prefer the SP way.
But with all of that said, none of this is entirely accurate. You don't really know what your BMR is. It's a guess. So is the burn rate due to activity. The only thing that's probably even remotely accurate is your consumption assuming that you measure and log every single bite.
2192 days ago
• YOGAYARNIE
I know the last time I used SparkPeople I used a calculator from another website to determine my calorie and macronutrient ranges because I as working out a lot.
2192 days ago
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# How to Calculate Implicit Interest Rate
An implicit interest rate is the nominal interest rate implied by borrowing a fixed amount of money and returning a different amount of money in the future. For example, if you borrow \$100,000 from your brother and promise to pay him back all the money plus an extra \$25,000 in 5 years, you are paying an implicit interest rate. There are other situations in every day life where you will encounter implicit interest rates.
Method 1
Method 1 of 3:
### Calculating Implicit Interest Manually
1. 1
Define implicit interest. If you borrow money from someone and agree to pay it back with an additional amount, you are not specifying any interest or interest rate. Let's use the example that you borrow \$100,000 from your brother and promise to pay him back in 5 years plus an extra \$25,000. In order to find the interest rate that is "implicit" or "implied" in this agreement, you need to do a mathematical calculation.[1]
• The formula you will use is total amount paid/amount borrowed raised to 1/number of periods = x. Then x-1 x100 = implicit interest rate.
2. 2
Calculate the implicit interest amount. For the example in Step 1, first divide the total payback amount by the borrowed amount. In this example, you borrowed \$100,000 and pay back a total of \$125,000, so \$125,000 divided by \$100,000 is 1.25.[2]
3. 3
Determine the number of years to repay. Raise the result of the first step to the power of 1/n, where n is the number of periods interest is paid. For simplicity, we can use n=5 for 5 years to calculate the implied annual interest rate. Thus, 1.25^(1/5) = 1.25^0.2 = 1.0456.[3]
4. 4
Calculate the implied interest percent. Subtract 1 from the above result. Thus 1.0456-1 = 0.0456. Then multiply the above result by 100, to arrive at 4.56%, which is the implicit interest rate per year.[4]
Method 2
Method 2 of 3:
### Calculating Implicit Interest Using a Spreadsheet
1. 1
Collect information needed for the implicit interest spreadsheet formula. This includes the number of periods such as months, total amount borrowed, monthly payment, and total number of years. You can find this information in your loan agreement.[5]
2. 2
Launch a computer spreadsheet application to help you calculate the implicit interest. Common spreadsheet programs include Microsoft Excel and iWork Numbers. You will be entering the data from Step 1 into a formula bar on the spreadsheet.[6]
3. 3
Click on cell A1 and then on the formula bar located above the column names. If you are taking out a \$300,000 real estate mortgage with monthly payments of \$2,000 for 30 years, enter this function formula in the formula bar: =RATE(30*12,-2000,300000). Then hit return.[7]
• The function calculates the value at .59%, which is a monthly interest rate. To annualize this monthly rate, multiply it by 12, and you get an implicit annual interest rate of 7.0203%.[8]
Method 3
Method 3 of 3:
### Using Implied Interest
1. 1
Determine implicit interest for leases. Many times business owners lease rather than purchase equipment. While lenders do not have to charge an explicit rate in a lease agreement in the U.S., finance firms making the loan are required to calculate the cost of borrowing for you.
• For example, a food products company needs to lease a large pasteurizing machine. They decide to lease rather than purchase it. If the total cost of the lease is \$1,000 and the company makes 12 payments of \$100 per month, then the lease agreement has an implicit interest rate of 20%.
2. 2
Determine implicit interest for bond purchases. When purchasing bonds, an implicit interest rate is the difference between the current yield (dividend) paid on a bond and the rate that the bondholder will receive at a fixed point in the future. The implicit rate may change from the rate stated in the bond contract at the time of purchase, since bonds can rise or fall in value during the bond term.
• For example, you purchase bonds with a promised dividend of \$5.00 per share to be paid in one year. Due to fluctuations in the marketplace, you receive \$10.00 per share on the one-year due date. The implicit interest rate earned is 50%.
3. 3
Calculate implicit interest before borrowing or leasing. If there is not an explicit interest stated, you should always calculate the implied interest rate before signing a lease or taking out a loan. This rate will determine your total financing expense. Do not rely only on monthly payment amounts or short-term yields on bonds before making financing decisions.
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Co-authored by:
This article was co-authored by Michael R. Lewis. Michael R. Lewis is a retired corporate executive, entrepreneur, and investment advisor in Texas. He has over 40 years of experience in business and finance, including as a Vice President for Blue Cross Blue Shield of Texas. He has a BBA in Industrial Management from the University of Texas at Austin. This article has been viewed 256,544 times.
Co-authors: 11
Updated: November 23, 2019
Views: 256,544
Categories: Lending
Article SummaryX
Implicit interest rate is the interest rate implied when borrowing a fixed amount of money and returning a different amount of money in the future. To calculate the implicit interest rate, divide the amount you’ll pay back by the amount you borrowed. Then, raise the result by the power of 1 divided by the number of periods, in this case years. So, if you borrow 100,000 dollars and promise to pay back 125,000 in 5 years, you’d divide 125,000 by 100,000 to give you 1.25. Then, raise this to the power of 1 divided by 5, which is 0.2, to give you 1.0456. Finally, subtract this by 1, then multiply it by 100 to give you the percentage. 1.0456 minus 1 would give you 0.456. Then, you'd multiply that by 100 and get 4.56. Therefore, the implicit interest rate is 4.56 percent. For more tips from our Financial co-author, including how to use a spreadsheet to calculate implicit interest rates, read on!
Thanks to all authors for creating a page that has been read 256,544 times.
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# How to Create Stunning SNS Histogram Plots with Code Examples: A Comprehensive Tutorial
## Table of content
### Introduction
Histogram plots are a powerful tool for visualizing patterns in data. They are especially useful for social media analytics, where visualizing user behavior over time can be a key part of understanding user engagement. This tutorial will provide a comprehensive guide to creating stunning histogram plots using Python. We will start with a brief overview of Python's data visualization packages and then dive into the key features of creating histogram plots. Along the way, we will provide detailed explanations of the code and examples, so that even those with limited Python programming experience can follow along. By the end of this tutorial, you'll have a solid foundation for creating stunning SNS histogram plots, which can be used to analyze social media data and improve your understanding of user behavior.
### Understanding Histogram Plots
Histogram plots are an essential tool for visualizing data distributions in SNS (Seaborn) Python library. In simple terms, a histogram is a graph that shows the frequency of data within defined intervals, known as bins. is crucial as it helps you interpret the data distribution and detect any patterns or outliers present in the dataset.
Histogram plots in SNS display the data distribution through color-coded bars that represent the frequency of data in each bin along the x-axis. The y-axis shows the count or density of data in each bin. The histogram plot allows you to adjust the number of bins, the color of the bars, and other aesthetical parameters to improve the visualization of your data.
One of the critical features of a histogram plot is the 'if statement with "name" parameter,' which helps you determine which column in your dataset to use for creating the histogram plot. This parameter works by selecting a specific column in the dataset by its name and using it to create the histogram plot in SNS.
Overall, understanding the visualization of data distribution through histogram plots is a fundamental step in using SNS Python library to analyze and visualize data. By learning how to use histogram plots, you can gain insights into your data's distribution and convey your findings to others more effectively.
### Data Preparation for SNS Histogram Plots
To create stunning SNS histogram plots, it is important to first prepare the data that will be used in the plot. Data preparation involves cleaning and structuring the data so that it can be easily visualized in the histogram plot.
One common way to prepare data for an SNS histogram plot is to use the Pandas library, which is a powerful tool for data manipulation and analysis in Python. The data can be imported into Pandas as a DataFrame, which is essentially a table of data with labeled columns and rows.
Once the data is in a Pandas DataFrame, it can be cleaned and filtered as needed. This may involve removing any missing or incorrect values, converting data types, and merging or grouping data in certain ways.
After the data has been properly cleaned and structured, it can be visualized using the Seaborn library's histogram function, which creates the histogram plot. The histogram function takes in the DataFrame, as well as other parameters such as the number of bins to use and the color palette to apply.
Overall, proper data preparation is crucial for creating stunning SNS histogram plots that accurately represent the underlying data. By using the Pandas and Seaborn libraries in Python, data can be quickly and easily prepared and visualized in a way that is both informative and visually appealing.
### Creating SNS Histogram Plots
is an essential skill for data analysts and scientists who want to visualize data distribution. Seaborn (SNS) is a popular data visualization library in Python that helps in creating visually appealing plots with a few lines of code. Here are the basics of creating histogram plots with SNS.
Firstly, you need to import the necessary libraries, including NumPy, Matplotlib, and Seaborn. Then, load the dataset you want to plot using NumPy or Pandas. You can then pass this data into the SNS histplot method to create a histogram plot. SNS allows you to customize various parameters like color, edge color, and bin width.
Another useful feature in SNS histogram plots is the ability to overlay multiple histograms on the same plot. This can be particularly helpful when comparing the distribution of different variables in your dataset.
One thing to keep in mind when creating histogram plots is to choose the appropriate bin size. A large bin size can oversimplify the distribution and hide important details, while a small bin size can make the plot difficult to read. SNS offers a feature that automatically calculates the ideal bin size based on the data range and distribution.
To summarize, involves importing the necessary libraries, loading the dataset, passing it to the histplot method, customizing the plot parameters, and choosing an appropriate bin size. By following these steps, you can easily create stunning histogram plots in Python using SNS.
### Customizing SNS Histogram Plots
SNS histogram plots can be customized in various ways to suit your needs. Here are some tips on how to customize your SNS histogram plots:
The bin size controls the width of the bars in a histogram plot. By default, SNS automatically calculates the bin size based on the distribution of the data. However, you can manually adjust the bin size by specifying the number of bins or the width of each bin.
sns.histplot(data = df, x = 'column_name', bins = 10)
This code will create a histogram with 10 bins. If you want to specify the width of each bin, you can use the bins argument like this:
sns.histplot(data = df, x = 'column_name', bins = [0, 10, 20, 30, 40, 50])
This code will create a histogram with bins of width 10 between 0 to 50.
You can add colors to your SNS histogram plot by specifying the color argument. You can also use the palette argument to specify a color palette.
sns.histplot(data = df, x = 'column_name', color = 'red')
This code will create a histogram with bars in red color. If you want to use a specific color palette, you can use the palette argument like this:
sns.histplot(data = df, x = 'column_name', palette = 'rocket')
This code will create a histogram with a color palette named rocket.
1. Changing bar style
You can change the style of the bars in your SNS histogram plot by using the element argument. The element argument can take values like 'bars', 'step', and 'poly'.
sns.histplot(data = df, x = 'column_name', element = 'step')
This code will create a histogram with a step-style bar. If you want to use a polygon-style bar, you can use the element argument like this:
sns.histplot(data = df, x = 'column_name', element = 'poly')
This code will create a histogram with a polygon-style bar.
Overall, is easy with the right code and arguments. By adjusting the bin size, adding colors, and changing bar styles, you can create stunning histogram plots that convey your data clearly and precisely.
### Code Examples
:
To create stunning SNS histogram plots, you need to use the Seaborn library in Python. The following code example shows how to import Seaborn and load data into a pandas data frame:
```import seaborn as sns
import pandas as pd
# Load data into pandas data frame
```
Once the data is loaded, you can create a histogram plot using the Seaborn `distplot()` function. The following code example shows how to create a histogram plot of a column named "age" in the data frame:
```# Create histogram plot of column "age"
sns.distplot(df['age'])
```
You can customize the plot by adding labels and changing the color of the plot. The following code example shows how to create a labeled plot with a blue color:
```# Create labeled plot with blue color
sns.distplot(df['age'], kde=False, label='Age', color='blue')
# Add plot title and x-axis label
plt.title('Age Distribution')
plt.xlabel('Age')
```
You can also create multiple histogram plots on the same plot to compare multiple columns. The following code example shows how to create a plot with two histograms side by side for columns named "age" and "income":
```# Create plot with two histograms side by side
sns.subplot(1, 2, 1)
sns.distplot(df['age'], kde=False, label='Age', color='blue')
sns.subplot(1, 2, 2)
sns.distplot(df['income'], kde=False, label='Income', color='green')
# Add plot titles and x-axis labels
plt.subplot(1, 2, 1)
plt.title('Age Distribution')
plt.xlabel('Age')
plt.subplot(1, 2, 2)
plt.title('Income Distribution')
plt.xlabel('Income')
```
In summary, creating stunning SNS histogram plots with is easy with the Seaborn library in Python. By following these simple steps and customizing the plot to your liking, you can create impressive visuals that showcase your data.
### Conclusion
In , creating stunning SNS histogram plots with Python is a fun and challenging task. It allows you to display data in a visually appealing way and convey information accurately. We hope that this comprehensive tutorial has provided you with the necessary knowledge to create your own SNS histogram plots using the Seaborn library. Remember to choose appropriate colors, adjust bin size, and add titles and axis labels to your plots to create a polished final product. By following the code examples provided in this tutorial and experimenting with different options, you will be able to create impressive SNS histogram plots that showcase your data in a compelling way. Happy plotting!
##### Surarchith Midhunakula
My passion for coding started with my very first program in Java. The feeling of manipulating code to produce a desired output ignited a deep love for using software to solve practical problems. For me, software engineering is like solving a puzzle, and I am fully engaged in the process. As a Senior Software Engineer at PayPal, I am dedicated to soaking up as much knowledge and experience as possible in order to perfect my craft. I am constantly seeking to improve my skills and to stay up-to-date with the latest trends and technologies in the field. I have experience working with a diverse range of programming languages, including Ruby on Rails, Java, Python, Spark, Scala, Javascript, and Typescript. Despite my broad experience, I know there is always more to learn, more problems to solve, and more to build. I am eagerly looking forward to the next challenge and am committed to using my skills to create impactful solutions.
Posts created 3177
## how to concatenate text from multiple rows into a single text string in sql server with code examples
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# How many college credits equals a high school credit?
Contents
Difference between high school and college credits: High school credits and college credits are not evaluated in the same way. Five college credits (quarter hours) are equivalent to one high school credit. Three college credits (quarter hours) are equivalent to .
## How do I convert college credits to high school credits?
How many college credits equals a high school credit? One whole college class is equivalent to one high school credit. If your child is taking one whole college class, worth 4, 5, or 6 credits, then it is one whole high school credit. If the college class is 1, 2, or 3 credits, I suggest calling it a half credit class.
## What equals a high school credit?
Traditionally, 1 credit in high school equals 120 hours of classwork, or 160 45-minute periods. Labs and projects, field trips, and independent reading can all count as classwork.
## How do you calculate college credits in high school?
The formula is to add all units together and all CPU’s together and divide the total CPU’s by the total units. So I divided 19.875 by 5.5 and got the GPA for this semester of 3.61. You can use this formula to get cumulative GPA’s for more than one semester.
## Can you still graduate high school with an F?
You can still finish college with one F on your transcript as long as you make up those lost credits, either by retaking the class or taking another class in its stead. As long as you have all the required credits to graduate, both in your major/program and in your electives, then you can graduate.
## What is a good GPA?
What is a good GPA – key takeaways. … Usually, a GPA of 3.0 – 3.5 is considered good enough at many high schools, colleges, and universities. Top academic institutions usually require GPAs higher than 3.5.
## How many credits should you have after freshman year in high school?
You will remain a Freshman until you reach 5.5 credits. You must earn 11.5 credits to become a Junior and 17.5 credits to become a Senior. If you pass everything during the school year you will have earned 7 credits each year giving you 28 credits.
## What is a half credit in high school?
A half-credit is awarded for one semester or one half-year of study. For classes such as physical education, students typically work three hours per week to fulfill one semester’s worth of work, or about 54 hours per semester.
## How many credits should a freshman have?
How many credits should I take as a freshman? While it might seem strange, for many students it’s better to take about 15 credits in their first semester. This is recommended because 12 credits are usually the minimum to be considered a full-time student at the college. It can even affect tuition in some cases.
IT\'S INTERESTING: Best answer: What degree is Air Command and Staff College?
## How are school credits calculated?
If your school determines your credit based on the number of classes you take, the calculation is fairly easy. All you do is add up all the classes you take and it will give you the total number of your credit. For this situation, count the number of classes you have taken and multiply that number by 5.
## What happens if I fail one semester in high school?
If you fail one semester of high school, those grades are permanently placed on your high school transcript and will be averaged as such. Even if you did earn all A’s subsequently, failing the first semester will definitely hurt you in the long run.
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Synchronous Motors MCQ Test - Set 06 - ObjectiveBooks
# Practice Test: Question Set - 06
1. A synchronous machine with large air gap has
(A) A higher value of stability limit
(B) A small value of inherent regulation
(C) A higher synchronizing power which makes the machine less sensitive to load variations
(D) All of the above
2. If in a synchronous motor, driving mechanical load and drawing current at lagging power factor from constant voltage supply, its field excitation is increased, then its power factor
(A) Become more
(B) Become less
(C) Remain constant
(D) None of the above
3. If one-phase of a 3-phase synchronous motor is short-circuited, motor
(A) Will refuse to start
(B) Will overheat in spots
(C) Will not come up to speed
(D) Will fail to pull into step
4. A 3-phase synchronous motor is running clockwise. If the direction of its field current is reversed
(A) The motor will stop
(B) The motor continue to run in the same direction
(C) The winding of the motor will burn
(D) The motor will run in the reverse direction
5. The back e.m.f. in the stator of a synchronous motor depends on
(A) Number of poles
(B) Flux density
(C) Rotor speed
(D) Rotor excitation
6. If the field winding of an unloaded salient pole synchronous motor is open circuited, the motor will
(A) Stop
(B) Run as induction motor
(C) Function as static condenser
(D) Burn with dense smoke
7. A synchronous motor which works on a leading power factor and does not drive a mechanical load is called as
(A) Static condenser
(B) Condenser
(C) Synchronous condenser
(D) None of the above
8. A pony motor is basically a
(A) Small induction motor
(B) D.C. series motor
(C) D.C. shunt motor
(D) Double winding A.C./D.C. motor
9. Mostly, synchronous motors are of
(A) Alternator type machines
(B) Induction type machines
(C) Salient pole type machines
(D) Smooth cylindrical type machines
10. In a synchronous motor which loss varies with load?
(A) Windage loss
(B) Bearing friction loss
(C) Copper loss
(D) Core loss
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# Introduction
This package contains a set of tools to classify the pixels of digital images into colour categories arbitrarily defined by the user. It contains functions to
• visualize the distribution of the pixel colours in the images,
• define classification rules
• classify the pixels and to store this information in R objects,
• save these as image files.
It is a simple version of the multivariate technique known as Support Vector Machine (Cortes and Vapnik, 1995; Bennet and Campbell, 2000), adapted to this particular use.
### The procedure
The basic steps of the procedure are the following:
• One or more digital images in JPEG or TIFF format is imported into R. The categories to identify are represented in this set (the test set).
• The values of the three three colour variables (or bands) that compose each image (R, G, and B) are transformed into proportions (r, g and b).
• The pixels of the image are plotted in the plane defined by two of the transformed variables (the user can select them arbitrarily) and, hopefully, they would form separate clusters (pixel categories).
• The user then traces straight lines that separate the pixel clusters. Using the mathematical expression for these rules and the rgb values, each pixel can be tested for membership in each category (see below).
• Recording the results of the tests as 1 or 0 (pass/fail), an incidence matrix is build for that rule. This is the result of the procedure, which can be submitted to posterior analysis or used to create a new version of the original image showing the category of each pixel.
The second step simplifies the problem because it makes one of the variables dependent on the other two (as r + g + b = 1). Moreover, the transformation eliminates colour variations due to differences in illumination.
The expressions for classification rules are the same as the expression for a straight line but using one of the comparison operators $$<$$, $$\leq$$, $$>$$ or $$\geq$$. For example: $$r \geq a g +c$$, being $$a$$ and $$c$$ the slope and intercept of the line, and $$r$$ and $$g$$ the colour variables selected for the classification. A single line can produce two classification rules.
### Using several rules per category
When there are more than two categories, or when the cluster of points has a complex shape, a single rule is not enough. In these cases the procedure has additional steps:
• several rules are defined for each category,
• incidence matrices are created for each rule,
• the incidence matrices are combined with the & operator to obtain the category incidence matrix.
The last step is equivalent to estimate the union of the incidence matrices, i e $$\mathbf{M} = \mathbf{M}_{1} \cap \mathbf{M}_{2} \cap \ldots \cap \mathbf{M}_{p}$$, being p the number of rules.
### Concave category shapes
A caveat of the method is that the rules must delimit a convex polygon to combine the individual rule results successfully (in a convex polygon, a line joining any two internal points is contained in the polygon). Not all clusters have convex shape. In these cases, the cluster must be divided in convex sub-polygons (subcategories) for which rules are defined as before. The incidence matrices of the subcategories are combined using the | operator, i.e. $$\mathbf{M} = \mathbf{M}_{1} \cup \mathbf{M}_{2} \cup \ldots \cup \mathbf{M}_{s}$$, being s the number of subcategories. Note that any polygon, convex or not, can be subdivided in triangles and, as triangles are convex polygons, it is always possible to solve this problem. Note that the goal is to obtain a minimal set of convex polygons, not a complete triangulation. The example presented below is one of such cases.
# The session
What follows is a sample session illustrating both the method and the use of the package functions. It uses an example image and a test set created by cutting small areas out of the example image. It is not a good test set, see below, but it is enough to show how the method works, and its problems.
library(pixelclasser)
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https://www.nag.co.uk/numeric/cl/nagdoc_latest/examples/source/f08usce.c.html
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```/* nag_zhbgst (f08usc) Example Program.
*
* Copyright 2017 Numerical Algorithms Group.
*
* Mark 26.1, 2017.
*/
#include <stdio.h>
#include <nag.h>
#include <nag_stdlib.h>
#include <nagf08.h>
int main(void)
{
/* Scalars */
Integer i, j, k1, k2, ka, kb, n, pdab, pdbb, pdx, d_len, e_len;
Integer exit_status = 0;
NagError fail;
Nag_UploType uplo;
Nag_OrderType order;
/* Arrays */
char nag_enum_arg[40];
Complex *ab = 0, *bb = 0, *x = 0;
double *d = 0, *e = 0;
#ifdef NAG_COLUMN_MAJOR
#define AB_UPPER(I, J) ab[(J-1)*pdab + k1 + I - J - 1]
#define AB_LOWER(I, J) ab[(J-1)*pdab + I - J]
#define BB_UPPER(I, J) bb[(J-1)*pdbb + k2 + I - J - 1]
#define BB_LOWER(I, J) bb[(J-1)*pdbb + I - J]
order = Nag_ColMajor;
#else
#define AB_UPPER(I, J) ab[(I-1)*pdab + J - I]
#define AB_LOWER(I, J) ab[(I-1)*pdab + k1 + J - I - 1]
#define BB_UPPER(I, J) bb[(I-1)*pdbb + J - I]
#define BB_LOWER(I, J) bb[(I-1)*pdbb + k2 + J - I - 1]
order = Nag_RowMajor;
#endif
INIT_FAIL(fail);
printf("nag_zhbgst (f08usc) Example Program Results\n\n");
/* Skip heading in data file */
scanf("%*[^\n] ");
scanf("%" NAG_IFMT "%" NAG_IFMT "%" NAG_IFMT "%*[^\n] ", &n, &ka, &kb);
pdab = ka + 1;
pdbb = kb + 1;
pdx = n;
d_len = n;
e_len = n - 1;
/* Allocate memory */
if (!(ab = NAG_ALLOC(pdab * n, Complex)) ||
!(bb = NAG_ALLOC(pdbb * n, Complex)) ||
!(d = NAG_ALLOC(d_len, double)) ||
!(e = NAG_ALLOC(e_len, double)) || !(x = NAG_ALLOC(n * n, Complex)))
{
printf("Allocation failure\n");
exit_status = -1;
goto END;
}
/* Read whether Upper or Lower part of A is stored */
scanf("%39s%*[^\n] ", nag_enum_arg);
/* nag_enum_name_to_value (x04nac).
* Converts NAG enum member name to value
*/
uplo = (Nag_UploType) nag_enum_name_to_value(nag_enum_arg);
/* Read A and B from data file */
k1 = ka + 1;
k2 = kb + 1;
if (uplo == Nag_Upper) {
for (i = 1; i <= n; ++i) {
for (j = i; j <= MIN(i + ka, n); ++j) {
scanf(" ( %lf , %lf ) ", &AB_UPPER(i, j).re, &AB_UPPER(i, j).im);
}
}
scanf("%*[^\n] ");
}
else {
for (i = 1; i <= n; ++i) {
for (j = MAX(1, i - ka); j <= i; ++j) {
scanf(" ( %lf , %lf ) ", &AB_LOWER(i, j).re, &AB_LOWER(i, j).im);
}
}
scanf("%*[^\n] ");
}
if (uplo == Nag_Upper) {
for (i = 1; i <= n; ++i) {
for (j = i; j <= MIN(i + kb, n); ++j) {
scanf(" ( %lf, %lf ) ", &BB_UPPER(i, j).re, &BB_UPPER(i, j).im);
}
}
scanf("%*[^\n] ");
}
else {
for (i = 1; i <= n; ++i) {
for (j = MAX(1, i - kb); j <= i; ++j) {
scanf(" ( %lf, %lf ) ", &BB_LOWER(i, j).re, &BB_LOWER(i, j).im);
}
}
scanf("%*[^\n] ");
}
/* Compute the split Cholesky factorization of B */
/* nag_zpbstf (f08utc).
* Computes a split Cholesky factorization of complex
* Hermitian positive-definite band matrix A
*/
nag_zpbstf(order, uplo, n, kb, bb, pdbb, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_zpbstf (f08utc).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
/* Reduce the problem to standard form C*y = lambda*y, */
/* storing the result in A */
/* nag_zhbgst (f08usc).
* Reduction of complex Hermitian-definite banded
* generalized eigenproblem Ax = lambda Bx to standard form
* Cy = lambda y, such that C has the same bandwidth as A
*/
nag_zhbgst(order, Nag_DoNotForm, uplo, n, ka, kb, ab, pdab, bb, pdbb,
x, pdx, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_zhbgst (f08usc).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
/* Reduce C to tridiagonal form T = (Q^T)*C*Q */
/* nag_zhbtrd (f08hsc).
* Unitary reduction of complex Hermitian band matrix to
* real symmetric tridiagonal form
*/
nag_zhbtrd(order, Nag_DoNotForm, uplo, n, ka, ab, pdab, d, e,
x, pdx, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_zhbtrd (f08hsc).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
/* Calculate the eigenvalues of T (same as C) */
/* nag_dsterf (f08jfc).
* All eigenvalues of real symmetric tridiagonal matrix,
* root-free variant of QL or QR
*/
nag_dsterf(n, d, e, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_dsterf (f08jfc).\n%s\n", fail.message);
exit_status = 1;
goto END;
}
/* Print eigenvalues */
printf(" Eigenvalues\n");
for (i = 0; i < n; ++i)
printf(" %8.4f", d[i]);
printf("\n");
END:
NAG_FREE(ab);
NAG_FREE(bb);
NAG_FREE(d);
NAG_FREE(e);
NAG_FREE(x);
return exit_status;
}
```
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https://www.toprankers.com/exams/niacl-assistant-2017-exam-analysis-23rd-april-slot-4/
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# NIACL Assistant 2017 Exam Analysis 23rd April (Slot 4)
NIACL Assistant 2017 Exam Analysis 23rd April (Slot 4)
#### NIACL Assistant Exam Analysis 2018
NIACL Assistant Slot 4 was conducted today from 4:30 – 5:30 PM. Check here Complete NIACL Assistant Prelims Exam Question Paper review of Slot 4, So, let us move forward with NIACL Assistant 23rd April Prelims Exam Analysis(Slot 4).
NIACL Assistant Prelims 2017 exam will have 100 questions for 3 sections English Knowledge, Reasoning and Numerical Ability for a time limit of 60 minutes.
Take FREE NIACL Assistant Mock Test
NIACL Assistant Prelims 2017 Exam Analysis 23rd April Slot 4 :
NIACL Assistant Prelims 2017 exam Slot 5 was Easy-Moderate and the number of safe attempts to clear the cut- off was 77-80.
Let us now move on to the section wise analysis, Cut off, Questions asked of NIACL Assistant 23rd April Slot 4 exam Question Paper review:
Reasoning Ability:
Good attempts for Reasoning Ability 30 -32, Level of Difficulty was Easy-Moderate.
Topics Difficulty level Number of questions Syllogism Easy 5 Puzzle & Seating Arrangement Easy-Modearte 15 Alphabet Series Easy 5 Alphanumeric Series Easy 5 Miscellaneous Easy 5
Click here for NIACL Assistant Exam Analysis – 23rd April 2017 (Slot 1)
Click here for NIACL Assistant Exam Analysis – 23rd April 2017 (Slot 2)
Click here for NIACL Assistant Exam Analysis – 23rd April 2017 (Slot 3)
Numerical Ability:
Good attempts for numerical Ability in Slot 4 exam is 21-23, Level of difficulty was Easy-moderate
Topics Difficulty level Number of questions Simplification and Approximation Easy-moderate 10 Number Series Easy 5 Data Interpretation Easy-moderate 5 Miscellaneous Easy-Moderate 15
English Language:
This Section was overall Easy, NIACL Assistant Pre exam – 23rd April 2017 Slot 4 English Language section Questions asked.
Topics Difficulty level Number of questions Reading Comprehension Easy 10 Error Correction Easy 10 Cloze Test Easy 10
Stay tuned for more updates, All the best!
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CC-MAIN-2023-50
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https://www.lmfdb.org/ModularForm/GL2/TotallyReal/2.2.85.1/holomorphic/2.2.85.1-1.1-b?display_eigs=False&numeigs=20
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# Properties
Base field $$\Q(\sqrt{85})$$ Weight [2, 2] Level norm 1 Level $[1, 1, 1]$ Label 2.2.85.1-1.1-b Dimension 4 CM no Base change yes
# Related objects
• L-function not available
## Base field $$\Q(\sqrt{85})$$
Generator $$w$$, with minimal polynomial $$x^{2} - x - 21$$; narrow class number $$2$$ and class number $$2$$.
## Form
Weight [2, 2] Level $[1, 1, 1]$ Label 2.2.85.1-1.1-b Dimension 4 Is CM no Is base change yes Parent newspace dimension 6
## Hecke eigenvalues ($q$-expansion)
The Hecke eigenvalue field is $\Q(e)$ where $e$ is a root of the defining polynomial:
$$x^{4} - 6x^{2} + 2$$
Norm Prime Eigenvalue
3 $[3, 3, w]$ $\phantom{-}e$
3 $[3, 3, w + 2]$ $\phantom{-}e$
4 $[4, 2, 2]$ $-e^{2} + 3$
5 $[5, 5, w + 2]$ $-e^{3} + 4e$
7 $[7, 7, w]$ $\phantom{-}e^{3} - 5e$
7 $[7, 7, w + 6]$ $\phantom{-}e^{3} - 5e$
17 $[17, 17, w + 8]$ $\phantom{-}2e^{3} - 14e$
19 $[19, 19, w + 1]$ $\phantom{-}2$
19 $[19, 19, w - 2]$ $\phantom{-}2$
23 $[23, 23, w + 9]$ $-2e^{3} + 11e$
23 $[23, 23, w + 13]$ $-2e^{3} + 11e$
37 $[37, 37, w + 11]$ $-2e^{3} + 10e$
37 $[37, 37, w + 25]$ $-2e^{3} + 10e$
59 $[59, 59, 3w + 10]$ $\phantom{-}6$
59 $[59, 59, 3w - 13]$ $\phantom{-}6$
73 $[73, 73, w + 15]$ $\phantom{-}e^{3} - 2e$
73 $[73, 73, w + 57]$ $\phantom{-}e^{3} - 2e$
89 $[89, 89, -w - 10]$ $\phantom{-}3e^{2} - 12$
89 $[89, 89, w - 11]$ $\phantom{-}3e^{2} - 12$
97 $[97, 97, w + 22]$ $-2e^{3} + 16e$
Display number of eigenvalues
## Atkin-Lehner eigenvalues
This form has no Atkin-Lehner eigenvalues since the level is $$(1)$$.
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en
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https://www.bankersadda.com/p/most-important-reasoning-questions-for_13.html
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Reasoning is a game of wits and presence of mind! Yes, it is true and it might seem as the greatest of the challenge after English Section’s surprises but yet this one can easily be dealt with. You just need correct practice and hardwire your brain to quickly make decisions of what to attempt and what to leave. And for same we are providing you questions of Reasoning Question and Answers. Solve these to Practice latest pattern reasoning question for bank exams.
Directions (1-5): Study the following information to answer the given questions.
Seven people A, B, C, D, E, F and G stay in a building that has seven floors. Only one person stays on one floor. The ground floor is numbered 1, the floor above it, number 2 and so on and the top most floor is numbered 7.
G stays on the ground floor, E stays on an odd numbered floor. B stays on a floor immediately below E’s floor. There are two floors between the floors on which E stays and the floor on which A stays. C stays on a floor immediately above D’s floor. D stays on any floor which is below the floor on which E stays.
Q1. Who amongst the following stays on the topmost floor?
(a) A
(b) E
(c) C
(d) D
(e) None of these
Q2. Who amongst the following stays on the fourth floor?
(a) B
(b) F
(c) D
(d) C
(e) None of these
Q3. Who amongst the following stays on the floor which is exactly between the floor on which A stays and the floor on which B stays?
(a) F
(b) C
(c) E
(d) D
(e) None of these
Q4. Which of the following is true as per the information provided?
(a) A stays on the floor which is immediately above the floor on which F stays.
(b) F stays on the floor which is immediately below the floor on which C stays.
(c) E stays on the fifth floor.
(d) C stays on the third floor.
(e) None of these
Q5. How many floors are there between the floors on which G stays and the floor on which D stays?
(a) Three
(b) None
(c) More than three
(d) One
(e) Two
Solution (1-5):
S1. Ans.(b)
Sol.
S2. Ans.(e)
Sol.
S3. Ans.(a)
Sol.
S4. Ans.(d)
Sol.
S5. Ans.(b)
Sol.
Direction (6-10): Study the following information carefully to answer the given questions.
A, B, C, D, E, F, G and H are eight people seated in a straight line but not necessarily in the same order. Some of them are facing south while some are facing north.
Only two people sit right of H. Two people sit between H and C. A sits second to right of C. Two people sit between A and G. G does not sit adjacent to C. B sits third to right of F. F does not sit adjacent to C. Immediate neighbor of B faces opposite direction (Opposite direction means if one neighbor faces north direction then other faces south and vice-versa). Those who sits on the extreme ends of the line faces opposite direction (Opposite direction means if one neighbor faces north direction then other faces south and vice-versa). D does not sit adjacent to E. Immediate neighbor of D faces same direction (Same direction means if one neighbor faces north then other also faces north and vice-versa). B sits second to left of D and both face same direction. Immediate neighbor of H faces opposite direction (Opposite direction means if one neighbor faces north direction then other faces south and vice-versa). B faces towards north direction.
6. How many people sits left of F?
(a)Three
(b) Four
(c) No one.
(d) Five
(e) One
7. Who sits second to the left of C?
(a) A
(b) B
(c) D
(d) F
(e) Cannot determined
8. How many person sit between H and C?
(a) Two
(b) Both (c) and (e)
(c) Five
(d) One
(e) Four
9. Four of the following five are alike in a certain way and hence they form a group. Which one of the following does not belong to that group?
(a) C
(b) E
(c) D
(d) F
(e) H
10. Who sits to the immediate right of E?
(a) G `
(b) H
(c) B
(d) A
(e) C
Solution (6–10):
S6. Ans. (c)
Sol.
S7. Ans. (d)
Sol.
S8. Ans. (a)
Sol.
S9. Ans. (c)
Sol.
S10. Ans. (a)
Sol.
Directions (11-12): Read the given information carefully and answer the given question.
Point N is 8m to the west of Point O. Point P is 4m to the south of Point O. Point Q is 4m to the east of Point P. Point R is 6m to the north of Point Q. Point S is 8m to the west of Point R. Point T is 2m to the south of Point S.
Q11. How far and in which direction is Point T with respect to Point N?
(a) 4m to the east
(b) 8m to the west
(c) 4m to the west
(d) 8m to the east
(e) 6m to the south
Q12. If point T is 4m to the north of point E, then what is the distance between E and Q?
(a) 11m
(b) 8m
(c) 15m
(d) 5m
(e) 9m
Solution(11-12):
S11. Ans.(a)
Sol.
S12. Ans.(b)
Sol.
Directions (13-14): Study the following information carefully and answer the given questions:
P, Q, R, S, T, U, V, W, X and Y are ten members of the family. P is married to Y. Q is daughter of X. Y and X are brothers. T is sister of Q. U is mother-in-law of P. X is the son of V. R is the mother of T. S is the son of R. W is the mother-in-law of Y.
Q13.How is Y related to T?
(a) Uncle
(b)Aunt
(c) Father
(d) Mother
(e) Brother
Q14.How is V related to Q?
(a) Grandfather
(b) Sister
(c) Grandmother
(d) Mother
(e) Father
Solution(13-14):
S13. Ans.(a)
Sol.
S14. Ans.(a)
Sol.
Q15. Shruti correctly remembers that her mother’s birthday is before twenty-third April but after nineteenth April, whereas her sister correctly remembers that their mother’s birthday is not on or after twenty-second April. On which day in April is definitely their mother’s birthday?
(a) Twentieth
(b) Twenty-first
(c) Twentieth or twenty-first
(d) Cannot be determined
(e) None of these
S15. Ans.(c)
Sol. According to Shruti, Shruti’s mother birthday on=20 April or 21 April or 22 April
According to Shruti’s sister birthday of their mother not on 22nd April or after 22nd April
So birthday is on= Either 20th April or 21st April
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https://www.geeksforgeeks.org/find-the-smallest-positive-number-missing-from-an-unsorted-array-set-2/?ref=lbp
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# Find the smallest positive number missing from an unsorted array | Set 2
Given an unsorted array with both positive and negative elements. Find the smallest positive number missing from the array in O(n) time using constant extra space. It is allowed to modify the original array.
Examples:
```Input: {2, 3, 7, 6, 8, -1, -10, 15}
Output: 1
Input: { 2, 3, -7, 6, 8, 1, -10, 15 }
Output: 4
Input: {1, 1, 0, -1, -2}
Output: 2
```
We have discussed an O(n) time and O(1) extra space solution in previous post. In this post another alternative solution is discussed.
We make the value at index corresponding to given array element equal to array element. For example: consider the array = {2, 3, 7, 6, 8, -1, -10, 15}. To mark presence of element 2 in this array, we make arr[2-1] = 2. In array subscript [2-1], 2 is element to be marked and 1 is subtracted because we are mapping an element value range [1, N] on index value range [0, N-1]. But if we make arr[1] = 2, we will loss data stored at arr[1]. To avoid this, we first store value present at arr[1] and then update it. Next we will mark presence of element previously present at arr[1], i.e. 3. Clearly this lead to some type of random traversal over the array. Now we have to specify a condition to mark the end of this traversal. There are three conditions that mark the end of this traversal:
1. If element to be marked is negative: No need to mark the presence of this element as we are interested in finding the first missing positive integer. So if a negative element is found, simply end the traversal as no more marking of presence of an element is done.
2. If element to be marked is greater than N : No need to mark the presence of this element because if this element is present then certainly it has taken a place of an element in range [1, N] in array of size N and hence ensuring that our answer lies in the range[1, N]. So simply end the traversal as no more marking of presence of an element is done.
3. If presence of current element is already marked: Suppose element to be marked present is val. If arr[val-1] = val, then we have already marked the presence of this element. So simply end the traversal as no more marking of presence of an element is done.
Also note that it is possible that all the elements of array in the range [1, N] are not marked present in current traversal. To ensure that all the elements in the range are marked present, we check each element of the array lying in this range. If element is not marked, then we start a new traversal beginning from that array element.
After we have marked presence of all array elements lying in the range [1, N], we check which index value ind is not equal to ind+1. If arr[ind] is not equal to ind+1, then ind+1 is the smallest positive missing number. Recall that we are mapping index value range [0, N-1] to element value range [1, N], so 1 is added to ind. If no such ind is found, then all elements in the range [1, N] are present in the array. So the first missing positive number is N+1.
How this solution works in O(n) time?
Observe that each element in range [1, N] is traversed at most twice in worst case. First while performing a traversal started from some other element in the range. Second when checking if a new traversal is required to be initiated from this element to mark the presence of unmarked elements. In worst case each element in the range [1, N] are present in the array and thus all N elements are traversed twice. So total computations are 2*n, and hence the time complexity is O(n).
Below is the implementation of above approach:
## C++
`/* CPP program to find the smallest ` ` ``positive missing number */` `#include ` `using` `namespace` `std; ` ` ` `// Function to find smallest positive ` `// missing number. ` `int` `findMissingNo(``int` `arr[], ``int` `n) ` `{ ` ` ``// to store current array element ` ` ``int` `val; ` ` ` ` ``// to store next array element in ` ` ``// current traversal ` ` ``int` `nextval; ` ` ` ` ``for` `(``int` `i = 0; i < n; i++) { ` ` ` ` ``// if value is negative or greater ` ` ``// than array size, then it cannot ` ` ``// be marked in array. So move to ` ` ``// next element. ` ` ``if` `(arr[i] <= 0 || arr[i] > n) ` ` ``continue``; ` ` ` ` ``val = arr[i]; ` ` ` ` ``// traverse the array until we ` ` ``// reach at an element which ` ` ``// is already marked or which ` ` ``// could not be marked. ` ` ``while` `(arr[val - 1] != val) { ` ` ``nextval = arr[val - 1]; ` ` ``arr[val - 1] = val; ` ` ``val = nextval; ` ` ``if` `(val <= 0 || val > n) ` ` ``break``; ` ` ``} ` ` ``} ` ` ` ` ``// find first array index which is ` ` ``// not marked which is also the ` ` ``// smallest positive missing ` ` ``// number. ` ` ``for` `(``int` `i = 0; i < n; i++) { ` ` ``if` `(arr[i] != i + 1) { ` ` ``return` `i + 1; ` ` ``} ` ` ``} ` ` ` ` ``// if all indices are marked, then ` ` ``// smallest missing positive ` ` ``// number is array_size + 1. ` ` ``return` `n + 1; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `arr[] = { 2, 3, 7, 6, 8, -1, -10, 15 }; ` ` ``int` `arr_size = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` ``int` `missing = findMissingNo(arr, arr_size); ` ` ``cout << ``"The smallest positive missing number is "` ` ``<< missing; ` ` ``return` `0; ` `} `
## Java
`/* Java program to find the smallest ` `positive missing number */` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ` ` ``// Function to find smallest positive ` ` ``// missing number. ` ` ``static` `int` `findMissingNo(``int` `[]arr, ``int` `n) ` ` ``{ ` ` ``// to store current array element ` ` ``int` `val; ` ` ` ` ``// to store next array element in ` ` ``// current traversal ` ` ``int` `nextval; ` ` ` ` ``for` `(``int` `i = ``0``; i < n; i++) { ` ` ` ` ``// if value is negative or greater ` ` ``// than array size, then it cannot ` ` ``// be marked in array. So move to ` ` ``// next element. ` ` ``if` `(arr[i] <= ``0` `|| arr[i] > n) ` ` ``continue``; ` ` ` ` ``val = arr[i]; ` ` ` ` ``// traverse the array until we ` ` ``// reach at an element which ` ` ``// is already marked or which ` ` ``// could not be marked. ` ` ``while` `(arr[val - ``1``] != val) { ` ` ``nextval = arr[val - ``1``]; ` ` ``arr[val - ``1``] = val; ` ` ``val = nextval; ` ` ``if` `(val <= ``0` `|| val > n) ` ` ``break``; ` ` ``} ` ` ``} ` ` ` ` ``// find first array index which is ` ` ``// not marked which is also the ` ` ``// smallest positive missing ` ` ``// number. ` ` ``for` `(``int` `i = ``0``; i < n; i++) { ` ` ``if` `(arr[i] != i + ``1``) { ` ` ``return` `i + ``1``; ` ` ``} ` ` ``} ` ` ` ` ``// if all indices are marked, then ` ` ``// smallest missing positive ` ` ``// number is array_size + 1. ` ` ``return` `n + ``1``; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `main (String[] args) ` ` ``{ ` ` ``int` `arr[] = { ``2``, ``3``, ``7``, ``6``, ``8``, -``1``, -``10``, ``15` `}; ` ` ``int` `arr_size = arr.length; ` ` ` ` ``int` `missing = findMissingNo(arr, arr_size); ` ` ` ` ``System.out.println( ``"The smallest positive"` ` ``+ ``" missing number is "` `+ missing); ` ` ``} ` `} ` ` ` `// This code is contributed by anuj_67. `
## Python 3
`# Python 3 program to find the smallest ` `# positive missing number ` ` ` `# Function to find smallest positive ` `# missing number. ` `def` `findMissingNo(arr, n): ` ` ` ` ``# to store current array element ` ` ` ` ``# to store next array element in ` ` ``# current traversal ` ` ``for` `i ``in` `range``(n) : ` ` ` ` ``# if value is negative or greater ` ` ``# than array size, then it cannot ` ` ``# be marked in array. So move to ` ` ``# next element. ` ` ``if` `(arr[i] <``=` `0` `or` `arr[i] > n): ` ` ``continue` ` ` ` ``val ``=` `arr[i] ` ` ` ` ``# traverse the array until we ` ` ``# reach at an element which ` ` ``# is already marked or which ` ` ``# could not be marked. ` ` ``while` `(arr[val ``-` `1``] !``=` `val): ` ` ``nextval ``=` `arr[val ``-` `1``] ` ` ``arr[val ``-` `1``] ``=` `val ` ` ``val ``=` `nextval ` ` ``if` `(val <``=` `0` `or` `val > n): ` ` ``break` ` ` ` ``# find first array index which is ` ` ``# not marked which is also the ` ` ``# smallest positive missing ` ` ``# number. ` ` ``for` `i ``in` `range``(n): ` ` ``if` `(arr[i] !``=` `i ``+` `1``) : ` ` ``return` `i ``+` `1` ` ` ` ``# if all indices are marked, then ` ` ``# smallest missing positive ` ` ``# number is array_size + 1. ` ` ``return` `n ``+` `1` ` ` `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` ` ``arr ``=` `[ ``2``, ``3``, ``7``, ``6``, ``8``, ``-``1``, ``-``10``, ``15` `] ` ` ``arr_size ``=` `len``(arr) ` ` ``missing ``=` `findMissingNo(arr, arr_size) ` ` ``print``( ``"The smallest positive"``, ` ` ``"missing number is "``, missing) ` ` ` `# This code is contributed ` `# by ChitraNayal `
## C#
`/* C# program to find the smallest ` `positive missing number */` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ``// Function to find smallest ` ` ``// positive missing number. ` ` ``static` `int` `findMissingNo(``int` `[]arr, ` ` ``int` `n) ` ` ``{ ` ` ``// to store current ` ` ``// array element ` ` ``int` `val; ` ` ` ` ``// to store next array element ` ` ``// in current traversal ` ` ``int` `nextval; ` ` ` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``{ ` ` ` ` ``// if value is negative or greater ` ` ``// than array size, then it cannot ` ` ``// be marked in array. So move to ` ` ``// next element. ` ` ``if` `(arr[i] <= 0 || arr[i] > n) ` ` ``continue``; ` ` ` ` ``val = arr[i]; ` ` ` ` ``// traverse the array until we ` ` ``// reach at an element which ` ` ``// is already marked or which ` ` ``// could not be marked. ` ` ``while` `(arr[val - 1] != val) ` ` ``{ ` ` ``nextval = arr[val - 1]; ` ` ``arr[val - 1] = val; ` ` ``val = nextval; ` ` ``if` `(val <= 0 || val > n) ` ` ``break``; ` ` ``} ` ` ``} ` ` ` ` ``// find first array index which ` ` ``// is not marked which is also ` ` ``// the smallest positive missing ` ` ``// number. ` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``{ ` ` ``if` `(arr[i] != i + 1) ` ` ``{ ` ` ``return` `i + 1; ` ` ``} ` ` ``} ` ` ` ` ``// if all indices are marked, ` ` ``// then smallest missing ` ` ``// positive number is ` ` ``// array_size + 1. ` ` ``return` `n + 1; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `Main (String[] args) ` ` ``{ ` ` ``int` `[]arr = {2, 3, 7, 6, ` ` ``8, -1, -10, 15}; ` ` ``int` `arr_size = arr.Length; ` ` ` ` ``int` `missing = findMissingNo(arr, arr_size); ` ` ` ` ``Console.Write(``"The smallest positive"` `+ ` ` ``" missing number is "` `+ ` ` ``missing); ` ` ``} ` `} ` ` ` `// This code is contributed ` `// by shiv_bhakt. `
## PHP
` ``\$n``) ` ` ``continue``; ` ` ` ` ``\$val` `= ``\$arr``[``\$i``]; ` ` ` ` ``// traverse the array until ` ` ``// we reach at an element ` ` ``// which is already marked ` ` ``// or which could not be marked. ` ` ``while` `(``\$arr``[``\$val` `- 1] != ``\$val``) ` ` ``{ ` ` ``\$nextval` `= ``\$arr``[``\$val` `- 1]; ` ` ``\$arr``[``\$val` `- 1] = ``\$val``; ` ` ``\$val` `= ``\$nextval``; ` ` ``if` `(``\$val` `<= 0 || ` ` ``\$val` `> ``\$n``) ` ` ``break``; ` ` ``} ` ` ``} ` ` ` ` ``// find first array index ` ` ``// which is not marked ` ` ``// which is also the smallest ` ` ``// positive missing number. ` ` ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++) ` ` ``{ ` ` ``if` `(``\$arr``[``\$i``] != ``\$i` `+ 1) ` ` ``{ ` ` ``return` `\$i` `+ 1; ` ` ``} ` ` ``} ` ` ` ` ``// if all indices are marked, ` ` ``// then smallest missing ` ` ``// positive number is ` ` ``// array_size + 1. ` ` ``return` `\$n` `+ 1; ` `} ` ` ` `// Driver code ` `\$arr` `= ``array``(2, 3, 7, 6, 8, ` ` ``-1, -10, 15); ` `\$arr_size` `= sizeof(``\$arr``) / ` ` ``sizeof(``\$arr``[0]); ` `\$missing` `= findMissingNo(``\$arr``, ` ` ``\$arr_size``); ` `echo` `"The smallest positive "` `. ` ` ``"missing number is "` `, ` ` ``\$missing``; ` ` ` `// This code is contributed ` `// by shiv_bhakt. ` `?> `
Output:
```The smallest positive missing number is 1
```
Time Complexity: O(n)
Auxiliary Space: O(1)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
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# How does the resistance force on a rolling ball depend on the ball radius?
A billiard ball set gently rolling on a billiard table slows and stops, because it is decelerated by resistance forces at the contact between the ball and table.
I expect the magnitude of the resistance force will depend on the ball radius, but what is the exponent on the radius in the equation that describes the force?
The exponent is ~3 if determined by the ball volume, ~2 if determined by the surface of the ball or contact area, and ~1 if determined by the ball perimeter. It would be ~0 if the resistance is independent of ball radius.
There may be an empirical constant of proportionality in the equation, but what I would like to know is how I could use physical reasoning to fix the exponent on the radius to an integer, and then fit this model to my experiment.
ADDENDUM: After reading some of the references suggested in the comments, I saw that my question was more naive than I realized. The subject of rolling resistance is large, complex, and with significant engineering applications. My requirements for an answer are likely more modest than what I now think might be offered here. Really, what I would like is a bit more understanding of the primary cause of rolling resistance, the underlying physics, and enough mathematics to be able to relate this to what I currently know about mechanics. Clearly experimentation is necessary to really understand this accurately, but some theoretical background would be useful for experimental design or explaining it to students.
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Hint: search for hertzian contact with friction for a ball on plane. When you also consider slipping and rolling the problem is actually rather complex. – ja72 Nov 28 '12 at 1:44
It is apparently far from being settled: en.wikipedia.org/wiki/Rolling_resistance#Depends_on_diameter – Jaime Nov 28 '12 at 3:58
ja72 and Jaime - thanks. Both comments were helpful in locating more information about this problem than I found on my own. Sometimes the first step is finding the right name or word for a thing. I've encountered the Hertzian contact fields in respect to hardness testing, and they certainly could come in handy here. Of course I should have recognized this is an instance of a wheel, a simple machine, and therefore a topic with lots of information available - Jaime's link answers my question. I will post an answer summarizing what I learn in a day or two. – Mark Rovetta Nov 28 '12 at 18:06
The assumption in the question is that the ball rolls without slipping. This may be a good approximation in many cases, but balls do slip when rolling after they are hit by the cue stick, rebound from a cushion, or are hit by another ball. – Ben Crowell Apr 29 '13 at 3:03
Here is my answer to my own question, based upon what I learned from the suggestions in comments.
The primary cause of the rolling resistance experienced by a billiard ball rolling across a billiard table is the transfer of (kinetic) energy from the ball to the table through hysteresis at the ball-table contact. A billiard table top consists of a cloth tightly stretched over a rigid base. To an approximation, the table-cloth-ball contact possesses characteristics of a linear viscoelastic material. As the ball rolls along the surface, its weight deforms the composite material (the ball and table-base deform little, the cloth much more,) the material rebounds upon unloading, but there is some dissipation of energy as heat (again mostly in the cloth.) As the ball loses kinetic energy, it slows.
If F is the rolling resistance force slowing the ball and N is the weight of the ball, a rolling resistance coefficient, C, can be defined as:
$$F=C\;N$$
The following is presented here and elsewhere as a physical formula for the rolling friction of a slow rigid wheel on a perfectly elastic surface, but I’ve so far been unable to derive it (or locate the original source.) Where z is the sinkage depth and R is the ball radius.
$$C = (\frac{z}{2 R})^\frac{1}{2}$$
The classical solution for the contact between a sphere and half-space gives the relation of sinkage depth to load (ball weight,) ball radius, and an effective modulus, E.
$$N = \frac{4}{3} E\, R^\frac{1}{2}\; z^\frac{3}{2}$$
Expressing the ball weight using radius and ball density, this analysis suggests that F may be proportional to $R^\frac{1}{3}$. So the exponent is not really close to any of the integer values I guessed at in the original question.
Some values for the physical properties of balls and tables can be found here: Physics of Pool and Billiards
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1.2: The Nature of Data
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# 1.2: The Nature of Data - PowerPoint PPT Presentation
CHS Statistics. 1.2: The Nature of Data. Objective: To understand the different types of data. Data. Data ( plural ) – observations (such as measurements, genders, and survey responses) that have been collected Datum ( singular )
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Presentation Transcript
CHS Statistics
### 1.2: The Nature of Data
Objective: To understand the different types of data
Data
• Data (plural)– observations (such as measurements, genders, and survey responses) that have been collected
• Datum (singular)
• Sometimes used to find statistics if the context of the data is randomly selected and/or representative of the population
Parameter vs. Statistic
• Parameter – a numerical measurement describing some characteristic of a population
• Statistic – a numerical measurement describing some characteristic of a sample
Parameter vs. Statistic – YOU DECIDE!
• A recent survey of a sample of MBAs reported that the average salary for an employee with an MBA is more than \$82,000.
• Starting salaries for the 667 MBA graduates for the University Of Chicago Graduate School Of Business increased 8.5% from the previous year.
• In a random check of a sample of retail stores, the Food and Drug Administration found that 34% of the stores were not storing fish at the proper temperature.
• When Lincoln was first elected to the presidency, he received 39.82% of the 1,865,908 votes cast.
Two Types of Data
• Quantitative Data – values that answer questions about the quantity or amount (with units) of what is being measured.
• Examples: income (\$), height (inches), weight (pounds)
• Categorical Data – (qualitative data) can be separated into different categories that are often distinguished by some nonnumeric characteristic
• Examples: sex, race, ethnicity, zip codes
• Wait? Hold up! Did I just see a zip codes as categorical data? I thought they were numbers…
Categorical vs. Quantitative - You Decide!
• Length of a song
• Responses in an opinion poll
• Telephone Number
• The genders (male/female) of college graduates
Discrete vs. Continuous Data
• Discrete Data– result when a number of possible values is either a finite number or a “countable” number (dealing with counts)
• Example: the number of students with blonde hair
• Continuous Data – result from infinitely many possible values that correspond to some continuous scale that covers a range of values without gaps, interruptions, or jumps (often times has units of measure attached)
• Example: the amount of rainfall in Zelienople this past month
Discrete vs. Continuous Data – YOU DECIDE!
• X represents the number of motorcycle accidents in one year in California.
• x represents the length of time it takes to get to work.
• x represents the volume of blood drawn for a blood test.
• x represents the number of rainy days in the month of July in Orlando, Florida.
• x represents the amount of snow (in inches) that fell in Nome, Alaska last winter.
Levels of Measurement
• Nominal – characterized by data that consist of names, labels, or categories only
• The data cannot be arranged in an ordering scheme (such as high to low)
• Example: survey responses of yes, no, and undecided
• Ordinal – can be arranged in some order, but the differences between the data values either cannot be determined or are meaningless
• Example: grade letters (A, B, C, D, F); movie ratings (1, 2, 3, 4, 5) – while you can find the difference between the ratings, it is meaningless. The difference of 1 or 2 is meaningless, because it cannot be compared to other similar differences.
Levels of Measurement (continued)
• Interval – similar to the ordinal level, but the difference between any two data values is meaningful. However, there is no natural zero starting point (where none of the quantity is present).
• Example: temperatures (while 0° F seems like a good starting point, it isn't necessarily)
• Ratio –similar to the interval, but has a natural zero starting point (where zero indicates none of the quantity is present)
• Differences and ratios are meaningful
• Example: weights of adult humans, prices of jeans
Levels of Measurement – YOU DECIDE!
• Body temperature in degrees Fahrenheit of a swimmer
• Collection of phone numbers
• Final standing for the football Northeastern Conference
• Heart rate (beats per minute) of an athlete.
1.2 Assignment
P. 10 #1 – 19 ALL
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# Quadratic Inequalities: Problems with Solutions
By Prof. Hernando Guzman Jaimes (University of Zulia - Maracaibo, Venezuela)
Problem 1
What is the solution to the inequality?
$x^{2}+2x-15>0$
Problem 2
Solve the inequality by factoring the expression on the left side.
$x^{2}-2x-3\leq 0$
Problem 3
$3x^{2}-x-2\leq 0$
Problem 4
Solve the inequality by factoring the expression on the left side.
$x^{2}-8x+12<0$
Problem 5
Solve the inequality by factoring the expression on the left side.
$x^{2}-5x\geq 0$
Problem 6
$3x^{2}-27<0$
Problem 7
$9x>2x^{2}-18$
Problem 8
$9x^{2}+30x>-25$
Problem 9
$4x^{2}-4x+1<0$
Problem 10
Solve the inequality by factoring the expression on the left side.
$x^{2}+6x\leq -9$
Problem 11
Solve the following inequiality
$-\left( x+1\right)\left( x+2\right) \left( x+3\right) < 0$
Problem 12
Solve the following inequiality
$-2\left( x-1\right)\left( x+\frac{1}{2}\right) \left( x-3\right) \leq 0$
Problem 13
$\left( x^{2}-1\right) \left(x^{2}-4\right) \leq 0$
Problem 14
$\left( x-1\right)^{2}\left( x+3\right) \left( x+5\right) >0$
Problem 15
$\left( x+3\right) ^{2}\left( x+4\right) \left( x-5\right)^{3}>0$
Problem 16
If $7$ times the square of a positive number is reduced by $3$ and the result is greater than $60$, what can the number be?
Problem 17
The number of diagonals $d$ of an n-sided polygon is given by the formula $d=\frac{1}{2}\left( n-1\right) n-n$.
Which polygon has the number of diagonals greater than $35$?
Problem 18
The number $t$ of dots, ordered as shown below is given by the formula $t=\frac{n(n+1)}{2}$, where n is the number of rows.
Find the range of rows if the number of dots is less than $5050$.
Problem 19
A rectangular garden should be twice as wide as it is long.
If the fenced area is greater than $98m^{2}$, what can we say about the width of the garden?
Correct:
Wrong:
Unsolved problems:
Feedback Contact email:
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# Math Puzzles Brain Teasers: Can You Solve This Rat And Cat Math Equation Puzzle?
Math Puzzles Brain Teasers: Can You Solve This Rat And Cat Math Equation Puzzle? can it be solved only if you have a vigilant mind with a penchant for solving puzzles. You can solve this picture equation with logical reasoning. Let us find out if we can solve this brain teaser.
by N Keerthana | Updated Sep 29, 2022
## Math Puzzles Brain Teasers: Can You Solve This Rat And Cat Math Equation Puzzle?
Brain teasers are an exciting form of a puzzle that needs thinking to solve. Brain teasers make you think out of the box and exploit your mind's potential. One of the most recent brain teasers trending on social media and boggling many minds is Math Puzzles Brain Teasers: Can You Solve This Rat And Cat Math Equation Puzzle?
Let us first have a look at what this puzzle is.
## Math Puzzles Brain Teasers: Can You Solve This Rat And Cat Math Equation Puzzle?- Solution
If you are still trying to get the answer to this, we have the answer to this puzzle. This brain teaser is a great way to test your observation skills and how sharp you are. If you still haven't gotten the answer, the picture given below will make you understand this solution better.
R=Rat
C=Cat
Equation 1
6R-9C=24
Equation 2
6R-8C=28
Equation 1- Equation 2
C=4
R=10
Disclaimer: The above information is for general informational purposes only. All information on the Site is provided in good faith, however we make no representation or warranty of any kind, express or implied, regarding the accuracy, adequacy, validity, reliability, availability or completeness of any information on the Site.
## Math Puzzles Brain Teasers: Can You Solve This Rat And Cat Math Equation Puzzle?-FAQ
1. How To Solve Brain Teasers?
Many people love to solve Brain teasers, analytical puzzles, and games. Brain teasers are like a trickier version of puzzles and riddles. You have to analyze these puzzles differently and think outside of the box. You must use your creative mind because the answer to Brain teasers cannot be found easily.
2. What are Brain Teasers?
A brain teaser is an exciting form of a puzzle that needs thinking to solve.
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• Brain Teaser: There Are Two Penguins That Do Not Fit Here. Can You Spot Them In 15 Secs?
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CC-MAIN-2022-49
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https://www.pegtop.net/delphi/articles/blendmodes/difference.htm
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pegtop.net > delphi section > articles > blend modes > difference mode
delphi section articles blend modes introduction normal mode average mode multiply mode screen mode darken mode lighten mode difference modes overlay mode hard light mode soft light mode dodge modes burn modes quadratic modes additive modes interpolation mode logical modes RGB modes HSL modes opacity dark modes bright modes final words components about me
difference mode
Description: Both parameters are subtracted from one another, the absolute value is taken. So the result shows the distance between both parameters, black stands for equal colors, white for opposite colors (one is black, the other white). The result looks a bit strange in many cases. This mode can be used to invert parts of the base image, and to compare two images (results in black if they are equal). An interesting note: Unlike XOR mode, this mode (generally) cannot be undone by applying it twice, but applying it three times, you get the same result as applying it once (and so on). Formula: f(a,b) = |a - b| Advantage: This mode is commutative (base and blend color can be swapped). Code: result := abs(a-b);
negation mode
Description: This one is the "opposite" of difference mode. Note that it is NOT difference mode inverted, because black and white return the same result, but colors between become brighter instead of darker. This mode can be used to invert parts of the base image, but NOT to compare two images. An interesting note: Unlike XOR mode, this mode (generally) cannot be undone by applying it twice, but applying it three times, you get the same result as applying it once (and so on). Since I didn't see this "forgotten" mode somewhere else, I gave it this (not that perfect) name. Formula: f(a,b) = 1 - |1 - a - b| Advantage: This mode is commutative (base and blend color can be swapped). Code: result := 255 - abs(255-a-b);
exclusion mode
Description: This mode is between difference and negation mode. Again black and white return the same result, but colors between become gray. This mode can be used to invert parts of the base image, but NOT to compare two images. Formula: f(a,b) = a + b - 2ab or f(a,b) = average(difference(a,b),negation(a,b)) Advantage: This mode is commutative (base and blend color can be swapped). Code: result := a + b - (a*b SHR 7);
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https://byjus.com/jee-questions/if-s-p-q-r-p-q-r-is-a-compound-statement-then-s-p-q-r-is/
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Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# If S (p, q, r) = (~ p) ∨ [~ (q ∧ r)] is a compound statement, then S (~ p, ~ q, ~ r) is
1) ~ S (p, q, r)
2) S (p, q, r)
3) p ∨ (q ∧ r)
4) p ∨ (q ∨ r)
5) S (p, q, ~ r)
Solution: (4) p ∨ (q ∨ r)
S (p, q, r) = (~ p) ∨ [~ (q ∧ r)]
⇒ S (~ p, ~ q, ~ r) = ~ (~ p) ∨ [~ (~ q ∧ ~ r)]
= p ∨ [~ (~ q) ∨ ~ (~ r)]
= p ∨ (q ∨ r)
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