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https://bifurcationkit.github.io/BifurcationKitDocs.jl/stable/branchswitching/ | 1,716,335,223,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00041.warc.gz | 106,753,754 | 16,548 | # Branch switching
The precise definition of the methods are given in Branch switching (branch point) and Branch switching (Hopf point).
## Summary of branching procedures
We collect in the following table the list of automatic branch switching (aBS) functions. Their detailed explanation follows in this page.
functionind-th bif. pointType Tdescription
continuation(br::ContResult{T}, ind::Int; kw...):bp, :ndEquilibriumContaBS from equilibria to equilibria
continuation(br::ContResult{T}, ind::Int, lens2::Lens; kw...):bp, :hopfEquilibriumContFold/Hopf continuation w.r.t. parameters getlens(br) and lens2
continuation(br::ContResult{T}, ind::Int; kw...):bt,:zh,:hhFoldCont,HopfContswitch to Fold/Hopf continuation from Hopf/Fold w.r.t. parameters of codim 2 br
continuation(br::ContResult{T}, ind_hopf::Int, ::ContinuationPar, prob::AbstractPeriodicOrbitProblem):hopfEquilibriumContBranch switching from Hopf point to periodic orbits
continuation(br::ContResult{T}, ind::Int, kw...):bp,:pdPeriodicOrbitContBranch switching from Branch / Period-doubling point of periodic orbits to curve of periodic orbits
continuation(br::ContResult{T}, ind::Int, kw...):gh,:zh,:hhTwoParamContBranch switching from Bautin / Zero-Hopf/ Hopf-Hopf point to curve of Fold/NS of periodic orbits
## From simple branch point to equilibria
You can perform automatic branch switching by calling continuation with the following options:
continuation(br::ContResult, ind_bif::Int, optionsCont::ContinuationPar; kwargs...)
where br is a branch computed after a call to continuation with detection of bifurcation points enabled. This call computes the branch bifurcating from the ind_bifth bifurcation point in br. An example of use is provided in 2d Bratu–Gelfand problem.
See Branch switching (branch point) precise method definition
### Simple example
using BifurcationKit, Setfield, Plots
# vector field of transcritical bifurcation
F(x, p) = [x[1] * (p.μ - x[1])]
# parameters of the vector field
par = (μ = -0.2, )
# problem (automatic differentiation)
prob = BifurcationProblem(F, [0.], par, (@lens _.μ); record_from_solution = (x, p) -> x[1])
# compute branch of trivial equilibria (=0) and detect a bifurcation point
opts_br = ContinuationPar(detect_bifurcation = 3)
br = continuation(prob, PALC(), opts_br)
# perform branch switching on one side of the bifurcation point
br1Top = continuation(br, 1, setproperties(opts_br; max_steps = 14) )
# on the other side
br1Bottom = continuation(br, 1, setproperties(opts_br; ds = -opts_br.ds, max_steps = 14))
scene = plot(br, br1Top, br1Bottom; branchlabel = ["br", "br1Top", "br1Bottom"], legend = :topleft)
### Algorithms
• for the pitchfork bifurcation, the normal form is computed and non-trivial zeros are used to produce guesses for points on the bifurcated branch.
## From non simple branch point to equilibria
We provide an automatic branch switching method in this case. The underlying method is to first compute the reduced equation (see Non-simple branch point) and use it to compute the nearby solutions. These solutions are then seeded as initial guess for continuation. Hence, you can perform automatic branch switching by calling continuation with the following options:
continuation(br::ContResult, ind_bif::Int, optionsCont::ContinuationPar;
kwargs...)
An example of use is provided in 2d Bratu–Gelfand problem.
See Branch switching (branch point) for the precise method definition
## From Hopf point to periodic orbits
In order to compute the bifurcated branch of periodic solutions at a Hopf bifurcation point, you need to choose a method to compute periodic orbits among:
Once you have decided which method to use, you use the following call:
continuation(br::ContResult, ind_HOPF::Int, _contParams::ContinuationPar,
prob::AbstractPeriodicOrbitProblem ;
δp = nothing, ampfactor = 1, kwargs...)
We refer to continuation for more information about the arguments. Here, we just say a few words about how we can specify prob::AbstractPeriodicOrbitProblem.
Several examples are provided in example ODE. In case of PDE, you can have a look at Brusselator or 2d Ginzburg-Landau equation.
See Branch switching (Hopf point) for the precise method definition
## From Branch / Period-doubling point of curve of periodic orbits
We do not provide (for now) the automatic branching procedure for these bifurcations of periodic orbits. As a consequence, the user is asked to provide the amplitude of the bifurcated solution.
We provide the branching method for the following methods to compute periodic orbits: PeriodicOrbitTrapProblem,PeriodicOrbitOCollProblem, ShootingProblem and PoincareShootingProblem. The call is as follows. Please note that a deflation is included in this method to simplify branch switching.
An example of use is provided in Lur'e problem.
continuation(br::AbstractBranchResult, ind_PD::Int, contParams::ContinuationPar;
δp = 0.1, ampfactor = 1, usedeflation = false, kwargs...)
## From Bogdanov-Takens (BT) point to Fold / Hopf curve
We provide an automatic branch switching method in this case (see for example Extended Lorenz-84 model or 2d Ginzburg-Landau equation (finite differences, codim 2, Hopf aBS)). Hence, you can perform automatic branch switching by calling continuation with the following options:
continuation(br::ContResult, ind_BT::Int,
options_cont::ContinuationPar = br.contparams;
nev = options_cont.nev,
detect_codim2_bifurcation::Int = 0,
start_with_eigen = false,
autodiff = false,
Teigvec = getvectortype(br),
scaleζ = norm,
kwargs...)
where ind_BT is the index of the BT point in br. Note that the BT has been detected during Fold or Hopf continuation. Calling the above method thus switches from Fold continuation to Hopf continuation (and vice-versa) automatically with the same parameter axis.
Check the docs of Fold / Hopf Continuation and particularly Setting the jacobian for improving the speed of computation for large scale systems.
## From Zero-Hopf (ZH) point to Fold / Hopf curve
We provide an automatic branch switching method in this case (see for example Extended Lorenz-84 model or 2d Ginzburg-Landau). Hence, you can perform automatic branch switching by calling continuation with the following options:
continuation(br::ContResult, ind_ZH::Int,
options_cont::ContinuationPar = br.contparams;
nev = options_cont.nev,
detect_codim2_bifurcation::Int = 0,
start_with_eigen = false,
autodiff = false,
Teigvec = getvectortype(br),
scaleζ = norm,
kwargs...)
where ind_ZH is the index of the ZH point in br. Note that the ZH has been detected during Fold or Hopf continuation. Calling the above method thus switches from Fold continuation to Hopf continuation (and vice-versa) automatically with the same parameter axis.
Check the docs of Fold / Hopf Continuation and particularly Setting the jacobian for improving the speed of computation for large scale systems.
## From Hopf-Hopf (HH) point to Fold / Hopf curve
We provide an automatic branch switching method in this case (see for example Extended Lorenz-84 model or 2d Ginzburg-Landau equation). Hence, you can perform automatic branch switching by calling continuation with the following options:
continuation(br::ContResult, ind_HH::Int,
options_cont::ContinuationPar = br.contparams;
δp = nothing, ampfactor::Real = 1,
nev = options_cont.nev,
detect_codim2_bifurcation::Int = 0,
start_with_eigen = false,
autodiff = false,
Teigvec = getvectortype(br),
scaleζ = norm,
kwargs...)
where ind_HH is the index of the HH point in br. Note that the HH has been detected during Hopf continuation. Calling the above method thus switches from Hopf continuation to another Hopf branch automatically with the same parameter axis.
Check the docs of Fold / Hopf Continuation and particularly Setting the jacobian for improving the speed of computation for large scale systems.
## From Bautin point to curve Folds of periodic orbits
From the Bautin normal form, we know that there is a curve of Fold of periodic orbits near the bifurcation point.
We provide an automatic branch switching method in this case which reads as follows:
continuation(br::HopfCont, ind_BAUTIN::Int,
_contParams::ContinuationPar,
prob::AbstractPeriodicOrbitProblem ;
δp = nothing, ampfactor = 1, kwargs...)
where prob is a method to compute periodic orbits (see From Hopf point to periodic orbits for more information).
Note that the two parameters in br will be used for the continuation of Fold points of periodic orbits.
See ODE for an example of use.
## From Zero-Hopf (ZH) point to curve NS of periodic orbits
From the Zero-Hopf normal form, we know that there is a curve of Neimark-Sacker (NS) bifurcations of periodic orbits near the bifurcation point.
We provide an automatic branch switching method in this case which reads as follows:
continuation(br::TwoParamCont, ind_ZH::Int,
_contParams::ContinuationPar,
prob::AbstractPeriodicOrbitProblem ;
δp = nothing, ampfactor = 1, kwargs...)
where prob is a method to compute periodic orbits (see From Hopf point to periodic orbits for more information).
Note that the two parameters in br will be used for the continuation of NS points of periodic orbits.
## From Hopf-Hopf (HH) point to curve NS of periodic orbits
From the Hopf-Hopf normal form, we know that there are two curves of Neimark-Sacker (NS) bifurcations of periodic orbits near the bifurcation point.
We provide an automatic branch switching method in this case which reads as follows:
continuation(br::TwoParamCont, ind_HH::Int,
_contParams::ContinuationPar,
prob::AbstractPeriodicOrbitProblem ;
δp = nothing, ampfactor = 1,
whichns = 1,
kwargs...)
where prob is a method to compute periodic orbits (see From Hopf point to periodic orbits for more information). The option whichns which belongs to {1,2} controls which NS curve you want to compute.
Note that the two parameters in br will be used for the continuation of NS points of periodic orbits.
See ODE for an example of use. | 2,424 | 9,985 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-22 | latest | en | 0.624808 |
https://aquamo.ch/09_02/28577.html | 1,659,903,171,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00043.warc.gz | 137,968,714 | 5,901 | ### Motor Torque Calculation
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A screw is a mechanism that converts rotational motion to linear motion, and a torque (rotational force) to a linear force. It is one of the six classical simple most common form consists of a cylindrical shaft with helical grooves or ridges called threads around the outside. The screw passes through a hole in another object or medium, with threads on the inside of the hole that ...
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Calculate Screw Conveyor Torque. calculate screw conveyor torque calculate screw conveyor torque. As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for any sizereduction requirements including quarry, aggregate, and different kinds of . | 1,137 | 5,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-33 | latest | en | 0.863129 |
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Class 17
# Class 17 - Center of Mass Reference Frames Physics 1301...
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1 Center of Mass, Reference Frames Physics 1301, Fall 2011: Class 17 Tuesday, October 11, 2011
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2 Objectives: Tuesday, October 11 Center of mass Reference frames
3 Demonstrations
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Center of Mass The response of an extended body to forces and torques depends on how the mass is distributed in the body A useful (but not complete) description is to characterize the body in terms of 3 parameters Mass (does not depend on distance) Center of Mass (depends on distance) Moment of Inertia (depends of distance squared) 4
Mass, CM, Moment of Inertia 5 M = m i i = 1 n M = ρ dV x cm = 1 M m i i = 1 n r i x cm = 1 M ρ r dV I = m i i = 1 n r i 2 I = ρ r 2 dV
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Example An object consists of three small spheres located at the centers of an equilateral triangle 10.0 cm on a side connected by very light rods. Each sphere has a mass of 1.00 kg. One of the spheres is located at the origin. What are the coordinates of the cm? 6 (0,0) (?,?) (?,?) (a) (0,0) (b) (10,0) (c) (10,10) (d) (5, 8.66) (e) (5,2.89)
Example An object consists of three small spheres located at the centers of an equilateral triangle 10.0 cm on a side connected by very light rods. Each sphere has a
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https://mashalscienceacademy.com/mcqs-ch-18-p12/ | 1,721,366,074,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00305.warc.gz | 320,010,682 | 15,518 | Each of the following questions is followed by four
answers. Select the correct answer in each case.
1. If the K.E of a free
electron doubles, its de Broglie wavelength changes by the factor
Solution: De Broglie wavelength is
Now let the K.E is doubled, i-e, 2K.E. Then the linear
momentum may be p. Therefore,
If λ is the new wavelength when the K.E is
doubled, then put this value from (2) in equation (1).
Thus the wavelength changes by a factor 1/√2.
2. Einstein’s photoelectric equation is Ek = hf – ϕ. In this equation Ek refers to
(a) K.E of all emitted electrons
(b) Mean K.E of emitted electrons
(c) Maximum K.E of emitted electrons
(d) Minimum K.E of emitted electrons
Explanation: The equation explains the conservation of energy concept in the case of photoelectric effect. The energy of the incident photon is partly used to escape the electron from the metal surface, called work function ϕ, and partly in accelerating the ejected electron from the surface of metal. Therefore, if hf is the energy of photon and ϕ is the work function of the metal, then their difference is the maximum K.E of the electron. See some details in comprehensive question 3.
3. De Broglie wave are associated with
(a) Moving charged particles only
(b) Moving neutral particles only
(c) All moving particles
(d) All particles whether in motion or at rest.
4. A perfect absorber must also be perfect
5. Pair production occurs only when energy of photon is at least equal to
Explanation: In the process of pair production, a
photon of energy hf disintegrates into the pair of electron and positron. The
excess energy of photon is used as the kinetic energies of electron and positron.
Since the rest masses of electron and positron are equal, therefore,
Now m0c2 = 0.51 MeV, this implies that
2m0c2 = 1.02 MeV.
Therefore, the energy of the photon must be at least equal to 1.2 MeV. For further details, see comprehensive question 5.
6. Pair production cannot take place in vacuum because
Explanation: Pair production takes place only when
there is a nucleus in the vicinity (surrounding). In this process the charge,
energy and momentum are conserved. The role of heavy nucleus is to share some
energy and momentum for their conservation. Therefore, pair production is not
possible in vacuum.
7. The positron has a charge which is in magnitude equal to the charge on
Explanation: Positron is the anti-particle of
electron. It has the same charge as the electron. All the given particles have
charge equal in magnitude (sign of the charge is not considered as per
condition of the question).
8. We can never accurately describe all aspects of subatomic particles simultaneously. It is correct according to
Explanation: Uncertainty principle of Heisenberg states that it is impossible to determine both position and momentum simultaneously. For further details, see comprehensive question 8.
1. An electron microscope
employs which to one of the following principles?
2. Electrons have a wave
nature
3. Electrons can be focused by
an electric field
4. Electrons can be focused by
a magnetic field
5. All of the above
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# What is the least common multiple of six and twenty-three?
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### What is the least common multiple of fifty six and eighty four?
168 is the least common multiple of 56 and 84
30
12
42
30
It is: 6
12
30
60
90
12 is.
30 | 173 | 592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-40 | latest | en | 0.904891 |
https://sysudoku.com/2015/12/15/hodokus-multiple-fin-krakens/ | 1,685,411,338,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644915.48/warc/CC-MAIN-20230530000715-20230530030715-00260.warc.gz | 633,512,432 | 28,483 | Hodoku’s Multiple Fin Krakens
This post finds multiple fin kraken swordfish, with additional removals, where Hodoku overlooks kraken analysis. This prompts an examination of the Hodoku kraken fish examples in its Last Resorts section, which offer no guidance for human solving.
The homework was to discover the two double fin swordfish defined by suset 146/1289 in the second example of Hodoku’s Finned/Sashimi Fish page. This one is a double fin kraken swordfish making additional eliminations. The c8 potentials see both fins. The SW potentials all turn on 8r9c9, turning off both fins. Only r3c2 escapes, and in an interesting way. It turns off the 8 group in c9 leaving only the two fins. One of them must be true. We don’t care which. The fish does not bite, regardless.
In the second one , we have the fish 146/1289 with two fins in c2. Two krakens are saved by confirming one of the fins. To be removed, a potential must see all of the fin cells. This is what makes multiple finned fish removals so rare.
While potential r9c9 could not be accused of seeing both fins, it did see enough to get into trouble, by stripping the North box of 8-candidates. It’s one of Andrew Stuart’s Unit Forcing Chains (Dec 2012).
The first swordfish was completely resolved, but this one is not. The available finned fish have left r8c9 unresolved. Has the 8-panel raised a question that it cannot answer? If so, we can look on the grid for forcing chains by which this candidate can see any of the toxic fin sets.
I did that, in effect, simply by doing a trial trace on a true 8r8c9. It ends with
8r8c9 => fin 8r6c2
and my simplest inference path looking like this:
Looking at that, I realized that I could have stopped short by recognizing that 8r8c9 sees 8r3c2. But that fact should have shown up on the 8-panel!
Well, guess what. It does. After the other removals, the 8-panel shows that, if 8r8c9 is true, then one of the two fins must be true also, and that 8r8c9 escapes the school of finned fish.
I didn’t have to show you how I stumbled into that, but it does make my point that the grid just may back you up when the X-panel fails to resolve a finned fish potential victim.
It also reminds us how removals accumulate. If a potential is removed by any fish it stays removed, even though its presence may enable a removal in another fish. However, potentials are subject to double jeopardy. If they escape one fish, they can still be eliminated on the next one.
Hodoku’s Last Resort page gives us a glimpse of the kraken analysis that a solver can do on today’s powerful computers. It demonstrates the use of multivalue forcing chains on the entire grid.
His first kraken fish example, a Franken sashimi 2-wing with two wing cells, is definitely computer solving technology parading as humanly practical technique. Would anyone dream up this Franken finned fish with the idea that a potential victim could see both fins?
So here is a sysudokie rendition of the computer printout. You can trace the grouped forcing chains through the ALS nodes from potential victim 2r8c7 to each fin.
Remember that this can be done without elaborate chain building by a trial trace toward the fin targets, as we saw above.
The removal claimed by the above result is obvious on the panel. But not in the way that Hobiger intends. Each potential, if true, removes 2r6c7, and therefore forces one of the fins to be true. The fish fails to remove any for that reason.
So how does 2r6c7 remove itself by seeing both fins via forcing chains? Simply by not being true, and starting no chain. Any potential that sees both fins also removes all 2’s from r6. It becomes a victim of an Andrew Stuart unit forcing chain.
What an elaborately staged and terribly flawed example of kraken fish.
The Hodoku comment on this example also misleading. Hobiger says:
• “If both fins are false, the fish is true and r1c7, r5c7, and r8c7 can be eliminated
• If fin r6c2 is true, 2 can be eliminated from r8c7 as proven by the first chain
• If fin r6c9 is true, 2 can be eliminated from r8c7 as proven by the second chain”
Just remain aware that successful removals by a finned fish do not mean that the fins are true. The first statement means nothing more than if all the fins are discovered later to be false, the fish becomes regular, and all previous free passes to potentials are revoked. The last two statements simply state that if a fin is later found to be true, then candidates that saw it are false. There is no justification in that for Hodoku’s example.
Hodoku labels his second kraken fish a Type 2. It puts a two-finned sashimi 4-wing to a new purpose. It’s not about eliminating 4’s!
Both fins see a 7-candidate. If it is true, both fins are false, and the sashimi 4-wing removes 4r3c2. But that makes 4r5c2 true and contradicts the premise. The conclusion is that 7r5c8 is false!
Finally, a case where Hodoku has a reason for a premise. Note the Medusa ANL created by this argument. It’s a new type of toxic set.
Type 2 may explain why Hobiger starts chains with the fins, rather than from potentials needing to see them.
But the idea of spending this much human solving effort on such a unlikely prospect is absurd.
In summary, Hodoku totally misses the most productive techniques for finned fishing. Single fin kraken analysis riddles his prime Franken and finned fish examples. X-panel and suset based kraken analysis reveals multiple fin examples that his solver missed. His finned fish techniques section should be revised, and the kraken analysis of the Hodoku Last Resort section, simply taken down.
Next time, we take a revealing second swipe at Hodoku’s suset-cooked Franken fish. | 1,371 | 5,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-23 | latest | en | 0.93726 |
https://www.thefreelibrary.com/Global+Energy+Solution+to+the+Schrodinger+Equation+Coupled+with+the...-a0578045224 | 1,638,176,036,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358702.43/warc/CC-MAIN-20211129074202-20211129104202-00526.warc.gz | 1,121,996,729 | 17,785 | # Global Energy Solution to the Schrodinger Equation Coupled with the Chern-Simons Gauge and Neutral Field.
1. Introduction
In this paper, we are interested in the Cauchy problem of the Chern-Simons-Schrodinger equations coupled with a neutral field (CSSn) in [R.sup.1+2]:
i[D.sub.0][psi] + [D.sub.j][D.sub.j][psi] = [[absolute value of ([psi])].sup.2] [psi] + 2N[psi], (1)
[[partial derivative].sub.00]N - [DELTA]N + N = -2[[absolute value of ([psi])].sup.2], (2)
[[partial derivative].sub.0][A.sub.1] - [[partial derivative].sub.1][A.sub.0] = 2 Im ([bar.[psi]][D.sub.2][psi]), (3)
[[partial derivative].sub.0][A.sub.2] - [[partial derivative].sub.2][A.sub.0] = -2 Im ([bar.[psi]][D.sub.1][psi]), (4)
[[partial derivative].sub.1][A.sub.2] - [[partial derivative].sub.2][A.sub.1] = [[absolute value of ([psi])].sup.2]. (5)
Here, [psi](t, x) : [R.sup.1+2] [right arrow] C is the matter field, N(t, x) : [R.sup.1+2] [right arrow] r is the neutral field, and [A.sub.[mu]](t, x) : [R.sup.1+2] [right arrow] R is the gauge field. [D.sub.[mu]] = [[partial derivative].sub.[mu]] - [A.sub.[mu]] is the covariant derivative, i = [square root of (-1)], [mathematical expression not reproducible], and [DELTA] = [[partial derivative].sub.j][[partial derivative].sub.j]. We use notation A = ([A.sub.0], [A.sub.j]) = ([A.sub.0], [A.sub.1], [A.sub.2]). From now on, Latin indices are used to denote T, 2 and the summation convention will be used for summing over repeated indices.
The CSSn system exhibits both conservation of the charge,
[mathematical expression not reproducible], (6)
and conservation of the total energy
[mathematical expression not reproducible], (7)
The CSSn system is invariant under the following gauge transformations:
[psi] [right arrow] [psi][e.sup.i[chi]],
N [right arrow] N,
[A.sub.[mu]] [right arrow] [A.sub.[mu]] + [[partial derivative].sub.[mu]][chi], (8)
where [chi] : [R.sup.1+2] [right arrow] R is a smooth function. Therefore, a solution to the CSSn system is formed by a class of gauge equivalent pairs ([psi], N, A). In this paper, we fix the gauge by adopting the Coulomb gauge condition [[partial derivative].sub.j][A.sub.j] = 0, which provides elliptic features for gauge fields A. Under the Coulomb gauge condition, the Cauchy problem of the CSSn system is reformulated as follows:
i[[partial derivative].sub.t][psi] + [DELTA][psi] = -[A.sub.0][psi] + [A.sup.2.sub.j][psi] + 2i[A.sub.j][[partial derivative].sub.j][psi] + [[absolute value of ([psi])].sup.2] [psi] + 2N[psi], (19)
[[partial derivative].sub.tt]N - [DELTA]N + N = -2[[absolute value of ([psi])].sup.2], (10)
[DELTA][A.sub.0] = 2 Im ([[partial derivative].sub.2][bar.[psi]][[partial derivative].sub.1][psi] - [[partial derivative].sub.1][bar.[psi]][[partial derivative].sub.2][psi]) + 2[[partial derivative].sub.2] ([A.sub.1] [[absolute value of ([psi])].sup.2]) (11)
-2[[partial derivative].sub.2] ([A.sub.2] [[absolute value of ([psi])].sup.2]), [DELTA][A.sub.1] = -[[partial derivative].sub.2] ([[absolute value of ([psi])].sup.2]), (12)
[DELTA][A.sub.2] = [[partial derivative].sub.1] ([[absolute value of ([psi])].sup.2]), (13)
with the initial data [psi](0, x) = [[psi].sub.0](x), N(0, x) = [n.sub.0](x), [[partial derivative].sub.t]N(0, x) = [n.sub.1](x). Note that [psi], N are dynamical variables and A are determined by [psi] through (11)-(13).
The CSSn system is derived from the nonrelativistic Maxwell-Chern-Simons model in [1] by regarding Maxwell term in the Lagrangian as zero. Compared with the Chern-Simons-Schrodinger (CSS) system which comes from the nonrelativistic Maxwell-Chern-Simons model by taking the Chern-Simons limit in [1], the CSSn system has the interaction between the matter field [psi] and the neutral field N. The CSS system reads as
[mathematical expression not reproducible] (14)
and has conservation of the total energy
[mathematical expression not reproducible] (14)
We remark that [mathematical expression not reproducible] has opposite sign in (7) compared with (14). In fact, this difference causes different global behavior of solution. The local well-posedness of the CSS system in [H.sup.2], [H.sup.1] was shown in [2, 3], respectively. We can prove the existence of a local solution of the CSSn system by applying similar argument. On the other hands, due to the nondefiniteness of total energy, the CSS system has a finite-time blow-up solution constructed in [2,4]. The CSSn system also has difficulty with nondefiniteness of N[[absolute value of ([psi])].sup.2] in the total energy, but we could obtain a global solution by controlling it with [H.sup.1]-norm.
Considering conservation of the energy (7), it is natural to study the Cauchy problem with the initial data [[psi].sub.0], [n.sub.0], [n.sub.1] [member of] [H.sup.1] x [H.sup.1] x [L.sup.2]. Our first result is concerned with a local solution in energy space.
Theorem 1. For the initial data ([[psi].sub.0], [n.sub.0], [n.sub.1]) [member of] [H.sup.1]([R.sup.2]) x [H.sup.1]([R.sup.2]) x [L.sup.2]([R.sup.2]), there are T > 0 and a unique local-in-time solution ([psi], N, A) to (9)-(13) such that
[mathematical expression not reproducible] (15)
where 2 < q < [infinity]. Moreover, the solution has continuous dependence on initial data.
Our second result is concerned with a global solution in energy space.
Theorem 2. For the initial data ([[psi].sub.0], [n.sub.0], [n.sub.1]) [member of] [H.sup.1]([R.sup.2]) x [H.sup.1]([R.sup.2]) x [L.sup.2] ([R.sup.2]), there exists a unique global solution ([psi], N, A) to (9)-(13) such that
[mathematical expression not reproducible] (16)
where 2 < q < [infinity]. Moreover, the solution has continuous dependence on initial data.
Note that, considering (11)-(13), [A.sub.j] can be determined by [psi] as
[A.sub.j] = [(-1).sup.j+1]/2[pi] ([x.sub.j']/[[absolute value of (x)].sup.2] * [[absolute value of ([psi])].sup.2]), (17)
and then [A.sub.0] can be determined as
[mathematical expression not reproducible] (18)
where j' = 2 if j = 1, and j' = 1 if j = 2. We present estimates for A and refer to [3, 5] for proof.
Proposition 3. Let [psi] [member of] [H.sup.1]([R.sup.2]) and let A be the solution of (11)-(13). Then, we have, for 2 < q [infinity],
[mathematical expression not reproducible] (19)
We will prove Theorems 1 and 2 in Sections 2 and 3, respectively. We conclude this section by giving a few notations. We use the standard Sobolev spaces [H.sup.s]([R.sup.2]) with the norm [mathematical expression not reproducible]. We will use c, C to denote various constants. When we are interested in local solutions, we may assume that T [less than or equal to] 1. Thus we shall replace smooth function of T, C(T) by C. We use A [??] B to denote an estimate of the form A [less than or equal to] CB.
2. Proof of Theorem 1
In this section we address the local well-posedness of solution to (9)-(13). We note that if we remove the gauge fields and the term [[absolute value of ([xi])].sup.2][psi] from the CSSn system, it is the same as the Klein-Gordon-Schodinger system with Yukawa coupling (KGS). There are many studies on the Cauchy problem of the KGS system in the Sobolev spaces [H.sup.s] [6-9]. Moreover, if we ignore the interaction with the neutral field N which does not cause any difficulty in obtaining a local solution, a local solution for the CSSn system can be obtained in a similar way to the CSS system. We could obtain a local regular solution by referring to [2, 8] and then construct a local energy solution by using the compactness argument introduced in [2, 3, 5, 6]. In other words, a local [H.sup.1]-solution is constructed by the limit of a sequence of more smooth solutions and it satisfies CSSn system in the distribution sense. For the proof, we follow the same argument as in [2]. So we omit the detail of the local existence here. Since the compactness argument does not guarantee the uniqueness and the continuous dependence on initial data of a local solution, we would rather contribute this section to show the uniqueness and the continuous dependence on initial data of a local solution.
Theorem 4. Let (f, N, A) and ([??], [??], [??]) be solutions to (9)-(13) on (0, T) x [R.sup.2] in the distribution sense with the same initial data ([[psi].sub.0], [n.sub.0], [n.sub.1]) [member of] [H.sup.1]([R.sup.2]) x [H.sup.1]([R.sup.2]) x [L.sup.2] ([R.sup.2]) satisfying
[mathematical expression not reproducible] (20)
for some M > 0. Then, we have
[mathematical expression not reproducible] (21)
for 0 [less than or equal to] t [less than or equal to] T. Moreover, the solution depends on initial data continuously.
Before beginning the proof, we gather lemmas used for the proof of Theorem 4. We use the following [L.sup.p] - [L.sup.p'] estimate proved in [10] which plays an important role to control the difference of solutions. It was used in [6] for the uniqueness of the KGS system.
Lemma 5. Let f(t, x) : [R.sup.1+2] [right arrow] R be a solution to
[[partial derivative].sub.tt] f - [DELTA]f + f = F, (t, x) [member of] [R.sup.1+2], f (0, x) = 0, (22)
[[partial derivative].sub.t] f (0, x) = 0, (23)
and T(t) be the Klein-Gordon propagator. Then, we have
f (t, x) = [[integral].sup.t.sub.0] T (t - s) F(s) ds, (24)
and
[mathematical expression not reproducible] (25)
The Hardy-Littlewood-Sobolev inequality is also used to control the difference of solutions. For the proof, we refer to Theorem 6.1.3 in [11].
Lemma 6. Let [I.sub.1] be the operator defined by
[mathematical expression not reproducible] (26)
If 1/q = 1/p - 1/2, 1 < p < 2, then we have
[mathematical expression not reproducible] (27)
The following Gagliardo-Nirenberg inequality with the explicit constant depending on q is used to show the uniqueness. It was proved in [12, 13] and used in [3, 5, 12, 13] to show the uniqueness of the nonlinear Schrodinger equations.
Lemma 7. For 2 [less than or equal to] q < [infinity], we have
[mathematical expression not reproducible] (28)
We need the following Groonwall type inequality.
Lemma 8. Let f(t) be a continuous nonnegative function defined on I = [0 ,a) and has zero only at 0. Suppose that f satisfies
[mathematical expression not reproducible] (29)
where [alpah], [beta] > 0 and q > 2. Then we have
[[integral].sup.t.sub.0] f (s) ds [less than or equal to] [([alpha][e.sup.2[beta]t/q - [alpha]/[beta]).sup.q/2] for t [member of] I. (30)
Proof. Define
h (t) = q/2 [([[integral].sup.t.sub.0] f(s) ds).sup.2/q] + q[alpha]/s[beta]. (31)
Then, the assumption (29) implies
[mathematical expression not reproducible] (32)
and the standard Gronwall' inequality gives
h (t) [less than or equal to] h(0) [e.sup.2[beta]t/q] = q[alpha]/2[beta] [e.sup.2[beta]t/q]. (33)
Considering the definition of h(t) in the above inequality, we have (30).
We also need the following inequality to show that the solution is continuously dependent on initial data. We refer to [14].
Lemma 9. Let q > 1 and a, b > 0. Let f : [0, [infinity]) [right arrow] [0, [infinity]) satisfy
f (t) [less than or equal to] a + b [[integral].sup.t.sub.0] [f.sup.1-1/q] (s)ds (34)
for all t [greater than or equal to] 0. Then, f(t) [less than or equal to] [([a.sup.1/q] + [bq.sup.-1]t).sup.q] for all t [greater than or equal to] 0.
Now we are ready to prove Theorem 4. The basic rationale is borrowed from [3, 5, 15]. Let ([psi], N, A) and ([??], [??], [??]) be solutions of (9)-(13) with the same initial data. If we set
[mathematical expression not reproducible] (35)
then the equations for u and v satisfy
[mathematical expression not reproducible] (36)
[[partial derivative].sub.tt]v - [DELTA]v + v = -2([[absolute value of ([psi])].sup.2] - [[absolute value of ([??])].sup.2]), (37)
where
u, v [member of] [L.sup.[infinity]] ([0, T); [H.sup.1] ([R.sup.2])) [intersection] C([0, T); [L.sup.2] ([R.sup.2])),
[[partial derivative].sub.t]v [member of] [L.sup.[infinity]] ([0, T); [L.sup.2] ([R.sup.2])). (38)
First of all, we will derive, for q > 2,
[mathematical expression not reproducible] (39)
where
[alpha] = [T.sup.2/q] [qM.sup.2] (1 + [M.sup.4/q] + [M.sup.2+4/q]) and [beta] = [M.sup.2]. (40)
Once we obtain (39), considering [mathematical expression not reproducible], Lemma 8 gives
[mathematical expression not reproducible] (41)
for 0 [less than or equal to] t [less than or equal to] T. We note that
[mathematical expression not reproducible] (42)
Let us take the time interval T' [less than or equal to] T satisfying (2 + [M.sup.2])(2[M.sup.2]T') < 1/2. Letting q [right arrow] [infinity] we have that [mathematical expression not reproducible] for 0 [less than or equal to] t [less than or equal to] T'. Using this argument repeatedly, we conclude that [mathematical expression not reproducible] for 0 [less than or equal to] t [less than or equal to] T.
To derive the estimate (39), multiplying (36) by u and integrating the imaginary part on [0, t] x [R.sup.2], we have
[mathematical expression not reproducible] (43)
Considering [[partial derivative].sub.j][[??].sub.j] = 0, we have (III) = 0. Except for the integral (VI), the right-hand side of (43) is bounded, by adopting the same manner described in [3, 5], as follows.
[mathematical expression not reproducible] (44)
We will provide, for instance, the bound for (II) and (IV). The rest can be proved in a similar way. Due to (17), Lemma 6 and Lemma 7 lead to
[mathematical expression not reproducible] (45)
and
[mathematical expression not reproducible] (46)
where p is determined by 1/q = 1/p-1/2. For 1/r + 1/q = 1/2, the Holder's inequality and Gagliardo-Nirenberg inequality yield
[mathematical expression not reproducible] (47)
Thus, the Holder's inequality gives
[mathematical expression not reproducible] (48)
For the integral (IV), similar estimate shows
[mathematical expression not reproducible] (49)
which implies
[mathematical expression not reproducible] (50)
For the integral (VI), we first apply Lemma 5 to (37) which leads to
[mathematical expression not reproducible] (51)
Then, we have
[mathematical expression not reproducible] (52)
Collecting these bounds (44), (52), we obtain (39) which implies
[mathematical expression not reproducible] (53)
On the other hand, multiplying (37) by [[partial derivative].sub.t]v and integrating over [0, t] x [R.sup.2], we have
[mathematical expression not reproducible] (54)
for 0 [less than or equal to] t [less than or equal to] T. The Holder's inequality and Gagliardo-Nirenberg inequality give us
[mathematical expression not reproducible] (55)
Finally, continuous dependence on initial data follows from the same estimates above and the same argument in [14]. Let ([psi], N, A) and ([??], [??], [??]) be solutions of (9)-(13) with the initial data ([[psi].sub.0], [n.sub.0], [n.sub.1]) and ([[??].sub.0], [[??].sub.0], [[??].sub.1]), respectively. If we set u = [psi] - [??] and [u.sub.0] = [[psi].sub.0] - [[??].sub.0], the above estimates show
[mathematical expression not reproducible] (56)
Applying Lemma 9 to (56), we have
[mathematical expression not reproducible] (57)
and this implies that the solution depends on initial data continuously in [L.sup.2] locally uniformly in time.
3. Proof of Theorem 2
In this section we study the existence of a global solution to (9)-(13). Firstly, we derive the conservation laws (6) and (7). Multiplying (1) by [bar.[psi]] and taking its conjugate, we have
[mathematical expression not reproducible] (58)
[mathematical expression not reproducible] (59)
Subtracting (59) from (58), we obtain
i[[partial derivative].sub.t] [[absolute value of ([psi])].sup.2] + [DELTA][psi][bar.[psi]] - [DELTA][bar.[psi]][psi] = 2i[[partial derivative].sub.j] ([A.sub.j] [[absolute value of ([psi]).sup.2]). (60)
Then, integration by parts on [R.sup.2] gives
[mathematical expression not reproducible] (61)
which implies (6).
Multiplying (1) by [[partial derivative].sub.t][bar.[psi]] and taking its conjugate, we have
[mathematical expression not reproducible] (62)
Summing the both sides and integrating by parts on [R.sup.2], we obtain
[mathematical expression not reproducible] (63)
Considering
[mathematical expression not reproducible] (64)
and
[mathematical expression not reproducible] (65)
the left side of (63) becomes
[mathematical expression not reproducible] (66)
On the other hands, multiplying (3), (4) by [[partial derivative].sub.t][A.sub.2], [[partial derivative].sub.t][A.sub.1], respectively, we have
[mathematical expression not reproducible] (67)
Adding the both sides, we have
[mathematical expression not reproducible] (68)
Replacing (iii) with this, integration by parts gives
[mathematical expression not reproducible] (69)
where (5) is used. Inserting (66) and (69) into (63), we have
[mathematical expression not reproducible] (70)
Now, multiplying (2) by [[partial derivative].sub.t]N and integrating on [R.sup.2] provide
[mathematical expression not reproducible] (71)
Adding (70) and (71), we finally obtain
d/dt E(t) = 0, (72)
Now we are ready to prove the existence of global solution. By the conservation laws (6) and (7), we have
[mathematical expression not reproducible] (73)
and
[mathematical expression not reproducible] (74)
Because we do not know the sign of the last term N[[absolute value of ([psi])].sup.2], the energy conservation (74) does not imply a global energy solution directly. Therefore we would find a bound for the last term and then a uniform bound for [H.sup.1-norm of solution which leads to global existence. We refer to [8].
Using the Holder's inequality, Lemma 7, and Young's inequality, we have
[mathematical expression not reproducible] (75)
From (75) and (74), it follows that
[mathematical expression not reproducible] (76)
which implies
[mathematical expression not reproducible] (77)
Referring to Proposition 3, the Holder's inequality and Young's inequality give
[mathematical expression not reproducible] (78)
which yields
[mathematical expression not reproducible] (79)
https://doi.org/10.1155/2018/3962062
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
Hyungjin Huh was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03028308).
References
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Hyungjin Huh (iD) and Bora Moon (iD)
Department of Mathematics, Chung-Ang University, Seoul 156-756, Republic of Korea
Correspondence should be addressed to Bora Moon; boramoon@cau.ac.kr
Received 26 March 2018; Accepted 8 May 2018; Published 4 June 2018 | 6,415 | 21,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-49 | latest | en | 0.765896 |
https://stackoverflow.com/questions/7855387/percentage-of-two-int | 1,716,394,520,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00869.warc.gz | 470,840,227 | 45,880 | # percentage of two int?
I want to get two ints, one divided by the other to get a decimal or percentage. How can I get a percentage or decimal of these two ints? (I'm not sure if it is right.. I'm probably way off...) for example:
``````int correct = 25;
int questionNum = 100;
float percent = correct/questionNum *100;
``````
This is how I thought I could do it, but it didn't work... I want to make the decimal (if there is one) into a percent out of 100 for example in this case it is %25. any ideas anyone?
Here is the correct code (thanks to Salvatore Previti!):
``````float correct = 25;
float questionNum = 100;
float percent = (correct * 100.0f) / questionNum;
``````
(btw, I am making a project using this for a quiz checking program that is why I need the percentage or decimal)
• Don't use float unless you aren't allowed to use double. You lose precision without much in the way of benefit. Oct 21, 2011 at 21:21
• percent should contain 25 in it. Not sure what your exact issue is, seems unclear. Please update your question, and post the function where you are using this code. Oct 21, 2011 at 21:23
• @Eric ah silly me :) i've been doing too much C all day hehe Java and strict types. Oct 21, 2011 at 21:40
• @HovercraftFullOfEels I am allowed to use double but how do i use it in this? Oct 21, 2011 at 21:56
• Everywhere you see `float`, substitute in `double`. Then do your math as `(double)n/v * 100;` Oct 21, 2011 at 21:59
If you don't add `.0f` it will be treated like it is an integer, and an integer division is a lot different from a floating point division indeed :)
``````float percent = (n * 100.0f) / v;
``````
If you need an integer out of this you can of course cast the `float` or the `double` again in integer.
``````int percent = (int)((n * 100.0f) / v);
``````
If you know your n value is less than 21474836 (that is (2 ^ 31 / 100)), you can do all using integer operations.
``````int percent = (n * 100) / v;
``````
If you get NaN is because wathever you do you cannot divide for zero of course... it doesn't make sense.
• 100F is sufficient. No need for the .0. Oct 21, 2011 at 21:42
• @SalvatorePreviti it gives me NAN as the answer when im using the float. with the int, it doesn't work at all.. Oct 21, 2011 at 21:58
• If v is zero, this calculation doesn't make sense, and is the only way you can get NaN (a number divided zero is not a number). Oct 21, 2011 at 22:00
• v is equal to 4, let me check it again. Oct 21, 2011 at 22:09
• no, still not working and i checked it and v is equal to 4 and n is equal to 1 Oct 21, 2011 at 22:15
Two options:
Do the division after the multiplication:
``````int n = 25;
int v = 100;
int percent = n * 100 / v;
``````
Convert an `int` to a `float` before dividing
``````int n = 25;
int v = 100;
float percent = n * 100f / v;
//Or:
// float percent = (float) n * 100 / v;
// float percent = n * 100 / (float) v;
``````
• I get NAN as an answer.. any ideas? Oct 21, 2011 at 22:02
• @Baruch: Both `n` and `v` must be 0, if you're getting `NaN`
– Eric
Oct 22, 2011 at 7:39
• Please, be careful with integer overflow. In case of overflow, you'll probably get negative numbers. Aug 31, 2016 at 16:26
One of them has to be a float going in. One possible way of ensuring that is:
``````float percent = (float) n/v * 100;
``````
Otherwise, you're doing integer division, which truncates the numbers. Also, you should be using `double` unless there's a good reason for the `float`.
The next issue you'll run into is that some of your percentages might look like 24.9999999999999% instead of 25%. This is due to precision loss in floating point representation. You'll have to decide how to deal with that, too. Options include a DecimalFormat to "fix" the formatting or BigDecimal to represent exact values.
• That's one of the many ways to do it. The important thing is that one of the numbers used in the division is a floating point number. Oct 21, 2011 at 21:33
• Yes ... you can do (n / (float)v) * 100; Oct 21, 2011 at 21:39
• No, the operation does not require any `float` values. See @Eric's answer. Oct 21, 2011 at 22:03
• @Loadmaster: That only works if `v` is evenly divisible by `n`. Otherwise you'll get rounding errors. The OP explicitly asked for a "decimal or percentage", and since this is for a quiz application, a non-fp answer won't work very well unless you want to set strict limits on the number of questions that can be on a quiz. Oct 21, 2011 at 22:21
``````float percent = (n / (v * 1.0f)) *100
`````` | 1,348 | 4,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-22 | latest | en | 0.903165 |
https://www.transtutors.com/homework-help/mathematics/probability-concepts/solved-problem.aspx | 1,516,729,688,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892059.90/warc/CC-MAIN-20180123171440-20180123191440-00237.warc.gz | 1,030,136,307 | 17,333 | Solved Problem Of Probability
When a coin is tossed, sample space S = {H, T}
Let E = the event of occurrence of a head. Then E = {H}
P(E) = Probability of occurrence of head = n(E) / n(S) = ½
(i) When a die is tossed, sample space S = (1, 2, 3, 4, 5, 6}
Let A = the event of occurrence of an odd number = {1, 3, 5}
B = the event of occurrence of a number greater than 4 = (5, 6).
(ii) When one ball is drawn at random from a bag containing 3 black and 4 red balls (balls of the same colour being identical or different), then sample space S = {B1, B2, B3, R1, R2, R3, R4}
\ n(S) = 7
Here the three black balls may be denoted by B|, B2 and B3 even if they are identical because while finding probability only number of black and red balls are to be taken into account.
Let E = the event of occurrence of a red ball.
Then E = {R,, R2, R3, R,) .-. n(E) = 4
(iii) When two coins are tossed, sample space S = (HH, HT, TH, TT}.
Let E = the event of occurrence of one head and one tail, then
E = {HT, TH}
Now P(E) = n(E) / n(S) = 2/4 = 1/2
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More Q&A | 413 | 1,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-05 | latest | en | 0.918575 |
https://www.mapleprimes.com/users/minhthien2016/questions?page=6 | 1,723,516,313,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00383.warc.gz | 671,615,622 | 24,596 | ## 350 Reputation
7 years, 213 days
## Are there two vectors a and b with integ...
Maple
With two vectors a and b, we know that
Norm(CrossProduct(a, b)) = Norm(a)* Norm(b) * sin(a,b).
I tried with a := <1, 2, -2>; b := <2, 10, 11>;
Note that a perpendicular to b and
Norm(CrossProduct(a, b)) = Norm(a)* Norm(b)
I tried
restart;
with(VectorCalculus);
SetCoordinates(cartesian[x, y, z]);
a := <1, 2, -2>;
b := <2, 10, 11>;
Norm(a);
Norm(b);
v := CrossProduct(a, b);
Norm(v);
Are there two vectors a and b with integer coordinates and not perpendicular, so that Norm(a), Norm(b), Norm(CrossProduct(a, b)) are interger numbers satisfying
Norm(CrossProduct(a, b)) = Norm(a)* Norm(b) * sin(a,b).
## How can I find three numbers a, b, c so ...
Maple
The system of equations x*y*z + y*z + y = 21, x*y*z + x*z + z = 30, x*y*z + x*y + x = 12
has three solutions, one of them is not an integer solution.
solve({x*y*z + y*z + y = 21, x*y*z + x*z + z = 30, x*y*z + x*y + x = 12}, {x, y, z})
How can I find three numbers a, b, c so that the system of equations
x*y*z + y*z + y = a, x*y*z + x*z + z = b, x*y*z + x*y + x = c
has three solutions (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3), where x1, y1, z1, x2, y2, z2, x3, y3, z3 are nine integer numbers.
## Tetrahedron with integer volume...
Maple
Tetrahedron with length of sides like this picture has volume is an integer number. Is there another tetrahedron like that?
## How to find six integer numbers a, b, c,...
Maple
I know that, the function f(x) = (5x^2 + 8x+ 2)/(2x^2 + 6x + 5) sastifying the conditions:
1. The solutions of the f'(x)=0 are -2 and -1;
2. f(-2) = 6 and f(-1) = -1.
How can I find six integer numbers a, b, c, d, e, m from 1 to 10 so that the function
f(x) = (a*x^2 + b*x + c)/(d*x^2 + e*x + m)
so that the equation f'(x)= 0 has two integer solutions x1, x2 and f(x1); f(x2) are also two integer numbers?
## For what the values of integer numbers k...
Maple
I find by my hand some equations have four integer solutions.
How can I tell Maple to do this? For what the values of integer numbers k, m, n, a, b, c, d so the equation
k/(x^2 + a x + b) + m/(x^2 + a x + c) + n/(x^2 + a x + d) = 0 have four integer solutions?
4 5 6 7 8 9 10 Last Page 6 of 16
| 823 | 2,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-33 | latest | en | 0.742185 |
https://www.jiskha.com/display.cgi?id=1306246134 | 1,516,538,354,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890582.77/warc/CC-MAIN-20180121120038-20180121140038-00701.warc.gz | 904,602,254 | 3,994 | # physics
posted by .
a rocket is launched from atop a 105-foot cliff with an initial velocity of 156 ft/s.The height of the rocket above the ground at time or t is given by h=-16tsquared+156t+105. When will the rocket hit the ground after it is launched?
• physics -
when is h= 0?
0=h(t)=-16t^2 + 156t+105
solve for t.
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https://www.codeproject.com/Questions/1081338/Rotate-image-in-loop-using-picture-box | 1,726,252,022,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00590.warc.gz | 643,554,252 | 33,627 | 15,996,848 members
1.00/5 (1 vote)
See more:
HI guys
I am building a Texas holdem game as my get into vb and learn project.
I have 3 picture boxes that hold the burn cards.
I found this neat code that after looking it over so much I kind of understand.
What it does is lets you set an angle to rotate your image.
Works great when i plug in the names of my picture boxes.
pbBurn1.image
pbBurn2.image
pbBurn3.image
I was trying to use the same method to loop thru all three images and rotate them all -45 degrees.
No errors in code seems to run thru instructions as expected but outputs nothing..can you please help...bellow I placed double comments on the code i tried adding to loop that never worked can someone tell me what i did wrong.
What I have tried:
VB
```Private Sub Button2_Click(sender As System.Object, e As System.EventArgs) Handles Button2.Click
''Dim BurnNum As Image = pbBurn1.Image
''Dim t As Integer = 1
''Dim count As Integer = 1
''For t = 1 To 3
'' If count = 2 Then BurnNum = pbBurn2.Image
'' If count = 3 Then BurnNum = pbBurn3.Image
' Copy the output bitmap from the source image.
Dim bm_in As New Bitmap(pbBurn2.Image) '' Dim bm_in As New Bitmap(BurnNum)
' Make an array of points defining the
' image's corners.
Dim wid As Single = bm_in.Width
Dim hgt As Single = bm_in.Height
Dim corners As Point() = { _
New Point(0, 0), _
New Point(wid, 0), _
New Point(0, hgt), _
New Point(wid, hgt)}
' Translate to center the bounding box at the origin.
Dim cx As Single = wid / 2
Dim cy As Single = hgt / 2
Dim i As Long
For i = 0 To 3
corners(i).X -= cx
corners(i).Y -= cy
Next i
' Rotate.
Dim theta As Single = Single.Parse(-45) * PI / 180.0
Dim sin_theta As Single = Sin(theta)
Dim cos_theta As Single = Cos(theta)
Dim X As Single
Dim Y As Single
For i = 0 To 3
X = corners(i).X
Y = corners(i).Y
corners(i).X = X * cos_theta + Y * sin_theta
corners(i).Y = -X * sin_theta + Y * cos_theta
Next i
' Translate so X >= 0 and Y >=0 for all corners.
Dim xmin As Single = corners(0).X
Dim ymin As Single = corners(0).Y
For i = 1 To 3
If xmin > corners(i).X Then xmin = corners(i).X
If ymin > corners(i).Y Then ymin = corners(i).Y
Next i
For i = 0 To 3
corners(i).X -= xmin
corners(i).Y -= ymin
Next i
' Create an output Bitmap and Graphics object.
Dim bm_out As New Bitmap(CInt(-2 * xmin), CInt(-2 * ymin))
Dim gr_out As Graphics = Graphics.FromImage(bm_out)
' Drop the last corner lest we confuse DrawImage,
' which expects an array of three corners.
ReDim Preserve corners(2)
' Draw the result onto the output Bitmap.
gr_out.DrawImage(bm_in, corners)
' Display the result.
pbBurn2.Image = bm_out '' Burnnum = bm_out
''count = count + 1
''Next
End Sub```
Posted
Updated 26-Feb-16 15:57pm
v3
Sergey Alexandrovich Kryukov 27-Feb-16 1:01am
Do it not using PictireBox.
—SA
## Solution 1
The reason the image doesn't animate is because your application starts on what is normally called the "UI Thread". This thread listens for messages coming into the message queue and processes these messages into events and handles everything UI related, like redrawing controls and windows.
The problem is that your code is hogging the UI thread and blocking it from processing the WM_PAINT messages that are coming in to tell your picturebox to repaint itself with the new image.
Instead of doing it in a loop it's better to use a timer and maintain a state of the image, rotating it and telling the picturebox to use the new image. Once this code ends for this one change in time, or tick of the timer, the UI thread can process the WM_PAINT message and the picturebox can repaint with the new image.
Really, you shouldn't be using PictureBox controls at all and just paint everything yourself on either the surface of the Form or on a Panel control using the Paint event for them.
Sergey Alexandrovich Kryukov 27-Feb-16 1:01am
5ed.
—SA | 1,069 | 3,866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.81042 |
http://mathhelpforum.com/advanced-statistics/162537-estimators.html | 1,481,278,838,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542693.41/warc/CC-MAIN-20161202170902-00507-ip-10-31-129-80.ec2.internal.warc.gz | 179,639,062 | 9,858 | 1. ## Estimators
Let x1,...xn be a sample random from pdf
f(x\t)=tx^(t-1) , 0<x<1 , t>0
1) to show that the pdf belongs to the exponential family
2)To find the lower limit of variance of two estimators unbiased to t
3)Find a enough statistics to t and its distribution
4)Suggest a estimator unbiased to t that be from unbiased function and check if it is efficient
Thank you
2. 1)
the function belongs to a exponential family if its pdf can be written as:
$f(x|t) = \bigg\{exp\bigg(c(t) \; T(x) \; + b(t)+S(x)\bigg)\bigg\}$
$t.x^{t-1} = \mbox{exp}(log(t)).\mbox{exp}((t-1)log(x)) = \mbox{exp}\bigg(logt+(t-1)log(x)\bigg) =...$ | 222 | 632 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2016-50 | longest | en | 0.783716 |
https://bookofproofs.github.io/branches/fun/dudeney/unclassified-problems/the-tiring-irons.html | 1,718,643,356,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861733.59/warc/CC-MAIN-20240617153913-20240617183913-00155.warc.gz | 124,573,666 | 4,585 | # Problem: The Tiring Irons
The illustration represents one of the most ancient of all mechanical puzzles. Its origin is unknown. Cardan, the mathematician, wrote about it in 1550, and Wallis in 1693; while it is said still to be found in obscure English villages (sometimes deposited in strange places, such as a church belfry), made of iron, and appropriately called "tiring-irons," and to be used by the Norwegians to-day as a lock for boxes and bags. In the toyshops, it is sometimes called the "Chinese rings," though there seems to be no authority for the description, and it more frequently goes by the unsatisfactory name of "the puzzling rings." The French call it "Baguenaudier."
The puzzle will be seen to consist of a simple loop of wire fixed in a handle to be held in the left hand, and a certain number of rings secured by wires which pass through holes in the bar and are kept there by their blunted ends. The wires work freely in the bar, but cannot come apart from it, nor can the wires be removed from the rings. The general puzzle is to detach the loop completely from all the rings, and then to put them all on again.
Now, it will be seen at a glance that the first ring (to the right) can be taken off at any time by sliding it over the end and dropping it through the loop; or it may be put on by reversing the operation. With this exception, the only ring that can ever be removed is the one that happens to be a contiguous second on the loop at the right-hand end. Thus, with all the rings on, the second can be dropped at once; with the first ring down, you cannot drop the second, but may remove the third; with the first three rings down, you cannot drop the fourth, but may remove the fifth; and so on. It will be found that the first and second rings can be dropped together or put on together, but to prevent confusion we will throughout disallow this exceptional double move, and say that only one ring may be put on or removed at a time.
We can thus take off one ring in $1$ move; two rings in $2$ moves; three rings in $5$ moves; four rings in $10$ moves; five rings in $21$ moves; and if we keep on doubling (and adding one where the number of rings is odd) we may easily ascertain the number of moves for completely removing any number of rings. To get off all the seven rings requires 85 moves. Let us look at the five moves made in removing the first three rings, the circles above the line standing for rings on the loop and those under for rings off the loop.
Drop the first ring; drop the third; put up the first; drop the second; and drop the first — $5$ moves, as shown clearly in the diagrams. The dark circles show at each stage, from the starting position to the finish, which rings it is possible to drop. After move $2$ it will be noticed that no ring can be dropped until one has been put on because the first and second rings from the right now on the loop are not together. After the fifth move, if we wish to remove all seven rings we must now drop the fifth. But before we can then remove the fourth it is necessary to put on the first three and remove the first two. We shall then have $7, 6, 4, 3$ on the loop, and may, therefore, drop the fourth. When we have put on $2$ and $1$ and removed $3, 2, 1,$ we may drop the seventh ring. The next operation then will be to get $6,$ $5, 4, 3, 2, 1$ on the loop and remove $4, 3, 2, 1,$ when $6$ will come off; then get $5, 4, 3, 2, 1$ on the loop, and remove $3, 2, 1,$ when $5$ will come off; then get $4, 3, 2, 1$ on the loop and remove $2, 1,$ when $4$ will come off; then get $3, 2, 1$ on the loop and remove $1,$ when $3$ will come off; then get $2, 1$ on the loop, when $2$ will come off; and $1$ will fall through on the 85th move, leaving the loop quite free. The reader should now be able to understand the puzzle, whether or not he has it in his hand in a practical form.
The particular problem I propose is simply this. Suppose there are altogether fourteen rings on the tiring-irons, and we proceed to take them all off in the correct way so as not to waste any moves. What will be the position of the rings after the $9,999$th move has been made?
Solutions: 1
Thank you to the contributors under CC BY-SA 4.0!
Github:
non-Github:
@H-Dudeney
### References
#### Project Gutenberg
1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917
This eBook is for the use of anyone anywhere in the United States and most other parts of the world at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this edition or online at http://www.gutenberg.org. If you are not located in the United States, you'll have to check the laws of the country where you are located before using this ebook. | 1,213 | 4,826 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-26 | latest | en | 0.971449 |
https://www.alexanderthamm.com/en/data-science-glossar/computational-statistics/ | 1,675,223,463,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499899.9/warc/CC-MAIN-20230201013650-20230201043650-00185.warc.gz | 669,167,825 | 36,562 | ## What are Computational Statistics?
The Computational Statistics field is the point of contact between information technology and statistics. Behind the term is an essential area of Data Sciencewhich currently enjoys a great deal of attention in a wide variety of application fields and will certainly continue to do so in the future, be it for Google PageRank, spam filters in e-mail inboxes or in the context of Big data analyses.
In addition to data science, computational statistics is also subordinate to simulation science; this generally involves recreating experiments in order to minimise the amount of work involved in research or to make experiments possible in the first place.
Computational statistics is often equated with statistical computing. In fact, the former is mainly about implementing algorithms in applications; in statistical computing it is the other way around and concepts from computer science are applied to statistics.
## Important methods:
• The Markov chain is a stochastic process that is used in a wide variety of fields: Economists use it to optimise traffic systems, in financial mathematics it is used to model share prices and online marketers use it to create texts; even the popular board game Monopoly can be understood as a Markov chain. In simplified terms, this mathematical method looks at the development of random systems over time. In other words, a sequence of dice rolls whose respective dice result is, of course, independent of the previous dice roll. Using Monopoly as an example, this process could now be used to determine how likely certain game scenarios are.
• The Monte Carlo simulation makes it possible to carry out statistical studies that would be impossible or very costly in other ways. If, for example, the average height of a person is to be determined, one could measure all the citizens of the earth and divide the sum by the world population - this is an impossible undertaking. In the Monte Carlo simulation, a smaller number of people are randomly selected, which keeps the workload low. The more measurements are made, the closer one gets to the real result - the reason for this is the law of large numbers. Monte Carlo simulation is also used in many areas: climate models predict the weather, for example, companies use it to weigh up risks and production processes in manufacturing plants are optimised with the help of this method.
• The Maximum likelihood method is a universally applicable estimation method - in bioinformatics it is considered a standard procedure. Like Monte Carlo simulation, the maximum likelihood method is used to keep the effort as low as possible. This means: If you want to try out different parameters for a statistic, but there are no measurements for them, the maximum likelihood method is used to determine the parameter that most likely leads to the desired result.
## What role do computational statistics play in the development of new technologies?
Computer-aided statistics is made up of various components. Based on the mathematical principles of probability, distribution, estimation and inference, methods (such as the Markov chain) are used to process data. Those who work in this field have mastered the procedures of statistics and their digital implementation.
In the future, work with computational statistics will play a role more than ever. Especially Areas of the Digitisation are mostly supplemented by computer-assisted statistics. In the field of autonomous driving For example, there is an urgent need for statistics; as safety is the primary concern in public road transport, computerised statistics are essential. The Nanotechnology and the medical sector in general will continue to rely on methods such as the maximum likelihood method to conduct research on DNA threads.
The Fields of technologisation require analysis by computer-aided statistics, be it the virtual reality, blockchain or the artificial intelligence.
An example of computerised statistics in the development of new technologies is an online platform for rental flats. Since the company was founded, there was the problem of the countless variables that make it difficult for landlords to set prices. Therefore, from the beginning, they relied on Data Science to calculate price suggestions for their clients. These suggestions reduce the workload for the landlord and thus make it less difficult to place an ad for the vacant flat. In turn, the resulting increase in turnover is processed statistically. Computer-aided statistics is closely interwoven with the development of new technologies; this can be seen in this example. | 849 | 4,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-06 | latest | en | 0.932518 |
https://www.mrexcel.com/board/threads/data-type-calculation-errors.328020/ | 1,680,345,466,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00107.warc.gz | 996,818,278 | 18,402 | # Data Type Calculation Errors
#### thorburnc
##### New Member
I have a workbook that calculates a sum of numbers for each week requested (a reporting tool) from 7-8 worksheets within this workbook.
My problem is that sometimes a cell does not contain a number so it is blank and when the formula looks at it to add the value to the sum, it is reading it as a different data type and can't add the "0" value to the ongoing sum calculation.
I hope that makes sense...we keep getting a debug error because it can't sum up all the numbers.
### Excel Facts
How to find 2nd largest value in a column?
MAX finds the largest value. =LARGE(A:A,2) will find the second largest. =SMALL(A:A,3) will find the third smallest
#### DonkeyOte
##### MrExcel MVP
you could possibly create an array or just use a handler:
=IF(ISNUMBER(cell1),cell1,0)+IF(ISNUMBER(cell2),cell2,0)
#### thorburnc
##### New Member
Well, it happens in two spots.
The first formula that I get this error is:
If c.Interior.ColorIndex = xlColorIndexNone Then dbl = dbl + c.Value
Then when I do the reporting for each of the sheets, the formula is
s = s + wksht.Cells(c + B, T).Value
#### DonkeyOte
##### MrExcel MVP
Same theory applies in VBA, you should test the variable before trying to add it, ie
Rich (BB code):
``If c.Interior.ColorIndex = xlColorIndexNone And IsNumeric(c.Value) Then dbl = dbl + c.Value``
Make sense ?
#### Norie
##### Well-known Member
Christine
An empty cell should be no problem when adding.
Care to post the rest of your code?
Perhaps you should be looking at some of the conversion functions like Val, CDbl, CInt etc.
#### thorburnc
##### New Member
Function sumWhite(rng As Range)
Dim c As Range
Dim dbl As Double
For Each c In rng
If c.Interior.ColorIndex = xlColorIndexNone Then dbl = Sum(dbl + c.Value)
Next
sumWhite = dbl
End Function
That is my code...
I will try adding in the ISNumberic(c.value) and hopefully that works! Thanks!
#### thorburnc
##### New Member
Adding the "And IsNumeric" statement worked!
Thank you!
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https://studyadda.com/solved-papers/jamia-millia-islamia/jamia-millia-islamia-solved-paper-2009/291 | 1,566,215,483,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314732.59/warc/CC-MAIN-20190819114330-20190819140330-00506.warc.gz | 654,763,047 | 49,786 | # Solved papers for JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2009
### done Jamia Millia Islamia Solved Paper-2009
• question_answer1) The magnitude of resultant of two equal forces is equal to either of two forces. The angle between forces is
A) $\frac{\pi }{3}$
B) $\frac{2\pi }{3}$
C) $\frac{\pi }{2}$
D) $\frac{3\pi }{4}$
• question_answer2) A particle projected vertically upwards attains a maximum height H. If the ratio of the times to attain a height$h(h<H)$is$\frac{1}{3}$then,
A) $4h=3H$
B) $3h=4H$
C) $3h=H$
D) $4h=H$
• question_answer3) A force vector applied on a body is given by $\overrightarrow{F}=6\hat{i}-8\hat{j}+10\hat{k}$and acquires an acceleration of$1\text{ }m{{s}^{2}}$. Then the mass of the body is
A) $10\sqrt{2}$kg
B) $2\sqrt{10}$
C) $10\,kg$
D) $20\,kg$
• question_answer4) In the adjoining figure, AB,BC and AC are light metallic rods hinged at B. The tension on rod BC is
A) 150 N
B) 180 N
C) 500 N
D) $166\frac{1}{3}N$
• question_answer5) A 2 kg block is dropped from a height of 0.4m on a spring of force constant$k=1960\text{ }N{{m}^{-1}}$. The maximum compression of the spring is
A) 0.1 m
B) 0.2 M
C) 0.3 M
D) 0.4 M
• question_answer6) A solid sphere of mass 2 kg rolls up a$30{}^\circ$incline with an initial speed$10\text{ }m{{s}^{-1}}$. The maximum height reached by the sphere is$(g=10\text{ }m{{s}^{-2}})$
A) 3.5m
B) 7.0m
C) 10.5m
D) 14.5m
• question_answer7) A body is moving in a vertical circle of radius r such that the string is just tout at its highest point. The speed of the particle when the string is horizontal is
A) $\sqrt{gr}$
B) $\sqrt{2gr}$
C) $\sqrt{3gr}$
D) $\sqrt{5gr}$
• question_answer8) In some region, the gravitational field is zero. The gravitational potential in this region
A) may be zero
B) cannot be zero
C) must be change
D) None of these
• question_answer9) Some gas at 300K is enclosed in a container. Now, the container is placed on a fast moving train. While the trains is in motion, the temperature of gas
A) rises above 300K
B) falls below 300K
C) remains unchanged
• question_answer10) The ratio of the adiabatic bulk modulus to the isothermal bulk modulus of a perfect gas is equal to (symbols have their usual meanings)
A) ${{C}_{p}}-{{C}_{V}}$
B) $\frac{{{C}_{P}}}{{{C}_{V}}}$
C) $\frac{{{C}_{V}}}{{{C}_{P}}}$
D) $\sqrt{\frac{{{C}_{P}}}{{{C}_{V}}}}$
• question_answer11) The volume of a block of metal changed by 0.12% when heated through$20{}^\circ C$. Then a is
A) $2.0\times {{10}^{-5}}/{}^\circ C$
B) $4.0\times {{10}^{-5}}/{}^\circ C$
C) $6.0\times {{10}^{-5}}/{}^\circ C$
D) $8.0\times {{10}^{-5}}/{}^\circ C$
• question_answer12) A simple pendulum has time period T. The pendulum is completely immersed in a non-viscous liquid whose density is one-tenth of that of the material of the bob. The time period of the pendulum immersed in liquid is
A) $T$
B) $\sqrt{\frac{9}{10}}T$
C) $\sqrt{\frac{10}{9}}T$
D) $\frac{T}{10}$
• question_answer13) How many times more intense is a 90 dB sound than a 40 dB sound?
A) 2.5
B) 5
C) 50
D) ${{10}^{5}}$
• question_answer14) The minimum distance of a reflector to hear the echo of monosyllabic sound is (speed of sound is$330\text{ }m{{s}^{-1}}$)
A) 16.5 m
B) 33 m
C) 165 m
D) 330m
• question_answer15) If$X=\frac{{{\varepsilon }_{0}}lV}{t},$where${{\varepsilon }_{0}}$is the permittivity of free space,$l$is light, V is potential difference and$t$is time. The dimensions of$X$are the same as that of
A) charge
B) resistance
C) voltage
D) current
• question_answer16) An electric dipole placed in a uniform electric field will have minimum potential energy when the dipole moment is inclined to the field at an angle
A) $\pi$
B) $\frac{\pi }{2}$
C) zero
D) $\frac{3\pi }{2}$
• question_answer17) In the figure, the equivalent capacitance between A and B is
A) $3.75\text{ }\mu F$
B) $5.25\text{ }\mu F$
C) $6.5\text{ }\mu F$
D) $10.5\text{ }\mu F$
• question_answer18) Potential at appoint at a distance r from the centre of uniformly charged sphere of radius a $a(<r)$ is proportional to
A) ${{a}^{3}}$
B) r
C) $\frac{1}{r}$
D) $\frac{1}{{{a}^{3}}}$
• question_answer19) If we add impurity to a metal those atoms also deflect electrons. Therefore,
A) the electrical and thermal conductivity both increase
B) the electrical and thermal conductivity both decrease
C) the electrical conductivity increases but thermal conductivity decreases
D) the electrical conductivity decreases but thermal conductivity increases
• question_answer20) The voltage V and current$I$graph for a conductor at two different temperatures${{T}_{1}}$and${{T}_{2}}$are shown in figure. The relations between${{T}_{1}}$and${{T}_{2}}$is
A) ${{T}_{1}}>{{T}_{2}}$
B) ${{T}_{1}}<{{T}_{2}}$
C) ${{T}_{1}}={{T}_{2}}$
D) ${{T}_{1}}\approx {{T}_{2}}$
• question_answer21) $1\,Wb/{{m}^{2}}$is equal to
A) $10{{\,}^{4}}G$
B) $10{{\,}^{2}}G$
C) $10{{\,}^{-2}}G$
D) $10{{\,}^{-4}}G$
• question_answer22) A bar magnet has coercively$4\times {{10}^{3}}A\text{ }{{m}^{-1}}$. It is desired to demagnetize it by inserting it inside a solenoid 12 cm long and having 60 turns. The current that should be sent through the solenoid is
A) 2 A
B) 4A
C) 6 A
D) 8 A
• question_answer23) A thyroidal solenoid with an air core has an average radius of 15 cm, are of cross-section 12cm2 and 1200 turns. Ignoring the field variation across the cross-section of the toroid the self-inductance of the toroid is
A) 4.6 mH
B) 6.9 mH
C) 2.3 mH
D) 9.2 mH
• question_answer24) If$\phi$is phase difference between current and voltage the wattless component of current is
A) $I\,\cos \phi$
B) $I\sin \phi$
C) $I\tan \phi$
D) $I{{\cos }^{2}}\phi$
• question_answer25) To double the covering range of a TV transmitter tower, its height should be made
A) 2 times
B) 4 times
C) V2 times
D) 8 times
• question_answer26) A bulb is placed between two plane mirrors inclined at an angle of$60{}^\circ$. The number of images formed is
A) 5
B) 6
C) 4
D) 3
• question_answer27) Two plane mirrors are placed perpendicular to each other. A ray strikes one mirror and after reflection from the second mirror will be
A) perpendicular to the original ray
B) parallel to the original ray
C) at$45{}^\circ$to the original ray
D) can be at any angle to the original ray
• question_answer28) An electron moving with velocity$2\times {{10}^{7}}m{{s}^{-1}}$ describes a circle in a magnetic field of strength$2\times {{10}^{-2}}T$.If$\left( \frac{e}{m} \right)$of electrons is$1.76\times {{10}^{11}}K\,k{{g}^{-1}},$then the diameter of the circle is nearly
A) 1.1 cm
B) 1.1 mm
C) 1.1 m
D) 11 cm
• question_answer29) A neutron is confined to a nucleus of size${{10}^{-14}}m$. The minimum momentum of the electron may be
A) $6.6\times {{10}^{-20}}kg\text{ }m{{s}^{-1}}$
B) $3.3\times {{10}^{-20}}kg\text{ }m{{s}^{-1}}$
C) $3.3\times {{10}^{-48}}kg\text{ }m{{s}^{-1}}$
D) $6.6\times {{10}^{-48}}kg\text{ }m{{s}^{-1}}$
• question_answer30) An electron makes transition inside a hydrogen atom. The orbital angular momentum of the electron may change by
A) $h$
B) $\frac{h}{3\pi }$
C) $\frac{h}{2\pi }$
D) $\frac{h}{4\pi }$
• question_answer31) If 1 mg of${{U}^{235}}$is completely annihilated, the energy liberated is
A) $9\times {{10}^{10}}J$
B) $9\times {{10}^{19}}J$
C) $9\times {{10}^{18}}J$
D) $9\times {{10}^{17}}J$
• question_answer32) The transfer ratio$\beta$of a transistor is 50. The input resistance of the transistor when used in the common emitter mode is$1\,k\Omega$. The peak value of the collector alternating current for an input peak voltage of 0.01 V is
A) $100\text{ }\mu A$
B) $\text{500 }\mu A$
C) $\text{0}\text{.01 }\mu A$
D) $\text{0}\text{.25 }\mu A$
• question_answer33) It is possible to observe total internal reflection when a ray travels from
A) air into water
B) air into glass
C) water into glass
D) glass into water
• question_answer34) A combination of convex and concave lenses has power 4D. If the convex lens has power 5D the focal length of the concave lens will be
A) 100 cm
B) $-100cm$
C) $-1\,cm$
D) $-\frac{100}{9}\,cm$
• question_answer35) A liquid of refractive index 1.62 is placed between two plano-convex identical lenses, the medium of which has refractive index 1.54. Two possible arrangement P and Q are shown. The system is
A) divergent in P, convergent in Q
B) convergent in P, divergent ion Q
C) convergent in both
D) divergent in both
• question_answer36) A scooter of mass 120 kg is moving with a uniform velocity of$108\text{ }km{{h}^{-1}}$. The force required to stop the vehicle in 10 s is
A) 360 N
B) 720 N
C) 180 N
D) $120\times 10.8\text{ }N$
• question_answer37) Energy required to accelerate a car from $10\text{ }m{{s}^{-1}}$to$20\text{ }m{{s}^{-1}}$compared with that. required to accelerate from 0 to$10\text{ }m{{s}^{-1}}$is
A) twice
B) three times
C) four times
D) same
• question_answer38) The moment of inertia of the body about on axis is$1.2\text{ }kg\text{ }{{m}^{2}}$. Initially the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25 rads-2 must be applied about the axis for the duration of
A) 2 s
B) 4 s
C) 8s
D) 10s
• question_answer39) A cooking pot should have
A) high specific heat and low conductivity
B) high specific heat and high conductivity
C) low specific heat and low conductivity
D) low specific heat and high conductivity
• question_answer40) A certain force gives a 2 kg object an acceleration of$0.5\text{ }m{{s}^{-2}}$. What acceleration would the same force give to a 10 kg object?
A) $0.1m{{s}^{-2}}$
B) $0.2m{{s}^{-2}}$
C) $0.5m{{s}^{-2}}$
D) $1.0m{{s}^{-2}}$
• question_answer41) A heat engine absorbs heat at$327{}^\circ C$and exhausts heat at$127{}^\circ C$. The efficiency of engine is$\eta$and the maximum amount of work performed by the engine per kilocalorie of heat input is W. Then
A) $\eta =0.38$
B) $\eta =0.88$
C) $W=1596J$
D) $W=1400J$
• question_answer42) A simple pendulum with length L and mass of the bob is vibrating with an amplitude a. Then the maximum tension in the string is
A) $mg$
B) $mg\left[ 1{{\left( \frac{a}{1} \right)}^{2}} \right]$
C) $mg{{\left[ 1+\frac{a}{2L} \right]}^{2}}$
D) $mg{{\left[ 1+\left( \frac{a}{L} \right) \right]}^{2}}$
• question_answer43) A sonometer wire is vibrating in the second overtone. In the wire there are
A) two nodes and two antinodes
B) one node and two antinodes
C) four nodes and three antinodes
D) three nodes and three antinodes
• question_answer44) Two charged conducting spheres of radii${{R}_{1}}$ and${{R}_{2}},$separated by a large distance, are connected by a long wire. The ratio of the charges on, them is
A) $\frac{{{R}_{1}}}{{{R}_{2}}}$
B) $\frac{{{R}_{2}}}{{{R}_{1}}}$
C) $\frac{R_{1}^{2}}{R_{2}^{2}}$
D) $\frac{R_{2}^{2}}{R_{1}^{2}}$
• question_answer45) A proton and an$\alpha -$particle, accelerated through the same potential difference, enter a region of uniform magnetic field normally. If the radius of the proton orbit is 10 cm the radius of$\alpha -$orbit is
A) 10cm
B) $10\sqrt{2}cm$
C) 20cm
D) $5\sqrt{2}cm$
• question_answer46) In a noiseless transformer an alternating current of 2 A is flowing in the primary coil. The number of turns on the primary and secondary coils are 100 and 20 respectively. The value of the current in the secondary coil is
A) 0.08 A
B) 0.4 A
C) 5 A
D) 10 A
• question_answer47) Two photons of energy 2.5 eV each are incident on a metal plate whose work function is 4.0 eV, then the number of electrons emitted from the metal surface will be
A) one
B) two
C) more than two
D) None of these
• question_answer48) The ratio of areas between the electron orbits for the first excited state to the ground state for hydrogen atom is
A) $2:1$
B) $4:1$
C) $8:1$
D) $16:1$
• question_answer49) The voltage gain of an amplifier without feedback is 100. If a negative feedback is introduced with a feedback fraction$p=0.1,$ then the gain of the feedback amplifier is
A) 9.09
B) 10
C) 100.1
D) 90.0
• question_answer50) The sun subtends an angle of half a degree at the pole of a concave mirror which has a radius of curvature of 15 m. Then the size (diameter) of the image of the sun formed by the concave mirror is
A) 7.5 cm
B) 6.55 cm
C) 3.7cm
D) 13.1cm
• question_answer51) Two bodies of equal masses are connected by a light inextensible string passing over a smooth frictionless pulley. The amount of mass that should be transferred from one to another so that both the masses mole 50 m in 5 s is
A) 30 %
B) 40 %
C) 70%
D) 50%
• question_answer52) In the following$p-V$diagram, two Aetobatus cut two isothermal at${{T}_{1}}$and${{T}_{2}}$The value of${{V}_{b}}/{{V}_{c}}$is
A) ${{V}_{a}}/{{V}_{d}}$
B) $<\frac{{{V}_{a}}}{{{V}_{d}}}$
C) $>\frac{{{V}_{a}}}{{{V}_{d}}}$
D) cannot say
• question_answer53) A copper wire of cross-sectional area $=2.0\text{ }m{{m}^{2}},$resistivity$=1.7\times {{10}^{-8}}\Omega m,$carries a current of 1 A. The electric field in the copper wire is
A) $8.5\times {{10}^{-5}}V{{m}^{-1}}$
B) $8.5\times {{10}^{-4}}V{{m}^{-1}}$
C) $8.5\times {{10}^{-3}}V{{m}^{-1}}$
D) $8.5\times {{10}^{-2}}V{{m}^{-1}}$
• question_answer54) A solenoid 30 cm long is made by winding 2000 loops of wire on an iron rod whose cross section is$1.5\text{ }c{{m}^{2}}$. If the relative permeability of the iron is 6000, what is the self-inductance of the solenoid?
A) 1.5 H
B) 2.5 H
C) 3.5 H
D) 0.5 H
• question_answer55) A luminous object is placed 20 cm from surface of a convex mirror and a plane mirror is set, so that virtual mirror is at a distance of 12 cm from object, then focal length of convex mirror is
A) 5 cm
B) 10 cm
C) 20cm
D) 40cm
• question_answer56) The optically active tartaric acid is named as$D-(+)-$tartaric acid because it has a positive
A) optical rotation and derived from D-glucose
B) pH in organic solvent
C) optical rotation and is derived from$D-(+)-$glyceraldehyde
D) optical rotation when substituted by deuterium
• question_answer57) Which one of the following pairs is not correctly matched?
A) Clemmnensen reduction
B) Wolff-Kishner reduction
C) $-COCl\to -CHO$Rosemond reduction
D) $-C\equiv N\to -CHO$Stephen reduction
• question_answer58) Which of the following radical is precipitated as sulphide when treated with hydrogen sulphide in ammonia Cal solution?
A) $B{{a}^{2+}}$
B) $N{{i}^{2+}}$
C) $M{{g}^{2+}}$
D) $C{{a}^{2+}}$
• question_answer59) Isomerism which arises due to the presence of two different atoms in the same ligand is called
B) hydrate
C) salt
D) Both (a) and (b)
• question_answer60) The black compound formed during the reaction between sodium thiosulphate and silver nitrate is
A) silver thiosulphate
B) silver sulphide
C) silver sulphate
D) silver sulphite
• question_answer61) Electrode potential data are given below $F{{e}^{3+}}(aq)++{{e}^{-}}F{{e}^{2+}}(aq)$ $E{}^\circ =+0.77V$ $A{{l}^{3+}}(aq)+3{{e}^{-}}\xrightarrow[{}]{{}}Al(s)$ $E{}^\circ =-1.66\,V$ $B{{r}_{2}}(aq)+2{{e}^{-}}\xrightarrow[{}]{{}}2B{{r}^{-}}(aq)$ $E{}^\circ =+1.08\,V$ Based on the data given above, reducing power of$F{{e}^{2+}},Al$and$B{{r}^{-}},$will increase, in the order
A) $B{{r}^{-}}<F{{e}^{2+}}<Al$
B) $F{{e}^{2+}}<Al<B{{r}^{-}}$
C) $Al<B{{r}^{-}}<F{{e}^{2+}}$
D) $Al<F{{e}^{2+}}<B{{r}^{-}}$
• question_answer62) Which of the following volume (V)$-$temperature (T) plots represent the behaviour of one mole of an ideal gas at one atmospheric pressure?
A)
B)
C)
D)
• question_answer63) Amongst the following, identify the species with an atom in$+6$oxidation state
A) $MnO_{4}^{-}$
B) $Cr(CN)_{6}^{3-}$
C) $NiF_{6}^{2-}$
D) $Cr{{O}_{2}}C{{l}_{2}}$
• question_answer64) What is the concentration of nitrate ions if equal-volumes of$0.1M\,AgN{{O}_{3}}$and$0.1\text{ }M\text{ }NaCl$are mixed together?
A) 0.1 N
B) 0.2 M
C) 0.05 M
D) 0.25 M
• question_answer65) Consider the following statements (I) A sigma$(\sigma )$ bond is formed when two$s-$ orbitals overlap (II) A pi$(\pi )$bond is formed when two$p-$ orbitals axially overlap (III) A$\sigma$ bond is weaker than$\pi -$bond Which of the above statements is/are correct?
A) II and III
B) I and II
C) I alone
D) II alone
• question_answer66) Speed of decomposition of${{H}_{2}}{{O}_{2}}$is reduced by
A) $N{{a}_{2}}C{{O}_{3}}$
B) $NaOH$
C) alcohol
D) $Pt$
A) $N<Be<B$
B) ${{F}^{-1}}<{{O}^{2-}}<{{N}^{3-}}$
C) $Na<Li<K$
D) $F{{e}^{3+}}<F{{e}^{2+}}<F{{e}^{4+}}$
• question_answer68) Amalgamation method is used for the extraction of
A) noble metals
B) alkali metals
C) alkaline earth metals
D) transition elements
• question_answer69) The alkali metal that reacts with nitrogen directly to form nitride is
A) $Li$
B) $Na$
C) K
D) $Rb$
A) a solution of magnesium
B) a solution of magnesium carbonate
C) a solution of magnesium bicarbonate
D) a solution of magnesium sulphate
• question_answer71) A compound with the molecular formula ${{C}_{3}}{{H}_{8}}O$on vigorous oxidation produces an acid ${{C}_{3}}{{H}_{6}}{{O}_{2}}$. It is
A) a tertiary alcohol
B) a secondary alcohol
C) a primary alcohol
D) not necessarily an alcohol
• question_answer72) The correct order of basic ties of the following compounds is (1) (2) $C{{H}_{3}}-C{{H}_{2}}-N{{H}_{2}}$ (3) ${{(C{{H}_{3}})}_{2}}NH$ (4) $C{{H}_{3}}-\overset{\begin{smallmatrix} O \\ |\,\,| \end{smallmatrix}}{\mathop{C}}\,-N{{H}_{2}}$
A) $2>1>3>4$
B) $1>3>2>4$
C) $3>1>2>4$
D) $1>2>3>4$
• question_answer73) $R-\overset{\begin{smallmatrix} O \\ |\,| \end{smallmatrix}}{\mathop{C}}\,-OH\xleftarrow[{}]{{{H}_{3}}{{O}^{+}}}X\xrightarrow[{}]{[H]}RC{{H}_{2}}N{{H}_{2}}$ Here X is
A) isonitrile
B) nitrile
C) nitrite
D) oxime
• question_answer74) Identify$X$${{C}_{6}}{{H}_{6}}\xrightarrow[reflux]{HN{{O}_{3}}/{{H}_{2}}S{{O}_{4}}}$Intermediate$\xrightarrow[heat]{Sn/HCl}X$
A)
B)
C)
D)
• question_answer75) Rutherfords a-particle dispersion experiment concludes
A) all positive ions are deposited at small part
B) all negative ions are deposited at small part
C) proton moves around the electron
D) neutrons are charged particles
• question_answer76) Identify the correct statement when following compounds are given$HF,HBr,{{H}_{2}}Se,{{H}_{2}}Te,{{H}_{3}}P$
A) HF is strong acid
B) ${{H}_{2}}Te$is strong alkali
C) $HBr$is strong acid
D) ${{H}_{3}}P$is strong alkali
• question_answer77) Calcium is obtained by
A) electrolysis of molten $CaC{{l}_{2}}$
B) electrolysis of aq solution of $CaC{{l}_{2}}$
C) reduction of$CaC{{l}_{2}}$with carbon
D) roasting of lime stone
• question_answer78) Heat of dissociation of benzene to elements in$5535\text{ }kJ\text{ }mo{{l}^{-1}}$. The bond enthalpies of$C-C,$ $C=C,$and$C-H$are 347.3, 615.0 and 416.2 kJ respectively resonance energy of benzene is
A) 1.51 kJ
B) 15.1 kJ
C) 151 kJ
D) 1511 kJ
• question_answer79) The rate constant for the reaction,$2{{N}_{2}}{{O}_{5}}\xrightarrow[{}]{{}}4N{{O}_{2}}+{{O}_{2}}$$3.0\times {{10}^{-5}}{{s}^{-1}}$. If the rate is$2.40\times {{10}^{-5}}$mol ${{L}^{-1}}{{s}^{-1}},$Then the concentration of${{N}_{2}}{{O}_{5}}$(in mol ${{L}^{-1}}$) is
A) 1.4
B) 1.2
C) 0.04
D) 0.8
• question_answer80) Which one of the following has highest pH?
A) Distilled water
B) $1M\,N{{H}_{3}}$
C) $1M\,NaOH$
D) Water saturated with chlorine
• question_answer81) Which of the following element of$IIIA$group form alum with aluminum like alkali metals?
A) $B$
B) $Ca$
C) $In$
D) $Te$
• question_answer82) Which of the following is used as a test for detecting the presence of carbon monoxide?
A) Reduction of metallic oxides to metals
B) Reduction of water to hydrogen
C) Reduction of$PdC{{l}_{2}}$to$Pd$(Black)
D) All of the above
• question_answer83) Nitrogen can exists in two forms which are correct about them? (i) $\alpha -$nitrogen with cubic crystalline structure (ii)$\beta -$nitrogen with cubic crystalline structure (iii)$\beta -$nitrogen with hexagonal crystalline structure
A) Both (i) and (iii)
B) Both (i) and (ii)
C) Both (ii) and (iii)
D) None of these
• question_answer84) Which of the following mixture Is called black ash?
A) ${{K}_{2}}C{{O}_{3}}+CuS$
B) $N{{a}_{2}}C{{O}_{3}}+CaS$
C) ${{K}_{2}}C{{O}_{3}}+N{{a}_{2}}S$
D) $N{{a}_{2}}C{{O}_{3}}+N{{a}_{2}}S$
A) 4 oxygen atom
B) 2 oxygen atom
C) 3 oxygen atom
D) 10 oxygen atom
• question_answer86) When glucose is warmed with dilute alkali solution converted into a mixture of
A) glucose and manose
B) glucose and fructose
C) manose and fructose
D) glucose and manose and fructose
• question_answer87) Methyl amine reacts with nitrous acid to form
A) methyl nitrile
B) dimethyle ether
C) Both (a) and (b)
D) None of these
A) ${{C}_{6}}{{H}_{5}}OH>{{C}_{6}}{{H}_{5}}COOH>C{{H}_{3}}COOH$
B) ${{C}_{6}}{{H}_{5}}COOH>C{{H}_{3}}COOH>{{C}_{6}}{{H}_{5}}OH$
C) $C{{H}_{3}}COOH>{{C}_{6}}{{H}_{5}}COOH>{{C}_{6}}{{H}_{5}}OH$
D) ${{C}_{6}}{{H}_{5}}OH>C{{H}_{3}}COOH>{{C}_{6}}{{H}_{5}}COOH$
• question_answer89) Which of the following is most reactive towards nucleophilic addition reaction?
A) $HCHO$
B) $C{{H}_{3}}CHO$
C) ${{C}_{2}}{{H}_{5}}CHO$
D) $C{{H}_{3}}.CO.C{{H}_{3}}$
• question_answer90) Meta formaldehyde is a
A) polymer
B) tetramer
C) trimer
D) dimer
• question_answer91) An aqueous solution of urea freezes at 272.8 K. An equimolar solution of acetic acid in water will freeze at
A) 272.8 K
B) 272.79 K
C) 272.81 K
D) 272.6 K
• question_answer92) A mixture of 0.3 mol of H2 and 0. 3 mole of 12 is allowed to react in a 10 L evacuated flask at$500{}^\circ C$. The reaction is${{H}_{2}}+{{I}_{2}}2HI$.The K is found to be 64. The amount of unreacted 12 at equilibrium is
A) 0.15 mol
B) 0.06 mol
C) 0.03 mol
D) 0.2 mol
• question_answer93) $IF{{N}_{2}}+3{{H}_{2}}2N{{K}_{3}}-K$and $2{{N}_{2}}+6{{H}_{2}}4N{{H}_{3}}-{{k}^{-1}}$ then${{k}^{-1}}$ will be
A) ${{k}^{2}}$
B) $\sqrt{k}$
C) $\frac{1}{\sqrt{k}}$
D) $\frac{1}{{{k}^{2}}}$
• question_answer94) With the rise in temperature, the surface tension of a liquid
A) increases
B) decreases
C) remain constant
D) first increase then decrease
• question_answer95) At STP a container has 1 mole of Ar, 2 mol of $C{{O}_{2}},$3 mol of${{O}_{2}}$and 4 mol of${{N}_{2}}$without changing the total pressure if one mole of${{O}_{2}}$is removed, the partical pressure of${{O}_{2}}$
A) is change by about 16%
B) is halved
C) is changed by 26%
D) is unchanged
• question_answer96) Due to Frenkel defect, the density of ionic solid
A) decreases
B) increases
C) does not change
D) charge
A) trivalent impurity
B) tetravalent impurity
C) pentavaient impurity
D) divalent impurity
• question_answer98) 0.45 g of acid of molecular weight 90 was neutralised by 20 mL of 0.5 N Caustic potash. The basicity of the acid is
A) 1
B) 2
C) 3
D) 4
• question_answer99) Which of the following cannot give iodometric titrations?
A) $F{{e}^{3+}}$
B) $C{{u}^{2+}}$
C) $P{{b}^{2+}}$
D) $A{{g}^{+}}$
• question_answer100) Oxidation state of Fe in$F{{e}_{3}}{{O}_{4}}$is
A) 2/3
B) 4/5
C) 5/4
D) 8/3
• question_answer101) A compound contains atoms of three elements AB and C. If the oxidation number of A is$+2,$B is$+5,$and that of C is$-2,$the possible formula of the compound is
A) ${{A}_{3}}{{(B{{C}_{4}})}_{2}}$
B) ${{A}_{3}}{{({{B}_{4}}C)}_{2}}$
C) $AB{{C}_{2}}$
D) ${{A}_{2}}{{(B{{C}_{3}})}_{2}}$
• question_answer102) For the redox reaction,$MnO_{4}^{-}+{{C}_{2}}O_{4}^{2-}+{{H}^{+}}\xrightarrow[{}]{{}}M{{n}^{2+}}+C{{O}_{2}}+{{H}_{2}}O$the correct coefficient of the reactants for the balanced reaction are $MnO_{4}^{-}$ ${{C}_{2}}O_{4}^{2-}$ ${{H}^{+}}$
A) 2 5 16
B) 16 5 2
C) 5 16 2
D) 2 16 5
• question_answer103) An example of Lewis acid is
A) $NaCl$
B) $MgC{{l}_{2}}$
C) $CC{{l}_{4}}$
D) $AlC{{l}_{3}}$
• question_answer104) The conjugate acid of$NH_{2}^{-}$is
A) $N{{H}_{3}}$
B) $N{{H}_{2}}OH$
C) $NH_{4}^{+}$
D) ${{N}_{2}}{{H}_{4}}$
• question_answer105) Which of the following is hot paramagnetic?
A) $NO$
B) ${{N}_{2}}$
C) $CO$
D) ${{O}_{2}}$
• question_answer106) The values of electronegativitys of atom A and B are 1.20 and 4.0 respectively. The percentage of ionic character of$A-B$bond is
A) 50%
B) 72.24%
C) 55.3%
D) 43%
• question_answer107) A sample of wood decayed to 1/16 of its, original value. What is the number of${{t}_{1/2}}$?
A) 3
B) 4
C) 8
D) 16
A) binary fission
B) nuclear fission
C) stable nuclei
D) decay of unstable nuclei
• question_answer109) The kinetic energy of an electron accelerated from rest through a potential difference of 5V will be
A) 5eV
B) 5J
C) 5 erg
D) 80 eV
• question_answer110) A 2.5 mol sample of hydrazine,${{N}_{2}}{{H}_{4}}$loses 25 mole of electrons in being converted to a new compound$X$. Assuming that all of the nitrogen appears in the new compound, what is the oxidation state of nitrogen in compound$X$?
A) $-1$
B) $-2$
C) $+3$
D) $+4$
• question_answer111) If the expression$\frac{\left[ \sin \left( \frac{x}{2} \right)+\cos \left( \frac{x}{2} \right)-i\tan (x) \right]}{\left[ 1+2i\sin \left( \frac{x}{2} \right) \right]}$is real the set of all possible value of$x$is
A) $n\pi +\alpha$
B) $2n\pi$
C) $\frac{n\pi }{2}+\alpha$
D) None of these
• question_answer112) If${{t}_{1}},{{t}_{2}}$and${{t}_{3}}$are distinct, points$({{t}_{1}},2a{{t}_{3}}+at_{1}^{3}),$$({{t}_{2}},2a{{t}_{2}}+at_{2}^{3})$and$({{t}_{3}},2a{{t}_{3}}+at_{3}^{3})$are collinear, if
A) ${{t}_{1}}{{t}_{2}}{{t}_{3}}=1$
B) ${{t}_{1}}+{{t}_{2}}+{{t}_{3}}={{t}_{1}}{{t}_{2}}{{t}_{3}}$
C) ${{t}_{1}}+{{t}_{2}}+{{t}_{3}}=0$
D) ${{t}_{1}}+{{t}_{2}}+{{t}_{3}}=-1$
• question_answer113) If the pair of straight lines$a{{x}^{2}}+2hxy-a{{y}^{2}}=0$ and $b{{x}^{2}}+2gxy-b{{y}^{2}}=0,$be such that each bisect the angle between the other, then
A) $hg+ab=0$
B) $ah+bg=0.$
C) ${{h}^{2}}-ab=0$
D) $ag+bh=0.$
• question_answer114) The locus of a points which moves such that the sum of the squares of its distance from three vertices of the triangle is constant is a/an
A) circle
B) straight line
C) ellipse
D) None of these
• question_answer115) AB is a chord of the parabola${{y}^{2}}=4ax$with vertex at A. BC is drawn perpendicular to AB meeting the axis at C. The projection of BC on the axis of the parabola is
A) 2
B) $2a$
C) $4a$
D) $8a$
• question_answer116) If$\tan {{\theta }_{1}},\tan {{\theta }_{2}}=-\frac{{{a}^{2}}}{{{b}^{2}}},$ then the chord joining two points${{\theta }_{1}}$and${{\theta }_{2}}$on the ellipse $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$ will subtend a right angle at
A) focus
B) centre
C) end of the major axis
D) end of the minor axis
• question_answer117) The largest interval for which${{x}^{2}}-{{x}^{9}}+{{x}^{4}}-x+1>0$is
A) $-4<x<0$
B) $0<x<1$
C) $-100<x<100$
D) $-\infty <x<\infty$
• question_answer118) The eccentricity of the conic represented by ${{x}^{2}}-{{y}^{2}}-4x+4y+16=0$is
A) 1
B) $\sqrt{2}$
C) 2
D) $1/2$
• question_answer119) Let$f$be a twice differentiable function such that$f(x)=-f(x)$and$f(x)=g(x)$. If$h(x)=[f{{(x)}^{2}}+g{{(x)}^{2}}]h(1)=8$and$h(0)=2,$then$h(2)$is equal to
A) 1
B) 2
C) 3
D) None of these
• question_answer120) The equation of$f$those tangents to $4{{x}^{2}}-9{{y}^{2}}=36$which are perpendicular to the straight line$5x+2y-10=0$are
A) $5(y-3)=2\left( x-\frac{\sqrt{117}}{2} \right)$
B) $2x-5y+10-2\sqrt{18}=0$
C) $2x-5y-10-2\sqrt{18}=0$
D) None of the above
• question_answer121) If$f(x)=a\log |x|+b{{x}^{2}}+x$has its extremum value at$x=-1$and$x=2,$then
A) $a=2,b=-1$
B) $a=2,b=\frac{-1}{2}$
C) $a=-2,b=\frac{1}{2}$
D) None of these
• question_answer122) If$\int{\frac{1}{(\sin x+4)(\sin x-1)}}dx$$=A=\frac{1}{\tan \frac{x}{2}-1}+B{{\tan }^{-1}}(f(x))+{{C}_{1}}$. Then
A) $A=\frac{1}{5},B=\frac{-2}{5\sqrt{15}},f(x)=\frac{4\tan x+3}{\sqrt{15}}$
B) $A=-\frac{1}{5},B=\frac{1}{\sqrt{15}},f(x)=\frac{4\tan x\left( \frac{x}{2} \right)+1}{\sqrt{15}}$
C) $A=\frac{2}{5},B=\frac{-2}{5},f(x)=\frac{4\tan x+1}{5}$
D) $A=\frac{2}{5},B=\frac{-2}{5\sqrt{15}},f(x)=\frac{4\tan \frac{x}{2}+1}{\sqrt{15}}$
• question_answer123) If${{a}_{1}},{{a}_{2}},.....,{{a}_{n}}$On are in arithmetic progression, where${{a}_{i}}>0$for all$i$. Then $\frac{1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}+\frac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}}}+...+\frac{1}{\sqrt{{{a}_{n-1}}}+\sqrt{{{a}_{n}}}}$is equal to
A) $\frac{{{n}^{2}}(n+1)}{2}$
B) $\frac{n-1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}}$
C) $\frac{n(n-1)}{2}$
D) None of these
• question_answer124) Let$f$be a positive function. Let${{I}_{1}}=\int_{1-k}^{k}{xf\{x(1-x)\}}dx,$${{I}_{2}}=\int_{1-k}^{k}{f\{x(1-x)\}}dx$where$2k-1>0$. Then, $\frac{{{I}_{1}}}{{{I}_{2}}}$is
A) 2
B) $k$
C) $\frac{1}{2}$
D) 1
• question_answer125) The area bounded by the curve$y=x|x|,$$x-$axis and the ordinates$x=1,\text{ }x=-1$is given by
A) 6
B) $\frac{1}{3}$
C) $\frac{2}{3}$
D) None of these
• question_answer126) The degree of the differential equation satisfying$\sqrt{1-{{x}^{2}}}+\sqrt{1-{{y}^{2}}}=a(x-y)$is
A) 1
B) 2
C) 3
D) None of these
• question_answer127) 7 relatives of a man comprises 4 ladies and 3 gentlemen his wife has also 7 relatives, 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of mans relative and 3 of the wifes relative?
A) 485
B) 500
C) 486
D) 102
• question_answer128) If$\overrightarrow{\alpha }=x(\overrightarrow{a}\times \overrightarrow{b})+y(\overrightarrow{b}\times \overrightarrow{c})+z(\overrightarrow{c}\times \overrightarrow{a})$and$[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]=\frac{1}{8},$then$x+y+z$is equal to
A) $8\overrightarrow{\alpha }.(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})$
B) $\overrightarrow{\alpha }.(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})$
C) $8(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})$
D) None of these
• question_answer129) The$xy-$plane divides the line joining the points$(-1,3,4)$and$(2,-5,6)$
A) internally in the ratio $2:3$
B) externally in the ratio$2:3$
C) internally in the ratio$3:2$
D) externally in the ratio$3:2$
• question_answer130) The plane$2x-(1+\lambda )y+3\lambda z=0$passes through the intersection of the plane
A) $2x-y=0$and $y+3z=0$
B) $2x-y=0$and $y-3z=0$
C) $2x+3z=0$and $y=0$
D) None of the above
• question_answer131) In a triangle ABC,$sin\text{ }A-cos\,B=cosC,$then angle B is
A) $\frac{\pi }{2}$
B) $\frac{\pi }{3}$
C) $\frac{\pi }{4}$
D) $\frac{\pi }{6}$
• question_answer132) Eight chair are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chair marked 1 to 4; and then the men select the chairs from amongst the remaining. The number of possible arrangements is
A) $^{6}{{C}_{3}}{{\times }^{4}}{{C}_{2}}$
B) $^{4}{{P}_{2}}{{\times }^{6}}{{P}_{3}}$
C) $^{4}{{C}_{2}}{{\times }^{4}}{{P}_{3}}$
D) None of these
• question_answer133) In a $\Delta ABC,\tan \frac{A}{2}=\frac{5}{6},\tan \frac{C}{2}=\frac{2}{5},$then
A) a, c, b are in AP
B) a, b, c are in AP
C) b, a, c are in AP
D) a, b, c are in GP
• question_answer134) $\Sigma {{a}^{3}}\cos (B-C)$is equal to
A) $3abc$
B) $3(cz+b+c)$
C) $abc(a+b+c)$
D) 0
• question_answer135) If$\alpha ,\beta ,\gamma \in \left[ 0,\frac{\pi }{2} \right],$ then the value of $\frac{\sin (\alpha +\beta +\gamma )}{sin\text{ }\alpha +sin\,\beta +sin\text{ }\gamma }$is
A) $<1$
B) $=-1$
C) $<0$
D) None of these
• question_answer136) If $f(x)=$$\left| \begin{matrix} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \\ \end{matrix} \right|$then$f(100)$is equal to
A) 0
B) 1
C) 100
D) $-100$
• question_answer137) The general value of 6 satisfying the equation $2si{{n}^{2}}\theta -3sin\theta -2=0$is
A) $n\pi +{{(-1)}^{n+1}}\frac{\pi }{6}$
B) $n\pi +{{(-1)}^{n}}\frac{\pi }{2}$
C) $n\pi +{{(-1)}^{n}}\frac{5\pi }{6}$
D) $n\pi +{{(-1)}^{n}}\frac{7\pi }{6}$
• question_answer138) In a right angled triangle the hypotenuse is$2\sqrt{2}$times the length of perpendicular drawn from the opposite vertex on the hypotenuse, then the other two angles are
A) $\frac{\pi }{3},\frac{\pi }{6}$
B) $\frac{\pi }{4},\frac{\pi }{4}$
C) $\frac{\pi }{8},\frac{3\pi }{8}$
D) $\frac{\pi }{12},\frac{5\pi }{12}$
• question_answer139) lf $x+y+z=xyz,$then $ta{{n}^{-1}}x+ta{{n}^{-1}}y+ta{{n}^{-1}}z$ is equal to
A) $\pi$
B) $\frac{\pi }{2}$
C) 1
D) None of these
• question_answer140) If two events A and B are such$P({{A}^{c}})=0.3$ $P(B)=0.4$and$P(A\cap {{B}^{c}})=0.5,$then $P[B/{{(A\cup B)}^{c}}]$is equal to
A) 1/2
B) 1/4
C) 0
D) None of these
• question_answer141) If${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}$are roots of the equation${{x}^{4}}-{{x}^{3}}$ $\sin 2\beta +{{x}^{2}}\cos 2\beta -x\cos \beta -\sin \beta =0,$ then ${{\tan }^{-1}}{{x}_{1}}+{{\tan }^{-1}}{{x}_{2}}+{{\tan }^{-1}}{{x}_{3}}+{{\tan }^{-1}}{{x}_{4}}$is equal to
A) $\beta$
B) $\frac{\pi }{2}-\beta$
C) $\pi -\beta$
D) $-\beta$
• question_answer142) The value of$\cos (2{{\cos }^{-1}}0.8)$is
A) 0.48
B) 0.96
C) 0.6
D) None of these
• question_answer143) ABC is a triangular park with$AB=AC=100\text{ }m$A clock tower is situated at the mid point of BC. The angle of elevation if the top of the tower at A and B are$co{{t}^{-1}}3.2$and$\cos e{{c}^{-1}}2.6$ respectively. The height of the tower is
A) 16 m
B) 25 m
C) 50 m
D) None of these
• question_answer144) The equation$\frac{3}{4}{{({{\log }_{2}}x)}^{2}}+{{\log }_{2}}x-\frac{5}{4}={{\log }_{x}}\sqrt{2}$has
A) at least one real solutions
B) exactly three real solutions
C) exactly one irrational solution
D) complex roots
• question_answer145) If G is the GM of the product of r set of observation with geometric means${{G}_{1}},{{G}_{2}},....{{G}_{r}}$respectively, then G is equal to
A) $\log {{G}_{1}}+\log {{G}_{2}}+....+\log {{G}_{n}}$
B) ${{G}_{1}}{{G}_{2}}....{{G}_{n}}$
C) $\log {{G}_{1}},\log {{G}_{2}},....,\log {{G}_{n}}$
D) None of the above
• question_answer146) If$Z=aX+bY$and r the correlation coefficient between$X$and$Y$,then$\sigma _{z}^{2}$is equal to
A) ${{a}^{2}}\sigma _{X}^{2}+{{b}^{2}}\sigma _{Y}^{2}+2abr\,{{\sigma }_{X}}{{\sigma }_{Y}}$
B) ${{a}^{2}}\sigma _{X}^{2}+{{b}^{2}}\sigma _{Y}^{2}-2abr\,{{\sigma }_{X}}{{\sigma }_{Y}}$
C) $2abr\,{{\sigma }_{X}}{{\sigma }_{Y}}$
D) None of the above
• question_answer147) If${{\log }_{2}}({{5.2}^{x}}+1),{{\log }_{4}}({{2}^{1-x}}+1)$and 1 are in AP, then$x$equals
A) $lo{{g}_{2}}5$
B) $1-lo{{g}_{2}}5$
C) $lo{{g}_{5}}2$
D) None of these
• question_answer148) If$A=\left[ \begin{matrix} \alpha & 0 \\ 1 & 1 \\ \end{matrix} \right]$and$B=\left[ \begin{matrix} 1 & 0 \\ 5 & 1 \\ \end{matrix} \right],$then value of a for which${{A}^{2}}=B$is
A) 1
B) $-1$
C) 4
D) no real value
• question_answer149) The maximum value of$|z|$when z satisfies the condition$\left| 2+\frac{2}{z} \right|=2$is
A) $\sqrt{3}-1$
B) $\sqrt{3}$
C) $\sqrt{3}+1$
D) $\sqrt{2}+\sqrt{3}$
• question_answer150) The value$1+\sum\limits_{k=0}^{14}{\left\{ \cos \frac{(2k+1)\pi }{15}+i\sin \frac{(2k+1)\pi }{15} \right\}}$is
A) 0
B) $-1$
C) 1
D) $i$
• question_answer151) The sum to n terms of the infinite series ${{1.3}^{2}}+{{2.5}^{2}}+{{3.7}^{2}}+...\infty$is
A) $\frac{n}{6}(n+1)(6{{n}^{2}}+14n+7)$
B) $\frac{n}{6}(n+1)(2n+1)(3n+1)$
C) $4{{n}^{3}}+4{{n}^{2}}+n$
D) None of the above
• question_answer152) If$f(x)={{\sin }^{2}}x+{{\sin }^{2}}\left( x+\frac{\pi }{3} \right)$$+\cos x\cos \left( x+\frac{\pi }{3} \right)$ and$g\left( \frac{5}{4} \right)=1,$then$gof(x)$is equal to
A) 0
B) 2
C) 1
D) 3
• question_answer153) If the roots of the quadratic equation${{x}^{2}}-4x-lo{{g}_{3}}a=0$are real, then the least value of a is
A) 81
B) 1/81
C) 1/64
D) None of these
• question_answer154) The exponent of 12 in 100 is
A) 48
B) 49
C) 96
D) None of these
• question_answer155) The value of$1.1!+2.2!+3.3!+...+n.n!$is
A) $(n+1)!$
B) $(n+1)!+1$
C) $(n+1)!-1$
D) None of these
• question_answer156) $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{r=1}^{2n}{\frac{r}{\sqrt{{{n}^{2}}+{{r}^{2}}}}}$equals
A) $1+\sqrt{5}$
B) $-1+\sqrt{5}$
C) $-1+\sqrt{2}$
D) $1+\sqrt{2}$
• question_answer157) If${{(1+2x+3{{x}^{2}})}^{10}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}$$+...+{{a}_{20}}{{x}^{20}},$then${{a}_{1}}$equals
A) 10
B) 20
C) 210
D) None of these
• question_answer158) The sum of series$\sum\limits_{n-1}^{\infty }{\frac{2n}{(2n+1)!}}$is
A) $e$
B) $-1$
C) $2e$
D) None of these
• question_answer159) If the value of the determinants$\left| \begin{matrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \\ \end{matrix} \right|$is positive, then
A) $abc>1$
B) $abc>-8$
C) $abc<-8$
D) $abc>-2$
• question_answer160) The function$f(x)=\frac{ln(1+ax)-ln(1-bx)}{x}$is not defined at$x=0$. The value which should be assigned to$f$at$x=0$so that it is continuous at$x=0,$is
A) $a-b$
B) $a+b$
C) $ln\text{ }a+ln\text{ }b$
D) None of these
• question_answer161) If$f(x+2y,x,x-2y)=xy,$then$f(x,y)$equals
A) $\frac{{{x}^{2}}-{{y}^{2}}}{8}$
B) $\frac{{{x}^{2}}-{{y}^{2}}}{4}$
C) $\frac{{{x}^{2}}+{{y}^{2}}}{4}$
D) $\frac{{{x}^{2}}-{{y}^{2}}}{2}$
• question_answer162) If$\underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{{{x}^{3}}+1}{{{x}^{2}}+1}-(ax+b) \right]=2,$then
A) $a=1,\text{ }b=-2$
B) $a=-1,\text{ }b=2$
C) $a=-1,\text{ }b=-2$
D) None of these
• question_answer163) $\underset{x\to \infty }{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+1}-\sqrt[3]{{{x}^{3}}+1}}{\sqrt{{{x}^{4}}+1}-\sqrt[3]{{{x}^{4}}+1}}$equals
A) 1
B) 0
C) $-1$
D) None of these
• question_answer164) The tangent at (1,7) to the curve${{x}^{2}}=y-6$touches the circle${{x}^{2}}+{{y}^{2}}+16x+12y+c=0$at
A) (6, 7)
B) $(-6,7)$
C) $(6,-7)$
D) $(-6,-7)$
• question_answer165) The set of points where the function$f(x)=x|x|$is differentiable is
A) $(-\infty ,\infty )$
B) $(-\infty ,0)\cup (0,\infty )$
C) $(0,\infty )$
D) $[0,\infty )$
• question_answer166) If$f(x)=|{{\log }_{e}}|x||,$then${{f}_{0}}(x)$equals
A) $\frac{1}{|x|},x\ne 0$
B) $\frac{1}{x}$for $|x|>1$and$\frac{-1}{x}$for $|x|<1$
C) $\frac{-1}{x}$for $|x|>1$and $\frac{1}{x}$for $|x|<1$
D) $\frac{1}{x}$for$x>0$and$-\frac{1}{x}$for $x<0$
• question_answer167) The equation of the tangent to the curve $y=(2x-1){{e}^{2(1-x)}}$at the points its maximum, is
A) $y-1=0$
B) $x-1=0$
C) $x+y-1=0$
D) $x-y+1=0$
• question_answer168) A function$f(x)=\left\{ \begin{matrix} 1+x,\,\,x\le 2 \\ 5-x,\,\,x>2 \\ \end{matrix} \right.$is
A) not continuous at $x=2$
B) differentiable at$x=2$
C) continuous but not differentiable at$x=2$
D) None of the above
• question_answer169) The interval of increase of the function$f(x)=x-{{e}^{x}}+\tan (2\pi /7)$is
A) $(0,\infty )$
B) $(-\infty ,0)$
C) $(1,\infty )$
D) $(-\infty ,-1)$
• question_answer170) Let$P(x)={{a}_{0}}+{{a}_{1}}{{x}^{2}}+{{a}_{2}}{{x}^{4}}+...+{{a}_{n}}{{x}^{2n}}$be a polynomial in a real variable$x$with $0<{{a}_{0}}<{{a}_{1}}<{{a}_{2}}<...<{{a}_{n}}$.The function$P(x)$has
A) neither a maximum nor a minimum
B) only one maximum
C) only one minimum
D) only one maximum and only one minimum | 14,914 | 39,326 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-35 | latest | en | 0.688578 |
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##### Math facts
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kidsrterrific Joined: Nov 2012 Posts: 992 Senior Member
kidsrterrific
Joined: Nov 2012
Posts: 992
Senior Member
Math facts
04-30-2019, 02:33 PM
#1
Hi all! My last five years teaching full-time I had second grade.
Right now I am tutoring a homeless child (through a volunteer community project) that is in second grade and struggling with her addition math facts.
I seem to remember a fun online free game called math blasters. Maybe I’m totally out of the loop because it has been four years. It was free unless you wanted to add more games. It was fun, it worked on each number’s math facts family.
This is not for our tutoring session but how she can practice for fun at home. The children are provided netbooks at school that they are able to bring home.
Any other ideas for fun practice other than flashcards? I have a memory game and a bingo that we can practice during our break. I’m thinking more interactive. I will take anything that you can offer.
iteachk2010 Joined: Jun 2011 Posts: 2,164 Senior Member
iteachk2010
Joined: Jun 2011
Posts: 2,164
Senior Member
05-01-2019, 03:48 AM
#2
What about working with dominoes? Use the ones that apply to the facts you are working on. This will give her a visual to go along with the fact. Hold up each domino. Have her tell how many in each group and how many altogether. Then have her sort all the dominoes by the sum. You can have her match the dominoes to the math fact (written on index cards), too.
We work with facts in groups.Once they master one group, we add the next group. i.e.the +0 facts and their turn arounds, the +1 facts and their turn arounds...doubles.
kidsrterrific Joined: Nov 2012 Posts: 992 Senior Member
kidsrterrific
Joined: Nov 2012
Posts: 992
Senior Member
Great idea
05-01-2019, 06:12 PM
#3
And...I just saw a set of dominos at the Center where we work today.
Thank you!
iluvmy3kids Joined: Oct 2009 Posts: 394 Senior Member
iluvmy3kids
Joined: Oct 2009
Posts: 394
Senior Member
math facts
07-21-2019, 01:39 PM
#4
If you have access to online try XtraMath. It is free. It is individualized practice. I personally used it with games and fun practice also, but it took the place of flash cards. Google XtraMath and look at the website. Most of my 2nd graders loved it, but I know a lot of the 1st grade teachers used it also.
Join the conversation! Post as a guest or become a member today. New members welcome!
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# Chapter 6 Section Reviews Pg 144 #1-4; pg 149 #1-3; pg 153 #1-3
## Presentation on theme: "Chapter 6 Section Reviews Pg 144 #1-4; pg 149 #1-3; pg 153 #1-3"— Presentation transcript:
Chapter 6 Section Reviews Pg 144 #1-4; pg 149 #1-3; pg 153 #1-3
Forces in Motion Chapter 6 Section Reviews Pg 144 #1-4; pg 149 #1-3; pg 153 #1-3
Pg 144 #1-4 1. How does air resistance affect the acceleration of falling objects? Air resistance slows or stops acceleration of falling objects. 2. Explain why an astronaut in an orbiting spaceship floats. An astronaut in an orbiting spaceship floats because both the astronaut and the spaceship are in free fall. Since both fall at the same rate, the astronaut floats inside. The astronaut has not sensation of falling.
Pg 144 #1-4 (cont.) 3. How is an orbit formed?
An orbit is formed by combining two motions: a forward motion and free fall toward Earth. The path that results is a curve that matches the curve of Earth’s surface. 4. Think about a sport you play that involves a ball. Identify at least four different instances in which an object is in projectile motion. Basketball – a player jumping to dunk the ball, a ball passed from one player to another, a ball shot toward the basket, a ball bounced on the floor.
Pg 149 #1-3 1. How is inertia related to Newton’s law of motion?
Newton’s first law says that matter resists any change in motion. Inertia is the tendency of objects (matter) to resist changes in motion. Newton’s first law is also known as the law of inertia. 2. Name two ways to increase the acceleration of an object. You can increase the acceleration of an object by increasing the force causing the acceleration or by reducing the object’s mass.
Pg 149 #1-3 (cont.) 3. If the acceleration due to gravity were somehow doubled to 19.6 m/s/s, what would happen to your weight? If the acceleration due to gravity were doubled, your weight would double. This is because of Newton’s second law: F = ma. Weight is the force due to the acceleration on mass. If acceleration is doubled and mass remains the same, the force (weight) is doubled, too.
Pg 153 #1-3 1. Name three action and reaction force pairs involved in doing your homework. Name what object is exerting and what object is receiving the forces. Using a pencil or pen (action: hand pushing on pencil; reaction: pencil pushing back on hand OR action: pencil pushing on paper; reaction: paper pushing on pencil).
Pg 153 #1-3 (cont.) 2. Which has more momentum, a mouse running at 1 m/s north or an elephant walking at 3 m/s east? Explain your answer. The elephant has both a greater mass and greater velocity.
Pg 153 #1-3 (cont.) 3. When a truck pulls a trailer, the trailer and truck accelerate forward even though the action and reaction forces are the same size but in opposite directions. Why don’t these forces balance each other out? The action and reaction forces do not balance each other because the forces are acting on two different objects. Because they act on two different objects, you cannot combine them to determine a net force.
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From: francogrex <francogrex_at_mail.com>
Date: Thu 11 Jan 2007 - 16:21:25 GMT
Hi guys again, it seems I haven't been doing the maximum likelihood estimation correctly. I quote below, can someone explain to me please what does it mean that the 2nd and 3rd derivatives of the function equals zero and how to compute that in R.
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--
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For questions related to the puzzle of islanders who can only leave when they deduce the colour of their own eyes.
The puzzle can be found here and is as follows.
A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
And lastly, the answer is not "no one leaves."
The answer is that on the 100th day, all 100 blue-eyed people leave the island. This tag is for questions about this puzzle or its many variants. | 453 | 2,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-10 | latest | en | 0.973467 |
https://www.airmilescalculator.com/distance/goh-to-jhs/ | 1,716,409,313,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00476.warc.gz | 569,423,582 | 13,110 | # How far is Sisimiut from Nuuk?
The distance between Nuuk (Nuuk Airport) and Sisimiut (Sisimiut Airport) is 200 miles / 322 kilometers / 174 nautical miles.
The driving distance from Nuuk (GOH) to Sisimiut (JHS) is 238 miles / 383 kilometers, and travel time by car is about 74 hours 38 minutes.
200
Miles
322
Kilometers
174
Nautical miles
## Distance from Nuuk to Sisimiut
There are several ways to calculate the distance from Nuuk to Sisimiut. Here are two standard methods:
Vincenty's formula (applied above)
• 200.070 miles
• 321.982 kilometers
• 173.856 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 199.502 miles
• 321.067 kilometers
• 173.362 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Nuuk to Sisimiut?
The estimated flight time from Nuuk Airport to Sisimiut Airport is 52 minutes.
## Flight carbon footprint between Nuuk Airport (GOH) and Sisimiut Airport (JHS)
On average, flying from Nuuk to Sisimiut generates about 54 kg of CO2 per passenger, and 54 kilograms equals 120 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Nuuk to Sisimiut
See the map of the shortest flight path between Nuuk Airport (GOH) and Sisimiut Airport (JHS).
## Airport information
Origin Nuuk Airport
City: Nuuk
Country: Greenland
IATA Code: GOH
ICAO Code: BGGH
Coordinates: 64°11′27″N, 51°40′41″W
Destination Sisimiut Airport
City: Sisimiut
Country: Greenland
IATA Code: JHS
ICAO Code: BGSS
Coordinates: 66°57′4″N, 53°43′45″W | 514 | 1,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-22 | latest | en | 0.831989 |
https://linearalgebra.me/mathematical-notes-the-solution-and-resultant-of-elimination/ | 1,716,579,874,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00024.warc.gz | 305,194,373 | 19,846 | # Mathematical Notes: The Solution and Resultant of Elimination
\]This form of the solution conforms most closely to common sense: since $x^{\prime}$ contains $b$ and is contained in $a^{\prime}$, it is natural that $x$ should be equal to the sum of $b$ and a part of $a^{\prime}$ (that is to say, the part common to $a^{\prime}$ and $x$). The solution is generally indeterminate (between the limits $a^{\prime}$ and $b$); it is determinate only when the limits are equal,$a^{\prime}=b,$for then$x=b+a^{\prime}x=b+bx=b=a^{\prime}.$Then the equation assumes the form$(ax+a^{\prime}x^{\prime}=0)=(a^{\prime}=x)$and is equivalent to the double inclusion\[(a^{\prime} indeterminate _.We shall reach the same conclusion if we observe that ($a+b$) is the superior limit of the function $ax+bx$ and that, if this limit is 0, the function is necessarily zero for all values of $x$,\[(ax+bx^{\prime} | 259 | 891 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-22 | latest | en | 0.803957 |
http://reference.wolfram.com/legacy/language/v10.4/ref/MeanShiftFilter.html | 1,508,647,784,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825141.95/warc/CC-MAIN-20171022041437-20171022061437-00497.warc.gz | 296,710,286 | 10,000 | # Wolfram Language & System 10.4 (2016)|Legacy Documentation
This is documentation for an earlier version of the Wolfram Language.
BUILT-IN WOLFRAM LANGUAGE SYMBOL
# MeanShiftFilter
MeanShiftFilter[image,r,d]
replaces each pixel with the mean of the pixels in a range-r neighborhood and whose value is within a distance d.
MeanShiftFilter[array,r,d]
applies mean-shift filtering to a data array.
## Details and OptionsDetails and Options
• MeanShiftFilter[image,r,d] finds the mean of the pixels in blocks centered on each pixel, taking into account only those pixels whose value is within a Euclidean distance d from the center pixel.
• MeanShiftFilter works with arbitrary 2D and 3D images, as well as data arrays of any rank.
• At the edges of an image, MeanShiftFilter uses smaller neighborhoods.
• MeanShiftFilter[image,{rrow,rcol},d] uses range in height, and in width.
• MeanShiftFilter[image,{rslice,rrow,rcol},d] uses range in height, in depth, and in width.
• The following options can be given:
• DistanceFunction EuclideanDistance how to compute the distance between values MaxIterations 1 maximum number of iterations to be performed
• For a complete list of possible settings for DistanceFunction, see the reference page for MeanShift.
• The possible range for the distance parameter d depends on the distance function as well as the dimension of the color space.
## ExamplesExamplesopen allclose all
### Basic Examples (3)Basic Examples (3)
Mean-shift filtering of a color image:
In[1]:=
Out[1]=
Mean-shift filtering of a grayscale image:
In[1]:=
Out[1]=
Mean-shift filtering of a 3D image:
In[1]:=
Out[1]= | 390 | 1,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-43 | latest | en | 0.740192 |
https://www.stackage.org/lts-22.17/package/csp-1.4.0 | 1,716,610,272,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058773.28/warc/CC-MAIN-20240525035213-20240525065213-00839.warc.gz | 848,880,965 | 5,787 | # csp
Discrete constraint satisfaction problem (CSP) solver.
LTS Haskell 22.23: 1.4.0 Stackage Nightly 2024-05-25: 1.4.0 Latest on Hackage: 1.4.0
See all snapshots `csp` appears in
LicenseRef-LGPL licensed and maintained by
This version can be pinned in stack with:`csp-1.4.0@sha256:051beb4864f737ea42377918e0c13a7aef4ee5c6f4f39993cc224f6f5bbf470d,1062`
#### Module documentation for 1.4.0
• Control
• Control.Monad
Depends on 4 packages(full list with versions):
# CSP
This package is available via Hackage where its documentation resides. It provides a solver for constraint satisfaction problems by implementing a `CSP` monad. Currently it only implements arc consistency but other kinds of constraints will be added.
Below is a Sudoku solver, project Euler problem 96.
``````import Data.List
import Control.Monad.CSP
mapAllPairsM_ :: Monad m => (a -> a -> m b) -> [a] -> m ()
mapAllPairsM_ f [] = return ()
mapAllPairsM_ f (_:[]) = return ()
mapAllPairsM_ f (a:l) = mapM_ (f a) l >> mapAllPairsM_ f l
solveSudoku :: (Enum a, Eq a, Num a) => [[a]] -> [[a]]
solveSudoku puzzle = oneCSPSolution \$ do
dvs <- mapM (mapM (\a -> mkDV \$ if a == 0 then [1 .. 9] else [a])) puzzle
mapM_ assertRowConstraints dvs
mapM_ assertRowConstraints \$ transpose dvs
sequence_ [assertSquareConstraints dvs x y | x <- [0,3,6], y <- [0,3,6]]
return dvs
where assertRowConstraints = mapAllPairsM_ (constraint2 (/=))
assertSquareConstraints dvs i j =
mapAllPairsM_ (constraint2 (/=)) [(dvs !! x) !! y | x <- [i..i+2], y <- [j..j+2]]
sudoku3 = [[0,0,0,0,0,0,9,0,7],
[0,0,0,4,2,0,1,8,0],
[0,0,0,7,0,5,0,2,6],
[1,0,0,9,0,4,0,0,0],
[0,5,0,0,0,0,0,4,0],
[0,0,0,5,0,7,0,0,9],
[9,2,0,1,0,8,0,0,0],
[0,3,4,0,5,9,0,0,0],
[5,0,7,0,0,0,0,0,0]]
solveSudoku sudoku3
``````
## Future
• Allow a randomized execution order for CSPs
• CSPs don’t need to use IO internally. ST is enough.
• Constraint synthesis. Already facilitated by the fact that constraints are internally nondeterministic
• Other constraint types for CSPs, right now only AC is implemented
• n-ary heterogeneous constraints | 751 | 2,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-22 | latest | en | 0.721121 |
http://spmaddmaths.onlinetuition.com.my/2014/01/function-example-2-and-3.html | 1,503,270,007,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106996.2/warc/CC-MAIN-20170820223702-20170821003702-00211.warc.gz | 387,627,095 | 21,563 | # Function Example 2 and 3
Example 2
Function f is defined as $f:x\to \frac{5}{2x-1},x\ne k$
Find the value of k.
Example 3
Given $g:x\to \frac{3x-5}{2x+7}$
Function g is defined for all values of x except x=a. Find the value of a. | 88 | 237 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-34 | longest | en | 0.542464 |
http://www.convertit.com/Go/Maps/Measurement/Converter.ASP?From=Roman+cubit&To=width | 1,670,359,751,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711114.3/warc/CC-MAIN-20221206192947-20221206222947-00807.warc.gz | 65,615,274 | 5,764 | Search Maps.com
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(800) 430-7532 | info@maps.com | Online Privacy Policy | 576 | 2,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-49 | latest | en | 0.690598 |
https://mcqslearn.com/cost-accounting/quizzes/quiz-questions-and-answers.php?page=5 | 1,708,875,243,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474617.27/warc/CC-MAIN-20240225135334-20240225165334-00143.warc.gz | 385,413,954 | 20,333 | BBA: Finance Courses
MBA Cost Accounting Certification Exam Tests
MBA Cost Accounting Practice Test 5
# Estimating Cost Functions Quiz Questions and Answers PDF - 5
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## Estimating Cost Functions Quiz Answers : Test 5
MCQ 21: A helpful technique, for accurate forecasts about costs to be incurred in future is a part of
A) unit estimation
B) production estimation
C) cost estimation
D) price estimation
MCQ 22: An approach used for choosing capacity level, having no beginning inventory, is classified as
A) write off variance approach
B) write in variance approach
MCQ 23: If the actual input price is \$150 and the budgeted input price is \$80, then the price variance will be
A) \$130
B) \$70
C) \$150
D) \$80
MCQ 24: The difference between the budgeted amounts and the actual results is classified as
A) standard deviation
B) variances
C) mean average
D) weighted average
MCQ 25: The standard input allows one unit, to be divided by standard cost per output unit, for variable direct cost input to calculate
A) standard price per input unit
B) standard price per output unit
C) standard cost per input unit
D) standard cost per output unit
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Business Statistics App (Android & iOS) | 497 | 2,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-10 | latest | en | 0.71944 |
http://www.dictionary.com/browse/asymptotically?qsrc=2446 | 1,500,737,877,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424079.84/warc/CC-MAIN-20170722142728-20170722162728-00167.warc.gz | 404,968,582 | 24,158 | Try Our Apps
asymptotic
or asymptotical
[as-im-tot-ik or as-im-tot-i-kuh l] /ˌæs ɪmˈtɒt ɪk or ˌæs ɪmˈtɒt ɪ kəl/
1.
of or relating to an asymptote.
2.
(of a function) approaching a given value as an expression containing a variable tends to infinity.
3.
(of two functions) so defined that their ratio approaches unity as the independent variable approaches a limit or infinity.
4.
(of a formula) becoming increasingly exact as a variable approaches a limit, usually infinity.
5.
coming into consideration as a variable approaches a limit, usually infinity:
asymptotic property; asymptotic behavior.
Origin of asymptotic
1665-1675
First recorded in 1665-75; asymptote + -ic
Related forms
Dictionary.com Unabridged
Based on the Random House Dictionary, © Random House, Inc. 2017.
Cite This Source
Examples from the Web for asymptotically
Historical Examples
• The theory is not a thing complete from the first, but a thing which grows, as it were asymptotically, towards certainty.
John Tyndall
British Dictionary definitions for asymptotically
asymptotic
/ˌæsɪmˈtɒtɪk/
1.
of or referring to an asymptote
2.
(of a function, series, formula, etc) approaching a given value or condition, as a variable or an expression containing a variable approaches a limit, usually infinity
Derived Forms
Collins English Dictionary - Complete & Unabridged 2012 Digital Edition
Publishers 1998, 2000, 2003, 2005, 2006, 2007, 2009, 2012
Cite This Source
Word Origin and History for asymptotically
asymptotic
1670s, see asymptote + -ic. Related: Asymptotical; asymptotically.
Online Etymology Dictionary, © 2010 Douglas Harper
Cite This Source
Difficulty index for asymptotic
Some English speakers likely know this word
Word Value for asymptotically
0
29
Scrabble Words With Friends | 494 | 1,774 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-30 | latest | en | 0.831474 |
https://www.tidy-finance.org/python/beta-estimation.html | 1,726,177,577,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00085.warc.gz | 956,298,861 | 26,013 | Note
You are reading Tidy Finance with Python. You can find the equivalent chapter for the sibling Tidy Finance with R here.
In this chapter, we introduce an important concept in financial economics: The exposure of an individual stock to changes in the market portfolio. According to the Capital Asset Pricing Model (CAPM) of Sharpe (1964), Lintner (1965), and Mossin (1966), cross-sectional variation in expected asset returns should be a function of the covariance between the excess return of the asset and the excess return on the market portfolio. The regression coefficient of excess market returns on excess stock returns is usually called the market beta. We show an estimation procedure for the market betas. We do not go into details about the foundations of market beta but simply refer to any treatment of the CAPM for further information. Instead, we provide details about all the functions that we use to compute the results. In particular, we leverage useful computational concepts: Rolling-window estimation and parallelization.
We use the following Python packages throughout this chapter:
import pandas as pd
import numpy as np
import sqlite3
import statsmodels.formula.api as smf
from regtabletotext import prettify_result
from statsmodels.regression.rolling import RollingOLS
from plotnine import *
from mizani.breaks import date_breaks
from mizani.formatters import percent_format, date_format
from joblib import Parallel, delayed, cpu_count
from itertools import product
Compared to previous chapters, we introduce statsmodels for regression analysis and for sliding-window regressions and joblib for parallelization.
## Estimating Beta Using Monthly Returns
The estimation procedure is based on a rolling-window estimation, where we may use either monthly or daily returns and different window lengths. First, let us start with loading the monthly CRSP data from our SQLite database introduced in Accessing and Managing Financial Data and WRDS, CRSP, and Compustat.
tidy_finance = sqlite3.connect(database="data/tidy_finance_python.sqlite")
sql="SELECT permno, date, industry, ret_excess FROM crsp_monthly",
con=tidy_finance,
parse_dates={"date"})
.dropna()
)
sql="SELECT date, mkt_excess FROM factors_ff3_monthly",
con=tidy_finance,
parse_dates={"date"}
)
crsp_monthly = (crsp_monthly
.merge(factors_ff3_monthly, how="left", on="date")
)
To estimate the CAPM regression coefficients $r_{i, t} - r_{f, t} = \alpha_i + \beta_i(r_{m, t}-r_{f,t})+\varepsilon_{i, t}, \tag{1}$ we regress stock excess returns ret_excess on excess returns of the market portfolio mkt_excess.
Python provides a simple solution to estimate (linear) models with the function smf.ols(). The function requires a formula as input that is specified in a compact symbolic form. An expression of the form y ~ model is interpreted as a specification that the response y is modeled by a linear predictor specified symbolically by model. Such a model consists of a series of terms separated by + operators. In addition to standard linear models, smf.ols() provides a lot of flexibility. You should check out the documentation for more information. To start, we restrict the data only to the time series of observations in CRSP that correspond to Apple’s stock (i.e., to permno 14593 for Apple) and compute $$\hat\alpha_i$$ as well as $$\hat\beta_i$$.
model_beta = (smf.ols(
formula="ret_excess ~ mkt_excess",
data=crsp_monthly.query("permno == 14593"))
.fit()
)
prettify_result(model_beta)
OLS Model:
ret_excess ~ mkt_excess
Coefficients:
Estimate Std. Error t-Statistic p-Value
Intercept 0.010 0.005 2.046 0.041
mkt_excess 1.384 0.109 12.696 0.000
Summary statistics:
- Number of observations: 516
- R-squared: 0.239, Adjusted R-squared: 0.237
- F-statistic: 161.196 on 1 and 514 DF, p-value: 0.000
smf.ols() returns an object of class RegressionModel, which contains all the information we usually care about with linear models. prettify_result() returns an overview of the estimated parameters. The output above indicates that Apple moves excessively with the market as the estimated $$\hat\beta_i$$ is above one ($$\hat\beta_i \approx 1.4$$).
## Rolling-Window Estimation
After we estimated the regression coefficients on an example, we scale the estimation of $$\beta_i$$ to a whole different level and perform rolling-window estimations for the entire CRSP sample.
We take a total of five years of data (window_size) and require at least 48 months with return data to compute our betas (min_obs). Check out the Exercises if you want to compute beta for different time periods. We first identify firm identifiers (permno) for which CRSP contains sufficiently many records.
window_size = 60
min_obs = 48
valid_permnos = (crsp_monthly
.dropna()
.groupby("permno")["permno"]
.count()
.reset_index(name="counts")
.query(f"counts > {window_size}+1")
)
Before we proceed with the estimation, one important issue is worth emphasizing: RollingOLS returns the estimated parameters of a linear regression by incorporating a window of the last window_size rows. Whenever monthly returns are implicitly missing (which means there is simply no entry recorded, e.g., because a company was delisted and only traded publicly again later), such a fixed window size may cause outdated observations to influence the estimation results. We thus recommend making such implicit missing rows explicit.
We hence collect information about the first and last listing date of each permno.
permno_information = (crsp_monthly
.merge(valid_permnos, how="inner", on="permno")
.groupby(["permno"])
.aggregate(first_date=("date", "min"),
last_date=("date", "max"))
.reset_index()
)
To complete the missing observations in the CRSP sample, we obtain all possible permno-month combinations.
unique_permno = crsp_monthly["permno"].unique()
unique_month = factors_ff3_monthly["date"].unique()
all_combinations = pd.DataFrame(
product(unique_permno, unique_month),
columns=["permno", "date"]
)
Finally, we expand the CRSP sample and include a row (with missing excess returns) for each possible permno-date observation that falls within the start and end date where the respective permno has been publicly listed.
returns_monthly = (all_combinations
.merge(crsp_monthly.get(["permno", "date", "ret_excess"]),
how="left", on=["permno", "date"])
.merge(permno_information, how="left", on="permno")
.query("(date >= first_date) & (date <= last_date)")
.drop(columns=["first_date", "last_date"])
.merge(crsp_monthly.get(["permno", "date", "industry"]),
how="left", on=["permno", "date"])
.merge(factors_ff3_monthly, how="left", on="date")
)
The following function implements the CAPM regression for a dataframe (or a part thereof) containing at least min_obs observations to avoid huge fluctuations if the time series is too short. If the condition is violated (i.e., the time series is too short) the function returns a missing value.
def roll_capm_estimation(data, window_size, min_obs):
"""Calculate rolling CAPM estimation."""
data = data.sort_values("date")
result = (RollingOLS.from_formula(
formula="ret_excess ~ mkt_excess",
data=data,
window=window_size,
min_nobs=min_obs,
missing="drop")
.fit()
.params.get("mkt_excess")
)
result.index = data.index
return result
Before we approach the whole CRSP sample, let us focus on a couple of examples for well-known firms.
examples = pd.DataFrame({
"permno": [14593, 10107, 93436, 17778],
"company": ["Apple", "Microsoft", "Tesla", "Berkshire Hathaway"]
})
It is actually quite simple to perform the rolling-window estimation for an arbitrary number of stocks, which we visualize in the following code chunk and the resulting Figure 1.
beta_example = (returns_monthly
.merge(examples, how="inner", on="permno")
.groupby(["permno"])
.apply(lambda x: x.assign(
beta=roll_capm_estimation(x, window_size, min_obs))
)
.reset_index(drop=True)
.dropna()
)
plot_beta = (
ggplot(beta_example,
aes(x="date", y="beta", color="company", linetype="company")) +
geom_line() +
labs(x="", y="", color="", linetype="",
title="Monthly beta estimates for example stocks") +
scale_x_datetime(breaks=date_breaks("5 year"), labels=date_format("%Y"))
)
plot_beta.draw()
## Parallelized Rolling-Window Estimation
Next, we perform the rolling window estimation for the entire cross-section of stocks in the CRSP sample. For that purpose, we can apply the code snippet from the example above to compute rolling window regression coefficients for all stocks. This is how to do it with the joblib package to use multiple cores. Note that we use cpu_count() to determine the number of cores available for parallelization but keep one core free for other tasks. Some machines might freeze if all cores are busy with Python jobs.
def roll_capm_estimation_for_joblib(permno, group):
"""Calculate rolling CAPM estimation using joblib."""
group = group.sort_values(by="date")
beta_values = (RollingOLS.from_formula(
formula="ret_excess ~ mkt_excess",
data=group,
window=window_size,
min_nobs=min_obs,
missing="drop"
)
.fit()
.params.get("mkt_excess")
)
result = pd.DataFrame(beta_values)
result.columns = ["beta"]
result["date"] = group["date"].values
result["permno"] = permno
return result
permno_groups = (returns_monthly
.merge(valid_permnos, how="inner", on="permno")
.groupby("permno", group_keys=False)
)
n_cores = cpu_count()-1
beta_monthly = (
pd.concat(
Parallel(n_jobs=n_cores)
(delayed(roll_capm_estimation_for_joblib)(name, group)
for name, group in permno_groups)
)
.dropna()
.rename(columns={"beta": "beta_monthly"})
)
## Estimating Beta Using Daily Returns
Before we provide some descriptive statistics of our beta estimates, we implement the estimation for the daily CRSP sample as well. Depending on the application, you might either use longer horizon beta estimates based on monthly data or shorter horizon estimates based on daily returns. As loading the full daily CRSP data requires relatively large amounts of memory, we split the beta estimation into smaller chunks. The logic follows the approach that we use to download the daily CRSP data (see WRDS, CRSP, and Compustat).
First, we load the daily Fama-French market excess returns and extract the vector of dates.
factors_ff3_daily = pd.read_sql_query(
sql="SELECT date, mkt_excess FROM factors_ff3_daily",
con=tidy_finance,
parse_dates={"date"}
)
unique_date = factors_ff3_daily["date"].unique()
For the daily data, we consider around three months of data (i.e., 60 trading days), require at least 50 observations, and estimate betas in batches of 500.
window_size = 60
min_obs = 50
permnos = list(crsp_monthly["permno"].unique().astype(str))
batch_size = 500
batches = np.ceil(len(permnos)/batch_size).astype(int)
We then proceed to perform the same steps as with the monthly CRSP data, just in batches: Load in daily returns, transform implicit missing returns to explicit ones, keep only valid stocks with a minimum number of rows, and parallelize the beta estimation across stocks.
beta_daily = []
for j in range(1, batches+1):
permno_batch = permnos[
((j-1)*batch_size):(min(j*batch_size, len(permnos)))
]
permno_batch_formatted = (
", ".join(f"'{permno}'" for permno in permno_batch)
)
permno_string = f"({permno_batch_formatted})"
crsp_daily_sub_query = (
"SELECT permno, date, ret_excess "
"FROM crsp_daily "
f"WHERE permno IN {permno_string}"
)
sql=crsp_daily_sub_query,
con=tidy_finance,
dtype={"permno": int},
parse_dates={"date"}
)
valid_permnos = (crsp_daily_sub
.groupby("permno")["permno"]
.count()
.reset_index(name="counts")
.query(f"counts > {window_size}+1")
.drop(columns="counts")
)
permno_information = (crsp_daily_sub
.merge(valid_permnos, how="inner", on="permno")
.groupby(["permno"])
.aggregate(first_date=("date", "min"),
last_date=("date", "max"),)
.reset_index()
)
unique_permno = permno_information["permno"].unique()
all_combinations = pd.DataFrame(
product(unique_permno, unique_date),
columns=["permno", "date"]
)
returns_daily = (crsp_daily_sub
.merge(all_combinations, how="right", on=["permno", "date"])
.merge(permno_information, how="left", on="permno")
.query("(date >= first_date) & (date <= last_date)")
.drop(columns=["first_date", "last_date"])
.merge(factors_ff3_daily, how="left", on="date")
)
permno_groups = (returns_daily
.groupby("permno", group_keys=False)
)
beta_daily_sub = (
pd.concat(
Parallel(n_jobs=n_cores)
(delayed(roll_capm_estimation_for_joblib)(name, group)
for name, group in permno_groups)
)
.dropna()
.rename(columns={"beta": "beta_daily"})
)
beta_daily_sub = (beta_daily_sub
.assign(
month = lambda x:
x["date"].dt.to_period("M").dt.to_timestamp()
)
.sort_values("date")
.groupby(["permno", "month"])
.last()
.reset_index()
.drop(columns="date")
.rename(columns={"month": "date"})
)
beta_daily.append(beta_daily_sub)
print(f"Batch {j} out of {batches} done ({(j/batches)*100:.2f}%)\n")
beta_daily = pd.concat(beta_daily)
## Comparing Beta Estimates
What is a typical value for stock betas? To get some feeling, we illustrate the dispersion of the estimated $$\hat\beta_i$$ across different industries and across time below. Figure 2 shows that typical business models across industries imply different exposure to the general market economy. However, there are barely any firms that exhibit a negative exposure to the market factor.
beta_industries = (beta_monthly
.merge(crsp_monthly, how="inner", on=["permno", "date"])
.dropna(subset="beta_monthly")
.groupby(["industry","permno"])["beta_monthly"]
.aggregate("mean")
.reset_index()
)
industry_order = (beta_industries
.groupby("industry")["beta_monthly"]
.aggregate("median")
.sort_values()
.index.tolist()
)
plot_beta_industries = (
ggplot(beta_industries,
aes(x="industry", y="beta_monthly")) +
geom_boxplot() +
coord_flip() +
labs(x="", y="",
title="Firm-specific beta distributions by industry") +
scale_x_discrete(limits=industry_order)
)
plot_beta_industries.draw()
Next, we illustrate the time-variation in the cross-section of estimated betas. Figure 3 shows the monthly deciles of estimated betas (based on monthly data) and indicates an interesting pattern: First, betas seem to vary over time in the sense that during some periods, there is a clear trend across all deciles. Second, the sample exhibits periods where the dispersion across stocks increases in the sense that the lower decile decreases and the upper decile increases, which indicates that for some stocks, the correlation with the market increases, while for others it decreases. Note also here: stocks with negative betas are a rare exception.
beta_quantiles = (beta_monthly
.groupby("date")["beta_monthly"]
.quantile(q=np.arange(0.1, 1.0, 0.1))
.reset_index()
.rename(columns={"level_1": "quantile"})
.assign(quantile=lambda x: (x["quantile"]*100).astype(int))
.dropna()
)
linetypes = ["-", "--", "-.", ":"]
n_quantiles = beta_quantiles["quantile"].nunique()
plot_beta_quantiles = (
ggplot(beta_quantiles,
aes(x="date", y="beta_monthly",
color="factor(quantile)", linetype="factor(quantile)")) +
geom_line() +
labs(x="", y="", color="", linetype="",
title="Monthly deciles of estimated betas") +
scale_x_datetime(breaks=date_breaks("5 year"), labels=date_format("%Y")) +
scale_linetype_manual(
values=[linetypes[l % len(linetypes)] for l in range(n_quantiles)]
)
)
plot_beta_quantiles.draw()
To compare the difference between daily and monthly data, we combine beta estimates to a single table. Then, we use the table to plot a comparison of beta estimates for our example stocks in Figure 4.
beta = (beta_monthly
.get(["permno", "date", "beta_monthly"])
.merge(beta_daily.get(["permno", "date", "beta_daily"]),
how="outer", on=["permno", "date"])
)
beta_comparison = (beta
.merge(examples, on="permno")
.melt(id_vars=["permno", "date", "company"], var_name="name",
value_vars=["beta_monthly", "beta_daily"], value_name="value")
.dropna()
)
plot_beta_comparison = (
ggplot(beta_comparison,
aes(x="date", y="value", color="name")) +
geom_line() +
facet_wrap("~company", ncol=1) +
labs(x="", y="", color="",
title="Comparison of beta estimates using monthly and daily data") +
scale_x_datetime(breaks=date_breaks("10 years"),
labels=date_format("%Y")) +
theme(figure_size=(6.4, 6.4))
)
plot_beta_comparison.draw()
The estimates in Figure 4 look as expected. As you can see, the beta estimates really depend on the estimation window and data frequency. Nevertheless, one can observe a clear connection between daily and monthly betas in this example, in magnitude and the dynamics over time.
Finally, we write the estimates to our database so that we can use them in later chapters.
(beta.to_sql(
name="beta",
con=tidy_finance,
if_exists="replace",
index=False
)
)
Whenever you perform some kind of estimation, it also makes sense to do rough plausibility tests. A possible check is to plot the share of stocks with beta estimates over time. This descriptive analysis helps us discover potential errors in our data preparation or the estimation procedure. For instance, suppose there was a gap in our output without any betas. In this case, we would have to go back and check all previous steps to find out what went wrong. Figure 5 does not indicate any troubles, so let us move on to the next check.
beta_long = (crsp_monthly
.merge(beta, how="left", on=["permno", "date"])
.melt(id_vars=["permno", "date"], var_name="name",
value_vars=["beta_monthly", "beta_daily"], value_name="value")
)
beta_shares = (beta_long
.groupby(["date", "name"])
.aggregate(share=("value", lambda x: sum(~x.isna())/len(x)))
.reset_index()
)
plot_beta_long = (
ggplot(beta_shares,
aes(x="date", y="share", color="name", linetype="name")) +
geom_line() +
labs(x="", y="", color="", linetype="",
title="End-of-month share of securities with beta estimates") +
scale_y_continuous(labels=percent_format()) +
scale_x_datetime(breaks=date_breaks("10 year"), labels=date_format("%Y"))
)
plot_beta_long.draw()
We also encourage everyone to always look at the distributional summary statistics of variables. You can easily spot outliers or weird distributions when looking at such tables.
beta_long.groupby("name")["value"].describe().round(2)
count mean std min 25% 50% 75% max
name
beta_daily 3327173.0 0.76 0.94 -44.85 0.21 0.69 1.24 61.64
beta_monthly 2104034.0 1.10 0.70 -8.96 0.64 1.03 1.47 10.35
The summary statistics also look plausible for the two estimation procedures.
Finally, since we have two different estimators for the same theoretical object, we expect the estimators to be at least positively correlated (although not perfectly as the estimators are based on different sample periods and frequencies).
beta.get(["beta_monthly", "beta_daily"]).corr().round(2)
beta_monthly beta_daily
beta_monthly 1.00 0.32
beta_daily 0.32 1.00
Indeed, we find a positive correlation between our beta estimates. In the subsequent chapters, we mainly use the estimates based on monthly data, as most readers should be able to replicate them and should not encounter potential memory limitations that might arise with the daily data.
## Exercises
1. Compute beta estimates based on monthly data using one, three, and five years of data and impose a minimum number of observations of 10, 28, and 48 months with return data, respectively. How strongly correlated are the estimated betas?
2. Compute beta estimates based on monthly data using five years of data and impose different numbers of minimum observations. How does the share of permno-date observations with successful beta estimates vary across the different requirements? Do you find a high correlation across the estimated betas?
3. Instead of using joblib, perform the beta estimation in a loop (using either monthly or daily data) for a subset of 100 permnos of your choice. Verify that you get the same results as with the parallelized code from above.
4. Filter out the stocks with negative betas. Do these stocks frequently exhibit negative betas, or do they resemble estimation errors?
5. Compute beta estimates for multi-factor models such as the Fama-French three-factor model. For that purpose, you extend your regression to $r_{i, t} - r_{f, t} = \alpha_i + \sum\limits_{j=1}^k\beta_{i,k}(r_{j, t}-r_{f,t})+\varepsilon_{i, t} \tag{2}$ where $$r_{i, t}$$ are the $$k$$ factor returns. Thus, you estimate four parameters ($$\alpha_i$$ and the slope coefficients). Provide some summary statistics of the cross-section of firms and their exposure to the different factors.
## References
Lintner, John. 1965. Security prices, risk, and maximal gains from diversification.” The Journal of Finance 20 (4): 587–615. https://doi.org/10.1111/j.1540-6261.1965.tb02930.x.
Mossin, Jan. 1966. Equilibrium in a capital asset market.” Econometrica 34 (4): 768–83. https://doi.org/10.2307/1910098.
Seabold, Skipper, and Josef Perktold. 2010. “Statsmodels: Econometric and Statistical Modeling with Python.” In 9th Python in Science Conference.
Sharpe, William F. 1964. Capital asset prices: A theory of market equilibrium under conditions of risk .” The Journal of Finance 19 (3): 425–42. https://doi.org/10.1111/j.1540-6261.1964.tb02865.x.
Team, Joblib Development. 2023. “Joblib: Running Python Functions as Pipeline Jobs.” https://joblib.readthedocs.io/. | 5,196 | 21,275 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-38 | latest | en | 0.792529 |
http://hackage.haskell.org/package/colour-2.3.5/docs/Data-Colour-RGBSpace-HSV.html | 1,606,727,084,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141211510.56/warc/CC-MAIN-20201130065516-20201130095516-00019.warc.gz | 43,346,109 | 3,055 | colour-2.3.5: A model for human colour/color perception
Data.Colour.RGBSpace.HSV
Synopsis
# Documentation
data RGB a Source #
An RGB triple for an unspecified colour space.
Instances
Source # Instance detailsDefined in Data.Colour.RGB Methodsfmap :: (a -> b) -> RGB a -> RGB b #(<\$) :: a -> RGB b -> RGB a # Source # Instance detailsDefined in Data.Colour.RGB Methodspure :: a -> RGB a #(<*>) :: RGB (a -> b) -> RGB a -> RGB b #liftA2 :: (a -> b -> c) -> RGB a -> RGB b -> RGB c #(*>) :: RGB a -> RGB b -> RGB b #(<*) :: RGB a -> RGB b -> RGB a # Eq a => Eq (RGB a) Source # Instance detailsDefined in Data.Colour.RGB Methods(==) :: RGB a -> RGB a -> Bool #(/=) :: RGB a -> RGB a -> Bool # Read a => Read (RGB a) Source # Instance detailsDefined in Data.Colour.RGB MethodsreadsPrec :: Int -> ReadS (RGB a) #readList :: ReadS [RGB a] #readPrec :: ReadPrec (RGB a) # Show a => Show (RGB a) Source # Instance detailsDefined in Data.Colour.RGB MethodsshowsPrec :: Int -> RGB a -> ShowS #show :: RGB a -> String #showList :: [RGB a] -> ShowS #
hsvView :: (Fractional a, Ord a) => RGB a -> (a, a, a) Source #
Returns the HSV (hue-saturation-value) coordinates of an RGB triple. See hue, saturation, and value.
hue :: (Fractional a, Ord a) => RGB a -> a Source #
The hue coordinate of an RGB value is in degrees. Its value is always in the range 0-360.
saturation :: (Fractional a, Ord a) => RGB a -> a Source #
Returns the saturation coordinate (range [0,1]) of an RGB triple for the HSV (hue-saturation-value) system. Note: This is different from saturation for the Data.Colour.RGBSpace.HSL
value :: (Fractional a, Ord a) => RGB a -> a Source #
Returns the value coordinate (raonge [0,1]) of an RGB triple for the HSV (hue-saturation-value) system.
hsv :: (RealFrac a, Ord a) => a -> a -> a -> RGB a Source #
Convert HSV (hue-saturation-value) coordinates to an RGB value. Hue is expected to be measured in degrees [0,360], while saturation and value are expected to be in the closed range [0,1]. | 573 | 2,009 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-50 | latest | en | 0.617874 |
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M 340L CS Homework Set 9 1. Let 1 2 3 3 2 1 5 3 , 2 , 1 , 3 . 0 1 4 1 u u u b a. Form the matrix 1 2 3 U u u u and confirm that the columns of U are orthogonal by computing . T U U b. Express b as a linear combination of 1 2 , u u and 3 . u (That is, solve . Ux b Be clever about using T U to do this.) 2. Let 1 2 2/ 3 1/ 3 1/ 3 , 2/ 3 2/ 3 0 u u a. Form the matrix 1 2 U u u and confirm that the columns of U are orthogonal by computing . T U U b. Normalize the columns and confirm that . T U U I 3. Answer true or false to the following. If false offer a simple counterexample. a. Every orthogonal set in n is linearly independent.
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Jill Tulane University ‘16, Course Hero Intern | 561 | 1,925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-26 | latest | en | 0.881103 |
http://stupas.ml/forum552-how-much-time-does-the-earth-actually-take-to-rotate-on-its-axis.html | 1,527,128,321,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865884.49/warc/CC-MAIN-20180524014502-20180524034502-00628.warc.gz | 279,065,537 | 7,563 | how much time does the earth actually take to rotate on its axis
# how much time does the earth actually take to rotate on its axis
By definition, the Earths rotation is the amount of time that it takes to rotate once on its axis. This is, apparently, accomplished once a day i.e. every 24 hours. However, there are actually two different kinds of rotation that need to be considered here. Greenwich. 7 Most time zones?Least time zones? Asia Australia. 8 International Date Line.Thursday in North America Friday in Asia. 10 How many time zones does the United States have? Earth rotates once/24 hrs 24 time zones around the earth. 1 hr between each time zone Philosophers, scientists and astronomers have been tackling lifes most pressing questions since the beginning of time. Theyve convinced us that the Earth is indeed round, that it revolves around the sun and that it rotates on its axis once approximately every 24 hours. A complete rotation of the earth on its axis takes approximately 23 hours, 56 minutes and four seconds with respect to the background stars.Aside fro How do you test a coil pack? Picture the Problem: The Earth rotates once on its axis every 24 hours.Because the fan slows down at a constant rate of acceleration, it takes exactly half the time for it to slow from 0.96 rev/s to 0.48 rev/s as it does to come to a complete stop. How do Earths Rotation and Revolution Work? We apologize but this resource is not available to you.SC.4.E.5.3 :Recognize that Earth revolves around the Sun in a year and rotates on its axis in a 24-hour day. Earths rotation is the rotation of Planet Earth around its own axis.
Earth rotates eastward, in prograde motion. As viewed from the north pole star Polaris, Earth turns counterclockwise. The North Pole, also known as the Geographic North Pole or Terrestrial North Pole This is the time it takes Earth to rotate 34. 7412 ms (milliseconds) or 0. Each day that goes by, A complete rotation of the earth on its axis takes approximately 23 hours, 56 minutes and four seconds with respect to the background stars. That giant flaming star in the sky does rotate, but moves at a much slower pace than the Earth.On average, the sun rotates on its axis once every 27 days. However, its equator spins the fastest and takes about 24 days to rotate, while the poles take more than 30 days. How fast does the earth rotate? Linear speed versus angular speed.
How long does it take the Earth to spin once on its axis? Whats this strange figure? Carry on readingIt actually takes the Earth slightly over 23 hours and 56 minutes to rotate once around this axis. The "circle" is actually a bit off, and a little like an oval despite this fact (that the earth is a little closer to the sun at some times than others), this is NOT what causes It is the fact that the earth is not "straight up and down" on its axis as it rotates around the sun that creates our seasons. Planet with nearly equal rotation and revolution time to the earth is .How many planets does our Solar System consist of? Which planet is called Double Planet? What is the direction of rotation of the earth on its axis? Like Earth, the Moon rotates on its axis. So why do we see only one view of its face?The moon does rotate on its axis. One rotation takes nearly as much time as one revolution around Earth. If the moon were to rotate quickly (several times each month) or not rotate at all, Earth would be This got me to wondering how much energy it would actually take to stop the world from spinning.Related. 3. How to know the time a disc takes to stop from rotating and the numbers of revolutions.How much effort would be required to fix the Earths rotation? 10. How much additional light does Did not find what you are looking for? The earth actually rotates on its axis once every 23 hours, 56 minutes, and 4 seconds (approximately). This is the time it takes for the earth toHow did Heracleides find out that Earth rotates around its axis once every 24 hours? Why do planets orbit stars? What tilted the Earth axis? I actually do stellar parallax as a researcher, so let me emphasize more strongly just how minisculeIn fact, if you take the difference between a Siderial (Earth spinning on its axis) day and a SolarDuring a year, the earth rotates 365.25 times around its axis (with each rotation taking exactly 24 How does the Earth continue to rotate around its axis? Where does the energy to keep it moving come from?As the nebula collapsed it began rotating, which may seem odd, but actually not rotating is far stranger than rotating. How long does Mars take to spin on its axis?A: It takes a total 1407.5 hours, or 58.646 Earth days, for Mercury to make a complete rotation on its axis. A day on Earth is only 23.934 hours long, which p Full Answer >. First, the Earth rotates on its axis once every 24 hours.How do we know these changes actually took place?The last time the Earths temperature was similar to todays is more than 100,000 years ago. Aryan, the Earths rotation on its axis (its day) is independent of its revolution about the Sun ( its year). So moving Earth closer to the Sun would have no affect on the length of its day.How much time does the earth take to complete one rotation?(with seconds). | 1,176 | 5,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-22 | latest | en | 0.940455 |
https://access.openupresources.org/curricula/our6-8math-v1/6/students/2/9.html | 1,716,456,512,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058614.40/warc/CC-MAIN-20240523080929-20240523110929-00815.warc.gz | 64,325,469 | 10,237 | Lesson 9: Constant Speed
Let’s use ratios to work with how fast things move.
9.1: Number Talk: Dividing by Powers of 10
Find the quotient mentally.
$30\div 10$
$34\div 10$
$3.4\div 10$
$34\div 100$
9.2: Moving 10 Meters
Your teacher will set up a straight path with a 1-meter warm-up zone and a 10-meter measuring zone. Follow the following instructions to collect the data.
1. The person with the stopwatch (the “timer”) stands at the finish line. The person being timed (the “mover”) stands at the warm-up line.
2. On the first round, the mover starts moving at a slow, steady speed along the path. When the mover reaches the start line, they say, “Start!” and the timer starts the stopwatch.
3. The mover keeps moving steadily along the path. When they reach the finish line, the timer stops the stopwatch and records the time, rounded to the nearest second, in the table.
4. On the second round, the mover follows the same instructions, but this time, moving at a quick, steady speed. The timer records the time the same way.
5. Repeat these steps until each person in the group has gone twice: once at a slow, steady speed, and once at a quick, steady speed.
row 1
1. After you finish collecting the data, use the double number line diagrams to answer the questions. Use the times your partner collected while you were moving.
Moving slowly:
Moving quickly:
1. Estimate the distance in meters you traveled in 1 second when moving slowly.
2. Estimate the distance in meters you traveled in 1 second when moving quickly.
3. Trade diagrams with someone who is not your partner. How is the diagram representing someone moving slowly different from the diagram representing someone moving quickly?
9.3: Moving for 10 Seconds
Lin and Diego both ran for 10 seconds, each at a constant speed. Lin ran 40 meters and Diego ran 55 meters.
1. Who was moving faster? Explain how you know.
2. How far did each person move in 1 second? If you get stuck, consider drawing double number line diagrams to represent the situations.
3. Use your data from the previous activity. How far could you travel in 10 seconds at your quicker speed?
4. Han ran 100 meters in 20 seconds at a constant speed. Is this speed faster, slower, or the same as Lin’s? Diego’s? Yours?
GeoGebra Applet ycJSQpXT
GeoGebra Applet ycJSQpXT
Summary
Suppose a train traveled 100 meters in 5 seconds at a constant speed. To find its speed in meters per second, we can create a double number line:
The double number line shows that the train’s speed was 20 meters per second. We can also find the speed by dividing: $100 \div 5 = 20$.
Once we know the speed in meters per second, many questions about the situation become simpler to answer because we can multiply the amount of time an object travels by the speed to get the distance. For example, at this rate, how far would the train go in 30 seconds? Since $20 \boldcdot 30 = 600$, the train would go 600 meters in 30 seconds.
Glossary
meters per second
meters per second
A unit to measure speed that tells you how many meters an object travels in one second. | 748 | 3,094 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2024-22 | latest | en | 0.919766 |
https://webraindor.info/qa/quick-answer-what-does-ph-stand-for-in-acidity.html | 1,624,235,453,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488259200.84/warc/CC-MAIN-20210620235118-20210621025118-00157.warc.gz | 538,214,016 | 8,546 | Quick Answer: What Does PH Stand For In Acidity?
What does Acidic mean in pH?
The pH scale measures how acidic or basic a substance is.
The pH scale ranges from 0 to 14.
A pH of 7 is neutral.
A pH less than 7 is acidic..
What pH is the blood?
The pH scale, ranges from 0 (strongly acidic) to 14 (strongly basic or alkaline). A pH of 7.0, in the middle of this scale, is neutral. Blood is normally slightly basic, with a normal pH range of about 7.35 to 7.45. Usually the body maintains the pH of blood close to 7.40.
Why pH is not more than 14?
One far end is not more than 1M of hydrogen ions, which results in a pH value of not more than 0. While on the other end is not more than 1M of hydroxide ions which results in a pH value of not more than 14.
What has a pH of 1?
The pH scaleIncreasing pH (Decreasing Acidity)Substances0 (most acidic)Hydrochloric acid (HCl)1Stomach acid2Lemon juice3Cola, beer, vinegar15 more rows
How do you know if titratable acidity?
The acid content of the must is determined by titrating a sample (a given volume) with a base such as sodium hydroxide solution to a phenolphthalein end point or alternatively, to a pH of 8.2. The titratable acidity is expressed as grams of tartaric acid per 100 ml.
How does acidity relate to pH?
If you add acid to a solution the concentration of hydrogen ions (acidity) increases and the pH decreases. Frequently people confuse pH with acidity—pH is the scale on which acidity is expressed, but it is not synonymous with acidity.
How is pH calculated?
To calculate the pH of an aqueous solution you need to know the concentration of the hydronium ion in moles per liter (molarity). The pH is then calculated using the expression: pH = – log [H3O+]. … Example: What is the pOH of a solution that has a hydroxide ion concentration of 4.82 x 10-5 M?
Is pH above 14 possible?
Thus, measured pH values will lie mostly in the range 0 to 14, though negative pH values and values above 14 are entirely possible. Since pH is a logarithmic scale, a difference of one pH unit is equivalent to a tenfold difference in hydrogen ion concentration.
Why is a pH of 7 neutral?
pH is a measure of the amount of Hydrogen ions (H+) in a solution. … Even in pure water ions tend to form due to random processes (producing some H+ and OH- ions). The amount of H+ that is made in pure water is about equal to a pH of 7. That’s why 7 is neutral.
What makes something acidic or alkaline?
An acid is a substance which donates hydrogen ions into solution, while a base or alkali is one which takes up hydrogen ions.
What is the pH of Coca Cola?
2.52pH ValuesPRODUCTpHSodaPepsi2.53Coca-Cola2.52Cherry Coke2.5266 more rows
What is the difference between acidity and pH?
Titratable acidity (also called total acidity) measures the total acid concentration in a food. … pH is defined as the negative log (base 10) of the hydrogen ion concentration and can span a range of 14 orders of magnitudes.
Why is lower pH more acidic?
High concentrations of hydrogen ions yield a low pH (acidic substances), whereas low levels of hydrogen ions result in a high pH (basic substances). … Therefore, the more hydrogen ions present, the lower the pH; conversely, the fewer hydrogen ions, the higher the pH.
Which solution is the most acidic?
What does it mean for a solution to be acidic or basic (alkaline)?pH ValueH+ Concentration Relative to Pure WaterExample010 000 000battery acid11 000 000gastric acid2100 000lemon juice, vinegar310 000orange juice, soda11 more rows
What happens when pH increases?
As the concentration of hydrogen ions in a solution increase, the more acidic the solution becomes. As the level of hydroxide ions increases the more basic, or alkaline, the solution becomes. Thus, as the hydrogen ion concentration increases hydroxide ion concentration falls, and vice versa. …
What happens if body pH is too alkaline?
An increase in alkaline causes pH levels to rise. When the levels of acid in your blood are too high, it’s called acidosis. When your blood is too alkaline, it is called alkalosis. Respiratory acidosis and alkalosis are due to a problem with the lungs.
What happens if the pH of blood changes?
If the body does not reset the pH balance, it can lead to more severe illness. For example, this can happen if the level of acidosis is too serious, or if the person’s kidneys are not working well. Depending on the cause, changes in blood pH can be either long lasting or brief.
Do acids have higher or lower pH?
Anything with a very low pH is acidic, while substances with a high pH are alkaline. … With that in mind, the pH scale made a lot more sense as a measure of acidity. Acids have a few different definitions, but overall they are substances that can generate hydrogen ions when in a solution.
How do you balance the pH in your body?
One of the ways you can work toward a healthy pH balance in your body is through diet. Take inventory of the foods you tend to eat and consider swapping out some acid-forming foods for acid-reducing foods and aiming for a diet with a balance of acidic and alkaline foods. Acid-Forming Foods.
How do you convert pH to acidity?
pH is the Measure of Solution AciditypH = -log[H3O+] … [H3O+][OH-] = Kw = 1 x 10-14 … pH + pOH = pKw = 14. … Weak acids and bases. … “If Ka is less than 0.001 (10-3) and c is at least 0.01 M, the approximation can be made. … More examples (The following values pertain to a weak acid, Ka=1.8 x 10-5)
What happens to pH as acidity decreases?
When an acidic solution is diluted with water the concentration of H + ions decreases and the pH of the solution increases towards 7. … This causes the pH of the alkali to fall towards 7, making the solution less alkaline as more water is added. | 1,434 | 5,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-25 | latest | en | 0.902288 |
https://www.coursehero.com/file/233777/HW2-Solution/ | 1,513,388,751,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948581033.57/warc/CC-MAIN-20171216010725-20171216032725-00213.warc.gz | 729,758,101 | 42,632 | HW2 Solution
# HW2 Solution - 3 and the 10-volt source we get –v 3 10 =...
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EE302 Homework #2 Chapter 2, Problem 12. In the circuit in Fig. 2.76, obtain v 1, v 2 , and v 3 . Chapter 2, Solution 12. + v 1 - + v 2 - + v 3 - 25v + + 10v - + 15v - + 20v - loop 1 loop 2 loop 3 For loop 1, -20 -25 +10 + v 1 = 0 v 1 = 35.0V For loop 2, -10 +15 -v 2 = 0 v 2 = 5.00V For loop 3, -v 1 +v 2 +v 3 = 0 v 3 = 30.0V
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Chapter 2, Problem 16. Determine V o in the circuit in Fig. 2.80. 6 Ω 2 Ω + _ + _ + _ V o 9 V 3 V y y Figure 2.80 For Prob. 2.16. Chapter 2, Solution 16. Apply KVL, -9 + (6+2)I + 3 = 0, 8I = 9-3=6 , I = 6/8 Also, -9 + 6I + V o = 0 V o = 9- 6I = 4.50 V
Chapter 2, Problem 17. Obtain v 1 through v 3 in the circuit Chapter 2, Solution 17. Applying KVL around the entire outside loop we get, –24 + v 1 + 10 + 12 = 0 or v 1 = 2.00V Applying KVL around the loop containing v 2 , the 10-volt source, and the 12-volt source we get, v 2 + 10 + 12 = 0 or v 2 = –22.0V Applying KVL around the loop containing v
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Unformatted text preview: 3 and the 10-volt source we get, –v 3 + 10 = 0 or v 3 = 10.0V Chapter 2, Problem 18. Find I and V ab in the circuit. Chapter 2, Solution 18. Applying KVL, -30 -10 +8 + I(3+5) = 0 8I = 32 I = 4.00A-V ab + 5I + 8 = 0 V ab = 28.0V Chapter 2, Problem 20. Determine i o in the circuit Chapter 2, Solution 20. Applying KVL around the loop, -36 + 4i + 5i = 0 i = 4.00A Chapter 2, Problem 22. Find V o in the circuit in Fig. 2.86 and the power dissipated by the controlled source. Chapter 2, Solution 22. KCL requires that v 2 10 4 v + + = 0 v = –4.44V The current through the controlled source is i = 2V = -8.888A v = (6 + 4) i (where i = v /4) = 10 111 . 11 4 − = v Hence, Power = (-8.888)(-11.111) = 98.8 W...
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HW2 Solution - 3 and the 10-volt source we get –v 3 10 =...
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Ask a homework question - tutors are online | 914 | 2,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-51 | latest | en | 0.71984 |
https://www.phrygane.tk/control-systems/graphs-in-general.html | 1,553,307,270,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202711.3/warc/CC-MAIN-20190323020538-20190323042538-00125.warc.gz | 871,845,842 | 5,369 | ## Graphs in General
A generalized graph is a set of vertices and edges. We usually represent vertices as dots and the edges as lines connecting the dots. The edges, the lines, also called links, connect the vertices. A more formal definition is a graph, G, is a non-empty set of elements, the vertices, and a list of unordered pairs of these elements, the edges. The set of vertices of graph G is the vertex set of G; we denote this set by V(G), and the list of edges is the edge set of
G, denoted E(G). If v and w are vertices of G, then the edge vw connects or joins v and w.
Why graphs are so useful is that graph theory treats only the number of elements in a network and their relationships to each other, both in terms of the characteristics of the edge set. First some terms.
0 0 | 184 | 791 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2019-13 | latest | en | 0.940642 |
https://mapleprimes.com/questions/200839-How-To-Replace-Dnotation-With-Standard | 1,713,140,837,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816904.18/warc/CC-MAIN-20240414223349-20240415013349-00019.warc.gz | 348,098,082 | 23,832 | # Question:How to replace D-notation with standard partial derivative notations
## Question:How to replace D-notation with standard partial derivative notations
Maple
I am trying to illustrate the chain rule for multivariet functions
diff(f(u(x,y),v(x,y),x)
The Maple responce is D1(f)(u(x,y),v(x,y)*(partial of u(x,y) wrt x) +..etc
I would like to replace the D- notation with the standard notation for the "partial of f wrt u" for obvious reasons - this is what students are familar with. The convert cmnd Doe Not Work in this case.
Similarly the cmnd diff(u(x,y),v(x,y),x,x) gives rise to D1,D11, D12 symbols which I would likee to convert to standard partial notation.
All this is a BIG DEAL when trying to illstrate the chain rule in Cal III.
Joe Salacuse
Mathematics
Kettering University
| 212 | 807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.86876 |
http://hobbyeleccircuits.blogspot.com/2014/06/input-trigger-synchronized-monostable.html | 1,553,427,977,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203438.69/warc/CC-MAIN-20190324103739-20190324125739-00543.warc.gz | 84,673,065 | 16,585 | ## Monday, June 30, 2014
### Input Trigger Synchronized Monostable Timer Circuit Using IC 555
Here we study a simple IC 555 based monostable circuit whose output monostable time duration starts only after the input trigger is released thus making sure that the trigger ON time duration is added with the monostable's pre-programmed ON time duration. The idea was requested by Mr. John Brogan
The Request
Hello,
I would like to know if I could hire you for a very simple project. This is to help me learn circuits.
I am looking for the following type of circuit (see below). Could you let me know what it would cost to design?
There will be 4 pins on the circuit board. 2 pins on the left side of the board, 2 pins on the right.
When someone closes the circuit of the LEFT side of the board, either momentarily, or for however long they keep the circuit closed, the pins on the RIGHT side of the board close *PLUS* 2 minutes after the time the circuit on the LEFT side of the board is opened. (that’s the part I’m stuck on – how to make a circuit stay closed for “n” minutes past the time another circuit is opened.
Please let me know what you would charge to diagram this and list the parts I need to buy to make this.
Thank you!
John Brogan
The Design
In other words, the above request demands a monostable whose output on state delay will initiate only once the input trigger is released, meaning suppose the monostable is designed to produce a delay of 2 minutes, and let's assume the input trigger hold time to be x minutes, the total delay at the output pin3 of the IC should be then = 2 minutes + "x" minutes.
The design may be simply configured by adding a PNP stage to a standard IC 555 monostable circuit.
Referring to the figure below we see a standard IC 555 monostable circuit which produces an output high for a time delay determined by R2 and C1. This initiates each time pin2 is grounded momentarily or may be for some relatively longer period of time.
However normally this would happen as soon as pin2 is grounded without considering the trigger ON duration, and we don't want this situation for the proposed design.
The issue is effectively remedied by the inclusion of the PNP device T1 across the shown position of the circuit.
As suggested in the request when the left pins are closed, T1 is allowed with a negative bias forcing it to conduct.
The above condition allows the output to go high but shorts the timing capacitor C1 via T1 emitter/cpllector so that it is unable to charge until the left pins are opened by the user.
Once the left pins are released, C1 is allowed to charge and initiate the monostable counting operation wherein the relay actuates and closes the right pins for a total duration of the set two minutes plus the duration for which the input was held closed. | 624 | 2,824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-13 | latest | en | 0.93489 |
https://gauravtiwari.org/tag/real-analysis/ | 1,685,645,778,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648000.54/warc/CC-MAIN-20230601175345-20230601205345-00736.warc.gz | 314,263,500 | 51,006 | # Real Analysis
## Dedekind’s Theory of Real Numbers
Let $\mathbf{Q}$ be the set of rational numbers. It is well known that $\mathbf{Q}$ is an ordered field and also the set $\mathbf{Q}$ is equipped with a relation called “less than” which is an order relation. Between two rational numbers there exists an infinite number of elements of $\mathbf{Q}$. Thus, the system…
## D’Alembert’s Ratio Test of Convergence of Series
In this article, we will formulate the D’ Alembert’s Ratio Test on convergence of a series. Let’s start. Statement of D’Alembert Ratio Test A series $\sum {u_n}$ of positive terms is convergent if from and after some fixed term $\dfrac {u_{n+1}} {u_n} < r < {1}$ , where r is a fixed…
## Derivative of x squared is 2x or x ? Where is the fallacy?
Derivative of x squared As we know, the derivative of x squared, i.e., differentiation of $x^2$ , with respect to $x$, is $2x$. i.e., $\dfrac{d}{dx} x^2 = 2x$ A Curious Case Suppose we write $x^2$ as the sum of $x$ ‘s written up $x$ times. i.e., \$ x^2… | 304 | 1,028 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-23 | latest | en | 0.879259 |
ligautamabet.net | 1,643,186,583,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304928.27/warc/CC-MAIN-20220126071320-20220126101320-00047.warc.gz | 401,584,166 | 9,926 | # A Brief History of the Metric System
Measurements conversion is one of the most important concepts when it comes to manufacturing procedures. Many products, like garments, food and drink, etc are made in different sizes. To ensure that the products can be manufactured in the required sizes, measurements conversion is required. Some measurements are so important that if these measurements are not taken, the products will not conform to international standards and can even cause damage to the users. The conversion factors between units can also cause inconvenience at times.
A good measurements conversion calculator can perform the task for you perfectly. All you need to do is to provide the data that you want to have converted into the units that are required. Once the data is entered on the unit conversion calculator, the results will be displayed immediately. The calculator will tell you whether the result that you have entered is correct or not. Also, you can easily see the data that has been input by using the handy buttons on the calculator unit conversion calculator.
There are different units conversions available for different purposes. Units of measurement are used for different purposes like measurements, measuring mass, measurements conversion, etc. A number of conversions are used in the world of science as well. For example, in measuring the velocity of light and the distance traveled by light, one may perform the conversions math by using the units of length, time and acceleration. Similarly, the www.measurementsconversion.com can be performed for different units like pressure, temperature, density and many more.
The most common conversions are the Fahrenheit and Celsius measurements. The Fahrenheit scale can be used to measure the temperatures of water and air. This is done by using the ‘atan formula’. In Celsius, it is followed by multiplying the square of the temperature by 1.3. The S-1 derivative also shows the values of temperatures of liquid and solid mass.
In order to perform the Fahrenheit conversion, one has to know about its conversions. The Fahrenheit and Celsius units are based on a standard which has been adopted by the United States and the United Kingdom. There are also some other conversions such as the English inch to metric yard, British yards to Metric yards, and meter to Nautical mile. In addition, the meter is also a conversion unit for measurements. These conversion systems allow the user to determine the distance and time that the measurements were taken from.
One can convert his measurements by using the metric system measurement calculator. There are many types of calculators available in the market, which include the English inch to Metric yard, the British yards to Metric yard, and the meter to Nautical mile. To get conversion results, the chosen conversion type should be entered.
One may also find the history of the metric system in the weights and measures conversion chart. This chart lists about the major units and their corresponding measurements which have been used since ancient times. Most countries use the English pound and the United States dollar as their measuring units. These days, the monetary units such as the US dollars and the Euro have replaced the English pounds and the British ounces.
Today, the use of the metric measurements dates back to the 16th century. This was followed by the decimal system, which was introduced by the French. This eventually replaced the English system of weights and measures with the metric measurements. The decimal system has been improved over the years. Today, the system of units for measuring weight and measures is used all over the world for its purposes including monetary calculations. | 690 | 3,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-05 | latest | en | 0.924787 |
https://www.assignmentexpert.com/homework-answers/economics/microeconomics/question-91563 | 1,653,764,745,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00446.warc.gz | 735,764,952 | 67,577 | 106 271
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Answer to Question #91563 in Microeconomics for swati
Question #91563
Assume your demand curve is P=60-5Q. Now draw your demand curve when you have an insurance policy that pays the amount of the doctor visit minus a copayment of $20. How much will you demand if the doctor charges$60? Draw the graph alongwith explaining your answer
1
2019-07-11T09:25:57-0400
If you have an insurance policy that pays the amount of the doctor visit minus a copayment of $20, then you pay$20 for every doctor visit and your demand curve is P = 20.
If the doctor charges \$60, then 60 - 5Q = 60,
Q = 0 visits.
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for any assignment or question with DETAILED EXPLANATIONS! | 218 | 806 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-21 | latest | en | 0.8928 |
https://www.aqua-calc.com/convert/density/stone-per-metric-tablespoon-to-slug-per-cubic-foot | 1,631,911,308,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00038.warc.gz | 715,137,666 | 11,168 | # Convert stones per (metric tablespoon) to slugs per (cubic foot)
## st/metric tbsp to sl/ft³ (st:stone, tbsp:tablespoon, sl:slug, ft:foot)
### stones per metric tablespoon to slugs per cubic foot conversion cards
• 1
through
20
stones per metric tablespoon
• 1 st/metric tbsp to sl/ft³ = 821.440197 sl/ft³
• 2 st/metric tbsp to sl/ft³ = 1 642.880394 sl/ft³
• 3 st/metric tbsp to sl/ft³ = 2 464.320591 sl/ft³
• 4 st/metric tbsp to sl/ft³ = 3 285.760788 sl/ft³
• 5 st/metric tbsp to sl/ft³ = 4 107.200985 sl/ft³
• 6 st/metric tbsp to sl/ft³ = 4 928.641182 sl/ft³
• 7 st/metric tbsp to sl/ft³ = 5 750.081379 sl/ft³
• 8 st/metric tbsp to sl/ft³ = 6 571.521576 sl/ft³
• 9 st/metric tbsp to sl/ft³ = 7 392.961773 sl/ft³
• 10 st/metric tbsp to sl/ft³ = 8 214.40197 sl/ft³
• 11 st/metric tbsp to sl/ft³ = 9 035.842167 sl/ft³
• 12 st/metric tbsp to sl/ft³ = 9 857.282364 sl/ft³
• 13 st/metric tbsp to sl/ft³ = 10 678.722561 sl/ft³
• 14 st/metric tbsp to sl/ft³ = 11 500.162758 sl/ft³
• 15 st/metric tbsp to sl/ft³ = 12 321.602955 sl/ft³
• 16 st/metric tbsp to sl/ft³ = 13 143.043152 sl/ft³
• 17 st/metric tbsp to sl/ft³ = 13 964.483349 sl/ft³
• 18 st/metric tbsp to sl/ft³ = 14 785.923546 sl/ft³
• 19 st/metric tbsp to sl/ft³ = 15 607.363743 sl/ft³
• 20 st/metric tbsp to sl/ft³ = 16 428.80394 sl/ft³
• 21
through
40
stones per metric tablespoon
• 21 st/metric tbsp to sl/ft³ = 17 250.244137 sl/ft³
• 22 st/metric tbsp to sl/ft³ = 18 071.684334 sl/ft³
• 23 st/metric tbsp to sl/ft³ = 18 893.124531 sl/ft³
• 24 st/metric tbsp to sl/ft³ = 19 714.564728 sl/ft³
• 25 st/metric tbsp to sl/ft³ = 20 536.004925 sl/ft³
• 26 st/metric tbsp to sl/ft³ = 21 357.445122 sl/ft³
• 27 st/metric tbsp to sl/ft³ = 22 178.885319 sl/ft³
• 28 st/metric tbsp to sl/ft³ = 23 000.325516 sl/ft³
• 29 st/metric tbsp to sl/ft³ = 23 821.765713 sl/ft³
• 30 st/metric tbsp to sl/ft³ = 24 643.20591 sl/ft³
• 31 st/metric tbsp to sl/ft³ = 25 464.646107 sl/ft³
• 32 st/metric tbsp to sl/ft³ = 26 286.086304 sl/ft³
• 33 st/metric tbsp to sl/ft³ = 27 107.526501 sl/ft³
• 34 st/metric tbsp to sl/ft³ = 27 928.966698 sl/ft³
• 35 st/metric tbsp to sl/ft³ = 28 750.406895 sl/ft³
• 36 st/metric tbsp to sl/ft³ = 29 571.847092 sl/ft³
• 37 st/metric tbsp to sl/ft³ = 30 393.287289 sl/ft³
• 38 st/metric tbsp to sl/ft³ = 31 214.727486 sl/ft³
• 39 st/metric tbsp to sl/ft³ = 32 036.167683 sl/ft³
• 40 st/metric tbsp to sl/ft³ = 32 857.60788 sl/ft³
• 41
through
60
stones per metric tablespoon
• 41 st/metric tbsp to sl/ft³ = 33 679.048077 sl/ft³
• 42 st/metric tbsp to sl/ft³ = 34 500.488274 sl/ft³
• 43 st/metric tbsp to sl/ft³ = 35 321.928471 sl/ft³
• 44 st/metric tbsp to sl/ft³ = 36 143.368668 sl/ft³
• 45 st/metric tbsp to sl/ft³ = 36 964.808865 sl/ft³
• 46 st/metric tbsp to sl/ft³ = 37 786.249062 sl/ft³
• 47 st/metric tbsp to sl/ft³ = 38 607.689259 sl/ft³
• 48 st/metric tbsp to sl/ft³ = 39 429.129456 sl/ft³
• 49 st/metric tbsp to sl/ft³ = 40 250.569653 sl/ft³
• 50 st/metric tbsp to sl/ft³ = 41 072.00985 sl/ft³
• 51 st/metric tbsp to sl/ft³ = 41 893.450047 sl/ft³
• 52 st/metric tbsp to sl/ft³ = 42 714.890244 sl/ft³
• 53 st/metric tbsp to sl/ft³ = 43 536.330441 sl/ft³
• 54 st/metric tbsp to sl/ft³ = 44 357.770638 sl/ft³
• 55 st/metric tbsp to sl/ft³ = 45 179.210835 sl/ft³
• 56 st/metric tbsp to sl/ft³ = 46 000.651032 sl/ft³
• 57 st/metric tbsp to sl/ft³ = 46 822.091229 sl/ft³
• 58 st/metric tbsp to sl/ft³ = 47 643.531426 sl/ft³
• 59 st/metric tbsp to sl/ft³ = 48 464.971623 sl/ft³
• 60 st/metric tbsp to sl/ft³ = 49 286.41182 sl/ft³
• 61
through
80
stones per metric tablespoon
• 61 st/metric tbsp to sl/ft³ = 50 107.852017 sl/ft³
• 62 st/metric tbsp to sl/ft³ = 50 929.292214 sl/ft³
• 63 st/metric tbsp to sl/ft³ = 51 750.732411 sl/ft³
• 64 st/metric tbsp to sl/ft³ = 52 572.172608 sl/ft³
• 65 st/metric tbsp to sl/ft³ = 53 393.612805 sl/ft³
• 66 st/metric tbsp to sl/ft³ = 54 215.053002 sl/ft³
• 67 st/metric tbsp to sl/ft³ = 55 036.493199 sl/ft³
• 68 st/metric tbsp to sl/ft³ = 55 857.933396 sl/ft³
• 69 st/metric tbsp to sl/ft³ = 56 679.373593 sl/ft³
• 70 st/metric tbsp to sl/ft³ = 57 500.81379 sl/ft³
• 71 st/metric tbsp to sl/ft³ = 58 322.253987 sl/ft³
• 72 st/metric tbsp to sl/ft³ = 59 143.694184 sl/ft³
• 73 st/metric tbsp to sl/ft³ = 59 965.134381 sl/ft³
• 74 st/metric tbsp to sl/ft³ = 60 786.574578 sl/ft³
• 75 st/metric tbsp to sl/ft³ = 61 608.014775 sl/ft³
• 76 st/metric tbsp to sl/ft³ = 62 429.454972 sl/ft³
• 77 st/metric tbsp to sl/ft³ = 63 250.895169 sl/ft³
• 78 st/metric tbsp to sl/ft³ = 64 072.335366 sl/ft³
• 79 st/metric tbsp to sl/ft³ = 64 893.775563 sl/ft³
• 80 st/metric tbsp to sl/ft³ = 65 715.21576 sl/ft³
• 81
through
100
stones per metric tablespoon
• 81 st/metric tbsp to sl/ft³ = 66 536.655957 sl/ft³
• 82 st/metric tbsp to sl/ft³ = 67 358.096154 sl/ft³
• 83 st/metric tbsp to sl/ft³ = 68 179.536351 sl/ft³
• 84 st/metric tbsp to sl/ft³ = 69 000.976548 sl/ft³
• 85 st/metric tbsp to sl/ft³ = 69 822.416745 sl/ft³
• 86 st/metric tbsp to sl/ft³ = 70 643.856942 sl/ft³
• 87 st/metric tbsp to sl/ft³ = 71 465.297139 sl/ft³
• 88 st/metric tbsp to sl/ft³ = 72 286.737336 sl/ft³
• 89 st/metric tbsp to sl/ft³ = 73 108.177533 sl/ft³
• 90 st/metric tbsp to sl/ft³ = 73 929.61773 sl/ft³
• 91 st/metric tbsp to sl/ft³ = 74 751.057927 sl/ft³
• 92 st/metric tbsp to sl/ft³ = 75 572.498124 sl/ft³
• 93 st/metric tbsp to sl/ft³ = 76 393.938321 sl/ft³
• 94 st/metric tbsp to sl/ft³ = 77 215.378518 sl/ft³
• 95 st/metric tbsp to sl/ft³ = 78 036.818715 sl/ft³
• 96 st/metric tbsp to sl/ft³ = 78 858.258912 sl/ft³
• 97 st/metric tbsp to sl/ft³ = 79 679.699109 sl/ft³
• 98 st/metric tbsp to sl/ft³ = 80 501.139306 sl/ft³
• 99 st/metric tbsp to sl/ft³ = 81 322.579503 sl/ft³
• 100 st/metric tbsp to sl/ft³ = 82 144.0197 sl/ft³
#### Foods, Nutrients and Calories
FRUIT SMOOTHIE, UPC: 028700147733 contain(s) 75 calories per 100 grams (≈3.53 ounces) [ price ]
11 foods that contain Choline, from sphingomyelin. List of these foods starting with the highest contents of Choline, from sphingomyelin and the lowest contents of Choline, from sphingomyelin
#### Gravels, Substances and Oils
CaribSea, Marine, CORALine, Aruba Puka Shell weighs 1 121.29 kg/m³ (69.99985 lb/ft³) with specific gravity of 1.12129 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Lithium nitride [Li3N] weighs 1 270 kg/m³ (79.28351 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-32, liquid (R32) with temperature in the range of -40°C (-40°F) to 65.56°C (150.008°F)
#### Weights and Measurements
A millimeter per minute squared (mm/min²) is a derived metric SI (System International) measurement unit of acceleration
The speed measurement was introduced to measure distance traveled by an object per unit of time,
lb/pt to mg/mm³ conversion table, lb/pt to mg/mm³ unit converter or convert between all units of density measurement.
#### Calculators
Compute cost of oils, fuels, refrigerants at different temperatures | 3,205 | 7,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-39 | latest | en | 0.200182 |
https://physicscatalyst.com/class-6/Important-questions_class6_science_electricity-1.php | 1,721,095,778,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514726.17/warc/CC-MAIN-20240716015512-20240716045512-00026.warc.gz | 397,963,689 | 16,007 | # Practice worksheets for Class 6 Science Chapter 9 Electricity and Circuits
In this page we have Practice worksheets for Class 6 Science Chapter 9 Electricity and Circuits . Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
State True or False
(a) Conductor allows the current to pass through it.
(b) Insulator does not allows the current to pass through it.
(c) Electric cell metal cap is a positive terminal.
(d) Electric cell metal disc is a negative terminal.
(e) A electric bulb with broken filament is called a fused bulb.
(f) An electric cell has four terminals.
(g) If the filament of the bulb is broken, the bulb will not glow.
(h) Air is good conductor of electricity.
(i) Themocol is an insulator.
(j) When the positive of bulb is connected with positive of the cell and negative of bulb is connected with negative of cell, the bulb glows.
(k) When the positive of bulb is connected with negative cell and negative of bulb is connected with positive of cell, the bulb does not glows.
1. True
2. True
3. True
4. True
5. True
6. False
7. True
8. False
9. True
10. True
11. False
Question 2
Match the column
Question 3
Crossword
Solve the below crossword
Across
1. A device that is used to either break the electric circuit or to complete it.
6. Materials which allow electric current to pass through them
7. The tiny coiled metal wire present inside the bulb
Down
2. A electric cell has two ______
3. Materials that do not allow electricity to pass through them
4. device which converts chemical energy into electric energy
5. the filament of a bulb breaks then it is called
6. The bulb glows only when _____ flows through the circuit.
1. Switch
2. terminals
3. insulators
4. Battery
5. Fused
6. current, Conductors
7. Filament
Question 4
FILL IN THE BLANKS.
1. ________ is a device used to break electric circuit and also join it when needed.
2. A bulb glows only when ________ flows through it.
3. An electric cell has ________ terminals.
4. ________ do not allow current to pass through them.
5. A ________. Bulb doesn't glow even when electric current pass through it.
6.
1. Switch
2. Current
3. Two
4. Insulators
5. Fused
Question 5
TRUE OR FALSE
1. Air is a conductor
2. Copper is a conductor.
3. Human body is a good conductor of electricity.
4. Sea water can conduct electricity better than distilled water.
5. The metal cap is the negative terminal of the electric cell.
6.
1. False
2. True.
3. True
4. True.
5. False
Question 6
UNSCRAMBLE THE FOLLOWING WORDS.
1. TILFNMESA
2. LRISASUTNO
3. ALETMINRS
4. UOCTNORCSD
5. CTECLRIE
1. FILAMENTS
2. INSULATORS
3. TERMINALS
4. CONDUCTORS
5. ELECTRIC
Question 7
1. Who invented the first electric bulb?
2. Which is the -ve terminal of a cell?
3. Who invented dry cell?
4. Name two good conductors.
5. Name two insulators.
(i)Thomas Alva Edison.
(ii)The side with the metal disc is the negative
(iii)Georges Leclanche.
(iv)copper & silver
(v) Wood & Plastic
### Practice Question
Question 1 What is $\frac {1}{2} + \frac {3}{4}$ ?
A)$\frac {5}{4}$
B)$\frac {1}{4}$
C)$1$
D)$\frac {4}{5}$
Question 2 Pinhole camera produces an ?
A)An erect and small image
B)an Inverted and small image
C)An inverted and enlarged image
D)None of the above | 900 | 3,270 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-30 | latest | en | 0.897088 |
https://brainmass.com/math/basic-algebra/93440 | 1,485,013,049,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281151.11/warc/CC-MAIN-20170116095121-00487-ip-10-171-10-70.ec2.internal.warc.gz | 791,259,151 | 17,840 | Share
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# Simplifying Expressions
Write
2 d^3 y^3 t
------------- - -----------
9ak 3a^3 d
as a single rational expression.
#### Solution Summary
An expression is simplified.
\$2.19 | 56 | 204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-04 | latest | en | 0.71605 |
https://studyres.com/doc/7747275/geocentric-solution-to-3-body-prob | 1,618,683,551,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038461619.53/warc/CC-MAIN-20210417162353-20210417192353-00124.warc.gz | 644,723,854 | 26,391 | Survey
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Transcript
```A Geocentric Solution to the 3-body Problem
1
A GEOCENTRIC SOLUTION TO THE
THREE-BODY PROBLEM
Prepared by
Gerard Bouw, Ph.D.
Introduction
In 2013 some humanistic, self-professed scientists proposed that, for the 3-body problem,
the theory of geocentricity should yield a different solution than what is observed. In
anticipation of that challenge Bob Sungenis invited me to derive a geocentric solution
within the framework of the theory of geocentricity. Our purpose was to derive a viable
geocentric framework for the 3-body problem. We were successful and this paper is the
fruit of our labors.
The Approach
The version of the 3-body problem we consider here is, as the name suggests, of three
bodies in mutual orbit about each other. Classically, the 3-body problem assumes that
two of the bodies are massive and the third body’s mass can be regarded as zero. Unlike
most modern dynamic derivations posted on the web which know the solutions ahead of
time, the approach in this paper is completely general and so could be used to locate any
additional critical points besides the five Lagrangian points. Geocentrically, the 3 bodies
are: the sun, the earth, and a third body, of negligible mass, that appear at rest relative to
those two bodies. Fundamentally, we are looking to answer this question: “At what
locations is the orbital speed of the negligible-mass body zero with respect to both the
sun and earth?” As a point of terminology, the negligible body is technically called the
“infinitesimal mass” or “the test mass.”
Figure 1 presents the geocentric geometry of the problem. The earth is located at point E
and the sun is located at point S. The angular velocity vector, D represents the diurnal
rotation of the firmament about the axis that runs from the center of the earth through its
poles. The angular velocity vector, A, is inclined at 23.5° to D, which is the angular
velocity of the earth-sun line about the earth in the course of the year. The geometry is
such that the sun is located at its southernmost point, namely the first day of winter, that
is, roughly 21 December. Thus the line from E to S goes clockwise around the point E as
seen from A, with a period of one year. Likewise, as seen from the tip of D, the entire
page rotates counterclockwise with a period of one sidereal day (23 hours 56 minutes).
We shall simplify where we can legitimately do so. In an earlier paper we have shown
specifically for the diurnal (daily) rotation case that the system holds together; that the
firmament’s gravitational field, which we call inertia, holds the stellar universe together
even though some may think that the orbital speeds would be immense. But rotation is
not the same as revolution, which is to say that if the entire universe rotates as one inertial
2
A Geocentric Solution to the 3-body Problem
field, the presence of that rotation cannot be felt by the atomic matter of the universe, or
by photons traveling through it.
We shall shortly demonstrate that neither the daily rotation of the firmament nor the
yearly orbital motion of the sun’s entire about the earth can disrupt the motions we
observe. When we are finished with that section, we can simplify our problem to that
presented in Figure 4 (pg. 7).
Figure 1: The Geocentric Situation
Derivation Of The Geocentric Equations For A Daily-Rotating Universe
By definition, physics deals with matter in motion. Mathematics is the language of
choice, used by physicists to describe motion. Usually physicists are well behaved in
their use of math, but at times, they introduce fudge factors to bridge what theory
demands and experiment lacks. Even then, the fudging is quite obvious from the names
given the fudge factors such as “guillotine factor,” for instance. But there are times when
reputations and careers are at stake and at those times, the fudging becomes quite subtle,
even mean-spirited at times.
A Geocentric Solution to the 3-body Problem
3
The mathematical language used to describe the gravitational forces of orbiting bodies,
and the behavior of spinning bodies is a case in point. When confronted by the mass of
evidence for the geocentric universe, physicists resort to sleight of hand to keep the earth
in orbit about the sun when all fundamental experimental results reveal earth to be
standing still in the firmament. In this case, they multiply one side of the generalized
equation of motion by the number one. Before multiplying by one, the equation is said to
be kinematic, describing the accelerations and velocities of the bodies but not taking the
masses of the bodies into consideration. For instance, consider this equation that
describes the velocity, v, of a body in circular motion with a rotational speed of at a
distance R from the center of the circle:
v = R.
(1)
This equation is said to be kinematic and even though it perfectly describes the velocity
and behavior of a body’s rotational and orbital motions, it is said to be unphysical.
Now suppose that we multiply the left-hand side of the equation by one, namely, by the
mass, m, divided by itself, i.e., m/m. This is equivalent to multiplying both sides of the
equation by the mass, m. Our equation (1) now looks as follows:
m v = m R.
(2)
This is said to be a dynamic description, that is to say, somehow this equation is more
“physical,” more “real,” than the kinematic equation (1) even though we can obviously
cancel out the m’s and simplify equation (2) back to equation (1). To hide this sleight of
hand, equation (2)’s left hand side is replaced by a single variable, p, called momentum.1
Thus equation (3), which is the same as equation (2) is rewritten as”
p = m R.
(3)
Since momentum is a dynamic concept, the mass is hidden and no physicist will cancel
its appearance on the right-hand side of the equation with its hidden counterpart in p.
But two can play at that game. Let us assume that God created the firmament with a
built-in set of reaction rules. These rules dictate the behaviors of accelerating bodies and
the set of all such reactions we group together under in the name of inertia.
Deriving the Geocentric Equations From First Principles
As seen from earth, its coordinates determine a star’s location. Just as our coordinates on
earth are specified by longitude and latitude, so a star’s coordinates are given by its right
ascension and declination. A star’s longitude is specified by its right ascension and its
latitude is measured by its declination north or south of the plane of earth’s equator.
Since the star’s coordinates are fixed to the celestial sphere, to model the rotation of the
firmament—carrying the star with it—we only need the star’s declination (see Figure 1).
1
Technically, it is more correct to say that p is the angular momentum, but that is irrelevant to the
argument at hand.
4
A Geocentric Solution to the 3-body Problem
Figure 2: The Geocentric View of the Daily Rotation
The following is a derivation of the dynamical equations for the universe rotating about
the earth in a daily rotation. In the derivation, we use the following notation:
F is the net gravitational force exerted on the star;
a is the net acceleration experienced by the star in its daily path about earth;
R is the shortest distance from the axis of rotation to the star;
D is the distance from earth to the star;
v is the velocity of the star;
m is the star’s mass;
is the declination (celestial latitude) of the star as measured from the equator; and
is the rotation rate of the firmament about the rotational axis that passes through
the north and south poles of earth: imagine it measured in degrees per second
although technically we use radians per second.2
The usual objection against geocentricity is that the earth is not massive enough to have
the universe “orbit” it once a day. In reality, neither the mass of the earth nor the mass of
the star enter into the force that holds the universe together during its rotation.
Acceleration is defined as a change in velocity per unit time. We can write this as:
d 2R
a 2
dt
2
There are 2 radians in the circumference of a circle, so a radian is roughly 57 degrees.
(4)
A Geocentric Solution to the 3-body Problem
5
Here, R is the distance to a moving object and t is time. This can be rewritten more
explicitly as:
a
d dR
dt dt
(5)
where dR/dt is the velocity, v, of the moving object, the star in our case. This equation
says, “Acceleration is the rate of change in velocity.”
But we’re not trying to model the speed and acceleration of an automobile here but that
of a distant star rotating about the earth once every 23 hours and 56 minutes. We must
thus add the rotational velocity (Equation (1)) into the mix. This requires us to rewrite
equation (5) as:
d dR
(6)
a
R
dt dt
where is the angular velocity (measured in degrees per second, for instance) and R is
the distance of the star from the axis of rotation.
Distributing the derivative (d/dt) through the terms in parentheses of Equation (6) gives
us equation (7):
d 2 R d
dR
(7)
a 2
R 2
R .
dt
dt
dt
Here the first term on the right-hand side of equation (7) is any acceleration that may be
d
imparted to the earth (the central point). The second term,
R , is the Euler force,
dt
which is not of interest here since it only kicks in if the length of the day changes
significantly over the course of a day. The third term (starting with the 2), is the Coriolis
force and the last term [R)] is the centrifugal force.
The Coriolis and centrifugal forces dominate the motion of the sun, planets, and stars in a
geocentric system. We shall thus ignore the Euler and local acceleration terms of
equation (7) and work only with:
a 2 v R
(8)
where v is the orbital speed of the star. Since the firmament rotates and not the earth, the
sign of v is in the opposite direction to the heliocentric system, and is thus negative. The
v in equation (8) is thus replaced by R .
After expanding v, equation (8) is now:
a 2 R R ;
or
a R .
Distributing the cross-product through the term in parentheses gives us:
(9)
6
A Geocentric Solution to the 3-body Problem
a R R .
(10)
Now the star is not located on the equator but at declination , whence
·R = D sin().
Our final equation for the geocentric system is thus:
a 2 R Dˆ sin .
(11)
Here ̂ is a unit vector pointing along the rotation axis, that is, in the direction of
which is perpendicular to the equator in general and here in the plane of the star’s circle
in Figure 1. This keeps the acceleration experienced by the star confined to the star’s
latitude, swept out by R and noted as the “Star’s daily path” in Figure 1.
Let’s Examine Our Results Thus Far
Equation (11) has two components, two vectors. They are pictured in Figure 2 where
they are shown as acceleration vectors. To make them dynamic, multiply each by the
star’s mass. The acceleration pictured by the sine term is aligned along the rotation
vector, , and serves to keep the star’s rotational
plane from “falling” up or down the rotational axis.
The second component is the cosine term. That
acceleration pulls the star towards the axis of
rotation. If multiplied by the star’s mass it becomes
a centripetal (non-fictitious) force, meaning that it
pulls the star towards the axis of rotation. The net
result of these two accelerations is to keep the star
in its place in the inertial field of the universe which
is the gravitational field of the firmament.
Of course, equation (11) is kinematic, not dynamic
and we have to show the geocentric model is
dynamically correct. To do that, all we have to do
is to multiply both sides by the star’s mass, m:
F ma m2 R Dˆ sin
(12)
Figure 3: Accelerations (Forces)
Although we assumed the body was a star in
deriving (12), it could just as well have been the sun, moon, any planet, artificial satellite,
or star circling the earth’s polar axis. Yet some will ask, “What about the speed of light?
Won’t the distant planets and stars orbit the earth way above the speed of light?”
The answer is, “No.” The speed of light is determined by the firmament. It is the
firmament that rotates on the polar axis once a day and so photons, which are transmitted
by the firmament, also participate in the daily rotation. Light, will also obey the above
A Geocentric Solution to the 3-body Problem
7
equations superimposed on its own motion. To object that it still exceeds the speed of
light we answer that the speed of light speed limit does not apply for rotation. In this
case it is equivalent to claiming that when a supersonic jet flies faster than the speed of
sound, you could not talk to the person in front of you because you were flying faster
than the speed of sound. But the air in the plane, too, was “flying” faster than the speed
of sound, so you can talk to the person in the seat in front of you because the soundbearing medium was carried with you, even as the light-bearing medium is carried along
with the sun, moon, and stars in the daily rotation of the firmament.
Conclusion
We have shown that the physics of the geocentric universe accounts perfectly for what
we see and measure of the daily rotation whether that rotation is of the earth within the
universe or the universe around the earth. In the final analysis, proofs based on
dynamical equations are not proofs of anything; nor are they proofs against the geocentric
universe.
By the same approach, we could show that the yearly orbital motion of the sun and
planets can be represented in the geocentric framework of the firmament. But for this
paper we shall only point out that every object in the universe obeys equation (12). Since
S, m, and even the c.m. in Figure 1 will all obey equation (12) for the rotational case, we
can simplify Figure 1 to Figure 4:
Figure 4: The Simplified Vector Diagram
.
Here: S is the sun;
E is the earth;
m is the infinitesimal mass (a.k.a. the test mass);
c.m. is the center of mass of the local system;
r1 is the vector representing the distance from sun to the c.m. (which is inside the
sun);
r2 is the vector representing the distance of the center of mass from the earth;
r is the distance of the test mass m from the c.m.
8
A Geocentric Solution to the 3-body Problem
There remains for us to derive the generalized equations of motion describing the path of
the test mass, m. Then we need to solve the generalized equations for locations where
m’s velocity is zero relative to both sun and earth.
Equations of Motion for the Infinitesimal Body
As a result of deriving Figure 4 the way we did by using the geocentric equations of
diurnal forces acting on distant masses, we can assume that the two finite masses revolve
in circles around their common center of mass. Since the test mass is assumed
infinitesimally small, it does not change the location of the earth-sun center of mass (i.e.,
the barycenter).
To further simplify our approach, let us assume the unit of mass such that the sum of the
3 masses = 1. This allows us to set the mass of one body equal to 1 - and the other
equal to. In our case, is the mass of the earth and 1 - is the mass of the sun. Here
we select the notation such that is less than or equal to ½.
Let the unit of length be the earth-sun distance, that is, the distance from E to S = 1.
Likewise, we select the unit of time to be such that k2 = 1, where k is such that
m
s k 2 F .
Set the origin of our coordinate system at the center of mass and let the -plane be
determined by their mutual rotation (i.e., the ecliptic plane). Set the coordinates of the
bodies so that 1 - and and the test mass be (and (
respectively, and
r1
r2
1
1 2
2
2 2 2 2
2
2
Then the differential equations of motion for m are:
1 2
d 2
1
'
2
dt
r13
r23
1 2
d 2
1
'
2
dt
r13
r23
d 2
1 3 3 .
2
dt
r1
r2
The choice of units makes the mean angular motion of the finite bodies be:
(13)
A Geocentric Solution to the 3-body Problem
k
9
1
a
3
2
= 1,
(14)
where a = r1 + r2 1.
If we now change coordinate systems so that the origin is still at the c.m. and the rotation
is still in the -plane in the direction that the bodies move with uniform angular velocity
unity see (14). The coordinates of the new system are defined by the equations:
x cos t y sin t ,
x sin t y cos t ,
z
(15)
and similar equations for letters with subscripts 1 and 2.
Taking the second derivatives of (15) we obtain and substituting into equation (13)
yields:
d 2 y
d 2 x
dy
dx
2 2 x cos t 2 2 y sin t
dt
dt
dt
dt
x x1 x x 2
y y1
y y 2
1
cos t 1 3
sin t ,
3
3
r1
r2
r1
r23
d 2 y
d 2 x
dy
dx
2
x
sin
t
2
2 2 y cos t
dt
dt
dt
dt
(16)
x x1 x x 2
y y1 y y 2
1
sin t 1
cos t ,
3
3
r1
r2
r13
r23
d 2z
z
z
1 3 3 .
2
dt
r1
r2
Multiplying the first two equations of the three by cos t and sin t respectively, and then by
–sin t and cos t, and adding; the results are:
10
A Geocentric Solution to the 3-body Problem
x x1 x x2
d 2x
dy
2
x 1
,
2
dt
dt
r13
r23
y y1 y y2
d2y
dx
2
y 1
,
2
dt
dt
r13
r23
d 2z
z
z
1 3 3 .
2
dt
r1
r2
Assuming the x-axis as fixed to the centers of the finite bodies (earth and sun, for
instance), then y1=0 and y2=0 and the equations become:
x x1 x x2
d 2x
dy
2
x 1
,
2
dt
dt
r13
r23
d2y
dx
y
y
2
y 1 3 3 ,
2
dt
dt
r1
r2
(17)
d 2z
z
z
1 3 3 .
2
dt
r1
r2
We now have the differential equations describing the motion of the test mass, m with
respect to axes rotating in such a way that the finite bodies always lie on the x-axis
(earth-sun). The unique property of these expressions is that they do not explicitly
include time. When we started our analysis in equations (13); were
functions of time.
Equations 17 can be integrated by Jacobi’s Integral if we let
U
1 2
1 ;
x y2
2
r1
r2
(18)
so that equations 17 can be rewritten as:
d 2x
dy U
2
,
2
dt
dt
x
d2y
dx U
2
,
2
dt
y
dt
(19)
d 2 z U
.
z
dt 2
Multiplying these equations by 2(dx/dt), 2(dy/dt) and 2(dz/dt) respectively, and then
summing, the resulting equation can be integrated because U is a function of x, y, and z
alone. Doing so gives:
A Geocentric Solution to the 3-body Problem
2
11
2
2
dx dy dz
2
v 2U C
dt dt dt
where C is the constant of integration.
By definition this gives
1 2 C
dx dy dz
2
2
x y 2
r1
r2
dt dt dt
2
2
2
(20)
Since our solution to the problem is of 6th order instead of the usual approach which is of
5th order, we need five more equations to solve the problem. If we confine the motion of
the test mass to the xy-plane, we only need three more equations. Finding one equation,
we can find the remaining two by Jacobi’s last multiplier; so we actually need to find one
more equation.
Equation (20) is a relation between the square of the velocity and the coordinates of m
referenced by the rotating axes. The constant C can be found by initial conditions, so
(20) specifies the velocity of the test mass at all points of the rotating space; and
conversely, given a velocity, (20) gives the locus of points where only m can be. In
particular, if the velocity is set to zero in (20), it will define the surfaces at which the
velocity is zero. On the one side of these surfaces, the velocity will be real and on the
other side, imaginary. We can regard this as saying that it is possible for the body to
move on one side of the surface and impossible to move on the other side.
The equation of the zero surfaces of relative velocity is:
x2 y2 2
r1
r2
1 2
r1
r2
C,
x x y z ,
x x y z .
2
2
2
1
2
2
(21)
2
2
Since only the squares of y and z occur, the surfaces defined by (21) are symmetrical
with respect to the xy and xz-planes, and, for the case that = 0.5, with respect to the yzplane also. The surfaces for 0.5 can be regarded as deformations of those for = 0.5.
From the geometry of z it follows that a line parallel to the z-axis pierces the surfaces in
two or none real points. Also, the surfaces are contained within a cylinder whose axis is
parallel to the z-axis and whose radius is C , to which certain folds are asymptotic at
z2 = ; for, as z2 increases, the equation approaches as a limit:
x2 + y2 = C.
12
A Geocentric Solution to the 3-body Problem
The equation of the curves of intersection of the surfaces with the xy-plane is obtainable
by setting z = 0 in equation (21):
x2 y2
21
x x
2
1
y2
2
x x
2
2
y2
C.
(22)
There are two cases we can use to approximate (22).
Case 1: x and y are large.
If x and y are large, the 3rd and 4th terms in (22) are negligible and the equation can
be written as:
x2 y2 C
where = the sum of terms 3 and 4, which are negligible in this case. This is the
equation of a circle whose radius is C . The larger the value of C, the greater
the values of x and y which satisfy the equation. The smaller , the more circular the
curve and the more nearly it reaches the asymptotic cylinder.
Case 2: x and y are small.
For small values of x and y, the first two terms of equation (22) become relatively
unimportant and the equation may be rewritten as:
1
C
r1
r2
These curves plot the locus of points of equal potential energy for the two centers of
force, 1- and . For large values of C they consist of closed ovals around each of
the bodies E and S. For small values of C these ovals unite between the bodies
forming a dumbbell shaped figure in which the ends are of different size except
when = 0.5. And for still smaller values of C, the handle of the dumbbell enlarges
until the figure becomes an oval enclosing both of the bodies.
It thus follows that the approximate forms of the curves in which the surfaces intersect
the xy-plane are given in Figure 5. The curves C1, C2, C3, C4, and C5 are in order of
decreasing values of the constant C. (Figures 5, 6, and 7 were not computed numerically
but are intended to show qualitatively the characteristics of the curves.)
A Geocentric Solution to the 3-body Problem
Figure 5: xy-Plane Contours
13
14
A Geocentric Solution to the 3-body Problem
Figure 6: xz-Plane Contours
A Geocentric Solution to the 3-body Problem
Figure 7: yz-Plane Contours
15
16
A Geocentric Solution to the 3-body Problem
The equation of the curves of intersection of the surfaces and the xz-plane is obtained by
setting y = 0 in equation (21):
x2
21
x x
2
1
z
2
2
x x
2
2
z
2
C
(23)
Again we have two cases:
Case 1: large values of x and z
For large values of x and z the 2nd and 3rd terms are negligible and may be written as:
x2 = C -
which is the equation of a symmetrical pair of straight lines parallel to the z-axis.
The larger the value of C, the larger the value of x which, for a given value of z,
satisfies the equation and, therefore, the smaller is . Hence, the larger C, the closer
the lines are to the asymptotic cylinder.
Case 2: small x and z.
For small values of x and z, the first term in (23) becomes negligible and the
equation can be written:
1 C
r1
r2 2
Hence the forms of the curves in the xz-plane are qualitatively like those in Figure 6.
Again, C1, …, C5 are in order of decreasing values of the integration constant, C.
We can likewise compute the curves for the yz=plane by setting x=0 in equation (21).
y2
x
21
2
1
y2 z2
x
2
2
2
y2 z2
C
(24)
Case 1: y, z large
Following the same reasoning as done for the earlier cases, for large y and z, we can
write
y 2 = C - ,
which is near the asymptotic cylinder.
Case 2: small y and z
For small values of y and z, (24) may be written as
21
C
r1
A Geocentric Solution to the 3-body Problem
17
which is the equation of a circle which becomes larger as C decreases. Hence the
forms of the curves in the yz-plane are qualitatively as given in Figure 7.
From these 3 figures it is easy to infer their forms for the different values of the
integration constant. They may be roughly described as consisting of, for large values of
C, a closed fold approximately spherical in form around each of the finite bodies, and of
curtains hanging from the asymptotic cylinder symmetrically with respect to the xyplane. For smaller values of C, the folds expand and coalesce (Figure 5, curve C3); for
still smaller values if C the united folds coalesce with the curtains, the first points of
cntact being in every case in the xy-plane; and for sufficiently small values of C the
surfaces consist of two parts symmetrical with respect to the xy-plane but not intersecting
it. (Figure 6, curve C5 and Figure 7 curve C6.)
Now that we know the forms of the surfaces, we have to find where the space motion is
real and where it is imaginary.
The square of the velocity is:
v2 x2 y2
21 2
C
r1
r2
Assume C is so large that the ovals and curtains all are separate. The motion will be real
in those portions of relative space for which the right member of the velocity equation is
positive. If it is positive in one point in a closed fold, it is positive in every other point
within it because the function changes sign only at a surface of zero relative velocity.
From the velocity equation that x and y can be taken so large that the right member will
be positive, regardless of how great C may be. Therefore, the motion is real outside the
curtains. It is also clear that a point can be chosen so near to either 1- or (earth or
sun), that is, either r1 or r2 may be taken so small that the right expression will be positive
however great C may be. Therefore, the motion is real within the folds around the finite
bodies.
If the value of C is so large that the folds around the finite bodies were closed, and if the
infinitesimal body should be within one of these folds at the origin of time, it would
always remain there since it could not cross a surface of zero velocity.
If the sun’s motion about the earth is taken to be circular, and the mass of the moon
infinitesimal, we find that the contour of C3, is 40 times larger than the orbit of the moon.
This is so large that the fold around it and the earth is closed with the moon within it.
Therefore, the moon cannot escape earth’s gravity.
Points on the surfaces can be found by determining the curves in the xy-plane and then
finding by approximations the values of z which satisfy equation 20. Specifically, the
curves in the xy-plane are of interest because the first points of contact, as the various
18
A Geocentric Solution to the 3-body Problem
folds coalesce, occur in this plane, and, indeed, on the x-axis as are evident from the
symmetries of the surfaces.
The equation of the curves in the xy-plane is:
x2 y2
21
x x
2
1
y
2
2
x x
2
2
y2
C
If this equation is rationalized and cleared of fractions the result is a polynomial of the
16th degree in x and y. When the value of one of the variables is taken arbitrarily the
corresponding values of the other can be found.
This problem presents great practical difficulties because of the high degree of the
equation (16th order), and these difficulties are exacerbated by the presence of extraneous
(imaginary) solutions which are introduced by the process of rationalization.
Transforming to polar coordinates can significantly reduce the degree of the equation.
That is, points on the curves can be defined by giving their distances from two fixed
points on the x-axis. We could not use this method if the curves were not symmetrical
with respect to the axis on which the poles lie.
Let the centers of the bodies 1- and be taken as poles; the distances from these poles
are r1 and r2 respectively. To complete the transformation it is necessary to express x2+y2
in terms of these quantities.
Figure 8: Transformed x-y axes
Let P be a point on one of the curves; then OA = x, AP=y, and, since O is the center of
mass of 1- and , O 1 , and O1 .
It follows that:
y 2 r12 x r12 x 2 2 x 2
2
y 2 r22 x 1 r22 x 2 21 x 1 .
On eliminating the first power of x from these equations and solving for x2 + y2, we find
that:
2
2
A Geocentric Solution to the 3-body Problem
19
x 2 y 2 1 r12 r22 1 .
As a consequence of this equation, equation (22) becomes:
r1
1 r12 2 r22
2
C 1 C
r2
(25)
If an arbitrary value of r2 is assumed, r1 can be computed from this equation: the points of
intersection of the circles around 1- and as centers, with the computed and assumed
values respectively of r1 and r2 as radii will be points on the curves. As a result, we may
let equation (25) be written in the form:
r13 ar1 b 0,
a
C
1 1
2 2
r2 ,
r2
(26)
b 2.
Since b=2 is positive, there is at least one real negative root of the first part of (26)
whatever value a may have. But the only value of r1, which has any meaning in this
problem, is real and positive; hence, the condition for real positive roots must be
considered.
2
It follows from (25) that C is always greater than r22 for all real, positive values
r2
of r1 and r2; therefore, a is always negative.
From the Theory of Equations we know that a cubic equation of this form (top line of
(26)) has three distinct real roots if
27b 2 4a 3 0 ; or, since b=2, if
a + 3 < 0.
Given this inequality, we can find the cubic roots as:
(27)
20
A Geocentric Solution to the 3-body Problem
sin
b 27
, ,
3
2 a
2
r11 2
a
sin ,
3
3
r12 2
a
sin 60 ,
3
3
r13 2
(28)
a
sin 60 .
3
3
where r11, r12, r13 are the three roots of the cubic.
The limit of the inequality (27), or, in terms of he original quantities, is,
r23 a r2 b 0
C 31
a
,
b 2.
(29)
The solution of this equation gives the extreme values of r2 for which (26) has real roots.
Therefore, in actual computation equation (29) should first be solved for r21, and r22. The
values of r2 to be substituted in (26) should be chosen at convenient intervals between
these roots.
Equation (29) will not have real, positive roots for all values of a´, the condition for real,
positive roots being:
a 3 0 ;
the limiting value of which is, in the original quantities,
C 31
3,
whence C´ = 3.
Therefore, C´ must be equal to, or greater than, 3 in order that the curves shall have real
points in the xy-plane. For C´=3, the curves are just vanishing from the xy-plane and it
follows that equation (25) is satisfied by r1 = 1, and r2 = 1; i.e., the surfaces vanish from
the xy-plane at the points which form equilateral triangles with 1- and .
From the overall form of the surfaces that the pairs of points which appear as C decreases
are all in the xy-plane. Therefore, it is sufficient here to consider the equation of the
curves in the xy-plane.
A Geocentric Solution to the 3-body Problem
21
There are three pairs of points on the x-axis which appear when the ovals around the
finite bodies touch each other and when they touch the exterior curve enclosing the both
of them. Two more appear as the surfaces vanish form the xy-plane, at the two points
making equilateral triangles with the finite bodies. These points are critical points of
their respective contours and they are connected with important dynamical properties of
the system.
Let the equation of a contour be written as:
F x, y x 2 y 2
21 2
C 0.
r1
r2
(30)
Differentiating we get the conditions for the twin pairs:
x x1 x x2 0
1 F
x 1
2 x
r13
r23
1 F
y
y
y 1 3 3 0
2 y
r1
r2
(31)
The left members of these equations are the same as the right members of equations (17)
for z=0. The terms to the left of the equal signs in (31) are proportional to the direction
cosines of the normal at all ordinary points of the curves; and, since dx/dt and dy/dt are
zero at the surfaces of zero velocity, it follows from (17) that the directions of
acceleration, i.e., the lines of effective force are orthogonal to he surfaces of zero
velocity.
Thus, if the infinitesimal body is placed on a surface of zero relative velocity it will start
moving in the direction of the normal. But at the twin points, the sense (direction and
amount) of the normal becomes ambiguous. Hence, it might be conclude that if the
infinitesimal body were placed at one of these points it would remain relatively at rest.
dx 2
dy 2
and
, which are the
dt 2
dt 2
accelerations in the x and y direction respectably, must vanish as per (17). The result of
the latter constraint on accelerations is that if the infinitesimal body, m, is placed at a twin
point with zero velocity, its coordinates will identically fulfill the differential equations of
motion and it will remain forever at rest with respect to sun and earth unless an external
force is brought to bear upon it.
The conditions imposed by (30) and (31) require that
Consider constraints (31); the second of which is fulfilled if we set y = 0. The twin
points on the x-axis and the linear solutions of the problem statement are given by the
conditions:
22
A Geocentric Solution to the 3-body Problem
x 1
x x1
x x
1
3
2 2
x x2
x x
2
3
2 2
0,
y 0,
(32)
z 0.
The first term of the first equation in (32), when taken as a function of x, is positive in the
limit as x goes to +. It is negative for x = x2 - where is a very small positive
quantity. It is positive for x = x2 - ; and it is negative for x = x1 + It is positive for
x=x1-; and it is negative for x=-. Therefore there are three positions along the line
through the finite bodies at which the infinitesimal body can remain when placed there.
We now have three cases:
Case 1:
Let the distance from to the double point on the x-axis between + and x2 be
represented by . Then:
x-x2=
x-x1=r1=1+
x=1-
Therefore, after clearing fractions, the first equation of (32) then becomes:
5 3 4 3 2 3 2 2 0
(33)
This 5th order equation has one change of signs in its coefficients, and thus has only
one real positive root. The value of this root depends upon , the mass of the earth.
Consider the left member of the equation as a function of and For =0, the
equation becomes:
3 2 3 3 0.
This has 3 roots: one of them is zero, and the other two are imaginary.
It follows that for 0 but sufficiently small, 3 roots of equation (33) can be
1
3
expressed as a power series in . One root will be real, the others complex.
Therefore the real root has the form:
1
2
3
a1 3 a 2 3 a3 3 ...
Substituting this into equation (33) and setting the result to zero, the coefficients (a1,
a2, a3, …) of corresponding powers of 1/3, we find that:
A Geocentric Solution to the 3-body Problem
a1
And so:
23
2
3
1
3
1
3
3
, a3 , …
, a2
3
9
27
1
2
3
3 1 3 1 3
r2 ...
3 3
9 3
3
r1 1 .
(34)
The corresponding value of C´ is found by substituting these values of r1 and r2 in
equation (25).
Case 2:
Let the distance from to the twin point on the x-axis between x2 and x1 be
represented by . Then x - x2 = -, x - x1 = r1 = 1 - , x = (1 - ) -. Therefore, the
first of equation (32) becomes:
5 3 4 3 2 3 2 2 0
On solving as in Case 1, the values of r2 and r1 are
1
2
3
3 1 3 1 3
r2 ...
3 3
9 3
3
r1 1 .
(35)
C´ is found by substituting r1 and r2 into (25).
Case 3:
Let the distance form 1- to the double point on the x-axis between x1 and - be
represented by 1-. Then:
x x 2 2
x x1 1
x 1
and the first equation of (32) becomes
5 7 4 19 6 3 24 13 2 12 14 7 0
(36)
If =0, which is to say that both m and the lesser of the two bodies’ masses is
negligible, we obtain
5 7 4 19 3 24 2 12 0
24
A Geocentric Solution to the 3-body Problem
which has only one root, =0. Therefore, can be expressed as a power series in
which converges for sufficiently small values of this parameter and vanishes with it
when = 0. This root has the form:
c1 c 2 2 c3 3 c 4 4 ...
Substituting the expression for into (36) and equating to zero the coefficients of the
various powers of , we find that:
c1
7
23 7 2
,…
, c 2 0 , c3
12
12 4
Hence
7
23 7 2 3
...
12
12 4
r1 1 ,
(37)
r2 1 r1 2
It’s C´ can be found by substituting into equation (25).
To find the twin points not on the x-axis, we again turn to equation (31). They, or any
two independent functions of them, define the twin points. Since y is distinct from zero,
we may divide it into the second equation, which yields:
1
1
3 0.
r13
r2
Multiplying this equation by x - x2 and x - x1, and subtracting the products separately
from the first equation of (31) gives:
x 2 1
x1
x2 x1
r13
x1 x2
r23
0
0
z 0
But, since x 2 1 , x1 , and x 2 x1 1 ; we conclude:
A Geocentric Solution to the 3-body Problem
25
1
1
0
r13
1
1
0
r23
z0
The only real solutions are r1 r2 1 so these points form equilateral triangles with the
two finite bodies, whatever their masses may be. Since z = 0, these points—called
Trojans—are located where the surfaces vanish from the xy-plane.
Figure 9 plots the solutions on the xy-plane.
Figure 9: The Five Lagrange Points Plotted on the xy-plane
It is clear from the presence of the asymptotic cylinders that many other critical points
have to exist. Such a potential candidate would look like the C6 contours on the y-axis of
Figure 7, which coincide with the L4 and L5 points.
In any case, we have taken the long way around to demonstrate that a geocentric
derivation follows from the dynamic explanation of the geocentric system. Indeed, the
so-called “fictitious forces” are brought into play because a geocentric system is
considered “fictional.” Nevertheless, in a geocentric coordinate system they come into
play because they are real, gravitational forces. Dynamic derivations lament the need for
invoking the fictitious forces in order to represent their Lagrange points derivations.
26
A Geocentric Solution to the 3-body Problem
THE LAGRANGE POINTS: A MODERN APPROACH3
The derivation we have applied in the previous section is a rigorous one derived from the
definition of force and geometry. In this section, we use another approach which is based
on Newtonian gravitational force. We again start with the same initial conditions as we
did with Figure 4, here presented as Figure 10.
Figure 10: The restricted 3-body problem
There are five equilibrium points to be found in the vicinity of two orbiting masses. They
are called Lagrange Points in honor of the French-Italian mathematician Joseph
Lagrange, who discovered them while studying the restricted three-body problem. The
term “restricted” refers to the condition that two of the masses are very much heavier
than the third. Today we know that the full three-body problem is chaotic, and so cannot
be solved in closed form. Therefore, Lagrange had good reason to make some
approximations. Moreover, there are many examples in our solar system that can be
accurately described by the restricted three-body problem.
The procedure for finding the Lagrange points is straightforward: We seek solutions to
the equations of motion which maintain a constant separation between the three bodies.
If M1 and M2 are two masses, and r1 and r2 are their respective position, then the total
force exerted on a third mass m, at a position r , will be
GM 1 m GM 2 m
F
(r r1 ) 3 r r2
3
r r1
r r2
3
Source: Neil J. Cornish, & Jeremy Goodman,
http://www.physics.montana.edu/faculty/comish/lagrange.pdf
(38)
A Geocentric Solution to the 3-body Problem
27
The catch is that both r1 and r2 are functions of time since M1 and M2 are orbiting each
other. Undaunted, one may proceed and insert the orbital solution for r1 (t) and r2 (t)
(obtained by solving the two-body problem for M1 and M2) and look for solutions to the
equation of motion
d 2 r t
,
F t m
dt 2
(39)
that keep the relative positions of the three bodies fixed. It is these stationary4 solutions
that are known as Lagrange points.
The easiest way to find the stationary solutions is to adopt a co-rotating frame of
reference in which the two large masses hold fixed positions. The new frame of
reference has its origin at the center of mass, and an angular frequency given by
Kepler’s law:
2 R 3 GM 1 M 2
(40)
Here R is the distance between the two masses [earth and sun —GB]. The only drawback
of using a non-inertial frame of reference is that we have to append various pseudoforces to the equation of motion.5 [Emphasis & footnote added —GB.] The effective
force in a frame rotating with angular velocity is related to the inertial force
F according to the transformation
dr
F F 2m m r .
dt
(41)
The first “correction” is the Coriolis force and the second is the centrifugal force. The
effective force can be derived from the generalized potential
1
U U v r r r ,
2
(42)
d
F rU vU .
dt
(43)
The velocity dependent terms in the effective potential do not influence the positions of
the equilibrium points, but they are crucial in determining the dynamical stability of
4
I.e., stationary relative to both the earth and the sun, this is clearly geostatic, as per the theory of
Geocentricity.
5
This refers to the so-called fictitious forces which in a geocentric framework are real gravitational forces.
In short, this statement says that in a geostatic framework these forces cannot be dismissed as fictitious
since they are necessary to obtain the correct equations.
28
A Geocentric Solution to the 3-body Problem
motion about the equilibrium points. A plot of U with v =0, M1 = 10, M2=1 and R = 10
is shown in Figure 11. The extrema of the generalized potential are labeled L1 through
L5.
Figure 11: A contour plot of the generalized potential
Choosing a set of Cartesian coordinates originating from the center of the masses with the
z-axis aligned with the angular velocity, we have
k
r xt iˆ y (t ) ˆj
r1 Riˆ
r2 Riˆ
(44)
where
M2
M1
,
.
M1 M 2
M1 M 2
(45)
A Geocentric Solution to the 3-body Problem
29
To find the static equilibrium points we set the velocity v dr / dt to zero and seek
solutions to the equation F 0 , where
x R R 3
x R R 3 ˆ
F 2 x
i
2
2
2 3/ 2
2 3/ 2
x R y
x R y
yR 3
2
y
x R 2 y 2
3/ 2
yR
x R
2
3
y2
3/ 2
(46)
ˆj
Here the mass m has been set equal to unity without loss of generality. The brute-force
approach for finding the equilibrium points would be to set the magnitude of each force
component to zero, and solve the resulting set of coupled, fourteenth order equations for x
and y. A more promising approach is to think about the problem physically, and use the
symmetries of the system to guide us to the answer [which is what we did in the first
section —GB].
Since the system is reflection-symmetric about the x-axis, the y component of the force
must vanish along this line. Setting y = 0 and writing x=R(u+ ) (so that u measures the
distance from M2 in units of R), the condition for the force to vanish along the x-axis
reduces to finding solutions to the three fifth-order equations.
u 2 1 s1 3u 3u 2 u 3 s0 2 s0 u 1 s0 s1 u 2 2u 3 u 4 ,
(47)
where s0 is the sign(u) and s1 is the sign(u+1). The three cases we need to solve have
(s0,s1) equal to (-1,1), (1,1), and (-1,-1). The case (1,-1) cannot occur. In each case there
is one real root to the quintic equation, giving us the positions of the first three Lagrange
points. We are unable to find closed-form solutions to equation (47) for general values of
<<1. To lowest order in , we find the first three Lagrange points to be positioned at
1
3
L1 : R 1 ,0 ,
3
1
3
L2 : R 1 ,0 ,
3
(48)
5
L3 : R 1 ,0 .
12
For the earth-sun system 3 10-6, R 1.5 108 km, and the first and second Lagrange
points are located approximately 1.5 million kilometers from earth. The third Lagrange
point orbits the sun just a bit further out than one astronomical unit [on the far side of the
sun from the earth].
30
A Geocentric Solution to the 3-body Problem
Identifying the remaining two Lagrange points requires a little more thought. We need to
balance the centrifugal force, which acts in a direction radially outward from the center of
mass, with the gravitational force exerted by the two masses. Clearly, force balance in
the direction perpendicular to centrifugal force will only involve gravitational forces.
This suggests that we should resolve the force into directions parallel and perpendicular
to r . The appropriate projection vectors are xiˆ yˆj and yiˆ xˆj . The perpendicular
projection yields
1
F y 2 R 3
x R 2 y 2
2/3
1
x R
2
.
2 2/ 3
y
(49)
Setting F 0 and y 0 tells us that the equilibrium points must be equidistant from
the two masses. Using this fact, the parallel projection simplifies to read
F|| 2
x2 y2
R
1
1
3
R
x R 2 y 2
3/ 2
(50)
Demanding that the parallel component of the force vanish leads to the condition that the
equilibrium points are at a distance R from each mass. In other words, L4 is situated at
the vertex of an equilateral triangle, with the two masses forming the other vertices. L5 is
obtained by mirror reflection of L4 about the x-axis. Explicitly, the fourth and fifth
Lagrange points have coordinates
R M M2 3
,
L4 : 1
R ,
2
M
M
2
2
1
.
R M1 M 2
3
,
L5 :
R
2 M1 M 2 2
(51)
Interestingly, the last two Lagrange points are stable because any deflecting of a particle
invokes a Coriolis force that brings it back to the point.
Since the only reason why the center of mass of the earth-sun balance was used only to
simplify the derivations, the earth could just as well have been taken as the center:
therefore, we expect no difference will be discovered for these points by any
geocentrically-based coordinate system. The rule that it is “six-to-one, half-dozen of the
other” when it comes to mathematical or physical proofs or disproof of heliocentric and | 17,177 | 48,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-17 | latest | en | 0.898797 |
https://collegephysicsanswers.com/openstax-solutions/current-river-exerts-force-90-n-balloon-floating-river-wind-exerts-force-130-n | 1,638,198,011,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358774.44/warc/CC-MAIN-20211129134323-20211129164323-00301.warc.gz | 253,971,993 | 13,347 | Change the chapter
Question
The current in a river exerts a force of 9.0 N on a balloon floating in the river. A wind exerts a force of 13.0 N on the balloon in the opposite direction. Draw a free-body diagram to show the forces acting on the balloon. Use your free-body diagram to predict the effect on the balloon.
The balloon will accelerate upstream.
Solution Video
OpenStax College Physics for AP® Courses Solution, Chapter 4, Problem 9 (Test Prep for AP® Courses) (0:36)
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Video Transcript
This is College Physics Answers with Shaun Dychko. This balloon is floating on the water so that means there is a buoyant force upwards of equal magnitude to the gravity downwards. There is a current moving to the right and it's applying a force due to the current of nine newtons and we're told that the wind or the air is blowing in the opposite direction to the current with a force of 13 newtons. Now, since the air is exerting a force greater than the current, there's going to be a net force in the direction of the air which is upstream and so this balloon will accelerate upstream. | 263 | 1,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-49 | latest | en | 0.931861 |
https://www.haroldserrano.com/blog/showcasing-beta-v006-of-the-game-engine | 1,550,486,582,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484928.52/warc/CC-MAIN-20190218094308-20190218120308-00498.warc.gz | 864,288,617 | 27,156 | # Game Engine Beta v0.0.6
Hi there,
These past two weeks I've been working on implementing a Frustum Culling algorithm for the engine. Here is a video of beta v0.0.6 of the engine with Frustum Culling.
In the previous beta version of the game engine, I had 20 soccer players on the screen. And I noticed that the game was a bit choppy. I recall the Frames-Per-Second (fps) being below 30. The engine was rendering all 20 players, the two goals, and soccer field at every frame without analyzing if the camera was able to see these models or not. You can agree that this is a problem.
The logic behind a Frustum Culling algorithm is essentially this:
"If the camera does not see the 3D model, then the engine should not render it".
Implementing this logic, allows you to have 100 models in a game, but only 10 or so many models being rendered at any time. Thus, improving the game experience.
For example, the minimum acceptable fps in a game is 30 fps. With Frustum Culling, the engine was able to render the entities at 60 fps and the lowest it got was 49 fps.
If you are entirely new to Frustum Culling, keep reading. I'm going to give you a brief tutorial on how Frustum Culling works.
## What is a Frustum
A game engine requires a camera. During initialization of a game, the engine computes a Frustum. A frustum is a chopped pyramid. And 3D models within the frustum are visible by the camera and by YOU.
Notice how the frustum is composed of six planes.
## What is Frustum Culling
A Frustum Culling Algorithm tests which 3D models lie within the frustum. If the 3D model lies outside the frustum, then it is ignored by the rendering engine. If the 3D model lies within the frustum, then it is rendered.
## Frustum Culling Test
Testing if a 3D model lies within a frustum is quite simple. First, the 3D model is wrapped within a box, in this case, with an AABB (Axis-Aligned Boundary Box).
Then, the eight vertices of the AABB are tested against a plane. In a Frustum Culling algorithm, the AABB's vertices are tested against all six frustum's planes.
If all vertices lie on the negative side of the plane, then the model is considered being outside the frustum. Otherwise, it lies within the frustum.
And that is it. You repeat the process for all game entities in your game.
## Improving the Frustum Culling Algorithm
Frustum Culling improves the efficiency of a game engine, but it is not enough. If your game only has ten 3D models, the engine can quickly do a Frustum Culling on all these entities. However, if your game has 200 or 1000 objects, then Frustum Culling by itself may not improve the efficiency of the engine.
What you need is an algorithm that can analyze the spatial area seen by the camera. For example, if the camera does not see the upper-left quadrant, then the engine should not perform Frustum Culling on the entities that lie on this quadrant. There are many algorithms that you can use. In the engine, I implemented a Boundary Volume Hierarchy (BVH) tree.
The image below shows the idea. I recursively wrapped the game entities in an AABB, until there is only one model per AABB. Then, starting from the root node, the engine test if the camera sees the AABB node or not. If the camera does not see the AABB node, then all the children node, containing the 3D models, are not rendered.
If the camera does see the AABB node, then it keeps performing Frustum Culling on the next AABB child node, until it reaches the leaf node.
In the example shown above, the algorithm rejects the red AABB box and does not need to test if the red and gray car is seen by the camera.
The camera does see the blue AABB box. Thus it tests the children nodes of the blue AABB box. It does this on and on until it reaches the leaf nodes. In this case, it sees the blue and orange car, but not the yellow car.
And that is Frustum Culling. I hope you learned something. | 915 | 3,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-09 | latest | en | 0.949033 |
https://www.pinterest.co.uk/explore/symmetric-group/ | 1,503,057,097,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104634.14/warc/CC-MAIN-20170818102246-20170818122246-00644.warc.gz | 929,551,888 | 61,999 | Symmetric group - Wikipedia
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These are drawings of some simple groups: the “lightswitch group” ℤ₂, the “hi-lo-off lightbulb” ℤ₃, and the symmetric group ₃=ℤ₂×ℤ₃, which is how the letters {A,B,C} or the boyfriends {Pankaj, Nadir, Ajay} permute.
### Paul McCobb Symmetric Group Sofa for Widdicomb
Paul McCobb Symmetric Group Sofa for Widdicomb image 8
These are drawings of some simple groups: the “lightswitch group” ℤ₂, the “hi-lo-off lightbulb” ℤ₃, and the symmetric group ₃=ℤ₂×ℤ₃, which is how the letters {A,B,C} or the boyfriends {Pankaj, Nadir, Ajay} permute.
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### Properties of Numbers Card Game
This properties of numbers card game is modeled after the traditional card game, Old Maid. Students play Old Math Guy in groups. Students will demonstrate their ability to match names and symbols of properties of numbers. Includes: reflexive, symmetric, transitive, substitution, additive identity, additive inverse, multiplicative identity, multiplicative inverse, multiplicative property of zero, commutative properties, associative properties, and properties of equality. Created by Free to…
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Pinterest | 670 | 2,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-34 | longest | en | 0.787731 |
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This is more of a philosophical or historical question, and I can be totally wrong in what I am about to write next. It looks to me, that complex-analytic geometry has lost its relative positions ...
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### A continuous inverse function problem: search for an inconsistent theory approach to return a 'false' metric [on hold]
My problem is about metrizable space definition: I need to find a topology that can be described by a metric without this being, in fact, a 'true metric'. I search an alternative approach to avoid to ...
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### Number of compositions of a positive number n, with factors between 1 and a certain number m
I'm trying to find the number of compositions of a positive number n, with factors between 1 and a certain number m. That is, all the combinations of limited numbers that add up to n $f(n, m)$ For ...
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### In search of a new isometric twisting invariant $T= \tau_1.\tau_2$
A curved line in $\mathbb R^3$ has properties of curvature and torsion, and, on an $\mathbb R^2$ possesses surface scalar properties of normal curvature and geodesic torsion $(\kappa_n, \tau_g).$ ...
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### analyze existence limit of sin(sqrt(x)) in infintiy [on hold]
i'm studying to test and i found a problem from last years: analyze the existence of limit $$sin(\sqrt x)$$ in infinity. Got some ideas about argument tending to infinity, but its especially hard for ...
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### On certain 2 dimensional foliation of $Gl(2,\mathbb{R})$ deleted by scalar matrices
We consider the following 4 dimensional open manifold $$M=Gl(2,\mathbb{R})\setminus \{\lambda I_2 \mid \lambda \in \mathbb{R}\}$$ where $I_2$ is the identity matrix. We consider the $2$ dimensional ...
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### Is a matrix similar to its transpose over $\mathbb{Z}_p$?
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### My question is about an article concerning p and t [on hold]
you say concerning p and t: An example of an element of p is the family of sets (indexed by k∈ℕ) defined by {m to the power of k :m∈ℕ}. But the second condition is not met because the set {2 to the ...
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### Relate solutions to a polynomial system in complex numbers to solutions in a finite field
Suppose I have a system of polynomials which are homogeneous but of distinct degrees that I want to solve simultaneously: $$F_1(z_1,\ldots,z_n)=\cdots=F_m(z_1,\ldots,z_n)=0.$$ Let $X(\mathbb F)$ ...
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### Fargues's Theorem for $Spa(C,C^+)$ (rather than $Spa(C,O_C)$
$\DeclareMathOperator\Spa{Spa}$Fargues's Theorem for $\Spa(C,O_C)$ states that the category of (mixed characteristic) shtukas with one paw at $x_C$ is equivalent to the category of Breuil-Kisin-...
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### Growth rate of bounded Lipschitz functions on compact finite-dimensional space
Let $\mathcal X$ be a metric space of diameter $D$ and "dimension" (e.g doubling dimension) $d$. Let $L \in [0, \infty]$ and $M \in [0, \infty)$ and consider the class $\mathcal H_{M,L}$ of $L$-...
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### Products of Cyclotomic Polynomials with Nonnegative Coefficients
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### Is there a short proof of the decidability of Kepler's Conjecture?
I've believed that the answer is "yes" for years, as suggested in various sources with reference to Tóth's work. For example, the Wikipedia article for Kepler Conjecture says: The next step toward ...
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### Potential for a Monotone Operator
[Cross-posted from math.stackexchange] I have a question about understanding the proof of Theorem 4.11 in the paper A Potential Theory for Monotone Multivalued Operators (accessible here). The ...
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### Reference for Shalika germs of GL(n)
I was reading the two Repka papers where he computes the leading and subleading Shalika germs for $GL_n$ and I was wondering, where are we since then? Have these germs (and the integrals) been ...
I am interested in the question above. I know that the answer is NO if the base field is for instance $\mathbb{Q}$ (the intersection of $\mathbb{Q}(x^2)$ and $\mathbb{Q}((x-1)^2)$ is $\mathbb{Q}$ ... | 1,180 | 4,529 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-51 | latest | en | 0.854894 |
https://www.mtholyoke.edu/academics/find-your-program/statistics | 1,721,457,098,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515020.57/warc/CC-MAIN-20240720052626-20240720082626-00803.warc.gz | 786,369,056 | 17,703 | # Statistics
Statistics is a branch of mathematics that deals with the gathering, organizing, analyzing and drawing conclusions from data.
## Program Overview
Statistics is used in most of the professions, in most sectors of the economy, and in a great many academic areas.
Courses in the Department of Mathematics and Statistics are designed to:
• teach the language of the mathematical sciences
• provide a command of powerful mathematical tools
• develop problem-solving skills
• foster the ability to ask questions and make independent discoveries
• encourage the experience of mathematics as a distinctively rigorous way of knowing.
In addition, especially through the statistics offering, the rich interplay between an applied context and mathematical models in working with numerical data is an area of particular emphasis.
Each year some of our students enter graduate programs in statistics, or in allied fields (business, economics, education, or psychology). Many enter the job market and later pursue advanced degrees.
Community Voices
## Spotlight on Math and Statistics students and alums
she/her
Kylee Miller ’24
she/her
Lupe Antonio Lopez ’23
she/her
## Learning Goals
We welcome all students into the Statistics major, and we aim to create an inclusive, supportive environment for everyone. Statistics majors have the following learning goals:
• Communicate statistical concepts and their interpretation, with statistical colleagues and with a wider audience
• Describe key statistical methods and the reasoning behind them
• Use appropriate statistical and computational methods to explore data and draw conclusions
• Ask questions about new methods and applications, learn new techniques, and make new discoveries
• Incorporate “big picture” reasoning, including ethics, practicality, and creativity, into statistical practice
• Develop the independence to approach new problems, and the ability to collaborate effectively.
## Requirements for the Major
A minimum of 36 credits:
As a prerequisite for MATH-203:
MATH-101
Calculus I
MATH-102
Calculus II
MATH-203Calculus III4
MATH-211Linear Algebra4
As a prerequisite for STAT-242:
STAT-140
Introduction to the Ideas and Applications of Statistics
STAT-242Intermediate Statistics4
STAT-340Applied Regression Methods4
MATH-342Probability4
STAT-343Mathematical Statistics4
12 additional credits in mathematics or statistics at the 200-level or above 112
Total Credits36
1
A 300-level course that contains substantial mathematical or statistical content in another discipline may be used to fulfill at most 4 credits toward the major with prior departmental approval.
Students considering developing a special major in statistics and economics should consult the Special Major chapter.
## Requirements for the Minor
A minimum of 16 credits:
At least one 200-level course in statistics4
At least one 300-level course in statistics4
Two additional courses in mathematics or statistics at the 200 level or above8
Total Credits16
Substitutions are possible with the permission of the department. Students planning a minor in statistics should consult one of the statistics advisors.
With departmental permission, students who have already completed one 100-level exploration course may elect to enroll in a second exploration course at the 200-level so that it may be counted toward the minor.
### Beginning the Study of Statistics
A natural way to begin if you have not studied statistics is with STAT-140, Introduction to the Ideas and Applications of Statistics.
A 200-level course in statistics is a good choice if you have taken an advanced placement statistics course or have taken the equivalent of a 100-level statistics course.
### Advice to Students with Special Interests
#### Actuarial science
Students interested in this area should plan to cover the material that is included in the first two actuarial exams as part of their undergraduate program. This material is included in Calculus I (MATH-101), Calculus II (MATH-102), Calculus III (MATH-203), Probability (MATH-342), and Mathematical Statistics (STAT-343), along with Macroeconomic Theory (ECON-211), Microeconomic Theory (ECON-212), and Economics of Corporate Finance (ECON-215). Students are also encouraged to obtain experience through an internship.
#### Biostatistics, public health, or natural resources
Students interested in these areas should include substantial work in biology, chemistry, geology, and/or environmental studies in their programs.
Many students with these interests design a special major in mathematics and economics or a special major in statistics and economics.
#### Engineering
Students interested in engineering often double-major in mathematics and physics and/or participate in one of the College’s five-year, dual-degree programs with Dartmouth’s Thayer School of Engineering, or California Institute of Technology, or the University of Massachusetts (see the Other Degree and Certificate Programs chapter).
Students preparing for graduate school in statistics or mathematics often participate in an undergraduate research program in the summer after the junior year and continue with an honors thesis in the senior year. Students considering graduate work in statistics at the level of a Ph.D. are encouraged to include abstract algebra and especially MATH-301.
#### Teaching certification
Students interested in pursuing certification for middle school or secondary school should major in mathematics rather than statistics. However, there is increasing emphasis on statistics in secondary school, and any of the applied courses would provide good preparation.
## Course Offerings
STAT-140 Introduction to the Ideas and Applications of Statistics
Fall and Spring. Credits: 4
This course provides an overview of statistical methods, their conceptual underpinnings, and their use in various settings taken from current news, as well as from the physical, biological, and social sciences. Topics will include exploring distributions and relationships, planning for data production, sampling distributions, basic ideas of inference (confidence intervals and hypothesis tests), inference for distributions, and inference for relationships, including chi-square methods for two-way tables and regression.
Applies to requirement(s): Math Sciences
T. Day, M. Ozanne
Advisory: 2 years of high school algebra
STAT-242 Intermediate Statistics
Fall and Spring. Credits: 4
In this course, students will learn how to analyze data arising from a broad array of observational and experimental studies. Topics covered will include exploratory graphics, description techniques, the fitting and assessment of statistical models, hypothesis testing, and communication of results. Specific topics may include multiple regression, ANOVA, and non-linear regression. Statistical software will be used.
Applies to requirement(s): Math Sciences
I. Beaudry, C. Hosman
Prereq: STAT-140 or equivalent.
STAT-244 Intermediate Topics in Statistics
STAT-244MP Intermediate Topics in Statistics: 'Survey Sampling'
Fall. Credits: 4
In this course, students will explore statistical techniques for designing and analyzing complex survey designs. Sample surveys are used to obtain data on demography, health, and development; to measure attitudes and beliefs; to estimate natural resources; to evaluate the impact of social programs; along with many other uses. The proper design and analysis of these surveys is crucial to their utility. We will cover topics including survey design, estimation, poststratification, imputation, and survey error. We will also apply these methods by frequently using real (and often messy) survey data through assignments and projects. The analysis of the data will be performed through R programming. Background should include estimation, confidence intervals and hypothesis testing.
Applies to requirement(s): Math Sciences
I. Beaudry
Prereq: STAT-140.
Notes: No prior knowledge of the statistical software R is required.
STAT-244NF Intermediate Topics in Statistics: 'Infectious Disease Modeling'
Spring. Credits: 4
Infectious disease has plagued humanity since time immemorial. Statistical models serve a critical role in improving understanding of the progression and proliferation of infection in a population, as well as the impact of interventions in stopping the spread of disease. In this course, we will explore regression and compartmental model-based approaches, which will be motivated by some of the most impactful epidemics and pandemics in recent history, including HIV/AIDS, Ebola, Zika, and COVID-19. R statistical software will be used.
Applies to requirement(s): Math Sciences
M. Ozanne
Prereq: STAT-140.
STAT-244NP Intermediate Topics in Statistics: 'Nonparametric Statistics'
Not Scheduled for This Year. Credits: 4
The methods taught in traditional statistics courses are based on assumptions that are often not satisfied by real data sets. In this course we will learn about approaches that require fewer assumptions, known as nonparametric methods. After taking this course, students will be able to examine assumptions for different approaches to statistical inference, compare nonparametric statistical tests such as sign and Wilcoxon tests to their parametric equivalents, and implement non-parametric approaches using R. In addition, the course will incorporate computational techniques for statistical analysis, including simulation, permutation tests, and bootstrapping.
Applies to requirement(s): Math Sciences
The department
Prereq: STAT-140.
Advisory: Students should have experience with the programming language R.
STAT-295 Independent Study
Fall and Spring. Credits: 1 - 4
The department
Instructor permission required.
Advisory: The permission of the department is required for independent work to count toward the major or minor.
STAT-331 Design of Experiments
Not Scheduled for This Year. Credits: 4
How do you get informative research results? By doing the right experiment in the first place. We'll look at the techniques used to plan experiments that are both efficient and statistically sound, the analysis of the resulting data, and the conclusions we can draw from that analysis. Using a framework of optimal design, we'll examine the theory both of classical designs and of alternatives when those designs aren't appropriate. On the applied side, we'll use R to explore real-world experimental data from science, industry, and everyday life; and we'll discuss key principles for working with expert (and not-so-expert) collaborators to help them set up the experiments they need.
Applies to requirement(s): Math Sciences
The department
Prereq: STAT-242.
STAT-340 Applied Regression Methods
Fall and Spring. Credits: 4
This course includes methods for choosing, fitting, evaluating, and comparing statistical models; introduces statistical inference; and analyzes data sets taken from research projects in the natural, physical, and social sciences.
Applies to requirement(s): Math Sciences
I. Beaudry, L. Tupper
Prereq: MATH-211 and STAT-242.
STAT-343 Mathematical Statistics
Spring. Credits: 4
This course is an introduction to the mathematical theory of statistics and to the application of that theory to the real world. Topics include probability, random variables, special distributions, introduction to estimation of parameters, and hypothesis testing.
Applies to requirement(s): Math Sciences
L. Tupper
Prereq: MATH-102 and MATH-342.
STAT-344 Seminar in Statistics and Scientific Research
STAT-344TM Seminar in Statistics and Scientific Research: 'Time Series Analysis'
Spring. Credits: 4
Time series -- data collected across time -- show up in a vast range of application areas, from climate to economics to music. But they can have special behaviors, like seasonality, memory, and directionality, that require some special treatment! We'll look at tools for describing, modeling, and predicting time series behavior. Topics include decomposition, ARIMA, exponential smoothing, dynamic regression, and a foray into the frequency domain. We'll apply these concepts using real datasets in R.
Applies to requirement(s): Math Sciences
L. Tupper
Prereq: STAT-242 and MATH-211.
Advisory: Previous or concurrent registration in STAT-340 (or other multiple regression experience) is recommended.
STAT-395 Independent Study
Fall and Spring. Credits: 1 - 8
The department
Instructor permission required.
Advisory: The permission of the department is required for independent work to count toward the major or minor.
The Department of Mathematics and Statistics strives to teach the language of the mathematical sciences, provide a command of mathematical tools, develop problem-solving skills, foster an ability to ask questions and make independent discoveries. | 2,521 | 12,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-30 | latest | en | 0.916059 |
https://www.physicsforums.com/threads/gravitational-acceleration-from-moon.564204/ | 1,544,997,807,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827998.66/warc/CC-MAIN-20181216213120-20181216235120-00221.warc.gz | 1,011,009,621 | 12,874 | Gravitational acceleration from moon
1. Jan 1, 2012
Irishwolf
Hi ,
How could I calculate the acceleration of an object on the earths surface , say at the equator , from the influence of the moon , when the moon is directly overhead?
I have calculated the force of attraction from the moon to be 1.99 x 10^(20) N.
Now to find the gravitaional acceleration do I use ==> a=F/M
Where big M = mass earth?
I do I use the mass (m) of the moon?
Or is this not the way to calculate it?
-------------------------------------------------------
Another attempt of solving this was i used the formula for influence of moon on tides:
a= GMm/(R-r)^(2) - GMm/(R^(2))
and grav acc due to moon is 0.000001131 m/s^(2)
But someone said I should solve it with the above equation?
2. Jan 1, 2012
Neandethal00
I'm not sure I understand your question. If your question is what is the net acceleration of an object on earth considering the gravitational effect of the moon, you may look at it with this formula:
a = GMearth/R2 - GMmoon/D2
Mearth = mass of earth
Mmoon = mass of moon
D = distance from center of moon to the object.
3. Jan 1, 2012
Irishwolf
Hi yes that was my question thanks,
But why is that a different formula than the one used to calculate the moons influence on tides?
4. Jan 1, 2012
Neandethal00
Not sure about tide formula. What does that do?
Note, mass of the object is irrelevant. For tide, is there a
well defined mass? You can treat all ocean water as one massive object,
but then acceleration of what water? Our simple formula then may not be
very accurate. | 409 | 1,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-51 | latest | en | 0.901994 |
https://www.teacherspayteachers.com/Browse/Search:math%20station%20games | 1,537,659,282,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158766.53/warc/CC-MAIN-20180922221246-20180923001646-00399.warc.gz | 907,042,795 | 54,381 | showing 1-24 of 13,853 results
This bundle includes the PDF version of Math Stations for Middle Grades (upper elementary), EDITABLE templates in both PowerPoint and Publisher format, and all of my Just Add Task Cards games. These games are easy to customize by adding your own task cards, so you can use them over and over again th
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\$21.95
189 Ratings
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Bundle
Your students will love adding money with Ice Cream Money Bump! Spin the money bump spinner twice. Add the money together. Place a counter on that value. If another player’s counter is on that number, BUMP it off. If your counter is on that number, stack the counters together and it freezes that
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\$2.00
66 Ratings
4.0
PDF (973.97 KB)
UPDATED 9/21/13--Added more addition and subtraction cards--also added more even/odd cookies. Second Grade Math Station Game Bundle (also sold individually) If you are looking for some cute games to keep your students motivated during math stations/centers, these cute matching and sorting cards ar
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\$4.00
55 Ratings
4.0
PDF (8.34 MB)
This place value game pack includes differentiated games to meet the needs of each student in your classroom. The games are aligned with the Common Core Standards for grades 1-2. Once you have leveled your students into math groups, they can play their leveled game in their math station, small gui
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\$5.95
73 Ratings
4.0
PDF (2.86 MB)
Your students will love this money math station! Students put all the coins in a pile. Take turns rolling the number cube and placing that number of pennies on the penny shapes on your game board. When you get 5 pennies, put them back in the pile and trade them for a nickel. Put your nickel in t
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\$1.50
31 Ratings
4.0
PDF (6.98 MB)
I created this Triangler Puzzle game for my common core math stations.The focus is on equivalent expressions created with the commutative, associative, distributive, and identity properties, as well as combining very simple like terms (no exponents). It is great for stations as it can be easily di
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\$2.00
51 Ratings
4.0
PDF (1.79 MB)
Fraction Action: Comparing Fractions The students will gain hands on experience comparing fractions. The students race around the game board with multiple opportunities to compare fractions and collaborate with their opponent to check each others work. I have also included a printable Follow-Up pr
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\$2.00
30 Ratings
4.0
PDF (221.09 KB)
This item includes 6 measurement activities/games that are perfect for math stations! Directions are included for each station. There are perfectly aligned with the 3rd and 4th grade measurement standards: 3.MD.A.1 3.MD.A.2 4.MD.A.1 4.MD.A.2 4.MD.A.3 These activities can also be used for 5th or 6t
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\$7.75
11 Ratings
4.0
PDF (18.41 MB)
This is a new station for practicing computation.This set focuses on dividing fractions and includes four different Roundabouts, each with eight problems. It is self checking in that the pieces fit together, linking problems with correct answers to form a circle. Print a copy and leave it together t
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\$2.00
35 Ratings
4.0
PDF (5.87 MB)
Fraction Action- Equivalent Fraction Game This fun game kit gives students practice finding equivalent fractions while incorporating a little friendly competition for motivation. Players race around the gameboard solving equivalent fraction problems and collaborate to check each others answers. A
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\$2.00
26 Ratings
4.0
PDF (232.46 KB)
This place value game pack includes differentiated games to meet the needs of each student in your classroom. The games are aligned with the Common Core Standards for grades 1-2. Once you have leveled your students into math groups, they can play their leveled game in their math station, small guide
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\$5.95
30 Ratings
4.0
PDF (3.24 MB)
Your class will love to practice adding and subtracting with these two fall themed bump games! For the addition bump game, students roll 3 dice and add the numbers together. For the subtraction game, they add the three numbers together and then subtract that number from 20. The student then places a
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\$2.00
28 Ratings
3.9
PDF (2.42 MB)
Your class will love to practice adding and subtracting with these two winter themed bump games! For the addition bump game, students roll 3 dice and add the numbers together. For the subtraction game, they add the three numbers together and then subtract that number from 20. The student then places
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\$2.00
26 Ratings
4.0
PDF (2.08 MB)
This activity was created for my middle school math stations on greatest common factor and least common multiple. It includes two activities, one for GCF and the other for LCM, as well as the four spinners you will need. I like to just use paper clips as the spinners, but you can use the arrows that
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\$3.00
23 Ratings
3.9
PDF (386.47 KB)
Reusable Primary Math Stations/Centers and Games Perfect for math differentiaton, workstations, & BUILD math. Games and activites that can be printed and used throughout the year for practicing Common Core skills. Games and activities are appropriate for primary level students. All printables
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\$8.00
24 Ratings
4.0
PDF (22.64 MB)
I created this game for my common core math stations. It can be played individually, in partners, or groups of three or four. These cards match word problems to inequalities. It works perfectly in stations as it is easily differentiated. As my students are rotating from one station to the next, I c
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\$2.00
23 Ratings
3.9
PDF (740.63 KB)
This is a new station for practicing computation.This set focuses on surface area and volume of rectangular prisms and includes four different Roundabouts, each with eight problems. It is self checking in that the pieces fit together, linking problems with correct answers to form a circle. Print a c
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\$2.00
22 Ratings
4.0
PDF (1.05 MB)
Addition Fluency, Make Ten, Addition Doubles, Add Ten, Add Three Numbers This game is a fun and engaging way for students to build fluency with addition facts. Choose from 8 boards featuring bunnies, bugs, umbrellas, baby chicks, birds, bees, flowers, and colorful worms. Color and B&W 8 Differ
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\$3.00
2 Ratings
4.0
PDF (3.99 MB)
This math station game will help students practice their division facts. This station only includes division facts without remainders. The following items are included in the file: Game Board Game Cards (facts 0-9) Print, Laminate, and store inside a Ziplock bag! I use colored counting blocks as
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\$1.99
18 Ratings
4.0
PDF (528.69 KB)
Are you looking for an engaging Hundreds Chart Math Game? Students use the number cards to make a hundreds chart (or chart to 120) on a table top or on the floor. Students practice finding one more, one less, ten more, ten less, five more, and five less of a given number on a hundreds or 120 ch
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Are you looking for a station to cover the Financial Literacy Standard for your students? Well look no more! This game is perfect for students to understand the concepts of needs, wants, saving, earning, and donating! Everything you need except a paper clip for the spinners are included! Please fe
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\$4.50
15 Ratings
4.0
PDF (2.89 MB)
This is a "Go Fish" Game focuses on equivalent ratios and unit rates. It includes 36 cards. This activity is designed in my classroom stations for groups of four. However, students could play in pairs or in larger groups. It works perfectly in stations as it is easily differentiated. As my students
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\$2.00
18 Ratings
3.9
PDF (1.14 MB)
This activity was created for middle school math stations on probability and statistics. As students rotate through stations, they can be challenged in individual, partner, and group activities and games. This is an individual practice station. Here, students work independently on a guided problem s
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CCSS:
\$2.00
18 Ratings
4.0
PDF (414.75 KB)
This is a "Go Fish" Game for modeling dividing fractions. Students match division problems to models. It includes 36 cards. This activity is designed in my classroom stations for groups of four. However, students could play in pairs or in larger groups. It works perfectly in stations as it is easily
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\$2.00
15 Ratings
4.0
PDF (1.43 MB)
showing 1-24 of 13,853 results
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,093 | 8,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-39 | latest | en | 0.908239 |
http://mathhelpforum.com/differential-geometry/186189-showing-sn-sm-homotopy-equivalent-sn-m-1-a.html | 1,527,132,104,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865884.49/warc/CC-MAIN-20180524014502-20180524034502-00176.warc.gz | 184,457,300 | 9,902 | Thread: Showing Sn\Sm is homotopy equivalent to Sn-m-1
1. Showing Sn\Sm is homotopy equivalent to Sn-m-1
Consider $\displaystyle S^m$ embedded in $\displaystyle S^n (m < n)$ as the subspace $\displaystyle \{(x_1,x_2,...,x_{m+1},0,....,0) | {\sum x_i^2 = 1}\}$. Show $\displaystyle S^n$ \ $\displaystyle S^m$ is homotopy equivalent to $\displaystyle S^{n-m-1}$.
If we consider m=1, n=2, then according to the theorem we're trying to prove, our quotient space should be homotopy equivalent to a point. However from our definition, $\displaystyle S^1$ embedded in $\displaystyle S^2$ is $\displaystyle \{(x_1,x_2,0) | {\sum x_i^2 = 1}\}$, i.e. the circle on the x-y plane. I thought the quotient space meant "shrinking all the points in the same equivalence class into a point", so our quotient space ends up more like $\displaystyle S^2 \wedge S^2$? So my first question is, is the question giving the right definition of embedding for us to prove the proposition?
In the case of m=1, n=2, if we use an embedding definition based on rotation about the z-axis (e.g. using spherical polars, points are in the same equivalence class if they have the same radius and elevation angle from x-y plane), we can see very clearly that $\displaystyle S^2$ \ $\displaystyle S^1$ is [-1,1]. Even if the embedding definition is correct in question, maybe there is an easier one to work with?
I probably have completely misunderstood the question, would love to know my mistake. Any comments are welcome.
2. Re: Showing Sn\Sm is homotopy equivalent to Sn-m-1
Wow, it took me so long to find out what is wrong with this - I thought I was taking a quotient space, but I actually need to remove the points of the embedding, and then can explicitly construct a homotopy. Well that was dumb of me.... | 484 | 1,785 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-22 | latest | en | 0.906403 |
http://booksbw.com/index.php?id1=4&category=mathematical&author=vantress-h&book=2001&page=15 | 1,566,781,070,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330913.72/warc/CC-MAIN-20190826000512-20190826022512-00433.warc.gz | 35,172,225 | 4,677 | Books in black and white
Books Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics
# Detection, Estimation modulation theory part 1 - Vantress H.
Vantress H. Detection, Estimation modulation theory part 1 - Wiley & sons , 2001. - 710 p.
ISBN 0-471-09517-6
Previous << 1 .. 9 10 11 12 13 14 < 15 > 16 17 18 19 20 21 .. 254 >> Next
Fig. 2.4 Decision regions.
Decision Criteria 25
We can now write the expression for the risk in terms of the transition probabilities and the decision regions:
^ = CqqPq f /7r|H0(R|#o)</R
JzQ
4- C10P0 f /?r|Ho(R|/f0) dR
Jz±
+ Ñã1б f pr^tRlHJdR
Jz1
+ ñ0ëà pr^mdR. (5)
Jz0
For an ^-dimensional observation space the integrals in (5) are N-fold integrals.
We shall assume throughout our work that the cost of a wrong decision is higher than the cost of a correct decision. In other words,
Ñþ > CQî,
r > r w
<-01 > '-n-
Now, to find the Bayes test we must choose the decision regions Z0 and
Zx in such a manner that the risk will be minimized. Because we require
that a decision be made, this means that we must assign each point R in the observation space Z to Z0 or Zx.
Thus
Z = Z0 + Z, A z0 u Zx. (7)
Rewriting (5), we have
31 = PoCqq f /?r|H0(R|^o) dR + PoCiq f /7r|Ho(R|^o) dR
J Zq J Z — Zq
+ PiCoi f ^ã³íÄ^²^³) dR + гÑö à pT\Hl(R\H1) dR. (8)
J Zq J Z — Zq
Observing that
à PrlHo(R\H0)dR = à pr^WHJdR = 1, (9)
Jz Jz
(8) reduces to
& = P0C10 + гÑö
+ f {[Ë(Ñ01 - Cu)pr|Jfl(R|^)]
Jz0
- [Po(C10 - C00)p,|Ho(R|H0)]} dR. (10)
26 2.2 Simple Binary Hypothesis Tests
The first two terms represent the fixed cost. The integral represents the cost controlled by those points R that we assign to Z0. The assumption in
(6) implies that the two terms inside the brackets are positive. Therefore all values of R where the second term is larger than the first should be included in Z0 because they contribute a negative amount to the integral. Similarly, all values of R where the first term is larger than the second should be excluded from Z0 (assigned to Zx) because they would contribute a positive amount to the integral. Values of R where the two terms are equal have no effect on the cost and may be assigned arbitrarily. We shall assume that these points are assigned to Hx and ignore them in our subsequent discussion. Thus the decision regions are defined by the statement: If
^i(Coi “ ѳ³)Ë,ß³äà|ß³) > Pq(Cw — ^oo)/7r|H0(R|^o), (11)
assign R to Zx and consequently say that H± is true. Otherwise assign R to Z0 and say H0 is true.
Alternately, we may write
ðã^Ø11Ðî(Ñ10 - Coo) n~
ðã,ÿ0(Ê|ßî)?Ë(Ñî³-Ñè)‘ {lZ)
The quantity on the left is called the likelihood ratio and denoted by A(R)
Ë(Þ Ä Pr|H,(R|#l)
~ Pt\h0(R\Ho)
(13)
Because it is the ratio of two functions of a random variable, it is a random variable. We see that regardless of the dimensionality of R, A(R) is a one-dimensional variable.
The quantity on the right of (12) is the threshold of the test and is denoted by rj:
Ä Wio ~ Cpo)
1 ~ Ë(Ñî³ - C„)'
Thus Bayes criterion leads us to a likelihood ratio test (LRT)
Hx
A(R) ^ v
Ho
(15)
We see that all the data processing is involved in computing A(R) and is not affected by a priori probabilities or cost assignments. This invariance of the data processing is of considerable practical importance. Frequently the costs and a priori probabilities are merely educated guesses. The result in (15) enables us to build the entire processor and leave 77 as a variable threshold to accommodate changes in our estimates of a priori probabilities and costs.
Likelihood Ratio Tests 27
Because the natural logarithm is a monotonic function, and both sides of (15) are positive, an equivalent test is
Two forms of a processor to implement a likelihood ratio test are shown in Fig. 2.5.
Before proceeding to other criteria, we consider three simple examples.
Example 1. We assume that under HY the source output is a constant voltage m. Under H0 the source output is zero. Before observation the voltage is corrupted by an additive noise. We sample the output waveform each second and obtain N samples. Each noise sample is a zero-mean Gaussian random variable n with variance a2. The noise samples at various instants are independent random variables and are independent of the source output. Looking at Fig. 2.6, we see that the observations under the two hypotheses are
Previous << 1 .. 9 10 11 12 13 14 < 15 > 16 17 18 19 20 21 .. 254 >> Next | 1,333 | 4,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-35 | latest | en | 0.763659 |
https://calculator.academy/ponderal-index-calculator/ | 1,696,028,611,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510529.8/warc/CC-MAIN-20230929222230-20230930012230-00785.warc.gz | 177,761,602 | 54,314 | Enter the mass and height of the individual into the calculator to determine the Ponderal Index; this calculator can also evaluate the mass or height given the Ponderal Index is known.
## Ponderal Index Formula
The following formula is used to calculate the Ponderal Index:
PI = m/{h^3}
Variables:
• PI is the Ponderal Index
• m is the mass of the individual
• h is the height of the individual
To calculate the Ponderal Index, divide the mass of the individual by the cube of their height.
## What is a Ponderal Index?
The Ponderal Index (PI) is a calculation used to assess a person’s leanness or corpulence. It is similar to the Body Mass Index (BMI), but it is considered to provide a more accurate representation of an individual’s leanness. The PI is calculated by dividing the individual’s mass by the cube of their height. The result is then compared to a reference range to determine whether the individual is underweight, normal weight, overweight, or obese. The PI is often used in medical and health-related research to study the relationship between body size, health, and disease risk. It is particularly useful in studies involving children and adolescents, as it takes into account the changes in body composition that occur with growth and development.
## How to Calculate Ponderal Index?
The following steps outline how to calculate the Ponderal Index.
1. First, determine the weight of the individual (kg).
2. Next, determine the height of the individual (m).
3. Next, calculate the body mass index (BMI) using the formula: BMI = weight (kg) / height^2 (m^2).
4. Finally, calculate the Ponderal Index using the formula: Ponderal Index = weight (kg) / height^3 (m^3).
5. After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem:
Use the following variables as an example problem to test your knowledge.
weight (kg) = 70
height (m) = 1.75 | 435 | 1,932 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-40 | latest | en | 0.903791 |
https://www.answersarena.com/ExpertAnswers/how-to-solve-5-a-sinusoidal-current-has-an-average-value-of-4-5-mathrm-ma-determine-the-f-pa736 | 1,718,405,037,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00145.warc.gz | 590,747,416 | 5,868 | # (Solved): how to solve 5. A sinusoidal current has an average value of $$4.5 \mathrm{~mA}$$. Determine the f ...
how to solve
5. A sinusoidal current has an average value of $$4.5 \mathrm{~mA}$$. Determine the following values: $\begin{array}{l} \mathrm{rms}= \\ \text { peak-to-peak }= \\ \text { peak }= \end{array}$
We have an Answer from Expert | 108 | 353 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-26 | latest | en | 0.779309 |
http://www.nzmaths.co.nz/resource/estimating-fractions | 1,467,210,843,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397748.48/warc/CC-MAIN-20160624154957-00086-ip-10-164-35-72.ec2.internal.warc.gz | 765,089,335 | 8,870 | Te Kete Ipurangi
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# Estimating with Fractions
Keywords:
Achievement Objectives:
Achievement Objective: NA4-4: Apply simple linear proportions, including ordering fractions.
AO elaboration and other teaching resources
Achievement Objective: NA4-2: Understand addition and subtraction of fractions, decimals, and integers.
AO elaboration and other teaching resources
Specific Learning Outcomes:
Solve problems that involve adding and subtracting fractions with related denominators.
Find equivalent fractions and order fractions.
Description of mathematics:
Number Framework Stage 7
Required Resource Materials:
Activity:
An important part of number sense is to have a concept of the size of fractions independent of doing calculations. For example, a student with a good sense of number will know immediately that 11/21 is more than 1/2 because 11 is over half the available 21 pieces.
#### Using Materials
Problem: Show 4/9 using the two different-coloured circle pieces and discuss whether it is more or less than 1/2.
Each student rotates the circles until one colour circle shows as near to 4/9 as they can. Discuss why it is 4/9.Look at the other colour and discuss what fraction it is.
(Answer: 5/9 because there are 9/9 in 1 whole.)
Examples: Show these fractions on the circles and work out the other fraction that adds up to 1 whole: 3/7 4/5 5/9 5/11 11/21...
#### Using Imaging
Problem: “I have made 15/31, but you cannot see it. Describe in words what this fraction looks like.”
(Possible answer: the fraction is just a little bit less than 1/2.)
“What is the other fraction?” (Answer: 16/31.)
Examples: Describe these fractions without using the circles and work out the other fraction that adds up to 1 whole: 5/11 2/5 4/7 8/17 6/13
#### Using Number Properties
Problem: “Maurice eats 2/5 of a cake, and Norris eats 3/7 of a same sized cake. In total, do they eat more or less than 1 cake?”
Discuss the answer. (Answer: Both are a little less than 1/2 so the total is less than 1 whole.)
Examples: Without calculating, determine whether these are more or less than 1:
2/5 + 12/23 17/33 + 6/11 51/100 + 6/12 24/48 + 29/60 24/47 + 61/120 ...
Examples: Is 1 – 21/43 more or less than 1/2? Is 11/45 more or less than 1/4? Is 3/4 – 11/40 more or less than 1/2? Is 3/4 + 15/62 more or less than 1?
Hard example: Is 3/7 + 14/27 more or less than 1? (Answer: Here the gap between 3/7 and 1/2 is more than the gap between 14/27 and 1/2. So 3/7 + 14/27 < 1.)
#### Understanding Number Properties:
Make up two fractions that both have denominators greater than 30 and that add up to just a little bit less than 1. | 739 | 2,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2016-26 | latest | en | 0.888337 |
https://samacheerkalviguru.com/samacheer-kalvi-10th-maths-chapter-2-ex-2-1/ | 1,618,543,572,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088471.40/warc/CC-MAIN-20210416012946-20210416042946-00447.warc.gz | 469,571,762 | 9,594 | ## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.1
Exercise 2.1 Class 10 Maths Samacheer Question 1.
Find all positive integers which when divided by 3 leaves remainder 2.
The positive integers when divided by 3 leaves remainder 2.
By Euclid’s division lemma a = bq + r, 0 ≤ r < b.
Here a = 3q + 2, where 0 ≤ q < 3, a leaves remainder 2 when divided by 3.
∴ 2, 5, 8, 11 ……………..
10th Maths Exercise 2.1 Samacheer Kalvi Question 2.
A man has 532 flower pots. He wants to arrange them in rows such that each row contains 21 flower pots. Find the number of completed rows and how many flower pots are left over?
Here a = 532, b = 21
Using Euclid’s division algorithm
a = bq + r
532 = 21 × 25 + 7
Number of completed rows = 21
Number of flower pots left over = 7
Ex 2.1 Class 10 Samacheer Question 3.
Prove that the product of two consecutive positive integers is divisible by 2.
Solution:
Let n – 1 and n be two consecutive positive integers. Then their product is (n – 1)n.
(n – 1)(n) = n2 – n.
We know that any positive integer is of the form 2q or 2q + 1 for some integer q. So, following cases arise.
Case I. When n = 2q.
In this case, we have
n2 – n = (2q)2 – 2q = 4q2 – 2q = 2q(2q – 1)
⇒ n2 – n = 2r, where r = q(2q – 1)
⇒ n2 – n is divisible by 2.
Case II. When n = 2q + 1
In this case, we have
n2 – n = (2q + 1)2 – (2q + 1)
= (2q + 1)(2q + 1 – 1) = 2q(2q + 1)
⇒ n2 – n = 2r, where r = q (2q + 1).
⇒ n2 – n is divisible by 2.
Hence, n2 – n is divisible by 2 for every positive integer n.
Hence it is Proved
10th Maths Exercise 2.1 Question 4.
When the positive integers a, b and c are divided by 13, the respective remainders are 9,7 and 10. Show that a + b + c is divisible by 13.
Let the positive integer be a, b, and c
We know that by Euclid’s division lemma
a = bq + r
a = 13q + 9 ….(1)
b = 13q + 7 ….(2)
c = 13q + 10 ….(3)
a + b + c = 13q + 9 + 13q + 7 + 13q + 10
= 39q + 26
a + b + c = 13 (3q + 2)
This expansion will be divisible by 13
∴ a + b + c is divisible by 13
10th Maths 2.1 Exercise Question 5.
Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4.
Solution:
Let x be any integer.
The square of x is x2.
Let x be an even integer.
x = 2q + 0
then x2 = 4q2 + 0
When x be an odd integer
When x = 2k + 1 for some interger k.
x2 = (2k + 1 )2
= 4k2 + 4k + 1
= 4k (k + 1) + 1
= 4q + 1
where q = k(k + 1) is some integer.
Hence it is proved.
10th Maths Exercise 2.1 In Tamil Question 6.
Use Euclid’s Division Algorithm to find the Highest Common Factor (H.C.F) of
(i) 340 and 412
To find the HCF of 340 and 412 using Euclid’s division algorithm. We get
412 = 340 × 1 + 72
The remainder 72 ≠ 0
Again applying Euclid’s division algorithm to the division of 340
340 = 72 × 4 + 52
The remainder 52 ≠ 0
Again applying Euclid’s division algorithm to the division 72 and remainder 52 we get
72 = 52 × 1 + 20
The remainder 20 ≠ 0
Again applying Euclid’s division algorithm
52 = 20 × 2 + 12
The remainder 12 ≠ 0
Again applying Euclid’s division algorithm
20 = 12 × 1 + 8
The remainder 8 ≠ 0
Again applying Euclid’s division algorithm
12 = 8 × 1 + 4
The remainder 4 ≠ 0
Again applying Euclid’s division algorithm
8 = 4 × 2 + 0
The remainder is zero
∴ HCF of 340 and 412 is 4
(ii) 867 and 255
To find the HCF of 867 and 255 using
Euclid’s division algorithm. We get
867 = 255 × 3 + 102
The remainder 102 ≠ 0
Using Euclid’s division algorithm
255 = 102 × 2 + 51
The remainder 51 ≠ 0
Again using Euclid’s division algorithm
102 = 51 × 2 + 0
The remainder is zero
∴ HCF = 51
∴ HCF of 867 and 255 is 51
(iii) 10224 and 9648
Find the HCF of 10224 and 9648 using Euclid’s division algorithm. We get
10224 = 9648 × 1 + 576
The remainder 576 ≠ 0
Again using Euclid’s division algorithm
9648 = 576 × 16 + 432
The remainder 432 ≠ 0
Using Euclid’s division algorithm
576 = 432 × 1 + 144
The remainder 144 ≠ 0
Again using Euclid’s division algorithm
432 = 144 × 3 + 0
The remainder is 0
∴ HCF = 144
The HCF of 10224 and 9648 is 144
(iv) 84,90 and 120
Find the HCF of 84, 90 and 120 using Euclid’s division algorithm
90 = 84 × 1 + 6
The remainder 6 ≠ 0
Using Euclid’s division algorithm
4 = 14 × 6 + 0
The remainder is 0
∴ HCF = 6
The HCF of 84 and 90 is 6
Find the HCF of 6 and 120
120 = 6 × 20 + 0
The remainder is 0
∴ HCF of 120 and 6 is 6
∴ HCF of 84, 90 and 120 is 6
Numbers And Sequences 10th Class Question 7.
Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.
Solution:
The required number is the H.C.F. of the numbers.
1230 – 12 = 1218,
1926 – 12 = 1914
First we find the H.C.F. of 1218 & 1914 by Euclid’s division algorithm.
1914 = 1218 × 1 + 696
The remainder 696 ≠ 0.
Again using Euclid’s algorithm
1218 = 696 × 1 + 522
The remainder 522 ≠ 0.
Again using Euclid’s algorithm.
696 = 522 × 1 + 174
The remainder 174 ≠ 0.
Again by Euclid’s algorithm
522 = 174 × 3 + 0
The remainder is zero.
∴ The H.C.F. of 1218 and 1914 is 174.
∴ The required number is 174.
10th Maths Ex 2.1 Question 8.
If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y.
Find the HCF of 32 and 60
60 = 32 × 1 + 28 ….(1)
The remainder 28 ≠ 0
By applying Euclid’s division lemma
32 = 28 × 1 + 4 ….(2)
The remainder 4 ≠ 0
Again by applying Euclid’s division lemma
28 = 4 × 7 + 0….(3)
The remainder is 0
HCF of 32 and 60 is 4
From (2) we get
32 = 28 × 1 + 4
4 = 32 – 28
= 32 – (60 – 32)
4 = 32 – 60 + 32
4 = 32 × 2 -60
4 = 32 x 2 + (-1) 60
When compare with d = 32x + 60 y
x = 2 and y = -1
The value of x = 2 and y = -1
10th Samacheer Maths Exercise 2.1 Question 9.
A positive integer when divided by 88 gives the remainder 61. What will be the remainder when the same number is divided by 11?
Solution:
Let a (+ve) integer be x.
x = 88 × y + 61
61 = 11 × 5 + 6 (∵ 88 is multiple of 11)
∴ 6 is the remainder. (When the number is divided by 88 giving the remainder 61 and when divided by 11 giving the remainder 6).
10th Maths 2.1 Question 10.
Prove that two consecutive positive integers are always coprime. | 2,239 | 6,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2021-17 | latest | en | 0.859589 |
https://help.scilab.org/docs/5.3.3/en_US/kron.html | 1,653,122,221,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539049.32/warc/CC-MAIN-20220521080921-20220521110921-00412.warc.gz | 347,674,472 | 6,601 | Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange
Scilab 5.3.3
Change language to: Français - Português - 日本語
See the recommended documentation of this function
# kron
Kronecker product (.*.)
kron(A,B)
A.*.B
### Description
kron(A,B) or A.*.B returns the Kronecker tensor product of two matrices A and B. The resulting matrix has the following block form:
If A is a m x n matrix and B a p x q matrix then A.*.B is a (m*p) x (n*q) matrix.
A and B can be sparse matrices.
### Examples
A=[1,2;3,4];
kron(A,A)
A.*.A
sparse(A).*.sparse(A)
A(1,1)=%i;
kron(A,A)
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Scilab EnterprisesCopyright (c) 2011-2017 (Scilab Enterprises)Copyright (c) 1989-2012 (INRIA)Copyright (c) 1989-2007 (ENPC)with contributors Last updated:Wed Oct 05 12:09:40 CEST 2011 | 274 | 832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-21 | latest | en | 0.604039 |
https://quizizz.com/admin/quiz/5bd242299815bc001a38b532/constant-of-proportionality | 1,721,067,322,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00330.warc.gz | 427,918,462 | 38,120 | # Constant of Proportionality
## 15 questions
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• 1. Multiple Choice
5 minutes
1 pt
Is the given table a proportional relationship? If so, what is the constant?
Yes; 1/3
Yes; 3
Yes; 5
No
• 2. Multiple Choice
5 minutes
1 pt
Write the equation for the table given.
y = 16x
y=8x
y=12x
• 3. Multiple Choice
5 minutes
1 pt
What is the constant of proportionality for this table?
\$1/9 per hour
\$9 per hour
\$8 per hour
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Already have an account? | 169 | 550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-30 | latest | en | 0.66934 |
http://de.metamath.org/mpeuni/cbvabv.html | 1,718,881,869,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00718.warc.gz | 8,342,231 | 4,107 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > cbvabv Structured version Visualization version GIF version
Theorem cbvabv 2734
Description: Rule used to change bound variables, using implicit substitution. (Contributed by NM, 26-May-1999.)
Hypothesis
Ref Expression
cbvabv.1 (𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
cbvabv {𝑥𝜑} = {𝑦𝜓}
Distinct variable groups: 𝜑,𝑦 𝜓,𝑥
Allowed substitution hints: 𝜑(𝑥) 𝜓(𝑦)
Proof of Theorem cbvabv
StepHypRef Expression
1 nfv 1830 . 2 𝑦𝜑
2 nfv 1830 . 2 𝑥𝜓
3 cbvabv.1 . 2 (𝑥 = 𝑦 → (𝜑𝜓))
41, 2, 3cbvab 2733 1 {𝑥𝜑} = {𝑦𝜓}
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 195 = wceq 1475 {cab 2596 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-clab 2597 df-cleq 2603 This theorem is referenced by: cdeqab1 3394 difjust 3542 unjust 3544 injust 3546 uniiunlem 3653 dfif3 4050 pwjust 4109 snjust 4124 intab 4442 intabs 4752 iotajust 5767 wfrlem1 7301 sbth 7965 cardprc 8689 iunfictbso 8820 aceq3lem 8826 isf33lem 9071 axdc3 9159 axdclem 9224 axdc 9226 genpv 9700 ltexpri 9744 recexpr 9752 supsr 9812 hashf1lem2 13097 cvbtrcl 13579 mertens 14457 4sq 15506 isuhgr 25726 isushgr 25727 isupgr 25751 isumgr 25761 nbgraf1olem5 25974 dispcmp 29254 eulerpart 29771 ballotlemfmpn 29883 bnj66 30184 bnj1234 30335 subfacp1lem6 30421 subfacp1 30422 dfon2lem3 30934 dfon2lem7 30938 frrlem1 31024 f1omptsn 32360 ismblfin 32620 glbconxN 33682 eldioph3 36347 diophrex 36357 cbvcllem 36934 ssfiunibd 38464 isuspgr 40382 isusgr 40383 isconngr 41356 isconngr1 41357 isfrgr 41430
Copyright terms: Public domain W3C validator | 1,050 | 2,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-26 | latest | en | 0.112743 |
https://www.justfreetools.com/en/calculators/lighting-calculators/candela-to-lumen-calculator | 1,632,637,565,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057830.70/warc/CC-MAIN-20210926053229-20210926083229-00050.warc.gz | 838,922,715 | 11,891 | # Candela to lumens free online calculator
Candela (cd) to lumens (lm) calculator and how to calculate.
## Candela to lumens calculator
Enter the luminous flux in lumens, apex angle in degrees and press the Calculate button
to get the luminous intensity in candela:
cd
°
lm
Lumens to candela calculator »
### Candela to lumens calculation
For uniform, isotropic light source, the luminous flux Φv in lumens (lm) is equal to the luminous intensity Iv in candela (cd),
times the solid angle Ω in steradians (sr):
Φv(lm) = Iv(cd) × Ω(sr)
The solid angle Ω in steradians (sr) is equal to 2 times pi times 1 minus cosine of half the cone apex angle θ in degrees (°):
Ω(sr) = 2π(1 - cos(θ/2))
The luminous flux Φv in lumens (lm) is equal to the luminous intensity Iv in candela (cd),
times 2 times pi times 1 minus cosine of half the apex angle θ in degrees (°):
Φv(lm) = Iv(cd) × ( 2π(1 - cos(θ/2)) )
So
lumens = candela × ( 2π(1 - cos(degrees/2)) )
Or
lm = cd × ( 2π(1 - cos(°/2)) )
Candela to lumens calculation »
Currently, we have around 1975 calculators, conversion tables and usefull online tools and software features for students, teaching and teachers, designers and simply for everyone. | 351 | 1,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-39 | latest | en | 0.711432 |
https://math.stackexchange.com/questions/53691/4-manifolds-of-which-there-exist-no-kirby-diagrams | 1,566,614,850,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319470.94/warc/CC-MAIN-20190824020840-20190824042840-00356.warc.gz | 557,792,715 | 31,618 | # 4-Manifolds of which there exist no Kirby diagrams
In 4-Manifold theory one makes often the use of Kirby Diagrams to construct 4-manifolds (compact or non-compact) with specific gauge and topological properties (for example small betti numbers, spin structure, etc.).
This raises a couple of questiona:
1.Is any compact or non-compact 4-manifold obtainable as a (finite or infinite) handle diagram ?
2.What are the properties needed for a compact or non-compact 4-manifold to be represented as a handle diagram ?
3.What are examples of 4-manifolds with no handle diagram ?
The diagrams can be as complicated as you want (so 0-, 2-, 3-, 4-) handles can be present. I do not know if you can get rid of all the 3-handles in the non-compact case.
This question came forth from the discussion explicit "exotic" charts . I am trying to get help of more people on that, then putting those things in comments (the question of explicit charts of an $\mathbb{E}\mathbb{R}^4$ is another one, albeit interesting in it's own right).
The question is answered by Bob Gompf by email, see my comment for the main part of his answer.
• A related question on math overflow: mathoverflow.net/questions/54143/… – Cheerful Parsnip Jul 25 '11 at 16:06
• Usually one finds a handle decomposition via a Morse function, but these don't behave so well in the noncompact case, so something subtle is going on. A quick Google search turned up this reference books.google.com/… which states that there are Morse functions on every noncompact manifold with no critical points.! – Cheerful Parsnip Jul 25 '11 at 16:24
• I have received a mail from Bob Gompf on this, seems that indeed every 4-manifold has a Kirby diagram, so my question is answered by external source. – Willem Noorduin Jul 27 '11 at 12:31
• The problem is that one cannot draw such a thing easily (but is does exist). There is also the problem of attaching infinitely many handles to its 0-handle, which can be fixed either by adding a collar to the boundary along with each handle, or introduceing canceling handle pairs so that there are infinitely many 0-handles. – Willem Noorduin Jul 27 '11 at 12:41
• In particular, Bob doesm't know any handle diagrams of large exotic R^4's. All known examples of these require infinitely many 3-handles in their handle decomposition – Willem Noorduin Jul 27 '11 at 12:47 | 585 | 2,359 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-35 | latest | en | 0.931972 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-2-linear-equations-and-functions-2-1-represent-relations-and-functions-2-1-exercises-skill-practice-page-76/5 | 1,701,379,475,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100232.63/warc/CC-MAIN-20231130193829-20231130223829-00566.warc.gz | 898,005,816 | 15,880 | ## Algebra 2 (1st Edition)
Domain = $-2,1,6$ Range = $-3, -1,5,8.$
The domain consists of all the $x$-coordinates of the ordered pairs: $\{-2,1,6\}.$ The range consists of all the $y$-coordinates of the ordered pairs: $\{-3, -1,5,8\}.$ The relation is graphed by plotting all the given ordered pairs as points. A mapping diagram shows (with arrows) which output(s) are associated with the inputs. | 122 | 397 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-50 | latest | en | 0.908059 |
http://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Parallel_postulate | 1,603,143,414,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107866404.1/warc/CC-MAIN-20201019203523-20201019233523-00633.warc.gz | 188,156,348 | 12,890 | # Parallel postulate
In geometry, the parallel postulate, also called Euclid's fifth postulate because it is the fifth postulate in Euclid's Elements, is a distinctive axiom in Euclidean geometry. It states that, in two-dimensional geometry:
If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.
This postulate does not specifically talk about parallel lines;[1] it is only a postulate related to parallelism. Euclid gave the definition of parallel lines in Book I, Definition 23[2] just before the five postulates.[3]
Euclidean geometry is the study of geometry that satisfies all of Euclid's axioms, including the parallel postulate. A geometry where the parallel postulate does not hold is known as a non-Euclidean geometry. Geometry that is independent of Euclid's fifth postulate (i.e., only assumes the modern equivalent of the first four postulates) is known as absolute geometry (or, in other places known as neutral geometry).
## Equivalent properties
Probably the best known equivalent of Euclid's parallel postulate, contingent on his other postulates, is Playfair's axiom, named after the Scottish mathematician John Playfair, which states:
In a plane, given a line and a point not on it, at most one line parallel to the given line can be drawn through the point.[4]
This axiom by itself is not logically equivalent to the Euclidean parallel postulate since there are geometries in which one is true and the other is not. However, in the presence of the remaining axioms which give Euclidean geometry, each of these can be used to prove the other, so they are equivalent in the context of absolute geometry.[5]
Many other statements equivalent to the parallel postulate have been suggested, some of them appearing at first to be unrelated to parallelism, and some seeming so self-evident that they were unconsciously assumed by people who claimed to have proven the parallel postulate from Euclid's other postulates. These equivalent statements include:
1. There is at most one line that can be drawn parallel to another given one through an external point. (Playfair's axiom)
2. The sum of the angles in every triangle is 180° (triangle postulate).
3. There exists a triangle whose angles add up to 180°.
4. The sum of the angles is the same for every triangle.
5. There exists a pair of similar, but not congruent, triangles.
6. Every triangle can be circumscribed.
7. If three angles of a quadrilateral are right angles, then the fourth angle is also a right angle.
8. There exists a quadrilateral in which all angles are right angles, that is, a rectangle.
9. There exists a pair of straight lines that are at constant distance from each other.
10. Two lines that are parallel to the same line are also parallel to each other.
11. In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides (Pythagoras' Theorem).[6][7]
12. The Law of cosines, a general case of Pythagoras' Theorem.
13. There is no upper limit to the area of a triangle. (Wallis axiom)[8]
14. The summit angles of the Saccheri quadrilateral are 90°.
15. If a line intersects one of two parallel lines, both of which are coplanar with the original line, then it also intersects the other. (Proclus' axiom)[9]
However, the alternatives which employ the word "parallel" cease appearing so simple when one is obliged to explain which of the four common definitions of "parallel" is meant – constant separation, never meeting, same angles where crossed by some third line, or same angles where crossed by any third line – since the equivalence of these four is itself one of the unconsciously obvious assumptions equivalent to Euclid's fifth postulate. In the list above, it is always taken to refer to non-intersecting lines. For example, if the word "parallel" in Playfair's axiom is taken to mean 'constant separation' or 'same angles where crossed by any third line', then it is no longer equivalent to Euclid's fifth postulate, and is provable from the first four (the axiom says 'There is at most one line...', which is consistent with there being no such lines). However, if the definition is taken so that parallel lines are lines that do not intersect, or that have some line intersecting them in the same angles, Playfair's axiom is contextually equivalent to Euclid's fifth postulate and is thus logically independent of the first four postulates. Note that the latter two definitions are not equivalent, because in hyperbolic geometry the second definition holds only for ultraparallel lines.
## History
For two thousand years, many attempts were made to prove the parallel postulate using Euclid's first four postulates. The main reason that such a proof was so highly sought after was that, unlike the first four postulates, the parallel postulate is not self-evident. If the order the postulates were listed in the Elements is significant, it indicates that Euclid included this postulate only when he realised he could not prove it or proceed without it.[10] Many attempts were made to prove the fifth postulate from the other four, many of them being accepted as proofs for long periods until the mistake was found. Invariably the mistake was assuming some 'obvious' property which turned out to be equivalent to the fifth postulate (Playfair's axiom). Although known from the time of Proclus, this became known as Playfair's Axiom after John Playfair wrote a famous commentary on Euclid in 1795 in which he proposed replacing Euclid's fifth postulate by his own axiom.
Proclus (410-485) wrote a commentary on The Elements where he comments on attempted proofs to deduce the fifth postulate from the other four, in particular he notes that Ptolemy had produced a false 'proof'. Proclus then goes on to give a false proof of his own. However he did give a postulate which is equivalent to the fifth postulate.
Ibn al-Haytham (Alhazen) (965-1039), an Arab mathematician, made an attempt at proving the parallel postulate using a proof by contradiction,[11] in the course of which he introduced the concept of motion and transformation into geometry.[12] He formulated the Lambert quadrilateral, which Boris Abramovich Rozenfeld names the "Ibn al-Haytham–Lambert quadrilateral",[13] and his attempted proof contains elements similar to those found in Lambert quadrilaterals and Playfair's axiom.[14]
The Persian mathematician, astronomer, philosopher, and poet Omar Khayyám (1050–1123), attempted to prove the fifth postulate from another explicitly given postulate (based on the fourth of the five principles due to the Philosopher (Aristotle), namely, "Two convergent straight lines intersect and it is impossible for two convergent straight lines to diverge in the direction in which they converge."[15] He derived some of the earlier results belonging to elliptical geometry and hyperbolic geometry, though his postulate excluded the latter possibility.[16] The Saccheri quadrilateral was also first considered by Omar Khayyám in the late 11th century in Book I of Explanations of the Difficulties in the Postulates of Euclid.[13] Unlike many commentators on Euclid before and after him (including Giovanni Girolamo Saccheri), Khayyám was not trying to prove the parallel postulate as such but to derive it from his equivalent postulate. He recognized that three possibilities arose from omitting Euclid's fifth postulate; if two perpendiculars to one line cross another line, judicious choice of the last can make the internal angles where it meets the two perpendiculars equal (it is then parallel to the first line). If those equal internal angles are right angles, we get Euclid's fifth postulate, otherwise, they must be either acute or obtuse. He showed that the acute and obtuse cases led to contradictions using his postulate, but his postulate is now known to be equivalent to the fifth postulate.
Nasir al-Din al-Tusi (1201–1274), in his Al-risala al-shafiya'an al-shakk fi'l-khutut al-mutawaziya (Discussion Which Removes Doubt about Parallel Lines) (1250), wrote detailed critiques of the parallel postulate and on Khayyám's attempted proof a century earlier. Nasir al-Din attempted to derive a proof by contradiction of the parallel postulate.[17] He also considered the cases of what are now known as elliptical and hyperbolic geometry, though he ruled out both of them.[16]
Nasir al-Din's son, Sadr al-Din (sometimes known as "Pseudo-Tusi"), wrote a book on the subject in 1298, based on his father's later thoughts, which presented one of the earliest arguments for a non-Euclidean hypothesis equivalent to the parallel postulate. "He essentially revised both the Euclidean system of axioms and postulates and the proofs of many propositions from the Elements."[17][18] His work was published in Rome in 1594 and was studied by European geometers. This work marked the starting point for Saccheri's work on the subject[17] which opened with a criticism of Sadr al-Din's work and the work of Wallis.[19]
Giordano Vitale (1633-1711), in his book Euclide restituo (1680, 1686), used the Khayyam-Saccheri quadrilateral to prove that if three points are equidistant on the base AB and the summit CD, then AB and CD are everywhere equidistant. Girolamo Saccheri (1667-1733) pursued the same line of reasoning more thoroughly, correctly obtaining absurdity from the obtuse case (proceeding, like Euclid, from the implicit assumption that lines can be extended indefinitely and have infinite length), but failing to refute the acute case (although he managed to wrongly persuade himself that he had).
In 1766 Johann Lambert wrote, but did not publish, Theorie der Parallellinien in which he attempted, as Saccheri did, to prove the fifth postulate. He worked with a figure that today we call a Lambert quadrilateral, a quadrilateral with three right angles (can be considered half of a Saccheri quadrilateral). He quickly eliminated the possibility that the fourth angle is obtuse, as had Saccheri and Khayyám, and then proceeded to prove many theorems under the assumption of an acute angle. Unlike Saccheri, he never felt that he had reached a contradiction with this assumption. He had proved the non-Euclidean result that the sum of the angles in a triangle increases as the area of the triangle decreases, and this led him to speculate on the possibility of a model of the acute case on a sphere of imaginary radius. He did not carry this idea any further.[20]
Where Khayyám and Saccheri had attempted to prove Euclid's fifth by disproving the only possible alternatives, the nineteenth century finally saw mathematicians exploring those alternatives and discovering the logically consistent geometries which result. In 1829, Nikolai Ivanovich Lobachevsky published an account of acute geometry in an obscure Russian journal (later re-published in 1840 in German). In 1831, János Bolyai included, in a book by his father, an appendix describing acute geometry, which, doubtlessly, he had developed independently of Lobachevsky. Carl Friedrich Gauss had also studied the problem, but he did not publish any of his results. Upon hearing of Bolyai's results in a letter from Bolyai's father, Farkas Bolyai, Gauss stated:
"If I commenced by saying that I am unable to praise this work, you would certainly be surprised for a moment. But I cannot say otherwise. To praise it would be to praise myself. Indeed the whole contents of the work, the path taken by your son, the results to which he is led, coincide almost entirely with my meditations, which have occupied my mind partly for the last thirty or thirty-five years."[21]
The resulting geometries were later developed by Lobachevsky, Riemann and Poincaré into hyperbolic geometry (the acute case) and elliptic geometry (the obtuse case). The independence of the parallel postulate from Euclid's other axioms was finally demonstrated by Eugenio Beltrami in 1868.
## Converse of Euclid's parallel postulate
Euclid did not postulate the converse of his fifth postulate, which is one way to distinguish Euclidean geometry from elliptic geometry. The Elements contains the proof of an equivalent statement (Book I, Proposition 27): If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another. As De Morgan[22] pointed out, this is logically equivalent to (Book I, Proposition 16). These results do not depend upon the fifth postulate, but they do require the second postulate[23] which is violated in elliptic geometry.
## Criticism
Attempts to logically prove the parallel postulate, rather than the eighth axiom,[24] were criticized by Arthur Schopenhauer. However, the argument used by Schopenhauer was that the postulate is evident by perception, not that it was not a logical consequence of the other axioms.
## Notes
1. Euclid's Elements, Book I, Definition 23
2. Euclid's Elements, Book I
3. Euclid's Parallel Postulate and Playfair's Axiom
4. Henderson & Taimiņa 2005, pg. 139
5. Eric W. Weisstein (2003), CRC concise encyclopedia of mathematics (2nd ed.), p. 2147, ISBN 1-58488-347-2, The parallel postulate is equivalent to the Equidistance postulate, Playfair axiom, Proclus axiom, the Triangle postulate and the Pythagorean theorem.
6. Alexander R. Pruss (2006), The principle of sufficient reason: a reassessment, Cambridge University Press, p. 11, ISBN 0-521-85959-X, We could include...the parallel postulate and derive the Pythagorean theorem. Or we could instead make the Pythagorean theorem among the other axioms and derive the parallel postulate.
7. Bogomolny, Alexander. "Euclid's Fifth Postulate". Cut The Knot. Retrieved 30 September 2011.
8. Weisstein, Eric W. "Proclus' Axiom – MathWorld". Retrieved 2009-09-05.
9. Florence P. Lewis (Jan 1920), "History of the Parallel Postulate", The American Mathematical Monthly, The American Mathematical Monthly, Vol. 27, No. 1, 27 (1): 16–23, doi:10.2307/2973238, JSTOR 2973238.
10. Katz 1998, pg. 269
11. Katz 1998, p. 269:
In effect, this method characterized parallel lines as lines always equidistant from one another and also introduced the concept of motion into geometry.
12. Rozenfeld 1988, p. 65
13. Smith 1992
14. Boris A Rosenfeld and Adolf P Youschkevitch (1996), Geometry, p.467 in Roshdi Rashed, Régis Morelon (1996), Encyclopedia of the history of Arabic science, Routledge, ISBN 0-415-12411-5.
15. Boris A. Rosenfeld and Adolf P. Youschkevitch (1996), "Geometry", in Roshdi Rashed, ed., Encyclopedia of the History of Arabic Science, Vol. 2, p. 447-494 [469], Routledge, London and New York:
Khayyam's postulate had excluded the case of the hyperbolic geometry whereas al-Tusi's postulate ruled out both the hyperbolic and elliptic geometries.
16. Katz 1998, pg.271:
But in a manuscript probably written by his son Sadr al-Din in 1298, based on Nasir al-Din's later thoughts on the subject, there is a new argument based on another hypothesis, also equivalent to Euclid's, [...] The importance of this latter work is that it was published in Rome in 1594 and was studied by European geometers. In particular, it became the starting point for the work of Saccheri and ultimately for the discovery of non-Euclidean geometry.
17. Boris A. Rosenfeld and Adolf P. Youschkevitch (1996), "Geometry", in Roshdi Rashed, ed., Encyclopedia of the History of Arabic Science, Vol. 2, p. 447-494 [469], Routledge, London and New York:
In Pseudo-Tusi's Exposition of Euclid, [...] another statement is used instead of a postulate. It was independent of the Euclidean postulate V and easy to prove. [...] He essentially revised both the Euclidean system of axioms and postulates and the proofs of many propositions from the Elements.
18. MacTutor's Giovanni Girolamo Saccheri
19. O'Connor, J.J.; Robertson, E.F. "Johann Heinrich Lambert". Retrieved 16 September 2011.
20. Faber 1983, pg. 161
21. Heath, T.L., The thirteen books of Euclid's Elements, Vol.1, Dover, 1956, pg.309.
22. Coxeter, H.S.M., Non-Euclidean Geometry, 6th Ed., MAA 1998, pg.3
23. Schopenhauer is referring to Euclid's Common Notion 4: Figures coinciding with one another are equal to one another. | 3,854 | 16,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-45 | latest | en | 0.956753 |
https://blogs.accu.org/category/calculus/page/2/ | 1,696,111,495,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00205.warc.gz | 155,463,878 | 36,383 | ## A Decent Borel Code – a.k.
A few posts ago we took a look at how we might implement various operations on sets represented as sorted arrays, such as the union, being the set of every element that is in either of two sets, and the intersection, being the set of every element that is in both of them, which we implemented with `ak.setUnion` and `ak.setIntersection` respectively.
Such arrays are necessarily both finite and discrete and so cannot represent continuous subsets of the real numbers such as intervals, which contain every real number within a given range. Of particular interest are unions of countable sets of intervals Ii, known as Borel sets, and so it's worth adding a type to the `ak` library to represent them.
## Finally On A Calculus Of Differences – student
My fellow students and I have spent much of our spare time this past year investigating the similarities between the calculus of functions and that of sequences, which we have defined for a sequence sn with the differential operator
Δ sn = sn - sn-1
and the integral operator
n Δ-1 sn = Σ si i = 1
where Σ is the summation sign, adopting the convention that terms with non-positive indices equate to zero.
We have thus far discovered how to differentiate and integrate monomial sequences, found product and quotient rules for differentiation, a rule of integration by parts and figured solutions to some familiar-looking differential equations, all of which bear a striking resemblance to their counterparts for functions. To conclude our investigation, we decided to try to find an analogue of Taylor's theorem for sequences.
## Further Still On A Calculus Of Differences – student
For some time now my fellow students and I have been whiling away our spare time considering the similarities of the relationships between sequences and series and those between the derivatives and integrals of functions. Having defined differential and integral operators for a sequence sn with
Δ sn = sn - sn-1
and
n Δ-1 sn = Σ si i = 1
where Σ is the summation sign, we found analogues for the product rule, the quotient rule and the rule of integration by parts, as well as formulae for the derivatives and integrals of monomial sequences, being those whose terms are non-negative integer powers of their indices, and higher order, or repeated, derivatives and integrals in general.
We have since spent some time considering how we might solve equations relating sequences to their derivatives, known as differential equations when involving functions, and it is upon our findings that I shall now report. | 526 | 2,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2023-40 | latest | en | 0.960969 |
https://numberworld.info/518 | 1,685,853,731,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00496.warc.gz | 470,677,260 | 3,763 | # Number 518
### Properties of number 518
Cross Sum:
Factorization:
2 * 7 * 37
Divisors:
1, 2, 7, 14, 37, 74, 259, 518
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 32:
g6
sin(518)
0.35488199371552
cos(518)
-0.93491110301274
tan(518)
-0.37958902463766
ln(518)
6.2499752422595
lg(518)
2.7143297597452
sqrt(518)
22.759613353482
Square(518)
### Number Look Up
Look Up
518 (five hundred eighteen) is a very special number. The cross sum of 518 is 14. If you factorisate the number 518 you will get these result 2 * 7 * 37. The figure 518 has 8 divisors ( 1, 2, 7, 14, 37, 74, 259, 518 ) whith a sum of 912. The figure 518 is not a prime number. The figure 518 is not a fibonacci number. The number 518 is not a Bell Number. The figure 518 is not a Catalan Number. The convertion of 518 to base 2 (Binary) is 1000000110. The convertion of 518 to base 3 (Ternary) is 201012. The convertion of 518 to base 4 (Quaternary) is 20012. The convertion of 518 to base 5 (Quintal) is 4033. The convertion of 518 to base 8 (Octal) is 1006. The convertion of 518 to base 16 (Hexadecimal) is 206. The convertion of 518 to base 32 is g6. The sine of 518 is 0.35488199371552. The cosine of the number 518 is -0.93491110301274. The tangent of the number 518 is -0.37958902463766. The square root of 518 is 22.759613353482.
If you square 518 you will get the following result 268324. The natural logarithm of 518 is 6.2499752422595 and the decimal logarithm is 2.7143297597452. that 518 is amazing figure! | 591 | 1,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-23 | latest | en | 0.745706 |
https://cmatutors.com/correlation-analysis-class-11-ahsec/ | 1,685,915,028,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650264.9/warc/CC-MAIN-20230604193207-20230604223207-00463.warc.gz | 205,818,404 | 60,181 | # Correlation Analysis, AHSEC Class 11 Economics Chapterwise Notes
## AHSEC Class 11 Economics Chapterwise Notes
Q.N.1. Define Correlation analysis. What are its various kinds?
Ans: – Definition: – Correlation is the degree of the relationship between two or more variables. It does not explain the cause behind the relationship. Kinds of correlation may be studied on the basis of:
I. Change in proportion.
II. Number of variation.
III. Change in direction.
(I) Basis of change in proportion:-There are two important correlations on the basis of change in proportion. They are:
(a) Linear correlation (b) Non-linear correlation
(a) Linear correlation: – Correlation is said to be linear when one variable move with the other variable in fixed proportion
(b) Non-linear correlation: – Correlation is said to be non-linear when one variable move with the other variable in changing proportion.
(II) On the basis of number of variables: On the basis of number of variables, correlation may be:
(a) Simple (b) Partial (c) Multiple
(a) Simple correlation: – When only two variables are studied it is a simple correlation.
(b) Partial correlation: – When more than two variables are studied keeping other variables constant, it is called partial correlation.
(c) Multiple correlations: – When at least three variables are studied and their relationships are simultaneously worked out, it is a case of multiple correlations
(III) On the basis of Change in direction: On the basis of Chang in direction, correlation may be
(a)Positive Correlation (b) Negative Correlation
(a) Positive Correlation: – Correlation is said to be positive when two variables move in same direction.
(b) Negative Correlation: – Correlation is said to be negative when two variables moves in opposite direction.
Q.N.2.What are the uses and limitations of Correlation?
Ans: – Following are the main advantages of correlation:
a) It gives a precise quantitative value indicating the degree of relationship existing between the two variables.
b) It measures the direction as well as relationship between the two variables.
c) Further in regression analysis it is used for estimating the value of dependent variable from the known value of the independent variable
d) The effect of correlation is to reduce the range of uncertainty in predictions.
Following are the main limitations of correlation:
a) Extreme items affect the value of the coefficient of correlation.
b) Its computational method is difficult as compared to other methods.
c) It assumes the linear relationship between the two variables, whether such relationship exist or not.
Q.N.3. What are the different degrees of Correlation? Or When correlation is positive or negative or zero
Ans: The different degrees of correlation are:
a) Perfect Correlation: – It two variables vary in same proportion, and then the correlation is said to be perfect correlation.
b) Positive Correlation: If increase (decrease) in the values of one variable result a corresponding increase (decrease) in the values of another variable then the variables are said to be positively correlated. For example heights and weights, income and expenditure etc. are positively correlated.
c) Negative Correlation: If increase (decrease) in the values of one variable result a corresponding decrease (increase) in the values of another variable then the variables are said to be negatively correlated. For example, demand and price of commodities.
d) Zero Correlation: Two variables are said to have Zero correlation between them if they tend to change with no connection to each other. For example one should expect zero correlation between the heights of the students and the marks obtained by them.
Q.N.4. What are the different methods of studying correlation?
Ans: – The different methods of studying relationship between two variables are:
a) Scatter diagram method.
b) Graphic method
c) Karl Pearson’s coefficient of correlation
d) Rank correlation method
a) Scatter Diagram Method: – It is a graphical representation of finding relationship between two or more variables. Independent variable are taken on the x-axis and dependent variable on the y-axis and plot the various values of x and y on the graph. If all values move upwards then there is positive correlation, if they move downwards then there is negative correlation.
Merits:
i) It is easy and simple to use and understand.
ii) Relation between two variables can be studied in a non-mathematical way.
Demerits:-
i) It is non-mathematical method so the results are non-exact and accurate.
ii) It gives only approximate idea of the relationship.
b) Graphic Method: – This is an extension of linear graphs. In this case two or more variables are plotted on graph paper. If the curves move in same direction the correlation is positive and if moves in opposite direction then correlation is negative. But if there is no definite direction, there is absence of correlation. Although it is a simple method, but this shows only rough estimate of nature of relationship.
Merits: –
i) It is easy and simple to use and understand.
ii) Relation between two variables can be studied in a non-mathematical way.
Demerits:-
i) It is non-mathematical method so the results are non-exact and accurate.
ii) It gives only approximate idea of the relationship.
c) Karl Pearson’s Coefficient of correlation: – Correlation coefficient is a mathematical and most popular method of calculating correlation. Arithmetic mean and standard deviation are the basis for its calculation. The Correlation coefficient (r), also called as the linear correlation coefficient measures the strength and direction of a linear relationship between two variables. The value of r lies between -1 to +1.
Properties of r:-
a) r is the independent to the unit of measurement of variable.
b) r does not depend on the change of origin and scale.
c) If two variables are independent to each other, then the value of r is zero.
Merits:-
a) The co-efficient of correlation measures the degree of relationship between two variables.
b) It also measures the direction.
c) It may be used to determine regression coefficient provided s.d. of two variables are known.
Demerits:-
a) It assumes always the linear relationship between the variables even if this assumption is not correct.
b) It is affected by extreme values.
c) It takes a lot of time to compute.
### AHSEC CLASS 11 CHAPTER-WISE NOTES
Part A: Introductory Micro Economics
Introduction to Micro Economics
Consumer Behaviour and Demand
* Law of Demand
* Elasticity of Demand
Producer Behaviour and Supply
Forms of Market and Price Determination:
* MonopolyÂ
* Perfect Comptetition
Simple Applications of Tools of Demand & Supply
Part B: Statistics for Economics
Introduction to Statistics
Collection, Organisation and Presentation of Data
Statistical Tools & InterpretationÂ
* Measure of Central Tendency
* Dispersion
* Index Number
Correlation analysis
Summary Notes of Statistics for Economics available here
d) Spearman’s rank Coefficient of correlation: – This is a qualitative method of measuring correlation co-efficient. Qualities such as beauty, honesty, ability, etc. cannot be measured in quantitative terms. So, ranks are used to determine the correlation coefficient.
Merits:-
a) It is easy and simple to calculate and understand.
b) This method is most suitable if the data are qualitative.
c) This is the only method that can be used where we are given the ranks not the actual data.
Demerits:-
a) This method is not accurate as Karl Pearson’s correlation coefficient.
b) This method cannot be used in case of grouped frequency distribution.
c) Where the number of items exceeds 30 the calculations become quite tedious and require a lot of time.
The rank method has two principal uses:
a) The initial data are in the form of ranks or qualitative in nature.
b) If N is fairly small (say, not larger than 25 or 30) rank method is sometimes applied to interval data as on approximation to the more time consuming R. This requires that the interval data be transferred to rank orders for both variables. If N is much in excess of 30, the labour required in ranking the scores, becomes greater than is justified by the anticipated saving of time through the rank formula.
Difference between Karl Pearson’s coefficient and Spearmen’s Rank correlation
                                     Karl Pearson Rank Correlation 1. This method is used when actual data is given. 1. This method is used when we are given rank not the actual data. 2. It measures only linear relationship between two variables. 2. It measures nonlinear monotonic relationship between two variables. 3. It is quantitative in nature. 3. It is qualitative in nature.
error: Content is protected !! | 1,888 | 8,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-23 | longest | en | 0.920431 |
http://www.cengage.com/aushed/instructor.do?disciplinenumber=1008&product_isbn=9780538476126&courseid=BS01&codeid=2786 | 1,418,963,893,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802768208.73/warc/CC-MAIN-20141217075248-00009-ip-10-231-17-201.ec2.internal.warc.gz | 424,675,177 | 7,090 | #### Online learning resources
Data Analysis and Decision Making, 4th Edition
Includes Online Content Printed Access Card.
ISBN-10: 0538476125 ISBN-13: 9780538476126
1080 Pages Casebound
Preface
1. Introduction to Data Analysis and Decision Making.
1.1. Introduction.
1.2. An Overview of the Book.
1.3. Modeling and Models.
1.4. Conclusion.
PART I: EXPLORING DATA.
2. Describing the Distribution of a Single Variable.
2.1 Introduction.
2.2 Basic Concepts.
2.3 Descriptive Measures for Categorical Variables.
2.4 Descriptive Measures for Numerical Variables.
2.5 Time Series Data.
2.6 Outliers and Missing Values.
2.7 Excel Tables for Filtering, Sorting, and Summarizing.
2.8 Conclusion.
3. Finding Relationships Among Variables.
3.1 Introduction.
3.2 Relationships Among Categorical Variables.
3.3 Relationships Among Categorical Variables and a Numerical Variable.
3.4 Relationships Among Numerical Variables.
3.5 Pivot Tables.
3.6 An Extended Example.
3.7 Conclusion.
PART II: PROBABILITY AND DECISION MAKING UNDER UNCERTAINTY
4. Probability and Probability Distributions.
4.1. Introduction.
4.2. Probability Essentials.
4.3. Distribution of a Single Random Variable.
4.4. An Introduction to Simulation.
4.5. Distribution of Two Random Variables: Scenario Approach.
4.6. Distribution of Two Random Variables: Joint Probability Approach.
4.7. Independent Random Variables.
4.8. Weighted Sums of Random Variables.
4.9. Conclusion.
5. Normal, Binomial, Poisson, and Exponential Distributions.
5.1. Introduction.
5.2. The Normal Distribution.
5.3. Applications of the Normal Distribution.
5.4. The Binomial Distribution.
5.5. Applications of the Binomial Distribution.
5.6. The Poisson and Exponential Distributions.
5.7. Fitting a Probability Distribution to Data with @RISK.
5.8. Conclusion.
6. Decision Making Under Uncertainty.
6.1. Introduction.
6.2. Elements of a Decision Analysis.
6.4. Bayes’ Rule.
6.5. Multistage Decision Problems.
6.6. Incorporating Attitudes Toward Risk.
6.7. Conclusion.
PART III: STATISTICAL INFERENCE.
7. Sampling and Sampling Distributions.
7.1. Introduction.
7.2. Sampling Terminology.
7.3. Methods for Selecting Random Samples.
7.4. An Introduction to Estimation.
7.5. Conclusion.
8. Confidence Interval Estimation.
8.1. Introduction.
8.2. Sampling Distributions.
8.3. Confidence Interval for a Mean.
8.4. Confidence Interval for a Total.
8.5. Confidence Interval for a Proportion.
8.6. Confidence Interval for a Standard Deviation.
8.7. Confidence Interval for the Difference Between Means.
8.8. Confidence Interval for the Difference Between Proportions.
8.9. Controlling Confidence Interval Length.
8.10. Conclusion.
9. Hypothesis Testing.
9.1. Introduction.
9.2. Concepts in Hypothesis Testing.
9.3. Hypothesis Tests for a Population Mean.
9.4. Hypothesis Tests for Other Parameters.
9.5. Tests for Normality.
9.6. Chi-Square Test for Independence.
9.7. One-Way ANOVA.
9.8. Conclusion.
PART IV: REGRESSION ANALYSIS AND TIME SERIES FORECASTING.
10. Regression Analysis: Estimating Relationships.
10.1. Introduction.
10.2. Scatterplots: Graphing Relationships.
10.3. Correlations: Indicators of Linear Relationships.
10.4. Simple Linear Regression.
10.5. Multiple Regression.
10.6. Modeling Possibilities.
10.7. Validation of the Fit.
10.8. Conclusion.
11. Regression Analysis: Statistical Inference.
11.1. Introduction.
11.2. The Statistical Model.
11.3. Inferences About the Regression Coefficients.
11.4. Multicollinearity.
11.5. Include/Exclude Decisions.
11.6. Stepwise Regression.
11.7. The Partial F Test.
11.8. Outliers.
11.9. Violations of Regression Assumptions.
11.10. Prediction.
11.11. Conclusion.
12. Time Series Analysis and Forecasting.
12.1. Introduction.
12.2. Forecasting Methods: An Overview.
12.3. Testing for Randomness.
12.4. Regression-Based Trend Models.
12.5. The Random Walk Model.
12.6. Autoregression Models.
12.7. Moving Averages.
12.8. Exponential Smoothing.
12.9. Seasonal Models.
12.10. Conclusion.
PART V: OPTIMIZATION AND SIMULATION MODELING.
13. Introduction to Optimization Modeling.
13.1. Introduction.
13.2. Introduction to Optimization.
13.3. A Two-Variable Product Mix Model.
13.4. Sensitivity Analysis.
13.5. Properties of Linear Models.
13.6. Infeasibility and Unboundedness.
13.7. A Larger Product Mix Model.
13.8. A Multiperiod Production Model.
13.9. A Comparison of Algebraic and Spreadsheet Models.
13.10. A Decision Support System.
13.11. Conclusion.
14. Optimization Models.
14.1. Introduction.
14.2. Worker Scheduling Models.
14.3. Blending Models.
14.4. Logistics Models.
14.5. Aggregate Planning Models.
14.6. Financial Models.
14.7. Integer Programming Models.
14.8. Nonlinear Programming Models.
14.9. Conclusion.
15. Introduction to Simulation Modeling.
15.1. Introduction.
15.2. Probability Distributions for Input Variables.
15.3. Simulation and the Flaw of Averages.
15.4. Simulation with Built-In Excel Tools.
15.5. Introduction to the @RISK Add-in.
15.6. The Effects of Input Distributions on Results.
15.7. Conclusion.
16. Simulation Models.
16.1. Introduction.
16.2. Operations Models.
16.3. Financial Models.
16.4. Marketing Models.
16.5. Simulating Games of Chance.
16.6. An Automated Template for @RISK Models.
16.7. Conclusion.
PART VI: BONUS ONLINE MATERIAL
2 Using the Advanced Filter and Database Functions.
17. Importing Data into Excel.
17.1 Introduction.
17.2 Rearranging Excel Data.
17.3 Importing Text Data.
17.4 Importing Relational Database Data.
17.5 Web Queries.
17.6 Cleansing the Data.
17.7 Conclusion. | 1,465 | 5,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2014-52 | latest | en | 0.548975 |
http://www.physicsforums.com/showthread.php?s=09a08d8488291a280dc478d8f906e24a&p=3838546 | 1,394,335,457,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999671301/warc/CC-MAIN-20140305060751-00069-ip-10-183-142-35.ec2.internal.warc.gz | 493,418,339 | 6,761 | # Help with indirect logic proof please!
by LCharette
Tags: indirect, logic, proof
P: 9 Using the five axioms below prove: p→q A1: p→~y A2: ~r→q A3: p→~z A4: x→ q or z A5: r→x or y Do I have to take the contrapositive of some of the axioms to begin this proof?
Math Emeritus Sci Advisor Thanks PF Gold P: 38,706 Yes, that would be the simplest thing to do. The very first "axiom" gives you p-> ~y but there is no "~y-> " so you cannot continue directly. However, you do have "A5: r->x or y which has contrapositive ~(x or y)= (~x) and (~y)->~r and then both "A2: ~r-> q" and "A4: x-> q or z".
P: 9 Am I on the right track with this? Conclusions Justifications 1. p Given 2. ~z or ~y All cases 3. ~z Case 1 4. ~x A4 5. ~r A5 6. q A2
Related Discussions Set Theory, Logic, Probability, Statistics 1 Precalculus Mathematics Homework 2 Calculus & Beyond Homework 5 Set Theory, Logic, Probability, Statistics 6 General Discussion 1 | 304 | 932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2014-10 | longest | en | 0.855958 |
https://gyires.inf.unideb.hu/KMITT/a53/ch06s06.html | 1,582,713,276,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146341.16/warc/CC-MAIN-20200226084902-20200226114902-00074.warc.gz | 383,095,204 | 4,594 | ## 29.6. 29.6 Construction of growing arrays using colouring
In this section particular constructions are presented.
### 29.6.1. 29.6.1 Construction of growing sequences
As the first example let and . Cellular calculates and Martin produces the cellular (2,1,2,4)-perfect sequence .
Since is symmetric, . Now Growing chooses multiplication coefficient , extension vector and uses Colour to construct a 4-ary perfect sequence.
Colour arranges copies into a blocks sized array receiving
Colour receives the indexing scheme and the colouring matrix transforming the elements of into digit length -ary numbers: .
Finally we colour the matrix using – that is multiply the elements of by and add the -th block of to both cells of the -th copy in :
Since , we use Colour again with and get the (8,1,2,64)-perfect sequence repeating 4 times, using the same indexing array and colouring array .
Another example is and . To guarantee the cellular property now we need a new alphabet size . Martin produces a (6,1,2,36)-perfect sequence , then Colour results a (12,1,2,144)-perfect sequence .
### 29.6.2. 29.6.2 Construction of growing squares
As the first example let and . Then . We construct the even sequence using Even and the symmetric perfect array in Figure 29.1.a using the meshing function (29.5). Since is symmetric, it can be used as . Now the greatest common divisor of and is 2, therefore indeed .
Growing chooses and Colour returns the array repeating the array times.
Colour uses the indexing scheme containing indices in the same arrangement as it was used in . Figure 29.1.b shows .
Transformation of the elements of into 4-digit -ary form results the colouring matrix represented in Figure 29.2.
Colouring of array using the colouring array results the (4,2,2,16)-square represented in Figure 29.3.
In the next iteration Colour constructs an 8-ary square repeating times, using the same indexing scheme and colouring by . The result is , a -perfect square.
### 29.6.3. 29.6.3 Construction of growing cubes
If , then the necessary condition (29.4) is for double cubes, implying is a cube number or is a multiple of 3. Therefore, either and then , or and so , that is, the smallest possible perfect double cube is the (8, 3, 2, 256)-cube.
As an example, let and . Cellular computes , Mesh constructs the -perfect square in Figure 29.1.a, then Shift uses Optimal-Martin with and to get the shift sizes for the layers of the -perfect output of Cellular , where . Shift uses as zeroth layer and the th layer is generated by cyclic shifting of the previous layer downwards by (div 4) and right by (mod 4), where . 16 layers of are shown in Figure 29.5.
Let be a sized perfect, rectangular matrix, whose 0th layer is the matrix represented in Figure 29.1, and the -perfect array in Figure 29.5, where and .
Growing uses Colour to retrieve a doubly symmetric cube. , thus and , that is we construct the matrix repeating times.
has the size and . Colour gets the colouring matrix by transforming the elements of into 8-digit -ary numbers – and arrange the elements into sized cubes in lexicographic order – that is in order (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), (1,1,1). Finally colouring results a double cube .
contains elements therefore it is presented only in electronic form (on the home page of the first author).
If we repeat the colouring again with , then we get a 64-ary sized double cube .
### 29.6.4. 29.6.4 Construction of a four-dimensional double hypercube
In 4 dimensions the smallest 's satisfying (29.3) are and . But we do not know algorithm which can construct -perfect or -perfect hypercube. The third chance is the -perfect hypercube. Let and . Cellular calculates , then calls Optimal-Martin receiving the cellular -perfect sequence . Then Cellular calls Mesh which constructs the cellular -perfect square shown in Figure 29.4.
Now Shift calls Optimal-Martin with and to get the shift sizes for the layers of the -perfect output of Cellular , where . Shift uses as zeroth layer and the th layer is generated by cyclic shifting of the previous layer downwards by (div 4) and right by (mod 4), where . The layers of the -perfect array are shown in Figure 29.5.
Up to this point the construction is the same as in [], but now , therefore we use Shift again to get a -perfect prism, then we fill an empty cube with -sized prisms and finally colouring results the required 4-dimensional hypercube.
Exercises
29.6-1 Explain the construction of Figure 29.5. | 1,099 | 4,528 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-10 | latest | en | 0.827137 |
http://tvtropes.org/pmwiki/pmwiki.php/Main/AbstractScale | 1,516,648,364,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891530.91/warc/CC-MAIN-20180122173425-20180122193425-00321.warc.gz | 344,515,832 | 26,703 | • Show Spoilers
• Night Vision
Abstract Scale
A system of measurement for something that doesn't seem like it could be measured in discrete units in the first place. Almost always used for humor. Broke the Rating Scale may invoke this when it gets silly. This is presumably how you tell if something is 20% More Awesome.
Compare Hiroshima as a Unit of Measure, which is when someone measures a quantifiable thing in a strange way.
Examples:
open/close all folders
Anime and Manga
• Scouter readings for Power Levels in Dragon Ball don't seem to just quantifying stored energy, but seems to be able to quantify fighting effectiveness, as evidenced by Goku's and Picolo's power levels registering higher after they take off their weighted clothing. The author states they are just supposed to be measurements of raw power, and thus an unreliable oversimplification for determining fighting ability, but what's shown in-series speaks to the contrary.
Blogs
• Deadcoders Reviews used Aelita as a measurement of Trauma and Hollywood as a measure of makeup.
Film
• At the end of the first Men in Black movie, J quantifies the battle with the Bug as ranking 9.5 on the "Weird-Shit-O-Meter".
• In Team America: World Police, terrorist acts are measured in multiples of the September 11 World Trade Center attack. One upcoming terrorist event, if not stopped, would be "Nine Eleven times a hundred" ("basically all the worst parts of The Bible"). Another is "Nine Eleven times a thousand." Kim Jong-Il described his ultimate plan, which involves simultaneous terrorist attacks with weapons of mass destruction across the world as "Nine Eleven times 2356."
• In The Abyss, Hippie measures one of the encounters with the aliens as a 9.5 on the Sphincter Scale.
Literature
• Beauty has been measured in milliHelens (the amount of beauty needed to launch a single ship) in The Rebel Angels, a novel by Robertson Davies. This system was invented by Cambridge mathematician W.A.H. Rushton; the term was possibly first suggested by Isaac Asimov. Irregular Webcomic!, however, pointed out that you shouldn't mix metric prefixes with Troy units.
• Note that Helen herself scores 1.186 Helens.
• Discworld has the thaum, a basic unit of magical strength "universally established as the amount of magic needed to create one small white pigeon or three normal-sized billiard balls." Naturally, it is measured by thaumometers. Some early books measure magic in "Primes", the amount of magic needed to move one pound of lead one foot. The Companion eventually explained that the Prime was the metric measurement, and the thaum is the "imperial" one. Wizards tending to be hidebound traditionalists, the Prime never caught on.
• A few straighter examples are found in Moving Pictures. "Numbers" Riktor was a mad wizard who believed everything could be measured, and created such devices as the "swamp meter".
• In later books, the thaum is regarded as a fundamental particle of magic rather than a unit, though they might happen to coincide.
• Good Omens gives us the alp as a way of measuring belief (in reference to the notion that "faith moves mountains"). Most people are only able to generate millialps, while more passionate believers like Anathema Device able to raise about half an alp. Adam's belief is measured in Everests.
• An amateur sci-fi writer group on LiveJournal attempted to come up with "Brian's Scale" to measure the fame of sci-fi authors, based on number of non-self-publications, with units ranging from the Trout to the Scalzi to finally the Gaiman.
• In The Schrodinger's Cat Trilogy, the sequel to The Illuminatus! Trilogy, Dr. Dashwood is a sex researcher, whose scales are named after pornographic stars:
"Sincerity we measure in Spelvins on a scale of zero to ten," Dashwood went on, totally absorbed in his subject. "Hedonism in Lovelaces-we've been lucky there; subjects are able to distinguish sixteen graduations. Finally, there's the dimension of Tenderness-we find zero to seven covers that, so that the perfect Steinem Job, if I may use the vernacular, would consist of ten Spelvins of Sincerity, sixteen Lovelaces of Hedonism, and seven Havens of Tenderness."
• In Stanislaw Lem's short story Experimenta Felicitologica, the protagonist uses a unit he calls "hedones" to measure the happiness of a being at a given time. His professor uses a unit called "bromeons" for the same purpose.
• America (The Book) book gauged the value of a news story in Buttafuocos.
• Words of Radiance: According to Wit, Shalan Davar is exactly 77% more agreeable than her mistress, Jasnah Kholin. He took a poll.
Jasnah: You're lying.
Wit: Okay, so it was a rather informal poll. But the ugly lizard-crab-thing gave you really poor marks for—
• In The Bible Jesus refers to "faith as small as a mustard seed", implying faith is measured by volume.
Live-Action TV
• Mystery Science Theater 3000: The Angels' Revenge episode had Tom Servo's "shame-o-meter", for measuring the amount of shame emanating from washed-out actors in a 70s Charlie's Angels rip-off movie. It measured in (Peter) Lawfords. Jim Backus registered well into the giga-Lawford range.
• One episode of The Colbert Report had Stephen rating Nazis on a scale of 1 to 10 Hitlers. Adolf Hitler himself got only 9 Hitlers because "nobody gets 10 Hitlers."
• The Daily Show with Trevor Noah has "What the Actual Fact?", a recurring segment in which correspondent Desi Lydic rates the Democrats' and Republicans' statements, based on how verifiably true they are. She's rated claims with "a Teletubby", "Chinese character tattoo at a strip mall", and "a hotel porn channel".
Tabletop Games
• Rather common in roleplaying games featuring magic or anything like it. In Dungeons & Dragons your basic magical ability can come from Intelligence, which is at least something people seriously attempt to measure in real life, but also Wisdom, which has something to do with good moral sense, or maybe attunement to nature if you're a druid, or the ability to see the broad picture - anyway divination spells are based on it. The third potential stat tied to casting spells is Charisma, which is both the measure of one's force of personality and how pretty they are.
• Certain spells, as well as the creation of magic items, require using materials "worth" a certain number of gold pieces, despite the value of resources (as well as the gold they are equivalent to) being relative and subjective. For instance, a Raise Dead spell requires 5000 gp worth of diamonds, so better hope a new diamond mine opening doesn't cause prices to drop! note
• Genius: The Transgression has several different unit systems (most likely a reference to all the different temperature scales) to measure "mania".
Video Games
• In Super Paper Mario, when Dimentio first brings the heroes to Dimension D, He believes that his power has increased by 256 times, though it actually increases everyone's power by that amount and he claims that he could obliterate the heroes with the amount of power it would take to lift an eyebrow.
Webcomics
• An archon in The Order of the Stick measured evil in terms of kilo-nazis, with a baseline of a hypothetical offspring of Cruella de Vil and Sauron clocking in at an average of 5 (that means as evil as 5000 nazis). The person being discussed, Belkar, for comparison had his worst at around 3400 nazis before meeting his new friend and boss Roy.
• Relating to the Dungeons & Dragons example above, an apprentice was sent to get some diamonds, and proudly reveals that they were on sale...only to be sent back for more, because the sale means they aren't "worth" enough to be a spell component.
• Early on in Schlock Mercenary, there is frequent mention of an absolute system of measurement for pain in Kill-o-Hurtz. Various medical instruments are rated according to how much pain they inflict, as measured by the Ouchdammitometer. The concept hasn't been mentioned in a long time, however, as the Schlock Mercenary universe (if not the actual story) has become somewhat more "serious".
• In Casey and Andy, Andy names the "fundamental unit of time travel" after himself (and the fundamental unit of stupid after Casey).
• This Saturday Morning Breakfast Cereal measures anger in miliHulks, fear in deciPantscraps and awkwardness in microWeiners (all in the "votey", the red button).
• MegaTokyo features a Magical Girl Detector that's calibrated in sparklogems, of course.
• El Goonish Shive:
• Questionable Content:
• Gets this now and again with its odd Twenty Minutes In The Future setting; usually coming from Hannelore, who grew up on a space station. For example, the current arc's "Fournier-Goldman Happiness Transforms", which measure happiness (Marten makes an attempt at calculating Hannelore's happiness for her father's benefit, but he couldn't follow the material after it brought up Irrational Birthday Integers).
• Faye in this installment claims to be a unit of measurement, but the formula is rather complex.
• Paranatural has the international scale of creepy to measure PJ's smile.
• One Achewood comic cites the made-up unit "the fermule" as "the basic unit of physics." One reader lampshades the silliness of this in the comments section: "That's right. Losing control of a 200 kg van on an icy road while traveling at 45 mph, skidding off a 45 degree turn and wrapping that van three times around a tree takes a total of 67 fermules of physics."
• In Ansem Retort, a plan devised to get Zexion elected governor of Pennsylvania by getting his opposition murdered is measured by Zexion in Michael Corleones.
Web Original
Western Animation
• Futurama once had a device that measured coolness in mega-Fonzies. There was also the funkometer for smells and a device that measured the musical talents of the Beastie Boys.
Leela: Impressive! They're laying down mad rhymes with an 80% success rate.
Bender: I believe that qualifies as "ill," at least from a technical standpoint.
• The What-If Machine can answer any what-if question accurate to within one tenth of a plausibility unit.
• In "Xmas Story", the department store sells a jolly-seeking missile launcher as an anti-Santa defense.
• It's stated in Bender's Big Score that a normal person emits about 10 millidooms per second. It becomes a plot point that duplicates created by the Time Sphere emit much larger quantities in order to prevent/correct paradoxes.
• Phineas and Ferb has devices that cause similar effects, such as Phineas' Cute-Tracker (Isabella made it overload). Santa's elves also carry meters that measure people's relative niceness or naughtiness.
Real Life
• The Hovind Scale measures the craziness of creationists. It was, of course, meant largely as a joke.
• Fame is sometimes measured in Warhols. Someone who is famous for 15,000 minutes would have one kiloWarhol.
• The Helen is a metric measurement of beauty, with 1 Helen of beauty being the amount of beauty required to launch a thousand ships. By extension, we can have a Kilohelen, which would be the amount of beauty required to launch a million ships, and a Millihelen - the amount of beauty required to launch a single ship.
• Adolf Hitler used to joke: "Do you know what is the difference between a goering and a goebbels? A goering is the amount of metal a man could pin on his chest and a goebbels is the amount of nonsense man could say in an hour." He often said the joke with both Joseph Goebbels and Hermann Goering present, to their great embarrassment.
• NASA measures resistance to Space Adaptation Sickness in Garns, where a person who rates one Garn being totally useless in microgravity.
• UNIX system load average is measured in Vastons, even though there's no meaningful or objective way to measure or compare load averages between systems.
• When a Linux computer boots, it measures the performance of the processor it is running on in "BogoMIPS", defined as "the number of million times per second a processor can do absolutely nothing". This unit is explicitly meaningless for any kind of comparison between computers (although that doesn't stop people from boasting with their values); its only purpose is to calibrate the kernel's internal busy-loop.
• The quality-adjusted life year (QALY) measures the total happiness over a span of time. It's nigh impossible to measure, but very important if you want to make sure your charity does a lot of good.
• In an inversion; there are lots of legitimate units you could use to say how fast somebody can run but in UK sport most commentators go with the 'yard of pace'.
• There's a common Facebook image macro that claims that the pain of a Groin Attack on a male is equivalent to the pain of many, many childbirths for a female, even offering an exact unit of measurement, e.g. 9000 del. Even though a "del" is not a recognized unit of measurement (pain isn't quantifiable except very generally, a la the 1-10 "scale") and even though the macro wasn't even remotely intended as serious, that hasn't stopped a lot of people from taking it as fact.
• Quality rating scales for films, games, restaurants, video games, apps, products, and so on.
• In the field of psychology, it's possible to create a rating scale for pretty much any human trait, so researchers sometimes end up creating unusual and offbeat scales for things that they want to study but few others attempt to measure. There are widely used and thoroughly validated scales for things like personality and emotion, of course, and slightly more offbeat measures like how much of a morning person you are or how much you conform to your designated gender role, but also weirder things like enjoyment of some random consumer product, experiences with alien abduction, obedience to political leaders, correlations between creativity and psychoticism, etc. Psychologists who measure such bizarre constructs sometimes get the Hard on Soft Science treatment and/or It Will Never Catch On.
Other
• The (sometimes) clever bits of wordplay known as "Sniglets" (as introduced by comedian Rich Hall on the '80s HBO program Not Necessarily the News) have featured a few of these, such as:
BEVAMETER: The distance a drink coaster, attached to the bottom of a wet glass, will travel before it falls back to earth.
CRAVAMETER: 3.72 inches, the proper distance between the ends of a tied tie.
http://tvtropes.org/pmwiki/pmwiki.php/Main/AbstractScale | 3,226 | 14,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-05 | longest | en | 0.952116 |
https://calculat.io/en/temperature/celsius-to-fahrenheit/270 | 1,726,328,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.38/warc/CC-MAIN-20240914125424-20240914155424-00786.warc.gz | 136,210,512 | 24,690 | # Convert 270 Celsius to Fahrenheit
## What is 270 Celsius in Fahrenheit?
Answer: 270 degrees Celsius is equal to 518 degrees Fahrenheit
270°C = 518°F
## Explanation of 270 Celsius to Fahrenheit Conversion
Celsius to Fahrenheit Conversion Formula: °F = (°C × 9/5) + 32
According to Celsius to Fahrenheit Conversion formula if you want to convert 270 (two hundred seventy) degrees Celsius (°C) to degrees Fahrenheit (°F) you have to multiply 270 by 9/5 (you can first multiply 270 by 9 and then divide the result by 5). And finally you have to add 32 to it.
Here is the complete solution:
((270 × 9) ÷ 5) + 32
=
(2430 ÷ 5) + 32
=
486 + 32
=
518
(five hundred eighteen degrees Fahrenheit )
## Celsius to Fahrenheit Conversion Table
Celsius degreesFahrenheit degrees
491°F
256°C492.8°F
257°C494.6°F
258°C496.4°F
259°C498.2°F
261°C501.8°F
262°C503.6°F
263°C505.4°F
264°C507.2°F
265°C
266°C510.8°F
267°C512.6°F
268°C514.4°F
269°C516.2°F
518°F
271°C519.8°F
272°C521.6°F
523.4°F
274°C525.2°F
276°C528.8°F
277°C530.6°F
278°C532.4°F
279°C534.2°F
281°C537.8°F
282°C539.6°F
283°C541.4°F
284°C543.2°F
## About "Celsius to Fahrenheit" Calculator
This converter will help you to convert degrees Celsius to degrees Fahrenheit (c to f). For example, it can help you find out what is 270 Celsius in Fahrenheit? (The answer is: 518). Enter the number of degrees Celsius (e.g. '270') and hit the 'Convert' button.
## FAQ
### What is 270 Celsius in Fahrenheit?
270 degrees Celsius is equal to 518 degrees Fahrenheit | 479 | 1,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-38 | latest | en | 0.471219 |
https://www.ams.org/publicoutreach/feature-column/fcarc-tidesi2 | 1,537,519,706,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156901.47/warc/CC-MAIN-20180921072647-20180921093047-00132.warc.gz | 679,452,025 | 9,278 | Fourier Analysis of Ocean Tides I
## 2. Tides and gravitation
From George H. Darwin's article "Tides," Encyclopaedia Britannica, IX Ed.,Chicago, 1890. Note that in this graph, time increases to the left. Larger image.
Tides. This figure shows the tidal record for two weeks (January 1-14, 1884) at Bombay. The tide was recorded on a cylindrical sheet that turned once every 24 hours. Each daily curve is labelled with its date. Some obvious features:
• there are usually two high tides and two low tides each day;
• the tides come about 50 minutes later each day;
• during the two-week period there is considerable variation in the daily pattern of highs and lows;
• there is usually a difference in height between two consecutive high tides (the diurnal inequality).
Gravitation. The tides are caused by the gravitational fields of of the Sun, the Moon and the Earth, acting on the Earth's oceans.
The yearly travel of the Earth around the sun, the monthly travel of the Moon around the Earth, and the daily rotation of the Earth on its axis are the three main rythms that govern the tides. (It is impossible to draw the Sun-Earth-Moon system to scale. To scale, the Sun is a basketball in the center of a football field, the Earth is the eraser on a pencil held on the sidelines, and the Moon is the tip of a straightened-out paper clip stuck in the eraser.)
In fact the "tidal force" depends on the gradient of the gravitational field. Because (in the case of the lunar tide) what creates the bulge in the Earth's oceans on the side facing the Moon is the fact that the surface of the Earth is closer to the Moon than the center, and is therefore attracted more strongly. This also explains the bulge on the opposite side: there the center is closer than the surface. In general a variable gravitational field will stretch bodies along its gradient; our ocean tides are a special case of this effect.
Why the gradient? The gravitational force exerted by a body of mass m on a piece of matter at a distance L centimeters gives it an acceleration Gm/L2 cm/sec2, where G= 6.67x10-8dyne-cm2/g2 is Newton's gravitational constant.
Suppose another piece of matter is k centimeters farther in the same direction. Then it undergoes a smaller acceleration: Gm/(L+k)2 cm/sec2. The difference between these two accelerations can be ascribed to a tidal force" pulling the two pieces apart. The estimation of this difference is a classic example of the linear approximation f(x+k) - f(x) ~ k f'(x). It yields
-2 G m k --------- cm/sec2. L3
• This calculation explains why the Sun (m = 2x1033, L = 1.5x1013, m/L3= 6x10-7) and the Moon (m = 7.3x1025, L = 3.8x1010, m/L3=13.3x10-7) exert comparable tidal forces at the surface of the Earth.
Welcome to the
Feature Column!
These web essays are designed for those who have already discovered the joys of mathematics as well as for those who may be uncomfortable with mathematics. | 724 | 2,995 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-39 | latest | en | 0.92928 |
https://www.jiskha.com/questions/22256/I-have-a-problem-that-I-cant-seem-to-figure-out-Could-you-show-me-how-its-done | 1,571,072,520,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653876.31/warc/CC-MAIN-20191014150930-20191014174430-00389.warc.gz | 973,414,824 | 5,141 | # algebra
I have a problem that I can't seem to figure out. Could you show me how it's done? Thank You :)
5sqrt(3x) + 3sqrt(3x)
What is 5a + 3a? Answer: 8a
1. 👍 0
2. 👎 0
3. 👁 80
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https://brainly.com/question/317027 | 1,485,220,123,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560283475.86/warc/CC-MAIN-20170116095123-00026-ip-10-171-10-70.ec2.internal.warc.gz | 781,182,048 | 9,358 | 2015-02-22T14:46:07-05:00
To do this you find out 10% of £68.
To find out 10% you divide by 10.
68/10=6.8.
So, 10% of £68 is 6.8
To find 40% you times 6.8 by 4.
6.8*4=27.2
40% of £68 is 27.2
To reduce something you're taking away something, in this case, 40%
£68-27.2=40.8
2015-02-22T14:58:10-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
The price of the jacket before the sale 68.0
The amount of reduction = 68.0 * 40/100 = 27.2
The price of the jacket after the reduction = 68.0 - 27.2 = 40.80 | 259 | 767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-04 | latest | en | 0.932112 |
https://convertoctopus.com/551-kilometers-to-meters | 1,657,066,888,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104655865.86/warc/CC-MAIN-20220705235755-20220706025755-00406.warc.gz | 231,749,032 | 7,399 | Conversion formula
The conversion factor from kilometers to meters is 1000, which means that 1 kilometer is equal to 1000 meters:
1 km = 1000 m
To convert 551 kilometers into meters we have to multiply 551 by the conversion factor in order to get the length amount from kilometers to meters. We can also form a simple proportion to calculate the result:
1 km → 1000 m
551 km → L(m)
Solve the above proportion to obtain the length L in meters:
L(m) = 551 km × 1000 m
L(m) = 551000 m
The final result is:
551 km → 551000 m
We conclude that 551 kilometers is equivalent to 551000 meters:
551 kilometers = 551000 meters
Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 meter is equal to 1.8148820326679E-6 × 551 kilometers.
Another way is saying that 551 kilometers is equal to 1 ÷ 1.8148820326679E-6 meters.
Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that five hundred fifty-one kilometers is approximately five hundred fifty-one thousand meters:
551 km ≅ 551000 m
An alternative is also that one meter is approximately zero times five hundred fifty-one kilometers.
Conversion table
kilometers to meters chart
For quick reference purposes, below is the conversion table you can use to convert from kilometers to meters
kilometers (km) meters (m)
552 kilometers 552000 meters
553 kilometers 553000 meters
554 kilometers 554000 meters
555 kilometers 555000 meters
556 kilometers 556000 meters
557 kilometers 557000 meters
558 kilometers 558000 meters
559 kilometers 559000 meters
560 kilometers 560000 meters
561 kilometers 561000 meters | 404 | 1,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-27 | latest | en | 0.840288 |
https://anewsstory.com/how-to-play-satta-matka-online/ | 1,722,906,538,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640461318.24/warc/CC-MAIN-20240806001923-20240806031923-00360.warc.gz | 72,450,266 | 15,378 | All
# How to Play Satta Matka Online?
Satta Matka Online has become the talk of the town these days. It has rapidly gained popularity
and is in great demand. The people love the simplicity and less complexity of the game and
focus on playing the suitable match to earn good money. However, it is widely known that lottery
games do not always offer a win-win situation, but there has been incredible hype about it. The
websites and applications are striving hard to make the Online Matka experience an excellent
hit for everyone around the globe.
Let’s have a look at what Satta Matka is and how it is played!
## What is Satta Matka?
Satta Matka had been widely known as the Ankara Play in the 1950s. With the increase in the
number of players and innovations in the game, it received a considerable push and evolved
into something extraordinary which no one thought of before. As the game changed, only the
name of the game remained as Matka. The Matka game being played these days is a simple
lottery game dependent on the correct choices about the numbers when putting the bets.
The winning numbers and the winning game are dependent on the skills and analysis of the
player. There are different choices of numbers which you can make in this kind of numbering
game, such as open and close, jackpot, Jodi, Sangam panel, and lots of others. There is not
just one set of numbers to follow; you can choose any number in any way. All you have to do is
stick to one secret strategy in the game to ace all bets and make a win. If you follow the winning
rules and numbers, you are going to succeed.
## How to Play Satta Matka Online?
For playing the Online Satta Matka in India, there are multiple open and legit options available in
the market. If you wish, you can play Satta Matka safely at Lottoland but the first thing you
should do is learn the games techniques and basics.
1. In the first step, you will have to pick three numbers from the standard range of 0-9. For
instance, you choose the numbers like 3,5, and 6 in the Matka Open range.
2. To add more thrill to the game, the sum of chosen numbers is taken out. In this case, the
sum will be 14.
3. From the sum, you can either choose the first digit or the second digit. Once you have
selected the number, your first draw will look like 3,6, and 5*4. Now, your first draw is
4. You will now repeat the whole process for the second draw.
5. When both of your draws are ready, you can place the bet.
## What must you keep in mind?
When you begin to play the Matka Online game, you must focus on some of the significant
points which are mentioned as follows:
1. To stay in the game for long and play safely, you must follow the rules and regulations of
the game. If you fail to understand the rules or have trouble understanding the basics,
you must opt for the introductory classes, which will walk you through the whole process
of the lottery game.
2. The game's primary focus is on how fast you respond to the numbers and how quickly
you place the bets. Everything depends on your fast action, so there are equal chances
of winning or losing a bet.
3. Online Matka lottery games are highly addictive and engaging, so make sure to play
within limits and make the best use of your time and money.
4. Don’t get lured by huge cash prizes being offered. Play safe and invest small portions of
money and first learn the game and gain experience. Then only you should opt for
increasing the amounts.
5. In India, you will find different sources and websites that offer the Matka game but first,
do your research before investing in money. You might come across fake websites or
frauds in the process who will provide you with bogus numbers and run away with your
hard-earned money. | 873 | 3,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.965919 |
https://www.learncram.com/maharashtra-board/class-7-maths-solutions-chapter-3-practice-set-10/ | 1,722,988,624,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640523737.31/warc/CC-MAIN-20240806224232-20240807014232-00421.warc.gz | 675,927,950 | 12,313 | # Maharashtra Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 10
## Maharashtra State Board Class 7 Maths Solutions Chapter 3 HCF and LCM Practice Set 10
Question 1.
Which number is neither a prime number nor a composite number?
Solution:
1
Question 2.
Which of the following are pairs of co-primes?
i. 8,14
ii. 4,5
iii. 17,19
iv. 27,15
Solution:
i. Factors of 8: 1, 2, 4, 8
Factors of 14: 1, 2, 7, 14
∴ Common factors of 8 and 14: 1,2
∴ 8 and 14 are not a pair of co-prime numbers.
ii. Factors of 4: 1, 4
Factors of 5: 1, 5
∴ Common factors of 4 and 5: 1
∴ 4 and 5 are a pair of co-prime numbers.
iii. Factors of 17: 1, 17
Factors of 19: 1, 19
∴ Common factors of 17 and 19: 1
∴ 17 and 19 are a pair of co-prime numbers.
iv. Factors of 27: 1, 3, 9, 27
Factors of 15: 1, 3, 5, 15 .
∴ Common factors of 27 and 15 : 1,3
∴ 27 and 15 are not a pair of co-prime numbers.
Question 3.
List the prime numbers from 25 to 100 and say how many they are.
Solution:
29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
There are 16 prime numbers from 25 to 100.
Question 4.
Write all the twin prime numbers from 51 to 100.
Solution:
1. 59 and 61
2. 71 and 73
Question 5.
Write 5 pairs of twin prime numbers from 1 to 50.
Solution:
1. 3,5
2. 5,7
3. 11,13
4. 17,19
5. 29,31
6. 41,43
Question 6.
Which are the even prime numbers?
Solution:
2
Maharashtra Board Class 7 Maths Chapter 3 HCF and LCM Practice Set 10 Intext Questions and Activities
Question 1.
Answer the following questions. (Textbook pg. no. 15)
i. Which is the smallest prime number?
ii. List the prime numbers from 1 to 50. How many are they?
iii. Identify the prime numbers in the list below.
17, 15 ,4, 3, 1, 2, 12, 23, 27, 35, 41, 43, 58, 51, 72, 79, 91, 97.
Solution:
i. 2 is the smallest prime number.
ii. There are 15 prime numbers from 1 to 50.
They are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
iii. [17], 15 ,4, [3], 1, [2], 12, [23], 27, 35, [41], [43], 58, 51, 72, [79], 91, [97]. | 816 | 1,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2024-33 | latest | en | 0.861093 |
https://www.thestudentroom.co.uk/showthread.php?page=34&t=3054709 | 1,544,919,991,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00463.warc.gz | 1,049,549,493 | 46,787 | You are Here: Home >< Maths
# Intermediate Maths Challenge - UKMT - 5th February 2015 watch
1. My teacher was sent the answers by UKMT.
They are:
C
B
A
E
E
E
C
B
C
C
D
C
A
B
E
E
E
B
C
D
C
A
B
C
B
I got 72, pretty happy with my result.
2. (Original post by alpegu66)
My teacher was sent the answers by UKMT.
They are:
C
B
A
E
E
E
C
B
C
C
D
C
A
B
E
E
E
B
C
D
C
A
B
C
B
I got 72, pretty happy with my result.
Do you know what you need to get a merit?
Posted from TSR Mobile
3. (Original post by theworkkid)
Do you know what you need to get a merit?
Posted from TSR Mobile
Nope, it's the top 25% of participants. What score did you get?
4. Do you know when those are told?
And 82 I think
Posted from TSR Mobile
5. Hopefully that was pink, i really wanna know what the solutions are for the grey kangaroo.... -_-
6. (Original post by theworkkid)
Do you know when those are told?
And 82 I think
Posted from TSR Mobile
It's supposed to be "the next month", though I'm not sure if it's April the first or the 19th
7. Asks just realised to meant April and not march
8. If those are the answers then I've done much worse than I thought:/ I'd have gotten 59, which I guess isn't bad, but I'd've thought I'd done a little better
Posted from TSR Mobile
9. The UKMT has uploaded the solutions to the Olympiad. It's on their website. I took the Maclaurin.
My solutions to 1,2 and 3 were correct, but I'm not sure the proofs were fully explained. My solution to four, was wishy-washy, so I'm not expecting many marks there, and as for question five, I only worked out four solutions, when there were seven. Do you think I'll get any marks for question five?
10. ''The 2015 materials will be added to this page on or soon after 20 Apr 15; Kangaroo organisation rules preclude publication within a calendar month of the event.''
(this is for the pink kangaroo)
11. (Original post by Pizzaa<3)
''The 2015 materials will be added to this page on or soon after 20 Apr 15; Kangaroo organisation rules preclude publication within a calendar month of the event.''
(this is for the pink kangaroo)
and grey.....
Ugh I can't wait that long
12. (Original post by alpegu66)
My teacher was sent the answers by UKMT.
They are:
C
B
A
E
E
E
C
B
C
C
D
C
A
B
E
E
E
B
C
D
C
A
B
C
B
I got 72, pretty happy with my result.
Looks like I got 89 with the potential of more. Is that a merit?
Posted from TSR Mobile
13. 100% a merit! That's awesome!
Posted from TSR Mobile
14. (Original post by theworkkid)
100% a merit! That's awesome!
Posted from TSR Mobile
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15. i failed sooooo badly and i'm even in the fast track maths set..
Posted from TSR Mobile
16. We find out next week I think
Posted from TSR Mobile
17. Don't be disheartened if you didn't do well. The achievement is that you managed to get a high enough score in the challenge to qualify for a follow-on round. That in itself is an accomplishment!
18. Got a merit in hamilton despite the fact that I only got 1 solution fully correct
so gassed right now
19. Wooooaahhh
20. I got 73 in Pink Kangaroo, which I'm happy about! I guessed the last five and got none of those ones right though😂 do you think I'll miss out on a merit?
Posted from TSR Mobile
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http://docplayer.net/21605261-Thermodynamics-of-mixing.html | 1,531,949,141,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.25/warc/CC-MAIN-20180718193656-20180718213656-00143.warc.gz | 99,628,660 | 25,861 | # Thermodynamics of Mixing
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## Transcription
1 Thermodynamics of Mixing Dependence of Gibbs energy on mixture composition is G = n A µ A + n B µ B and at constant T and p, systems tend towards a lower Gibbs energy The simplest example of mixing: What is the Gibbs free energy if we take two pure ideal gases and mix them together? Consider gas A and gas B, both in separate containers at pressure p at temperature T. The chemical potentials are at their pure values at this point. Gibbs energy is G = n A µ A + n B µ B = n A µ o A + RT ln p p o + n µ o + RT ln p B B p o We can simplify things by letting p denote the pressure relative to p o, writing G = n A ( µ o A + RT ln p)+ n B ( µ o B + RT ln p) Prof. Mueller Chemistry Fall 2003 Lecture 16-1
2 Gibbs Energy of Mixing After mixing, the partial pressures of the gases are p A and p B, where the total pressure is p = p A + p B. The total Gibbs energy is then G = n A ( µ o A + RT ln p A )+ n B µ o B + RT ln p B ( ) The difference in Gibbs energies, G f - G i, is the Gibbs energy of mixing G = n A RT ln p A + n B RT ln p B p p We use mole fractions, replacing n J with x J n : G = nrt( x A ln x A + x B ln x B ) Since the mole fractions are never greater than 1, the ln terms are negative, and G < 0 This allows is to conclude that mixing processes are spontaneous, and gases mix spontaneously in all proportions Prof. Mueller Chemistry Fall 2003 Lecture 16-2
3 Calculating Gibbs Energy of Mixing Two containers of equal volume are partitioned from one another, with one containing 3.0 mol H 2 and the other 1.0 mol N 2 at 25 o C. Calculate the Gibbs energy of mixing when the partition is removed We assume ideal gas behavior, with pressure of N 2 being p, pressure of H 2 being 3p. Volume of each gas will double, and therefore partial pressures fall by a factor of 2 Prof. Mueller Chemistry Fall 2003 Lecture 16-3
4 Calculating Gibbs Energy of Mixing Two containers of equal volume are partitioned from one another, with one containing 3.0 mol H 2 and the other 1.0 mol N 2 at 25 o C. Calculate the Gibbs energy of mixing when the partition is removed o o G i = ( 3.0 mol ( )µ H 2 + RT ln 3p)+ ( 1.0 mol) ( µ N2 + RT ln p) o G f = ( 3.0 mol) µ H 2 + RT ln 3p o + ( 1.0 mol) µ 2 N2 + RT ln p 2 Gibbs energy of mixing is the difference between the above quantities 3p G = ( 3.0 mol) RT ln mol 3p = ( 3.0 mol)rt ln2 ( 1.0 mol)rt ln2 = ( 4.0 mol)rt ln2 = 6.9 kj ( ) RT ln p 2 p Prof. Mueller Chemistry Fall 2003 Lecture 16-4
5 Since Other Thermodynamic Mixing Functions G T p,n = S, and G = nrt( x A ln x A + x B ln x B ) it follows that for a mixture of perfect gases, the entropy of mixing is S = G T p,n A,n B Since ln x < 0, then S > 0 for mixtures of all compositions We expect this increase IF WE SPEAK ABOUT DISORDER, since dispersal of one gas into another implies greater disorder in the system. What is the preferred interpretation? Example: = nr( x A ln x A + x B ln x B ) G = ( 4.0 mol)rt ln2 = 6.9 kj S = ( 4.0 mol)rln2 =+23 J K -1 The isothermal isobaric enthalpy of mixing of two gases is found from G = H T S, and we find H = 0 This is expected for a system in which there are no interactions between the molecules in the mixture. Prof. Mueller Chemistry Fall 2003 Lecture 16-5
6 Chemical Potentials of Liquids Some nomenclature: Quantities relating to pure substances will be denoted with asterisks, so the chemical potential of pure liquid A is µ A* (l). Since the vapor pressure of pure liquid A is p A*, the chemical potential of A in the vapor is µ o + RT ln p A * (p A * as relative pressure p A* /p o ). Chemical potentials are at equilibrium: µ A = µ A o + RT ln p A If another substance is present (e.g., a solute in the liquid), chemical potential of A in the liquid is µ A and vapor pressure is p A µ A = µ A o + RT ln p A Combine to eliminate the standard potential: µ A = µ A + RT ln p A p A The chemical potential of A depends on its partial vapor pressure. It follows that chemical potential of liquid A is related to its partial pressure Prof. Mueller Chemistry Fall 2003 Lecture 16-6
7 Raoult s Law After experimenting with mixtures of similar liquids, the French chemist Francois Raoult found that the ratio of the partial vapor pressure of each component to its vapor pressure as a pure liquid, p A /p A*, is approximately equal to the mole fraction of A in the liquid mixture p A = x A p A Some mixtures obey Raoult s law very closely, especially when they are structurally similar these are called ideal solutions, and they obey µ A = µ A + RT ln x A Prof. Mueller Chemistry Fall 2003 Lecture 16-7
8 Molecular Interpretation of Raoult s Law We have to consider the rate at which molecules leave and enter solution to understand Raoult s law The presence of a second component actually hinders molecules from leaving the solution, but does not inhibit the rate at which they return The rate at which A molecules leave the surface is proportional to the number of them at the surface, which is proportional to their mole fraction rate of vaporization = k x A where k is a proportionality constant. The rate at which molecules return is proportional to their gas phase concentration, which is proportional to their partial pressure rate of condensation = k p A At equilibrium, rates of vaporization and condensation are equal, so For a pure liquid, x A = 1, so kx A = k p A p A = kx A k p A = k k Prof. Mueller Chemistry Fall 2003 Lecture 16-8
9 Departure from Raoult s Law, Dilute Solutions Some solution mixtures behave very differently from Raoult s law, notably with mixtures of structurally and chemically dissimilar liquids Mixture of CS 2 and acetone: When the solute is dilute, and solvent obeys Raoult s Law very closely When the solvent is nearly pure, it has a vapor pressure proportional to the mole fraction with slope p B * When it is the minor component (the solute) the vapor pressure has a different constant of proportionality Prof. Mueller Chemistry Fall 2003 Lecture 16-9
10 Henry s Law The English chemist William Henry found that for real solutions at low concentrations, although the vapor pressure of the solute is proportional to the mole fraction, the constant of proportionality is not the vapor pressure of the pure substance p B = x B K B Here x B is the mole fraction of the solute B, and K B is an empirically determined constant with pressure dimensions. K B is chosen so that a plot of vapor pressure of B against mole fraction is tangent to the experimental curve at x B = 0 Such solutions are called ideal-dilute solutions Solvent behaves like a slightly modified pure liquid Solute behaves entirely differently from its pure state (since its molecules are surrounded by solvent molecules), unless the molecules of the two components are very similar Prof. Mueller Chemistry Fall 2003 Lecture 16-10
11 The Validity of Raoult s & Henry s Laws Consider an acetone (A) and chloroform (C) mixture at 35 o C x C p C (Torr) p A (Torr) K = 175 Torr for (A) and K = 165 Torr for (C) When mole fractions of (A) or (C) are near 1, Raoult s Law predicts the partial pressures When mole fractions of (A) or (C) are near 0, Henry s Law predicts the partial pressures In both cases, there are some deviations from this predicted behavior Prof. Mueller Chemistry Fall 2003 Lecture 16-11
12 Using Henry s Law Estimate the molar solubility of oxygen in water at 25 o C and partial pressure of 120 Torr, its partial pressure at sea level The amount of O 2 dissolved is very small, and x J = p J /K J, so x( O 2 )= ( ) ( ) n ( O 2) n( H 2 O) n O 2 n( O 2 )+ n H 2 O n( O 2 )= x( O 2 )n( H 2 O)= p O 2 ( )n H 2 O K ( ) = ( ) Torr ( 160 Torr) 55.5 mol kg -1 = mol kg -1 So the molar solubility of oxygen is 2.69 x 10-4 mol kg -1 corresponding to concentration 2.7 x 10-4 mol L -1 Henry s law constants are very well known by biochemists for treatment of gas behavior in fats and lipids - very important in understanding respiration processes: especially when partial pressure of O 2 is abnormal, such as in mountaineering, scuba diving, and in the use of gaseous anaesthetics. Prof. Mueller Chemistry Fall 2003 Lecture 16-12
### Thermodynamics and Kinetics. Lecture 14 Properties of Mixtures Raoult s Law Henry s Law Activity NC State University
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### Chapter 12 Solutions
Chapter 12 Solutions 12.1 Ideal solutions Solutions are arguably the most important kind of system studied by chemical engineers, whether in large chemical processing or cellular-level molecular phenomena. | 8,583 | 34,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2018-30 | longest | en | 0.86376 |
https://se.mathworks.com/help/stats/nancov.html | 1,675,561,619,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500158.5/warc/CC-MAIN-20230205000727-20230205030727-00570.warc.gz | 512,863,248 | 19,037 | # nancov
(Not recommended) Covariance ignoring `NaN` values
`nancov` is not recommended. Use the MATLAB® function `cov` instead. With the `cov` function, you can specify whether to include or omit `NaN` values for the calculation. For more information, see Compatibility Considerations.
## Syntax
```Y = nancov(X) Y = nancov(X1,X2) Y = nancov(...,1) Y = nancov(...,'pairwise') ```
## Description
`Y = nancov(X)` is the covariance `cov` of `X`, computed after removing observations with `NaN` values.
For vectors `x`, `nancov(x)` is the sample variance of the remaining elements, once `NaN` values are removed. For matrices `X`, `nancov(X)` is the sample covariance of the remaining observations, once observations (rows) containing any `NaN` values are removed.
`Y = nancov(X1,X2)`, where `X1` and `X2` are matrices with the same number of elements, is equivalent to `nancov(X)`, where `X = [X1(:) X2(:)]`.
`nancov` removes the mean from each variable (column for matrix `X`) before calculating `Y`. If n is the number of remaining observations after removing observations with `NaN` values, `nancov` normalizes `Y` by either n – 1 or n , depending on whether n > 1 or n = 1, respectively. To specify normalization by n, use ```Y = nancov(...,1)```.
`Y = nancov(...,'pairwise')` computes `Y(i,j)` using rows with no `NaN` values in columns `i` or `j`. The result `Y` may not be a positive definite matrix.
## Examples
Generate random data for two variables (columns) with random missing values:
```X = rand(10,2); p = randperm(numel(X)); X(p(1:5)) = NaN X = 0.8147 0.1576 NaN NaN 0.1270 0.9572 0.9134 NaN 0.6324 NaN 0.0975 0.1419 0.2785 0.4218 0.5469 0.9157 0.9575 0.7922 0.9649 NaN```
Establish a correlation between a third variable and the other two variables:
```X(:,3) = sum(X,2) X = 0.8147 0.1576 0.9723 NaN NaN NaN 0.1270 0.9572 1.0842 0.9134 NaN NaN 0.6324 NaN NaN 0.0975 0.1419 0.2394 0.2785 0.4218 0.7003 0.5469 0.9157 1.4626 0.9575 0.7922 1.7497 0.9649 NaN NaN```
Compute the covariance matrix for the three variables after removing observations (rows) with `NaN` values:
```Y = nancov(X) Y = 0.1311 0.0096 0.1407 0.0096 0.1388 0.1483 0.1407 0.1483 0.2890```
## Version History
Introduced before R2006a
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# b= 11/13e what % is b less than the combined of b and e
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b= 11/13e what % is b less than the combined of b and e
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29 May 2006, 15:35
[quote="withme"]b= 11/13e what % is b less than the combined of b and e[/quote]
What exactly does "combined" mean here? Is it is the sum, product? Also is the question; b = (11/13)e or 11/13e
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Re: Maths Question - % - PS question [#permalink] 29 May 2006, 15:35
Display posts from previous: Sort by | 493 | 1,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2017-43 | latest | en | 0.884601 |
https://dsp.stackexchange.com/questions/29625/combination-of-accelation-signals-from-two-devices | 1,632,615,772,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057787.63/warc/CC-MAIN-20210925232725-20210926022725-00672.warc.gz | 286,398,330 | 39,051 | # Combination of accelation signals from two devices
I have $x_A$, $y_A$, and $z_A$ acceleration signals collected from a smartphone and I call it: set $A$ and I also have $x_B$, $y_B$, and $z_B$ same acceleration signals from smartwatch: let's call this set $B$. I need to combine them together to obtain set $C$, but since I don't have good knowledge in signal processing I am not sure if it is this logically correct to combine them and take the average of them, or any other way should I follow to have one set of signals instead of having two separate sets?
How to select the set if it is possible? The signals are raw data.
• One search term you could try is "sensor fusion". Mar 22 '16 at 7:12
• What are you trying to achieve? You have the movement of the watch and the movement of the phone. You try combine them to get the movement of what exactly? Mar 22 '16 at 12:37
• Do you try to combine them online, or does offline suffice? Mar 22 '16 at 19:43
• I am trying to find which combination of devices give optimal result in regard to detect the subject's activities. I have more than these two sensors but I just ask about these two and then if I got an answer then apply it on all. I will combine smartphone with smartwatch, smartphone with sensor placed on upper arm and etc. Regarding the online or offline not sure which to use, but after collecting the raw data I want to know if it is better to do the combination after or before extracting features.
Mar 23 '16 at 8:06
Mar 25 '16 at 4:45
May be you can find variance of acceleration along x,y, and z direction in SET A and in set B separately and choose the the device which has least variance.
As this Wikipedia page suggests, you can combine the two by weighting by their respective variances.
So, to get $x_C$ simply: $$x_C = \frac{1}{\sigma_1^{-2} + \sigma_2^{-2}} \left( \frac{x_A}{\sigma_1^2} + \frac{x_B}{\sigma_2^2}\right)$$
Putting some numbers on this, suppose $\sigma^2_1 = 1$ and $\sigma^2_2 = 0.5$, then $$\sigma_3^2 = \frac{1}{\sigma_1^{-2} + \sigma_2^{-2}} = \frac{1}{1 + 2} = \frac{1}{3}$$ which is a lower variance than either of the individual measurements.
Use the same process to find $y_C$ and $z_C$.
• thank you for your comment, but I need to have both and I want to find a way to combine them and to see how can they give me same result when they are combined comparing to the individual results. What I'm doing is that I am detecting activities from both the smartphone and the smartwatch, and I am able to detect the right activity when using each device's data separate and now I want to know how to combine their signals and run my algorithm to see if it can detect the right activities or not. | 699 | 2,699 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-39 | latest | en | 0.936168 |
https://www.kidport.com/Grade7/Math/Lesson_6__Word_Problems__Percents.html | 1,561,448,362,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999814.77/warc/CC-MAIN-20190625072148-20190625094148-00357.warc.gz | 795,151,889 | 5,031 | << Back to Lessons Index
# 7th Grade Math / Lesson 6: Word Problems: Percents
Word Problems What will we be learning in this unit? In this lesson you will learn how to solve problems involving percents. Vocabulary words are found in this purple color throughout the lesson. Remember to put these in your notebook.
Percents Percentages are ratios that compare numbers to 100. The word percent means parts per hundred. Suppose you got a 7 out of 9 on a quiz and you wanted to know what percent that was. You would set it up as a proportion like this:7 = ? This is comparing 7 out of 9 to "how many"9 100 out of 100 (percent). Or in other words, if 9 was equivalent to 100, 7 would be equivalent to what?These proportions are set up using the following model: Portion that you have/want = % Total that you should have/want 100
Percents Percentages are ratios that compare numbers to 100. The word percent means parts per hundred. Suppose you got a 7 out of 9 on a quiz and you wanted to know what percent that was. You would set it up as a proportion like this:7 = ? This is comparing 7 out of 9 to "how many"9 100 out of 100 (percent). Or in other words, if 9 was equivalent to 100, 7 would be equivalent to what?These proportions are set up using the following model: Portion that you have/want = % Total that you should have/want 100
Percents Use the model to set up the problems. Portion = % Total 100 1) There were 34 students out of 146 that attended the dance. What percent of the students attended? 2) 55% of the people were from the south. There were 60 people altogether. How many were from the south?
Percents Use the model to set up the problems.Portion = %Total 100Examples1) There were 34 students out of 146 that attended the dance. What percent of the students attended?34 = % portion = 34, total = 146146 1002) 55% of the people were from the south. There were 60 people altogether. How many were from the south?Portion = 55 portion is unknown, total = 60 60 100
Percents To solve these proportions:If we go back to our quiz example. You got 7 out of 9. What percent is that?7 = ? Find two numbers on a diagonal that you have.9 100 The 7 and the 100.
Percents
Percents | 614 | 2,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2019-26 | latest | en | 0.952608 |
http://mathhelpforum.com/algebra/51825-solved-urgent.html | 1,524,613,057,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947421.74/warc/CC-MAIN-20180424221730-20180425001730-00218.warc.gz | 211,023,744 | 9,746 | 1. ## [SOLVED] Urgent!
Hey guys
This is on section 6 #6
At the first stop after after a school bus left school, 4 children got off the bus. One-quarter of the remaining children got off at the second stop. And 10 children got off at the third stop. If there were still 14 children on the bus after the third stop, how many children were on the bus as it left the school?
let x denote the children on the bus as it left the school
[1/4(x-4)]-10=24
2. Your mistake is that you were counting the children who left the bus rather than the children who remained. Look carefully here:
When bus left the school, x children were in the bus.
When bus reached the first stop, x - 4 children remained in the bus
When bus reached the second stop, (1/4)(x-4) left and (3/4)(x-4) remained.
When bus reached that third stop, (3/4)(x-4) - 10 remained.
(3/4)(x-4) - 10 = 14
Solve for x and we're done. | 254 | 892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-17 | latest | en | 0.976686 |
https://encyclopedia2.thefreedictionary.com/Iterated+integrals | 1,544,874,053,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826856.55/warc/CC-MAIN-20181215105142-20181215131142-00539.warc.gz | 588,927,769 | 11,010 | # Multiple Integral
(redirected from Iterated integrals)
## multiple integral
[′məl·tə·pəl ′int·ə·grəl]
(mathematics)
An integral over a subset of n-dimensional space.
## Multiple Integral
an integral of a function defined on some region in a plane and in three-dimensional or n -dimensional space. The corresponding multiple integrals are referred to as double integrals, triple integrals, and n-tuple integrals, respectively.
Let the function f(x, y ) be defined on some region D of the plane xOy. Let us divide D into n subregions di whose areas are equal to si, choose a point (ξi, ηi) in each subregion di (see Figure 1), and form the integral sum
If as the maximal diameter of the subregions d, decreases without bound the sums S have a limit independent of the choice of the points (ξi, ηi), then this limit is called the double integral of the function f(x, y) over the region D and is denoted by
∫ ∫Df (x,y) ds
A triple integral and, in general, an n -tuple integral are defined analogously.
Figure 1
In order for the double integral to exist, it is sufficient that, for example, the region D be a closed (Jordan) measurable region and that the function f(x, y) be continuous throughout D. Multiple integrals possess a number of properties similar to those of ordinary integrals. In order to calculate a multiple integral we reduce it to an iterated integral. Green’s formulas and the Green-Ostrogradskii theorem can be used in special cases to reduce multiple integrals to integrals of lower dimension. Multiple integrals find wide application. Volumes of bodies, as well as masses, static moments, and moments of inertia ( of bodies, for example) are expressed using multiple integrals.
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Open / Close | 414 | 1,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2018-51 | latest | en | 0.851892 |
http://www.sudokudaddy.com/printable-sudoku-puzzles/free-medium-puzzles/18-medium-sudoku-puzzles-menu.php | 1,369,147,029,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700107557/warc/CC-MAIN-20130516102827-00077-ip-10-60-113-184.ec2.internal.warc.gz | 719,492,244 | 6,946 | # 18 x Medium Sudoku Puzzles
## Free Medium Difficulty Printable Sudoku Puzzles + Answers
### 5 Approaches for Solving Sudoku Puzzles
Sudoku is one of the most trendy numeral puzzle games right now. This game entails placing in numbers into blank positions. Now, the challenge lies in the restrictions it imposes.
1. To start with, no identical digit from 1 to 9 should be repeated in the horizontal lines.
2. Secondly, no equivalent numbers should be seen in the vertical rows as well.
3. And lastly, none of the 1 to 9 digits must appear in the 3x3 sub-squares of the puzzle.
Since completing sudoku demands logic, then it is but understandable that a number of special techniques have been devised to help in quickly solving sudoku without making any mistakes along the way.
## Popular Techniques for Completing Sudoku Puzzles
### 1. The Simple Solving Techniques
These procedures are generally effective for beginner level sudoku puzzles. Given that the majority of the digits have already been supplied, the sudoku puzzle can then be easily concluded. Some examples of these easy solving sudoku tactics are as follows:
#### The Single Candidate System
Like the name suggests, this requires finding out accurately what figure should go in a specific cell. The most resourceful way to perform this method is to first recognize rows, columns, or sub-squares that have been complete with the most digits already. From there, you can quickly presume valid values to place in the particular unfilled cells.
#### The Single Position Approach
This technique will entail you to check rows, columns, or sub-squares that are previously partly filled. This way, you get to identify precisely which digits go in which vacant cells.
### 2. The Medium Solving Approaches
These sets of solving sudoku techniques are your next approach of attack if you are working on more difficult sudoku puzzles. Furthermore, these medium level approaches can be employed as well if your pencil markings have proven to be a little incompetent in placing valid numbers in the different cells.
A number of these methods are detailed below:
#### The Candidate Lines Attack
Much like the single candidate method, performing this approach to help you in completing sudoku puzzles requires checking of the encircling cells.
But instead of merely searching for helpful clues in the immediate cells, you will have to look meticulously at the cells that are situated farther than the cell you are attempting to crack. This way, you get to properly eliminate which number guesses have void placements so that you get to invariably trim down your options.
#### The Multi-Line Approach
This method will usually require you to take a glance at three units of closely located sub-squares so that you get to rapidly spot which lines can be inserted with exacting digits.
#### The Double Pair System
Much like the multi-line approach, this system for speedily solving sudoku puzzles requires you to come up with two number estimates for a cell, after which you use these to try and deduce the possible solutions in the nearest surrounding cells.
### 3. The Superior Methods for Solving Sudoku
If the first few techniques we described above do not aid you in distinguishing suitable number placements, then maybe it’s time for you to use much cleverer systems for completing sudoku puzzles.
#### The Naked Pairs or Triples Approach
Grab your pencil and begin annotating unfilled boxes of the potential values. Now, the secret here is to look for precise number estimates that appear in precise same combinations in other cells. Usually, these numeral guesses are either in two’s or in three’s.
Now, bear in mind that these pairs or triples can not be inserted in any other cells except the ones they have been annotated in. Because of this, you can then swiftly eradicate these pairs or triples from further annotated cells thereby aiding you in rapidly and methodically completing sudoku cells.
#### The Hidden Pairs or Triples Procedure
Now, this system is much more hard to utilize when completing sudoku puzzles. This is because instead of seeing straight away figures in pairs or triples, an extra value conceals them, hence the name “hidden”.
### 4. Master Level Approaches
Of course, highly smart techniques for solving sudoku puzzles are available as well. Plus, they have cunning names as well!
#### The X-Wing Approach
This practice demands you to recognize four cells in horizontal columns and/or vertical rows which hold the same number guesses. Once these are spotted, then it’s much simpler for you to rule out specific values in particular cells. Generally, the same figure guesses are positioned such that they shape an invisible letter X, consequently the name.
#### The Forcing Chains Practice
This approach first starts out with you identifying cells that have a pair each of the same figure guesses. Selecting only cells with candidate pairs is important as placing one or the other of the values inevitably “forces” the unused value into the other unfilled cell by this means solving the sudoku puzzle more rapidly.
### 5. Enormously Hard Solving Process
Much like any other puzzle game that demands cunning as well as logic, this puzzle also has one thoroughly tricky technique for successful solving. This approach is usually utilized if all the other less complicated tactics prove to be inadequate for solving sudoku puzzles.
#### The Nishio Method
This technique for solving sudoku puzzles is pretty much similar to guessing. But to improve the odds of getting the correct solutions, you will need to recognize cells that have only two potential values.
You can then select out one of these values to place in the cell so that you start working your way through the puzzle. However if you get to encounter some repetitions, then you will inevitably need to go back to the first guess you put in.
#### New Free to Print Sudoku Puzzles
We have uploaded several new printable sudoku puzzles for you to enjoy. You can locate the new puzzles by clicking the following link.
#### Free Online Tetris Game
Play the classic puzzle rotatation game Tetris. See if you can beat the highest score! Play Free Tetris
#### Free Maths Game
Random numerical questions, try and correctly solve the sums as swiftly as possible. Free Maths Game
#### Printable Evil Sudoku Puzzles
###### Sudoku Techniques and Tips
The appeal of Sudoku lies in the broad difficulty the game can present, from simple to annoying, which allows users of all skill levels to play at an suitable level. Sudoku also offers a wide choice of procedures that can be used to complete puzzles, from straightforward to complex.
And that's where the real appeal of the game can be found, as every board can be solved through nothing but logical deductions. There is no guesswork or trial and error necessary, though these are often employed tactics as well.
Sudoku Techniques and Tips | 1,370 | 6,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2013-20 | longest | en | 0.89553 |
https://physics.stackexchange.com/questions/134124/conservation-laws-and-continuity-equations | 1,726,602,766,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00452.warc.gz | 414,894,324 | 45,143 | # Conservation laws and continuity equations
I'm a bit messed up with conservation laws and continuity equations.
This is my understanding:
• A conservation law describes that a physical quantity $G$ is conserved with time. It does not prevent "quantity teleportation", as long as at a given time, the created quantity and the disappeared quantity are equal. In Wikipedia's wordings: "For example, it is true that "the total energy in the universe is conserved". But this statement does not immediately rule out the possibility that energy could disappear from Earth while simultaneously appearing in another galaxy."
• A continuity equation is stronger: It implies that there is no "quantity teleportation".
• A global conservation law describes that globally, a physical quantity $G$ is conserved with time: $$\dfrac{\mathrm{d}G}{\mathrm{d}t}=0$$ Using mathematics, this can be written as the sum of an integral over a volume and an integral over a surface, or as a single integral over a volume using Stokes theorem (and introducing a divergence)
• A local conservation law is the result of writing that the integrand of the global law are equal.
Questions:
1. Is "quantity transportation" possible in a local conservation law? If not is there any difference between the two?
2. In the equation (i.e. mathematically), where do you see the differences between continuity equations and conservation laws?
• 1. Do you know about flux? And how it comes into these kind of equations? 2. I would say, focus on the physical meaning of the terms in the equations, rather than the equation itself. Commented Sep 5, 2014 at 18:50
• @Bernhard: 1. Yes. I thought, but that might be where I'm missing something about continuity. 2. When I do so, I don't see how a local conservation law could include "quantity teleportation". Commented Sep 5, 2014 at 18:54
• Is it correct that writing the local conservation law requires continuity? Continuity would make Stokes theorem applicable, and transform the flux density with a divergence. Commented Sep 5, 2014 at 18:57
• I don't really understand what you mean with "quantity teleportation". Also, do you have a specific field in mind? Commented Sep 5, 2014 at 20:05
• @Bernhard By "quantity teleportation" I mean the following: Consider the case when the quantity (momentum, thermal energy, mass, etc.) we're looking at $G$ in a given domain at $t$ and at $t+\mathrm{d}t$ is the same: $G$ is conserved ($\dot G(t)=0$). Yet, it could be that some quantity $G_0$ vanished instantaneously in a small area, and appeared instantaneously in another area : $G$ would still be conserved: $G(t+\mathrm{d}t)=G(t)+G_0-G_0=G(t)$. Now, if continuity (of the density of $G$?) is imposed, this is not possible. Commented Sep 5, 2014 at 20:46
Continuity equations are an embodiment of local conservation laws, and they both reflect the fact that there is no 'quantity teleportation'. That said, the local transport of a quantity is perfectly possible within local conservation laws and it is precisely this that the continuity equation models.
Your distinction between global and local conservation laws could use some refinement, though. Consider a quantity $G$ whose local density is $g(\mathbf r)$, and whose flow density (i.e. flux) is $\mathbf j(\mathbf r)$. With this notation, a global conservation law establishes only that the total amount of $G$, i.e. $$G=\int g(\mathbf r)\text d\mathbf r,$$ where the integral is over all space, is constant over time.
Saying that $G$ additionally obeys a local conservation law is a stronger statement, and it is exactly the statement that $g$ and $\mathbf j$ obey the continuity equation. This one comes in two flavours:
• differential, $$\frac{\partial g}{\partial t}+\nabla\cdot\mathbf j=0,$$
• and integral, $$\frac{d}{dt}\int_Vg(\mathbf r)\text d\mathbf r +\bigcirc \!\!\!\!\!\!\!\!\!\iint_{\partial V}\mathbf j(\mathbf r)\cdot \text d\mathbf a.$$
It is important to note that both of these forms are completely equivalent (modulo technical assumptions on point, line and surface charges). The differential form holds at every point $\mathbf r$, and the integral form holds for every volume $V$, and one can use appropriate calculations to translate between both forms and therefore between both freedoms.
The reason that we say continuity equations embody local conservation laws is that they make precise the intuition that all the $g$ that "comes out" of some region can be "seen crossing the boundary", which is measured by the surface integral / the divergence term.
This is as opposed to, for example, a quantity with a density of the form $$g(\mathbf r,t)=g_0 \cos^2(\omega t)e^{-(\mathbf r-\mathbf r_1)^2/\sigma^2} + g_0 \sin^2(\omega t)e^{-(\mathbf r-\mathbf r_2)^2/\sigma^2}$$ where $\mathbf r_1$ and $\mathbf r_2$ are in principle far apart. Here $G$ stays constant, but between $t=0$ and $\pi/2\omega$, all of the $G$ near $\mathbf r_1$ has disappeared without there being any flux through the plane between it and $\mathbf r_2$. Here $G$ obeys a global conservation law, but not a local one.
For clarity, I should note that your statement that "A local conservation law is the result of writing that the integrand of the global law are equal" is incorrect, and depending on exactly what you mean by it, there may be exceptionally few systems that obey that.
• That's what I was asking for, TY. Two remarks though: 1) when you say "the integral form holds for every volume", do you mean for any volume with any velocity, or for any volume attached to matter (material velocity field)? 2) Answer to your last paragraph: I meant $\forall\mathcal{D},\ \int_\mathcal{D}f=0\Longrightarrow f=0$. I now understand that in order to apply this, all the terms must be gathered in an integral over a volume (integrated on $\mathcal{D}$, which requires continuity. Still have to assimilate all your answer though. Commented Sep 5, 2014 at 21:05
• Your definition is not that useful unless you specify what D and f are. The volumes in the integral continuity equation are fixed in space (i.e. in an inertial frame) and do not move, either with the fluid or on their own. Commented Sep 6, 2014 at 8:23
• "all of the G near r1 has disappeared without there being any flux through the plane between it and r2." Can you explain why? Assuming $\partial_t g + \nabla \cdot \mathbf j = 0$ and some boundary conditions on $\mathbf j$, appropriate $\mathbf j$ might be found. Commented Sep 6, 2014 at 8:43
• @Ján yes, that's true, particularly if $g$ is a signed measure. Nevertheless, the lack of $g$ anywhere in between does rule out many physical situations. Commented Sep 8, 2014 at 9:37
• I do not follow. Why is sign of any importance? What do you mean by lack of g in between? Commented Sep 8, 2014 at 18:42
In the equation (i.e. mathematically), where do you see the differences between continuity equations and conservation laws?
The continuity equation is not sufficient to derive conservation of something. For example, continuity equation for fluid flow in non-relativistic theory is
$$\partial_t \rho + \nabla \cdot (\rho \mathbf v) = 0$$
wherer $\rho$ is density of fluid and $\mathbf v$ is its velocity. Integrating this equation over some region $V$ with boundary surface $\Sigma$, switching the order of integration and differentiation and using the Gauss-Ostrogradskii theorem we obtain $$\frac{d}{dt}\int_V \rho \,dV = -\oint_\Sigma \rho \mathbf v \cdot d\boldsymbol{\Sigma}.$$ As the region $V$ is expanded to contain all space in a limit, the left-hand side gives rate of range of mass in the whole space. This is not necessarily zero, for the right hand side may be generally have non-zero limit.
To get conservation, we have to impose additional condition that there is a surface $\Sigma$ such that value of the surface integral is zero or at least goes to zero as the surface is expanded to infinity. This may not be always possible.
For example, the Poynting theorem for current-free region
$$\partial_t \left(\frac{1}{2}\epsilon_0E^2 + \frac{1}{2\mu_0}B^2 \right) + \nabla \cdot (\mathbf E\times \mathbf B/\mu_0) = 0$$ has the same form as the above equation of continuity, but the integral version $$\frac{d}{dt}\int_V \left(\frac{1}{2}\epsilon_0E^2 + \frac{1}{2\mu_0}B^2 \right)\,dV = -\oint_\Sigma \mathbf E\times \mathbf B/\mu_0 \cdot d\boldsymbol \Sigma$$ may have nonvanishing surface integral on the right-hand side, if there are EM waves present at infinity. From this it follows that the rate on the left-hand side is not zero and there is no conservation of value of the integral there (Poynting energy).
So continuity does not automatically mean conservation.
Similarly, conservation does not imply continuity - mass or energy can jump from place to place suddenly while the net value remains the same.
• "So continuity does not automatically mean conservation." It seems to me you're talking about global conservation, because you take the limit of $\Sigma$ to infinity. Do you agree with "So continuity does not automatically mean global conservation" but "continuity does automatically mean local conservation"? Helpful answer also by the way, thank you very much. Commented Sep 5, 2014 at 21:23
• I think "continuity does automatically mean local conservation" is wrong, sorry. What is true (see Emilio's answer) is that global conservation + continuity implies local conservation. Additionally, you showed that continuity does not imply global conservation and that global conservation does not imply continuity. Agree? Commented Sep 5, 2014 at 21:28
• "Do you agree with <So continuity does not automatically mean global conservation> but <continuity does automatically mean local conservation>?" Yes, but I think it is just a definition - local conservation seems to be defined by the validity of continuity equation. Commented Sep 6, 2014 at 8:45
• for context ,that poyenting theorem ,means energy transferd is connected to how much momentum flows one surface to another.left hand side is energy term and right hand side is telling me the momentum of the electromagmetic field.this is also called a conservation ,does not? Commented Mar 26, 2023 at 23:45
• The Poynting theorem is interpreted in terms of energy, not momentum. For momentum, there is another similar set of equations, involving the Maxwell stress tensor. Commented Mar 26, 2023 at 23:59 | 2,586 | 10,388 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-38 | latest | en | 0.931487 |
https://www.myaccountingcourse.com/accounting-dictionary/free-cash-flow-yield | 1,553,212,641,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202588.97/warc/CC-MAIN-20190321234128-20190322020128-00449.warc.gz | 831,105,653 | 12,502 | What is Free Cash Flow Yield?
Definition: Free cash flow yield, or FCF yield, is a valuation metric to measure the yield of a firm’s free cash compared to its size.
What Does Free Cash Flow Yield Mean?
What is the definition of free cash flow yield? FCF is a useful valuation metric to determine a firm’s operating performance. Generally, firms with strong cash flows are financially healthy as opposed to firms with weak FCFs may be unable to meet their short-term obligations.
The FCFY provides an additional indication of a firm’s operating strength by comparing the free cash flows to the market cap. As expected, a low FCFY indicates a firm in which investors are putting more money than the returns it generates. Financial analysts investigate FCF yield along with a firm’s capital expenditures to determine whether potentially weak FCFs can be justified by investments in fixed assets.
Let’s look at an example.
Example
Mary works as an accountant in a retail company. She is asked to calculate the free cash flows formula of the company for the period 2010 – 2015 and determine whether the FCFY is attractive.
Mary calculates the FCFs as follows:
Then, she finds the stock’s historical prices for the relevant years and the number of shares outstanding to calculate the market cap.
So, based on Mary’s calculations, in 2010, the company had 750,000 shares outstanding, which traded at \$35.85, hence a market cap of \$26.9 million. As it can be seen, the market cap decreases in 2012 and in 2013, thereby affecting the firm’s cash flows. Apparently, the firm had released negative results or had failed in a project to the extent that it could not generate strong cash flows as in 2010.
Mary calculates the FCFY of each year by dividing the free cash flow over the market cap.
However, to get a firm idea of how strong or weak is the FCFY, financial analysts compare it to a benchmark firm in the industry.
Summary Definition
Define Free Cash Flow Yield: FCF Yield means a financial measurement designed to evaluate a firm’s profitability and efficiency. | 449 | 2,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-13 | latest | en | 0.937906 |
https://community.gamedev.tv/t/mining-quest-keep-your-enemies-clo-farther-solutions/167069 | 1,701,787,750,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100551.2/warc/CC-MAIN-20231205140836-20231205170836-00123.warc.gz | 208,975,415 | 6,095 | # MINING QUEST: ‘Keep Your Enemies Clo...Farther’ - Solutions
Quest: Mining Quest
Feel free to share your solutions, ideas and creations below. If you get stuck, you can find some ideas here for completing this challenge.
Making progress on the challenges. This one took me a while.
There is probably a simpler solution, but it is working.
In Start() I calculate the player starting position (center). This will be needed, once the map size gets variable too.
`````` private int centerLocationX;
private int centerLocationY;
void Start()
{
centerLocationX = (xSize / 2); // integer division
centerLocationY = (ySize / 2);
}
``````
In SpawnEnemies() I start by calculating a List of valid locations (each an array with coordinates). If at least one coordinate x/y is not in the safe zone, the location is valid. Later I pick a random location from that list for each enemy.
``````private void SpawnEnemies()
{
// Challenge 5, safe space 3x3 around player
List<int[]> validLocations = new List<int[]>();
for (int x = 0; x < xSize; x++)
{
for (int y = 0; y < ySize; y++)
{
if (x < centerLocationX - 1 || x > centerLocationX + 1 ||
y < centerLocationY - 1 || y > centerLocationY + 1)
{
int[] location = { x, y };
}
}
}
for (int i = 0; i < amountOfEnemies; i++)
{
int[] randomLocation = validLocations[Random.Range(0, validLocations.Count - 1)];
Vector2 spawnLocation = new Vector2(randomLocation[0], randomLocation[1]);
Instantiate(enemyPrefab, spawnLocation, Quaternion.identity);
}
}
``````
3 Likes
Because the map changes sizes at runtime, I first had to ensure the player ended up generally in the middle. It’s a bit brute force, but it works. I then iterate through a 3x3 grid around the player and feed those to my Blocked Coordinates method. That method takes the x/y coordinates and drops it into a list of Vector2s that will be referenced later to tell the rest of the items and enemies what coordinates they cannot spawn on:
`````` void Awake()
{
gridLength = Random.Range(gridMin,gridMax);
playerSpawn = gridLength / 2;
PlacePlayer();
blockedCoordinates = new List<Vector2>();
}
void Start()
{
SetPlayerSafeZone();
GenerateEnviron();
SpawnEnemies();
}
private void PlacePlayer()
{
playerPrefab.transform.position = new Vector2(playerSpawn,playerSpawn);
}
private void SetPlayerSafeZone()
{
int playerXPosOffset = (int)playerPrefab.transform.position.x -1;
int playerYPosOffset = (int)playerPrefab.transform.position.y -1;
for (int x = (int)playerPrefab.transform.position.x - 1; x < playerXPosOffset + 3; x++)
{
for (int y = (int)playerPrefab.transform.position.y - 1; y < playerYPosOffset + 3; y++)
{
BlockCoordinates(x,y);
}
}
}
``````
And then in my SpawnEnemies() method I first divide the length of the grid by an int (currently set to 3) to determine how many enemies will spawn, so that the number of enemies increases as the grid does. I then iterate through the for loop, and while it still picks a random Vector2 to instantiate on, before actually instantiating it checks my list of Blocked Coordinates. If the Vector2 is not on the list, the enemy is instantiated, otherwise I subtract 1 from “I” in order to rerun the check and ensure that all of the enemies spawn.
``````private void SpawnEnemies()
{
int amountOfEnemies = gridLength/enemyDivisor;
for (int i = 0; i < amountOfEnemies; i++)
{
Vector2 spawnLocation = new Vector2(Random.Range(1, gridLength), Random.Range(1, gridLength));
if(CanSpawn((int)spawnLocation.x,(int)spawnLocation.y))
{
GameObject newEnemy = Instantiate(enemyPrefab, spawnLocation, Quaternion.identity);
}
else
{
i--;
}
}
}
`````` | 904 | 3,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-50 | latest | en | 0.771632 |
mommytanya.blogspot.com | 1,531,860,009,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589902.8/warc/CC-MAIN-20180717203423-20180717223423-00104.warc.gz | 242,652,050 | 17,448 | ## Wednesday, March 14, 2012
### Pi Day
Let your nerd flag fly high kids because today is Pi Day!!
π (sometimes written pi) is a mathematical constant that is the ratio of any Euclidean[1] circle's circumference to its diameterπ is approximately equal to 3.14. Many formulae in mathematics, science, and engineering involve π, which makes it one of the most important mathematical constants.[2] For instance, the area of a circle is equal to π times the square of the radius of the circle. (http://en.wikipedia.org/wiki/Pi)
And because we are proud nerds in this family March 14th, or 3/14, or 3.14 is the day we eat pie!
For our main course I busted out a Steak and Mushroom Guinness pie. this is also a nod to The Eagle and Child, where I had my first steak pie when I was at Oxford, and was also Tolkein and Lewis' old haunt.
I got fancy and carved a π into the puff pastry. Unfortunately because of how soupy the filling was it didn't puff the way it was supposed to. No worries it still tasted amazing!
And you can't celebrate properly with out desert so I made a delicious chocolate pie! I baked up a crust, made some delicious pudding and topped it with cool whip and chocolate shavings!!
ETA (thanks Angi)
#### 1 comment:
1. awww i love the pies. and T is so cute! | 334 | 1,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-30 | latest | en | 0.923512 |
http://www.docstoc.com/docs/104677050/Chapter-14---Download-as-PowerPoint | 1,386,665,934,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164014082/warc/CC-MAIN-20131204133334-00029-ip-10-33-133-15.ec2.internal.warc.gz | 320,621,340 | 20,282 | Document Sample
``` Chapter 14
Rates of Reactions
Kinetics
I. Introduction
A) Demonstrations
B) Chemical Kinetics is the study
of the rates (speeds) of chemical
reactions and the mechanisms of
chemical reactions.
C) The rate of a chemical reaction
is a measure of how fast reactants
are consumed and/or how fast
D) The mechanism of a reaction is a
detailed description of the way a
reaction occurs. It is a sequence of
reactants to products.
Mechanisms can be proven
wrong through _____________,
but they can never be called
_________________________
since they are, in general,
educated guesses.
Practical reasons for studying kinetics:
Some reactions we would like to speed
up: drug delivery, paint drying,
destruction of air pollutants in auto
exhaust breakdown of materials in
landfills.
Some reactions we would like to slow
down: food decay, rubber decay, human
aging, destruction of the ozone layer,
rusting, corrosion
E) Some reactions take place in
a fraction of a second and other
take many years. What
variables affect the reaction
rate?
1) The characteristics of the
reactants and the products.
2) The concentration of the
reactants – in some reactions
the rate is unaffected by the
concentration of one of the
reactants as long as it is there
in some amount.
3) The presence of a catalyst, a
substance that …
4)The temperature at which the
reaction occurs. Increasing the
temperature usually increases
the rate. A general rule is …
5) The surface area of a solid
reactant or catalyst affects the
rate.
G) Reaction rate is the increase
in molar concentration of
product of a reaction per unit
time or the decrease in molar
concentration of a reactant per
unit time.
For example, for the reaction:
2 N2O5 4 NO2 + O2
The following table shows the
concentration of N2O5 as a
function of time at 45 oC.
Time in min [N2O5] mol/L
0 0.01756
20 0.0093
40 0.00531
60 0.00295
80 0.00167
100 0.00095
160 0.00014
1) The rate of reaction can be
written in the form:
2) Usually the rate is a rapidly
changing quantity, as the reaction
proceeds the reactants are used up
and there remains less and less
material to undergo reaction.
3)Generally we obtain for a
reactant a curve which resembles
the one below.
If we take the first 2 points from
the table above, we can find as
avg. rate of decomposition of
N2O5.
You should be able to see that the
avg. rate is decreasing, hence the
curve.
Even though we obtain a
negative value for the rate, since
N2O5 is decreasing, rates are
reported as positive values.
Look at the graph again,
And make the time interval
smaller and smaller, we can
obtain an instantaneous rate.
The instantaneous rate is equal
to the slope of the line at that
point. Calculus??? To what is the
slope of the line at that point
equal?
4) How is data obtained for a
concentration curve?
a) Monitor a color change.
b) Measure pressure if a gas is
produced.
c) Monitor a change in pH if an
acid or base reaction.
5) A look at the change in rate over
time for another reaction:
2 NO2(g) 2 NO(g) + O2(g) at 300 oC
H) What does the balanced
The equation we will look at is:
H2 (g) + I2 (g) ----> 2 HI(g)
H2 and I2 must disappear at the
same rate since 1 molecule of H2
reacts with 1 molecule of I2.
In the same amount of time, 2
molecules of HI must appear.
The rate of appearance of HI
must equal twice the rate of
disappearance of H2 (g) and I2 (g).
The rate of disappearance of
H2(g) = the rate of disappearance
of I2(g) = ½ the rate of
appearance of HI.
In general, for the equation:
aA + bB -----> cC + dD
To obtain a rate equation of the
rates of the substances in relation
to each other we divide through by
the coefficients.
But we usually want the first
reagent in terms of all the others,
so we multiply by the coefficient of
the first reagent.
II. The Rate Law (Rate
expression), Rate Constant
Order of Reaction
A) The following equation has
been studied in the gaseous state
and the data at 250 K may be
summarized as follows:
F2(g) + 2 ClO2(g) ---> 2 ClO2F(g)
B) From this data the answers to
the following questions can be
obtained:
1) What is the rate law of the
reaction?
2) What is the order of the
reaction?
3) What is the value of the rate
constant k?
4) What is the rate of formation
of ClO2F when [F2]is 0.010 mol/L
and[ClO2] is 0.020 mol/L?
C) What is a rate law?
1) In 1864 it was discovered that
the rate of a reaction is
proportional to some power of
the concentration of reactants at
constant temperature.
2) In general, for the equation:
aA + bB cC + dD
rate law = rate equation = rate
expression
k is the specific rate constant
which is independent of
concentration.
k depends on the nature of the
reactants; fast reactions have
large k's and slow reactions
have small k's ordinarily k
________________ with
temperature.
THE EXPONENTS x AND y
MUST BE EXPERIMENTALLY
DETERMINED.
They are not automatically
obtained from the balanced
equation.
Some experimentally determined
rate laws for equations are as
follows:
a) 2 N2O5 (soln) 2 N2O4 (soln) + O2 (g)
Notice that the exponent is NOT 2, the
coefficient in the balanced equation, but
has been experimentally determined to
be 1.
The reaction order with respect to a given
species equals the exponent of the
concentration of that species in the rate
law as determined experimentally.
The order of the above reaction with
respect to N2O5 is 1.
The order of a reaction is the sum of the
exponents of the reactants in a rate law.
The order of the above reaction, since 1
is the only exponent, is 1 as well. It is a
first order reaction.
This means that when we double the
concentration of N2O5, we double the rate
of reaction OR if we halve the
concentration, we halve the rate.
b) For the reaction: 2 NO + O2 2 NO2
What is the situation here?
D) The answers to the questions then are
obtained in the following manner.
The rate law for
F2(g) + 2 ClO2(g) ---> 2 ClO2F(g)
will look like the following.
Your job is to find the values of x and y
from the experimental data.
You need to do two division problems to
find x and y.
What is the order of the reaction with
respect to F2? _______
What is the order of the reaction with
respect to ClO2? ______
What is the overall order of the
reaction? __________
To find k, the rate constant, we take the
experimentally determined rate law,
put the data in from one of the
experiments and solve for k. The units
of k are important.
k = _________________
To find the rate of formation of ClO2F
when [F2] is 0.010 M and the [ClO2] is
0.020 M, we need to look at the
relationship between the rate of
formation of [ClO2], and the rate of
disappearance of F2.
Rate = __________________________
Transparency – in class problem –
collect it
IV. A graphical method is often used to
show the order of a reaction, or from a
graph we can obtain the order of a
reaction.
A)For the general reaction of A --->
products
B) If the reaction is first order, we can
write
When we divide both sides by [A] and
multiply both sides by dt we obtain the
following:
Those of you who have had calculus
should recognize this as:
Which can be changed to the following
in log to the base 10.
The above equations can be rewritten in
a more familiar form.
THIS MEANS THAT IF WE PLOT ln
[A] vs. t AND OBTAIN A STRAIGHT
LINE THE RATE IS FIRST ORDER.
Or plot log [A] vs. t.
Slope of the line is -k/2.303. The rate
constant k then is = to -slope of the line
times 2.303.
If we plot ln[A] vs.time then the slope of
the line is equal to -k.
C) Half-life - t½ is the time required
for half of the original concentration
of the limiting reactant to be
consumed.
Half-life is _________________ to the
rate constant k.
A fast reaction with a large k has a
_________________ half-life, a slow
reaction with a small k has a
___________________ half-life.
D) For the reaction :
A ----> B + C
We obtain the following data.
Prove that the reaction is first order in
A. Calculate the half-life of the reaction,
t½ .
What is our criterion for a first order
reaction?
We make the following table.
When we plot log [A] vs. t we obtain a
_________________, therefore the
reaction is first order.
The equation for the line is:
V. Second order reaction for a single
reactant
A) A ----> products
B) The rate law is: rate = k[A]2
C) The integrated form of the rate law
gives us the following:
D) A straight line is obtained with slope
= ____.
E) This means that we can tell the
difference between a first order and
second order reaction by
________________________________.
G) The half-life of a second order reaction
depends on the initial concentration of the
reactant as can be seen from the following
derivation:
H) When we compare this to the formula
for the half-life of a first order reaction,
we notice that the t½ for the first order
reaction is __________________of
the initial concentration of the reactant.
The t½ for the second order reaction is
________________________to the initial
concentration of the reactant.
I) The following equations will be given
to you on an exam or quiz for you to
apply to questions in an appropriate
manner.
VI. Collision Theory
A)Consider the elementary process
A2(g) + B2(g) ---> 2AB
B) Collision theory assumes that for gas
molecules to react they must collide.
C) Binary collisions per unit time at
room temperature and 1 atm pressure
are extremely large, approx. 1 x 1031
collisions per Liter-sec.
D) With this many collisions, why isn't
there an immediate explosion when we
put H2(g) and O2(g) together in a balloon?
E) Why isn't every collision an effective
collision; i.e. one that gives product?
F) For a reaction to occur reactant
molecules must collide with an energy
greater than some minimum value and
with the proper orientation.
G) The minimum energy of collision
required for two molecules to react is
called the activation energy, Ea. The
value of Ea depends on a particular
reaction.
H) Effect of temperature
1) The rates of almost all chemical
reactions_________ when T is raised.
Generally speaking for a _________
increase, the reaction rate ________.
2) What is the relationship between
temperature and collision theory?
a) Approx. ___ of the reaction rate
increase due to an increase in
temperature is accounted for by the
increase in the number of___________.
b) Approx. ____of the reaction rate
increase is accounted for by the
increase in the number of high energy
_______________.
3) Recall our graph of the number of
molecules vs. speed in a gas sample.
J) Arrhenius in 1889 proposed the
following equation which relates the rate
constant k to the temperature:
where k is the rate constant from the
rate equation ( rate = k[A])
A is the frequency factor (the collision
factor and the orientation factor) .
e is the base of the natural logarithms
2.718
Ea is the activation energy in Joules per
mole.
R is the molar gas constant, 8.3145 J/K
mol.
T is the temperature in Kelvins.
J) The most valuable use for this
equation is in determining the activation
energy of a reaction from rate
experiments at different temperatures.
L) Ea is best obtained graphically.
This is of the form y = mx + b
If we plot the log of k vs 1/T we obtain
the slope of the line which is equal to
- Ea/ 2.303 R. then Ea = - 2.303 R slope.
L) The equation can be used to compare
rates at various temperatures when Ea is
known, OR the Ea can be obtained from
a comparison of two sets of rate data.
r1 = k1 [A]x at T1
r2 = k2 [A]x at T2
We divide equation 1 by equation 2,
take the log of both sides, and substitute
for log k –Ea/2.303RT. This should give:
We remove the negative sign and we
get the equation which we will use and
which will be given to you on the exam
or quiz.
1) An example: The activation energy
for the reaction involved in the souring
of raw milk is 75 kJ. Milk will sour in
about 8 hours at 21oC (room temp-
70oF). How long will raw milk last in a
refrigerator at 5oC?
2) The gas phase reaction between
methane and diatomic sulfur is given by
the equation:
CH 4(g) + 2 S2(g) ---> CS2(g) + 2 H2S(g)
At 550 oC the rate constant, k1, for the
reaction is 1.1 L/mol sec. At 625 oC the
rate constant, k2, is 6.4 L/mol sec. What
is the activation energy of this reaction?
N) Transition State Theory
1) The hypothetical elementary process
of:
A2 + B2 --> 2 AB
can be represented as the following:
is an activated complex. It is a
transition state species that
__________________________.
In the past, these could not be detected only
postulated. Now with laser technology we
can detect species which exist for only a
femto second(1x10-15), and evidence for the
existence of some intermediates is available.
2) We can graph the potential energy
changes which occur during the course of
a reaction.
3) We have shown reactions in which
DH is - (negative). The reactions are
exothermic.
DHrxn = Eaf - Ear
4) Potential energy diagrams are helpful
in visualizing why it is that highly
exothermic reactions can be very slow.
If Ea is large then __________________ .
VII. The mechanism of a reaction
A) The mechanism of a reaction is the
detailed pathway followed when
reactants are converted to products.
B) It must agree with
1) ____________________ .
2) ____________________ .
3) ____________________ .
C) What do we mean by the rate
determining step? Let's look at a factory
where toasted corn puffs are made.
What is the rate determining step? ___
How fast can the factory produce
packages of corn puffs with a prize?
_________________
What does the factory need to do to
improve the rate of production?
__________________
D) For the reaction:
2 NO + F2 ---> 2 NOF
nitric oxide plus fluorine gives nitrosyl
fluoride
1) The experimentally determined rate
law is: rate = k [NO][F2]
2) A mechanism is proposed on the
basis of the rate equation and other
information gained from a study of
similar reactions over a period of years.
3) Remember that the mechanism is ___ .
4) A suggested mechanism for this
reaction is:
NO + F2 --> NOF + F (slow step)
NO + F --> NOF (fast step)
The sum of these two reactions gives:
2 NO + F2 + F --> 2 NOF + F
Subtracting F from both sides gives the
5) The rate determining step is the _____ .
6) Each step is an elementary reaction
and the rate equation can be written
directly from a _________________
elementary reaction.
7) The rate law of our reaction is rate = k
[NO][F2]. The proposed mechanism
satisfies the requirements ___________
__________________________________ .
E) Let's look at another equation its
experimentally determined rate law and
the proposed mechanism to see if they
are in agreement.
2 O3 -->3 O2
1) The experimentally determined rate
law is:
2) The proposed mechanism is:
The rate of the overall reaction is simply
the rate of step ______ .
Rate = k2[O3][O]
There is a problem. We do NOT report
rate laws with the concentrations of
species which are not in the _________ .
How can we convert Rate = k2[O3][O]
to
3) We need to use a combination of
knowledge from chemistry and algebra to
change one rate law to another.
The chemical knowledge we bring to
the situation is that at equilibrium,
since both step 1 and step 2 are fast and
2 is the reverse of 1 we will make their
rates equal.
k1[O3] = k-1[O2][O]
Then we can use algebra to solve for [O],
since that is what we want to __________
We can then substitute this relationship
of [O] into the rate law we obtained from
____________ to obtain
F) Catalysts
1) The rates of many reactions are
increased by the presence of a catalyst, a
substance which increases the rate of
reaction without being consumed by it.
2) How is this possible?
3) NO (nitric oxide) catalyzes the
decomposition of O3. The following
mechanism is proposed for this reaction:
What should you notice about NO?
4) A catalyst acts by making available a
new reaction mechanism with a lower
activation energy, Ea.
Sometimes is orients the molecules for a
successful collision, for example the
lock and key model mechanism of
enzymes.
Other times it provides an alternate
pathway for the reaction.The depletion of
ozone in the stratosphere by Cl atoms
provides an example of the lowering of
activation energy by a catalyst. The
uncatalyzed reaction has such a large
activation energy as can be seen from the
following diagram that its rate is
extremely slow.
VIII. Review - Factors which affect the
rate of reaction:
A) Nature of the reactants -
B) Concentration -
C) Temperature -
D) Catalysts -
```
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How are you planning on using Docstoc? | 4,377 | 16,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2013-48 | longest | en | 0.905021 |
http://free-apps-android.com/calcnote/ | 1,521,436,514,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646375.29/warc/CC-MAIN-20180319042634-20180319062634-00784.warc.gz | 119,560,542 | 17,646 | # CalcNote
## [App] CalcNote – Notepad Calculator
CalcNote is the the notepad calculator designed for the smart phone.
You can immediately obtain a calculation result while allowing recognition of a calculation process.
CalcNote calculates instantly as you type. No need to tap equals.
You can do multiple calculations over multiple lines, and even connect the lines together using line references. When one line changes, the other lines update.
If you make a mistake, you don’t need to start again, you can edit your expression like in a text editor.
Of course, you can write a memo for your problems.
How to use
Result will be shown realtime to right pane when you enter the expression to left pane.
Tap the result to show the result as hex, binary, tax included and tax excluded. Then you can copy the result to tap it.
Tap the line number to refer to the previous answer.
If expression has syntax error, the result won’t shown.
You can edit your expression like in a text editor. Also you can copy and paste the expression.
Your expression and result will be saved automatically.
Result will be exponential notation if result is over 10^9.
The button can have sub buttons. You can perform the assigned operation to sub buttons by swipe to the up or down.
Swipe right on the display to show the side menu.
Swipe left or right on the numeric keypad to reveal mathematical keypad.
You can customize the appearance of editor. (Background color, Text color, Show line number)
You can refer to the previous answer by specifying the line number with ‘\$’.
Ex) When you want to calculate the consumption tax.
2000 * 4
1900 * 2
(\$1+\$2)*1.08 # =(8000+3800)*1.08
Please refer to online manual for more details
https://github.com/burton999dev/CalcNoteHelp/blob/master/documents/en/index.md
Main Functions
Four arithmetic operations
Percentage calculation
Modulo operation
Rounded, Truncated, Rounded-up
PI
Square root, Cube root
Power, Factorial
Permutations, Combinations
Trigonometric function sin,asin,sinh, cos,acos,cosh, tan,atan,tanh
Logarithm, Natural logarithm, Natural exponential function
Customizable fonts
Customizable colors
Customizable display format (European style and Indian style)
Absolute value
Saving a expression
Exporting a expression
If you have any concerns, feedback, or questions email me at burton9999dev@gmail.com
Thank you.
CalcNote user reviews :
Awesome No comment just loved it
Love it. Thanks for creating it. I find it more useful and use it most of the times. Reason – it will let me compare two or more computations right there. Also its strength is in using \$lineNo. …Since it has two columns, one for computing the other for result, it makes sense to have a 3rd column for brief note per line. Revised. History is another essential feature it offers. Built in unit conversion makes it marvelous app that I discovered in few days. Thank you for so useful app. Plz do consider a 3rd column.
• burton999 January 14, 2017
• Thank you for your positive review. You can include notes in your expression. Everything to the left of the colon will not be evaluated. eg. `Salary for 10/2016: 950 JPY * 120 hours` Also everything to the right of the semicolon will not be evaluated. eg. `950 JPY * 120 hours; Salary for 10/2016` Best regards,
Contact developer :
Email
Thanks to Noriyoshi for sharing CalcNote | 768 | 3,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-13 | latest | en | 0.843895 |
http://tuxar.uk/about-computers/alan-turing/turing-machines/ | 1,416,645,122,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400376728.38/warc/CC-MAIN-20141119123256-00169-ip-10-235-23-156.ec2.internal.warc.gz | 345,748,698 | 11,537 | # Turing Machines
As a mathematician, Alan Turing was interested in mathematical logic, and in particular with the foundations of mathematics. In 1935 he learnt from M.H.A. Newman that Hilbert’s Entsceidungsproblem (decision problem) was open: could there be a definite method to determine if any given mathematical statement was provable? To answer this, he analysed what a human ‘computer’ would do, and derived a simplified model that seemed capable of performing autmomatic calculation.
A Turing machine is a device that manipulates symbols on a strip of tape according to a table of rules. Despite its simplicity, a Turing machine can be adapted to simulate the logic of any computer algorithm, and is particularly useful in explaining the functions of a CPU inside a computer. The “Turing” machine was described by Alan Turing in 1936, who called it an “a(utomatic)-machine”:
We may compare a man in the process of computing a real number to a machine which is only capable of a finite number of conditions q1, q2, …, qR which will be called “m-configurations”. The machine is supplied with a “tape”, (the analogue of paper) running through it, and divided into sections (called “squares”) each capable of bearing a “symbol”. At any moment there is just one square, say the r-th, bearing the symbol S(r) which is “in the machine”. We may call this square the “scanned square”. The symbol on the scanned square may be called the “scanned symbol”. The “scanned symbol” is the only one of which the machine is, so to speak, “directly aware”. However, by altering its m-configuration the machine can effectively remember some of the symbols which it has “seen” (scanned) previously. The possible behaviour of the machine at any moment is determined by the m-configuration qn and the scanned symbol S(r). This pair qnS(r) will be called the “configuration”: thus the configuration determines the possible behaviour of the machine. In some of the configurations in which the scanned square is blank (i.e. bears no symbol) the machine writes down a new symbol on the scanned square: in other configurations it erases the scanned symbol. The machine may also change the square which is being scanned, but only by shifting it one place to right or left. In addition to any of these operations the m-configuration may be changed. Some of the symbols written down will form the sequence of figures which is the decimal of the real number which is being computed. The others are just rough notes to “assist the memory”. It will only be these rough notes which will be liable to erasure.
The Turing machine is not intended as a practical computing technology, but rather as a hypothetical device representing a computing machine. Turing machines help computer scientists understand the limits of mechanical computation.
There are many virtual implementations of the Turing machine, in various languages, and even some in hardware that actually read and write on moving tape. (The theoretical machine requires infinite tape, serving as memory). As implied just above, these aren’t practical devices but serve a pedagocial use. So here is a very elementary example in Python:
```# Add unary numbers (e.g. 2+1)
table={'1': {'0': '1>2', '1': '1>1'},'2':{'0':'0<3','1':'1>2'},'3':{'0':'0>3','1':'1>0'}}
tape='01101000'
state='1'
while state != '0':
print tape
next=table[state]
tuple=next[symbol] | 757 | 3,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2014-49 | latest | en | 0.941934 |
https://oeis.org/A194119 | 1,719,104,566,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00517.warc.gz | 376,248,342 | 3,941 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A194119 Triangular array: T(n,k)=C(n+3,k)+C(n+3,k+3), 0<=k<=n. 4
2, 5, 5, 11, 10, 11, 21, 21, 21, 21, 36, 42, 42, 42, 36, 57, 78, 84, 84, 78, 57, 85, 135, 162, 168, 162, 135, 85, 121, 220, 297, 330, 330, 297, 220, 121, 166, 341, 517, 627, 660, 627, 517, 341, 166, 221, 507, 858, 1144, 1287, 1287, 1144, 858, 507, 221 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,1 LINKS Table of n, a(n) for n=0..54. EXAMPLE Northwest corner: 2 5...5 11...10...11 21...21...21...21 36...42...42...36 57...78...84...84...78...56 MATHEMATICA T[n_, k_] := Binomial[n + 3, k] + Binomial[n + 3, k + 3] Flatten[Table[T[n, k], {n, 0, 10}, {k, 0, n}]] (* A194119 as a sequence *) TableForm[Table[T[n, k], {n, 0, 10}, {k, 0, n}]] (* A194119 as an array *) CROSSREFS Cf. A007318, A194118, A194120. Sequence in context: A086651 A074495 A081467 * A084721 A273041 A274108 Adjacent sequences: A194116 A194117 A194118 * A194120 A194121 A194122 KEYWORD nonn,tabl AUTHOR Clark Kimberling, Aug 16 2011 STATUS approved
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Last modified June 22 20:54 EDT 2024. Contains 373608 sequences. (Running on oeis4.) | 591 | 1,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.632806 |
http://www.lsa.umich.edu/saa/publications/courseguide/winter/archive/Winter90.cg/statistics.html | 1,511,456,249,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806844.51/warc/CC-MAIN-20171123161612-20171123181612-00499.warc.gz | 436,754,074 | 5,713 | # Courses in Statistics (Division 489)
100(300). Introduction to Statistical Reasoning. (4). (NS).
This course is designed to provide an overview of the field of statistics. Course topics include methods of collecting, analyzing and summarizing data (with special emphasis on graphical techniques), statistical reasoning as a means of learning from observations, and techniques for dealing with uncertainty in drawing conclusions from collected data. Basic fallacies in common statistical analyses and reasoning are discussed. Alternative approaches to statistical inference are also discussed. The course emphasis is on presenting basic underlying concepts rather than on covering a wide variety of different methodologies. The course format is lecture, with a weekly one-hour laboratory. Evaluation is based on three evening midterms and a final examination. (Section 001 – Gunderson. Section 002 – Muirhead. Section 003 – Keener.)
170(270). The Art of Scientific Investigation. (4). (NS).
A Collegiate Fellows course; see page 3 of this Course Guide fora complete list of Collegiate Fellows courses. The objective of this course is to introduce students to the learning process in a non-deterministic environment. An appreciation for measurement, bias and variation is essential to formulate questions and learn about things. Underlying this course is the Edwards Deming philosophy. Deming, an American statistician, was invited to Japan in the early 1950's to help improve the quality of mass produced items. His success in Japan is, in part, responsible for our current balance of trade deficit; and here, the Ford Motor Co., has also attained a larger market share as a result of his ideas. Implementation of the Deming message requires a critical appreciation of variation and the scientific method. Specifically, we will discuss: (1) Historical attempts to learn and the advent of the modern scientific method. (2) The differences between special or assignable causes and common causes of variation. Before we can learn how a process operates, the process must be stable. (3) Differences between observational and controlled randomized studies and associated ethical issues. (4) The 'what' and 'how' of measurement and the quantification of uncertainty-subjective and frequency notions of probability. (5) Understanding bias and variation. (6) How to use bias to design efficient studies. (7) Differences between enumerative and analytic studies. Many of the ideas will be introduced through experimentation (e.g., the red bead and funnel experiments) and the mathematical level will not require more than a modest background in high school algebra. The course format includes 3 lectures and a laboratory (1.5 hours per week). (Rothman)
311/I.O.E. 365. Engineering Statistics. Math. 215 and Eng. 102, or equivalent. No credit granted to those who have completed 412. (4). (Excl).
Analysis of engineering data associated with stochastic industrial processes. Topics include: fundamentals of distribution analyses; process model identification, estimation, testing of hypothesis, validation procedures, and evaluation of models by regression and correlation. Students are required to use the MTS computer system for problem solving. (Aras)
402. Introduction to Statistics and Data Analysis. No credit granted to those who have completed 412. (4). (NS).
In this course students are introduced to the concepts and applications of statistical methods and data analysis. Statistics 402 has no prerequisite and has been elected by students whose mathematics background includes only high school algebra. Examples of applications are drawn from virtually all academic areas and some attention is given to statistical process control methods. The course format includes three lectures and a laboratory (l.5 hours per week). The laboratory section deals with the computational aspects of the course and provides a forum for review of lecture material. For this purpose, students are introduced to the use of a micro-computer package and the Macintosh computer. Course evaluation is based on a combination of three examinations GIVEN WEDNESDAY EVENINGS, a final examination and teaching fellow input. [Cost:2] [WL:3] (Rothman)
403. Introduction to Statistics and Data Analysis II. Stat. 402. (4). (Excl).
APPLIED REGRESSION. The course will also cover various topics associated with the general linear model emphasizing applications. Topics include: multiple regression, variable selection, stepwise regression, residual analysis, analysis of variance models, covariance analysis and principal components. OTHER TOPICS. As time allows, the course may cover some aspects of probit and logit analyses, analysis of time series data, reliability analysis, and topics in experimental design. Three hours of lecture and one and one-half hours of lab per week. [Cost:2] [[WL:3] (Smith)
405/Econ. 405. Introduction to Statistics. Math. 115 or permission of instructor. Juniors and seniors may elect concurrently with Econ. 201 and 202. No credit granted to those who have completed Econ. 404. (4). (Excl).
Principles of statistical inference, including: probability, experimental and theoretic derivation of sampling distributions, hypothesis testing, estimation, and simple regression. [Cost:2] [WL:3] (Hill)
412. Introduction to Probability and Statistics. Prior or concurrent enrollment in Math. 215 and either CS 283 or Engin. 102. No credit granted to those who have completed 311 or 402. (3). (Excl).
The objectives of this course are to introduce students to the basic ideas of probability and statistical inference and to acquaint students with some important data analytic techniques, such as regression and the analysis of variance. Examples will emphasize applications to the natural sciences and engineering. There will be regular homework, including assignments which require the use of MTS, two midterms, and a final exam. [Cost:2] [WL:1] (Jeganathan)
425/Math. 425. Introduction to Probability. Math. 215. (3). (N.Excl).
See Math 425.
426. Introduction to Mathematical Statistics. Stat. 425. (3). (NS).
This course covers the basic ideas of statistical inference, including sampling distributions, estimation, confidence intervals, hypothesis testing, regression, analysis of variance, nonparametric testing, and Bayesian inference. The sequence of Statistics 425/426 serves as a prerequisite for more advanced Statistics courses. Weekly problem sets, two hourly exams, and one final exam. [Cost:3] [WL:2; see the instructor for an override] (Sun)
466/IOE 466. Statistical Quality Control. Statistics 311 or IOE 365. (3). (Excl).
Design and analysis of procedures for forecasting and control of production processes. Topics include: attribute and variables; sampling plans; sequential sampling plans; rectifying control procedures; charting, smoothing, forecasting, and prediction of discrete time series.
470. The Design of Scientific Experiments. Stat. 311, 402, 412, or 426; or permission of instructor. (4). (Excl).
The objective of this course is to introduce students to the process of planning, designing and implementation of a study. Includes enumerative, Monte Carlo, observational and controlled randomized experimentation. Emphasis is on the conceptual framework not on the mathematical theory of design (e.g., Statistics 570).
501. Applied Statistics II. Stat. 500 or permission of instructor. (3). (Excl).
Topics in applied multivariate analysis including Hotelling's T2 multivariate ANOVA, discriminant functions, factor analysis, principal components, canonical correlations, and cluster analysis. Selected topics from: maximum likelihood and Bayesian methods, robust estimation and survey sampling. Applications and data analysis using the computer will be stressed. (Faraway)
511. Mathematical Statistics II. Stat. 510. (3). (Excl).
Topics covered will include: hypothesis testing and related topics such as size, power, similarity and optimality properties. Likelihood ratio tests, generalized likelihood ratio tests, decision theory and Bayes approaches. Sequential procedures, large sample theory and various other topics. [Cost:2] [WL:3] (Keaner)
525(510)/Math. 525. Probability Theory. Math. 450 or 451; or permission of instructor. Students with credit for Math. 425/Stat. 425 can elect Math. 525/Stat. 525 for only 1 credit. (3). (Excl).
See Math 525.
526/Math. 526. Discrete State Stochastic Processes. Math. 525, or Stat. 525, or EECS 501. (3). (Excl).
Generating functions: recurrent events and the renewal theorem; random walks. Markov chains; branching processes; limit theorems; Markov chains in continuos time with emphasis on birth and death processes and queueing theory. An introduction to Brownian motion, stationary processes and martingales. (Belisle)
531. Statistical Analysis of Time Series. Stat. 426. (3). (Excl).
The major topics include time- and frequency-domain characteristics of stationary discrete time series, autoregressive and moving average models, prediction theory, estimation and hypothesis testing, and computer applications. Special topics might include vector autoregression, cross-spectral analysis, causality testing or other issues of current interest. Statistics 511 or Economics 775 is the standard prerequisite. Student evaluation is based on exams, homework, and a term paper. Lectures and problem sets including computer exercises are the main methods of instruction. (Howrey)
576/Econ. 776. Econometric Theory II. Econ. 775 or equivalent. (3). (Excl).
This is a course in advanced econometrics. It includes a thorough treatment of the general linear model, a development of simultaneous equation techniques, and an introduction to nonlinear models. Maximum likelihood and generalized method-of-moments estimators are rigorously treated. [Cost:3] [WL:4] (Mackie-Mason)
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Trademarks of the University of Michigan may not be electronically or otherwise altered or separated from this document or used for any non-University purpose. | 2,123 | 10,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-47 | latest | en | 0.922619 |
https://web2.0calc.com/questions/quadratic_95503 | 1,653,030,437,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531762.30/warc/CC-MAIN-20220520061824-20220520091824-00124.warc.gz | 675,263,332 | 5,389 | +0
0
47
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What real value of t produces the smallest value of the quadratic t^2 -12t - 36 + 18t - 4?
Mar 29, 2022
#1
+36416
+2
t^2 +6t -40 will have minumum value at t = - b/2a = - (6)/(2*1) = - 3
Mar 29, 2022 | 103 | 226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-21 | longest | en | 0.469395 |
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