text: Flow Computation '' is the set V is the set of all, most of its cool are... Adobe Flash plugin is needed to view this content use in your PowerPoint Presentations the moment you need them O. )bD-.6, /ProcSet 2 0 R cuai3F2WgYk\U@:]Z4qHG?s-Ef7pTP>s4s6VCIcZSh;M[Gr%+1!A/a2Un\,EMDi4@ /F6 7 0 R ]nf4>N!YgGB_\ZmGP?a"F4-jAfknck@NF:c'0/0MCPT^#b5AW%4 38 0 obj aG. The following sections present Python and C# programs to find the maximum flow from the source (0) to the sink (4). Objectives and Agenda: 1 l+^UE4HN ) # _t27 Y ; Vi2- graph Theorem 00FK ( 0 of in... O *,6kb= ; t ( TdjAPK: XE3UNK\tAIRN6W1ZOfs0 '' & a loop while there an.! ] b) Incoming flow is equal to outgoing flow for every vertex except s and t. 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The path is indicated by green forward arcs and the flow on the path is denoted f . Lemma. 8Pbk ; ( ^ ( 3I @... Niqwi8Rftshre > ^lib! Most of its cool features are free and easy to use its rightful owner finding a feasible through... And Agenda: 1. 6 0 obj A key question is how self-governing owners in the network can cooperate with each other to maintain a reliable flow. Mkmoqu % _ (  IC Nl/3 * P/=g_H  e+C, hh+c $U-! This presentation, you 'll need to allow Flash '' 6HLYZNA? <> 6N... Algorithms by Kleinberg and Tardos denote capacities let c denote edge costs ;, 2J. !aRk)ISX+$1^a#.mgc2HXHq]GU2.Z/=8:.e )VqG-=/NRjY1i->ZL]TfY:]Y(h![l5Qb(V6?qu. 13 grudnia 2020 "?Qq/#\8>kK(s]1(CEX$#&cj8cuc86 /ProcSet 2 0 R /F7 17 0 R stream /Font *fD\"PrAqjLF[sX? Templatesâ from Presentations Magazine network would allow to flow from source to sink with positive capacity 2 [ & )! The minimum-cost flow problem (MCFP) is an optimization and decision problem to find the cheapest possible way of sending a certain amount of flow through a flow network.A typical application of this problem involves finding the best delivery route from a factory to a warehouse where the road network has some capacity and cost associated. endobj Max Flow Theorem. :3lYJ6P/!qY8==R%\Bn6\; =^>%56A_GEF_[? The first step in determining the maximum possible flow of railroad cars through the rail system is to choose any path arbitrarily from origin to destination and ship as much as possible on that path. Prerequisite : Max Flow Problem … congestion time. In every network, the maximum flow equals the cost of the st-mincut Max flow = min cut = 7 Next: the augmented path algorithm for computing the max-flow/min-cut Maxflow Algorithms Augmenting Path Based Algorithms 1. 60 0 obj !LLriEt4KF\/N:l&?nL+7Q'!/@]t4V1"WCaTKU.5UJfUsSHRrBBaN:nG;fHqNol endobj %WSU6n/-5\]KARhSnkcq(]H@0,6%=4LQ,elPe:Ia.k(iqPVKl-TI+"=Ums8C)K+F oW)Cj_6(PL.e>i=#(2M(?,)1hih#TET2>A76iZrSaT>3#(#0&<288$(7WZhRcR ?O)f#$"i%j#*!KM\cs=4h0c3&B)*lXmiID$2W;/WE:XY[H$'>?fP'#GNP\r? ⇐ Suppose max flow value is k. By integrality theorem, there exists {0, 1} flow f of value k. Consider edge (s,v) with f(s,v) = 1. ... Lec-22 Maximum Flow Problem - … • If t 6∈S, then S is a saturated cut and f is maximum. Uncertain conditions effect on proper estimation and ignoring them may mislead decision makers by overestimation. This study investigates a multiowner maximum-flow network problem, which suffers from risky events. The max-flow problem and min-cut problem can be formulated as two primal-dual linear programs. Vietnamese Books Pdf, 8Pbk ; ( ^ ( 3I ) @ Q3T the of! Ghm\Oq: = 00FK ( 0 loop! Max flow formulation: assign unit capacity to every edge. Ovation Award for âBest PowerPoint Templatesâ from Presentations Magazine maximum-flow algorithm for Large Small-World Graphs! endobj Cost flow Notations: directed graph G= ( V, Ef ) Ef ( u, V ) ZJW_. Solve practice problems for Maximum flow to test your programming skills. RaudiY^? 6fP9s;CSVHAYR[B&:CEKISe#1MU68%&4m4\Re]RW?ts4X!Z;8uHDPAP5g4]PWN7OZ "KT\!F2Q;Z2_MV)G:\X!4mi&;RNuaHimQ-T%RZV/;K^:bHZAgGInZn3jY4p8R+SS,mGJh7pJR@cKS) >> 7RuafU>)JklS\g;(R"#g3&HAqERr5\)Y4uuY'0BLk/!Ba#i)e"IIM[N^;s&HV;rtO 00FK(0. A ow of f(v;w) units on edge (v;w) contributes cost c(v;w)f(v;w) to the objective function. It is a classic academic question to assess whether you can see beyond the confusion and look at only what matters. Given as input a table that specifies which widgets and boxes can go together, find some way to fit all n widgets one to a box. Given a ﬂow f in a ﬂow network G0, constructed for a bipartite graph G = (V;E), consider M = f(a;b) 2E jf(a;b) = 1 g: Then, v(f) = #M and M is a matching in G. Corollary. It is defined as the maximum amount of flow that the network would allow to flow from source to sink. • Maximum flow problems find a feasible flow through a single-source, single-sink flow network that is maximum. Minimum Cost Flow Notations: Directed graph G= (V;E) Let u denote capacities Let c denote edge costs. All nodes reachable from s in Gf support us by buying something from amazon 3 Add an from! The main theorem links the maximum flow through a network with the minimum cut of the network. GJLiar_Jr!0.sA>B_ijjK*&OadkG]D1_7Ut2'\k5W4&-u":2LKjEd(;(Inso[ Is labeled with capacity, the size of the above graph is 23, memorable appearance - the kind problems! 8Pbk ; ( ^ ( 3I ) @?. 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Ford Fulkerson algorithm for Maximum Flow Problem Example - Duration: 13:13. 'dP % D &. There is another augmenting path in the graph, s → 1 → t , with both arcs used in the forward direction. gVQ3 # 5eE.EcYGe capacity uij of industry a reliable flow ; peHXe Add an edge from vertex... F, and let s be the residual graph Theorem M: kBtW$! RaudiY^? <>/Metadata 1117 0 R/ViewerPreferences 1118 0 R>> Uq7, @ % 5iHOc52SDb ] ZJW_ s is a special case of the interior of... 6Hlyzna? The maximum possible flow in the above graph is 23. Raspberry Symbolic Meaning. a) Flow on an edge doesn’t exceed the given capacity of the edge. * W\__F3L_/VAF4? 1 0 obj endobj ul. 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Finding maximum or minimum values is called optimisation K$ gY ; OL #? Strajku Łódzkich Studentów w 1981r. There are specialized algorithms that can be used to solve for the maximum flow. We run a loop while there is an augmenting path. Augmenting flow â¦ maximum flow a key question is how self-governing owners the. If the value of the ﬂow f in G0is maximal over all ﬂows, then the matching M corresponding to f is maximal as well. You have n widgets to put in n boxes, but the widgets and boxes are highly individualized and not all widgets will fit in all boxes. The minimum arc flow and arc capacities are specified as lower and upper bounds in square brackets, respectively. The bottleneck here is the arc B-E. 0/r?Y^M7+=/+5Ihf[n-eh+Tkqo9?os/McYD6ZaT1Ks(F#qD45O>jL /Length 28 0 R Or use it to upload your own PowerPoint slides so you can share them with your teachers, class, students, bosses, employees, customers, potential investors or the world. 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text:
# Pulling a box by a 10N force

1. Mar 8, 2007

### future_vet

1. The problem statement, all variables and given/known data
A 2.0 kg box is pulled along a horizontal plank by a 10N force parallel to the direction of motion. The rate of acceleration is 0.50 m/s^2. Find the coefficient of kinetic friction.

I think we use the formula: coefficient of friction = F (so 10) / N.
But I am not sure how to use the acceleration and the N variable.

2. Relevant equations
F - μk mg = ma
10 - 2(9.8)μk = 2 x 0.5

Would this be the correct way to go?

Thanks!

2. Mar 8, 2007

### PhanthomJay

Yes, but it contradicts what you said in the problem statement.

3. Mar 9, 2007

### future_vet

Yes, so I use the second equation? Is the data correctly plugged in?

4. Mar 9, 2007

### cristo

Staff Emeritus
Yes it is.

5. Mar 9, 2007

### future_vet

I am getting -0.005 as the answer... does it sound right? Where should the acceleration be used?

Thanks for all the help! :)

6. Mar 9, 2007

### PhanthomJay

you have the correct equation, with the given acceleration entered correctly, but your math is off.
$$10 -19.6\mu_k = 2(.5) = 1$$
$$-19.6\mu_k = -9$$
$$\mu_k = ?$$

7. Mar 9, 2007

### future_vet

oops, it's 0.46, right?

8. Mar 9, 2007

### PhanthomJay

yes, correct.

format: plain


text:
# naginterfaces.library.correg.coeffs_pearson_subset¶

naginterfaces.library.correg.coeffs_pearson_subset(x, kvar)[source]

coeffs_pearson_subset computes means and standard deviations, sums of squares and cross-products of deviations from means, and Pearson product-moment correlation coefficients for selected variables.

For full information please refer to the NAG Library document for g02bg

https://www.nag.com/numeric/nl/nagdoc_29/flhtml/g02/g02bgf.html

Parameters
xfloat, array-like, shape

must be set to , the value of the th observation on the th variable, for , for .

kvarint, array-like, shape

must be set to the column number in of the th variable for which information is required, for .

Returns
xbarfloat, ndarray, shape

The mean value, , of the variable specified in , for .

stdfloat, ndarray, shape

The standard deviation, , of the variable specified in , for .

sspfloat, ndarray, shape

is the cross-product of deviations, , for the variables specified in and , for , for .

rfloat, ndarray, shape

is the product-moment correlation coefficient, , between the variables specified in and , for , for .

Raises
NagValueError
(errno )

On entry, .

Constraint: .

(errno )

On entry, and .

Constraint: and .

(errno )

On entry, , and .

Constraint: .

Notes

No equivalent traditional C interface for this routine exists in the NAG Library.

The input data consist of observations for each of variables, given as an array

where is the th observation on the th variable, together with the subset of these variables, , for which information is required.

The quantities calculated are:

1. Means:

2. Standard deviations:

3. Sums of squares and cross-products of deviations from zero:

4. Pearson product-moment correlation coefficients:

If or is zero, is set to zero.


text:
# Defining a function for the construction of a bordered hessian

I think in Mathematica the omission of a function that can create a bordered matrix. This code works

m = {{a, b}, {c, d}};
k = {α, β}
l1 = Insert[ k, 0, 1];
c1 = Insert[Transpose[m], k, 1] // Transpose;
m1 = Insert[c1, l1, 1];
m1 // MatrixForm

and gives

$\begin{pmatrix} 0 & \alpha & \beta\\ \alpha & a & b\\ \beta & c & d \end{pmatrix}$

as expected.

So I have tried to design a function. But obviously it doesn't work

bordhess[mat_?MatrixQ[mat], vec_?VecQ[vec] ] :=
Block[{x = mat, y = vec},
l1 = Insert[ y, 0, 1];
c1 = Insert[Transpose[x], k, 1] // Transpose;
Insert[c1, l1, 1]
]

and even If I have been able to create such a function, I wonder how to generalize the problem to many vectors.

• ArrayFlatten[{{0, {k}}, {Transpose@{k}, m}}] Jan 18 '17 at 9:10
• Thanks too much simple for my complicated mind! Jan 18 '17 at 9:18
• @SimonRochester You should make this an answer :) Jan 18 '17 at 10:07

## 3 Answers

Also,

borderF = ArrayReshape[{0, #, Transpose[{##}]}, 1 + Dimensions@#2] &;

mat = Array[Subscript[a, ##] &, {4, 4}];
multipliers = Array[Subscript[λ, #] &, 4];
borderF[multipliers, mat] // MatrixForm

• Does this do the multi-row block-bordering shown in the last images of Simon's and my answers? I believe that's what the OP is asking about (when wondering "how to generalize the problem to many vectors"), but I could be wrong. Jan 19 '17 at 1:58
• @MichaelE2, no it doesn't handle multi-row borders. Still trying to figure out the appropriate argument for ArrayReshape to get multi-row borders.
– kglr
Jan 19 '17 at 2:14
• @cyrcyrille.piatecki , i suppose the Accept is a mistake:) considering the generality and elegance of Simon's and Michal's answers.
– kglr
Jan 19 '17 at 3:00
• Of course, I have checked yours as I could have checked an other because all were satisfying I should have checked the more elegant but I do not know <hich is the one. Perhaps we could ask for the possibility to multi check some time. Jan 19 '17 at 12:38

You can do this with ArrayFlatten. Here is a generalization to one or more vectors -- I'm not sure if it's what you're looking for:

bordhess[mat_?MatrixQ, mat1_?MatrixQ] := ArrayFlatten[{{0, mat1}, {Transpose@mat1, mat}}]

bordhess[mat_?MatrixQ, vec_?VectorQ] := bordhess[mat, {vec}]

Then you have

m = {{a, b}, {c, d}};
k = {α, β};
j = {γ, δ};

bordhess[m, k] // MatrixForm

bordhess[m, {k, j}] // MatrixForm

Here's an answer to the title question, about constructing a bordered Hessian, in case someone come looking for answer to it. It comes directly from calculus, instead of playing with matrices.

Basically thus:

D[f + λ.g, {{λ1, λ2,..., x1, x2,...}, 2}]

where

λ = {λ1, λ2,...}

are the Lagrange multipliers and

x = {x1, x2,...}

are the variables of f and g.

Example:

ClearAll[bH];
bH[f_, g_List, x_List, param_: K] := bH[f, g, x, Array[param, Length@g]];
bH[f_, g_List, x_List, λ_List] := D[f + λ.g, {Join[λ, x], 2}];

bH[w^4 x + x^3 y + y^2 z + z^4,
{w^2 + x^2 + y^2 + z^2, w^2 + 2 x^2 + 3 y^2 + 4 z^2},
{w, x, y, z}, {λ, μ}] // MatrixForm

• Bordered Hessian can also be used in checking quasi-concavity Jan 19 '17 at 12:40
• @cyrille.piatecki Yes, second derivatives are useful. (I'm not sure why the comment, actually. I suppose you mean specializing at bH[f, {f}, {x, y,..}, {λ}] /. λ -> 0.) Jan 19 '17 at 13:19
• Michael E2 Not exactly. In economics, one is obliged to assume that every upper level set of the consumer's ordering is convex, which is equivalent to the condition that any function that represents the consumer's ordering is quasiconcave (for definition see mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/…). If the function is $C^2$ the condition of quasiconcavity is linked to the bordered hessian Jan 20 '17 at 6:01
• -> continued --- principal minors should be alternatively negative/ positive beginning with the second order --- by construction the first order is 0. If the function is strtictly quasi-convex then on has a maximum. Jan 20 '17 at 6:01
• @cyrille.piatecki It seems the same thing, no? At least the formula for bordered Hessian in the ref. link in your comment is the same. Jan 20 '17 at 10:34


text:
# Gay Lussac's Law — Avogadro's Law — Mole Concept

## Lussac's Law

#### Question 1

Nitrogen reacts with hydrogen to give ammonia. Calculate the volume of the ammonia gas formed when nitrogen reacts with 6 litres of hydrogen. All volumes measured at s.t.p.

[By Lussac's law]

$\begin{matrix} \text{N}_2 & + & 3\text{H}_2 & \longrightarrow & 2 \text{NH}_3 \\ 1 \text{ vol.} & : & 3 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}$

To calculate the volume of ammonia gas formed.

$\begin{matrix} \text{H}_2 & : & \text{NH}_3 \\ 3 & : & 2 \\ 6 & : & x \end{matrix}$

Therefore,

$\dfrac{2}{3} \times 6 = x \\[0.5em] \Rightarrow x = 4 \text{ lts}$

Hence, volume of ammonia gas formed is 4 lts

#### Question 2

2500 cc of oxygen was burnt with 600 cc of ethane [C2H6]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed.

[By Lussac's law]

$\begin{matrix} \text{\scriptsize{2C}}_2\text{\scriptsize{H}}_6 & \text{\scriptsize{+}} & \text{\scriptsize{7O}}_2 & \longrightarrow & \text{\scriptsize{4CO}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{6H}}_2\text{\scriptsize{O}} \\ \text{\scriptsize{2 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{7 vol.}} & \longrightarrow & \text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{6 vol.}} \end{matrix}$

To calculate the volume of unused oxygen.

$\begin{matrix} \text{C}_2\text{H}_6 & : &\text{O}_2 \\ 2 & : & 7 \\ 600 & : & x \end{matrix}$

Therefore,

$\dfrac{7}{2} \times 600 = x \\[0.5em] \Rightarrow x = 2100 \text{cc}$

Therefore, volume of unused oxygen = 2500 - 2100 = 400 cc

To calculate the volume of carbon dioxide formed

$\begin{matrix} \text{C}_2\text{H}_6 & : & \text{CO}_2 \\ 2 & : & 4 \\ 600 & : & x \end{matrix}$

Therefore,

$\dfrac{4}{2} \times 600 = x \\[0.5em] \Rightarrow x = 1200 \text{cc}$

Therefore, volume of carbon dioxide formed is 1200 cc

#### Question 3

20 ml. each of oxygen and hydrogen and 10 ml. of carbon monoxide are exploded in an enclosure. What will be the volume and composition of the mixture of the gases when they are cooled to room temperature.

Given,

20 ml. each of oxygen and hydrogen and 10 ml. of carbon monoxide

[By Lussac's law]

$\begin{matrix} 2\text{CO}& + & \text{O}_2 & \longrightarrow & 2\text{CO}_2 & \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} & \\ \end{matrix}$

To calculate the amount of CO2 produced,

$\begin{matrix} \text{CO}& : & \text{CO}_2 & \\ 2 \text{ vol.} & : & 2 \text{ vol.} & \\ 10 \text{ ml} & : & x \text{ ml} & \\ \end{matrix}$

Therefore, CO2 produced is 10 ml

To calculate the amount of O2 used,

$\begin{matrix} \text{CO}& : & \text{O}_2 & \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \\ 10 \text{ ml} & : & x \text{ ml} & \\ \end{matrix}$

$\dfrac{1}{2} \times 10 = x \\[0.5em] \Rightarrow x = 5 \text{ ml}$

From relation,

$\begin{matrix} 2\text{H}_2 & + & \text{O}_2 & \longrightarrow & 2\text{H}_2 \text {O} & \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} & \\ \end{matrix}$

$\begin{matrix} \text{H}_2 & : & \text{O}_2 & \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \\ 20 \text{ ml} & : & x \text{ ml} & \\ \end{matrix}$

$\dfrac{1}{2} \times 20 = x \\[0.5em] \Rightarrow x = 10 \text{ ml}$

Therefore, total vol. of oxygen used = 10 + 5 = 15 ml

Hence, oxygen left = 20 - 15 = 5 ml

Therefore, oxygen left is 5 ml and CO2 produced is 10 ml.

#### Question 4

224 cm3 of ammonia undergoes catalytic oxidation in presence of Pt to given nitric oxide and water vapour. Calculate the volume of oxygen required for the reaction. All volumes measured at room temperature and pressure.

[By Lussac's law]

$\begin{matrix} \text{\scriptsize{4NH}}_3 & \text{\scriptsize{+}} & \text{\scriptsize{5O}}_2 & \longrightarrow & \text{\scriptsize{4NO}} & \text{\scriptsize{+}} & \text{\scriptsize{6H}}_2\text{\scriptsize{O}} \\ \text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{5 vol.}} & \longrightarrow &\text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{6 vol.}} \end{matrix}$

To calculate the volume of oxygen required.

$\begin{matrix} \text{NH}_3 & : & \text{O}_2 & \\ 4 & : & 5 \\ 224 & : & x \end{matrix}$

Therefore,

$\dfrac{5}{4} \times 224 = x \\[0.5em] \Rightarrow x = 280 \text{ cm}^3$

Therefore, volume of oxygen required is 280 cm3.

#### Question 5

Acetylene [C2H2] burns in air forming carbon dioxide and water vapour. Calculate the volume of air required to completely burn 50 cm3 of acetylene. [Assume air contains 20% oxygen].

[By Lussac's law]

$\begin{matrix} \text{\scriptsize{2C}}_2 \text{\scriptsize{H}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{5O}}_2 & \longrightarrow & \text{\scriptsize{4CO}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{2H}}_2\text{\scriptsize{O}} \\ \text{\scriptsize{2 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{5 vol.}} & \longrightarrow & \text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{2 vol.}} \end{matrix}$

To calculate the volume of air required,

$\begin{matrix} \text{C}_2 \text{H}_2 & : & \text{O}_2 & \\ 2 & : & 5 \\ 50 & : & x \end{matrix}$

Therefore, volume of oxygen (x),

$\dfrac{5}{2} \times 50 = x \\[0.5em] \Rightarrow x = 125 \text{ cm}^3$

When oxygen is 20% then air is 100%
Therefore when, oxygen is 125 cm3 then air is

$\dfrac{100}{20} \times 125 = x \\[0.5em] \Rightarrow x = 625 \text{ cm}^3$

Therefore, volume of air required is 625 cm3.

#### Question 6

On igniting a mixture of acetylene [C2H2] and oxygen, 200 cm3 of CO2 is collected at s.t.p. Calculate the volume of acetylene & O2 at s.t.p. in the original mixture.

[By Lussac's law]

$\begin{matrix} \text{\scriptsize{2C}}_2 \text{\scriptsize{H}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{5O}}_2 & \longrightarrow & \text{\scriptsize{4CO}}_2 & \text{\scriptsize{+}} & \text{\scriptsize{2H}}_2\text{\scriptsize{O}} \\ \text{\scriptsize{2 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{5 vol.}} & \longrightarrow & \text{\scriptsize{4 vol.}} & \text{\scriptsize{:}} & \text{\scriptsize{2 vol.}} \end{matrix}$

To calculate the volume of acetylene :

$\begin{matrix} \text{CO}_2 & : & \text{C}_2 \text{H}_2 & \\ 4 & : & 2 \\ 200 & : & x \end{matrix}$

Therefore,

$\dfrac{2}{4} \times 200 = x \\[0.5em] \Rightarrow x = 100 \text { cm}^3$

Hence, volume of acetylene in the original mixture is 100 cm3.

To calculate the volume of oxygen :

$\begin{matrix} \text{CO}_2 & : & \text{O}_2 & \\ 4 & : & 5 \\ 200 & : & x \end{matrix}$

Therefore,

$\dfrac{5}{4} \times 200 = x \\[0.5em] \Rightarrow x = 250 \text { cm}^3$

Hence, volume of oxygen in the original mixture is 250 cm3.

#### Question 7

Ammonia is formed from the reactants nitrogen and hydrogen in presence of a catalyst under suitable conditions. Assuming all volumes are measured in litres at s.t.p. Calculate the volume of ammonia formed if only 10% conversion has taken place.

[By Lussac's law]

$\begin{matrix} \text{N}_2 & + & 3\text{H}_2 & \longrightarrow & 2\text{NH}_3 & \\ 1 \text{ vol.} & : & 3 \text{ vol.} & \longrightarrow & 2 \text{ vol.} & \\ 1 \text{ lit.} & : & 3 \text{ lit.} & \longrightarrow & 2 \text{ lit.} & \end{matrix}$

To calculate the volume of ammonia formed when only 10% conversion takes place,

$\therefore 10\% \text{ of 2 lit} = \dfrac{10}{100} \times 2 = 0.2 \text{ lit}$

Hence, volume of ammonia formed is 0.2 lit or 20% or 1/5th of vol of N2 and H2

#### Question 8

100 cc. each of water gas and oxygen are ignited and the resultant mixture of gases cooled to room temp. Calculate the composition of the resultant mixture. [Water gas contains CO and H2 in equal ratio]

Given,

Vol of CO = 50 cc and H2 = 50 cc [As Water gas contains CO and H2 in equal ratio] and O2 = 100 cc

[By Lussac's law]

$\underset{\text{\scriptsize{1 vol.}}}{\text{\scriptsize{CO}}} \underset{:}{\text{\scriptsize{+}}} \underset{\text{\scriptsize{1 vol.}}}{\text{\scriptsize{H}}_2} \underset{:}{\text{\scriptsize{+}}} \underset{\text{\scriptsize{1 vol.}}}{\text{\scriptsize{O}}_2} \longrightarrow \underset{\text{\scriptsize{1 vol}}}{\text{\scriptsize{CO}}_2} \underset{:}{\text{\scriptsize{ +}}} \underset{\text{\scriptsize{1 vol}}}{\text{\scriptsize{ H}}_2 \text{\scriptsize{O}}}$

To calculate the amount of O2 used,

$\begin{matrix} \text{CO}& : & \text{O}_2 & \\ 1 \text{ vol.} & : & 1 \text{ vol.} & \\ 50 \text{ cc} & : & x \text{ cc} & \\ \end{matrix}$

As ratio between CO and O2 is same so, O2 used is 50 cc.

Hence, remaining O2 = 100 - 50 = 50 cc.

To calculate the amount of CO2 produced,

$\begin{matrix} \text{CO}& : & \text{CO}_2 & \\ 1 \text{ vol.} & : & 1 \text{ vol.} & \\ 50 \text{ cc} & : & y \text{ cc} & \\ \end{matrix}$

1 vol of CO produces 1 vol. of CO2
Hence, CO2 produced = 50 cc.

Therefore, the resultant mixture has 50 cc of O2 + 50 cc of CO2

## Mole Concept — Avogadro's Law and No.

#### Question 1

Calculate the mass of 2.8 litres of CO2. [C = 12, O = 16]

[1 mole = 1 gm mol. wt. and occupies 22.4 lit at s.t.p.]

gm mol. wt. of CO2 = 12 + (16 x 2) = 44 g

1 mole of CO2 = 1 gm mol. wt.

22.4 lit of CO2 has mass 44 g [s.t.p.]

Therefore, 2.8 lit of CO2 has mass

$= \dfrac{44}{22.4} \times 2.8 = 5.5 \text{ g}$

Therefore, mass of 2.8 litres of CO2 is 5.5 g

#### Question 2

Calculate the volume occupied by 53.5 g of Cl2. [Cl = 35.5]

[1 mole = 1 gm mol. wt. and occupies 22.4 lit at s.t.p.]

gm mol. wt. of Cl2 = (35.5 x 2) = 71 g

1 mole of Cl2 = 1 gm mol. wt. and occupies 22.4 lit.

71 g of Cl2 occupies 22.4 lit [s.t.p.]

∴ Vol. occupied by 53.5 g of Cl2

$= \dfrac{22.4}{71} \times 53.5 = 16.87 \text{ lit}$

Therefore, volume occupied by 53.5 g of Cl2 is 16.87 lit

#### Question 3

Calculate the number of molecules in 109.5 g of HCl. [H = 1, Cl = 35.5]

gm mol. wt. of HCl = 1 + 35.5 = 36.5 g

At s.t.p.,

36.5 g of HCl = 6.023 x 1023 number of molecules    [Avogadro's law]

Therefore, 109.5 g of HCl

$= \dfrac{6.023 \times 10^{23}}{36.5} \times 109.5 \\[0.5em] = \dfrac{109.5}{36.5} \times 6.023 \times 10^{23} \\[0.5em] = 3 \times 6.023 \times 10^{23} \\[0.5em]$

Hence, number of molecules in 109.5 g of HCl is 3 x 6.023 x 1023

#### Question 4

Calculate the number of :

(i) molecules [S = 32]

(ii) atoms in 192 g. of sulphur. [S8]

(i) 1 mole of any substance contains 6.023 x 1023 number of molecules.

S8 = 8 atoms = 8 x 32 = 256.

256 g = 6.023 x 1023 number of molecules

So, 192 g will have

$= \dfrac{6.023 \times 10^{23}}{256} \times 192 \\[0.5em] = \dfrac{192}{256} \times 6.023 \times 10^{23} \\[0.5em] = 0.75 \times 6.023 \times 10^{23} \text { molecules} \\[0.5em]$

(ii) Gram atomic mass of S = 32 g

32 g of S has 6.023 x 1023 number of atoms

Therefore, 192 g will have

$= \dfrac{6.023 \times 10^{23}}{32} \times 192 \\[0.5em] = \dfrac{192}{32} \times 6.023 \times 10^{23} \\[0.5em] = 6 \times 6.023 \times 10^{23} \text { atoms} \\[0.5em]$

#### Question 5

Calculate the mass of Na which will contain 6.023 × 1023 atoms. [Na = 23]

As, a mole of atoms contain 6.023 × 1023 atoms [Avogadro number] and has weight [mass] equal to gram atomic mass of the element.

Therefore, a mole of Na will contain 6.023 × 1023 atoms and will have mass equal to gram atomic mass of Na = 23 g

#### Question 6

Calculate the no. of atoms of potassium present in 117 g. of K. [K = 39]

Gram atomic mass of K = 39

At s.t.p.,

39 g of K = 6.023 x 1023 atoms of K    [Avogadro's law]

Therefore, 117 g of K

$= \dfrac{6.023 \times 10^{23}}{39} \times 117 \\[0.5em] = \dfrac{117}{39} \times 6.023 \times 10^{23} \\[0.5em] = 3 \times 6.023 \times 10^{23} \\[0.5em]$

Hence, number of atoms in 117 g of K = 3 x 6.023 x 1023

#### Question 7

Calculate the number of moles and molecules in 19.86 g. of Pb(NO3)2. [Pb = 207, N = 14, O = 16]

Gram molecular mass of Pb(NO3)2
= Pb + 2 [N x 3(O)]
= Pb + 2N + 6O
= 207 + (2 x 14) + (6 x 16)
= 207 + 28 + 96 = 331

As,

331 g of Pb(NO3)2 = 1 mole
Therefore, 19.86 g

$= \dfrac{1}{331} \times 19.86 = 0.06 \text { moles} \\[0.5em]$

Hence, number of moles in 19.86 g. of Pb(NO3)2 = 0.06 moles

As,

1 mole of Pb(NO3)2 weighs 331 g and has 6.023 x 1023 molecules

Therefore, 19.86 g of Pb(NO3)2 will have

$= \dfrac{ 6.023 \times 10^{23}}{331} \times 19.86 \\[0.5em] = \dfrac{19.86}{331} \times 6.023 \times 10^{23} \\[0.5em] = 0.06 \times 6.023 \times 10^{23} \text{molecules}$

Hence, number of molecules in 19.86 g. of Pb(NO3)2 = 0.06 x 6.023 x 1023 moles

#### Question 8

Calculate the mass of an atom of lead [Pb = 202].

As,

1 mole of Pb weighs 202 g and has 6.023 x 1023 atoms.

So, 6.023 x 1023 atoms of Pb has mass = 202 g

Therefore, 1 atom of Pb will have mass

$= \dfrac{202}{6.023 \times 10^{23}} = 33.53 \times 10^{-23} \text{g}$

Hence, mass of an atom of lead is 33.53 x 10-23 g

#### Question 9

Calculate the number of molecules in 1½ litres of water. [density of water 1.0 g./cc. ∴ mass of water = volume × density]

Given,

Vol of water = 1.5 lit = 1500 mL =1500 cc

Density of water 1.0 g./cc..

∴ Mass of water = volume × density

Hence, mass of water = 1500 x 1 = 1500 g

Gram molecular mass of water

= 2H + O

= (2 x 1) + 16 = 18 g

So, 18 g of water = 6.023 x 1023 molecules

Therefore, 1500 g of water will have

$\dfrac{6.023 \times 10^{23}}{18} \times 1500 \\[0.5em] = \dfrac{1500}{18} \times 6.023 \times 10^{23} \\[0.5em] = 83.33 \times 6.023 \times 10^{23}$

Hence, the number of molecules in 1½ litres of water is 83.33 x 6.023 x 1023 molecules.

#### Question 10

Calculate the gram-atoms in 88.75 g of chlorine [Cl = 35.5]

Gram atoms is the relative atomic mass of an element expressed in grams.

$\text{Gram atoms} = \dfrac{\text{Mass in grams}}{\text{Relative At. mass [At. wt.]}} \\[0.5em] = \dfrac{88.75}{35.5} \\[0.5em] = 2.5 \text { gram atoms} \\[0.5em]$

Hence, gram-atoms in 88.75 g of chlorine is 2.5 g. atoms

#### Question 11

Calculate the number of hydrogen atoms in 0.25 mole of H2SO4.

1 mole of hydrogen atom has 2 x 6.023 x 1023 atoms.

∴ 0.25 mole will have

$= \dfrac{2 \times 6.023 \times 10^{23}}{1} \times 0.25 \\[0.5em] = 0.5 \times 6.023 \times 10^{23} \text{ atoms}$

Hence, number of hydrogen atoms in 0.25 mole of H2SO4 = 0.5 x 6.023 x 1023 particles

#### Question 12

Calculate the gram molecules in 21 g of nitrogen [N = 14]

Gram molecules is the relative molecular mass of a substance expressed in grams.

Relative molecular mass of N2 = 2 x 14 = 28 g

$\text{Gram molecules} = \dfrac{\text{Mass in grams [of nitrogen]}}{\text{Rel. molecular mass [Mol. wt.]}} \\[0.5em] = \dfrac{21}{28} \\[0.5em] = 0.75 \text { gram molecules}$

Hence, gram molecules in 21 g of nitrogen = 0.75 gram molecules

#### Question 13

Calculate the number of atoms in 10 litres of ammonia [N = 14, H = 1]

1 mole of NH3 = N + 3 H = 4 atoms = 4 x 6.023 x 1023 atoms and occupies 22.4 lit at s.t.p.

If 22.4 lit of NH3 has 4 x 6.023 x 1023 atoms,

then, 10 lit will have

$= \dfrac{4 \times 6.023 \times 10^{23}}{22.4} \times 10 \\[0.5em] = \dfrac{40}{22.4} \times 6.023 \times 10^{23} \\[0.5em] = 1.786 \times 6.023 \times 10^{23} \text{ atoms}$

Hence, number of atoms in 10 litres of ammonia = 1.786 x 6.023 x 1023 atoms

#### Question 14

Calculate the number of atoms in 60 g of neon [Ne = 20]

Gram atomic mass of Ne = 20 g

As, 1 mole of Ne weighs 20 g and has 6.023 x 1023 atoms

So, 60 g of Ne will have

$\dfrac{ 6.023 \times 10^{23}}{20} \times 60 \\[0.5em] = \dfrac{60}{20} \times 6.023 \times 10^{23} \\[0.5em] = 3 \times 6.023 \times 10^{23} \text{ atoms}$

Hence, number of atoms in 60 g of Ne = 3 x 6.023 x 1023 atoms

#### Question 15

Calculate the number of moles of 'X' atoms in 93 g of 'X' [X is phosphorus = 31]

Gram atomic mass of X (phosphorus) = 31 g

As, 31 g of phosphorus = 1 mole

So, 93 g of phosphorus = $\dfrac{1}{31} \times 93 = 3 \text{ moles}$

Hence, number of moles in 93 g of X is 3 moles

#### Question 16

Calculate the volume occupied by 3.5 g of O2 gas at 27 °C and 740 mm pressure. [O = 16]

Gram molecular mass of O2 = 2 x 16 = 32 g

1 mole of O2 weighs 32 g and occupies 22.4 lit. vol.

∴ 3.5 g of O2 occupies = $\dfrac{22.4}{32} \times 3.5 = 2.45 \text{ lit.}$

Volume occupied by 3.5 g of O2 gas at 27°C and 740 mm pressure:

s.t.p.given values
P1 = 760 mm of HgP2 = 740 mm of Hg
V1 = 2.45 litV2 = x lit
T1 = 273 KT2 = 27 + 273 K

Using the gas equation,

$\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$

Substituting the values we get,

$\dfrac{760 \times 2.45}{273} = \dfrac{740 \times x}{300} \\[0.5em] x = \dfrac{760 \times 2.45 \times 300}{740 \times 273 } \\[0.5em] x = \dfrac{5,58,600}{2,02,020} \\ \\[0.5em] x = 2.76 \text{ lit}$

Hence, the volume occupied by 3.5 g of O2 gas at 27°C and 740 mm pressure is 2.76 lit.

#### Question 17

Calculate the moles of sodium hydroxide contained in 160 g of it. [Na = 23, O = 16, H = 1]

Gram molecular mass of sodium hydroxide
= Na + O + H
= 23 + 16 + 1
= 40 g

As, 40 g of sodium hydroxide = 1 mole

∴ 160 g of sodium hydroxide = $\dfrac{ 1}{40} \times 160 = 4 \text{ moles}$

Hence, number of moles in 160 g of sodium hydroxide is 4

#### Question 18

Calculate the weight in g. of 2.5 moles of ethane [C2H6]. [C = 12, H = 1]

Gram molecular mass of C2H6 = (2 x 12) + (6 x 1)
= 24 + 6
= 30 g

1 mole of C2H6 weighs 30 g

∴ 2.5 moles weighs = $\dfrac{ 30}{1} \times 2.5 = 75 \text{ g}$

Hence, weight of 2.5 moles of ethane [C2H6] is 75 g.

#### Question 19

Calculate the molecular weight of 2.6 g of a gas which occupies 2.24 lits. at 0°C and 760 mm press.

Given,

2.24 lit volume of gas weighs 2.6 g

∴ 22.4 lit volume of gas weighs = $\dfrac{2.6}{2.24} \times 22.4 = 26 \text{ g}$

As 22.4 lit of gas at s.t.p. weighs 26 g
∴ Molecular weight = 26 g

#### Question 20

Calculate the gram atoms in 46 g of sodium [Na = 23]

Gram atoms is the relative atomic mass of an element expressed in grams.

$\text{Gram atoms} = \dfrac{\text{Mass in grams [of sodium]}}{\text{Rel. atomic mass [At. wt.]}} \\[0.5em] = \dfrac{46}{23} = 2 \text { gram atoms}$

Hence, gram-atoms in 46 g of sodium is 2 g. atoms

#### Question 21

Calculate the number of moles of KClO3 that will be required to give 6 moles of oxygen.

2KClO3 ⟶ 2KCl + 3O2

$\begin{matrix} \text{O}_2 & : & \text{KClO}_3 & \\ 3 & : & 2 \\ 6 & : & x \end{matrix}$

$\dfrac{2}{3} \times 6 = x \\[0.5em] \Rightarrow x = 4 \text{ moles}$

Hence, 4 moles of KClO3 are required to give 6 moles of oxygen.

#### Question 22

Calculate the weight of the substance if it's molecular weight is 70 and in the gaseous form occupies 10 lits. at 27°C and 700 mm pressure.

Initial ConditionsFinal Conditions (s.t.p.)
P1 = 700 mm of HgP2 = 760 mm of Hg
V1 = 10 litV2 = x lit
T1 = 27 + 273 KT2 = 273 K

Using the gas equation,

$\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$

Substituting the values we get,

$\dfrac{700 \times 10}{300} = \dfrac{760 \times x}{273} \\[0.5em] x = \dfrac{700 \times 10 \times 273}{300 \times 760 } \\[0.5em] x = \dfrac{1911}{228 } \\[0.5em] x = 8.38 \text{ lit}$

1 gram molecular weight of the gas occupies 22.4 lit. at s.t.p.

∴ 70 g of the gas occupies 22.4 lit. at s.t.p.

If $y$ g of the gas occupies 8.38 lits, then

$y = \dfrac{70}{22.4} \times 8.38 \\[0.5em] = 26.18 \text{ g}$

Hence, weight of substance is 26.18 g

#### Question 23

State which has higher number of moles : 5 g. of N2O or 5 g. of NO [N = 14, O = 16]

Gram molecular mass of N2O
= (2 x 14) + 16
= 44 g

44 g of N2O = 1 mole

∴ 5 g of N2O = $\dfrac{1}{44} \times 5 = 0.11$ moles

Gram molecular mass of NO

= 14 + 16
= 30 g

30 g of NO = 1 mole

∴ 5 g of NO = $\dfrac{1}{30} \times 5 = 0.16$ moles

Hence, 5 g. of NO has higher moles than 5 g. of N2O

#### Question 24

State which has higher mass : 1 mole of CO2 or 1 mole of CO [C = 12, O = 16]

Gram molecular mass of CO2
= 12 + (2 x 16)
= 12 + 32
= 44 g

1 mole of CO2 = 44 g

Gram molecular mass of CO
= 12 + 16
= 28 g

1 mole of CO = 28 g

Therefore, 1 mole of CO2 has higher mass.

#### Question 25

State which has higher no. of atoms : 1 g of O2 or 1 g of Cl2 [O = 16, Cl = 35.5]

Molecular mass of O2 = 2 x 16 = 32 g

32 g of O2 = 2 x 6.023 x 1023 atoms

∴ 1 g of O2

$= \dfrac{2 \times 6.023 \times 10^{23}}{32} \\[0.5em] = 0.0625 \times 6.023 \times 10^{23} \text{ atoms}$

Molecular mass of Cl2 = 2 x 35.5 = 71 g

71 g of Cl2 = 2 x 6.023 x 1023 atoms

∴ 1 g of Cl2

$= \dfrac{2 \times 6.023 \times 10^{23}}{71} \\[0.5em] = 0.0281 \times 6.023 \times 10^{23} \text{ atoms}$

Hence, 1 g of O2 has more number of atoms.

## Vapour Density And Molecular Weight

#### Question 1

500 ml. of a gas 'X' at s.t.p. weighs 0.50 g. Calculate the vapour density and molecular weight of the gas. [1 lit. of H2 at s.t.p. weighs 0.09 g].

Given, 500 ml. of gas 'X' at s.t.p. weighs 0.50 g

Therefore, 1000 ml of gas 'X' at s.t.p. will weigh (0.50 x 2) g

Vapour density of gas X =

$\dfrac{\text{Wt. of 1000 ml of gas at s.t.p.}}{\text{Wt. of 1000 ml of H}_2 \text{ at s.t.p.}} \\[0.5em] = \dfrac{0.50 \times 2}{0.09} \\[0.5em] = 11.1$

Hence, vapour density of gas is 11.11

Molecular weight = 2 x Vapour density
= 2 x 11.1
= 22.2 g

Hence, molecular weight of gas is 22.2 g

#### Question 2

A gas cylinder holds 85 g of a gas 'X'. The same cylinder when filled with hydrogen holds 8.5 g of hydrogen under the same conditions of temperature and pressure. Calculate the molecular weight of 'X'.

Vapour density of gas X =

$\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \dfrac{85}{8.5} \\[0.5em] = 10$

Molecular weight = 2 x Vapour density
= 2 x 10 = 20 g

Hence, molecular weight of gas X is 20 g

#### Question 3

Calculate the relative molecular mass [molecular weight] of 290 ml. of a gas 'A' at 17 °C and 1520 mm pressure which weighs 2.73 g at s.t.p. [1 litre of hydrogen at s.t.p. weighs 0.09 g.]

Convert the volume to s.t.p. using gas equation

Initial ConditionsFinal Conditions (s.t.p.)
P1 = 1520 mm of HgP2 = 760 mm of Hg
V1 = 290 mlV2 = x lit
T1 = 17 + 273 KT2 = 273 K

Using the gas equation,

$\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$

Substituting the values we get,

$\dfrac{1520 \times 290}{290} = \dfrac{760 \times x}{273} \\[0.5em] x = \dfrac{1520 \times 273}{760} \\[0.5em] x = 546 \text{ ml}$

546 ml of gas at s.t.p. weighs 2.73 g

∴ Wt. of 1000 ml of gas = $\dfrac{2.73}{546} \times 1000 = 5 \text { g}$

Vapour density of gas X =

$\dfrac{\text{Wt. of 1 lit of gas X}}{\text {Wt. of 1 lit of H}_2} \\[0.5em] = \dfrac{5}{0.09} \\[0.5em] = 55.55$

Hence, vapour density of gas is 55.55

Molecular weight = 2 x Vapour Density
= 2 x 55.555
= 111.11 g

Hence, molecular weight of gas is 111.11 g

#### Question 4

State the volume occupied by 40 g of a hydrocarbon – CH4 at s.t.p. if it's V.D. is 8.

Vapour density of hydrocarbon = 8

∴ Molecular weight = 2 x Vapour density
= 2 x 8 = 16 g

16 g of hydrocarbon at s.t.p. occupy 22.4 lit

∴ 40 g of hydrocarbon will occupy = $\dfrac{22.4}{16} \times 40 = 56 \text{ lit.}$

Hence, volume occupied by 40 g of a hydrocarbon (CH4) at s.t.p. is 56 lit.

#### Question 5

Calculate the atomicity of a gas X [at. no. 35.5] whose vapour density is equal to it's relative atomic mass.

Given,

Vapour density = relative atomic mass = 35.5

Molecular weight = 2 x Vapour Density = 2 x 35.5 = 71 g

Atomicity is the number of atoms present in one molecule of that element.

Number of atoms

$= \dfrac{\text{Molecular weight}}{\text{Atomic weight}} \\[0.5em] = \dfrac{\text{71}}{\text{35.5}} \\[0.5em] = 2$

Hence, atomicity of a gas X is 2

#### Question 6

Calculate the relative molecular mass and vapour density of methyl alcohol [CH3OH] if 160 g. of the alcohol on vaporization has a volume of 112 litres at s.t.p.

Given,

Weight of 112 lit of CH3OH = 160 g [at s.t.p.]

∴ Weight of 1 lit of CH3OH = $\dfrac{160}{112}$ = 1.4286 g

Weight of 1 lit of H2 = 0.09 g [at s.t.p.]

Under similar temperature and pressure,

Vapour density of gas methyl alcohol =

$\dfrac{\text{Wt. of certain vol. of gas}}{\text {Wt. of same vol. of H}_2} \\[0.5em] = \dfrac{1.4286}{0.09} \\[0.5em] = 15.87 \approx 16$

Molecular weight = 2 x Vapour density
= 2 x 15.87 = 31.74 g ≈ 32 g

Hence, molecular weight of methyl alcohol is 32 g and vapour density of methyl alcohol is 16


text:
6 deleted 124 characters in body

# Brain-Flak, 1101, 1085, 1085981 bytes

{({}<>)(<>())(<>{}<>)<(({}))((((()()()()()){}){}){})({}[{}]<(())>){((<{}{}>))}{}{
{}>{}(<{}>)(((()))<>)(<><{}><>)}{}<(({
}))((((()()()()()){}){}){}())({}[{}]<(())>){((
<{}{}>))}{}{{}>{}(<{}>)(((()))<>)({}[{}](<()>)){
{}{}(<(())>)}{}{<>{{}}<>{{}}((<()
>))}{}(<><{}><>)}{}<(({}))(((((()()()()()){}){})){}{}
)({}[{}]<(())>){((<{}{}>))}{}{
{}>{}(<({}>)(((())<>)>()<>)(<><{}><>)}{}<(({}))(((((()()()()()){}){})(
)){}{})({}[{}]<(()
)>){((<{}{}>))}{}{{}>{}(<({}>)(((())<>)>()<>)({}[{}](<()>)){{}{}(<(())>)}{}{
<>{{}}<>{
{}}((<()>))}{}(<><{}><>)}{}<(({}))((((()()()){}){}()){({}[()])}{})({}[{}]<(()
)>){((
<{}{}>))}{}{{}>{}(<({}>)(((()()<>)>()()<>)(<><{}>)}{}<(({}))((((((()()()()()){})){}{}()
)){}{}
)({}[{}]<(())>){((<{}{}>))}{}{{}>{}(<({}>)(((()))()()<>)>()({}[{}](<()>)){{}{}(
<(())>)}{}{<>{{
}}<>{{}}((<()>))}{}(<><{}><>)}{}<(({}))((((((()()()()()){}){}){}()))
{}{})({}[{}]<(())>
){((<{}{}>))}{}{{}>{}(<({}>)(((()))()()()<>)(<><{}>()(<>)}{}<(({}))(((
(((()()()()()){}){}){}())()){}
{})({}[{}]<(())>){((<{}{}>))}{}{{}>{}(<({}>)(((()))(
)()()<>)>()({}[{}](<()>)){{}{}(<(())>)}{}{
<>{{}}<>{{}}((<()>))}{}(<><{}><>)}{}<{}>[()]){<>{{}}
<>(<()>)<>{{}}(<()>)}{}}<>([]<>)({}<
(())>){((<{}{}>))}{}

This is 1084980 bytes of source code, and +1 for the -a flag allowing ASCII input (but decimal output)

• First, we push a counter underneath the top character. We'll call it the "equals counter". This counter is initialized to 1, and as soon as the top character matches a bracket, we will set this to 0.

• Then weWe check each character to see if it matches any bracket. If it is an opening bracket, we push a number onto the other stack according to the following mapping:

( = 1
< = 2
[ = 3
{ = 4

• Then we check to see if it matches any closing bracket. If it does, we push the equivalent number onto the alternate stack, just like for opening brackets. Then, we check if the top two numbers are equal. If they are, the both get popped and the program continues as normal. If they are not, we clear both stacks (to stop the looping) and push a one onto the alternate stack. This is essentially a "break" statement.

• After checking the 8 bracket types, we poppush the character and look atvalue of this run through the "equals counter"loop. If it's truthySince we zero out most of it, the only snippets that means none ofhave any value are the conditionals when we compare against brackets. So if any bracket is matched, so we have an invalid characterthe whole loop has a value of 1. If none of them did, the whole loop has a value of 0. In this case, we will clear both stacks and push a 10 onto the alternate stack. Again, this is like a "break" statement.

• Removed some push pop redundancy

Removed some push pop redundancy

• Changed my zero counter logic

# Brain-Flak, 1101, 1085 bytes

{({}<>)(<>())(<>{}<>)(({}))((((()()()()()){}){}){})({}[{}]<(())>){((<{}{}>))}{}{
{}{}(<{}>)(((()))<>)(<><{}>)}{}(({}))((((()()()()()){}){}){}())({}[{}]<(())>){((
<{}{}>))}{}{{}{}(<{}>)(((()))<>)({}[{}](<()>)){{}{}(<(())>)}{}{<>{{}}<>{{}}((<()
>))}{}(<><{}>)}{}(({}))(((((()()()()()){}){})){}{})({}[{}]<(())>){((<{}{}>))}{}{
{}{}(<{}>)(((()))()<>)(<><{}>)}{}(({}))(((((()()()()()){}){})()){}{})({}[{}]<(()
)>){((<{}{}>))}{}{{}{}(<{}>)(((()))()<>)({}[{}](<()>)){{}{}(<(())>)}{}{<>{{}}<>{
{}}((<()>))}{}(<><{}>)}{}(({}))((((()()()){}){}()){({}[()])}{})({}[{}]<(())>){((
<{}{}>))}{}{{}{}(<{}>)(((()))()()<>)(<><{}>)}{}(({}))((((((()()()()()){})){}{}()
)){}{})({}[{}]<(())>){((<{}{}>))}{}{{}{}(<{}>)(((()))()()<>)({}[{}](<()>)){{}{}(
<(())>)}{}{<>{{}}<>{{}}((<()>))}{}(<><{}>)}{}(({}))((((((()()()()()){}){}){}()))
{}{})({}[{}]<(())>){((<{}{}>))}{}{{}{}(<{}>)(((()))()()()<>)(<><{}>)}{}(({}))(((
(((()()()()()){}){}){}())()){}{})({}[{}]<(())>){((<{}{}>))}{}{{}{}(<{}>)(((()))(
)()()<>)({}[{}](<()>)){{}{}(<(())>)}{}{<>{{}}<>{{}}((<()>))}{}(<><{}>)}{}{}{{{}}
<>(())<>}{}}<>([]<>)({}<(())>){((<{}{}>))}{}

Try it online!

This is 1084 bytes of source code, and +1 for the -a flag allowing ASCII input (but decimal output)

• First, we push a counter underneath the top character. We'll call it the "equals counter". This counter is initialized to 1, and as soon as the top character matches a bracket, we will set this to 0.

• Then we check to see if it matches any bracket. If it is an opening bracket, we push a number onto the other stack according to the following mapping:

( = 1
< = 2
[ = 3
{ = 4

• Then we check to see if it matches any closing bracket. If it does, we push the equivalent number onto the alternate stack, just like for opening brackets. Then, we check if the top two numbers are equal. If they are, the both get popped and the program continues as normal. If they are not, we clear both stacks (to stop the looping) and push a one onto the alternate stack. This is essentially a "break" statement.

• After checking the 8 bracket types, we pop the character and look at the "equals counter". If it's truthy, that means none of the brackets matched, so we have an invalid character. In this case, we clear both stacks and push a 1 onto the alternate stack. Again, this is like a "break" statement.

• Removed some push pop redundancy

# Brain-Flak, 1101, 1085, 981 bytes

{(<(({}))((((()()()()()){}){}){})({}[{}]<(())>){((<{}{}>))}{}>{({}<>)(<>)}{}<(({
}))((((()()()()()){}){}){}())({}[{}]<(())>){((<{}{}>))}{}>{({}<>)({}[{}](<()>)){
{}{}(<(())>)}{}{<>{{}}<>{{}}((<()>))}{}(<>)}{}<(({}))(((((()()()()()){}){})){}{}
)({}[{}]<(())>){((<{}{}>))}{}>{<({}()<>)>()(<>)}{}<(({}))(((((()()()()()){}){})(
)){}{})({}[{}]<(())>){((<{}{}>))}{}>{<({}()<>)>()({}[{}](<()>)){{}{}(<(())>)}{}{
<>{{}}<>{{}}((<()>))}{}(<>)}{}<(({}))((((()()()){}){}()){({}[()])}{})({}[{}]<(()
)>){((<{}{}>))}{}>{<({}()()<>)>()(<>)}{}<(({}))((((((()()()()()){})){}{}())){}{}
)({}[{}]<(())>){((<{}{}>))}{}>{<({}()()<>)>()({}[{}](<()>)){{}{}(<(())>)}{}{<>{{
}}<>{{}}((<()>))}{}(<>)}{}<(({}))((((((()()()()()){}){}){}())){}{})({}[{}]<(())>
){((<{}{}>))}{}>{<({}()()()<>)>()(<>)}{}<(({}))((((((()()()()()){}){}){}())()){}
{})({}[{}]<(())>){((<{}{}>))}{}>{<({}()()()<>)>()({}[{}](<()>)){{}{}(<(())>)}{}{
<>{{}}<>{{}}((<()>))}{}(<>)}{}<{}>[()]){<>{{}}(<()>)<>{{}}(<()>)}{}}<>([]<>)({}<
(())>){((<{}{}>))}{}

Try it online!

This is 980 bytes of source code, and +1 for the -a flag allowing ASCII input (but decimal output)

• We check each character to see if it matches any bracket. If it is an opening bracket, we push a number onto the other stack according to the following mapping:

( = 1
< = 2
[ = 3
{ = 4

• Then we check to see if it matches any closing bracket. If it does, we push the equivalent number onto the alternate stack, just like for opening brackets. Then, we check if the top two numbers are equal. If they are, the both get popped and the program continues as normal. If they are not, we clear both stacks (to stop the looping) and push a one onto the alternate stack. This is essentially a "break" statement.

• After checking the 8 bracket types, we push the value of this run through the loop. Since we zero out most of it, the only snippets that have any value are the conditionals when we compare against brackets. So if any bracket is matched, the whole loop has a value of 1. If none of them did, the whole loop has a value of 0. In this case, we will clear both stacks and push a 0 onto the alternate stack. Again, this is like a "break" statement.

• Removed some push pop redundancy

• Changed my zero counter logic

5 added 23 characters in body

# Brain-Flak, 1101 bytes(non-competing)1101, 1085 bytes

{({}<>)(<>())(<>{}<>)(({}))((((()()()()()){}){}){})({}[{}]<(())>){((<{}{}>))}{}{{
{}{}(<{}>)((()())(()<>)(<><{}>)}{}(({}))((((()()()()()){}){}){}())({}[{}]<(())>){((
<{}{}>))}{}{{}{}(<{}>)(((()))(()<>)({}[{}](<()>)){{}{}(<(())>)}{}{<>{{}}<>{{}}((<(
)
>))}{}(<><{}>)}{}(({}))(((((()()()()()){}){})){}{})({}[{}]<(())>){((<{}{}>))}{}{
{}{}(<{}>)(((()))(()()<>)(<><{}>)}{}(({}))(((((()()()()()){}){})()){}{})({}[{}]<((
)
)>){((<{}{}>))}{}{{}{}(<{}>)((()())(()()<>)({}[{}](<()>)){{}{}(<(())>)}{}{<>{{}}<
><>{
{}}((<()>))}{}(<><{}>)}{}(({}))((((()()()){}){}()){({}[()])}{})({}[{}]<(())>){(
(
<{}{}>))}{}{{}{}(<{}>)((()())(()()()<>)(<><{}>)}{}(({}))((((((()()()()()){})){}{}
()
)){}{})({}[{}]<(())>){((<{}{}>))}{}{{}{}(<{}>)((()())(()()()<>)({}[{}](<()>)){{}
{}(
<(())>)}{}{<>{{}}<>{{}}((<()>))}{}(<><{}>)}{}(({}))((((((()()()()()){}){}){}()
))
{}{})({}[{}]<(())>){((<{}{}>))}{}{{}{}(<{}>)((()())(()()()()<>)(<><{}>)}{}(({}))
(((
(((()()()()()){}){}){}())()){}{})({}[{}]<(())>){((<{}{}>))}{}{{}{}(<{}>)((()())
(()(
)()()<>)({}[{}](<()>)){{}{}(<(())>)}{}{<>{{}}<>{{}}((<()>))}{}(<><{}>)}{}{}{{
{}}
<>(())<>}{}}<>([]<>)({}<(())>){((<{}{}>))}{}

This is 11001084 bytes of source code, and +1 for the -a flag allowing ASCII input (but decimal output)

• Thanks to @WheatWizard for coming up with a good stack-clean "equals" and "logical not" snippet, and for regularly updating the github wiki with useful examples.

• Thanks to @ASCII-only for writing an online integer metagolfer which helped immensely in writing this program

revisions

• Removed some push pop redundancy

# Brain-Flak, 1101 bytes(non-competing)

{({}<>)(<>())(<>{}<>)(({}))((((()()()()()){}){}){})({}[{}]<(())>){((<{}{}>))}{}{{
}{}(<{}>)((()))(()<>)(<><{}>)}{}(({}))((((()()()()()){}){}){}())({}[{}]<(())>){((
<{}{}>))}{}{{}{}(<{}>)((()))(()<>)({}[{}](<()>)){{}{}(<(())>)}{}{<>{{}}<>{{}}((<(
)>))}{}(<><{}>)}{}(({}))(((((()()()()()){}){})){}{})({}[{}]<(())>){((<{}{}>))}{}{
{}{}(<{}>)((()))(()()<>)(<><{}>)}{}(({}))(((((()()()()()){}){})()){}{})({}[{}]<((
))>){((<{}{}>))}{}{{}{}(<{}>)((()))(()()<>)({}[{}](<()>)){{}{}(<(())>)}{}{<>{{}}<
>{{}}((<()>))}{}(<><{}>)}{}(({}))((((()()()){}){}()){({}[()])}{})({}[{}]<(())>){(
(<{}{}>))}{}{{}{}(<{}>)((()))(()()()<>)(<><{}>)}{}(({}))((((((()()()()()){})){}{}
())){}{})({}[{}]<(())>){((<{}{}>))}{}{{}{}(<{}>)((()))(()()()<>)({}[{}](<()>)){{}
{}(<(())>)}{}{<>{{}}<>{{}}((<()>))}{}(<><{}>)}{}(({}))((((((()()()()()){}){}){}()
)){}{})({}[{}]<(())>){((<{}{}>))}{}{{}{}(<{}>)((()))(()()()()<>)(<><{}>)}{}(({}))
((((((()()()()()){}){}){}())()){}{})({}[{}]<(())>){((<{}{}>))}{}{{}{}(<{}>)((()))
(()()()()<>)({}[{}](<()>)){{}{}(<(())>)}{}{<>{{}}<>{{}}((<()>))}{}(<><{}>)}{}{}{{
{}}<>(())<>}{}}<>([]<>)({}<(())>){((<{}{}>))}{}

Try it online!

This is 1100 bytes of source code, and +1 for the -a flag allowing ASCII input (but decimal output)

• Thanks to @WheatWizard for coming up with a good stack-clean "equals" and "logical not" snippet, and for regularly updating the github wiki with useful examples.

• Thanks to @ASCII-only for writing an online integer metagolfer which helped immensely in writing this program

# Brain-Flak, 1101, 1085 bytes

{({}<>)(<>())(<>{}<>)(({}))((((()()()()()){}){}){})({}[{}]<(())>){((<{}{}>))}{}{
{}{}(<{}>)(((()))<>)(<><{}>)}{}(({}))((((()()()()()){}){}){}())({}[{}]<(())>){((
<{}{}>))}{}{{}{}(<{}>)(((()))<>)({}[{}](<()>)){{}{}(<(())>)}{}{<>{{}}<>{{}}((<()
>))}{}(<><{}>)}{}(({}))(((((()()()()()){}){})){}{})({}[{}]<(())>){((<{}{}>))}{}{
{}{}(<{}>)(((()))()<>)(<><{}>)}{}(({}))(((((()()()()()){}){})()){}{})({}[{}]<(()
)>){((<{}{}>))}{}{{}{}(<{}>)(((()))()<>)({}[{}](<()>)){{}{}(<(())>)}{}{<>{{}}<>{
{}}((<()>))}{}(<><{}>)}{}(({}))((((()()()){}){}()){({}[()])}{})({}[{}]<(())>){((
<{}{}>))}{}{{}{}(<{}>)(((()))()()<>)(<><{}>)}{}(({}))((((((()()()()()){})){}{}()
)){}{})({}[{}]<(())>){((<{}{}>))}{}{{}{}(<{}>)(((()))()()<>)({}[{}](<()>)){{}{}(
<(())>)}{}{<>{{}}<>{{}}((<()>))}{}(<><{}>)}{}(({}))((((((()()()()()){}){}){}()))
{}{})({}[{}]<(())>){((<{}{}>))}{}{{}{}(<{}>)(((()))()()()<>)(<><{}>)}{}(({}))(((
(((()()()()()){}){}){}())()){}{})({}[{}]<(())>){((<{}{}>))}{}{{}{}(<{}>)(((()))(
)()()<>)({}[{}](<()>)){{}{}(<(())>)}{}{<>{{}}<>{{}}((<()>))}{}(<><{}>)}{}{}{{{}}
<>(())<>}{}}<>([]<>)({}<(())>){((<{}{}>))}{}

Try it online!

This is 1084 bytes of source code, and +1 for the -a flag allowing ASCII input (but decimal output)

• Thanks to @WheatWizard for coming up with a good stack-clean "equals" and "logical not" snippet, and for regularly updating the github wiki with useful examples.

• Thanks to @ASCII-only for writing an online integer metagolfer which helped immensely in writing this program

revisions

• Removed some push pop redundancy
4 replaced http://codegolf.stackexchange.com/ with https://codegolf.stackexchange.com/
• Thanks to @WheatWizard for coming up with a good stack-clean "equals" and "logical not" snippet, and for regularly updating the github wiki with useful examples.

• Thanks to @ASCII-only for writing an online integer metagolfer which helped immensely in writing this program

• Thanks to @WheatWizard for coming up with a good stack-clean "equals" and "logical not" snippet, and for regularly updating the github wiki with useful examples.

• Thanks to @ASCII-only for writing an online integer metagolfer which helped immensely in writing this program

3 added 16 characters in body
2 added 507 characters in body
1


text:
# Tag Info

41

No clipping or even-odd-rule required: \documentclass[tikz,border=5]{standalone} \tikzset{% v 0/.style={fill=white}, v 1/.style={fill=blue!30}, pics/venn/.style args={#1#2#3#4}{code={% \fill [v #1/.try] (-2,-1.5) rectangle (2,1.5); \fill [v #2/.try] (90:sin 60) arc (120:-120:1) arc (-60:60:1); \fill [v #3/.try] (90:sin 60) arc (60:300:1) ...

17

Well you can just remove the (C) and you are done \documentclass{article} \usepackage{tikz} \makeatletter \def\venn@strip#1#2\venn@STOP{\def\venn@next{#1}\gdef\venn@rest{#2}} \newcommand{\venn}[1]{% \begin{tikzpicture}[scale=0.2] \coordinate (A) at (0,0); \coordinate (B) at (2,0); \coordinate (S-SE) at (5,-3); \coordinate (S-NW) at (-3,{sqrt(3)+3}); \...

16

One possibility using TikZ for the first row and pgfplots together with its fillbetween library (requires an updated version of the package) for the second row. The third column is left as an exercise: \documentclass{article} \usepackage{pgfplots} \usepackage{subcaption} \pgfplotsset{compat=1.10} \usepgfplotslibrary{fillbetween} \pgfmathdeclarefunction{...

14

The areas in the diagrams are drawn with a combination of clip and fill commands. Each bit position encodes one of the four intersection free areas of the diagram. These areas are drawn independently. \documentclass{article} \usepackage{tikz} \begin{document} \begin{tikzpicture} \def\w{1.35} \def\h{1} \pgfmathsetmacro\rad{\h*.35} \def\sep{.2} \...

13

And here's a version in Metapost for comparison. This exploits a feature of buildcycle with two circular paths that was discussed in my answer to this quesion. prologues := 3; outputtemplate := "%j%c.eps"; color venn_color; venn_color = 0.8[blue,white]; vardef venn(expr p, q, r, s) = save a, b, c, d, f, u; path a,b,c,d, f[]; u = 1cm; f1 = fullcircle ...

12

A possible implementation with the new Blend Modes feature from TikZ v. 3.0.0 (see What are the new features in TikZ/pgf 3.0?). This allows to avoid using some clip operation: \documentclass[tikz,border=10pt]{standalone} \begin{document} \def\firstcircle{(0,0) circle (3.0cm)} \def\secondcircle{(360:3.5cm) circle (3.0cm)} \begin{tikzpicture}[blend group=...

11

My audacious mathematical theory about Euler diagrams seem to have failed to convince in the comments, so I revert back. (and my code did not compile due to a linebreak) Anyway, I borrow some of the techniques demo'd by @CarLaTeX here. Props to her ! Edits Fixed colors and fonts Added CSS-like padding syntax No more rules, added node for title ...

10

with symmetry in circle placing the drawing of all elements is simpler: \documentclass{article} \usepackage{tikz} \begin{document} \begin{center} \begin{tikzpicture} \fill[gray!50] (-0.75,0) circle (1cm); \fill[gray!50] ( 0.75,0) circle (1cm); \draw (-0.75,0) circle (1cm) node {$A$}; \draw ( 0.75,0) circle (1cm) node {$B$}; \draw (-...

10

A code with the dedicated package: \documentclass[svgnames]{article} \usepackage{venndiagram} \begin{document} \begin{venndiagram3sets}[labelA= ,labelB= , labelC= ,radius=2cm, overlap=2cm]% \setkeys{venn}{shade=OliveDrab!40} \fillACapBCapC \setpostvennhook {% \draw (labelA) ++(-120:1.8cm) node{$A$}; \draw (labelB) ...

10

You can use the vendiagram package. Its very easy to use. Do you mean something like this? \documentclass[]{article} \usepackage{tikz,venndiagram} \begin{document} \begin{venndiagram3sets} \fillOnlyC \fillBCapCNotA \fillACapCNotB \end{venndiagram3sets} $C\setminus (A \cap B \cap C)$ \end{document}

8

In venndiagram.sty on lines 138, 140, 147, 159, 151, 153, the arguments of atan2 are in the wrong order. When I reverse them, one gets the desired results. What I mean is that the original arguments to atan2 were entered as (dx,dy), but they should be entered as (dy,dx). Corrected passage: (lines 128-164) \def\fillOnlyA{% \begin{scope} \path[...

8

I don't know what it should look like (I'm colorblind so everything looks OK to me) but here is the second one. Ifyour viewer can handle it play with the blend mode parameter for different effects. \documentclass[tikz]{standalone} \begin{document} \begin{tikzpicture}[mys/.style={pink!80,fill opacity=0.5,draw=black }] \begin{scope}[transparency group] \...

8

If I may, given Gonzalo Medina's solution, this proposal provides a complement solution where clip technique within scope environment is used. Note: For those without updated version of the pgfplots package. Code \documentclass[border=10pt]{standalone}%{article} \usepackage{pgfplots} \pgfplotsset{compat=1.8} \pgfmathdeclarefunction{gauss}{2}{% \...

8

Here is a nice introduction I found to drawing Venn diagrams: http://users.ju.edu/hduong/math220/venn_diagrams.pdf I put it here, since this question is one of the first results google gives when looking for "how to draw Venn diagrams" (naturally), and also other places refer here, but I feel none of the answers give an introduction to drawing Venn ...

8

Another solution with nodes placed on a matrix (you could also use positioning library) and color boxes drawn as fit nodes on a background layer. \documentclass[tikz,border=2mm]{standalone} \usetikzlibrary{matrix, backgrounds, fit} \usepackage{lmodern} \begin{document} \begin{tikzpicture}[mytext/.style={text=white, font=\sffamily\large}] \matrix (N) [ ...

8

A solution with "plain tikz": \documentclass{article} \usepackage{tikz} \usetikzlibrary{backgrounds,positioning} \usepackage{float} \usepackage{showframe} \begin{document} \begin{figure}[H] \centering \begin{minipage}[t]{0.45\textwidth} \begin{tikzpicture}[scale=0.6] \draw (-4,-3)rectangle (6,3); \fill[gray,...

7

You tagged tikz-pgf, so here goes. The shadings library is used to shade the ring similar to your image. The even odd rule makes it possible to use another circle to not-fill the inner part. As both circular paths use different line option (one is blue-ish, the other is one black) we cannot use the same \path (implicitly called by \shade) but use two ...

7

Another option could be to use xpatch to patch those two environments, either removing the line altogether, or adding a style to the path, like this: \documentclass[border=5mm]{standalone} \usepackage{venndiagram} \usepackage{xpatch} \tikzset{ vennframe/.style={draw=none} % define a new style for the frames } \makeatletter \xpatchcmd{\endvenndiagram3sets} ...

7

Welcome to TeX.SE! It is good that you included your code. In terms of drawing a rectangle around the venn diagram, you could import the fit library from Tikz using \usetikzlibrary{fit}. I have completed an example to illustrate. Also, I have changed the code a little bit. You now have one tikzpicture (with no minipage). The first venn diagram has its own ...

7

Here is a different approach. We fill the second and third circles with white, then fill the first circle with yellow. Finally, we color the intersection of the three circles white. \documentclass{article} \usepackage{tikz} \usetikzlibrary{shapes,backgrounds} \begin{document} \def\firstcircle{(0,0) circle (1.5 cm)} \def\secondcircle{(0: 2 cm) circle (1....

7

From your code with minor changes: using \draw[fill=orange!60] (0.9,1.5) circle(1.5); and removing \clip (0.9,1.5) circle(1.5cm); \documentclass[tikz,border=0.2cm]{standalone} \begin{document} \begin{tikzpicture}[thick] \draw[fill=black!30] (0,0) circle(1.5) +(135:1.8) node{$A$} (1.8,0) circle(1.5) +(45:1.8) node{$B$}; % Intersection \begin{scope} ...

6

This is now fixed in venndiagram v1.1 which checks \pgfversion to determine what syntax to use for atan2, so venndiagram should now work with both pgf v2.0 and pgf v3.0.

6

There is a simple package to Venn diagrams, maybe somebody likes this: package venndiagram. I'm using it, it's enough for me.

6

The following code should do what you want (based on http://www.texample.net/tikz/examples/venn-diagram/): \documentclass[border=2mm]{standalone} \usepackage{tikz} \begin{document} \begin{tikzpicture} \fill[gray] (0.6,-1.04) circle (1); \begin{scope} \clip (0.6,-1.04) circle (1); \clip (1.2,0) circle (1); \fill[white] (0,0) circle (1); \end{scope} \...

6

With path picture you can fill a path with whatever you want and keep it clipped to the original path. Does it help? \documentclass[tikz,border=2mm]{standalone} \usetikzlibrary{positioning, shapes.geometric} \begin{document} \begin{tikzpicture} \draw[blue] (0,0) ellipse(2cm and 3cm) [path picture={ \node[draw=blue, minimum width=4cm, minimum ...

6

Something like this (it's just a rough proposition): \documentclass[border=5pt,tikz]{standalone} \usetikzlibrary{patterns} %\usepackage{venndiagram} % \begin{document} %\begin{venndiagram2sets}[shade=orange,tikzoptions={red}] % \fillA %\end{venndiagram2sets} \begin{tikzpicture}[red,every node/.style={font=\Large}] \draw[pattern=north west lines] (0,...

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text:
# A hiker travels 300 m [N], turns around and hikes 550 m [S], and then hikes back another 50.0 m[N]. He manages to do this in 7200 seconds. What is the hiker's displacement? What is the hiker's average speed? What is the hiker's velocity?

Jul 20, 2015

Displacement: 200 m
Average speed: 0.125 m/s
Average velocity: 0.0278 m/s

#### Explanation:

Let's start with the hiker's displacemnt, which is simply the distance between its starting point and its finish point.

Displacement does not take into account the path taken from start to finish, it just deals with the shortest distance between these two points.

So, you know that your hiker first travels 300 m north, then turns around and hikes 550 m south.

At this point, he will be 250 m south of his starting point, since hiking 300 m south would have taken him right back to his starting point.

After this, he walks 50 m north again, towards the starting point.

This means that the distance between the starting point and the finish point will be equal to 200 m. The hiker's displacement will thus be

d_"hiker" = color(green)("200 m")

Now for his average speed. Average speed is defined as the total distance covered by the hiker divided by the time total time needed to cover this distance.

The total distance covered by the hiker will be

${d}_{\text{total" = 300 + 550 + 50 = "900 m}}$

This means that his average speed was

v_"avg" = d_"total"/t_"total" = "900 m"/"7200 s" = color(green)("0.125 m/s")

Velocity is actually a vector quantity, so you need to determine both the magnitude and the direction of the vector.

Because your hiker changes direction, you can't have a constant velocity. This means that you have to use average velocity, which, for straight line motions, is simply the ratio between the displacement of the hiker and the total time of the motion.

bar(v) = d/t_"total" = "200 m"/"7200 s" = color(green)("0.0278 m/s")


text:
Help with Integration

1. Apr 14, 2008

elitespart

1. How would I do this: $$\int$$$$\frac{x^{2} - 1}{x}$$

I don't need the answer, only the method I need to use (integration by parts, partial fractions etc.)

Thanks.

2. Apr 14, 2008

quasar987

Just observe that

$$\frac{x^2-1}{x}=\frac{x^2}{x}-\frac{1}{x}$$

3. Apr 15, 2008

elitespart

ahh now i feel dumb. Thanks man.


text:
# What was the role of mr. william bishop in williams' life?

###### Question:

What was the role of mr. william bishop in williams' life?

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text:
# Derivative of cos(50)

## Derivative of cos(50). Simple step by step solution, to learn. Simple, and easy to understand, so dont hesitate to use it as a solution of your homework.

If it's not what You are looking for type in the derivative calculator your own function and let us solve it.

## Derivative of cos(50):

(cos(50))'0`
The calculation above is a derivative of the function f (x)

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text:
# Very-Short and Short-Answer Questions

Question:

Very-Short and Short-Answer Questions

How many three-digit natural numbers are divisible by 7?

Solution:

The three-digit natural numbers divisible by 7 are 105, 112, 119, ..., 994.

Clearly, these number are in AP.

Here, = 105 and d = 112 − 105 = 7

Let this AP contains n terms. Then,

$a_{n}=994$

$\Rightarrow 105+(n-1) \times 7=994 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 7 n+98=994$

$\Rightarrow 7 n=994-98=896$

$\Rightarrow n=128$

Hence, there are 128 three-digit numbers divisible by 7.