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https://ds4beginners.wordpress.com/2007/04/20/infosys-paper6/ | 1,524,761,838,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948426.82/warc/CC-MAIN-20180426164149-20180426184149-00620.warc.gz | 616,991,728 | 18,237 | # INFOSYS-PAPER6
Infosys Technologies Ltd. 1 . There are 4 political parties . Day Flight , Eat well , Good Sleep = Deposit Loss. The 3 statements are a. Either Day Flight or Eat Well will win the election. b. Day Flight cannot win election. c. Neither Deposit Loss nor Eat well can win the election. Only one statement is true while other two are false .who win the = election =85 Soln : Deposit Loss ( F T F ) 2. Javed , Roshan , Bharati and Sheela are supposed to have won prizes = in math , French , Logic , and Eng. But its not clear who won what. a. Sheela guessed that Roshan must have won logic. b. Roshan guessed that Sheela will never win Math. =20 c. Javed guessed that Bharati has won French. d. Bharati guessed Javed has got Eng. Of those guesses the winners of Math and logic prizes guessed correctly = while other two guessed wrong. Who won which prize. Soln Sheela —-English. Roshan —French Bharati—Logic Roshan-Maths. 3. there are 10 cig packs each conatining an unknown no of cigs( each = pack has between 50 and 100). One pack has all bad cigs . Each good cig = ( if u can agree that any cig is good) weighs 1gm. Each bad cig weighs = 1.1 gm or .99 gm but all bad cigs weigh same. Using a spring chemical = balance ( which gives u the exactweight) .Find out which pack is bad , = using the balace only once. 4. 24 apples were divided among 3 brothers . Each of them got the same = no of apples as their age 3 years ago. The youngest of them kept half = of his share and divided the rest of them equally between other two. = Other two did the same thing. They all ended up with 8 apples each . = Determine their ages. Ans 7 , 10 , 16. 5. a city is divided into 16 blocks with rows passing betwwen them = Find the no of ways one can move from A and C. ( if u can only move = upwards and to the right. Different paths may have common routes in = some parts ans 8 C 4 =3D 70. 6. A is murdered And the statements given by B , C , D are B – I am a lawyer. I haven’t killed. C – I am a layer I haven’t killed. D- I am not a lawyer. A lawyer is a murderer. Out of these 6 2 are are correct and two of them lawyer. Ans . first 2 are Truth . Last four are false . c has killed. 7 . Baron’s guide problem colur prob. 7 C E E=20 T E S S _______________ =20 9 9 9 9 8. From the north pole a flight takes up with full capacity of fuel = It can travel half of the globe. You canchange fuel and do whatever = you like in air. How many minimum flights will be reqd travel the = earth once through south pole. Ans 3. 9. Every day a cyclist meets a train at a particular crossing . The = road is straight. The cyclist travels with a speed of 10 kmph . one day = the cyclist comes late by 25 min and meets the train 5 km from the = crossing. What is the speed of the train? Ans assume that the man and the train normally meet at the crossing = at 8 a.m . he is 5 km behind at 7.30 . But when the cyclist is late he = arrives at the crossing at 8.25 and therefore 5 km behind at 7.55 am = . Since the train takes 5 min to travel 5 km , the speed of the train = is 60 kmph. 10.A frog starts climbing a 30 ft wall.Each hr he climbs 3 ft and slips back 2. How many days does it take him to reach the top and get out ? Ans: 28hrs At the end of 27 hrs hes 3ft from the top. And during the 28th hr he climbs the remaining 3ft And hes out. So 28 hrs (shankuntala Devis more puzzles to puzzle u, 197th question). 11. A truck and a car are moving in the same direction at 40 and 60km/hr respectively. Truck was 1500 m Ahead of the car at the start. After some time they both crash. What was distance one minute b4 the Crash. Convert them into m/sec and do it. The figures 40 and 60 hes not sure. Suppose one is 15m/s another is 10m/s. So in one sec car covers 5m. For 1500m it will take 300s =5 min. One min b4 I: e 4min I; e 240 sec … it will cover 240 * 5 = 1200. So the distance b/w them is 300. THE FIGURES ARE NOT THE SAME. It’s been just an example. 12. Professor Kittredge’s literature seminar includes students with varied tastes in poetry. All those in the seminar who enjoy the poetry of browning also enjoy the poetry of Eliot. Those who enjoy the poetry of Eliot despise the poetry of colridge. Some of those who enjoy the poetry of Eliot also enjoy the poetry of Auden. All those who enjoy the poetry of Coleridge also enjoy the poetry of Donne. Some of those who enjoy the poetry of donne also enjoy the poetry of Eliot. Some of those who enjoy the poetry of auden despise the poetry of cleridge All those who enjoy the poetry of donne also enjoy the poetry of frost. 1. Miss garfield enjoys the poetry of donne. Which of the following must be true? A) she may or may not enjoy the poetry of coleridge. B) She doesn’t enjoy the poetry of bropwning. C) She enjoys the poetry of auden. D) She doesnot enjoy the poetry of eliot. E) She enjoys the poetry of coleridge. 2. Mr. Huxtable enjoys the poetry of browning. He may also enjoy any of the following poets except A) auden B) coleridge C) donne D) eliot E) frost 3) Miss inaguchi enjoys the poetry of coleridge. Which of the fllwing must be false? A) she doesnot enjoy the poetry of auden. B) She enjoys the poetry of donne. C) She enjoys the poetry of frost. D) She doesnot enjoy the poetry of browning. E) She may enjoy the poetry of eliot. 4) Based on the info provide, which of the following stmnts concerning the members of the seminar must be true? a) all those who enjoy the poetry of eliot also enjoy the poetry of browning. b) None of those who despise the poetry of frost enjoy the poetry of auden. c) Some of those who enjoy the poetry of auden des-ise the poetry of coleridge. d) None of those who enjoy the poetry of browning despise the poetry of donne. e) Some of those who enjoy the poetry of frost despise the poetry of donne. Ans 13) The person meets one train at the crossing every day. One day hes 20 min late . He meets the train 5 miles from the place If the speed of the man is 12 mile/hr What is trains speed. 6 **)*****(*7* *77 ———– *** *7* ——— ** ** Ans 53)*****(971 I think this is the ans so that u can find the rest. Any way check it out again. ^ | 1,696 | 6,282 | {"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} | 3.734375 | 4 | CC-MAIN-2018-17 | latest | en | 0.96108 |
https://text.123doc.org/document/5623649-financial-information-for-managemetn-paper-1-2-2006-q2.htm | 1,566,347,013,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315695.36/warc/CC-MAIN-20190821001802-20190821023802-00190.warc.gz | 663,834,544 | 13,341 | # Financial information for managemetn paper 1 2 2006 q2
PART 1
FRIDAY 9 JUNE 2006
QUESTION PAPER
Time allowed 3 hours
This paper is divided into two sections
Section A
ALL 25 questions are compulsory and MUST be
Section B
ALL FIVE questions are compulsory and MUST be
Formulae Sheet is on page 13
Do not open this paper until instructed by the supervisor
This question paper must not be removed from the examination
hall
The Association of Chartered Certified Accountants
Paper 1.2
Financial
Information for
Management
Section A – ALL 25 questions are compulsory and MUST be attempted.
Please use the Candidate Registration Sheet provided to indicate your chosen answer to each multiple choice question.
Each question within this section is worth 2 marks.
1
A supplier of telephone services charges a fixed line rental per period. The first 10 hours of telephone calls by the
customer are free, after that all calls are charged at a constant rate per minute up to a maximum, thereafter all calls
in the period are again free.
Which of the following graphs depicts the total cost to the customer of the telephone services in a period?
AA
BB
£
£
0
Hours
0
Hours
CC
DD
£
£
0
2
Hours
0
Hours
Four vertical lines have been labelled P, Q, R and S at different levels of activity on the following profit-volume chart:
£
S
Output
0
P
Q
R
Which line represents the total contribution at that level of activity?
A
B
C
D
Line
Line
Line
Line
P
Q
R
S
2
3
A company manufactures a single product which it sells for £15 per unit. The product has a contribution to sales ratio
of 40%. The company’s weekly break-even point is sales of £18,000.
What would be the profit in a week when 1,500 units are sold?
A
B
C
D
4
£900
£1,800
£2,700
£4,500
The following production and total cost information relates to a single product organisation for the last three months:
Month
1
2
3
Production
units
1,200
1,900
1,400
Total cost
£
66,600
58,200
68,200
The variable cost per unit is constant up to a production level of 2,000 units per month but a step up of £6,000 in
the monthly total fixed cost occurs when production reaches 1,100 units per month.
What is the total cost for a month when 1,000 units are produced?
A
B
C
D
5
Which of the following is NOT a feasible value for the correlation coefficient?
A
B
C
D
6
£54,200
£55,000
£59,000
£60,200
+ 1·2
+ 0·6
+0
– 0·6
The following statements relate to responsibility centres:
(i) Return on capital employed is a suitable measure of performance in both profit and investment centres.
(ii) Cost centres are found in manufacturing organisations but not in service organisations.
(iii) The manager of a revenue centre is responsible for both sales and costs in a part of an organisation.
Which of the statements, if any, is true?
A
B
C
D
7
(i) only
(ii) only
(iii) only
None of them
The purchase price of a stock item is £25 per unit. In each three month period the usage of the item is 20,000 units.
The annual holding costs associated with one unit equate to 6% of its purchase price. The cost of placing an order
for the item is £20.
What is the Economic Order Quantity (EOQ) for the stock item to the nearest whole unit?
A
B
C
D
1,730
1,894
1,461
1,633
3
[P.T.O.
8
A company determines its order quantity for a raw material using the EOQ model.
What would be the effects on the EOQ and on the total annual stockholding cost of a decrease in the cost of
placing an order for the raw material?
EOQ
A
B
C
D
9
Increase
Decrease
Increase
Decrease
Total annual
stockholding cost
No effect
No effect
Increase
Decrease
A company uses standard absorption costing. The following data relate to last month:
Budget
Actual
Sales and production (units)
1,000
900
Standard
Actual
£
£
Selling price per unit
50
52
Total production cost per unit
39
40
What was the adverse sales volume profit variance last month?
A
B
C
D
£1,000
£1,100
£1,200
£1,300
10 A company operates a standard marginal costing system. Last month actual fixed overhead expenditure was 2%
below budget and the fixed overhead expenditure variance was £1,250.
What was the actual fixed overhead expenditure for last month?
A
B
C
D
£61,250
£62,475
£62,500
£63,750
11 An organisation’s stock records show the following transactions for a specific item during last month:
Date
4th
13th
20th
27th
Receipts
units
Issues
units
50
200
50
50
The stock at the beginning of last month consisted of 100 units valued at £6,700.
The receipts last month cost £62 per unit.
The value of the closing stock for last month has been calculated twice – once using a FIFO valuation and once using
a LIFO valuation.
Which of the following statements about the valuation of closing stock for last month is correct?
A
B
C
D
The
The
The
The
FIFO
LIFO
FIFO
LIFO
valuation
valuation
valuation
valuation
is
is
is
is
higher
higher
higher
higher
than
than
than
than
the
the
the
the
LIFO
FIFO
LIFO
FIFO
valuation
valuation
valuation
valuation
4
by
by
by
by
£250.
£250.
£500.
£500.
12 A company uses absorption costing with a predetermined hourly overhead absorption rate. The following situations
arose last month:
(i) Actual hours worked exceeded planned hours.
(ii) Actual overhead expenditure exceeded planned expenditure.
Which of the following statements is correct?
A
B
C
D
Situation (i) would cause overheads to be over absorbed and situation (ii) would cause overheads to be under
absorbed.
Situation (i) would cause overheads to be under absorbed and situation (ii) would cause overheads to be over
absorbed.
Both situations would cause overheads to be over absorbed.
Both situations would cause overheads to be under absorbed.
13 A factory consists of two production cost centres (G and H) and two service cost centres (J and K). The total overheads
allocated and apportioned to each centre are as follows:
G
H
J
K
£40,000
£50,000
£30,000
£18,000
The work done by the service cost centres can be represented as follows:
G
H
J
K
Percentage of service cost centre J to
30%
70%
Percentage of service cost centre K to
50%
40%
10%
The company apportions service cost centre costs to production cost centres using a method that fully recognises any
work done by one service cost centre for another.
What are the total overheads for production cost centre G after the reapportionment of all service cost centre
costs?
A
B
C
D
14 The
(i)
(ii)
(iii)
£58,000
£58,540
£59,000
£59,540
following statements refer to strategic planning:
It is concerned with quantifiable and qualitative matters.
It is mainly undertaken by middle management in an organisation.
It is concerned predominantly with the long term.
Which of the statements are correct?
A
B
C
D
(i) and (ii) only
(i) and (iii) only
(ii) and (iii) only
(i), (ii) and (iii)
15 The following statements refer to situations occurring in Process Q of an organisation which operates a series of
consecutive processes:
(i) Direct labour is working at below the agreed productivity level.
(ii) A machine breakdown has occurred.
(iii) Direct labour is waiting for work to be completed in a previous process.
Which of these situations could give rise to idle time?
A
B
C
D
(i) and (ii) only
(i) and (iii) only
(ii) and (iii) only
(i), (ii) and (iii)
5
[P.T.O.
16 The
(i)
(ii)
(iii)
following terms relate to computers:
Floppy disks
Operating systems
Which of these terms are examples of computer software?
A
B
C
D
(i) and (ii) only
(i) and (iii) only
(ii) and (iii) only
(i), (ii) and (iii)
17 A company operates a job costing system. Job number 506 requires £64 of direct materials and 7 hours of direct
labour. Direct labour is paid £8 per hour. Production overheads are absorbed at the rate of £20 per direct labour hour
and non-production overheads at a rate of 60% of prime cost.
What is the total cost of job number 506?
A
B
C
D
£332
£352
£416
£448
18 All of a company’s skilled labour, which is paid £8 per hour, is fully employed manufacturing a product to which the
following data refer:
£ per unit
Selling price
Less Variable costs:
Less Skilled labour
Less Others
Contribution
£ per unit
60
20
15
–––
(35)
–––
25
–––
The company is evaluating a contract which requires 90 skilled labour hours to complete. No other supplies of skilled
labour are available.
What is the total relevant skilled labour cost of the contract?
A
B
C
D
£720
£900
£1,620
£2,160
19 A company requires 600 kg of raw material Z for a contract it is evaluating. It has 400 kg of material Z in stock which
were purchased last month. Since then the purchase price of material Z has risen by 8% to £27 per kg. Raw material
Z is used regularly by the company in normal production.
What is the total relevant cost of raw material Z to the contract?
A
B
C
D
£15,336
£15,400
£16,200
£17,496
6
The following information relates to questions 20 and 21:
A company operates a process costing system using the first-in-first-out (FIFO) method of valuation. No losses occur in the
process. All materials are input at the commencement of the process. Conversion costs are incurred evenly through the
process.
The following data relate to last period:
Units
Degree of completion
Opening work in progress
12,000
60%
Total number of units completed
14,000
Closing work in progress
13,000
30%
£
Costs arising:
Materials
151,000
Conversion
193,170
20 What was the total number of units input during last period?
A
B
C
D
12,000
13,000
15,000
17,000
21 What was the value of the closing work in progress for last period?
A
B
C
D
£21,330
£21,690
£22,530
£22,890
22 A company is attempting to break into an existing market by launching a new product at an initially low selling price.
What pricing policy is the company following?
A
B
C
D
Price skimming
Price discrimination
Penetration pricing
23 A company has established the following equations for one of its products:
Selling price (£ per unit)
=
Marginal revenue (£ per unit)
=
Total cost per week (£)
=
Q in each case represents the number
40 – 0·008Q
40 – 0·016Q
2,500 + 8Q
of units produced and sold per week.
At what selling price per unit will weekly profits be maximised?
A
B
C
D
£8
£16
£24
£32
7
[P.T.O.
The following information relates to questions 24 and 25:
A company which manufactures and sells two products (X and Y) aims to maximise its profits. It holds no stocks. Product
X makes a contribution per unit of £4 and product Y makes a contribution per unit of £1.
Next period the company faces three ‘less than’ production constraints and these are shown as the lines labelled (1), (2)
and (3) on the following graph:
Product Y
units
11
’000
10
(2)
9
(3)
8
7
H
6
J
5
4
3
2
K
(1)
1
L
1
2
3
4
5
6
7
8
9
10
11
12
13
Product X
14 units
’000
24 Which of the following points shown on the graph is optimal for next period?
A
B
C
D
Point
Point
Point
Point
H
J
K
L
25 Which of the following constraint formulations is represented by the line labelled (2) on the graph?
A
B
C
D
10X
17X
17X
13X
+
+
+
+
17Y
10Y
13Y
1 7Y
70,000
70,000
91,000
91,000
(50 marks)
8
Section B – ALL FIVE questions are compulsory and MUST be attempted
1
Corcoran Ltd operates several manufacturing processes. In process G, joint products (P1 and P2) are created in the
ratio 5:3 by volume from the raw materials input. In this process a normal loss of 5% of the raw material input is
expected. Losses have a realisable value of £5 per litre. The company holds no work in progress. The joint costs are
apportioned to the joint products using the physical measure basis.
The following information relates to process G for last month:
Raw materials input
60,000 litres (at a cost of £381,000)
Abnormal gain
11,000 litres
Other costs incurred:
Direct labour
£180,000
Direct expenses
1£54,000
110% of direct labour cost.
Required:
(a) Prepare the process G account for last month in which both the output volumes and values for each of the
joint products are shown separately.
(7 marks)
The company can sell product P1 for £20 per litre at the end of process G. It is considering a proposal to further
process product P1 in process H in order to create product PP1. Process H has sufficient spare capacity to do this
work. The further processing in process H would cost £4 per litre input from process G. In process H there would
be a normal loss in volume of 10% of the input to that process. This loss has no realisable value. Product PP1 could
then be sold for £26 per litre.
(b) Determine, based on financial considerations only, whether product P1 should be further processed to create
product PP1.
(3 marks)
(c) In the context of process G in Corcoran Ltd, explain the difference between ‘direct expenses’ and ‘production
(2 marks)
(12 marks)
9
[P.T.O.
2
Buttercup Ltd manufactures and sells three products (R, S and T). These products are made using the same
machinery. The total machining time available each month is 10,500 hours but this is insufficient to produce all the
units of R, S and T required to meet maximum demands. No stocks of these products are held.
The following information is available:
Selling price per unit
Contribution to sales ratio
Machining minutes per unit
Maximum monthly demand (units)
Product R
£60
20%
40
9,000
Product S
£75
24%
54
6,000
Product T
£84
25%
75
3,000
Required:
(a) Calculate the monthly shortfall in machining hours.
(2 marks)
(b) Determine the monthly production plan in units that will maximise the company’s total contribution from
products R, S and T and calculate this total contribution.
(6 marks)
(8 marks)
3
Deadeye Ltd operates a standard costing system in which all stocks are valued at standard cost. The standard direct
material cost of one unit of product MS is £36, made up of 4·8 kg of material H at £7·50 per kg. Material H is used
only in the manufacture of product MS.
The following information relates to last month:
Material H:
Purchased
40,000 kg for £294,000
Issued into production
36,500 kg
Finished output of MS
17,200 units
Required:
(a) Calculate the direct material price and usage variances for last month.
(3 marks)
(b) Prepare a statement that reconciles the actual cost of material H purchased with the standard material cost
of actual production of MS for last month. The statement should incorporate the variances calculated in (a).
(3 marks)
(c) (i)
Suggest ONE possible cause for EACH of the variances calculated in (a).
(ii) Who should the direct material price variance be reported to, and why?
(4 marks)
(10 marks)
10
4
The management accountant at Josephine Ltd is trying to predict the quarterly total maintenance cost for a group of
similar machines. She has extracted the following information for the last eight quarters:
Quarter number
1
2
3
4
5
6
7
8
Total maintenance
cost (£’000)
265
302
222
240
362
295
404
400
Production units
(‘000)
20
24
16
18
26
22
32
30
The effects of inflation have been eliminated from the above costs.
The management accountant is using linear regression to establish an equation of the form y = a + bx and has
produced the following preliminary calculations:
Σ (total maintenance cost x production units)
= £61,250 million
Σ (total maintenance cost)2
= £809,598 million
Σ (production units)2
= 4,640 million
Required:
(a) Establish the equation which will allow the management accountant to predict quarterly total maintenance
costs for a given level of production. Interpret your answer in terms of fixed and variable maintenance costs.
(7 marks)
(b) Using the equation established in (a), predict the total maintenance cost for the next quarter when planned
production is 44,000 units. Suggest a major reservation, other than the effect of inflation, you would have
(3 marks)
(10 marks)
11
[P.T.O.
5
Pinafore Ltd manufactures and sells a single product. The budgeted profit statement for this month, which has been
prepared using marginal costing principles, is as follows:
£’000
£’000
Sales (24,000 units)
864
Less Variable production cost of sales:
Less Opening stock (3,000 units)
169
Less Production (22,000 units)
506
Less Closing stock (1,000 units)
1(23)
––––
(552)
––––
312
Less Variable selling cost
1(60)
––––
Contribution
252
Less Production
125
140
––––
(165)
––––
Net profit
187
––––
The normal monthly level of production is 25,000 units and stocks are valued at standard cost.
Required:
(a) Prepare in full a budgeted profit statement for this month using absorption costing principles. Assume that
fixed production overhead costs are absorbed using the normal level of activity.
(6 marks)
(b) Prepare a statement that reconciles the net profit calculated in (a) with the net profit using marginal costing.
(2 marks)
(c) Which of the two costing principles (absorption or marginal) is more relevant for short-run decision-making,
and why?
(2 marks)
(10 marks)
12
Formulae Sheet
End of Question Paper
13
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× | 4,446 | 16,895 | {"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} | 2.96875 | 3 | CC-MAIN-2019-35 | latest | en | 0.900631 |
http://www.algebra.com/algebra/homework/Linear-equations/Linear-equations.faq.question.105067.html | 1,369,539,458,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706578727/warc/CC-MAIN-20130516121618-00069-ip-10-60-113-184.ec2.internal.warc.gz | 322,805,918 | 5,124 | # SOLUTION: I was doing my assignment I got stuck on this problem. A repair service charges a fixed fee for a house call plus a fixed hourly rate. If a 2 hour call cost \$57 and a 4 1/2 hour
Algebra -> Algebra -> Linear-equations -> SOLUTION: I was doing my assignment I got stuck on this problem. A repair service charges a fixed fee for a house call plus a fixed hourly rate. If a 2 hour call cost \$57 and a 4 1/2 hour Log On
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Algebra: Linear Equations, Graphs, Slope Solvers Lessons Answers archive Quiz In Depth
Question 105067: I was doing my assignment I got stuck on this problem. A repair service charges a fixed fee for a house call plus a fixed hourly rate. If a 2 hour call cost \$57 and a 4 1/2 hour call costs \$97, what is the fixed fee for a house call. I have: 2/57=4.5/97 194=256.5 Now what?Answer by Earlsdon(6291) (Show Source): You can put this solution on YOUR website!Let x = the fixed fee and y = the hourly rate. For the 2-hour call, you can write: 1) ...and for the 4.5-hour call, you can write: 2) You need to solve this system of equations for x, the fixed fee. Solve equation 1) for y and substitute into equation 2) then solve it for x. 1) 2) Simplify and solve for x. Multiply both sides by 2. The fixed fee is \$25.00 The hourly fee is: Substitute x = 25. The hourly fee is \$16.00 | 416 | 1,509 | {"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} | 3.9375 | 4 | CC-MAIN-2013-20 | latest | en | 0.878958 |
https://sourceforge.net/p/matplotlib/mailman/message/13934181/ | 1,469,426,013,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824204.27/warc/CC-MAIN-20160723071024-00193-ip-10-185-27-174.ec2.internal.warc.gz | 863,138,473 | 8,454 | [Matplotlib-users] Irregularly sized data without interpolation
[Matplotlib-users] Irregularly sized data without interpolation From: Matthew Turk - 2007-05-31 21:06:42 ```Hi there. I'm investigating using matplotlib for plotting of Adaptive Mesh Refinement ( http://en.wikipedia.org/wiki/Adaptive_mesh_refinement ) data -- the primary characteristic of which is that it is of non-equal resolution. I've used the scipy/delaunay method, as mentioned in the cookbook, but unfortunately that provides a level of interpolation that is not always desirable; very often when plotting data, we want to be able to see clear cell boundaries, as well as boundaries between resolution levels. Essentially what I have are the following pieces of data: x, y, dx, dy, z. The simplest way is for me to sample this using a loop over points in the module where I handle the data; however, what I'd like to be able to do is hand it off to matplotlib, and the on-the-fly change the x,y (and z) bounds. (This seems as though it would be the more efficient manner of handling the data, anyway.) Is there a way to do this? If not, would it be terribly difficult for me to implement? I've browsed the code, and it seems that the best starting place would by pcolor in lib/matplotlib/pylab.py or src/_image.cpp. I very much would like to leverage the abilities of matplotlib -- specifically, I'm very excited about being able to plot this data, and then overplot contour or quiver plots (which I have done with my data using the delaunay method.) Any ideas? Thanks! -Matt ```
[Matplotlib-users] Irregularly sized data without interpolation From: Matthew Turk - 2007-05-31 21:06:42 ```Hi there. I'm investigating using matplotlib for plotting of Adaptive Mesh Refinement ( http://en.wikipedia.org/wiki/Adaptive_mesh_refinement ) data -- the primary characteristic of which is that it is of non-equal resolution. I've used the scipy/delaunay method, as mentioned in the cookbook, but unfortunately that provides a level of interpolation that is not always desirable; very often when plotting data, we want to be able to see clear cell boundaries, as well as boundaries between resolution levels. Essentially what I have are the following pieces of data: x, y, dx, dy, z. The simplest way is for me to sample this using a loop over points in the module where I handle the data; however, what I'd like to be able to do is hand it off to matplotlib, and the on-the-fly change the x,y (and z) bounds. (This seems as though it would be the more efficient manner of handling the data, anyway.) Is there a way to do this? If not, would it be terribly difficult for me to implement? I've browsed the code, and it seems that the best starting place would by pcolor in lib/matplotlib/pylab.py or src/_image.cpp. I very much would like to leverage the abilities of matplotlib -- specifically, I'm very excited about being able to plot this data, and then overplot contour or quiver plots (which I have done with my data using the delaunay method.) Any ideas? Thanks! -Matt ```
No, thanks | 699 | 3,055 | {"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-2016-30 | latest | en | 0.926116 |
http://mathhelpforum.com/math-challenge-problems/92286-ta-s-challenge-problem-3-a.html | 1,481,436,196,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544140.93/warc/CC-MAIN-20161202170904-00077-ip-10-31-129-80.ec2.internal.warc.gz | 182,374,798 | 10,110 | # Thread: TA’s Challenge Problem #3
1. ## TA’s Challenge Problem #3
If $x,y,z$ are non-negative real numbers, show that
$(x+y+z)^2+xy+yz+zx\ \geqslant\ 2\sqrt{xy}(x+y)+2\sqrt{yz}(y+z)+2\sqrt{zx}(z+x)$
2. Originally Posted by TheAbstractionist
If $x,y,z$ are non-negative real numbers, show that
$(x+y+z)^2+xy+yz+zx\ \geqslant\ 2\sqrt{xy}(x+y)+2\sqrt{yz}(y+z)+2\sqrt{zx}(z+x)$
we have $(x+y-2\sqrt{xy})^2 \geq 0,$ which gives us: $x^2 + y^2+6xy \geq 4\sqrt{xy}(x+y).$ similarly: $y^2 + z^2+6yz \geq 4\sqrt{yz}(y+z)$ and $x^2 + z^2+6xz \geq 4\sqrt{xz}(x+z).$
add these three inequalities together and then divide by 2 to get the result.
3. Very good! | 279 | 656 | {"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": 8, "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} | 3.84375 | 4 | CC-MAIN-2016-50 | longest | en | 0.589727 |
https://www.docme.ru/doc/4317300/jph04142108 | 1,566,055,234,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313428.28/warc/CC-MAIN-20190817143039-20190817165039-00145.warc.gz | 756,609,407 | 12,137 | Забыли?
?
# JPH04142108
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```Patent Translate
Notice
This translation is machine-generated. It cannot be guaranteed that it is intelligible, accurate,
complete, reliable or fit for specific purposes. Critical decisions, such as commercially relevant or
financial decisions, should not be based on machine-translation output.
DESCRIPTION JPH04142108
[0001]
BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to a
cosine equalizer device having cosine-shaped amplitude frequency characteristics, and in
particular, a wideband delay line of a cosine equalizer capable of varying the amount of delay by
changing current or voltage externally. It concerns the built-in. PRIOR ART FIG. 2 shows the
principle of a conventional cosine equalizer apparatus. In the figure, 7 is a first delay line, 8 is a
second delay line, 4 'is a first adder, 9 is an amplitude adjuster, 10 is a phase inverter, and 11 is a
second adder. Next, the principle of operation will be described. The input signal input to the
input terminal is Ee4 ′ ′ ′ ′. Assuming that the delay amount of the 1st second delay lines 7
and 8 is τ, the signals at the input end (point A) and points B and C are respectively A point = Ee
′ ′ ′ □ ′ B point ==: E e I 1 “at” point C = Ee ′ (1) (“2”). Then, the signal at point D
becomes point D = point A + point C E = 2 ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′
cos ωr and is multiplied by the amplitude adjuster and inverted by the phase inverter, the signal
at point E becomes point E = −2 k E e “” ”“ 'cos ωr. Therefore, the output signal output to
the output terminal is output signal = point B + point E = E (1-2 k cos ωr) e ′ ′ ′ ′ ′ ′ c ′
(1). From the equation (1), it can be seen that the output signal has a time delay of τ with
respect to the input signal, or there is no phase distortion at all. Further, the above equation can
be expressed as an amplitude frequency characteristic as shown in FIG. From equation (1) or FIG.
3, ωτ = a · π (a is 1. ３． In the case of an odd number of 5 ...), that is, assuming that a = 1, it
exhibits a cosine characteristic in which the amplitude becomes maximum at a frequency of f = 1
/ 2τ (ω = 2πf). Here, a usage example of the cosine equalizer device will be described. In
magnetic recording, in the recording and reproducing process, the amplitude of the lower side
wave is larger than the upper side wave and the reproduction is performed. This situation is
shown in FIG. By adjusting the value of k using a cosine equalizer to adjust a left-right
08-05-2019
1
without causing phase distortion as shown in FIG. Next, the actual operation will be described.
FIG. 6 shows a conventional cosine equalizer apparatus. In the figure, 12 is a matching resistor of
a delay line, 13 is a delay line, 14 is a high input impedance variable gain amplifier, and 15 is a
high input impedance difference. Next, the operation will be described. The delay line 13 is a
block filter configured by combining a capacitor, a coil and others, and there is no reflection
because the signal input to the point F has a matching resistance 12 or not, the point H is high
because it is not a matching resistance 12 Due to the impedance and the so-called oven state, the
signal is inverted 100% in phase.
Therefore, the point G is obtained from the signal D = A to C similar to the point -0 in the
operation principle diagram of FIG. This signal is determined by the variable gain amplifier 14
and is input to the difference unit 15. Since this differencer 15 is equivalent to (phase inverter 10
+ tenth adder 11) in FIG. 2, this device operates as a cosine equalizer. The delay line 13 is
configured as a discrete component, and the variable gain amplifier 14 and the difference unit 15
are configured as a discrete component or a semiconductor integrated circuit. [Problems to be
Solved by the Invention] Since the conventional cosine equalizer device is configured as
described above, when it is desired to vary the frequency at which the amplitude of the cosine
equalizer device is maximum, the delay amount is calculated according to the relationship of f = 1
/ 2τ. Because of the fixed delay amount of the discrete block filter as a delay line, it is necessary
to switch several kinds of filters to change f, and it is necessary to change the configuration or
complexity. In addition, the filter itself has an error in the amount of delay, and there is also a
problem in that f or the like changes slightly. SUMMARY OF THE INVENTION The present
invention has been made to solve the above-mentioned problems, and an object thereof is to
obtain a cosine equalizer device capable of incorporating the entire device into an integrated
circuit and continuously varying the frequency at which the amplitude is maximum. . According
to the present invention, there is provided a cosine equalizer device comprising: a first variable
delay line variable in delay, which delays a signal input from an input terminal; and an output of
the first variable delay line. , An adder for adding the output of the input signal and the output of
the second variable delay line, a variable gain amplifier for controlling the output of the adder,
and an output of the amplifier And a difference unit for extracting the difference between the
outputs of the first variable delay line, and an interface circuit for controlling the delay amounts
of the first and second variable delay lines from the outside so that they have almost the same
desired amount. It is provided in a semiconductor integrated circuit. [Operation] The cosine
equalizer device in the present invention uses two delay lines which can be incorporated as a
delay line in a semiconductor integrated circuit and whose delay amount can be varied by voltage
or current, so that the entire device is incorporated in a semiconductor integrated circuit. And the
frequency at which the amplitude is maximum can be easily varied freely. An embodiment of the
present invention will be described below with reference to the drawings. The first shows a
cosine equalizer apparatus according to an embodiment of the present invention. In the figure, 1
is a first variable delay line, 2 is a second variable delay line, 3 is an interface circuit, 4 is an
08-05-2019
2
adder, 5 is a variable gain amplifier, and 6 is a difference unit.
Next, the operation will be described. The difference from the diagram showing the operation
principle of FIG. 2 is that the first and second delay lines 78 are changed to variable type, and the
control voltage and current from the outside are controlled as desired to control the delay
amount. The only difference is that the interface circuit 3 for converting into a signal is added
and the combination of the phase inverter 10 and the adder 2 or the differencer 6 is changed.
Therefore, it is possible to equalize the input signal by the same operation as that of the
conventional device. At this time, in the present embodiment, the delay amount of the delay line
can be changed continuously or discretely by the interface circuit, and the amplitude based on f,
= 1 / 2τ by the delay amount τ1 determined by this delay control. The maximum frequency (=
f) and hence the entire amplitude frequency characteristic can be freely displaced in the
frequency axis direction. Therefore, even if the carrier frequencies of the cosine equalizer device
according to the present embodiment are different signals, such as VH3 signal and 5-VH3 signal
in VTR, the cosine equalizer can be equalized by changing the delay amount. Since the delay
amount is arbitrary, it can be applied to equalization of various other signals. Furthermore, the
present invention can be applied not only to VTRs, but also to general electromagnetic
conversion systems. As described above, the cosine equalizer device according to the
embodiment of the present invention responds to the control of the frequency at which the
amplitude of the device is maximum, in the case of the carrier frequency or a different mode, or
in the case of error correction of the delay amount of the delay line itself, etc. It can be freely
changed with a simple configuration. The first and second variable delay lines 1 and 2 are
generally gm'1liT variable control type filters consisting of an operational amplifier and a
capacitor, but the gain 2 group delay amount within the use band is flat. It is desirable. Also,
barrier pull capacitor control may be used. Further, although the adder 4 and the variable gain
amplifier 5 are separately provided, both processing may be simultaneously performed by the
adder 2 capable of amplitude control of the output. As described above, according to the cosine
equalizer device of the present invention, it is possible to use a delay-variable first variable delay
line for delaying a signal input from an input terminal, and a first variable delay line of the first
variable delay line. A variable delay second variable delay line for delaying an output, an adder
for adding the output of the input signal and the output of the second variable delay line, a
variable gain amplifier for amplitude controlling an output of the adder, A differencer for
extracting the difference between the output of the amplifier and the output of the first variable
delay line, and an interface for controlling the delay amounts of the first and second variable
delay lines from the outside so as to be substantially the same desired amount Since the circuit
and the circuit are provided in the semiconductor integrated circuit, control of the frequency at
which the amplitude of the cosine equalizer apparatus is maximized.
There is an effect that it can be freely varied with a simple configuration according to the case of
08-05-2019
3
the carrier frequency or a different mode, or the case of correcting the error of the delay amount
of the delay line itself.
[0002]
Brief description of the drawings
[0003]
1 shows a cosine equalizer according to an embodiment of the present invention, FIG. 2 shows
the operation principle of the conventional cosine equalizer, FIG. 3 shows a cosine characteristic,
and FIG. FIG. 5 is a diagram showing frequency characteristics at the time of reproduction of
recording, FIG. 5 is a diagram showing that correction is performed after passing through a
cosine equalizer device, and FIG. 6 is a diagram showing a conventional cosine equalizer device.
In the figure, 1 is a first variable delay line, 2 is a second variable delay line, 3 is an interface
circuit, 4 adders, 4 'is a first adder, 5 is a variable gain tank 1 and 6 is a differencer , 7 is a first
delay line, 8 is a second delay line, 9 is an amplitude adjuster, 10 is a phase inverter, and l is a
second adder. In the drawings, the same reference numerals denote the same or corresponding
parts.
08-05-2019
4
```
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Пожаловаться на содержимое документа | 2,555 | 10,977 | {"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} | 2.765625 | 3 | CC-MAIN-2019-35 | latest | en | 0.875189 |
http://brainly.com/question/136932 | 1,480,861,835,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541322.19/warc/CC-MAIN-20161202170901-00067-ip-10-31-129-80.ec2.internal.warc.gz | 40,178,784 | 9,867 | # How do you solve -6-6y = 6x 45-9x = 18y ?
1
by alexgcp
Is this just a system of equations? -6-6y=6x is the first equation and 45-9x=18y is the second?
yup! | 68 | 158 | {"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} | 3.5 | 4 | CC-MAIN-2016-50 | latest | en | 0.955333 |
https://speakerdeck.com/heyitsmohit/functional-programming-with-arrow | 1,702,325,692,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679516047.98/warc/CC-MAIN-20231211174901-20231211204901-00126.warc.gz | 573,300,255 | 38,171 | Mohit S
January 24, 2018
1.4k
Functional Programming with Arrow
Discussion of data types and type classes available in the library called Arrow for Kotlin.
January 24, 2018
Transcript
1. Functional Programming
with Λrrow
Mohit Sarveiya
2. Functional Programming with Λrrow
• Data Types
• Optics
• Error Handling
• Type Classes
• Types of Polymorphism
• Type classes proposal for Kotlin
3. Optics
Iso
Lens
4. Optics
Iso
Lens
5. data class Point2D
data class Cords
(val x: Int, val y: Int)
Isomorphism
(val x: Int, val y: Int)
6. data class Point2D
data class Cords
(val x: Int, val y: Int)
Isomorphism
(val x: Int, val y: Int)
7. data class Point2D
data class Cords
(val x: Int, val y: Int)
Isomorphism
(val x: Int, val y: Int)
fun point2DToCords (point2D: Point2D): Cords {
return Cords(point2D.x, point2D.y)
}
8. data class Point2D
data class Cords
(val x: Int, val y: Int)
Isomorphism
(val x: Int, val y: Int)
fun cordsToPoint2D (cord: Cord): Point2D {
return Point2D(cord.x, cord.y)
}
9. Isomorphism
point2DToCords: Point2D "-> Cords
cordsToPoint2D: Cords "-> Point2D
10. Isomorphism
Source:
Focus:
f ≇ g = IdA
g ≇ f = IdS
f: "->
g: "->
S
A
S A
A S
11. Isomorphism
interface Iso {
/**
* Get the focus of a [Iso]
"*/
fun get(s: S): A
/**
* Get the modified focus of a [Iso]
"*/
fun reverseGet(b: B): T
}
12. Isomorphism
val pointIsoCords : Iso =
Iso(
get = { point !-> Cords(point.x, point.y) },
reverseGet = { cords !-> Point2D(cords.x, cords.y) }
)
13. Isomorphism
val point = Point2D(6, 10)
val cords: Cords = pointIsoCords.get(point)
"// Cords(6, 10)
14. Isomorphism
val cords = Cords(6, 10)
val point2D: Point2D = pointIsoCords.reverseGet(cords)
"// Point2D(6, 10)
15. data class Point2D
data class Tuple2
(val x: Int, val y: Int)
Isomorphism
(val x: Int, val y: Int)
16. data class Tuple2(val a: A, val b: B)
Isomorphism
data class Tuple3(val a: A,
val b: B,
val c: C)
data class Tuple4("..)
17. val pointIsoTuple: Iso> =
Iso(
get = { point !-> point.x toT point.y },
reverseGet = { tuple !-> Point2D(tuple.a, tuple.b) }
)
Isomorphism
18. var point2D = Point2D(1, 2)
val tuple2 = pointIsoTuple.get(point2D)
"// Tuple2(a=1, b=2)
val point2D = pointIsoTuple.reverseGet(tuple2)
"// Point2D(x=1, y=2)
Isomorphism
19. val pointIsoTuple = Iso(
get = { point !-> Tuple2(point.x, point.y) },
reverseGet = { tuple !-> Point2D(tuple.a, tuple.b) }
)
Generating Isomorphism
data class Point2D (val x: Int, val y: Int)
20. @Retention(SOURCE)
@Target(CLASS)
annotation class isos
Generating Isomorphism
21. Generating Isomorphism
@isos data class Point2D (val x: Int, val y: Int)
22. Optics
Iso
Lens
23. Lens
Source: S
Purpose: Set, Get, Copy an immutable data structure
through functional references.
24. Lens
Lens
Get : S "-> A
Set : A "-> (S "-> S)
25. Lens
data class Person(val name: String, val age: Int)
26. Lens
data class Person(val name: String, val age: Int)
val personNameLens : Lens = Lens(
get = { person !-> person.name },
set = { newName !-> {foo !-> foo.copy(name = newName)} }
)
27. Lens
val person = Person("John Doe", 100)
val personNameLens : Lens = Lens(
get = { person !-> person.name },
set = { newName !-> {foo !-> foo.copy(name = newName)} }
)
28. Lens
val person = Person("John Doe", 100)
val newPerson = personNameLens.set(person, "John Doe Jr”)
// Person(name=John Doe Jr, age=100)
29. Lens
data class Employee(val name: String, val company: Company)
data class Address(val city: String, val street: Street)
data class Street(val number: Int, val name: String)
30. Lens
val street = Street(1, "hacker Way”)
val employee = Employee("John Doe", company)
31. Lens
employee.copy(
company = employee.company.copy(
)
)
)
)
32. Lens
data class Street(val number: Int, val name: String)
data class Employee(val name: String, val company: Company)
33. val company: Company)
Lens
data class Employee(val name: String, val company: Company)
val employeeCompany: Lens =
Lens(get = { it.company },
set = { company !->
{ employee !->
employee.copy(company = company)
}
}
)
Lens
Lens(
{ company !->
}
})
35. data class Address(val city: String, val street: Street)
Lens
Lens(
get = { it.street },
set = { street !-> {
}
})
val street: Street)
36. data class Street(val number: Int, val name: String)
Lens
val streetName: Lens = Lens(
get = { it.name },
set = { name !-> { street !-> street.copy(name = name) } }
)
val name: String)
37. Lens
val employeeStreetName = employeeCompany compose
streetName
val employee = employeeStreetName.modify(
employee,
String"::capitalize)
// Employee(name=John Doe, ... name=Hacker Way))))
38. val company: Company)
Generating Lenses
val employeeCompany: Lens =
Lens(get = { it.company },
set = { company !->
{ employee !->
employee.copy(company = company)
}
}
)
data class Employee(val name: String, val company: Company)
39. Generating Lenses
@Retention(SOURCE)
@Target(CLASS)
annotation class lenses
40. Generating Lens
@lenses
data class Employee(val name: String,
val company: Company)
41. Error Handling
Option
Either
Try
42. Error Handling
Option
Either
Try
43. val books : List? = getBooks()
44. val books : List? = getBooks()
books"?.map { ""... }
books"?.flatMap { ""... }
books"?.fold(""...)
45. val book : Book? = getBookForId(1000)
46. val book : Book? = getBookForId(1000)
book"?.map { ""... }
book"?.flatMap { ""... }
book"?.fold(""...)
47. Option
A
Some None
48. sealed class Option {
object None : Option()
data class Some(val t: T) : Option()
}
49. val option : Option = getBookForId(1000)
fun getBookForId(id : Int) : Option {
val book : Book? = getAllBooksFromDB().find {
it.id "== id
}
return if (book "!= null) {
Some(book)
} else {
None
}
}
50. val option : Option = getBookForId(1000)
fun getBookForId(id : Int) : Option {
val book : Book? = getAllBooksFromDB().find {
it.id "== id
}
}
return if (book "!= null) {
Some(book)
} else {
None
}
51. val option : Option = getBookForId(1000)
fun T?.toOption(): Option =
if (this != null) {
Some(this)
} else {
None
}
52. val option : Option = getBookForId(1000)
fun getBookForId(id : Int) : Option {
return getAllBooksFromDB()
.find { it.id "== id }
.toOption()
}
53. val option : Option = getBookForId(1000)
val totalPrice: Option = option.map { book !->
book.price + tax
}
54. val option : Option = getBookForId(1000)
val description = option.fold(
ifEmpty = {
"No Description"
},
some = {
it.description
}
)
55. Error Handling
Option
Either
Try
56. fun reciprocal(i: Int): Double {
if (i "== 0)
throw IllegalArgumentException("Can’t be 0")
return 1.0 / i
}
57. throw IllegalArgumentException(“Can’t be 0")
fun reciprocal(i: Int): Double {
if (i == 0)
return 1.0 / i
}
58. try {
val reciprocal = reciprocal(2)
compute(reciprocal)
} catch (e: IllegalArgumentException) {
log("Failed to take reciprocal”)
}
59. Either
A
Left Right
B
60. sealed class Either {
data class Left(val a: A) : Either()
data class Right(val b: B) : Either()
}
61. fun reciprocal(i: Int): Double {
if (i == 0)
throw IllegalArgumentException(“Can’t be 0")
return 1.0 / i
}
62. fun reciprocal(i: Int): {
if (i == 0)
throw IllegalArgumentException(“Can’t be 0")
return 1.0 / I
}
Either
63. fun reciprocal(i: Int): {
return if (i == 0) {
} else {
1.0 / i
}
}
Either.left(IllegalArgumentException(“Can’t be 0")
Either
64. fun reciprocal(i: Int): {
return if (i == 0) {
} else {
}
}
Either.left(IllegalArgumentException(“Can’t be 0")
Either
Either.right(1.0 / i)
65. fun reciprocal(i: Int): {
return if (i == 0) {
} else {
}
}
Either.left(IllegalArgumentException(“Can’t be 0")
Either
Either.right(1.0 / i)
66. val either: Either = reciprocal(2)
67. val either: Either = reciprocal(2)
when(either) {
is Either.Left "-> log("Failed to take reciprocal")
is Either.Right "-> compute(either.b)
}
68. Error Handling
Option
Either
Try
69. fun reciprocal(i: Int): Double {
if (i == 0)
throw IllegalArgumentException(“Can’t be 0")
return 1.0 / i
}
External Library
70. var reciprocal : Double = try {
reciprocal(0)
} catch (e: IllegalArgumentException) {
1.0
}
71. val reciprocal = Try {
reciprocal(2)
}.getOrDefault {
1.0
}
72. Try {
reciprocal(0)
}.fold(
{ t: Throwable !->
log("Failed to take reciprocal", t)
},
{ num: Double !->
compute(num)
})
73. sealed class Try {
data class Failure(val e: Throwable) : Try()
data class Success(val value: A) : Try()
}
74. Adhoc Polymorphism & Type Classes
75. Type Classes
Type Constructors
Option, Try, Either, IO, Ior, Reader, State
76. Polymorphism
77. Parametric Polymorphism
fun head(a : List) : A {
return a[0]
}
• Same implementation for all types.
fun combineAll(a: Matrix, b: Matrix) : Matrix {
return doMatrixComputation()
}
• Same name with different implementation.
• Type Dependent.
fun combineAll(a: String, b: String) : String {
return a + b
}
79. Ad-hoc Polymorphism with Type Class
inline fun combineAll(
a: F, b: F,
ev: Monoid = monoidal()
): F {
return ev.combine(a, b)
}
• Polymorphism with Type Classes.
• How do I achieve this?
80. Define Type Class
interface Monoid {
fun combine(a: F, b: F): F
}
81. Define Type Class
interface Monoid {
fun combine(a: F, b: F): F
}
val monoidMap = mutableMapOf, Monoid"<*">>()
82. Define Instances of Type Class
val monoidMap = mutableMapOf, Monoid"<*">>()
monoidMap[String"::class.java] = object: Monoid {
override fun combine(a: String, b: String): String {
return a + b
}
}
83. Define Instances of Type Class
val monoidMap = mutableMapOf, Monoid"<*">>()
monoidMap[Matrix"::class.java] = object: Monoid {
override fun combine(a: Matrix, b: Matrix): Matrix {
}
}
84. Getting Instances of Type Class
val monoidMap = mutableMapOf, Monoid"<*">>()
inline fun monoidal(): Monoid {
return monoidMap[F"::class.java] as Monoid
}
85. Ad-hoc Polymorphism with Type Class
inline fun combineAll(
a: F, b: F,
ev: Monoid = monoidal()
): F {
return ev.combine(a, b)
}
86. Ad-hoc Polymorphism with Type Class
inline fun combineAll(
a: F, b: F,
ev: Monoid = monoidal()
): F {
return ev.combine(a, b)
}
combineAll(matrix, matrix)
combineAll("Hello", "World")
87. Type Classes in Λrrow
@typeclass
interface Functor : TC {
fun map(fa: HK, f: (A) -> B): HK
}
88. Type Classes Instance in Λrrow
@instance(Option"::class)
interface OptionFunctorInstance : Functor {
override fun map(
fa: OptionKind, f: kotlin.Function1
): Option =
fa.ev().map(f)
}
89. Type Classes Proposal
typeclass Monoid {
fun A.combine(b: A): A
fun empty(): A
}
90. Type Classes Proposal
package intext
instance object IntMonoid : Monoid {
fun Int.combine(b: Int): Int = this + b
fun empty(): Int = 0
}
91. Type Classes Proposal
import intext.IntMonoid
1.combine(2) "// 3
Int.empty() "// 0
92. Type Classes Proposal
import intext.IntMonoid
fun add(a: A, b: A): A given Monoid = a.combine(b)
add("a", "b") "// does not compile: No `String: Monoid`
93. Swift Extension Protocols
extension Int : Monoid {
static func e() "-> Int {
return 0
}
}
extension Bool : Monoid {
static func e() "-> Bool {
return false
}
}
95. Thank you! | 3,341 | 10,952 | {"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} | 2.765625 | 3 | CC-MAIN-2023-50 | longest | en | 0.439746 |
http://math.stackexchange.com/questions/118619/subspace-of-real-valued-functions | 1,469,802,807,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830091.67/warc/CC-MAIN-20160723071030-00047-ip-10-185-27-174.ec2.internal.warc.gz | 151,342,377 | 20,194 | # Subspace of Real-valued Functions
I've been looking at the product topology, and came across this question.
Let X be the set of all real-valued functions which are zero outside of a countable subset of $\mathbb{R}$. Consider X as a subspace of $\mathbb{R}^{\mathbb{R}}$ with the product topology.
1) Is X separable?
2) Does X has the Souslin property? That is, every collection of pairwise disjoint non-empty open subsets of X is countable.
3) Show Y = $\{f \in X: |f(x)| \leq 1 \mbox{ for every } x \in \mathbb{R} \}$ $\thinspace$ is countably compact.
4) Is X Lindelof?
I believe that X is separable, but I am having difficulty formulating a proof. Also, I know that if X is separable, then it has the Souslin property.
-
It is not separable. Let $\{f_n\}$ be a countable set of such functions. We show it is not dense. For each $n$, let $C_n=\{x:f(x)\neq 0\}$. Each $C_n$ is countable and so is $C=\bigcup_n C_n$. Let $x\neq C$. The product set of all admissible functions $f$ such that $f(x)\in(1,\infty)$ is open, but doesn't contain an element from $\{f_n\}$. – Michael Greinecker Mar 10 '12 at 18:12
For 3, any countable set essentially "lives" on a countable index set (as the previous commenter used to show non-separability) and thus lives on a countable metric space $[-1,1]^N$ and thus has a limit point. – Henno Brandsma Mar 10 '12 at 19:06
Once you know it's countably compact, and non-compact (it's dense in the product) it cannot be Lindelöf anymore. – Henno Brandsma Mar 10 '12 at 19:08
Hint: it is ccc; it should follow using the fact that the whole product is. – Henno Brandsma Mar 10 '12 at 19:09
(1) $X$ is not separable. (That would make life too easy!) Let $D$ be a countable subset of $X$. For each $x\in D$ let $S(x)=\{\alpha\in\Bbb R:x(\alpha)\ne 0\}$, the support of $x$. Let $$S=\bigcup_{d\in D}S(x)\;;$$ each $S(x)$ is countable, so $S$ is countable. Now let $\alpha_0\in\Bbb R\setminus S$, and define $p\in X$ by $$p(\alpha)=\begin{cases}1,&\text{if }\alpha=\alpha_0\\0,&\text{otherwise}\;.\end{cases}$$ Let $B=\{x\in X:x(\alpha_0)\ne 0\}$; then $B$ is an open nbhd of $p$ disjoint from $D$, and $D$ is therefore not dense in $X$.
(2) $X$ is ccc (i.e., does have the Suslin property). Let $\mathscr{I}$ be the set of open intervals with rational endpoints. For each finite $F=\{\alpha_1,\dots,\alpha_n\}\subseteq\Bbb R$ and function $\varphi:F\to\mathscr{I}$ let $$B(F,\varphi)=\{x\in X:x(\alpha_k)\in\varphi(\alpha_k)\text{ for }k=1,\dots,n\}\;;\tag{1}$$ $X$ has a base of such open sets, so show that $X$ is ccc, it suffices to show that it has no uncountable pairwise disjoint family of open sets of the form $(1)$. Suppose, then, that $I$ is an uncountable index set, and $\mathscr{B}=\{B(F_i,\varphi_i):i\in I\}$ is a family of these basic open sets. By the $\Delta$-system lemma there are a finite $F\subseteq\Bbb R$ and an uncountable $I_0\subseteq I$ such that $F_i\cap F_j=F$ for every pair of distinct $i,j\in I_0$. There are only countably many finite collections of open intervals with rational endpoints, so there are a $\varphi:F\to\mathscr{I}$ and an uncountable $I_1\subseteq I_0$ such that for each $i\in I_1$ and each $\alpha\in F$, $\varphi_i(\alpha)=\varphi(\alpha)$. (In other words, the basic open sets $B(F_i,\varphi_i)$ for $i\in I_1$ all restrict the coordinates in $F$ in exactly the same way.) But then for any $i,j\in I_1$ we have $B(F_i,\varphi_i)\cap B(F_j\varphi_j)\ne\varnothing$, and $\mathscr{B}$ is not pairwise disjoint.
Added: Alternatively, if you know that $\Bbb R^{\Bbb R}$ is separable, which follows from the Hewitt-Marczewski-Pondiczery Theorem, then you know that $\Bbb R^{\Bbb R}$ is ccc, and you can easily show that any dense subspace must also be ccc. (But the $\Delta$-system lemma is a handy tool to have anyway.)
(3) Adapt the idea that I used in (1) to show that $Y$ does not contain an infinite closed discrete subset.
(4) Show that $Y$ is a closed subset of $X$ that is not compact. Conclude that $Y$ cannot be Lindelöf. What does this tell you about the Lindelöfness of $X$?
$X$ is an example of what is known as a $\Sigma$-product. More generally, let $\{X_\alpha:\alpha\in A\}$ be a family of spaces, and for each $\alpha\in A$ fix a point $p_\alpha\in X_\alpha$. Let $$p=\langle p_\alpha:\alpha\in A\rangle\in \prod_{\alpha\in A}X_\alpha\;,$$ and let $$X=\left\{x\in\prod_{\alpha\in A}X_\alpha:\{\alpha\in A:x_\alpha\ne p_\alpha\}\text{ is countable}\right\}\;;$$ $X$ is the $\Sigma$-product of the $X_\alpha$ with base point $p$. In your case each $X_\alpha$ is $\Bbb R$, and each $p_\alpha=0$.
-
Nice work! As an addendum to the addition to (3), you can also show that $\mathbb{R}^\mathbb{R}$ is ccc by using the fact that a Cartesian product $\prod_{i \in I} X_i$ is ccc iff $\prod_{i \in J} X_i$ is ccc for all finite $J \subseteq I$, and noting that the "finite subproducts" of $\mathbb{R}^\mathbb{R}$ are just the Euclidean spaces $\mathbb{R}^n$ which are easily seen to be ccc. – arjafi Mar 10 '12 at 21:13
I believe the separability of $\mathbb{R}^\mathbb{R}$ is easier than the H-M-P theorem. Take, for instance, the polynomials with rational coefficients. It is clear that every basic open set contains such a polynomial. – Nate Eldredge Mar 10 '12 at 22:16
@Arthur: Yes, another good use of the $\Delta$-system lemma. – Brian M. Scott Mar 11 '12 at 6:30
@Nate: That works, but I don’t think of it that way: I tend to view $\Bbb R^{\Bbb R}$ as a product, not as a function space. – Brian M. Scott Mar 11 '12 at 6:34
@Brian: Thank you for answering my question! I understand 1 and 2, and I am able to follow the hint given for 4. For 3, I was able to figure out that any countable subset of $Y$ has limit point since it is a subspace of $[-1, 1]^{\omega}$ (thanks to Henno). However, I'm not sure how this implies every infinite subset of $Y$ has a limit point in $Y$. I'm having a hard time adapting the idea from 1. – Maria Mar 11 '12 at 16:03 | 1,956 | 5,959 | {"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-2016-30 | latest | en | 0.832054 |
https://www.statisticshowto.com/probability-and-statistics/hypothesis-testing/standardized-test-statistic/ | 1,718,255,477,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.11/warc/CC-MAIN-20240613025523-20240613055523-00283.warc.gz | 916,968,400 | 14,636 | # Standardized Test Statistic: What is it?
## Standardized Test Statistic Formula
If you’re taking the AP Statistics test, you’ll need to know the general formula for a standardized test statistic. Standardized test statistics are used in hypothesis testing. The general formula is:
Standardized test statistic: (statistic-parameter)/(standard deviation of the statistic).
The formula by itself doesn’t mean much, unless you also know the three major forms of the equation for z-scores and t-scores.
How to use these formulas: All of the formulas require you to insert three pieces of information:
1. Your test statistic. For example, the median.
2. The known population parameter.
3. The standard deviation for the statistic.
### Standardized test statistic for z-scores
For easy steps on how to solve this formula, see: How to calculate a z-score.
### T-score (single population)
For easy steps on how to solve this formula, see: What is a T Score Formula?
## What does a Standardized Test Statistic mean?
Standardized test statistics are a way for you to compare your results to a “normal” population. Z-scores and t-scores are very similar, although the t-distribution is a little shorter and fatter than the normal distribution. They both do the same thing. In elementary statistics, you’ll start by using a z-score. As you progress, you’ll use t-scores for small populations. In general, you must know the standard deviation of your population and the sample size must be greater than 30 in order for you to be able to use a z-score. Otherwise, use a t-score. See: T-score vs. z-score.
## Calculating a Standardized Test Statistic: Example Problem
Need help with a homework question? Check out our tutoring page!
The mean life of a particular battery is 75 hours. A sample of 9 light bulbs is chosen and found to have a standard deviation of 10 hours and a mean of 80 hours. Find the standardized test statistic.
The population standard deviation isn’t known, so I’m going to use the t-score formula.
Step 1: Plug the information into the formula and solve:
x̄ = sample mean = 80
μ0 = population mean = 75
s = sample standard deviation = 10
n = sample size = 9
t = 80-75 / (10/√9) = 1.5.
This means that the standardized test statistic (in this case, the t-score) is 1.5.
Check out our YouTube channel for more stats help and tips!
## References
Everitt, B. S.; Skrondal, A. (2010), The Cambridge Dictionary of Statistics, Cambridge University Press.
Gonick, L. (1993). The Cartoon Guide to Statistics. HarperPerennial. | 593 | 2,547 | {"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} | 4.59375 | 5 | CC-MAIN-2024-26 | latest | en | 0.894827 |
https://brainly.ph/question/115935 | 1,487,597,897,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170562.59/warc/CC-MAIN-20170219104610-00011-ip-10-171-10-108.ec2.internal.warc.gz | 701,720,849 | 10,897 | # The steel plate to be used in making the hood of a certain utility vehicle is an isosceles trapezoid of altitude 100 cm. The slant height is equal to the shorter base while the longer base is one ad a half times the slant height. Find the area of the steel plate?
1
by patappl912
2015-05-13T20:46:48+08:00
### 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.
Refer to the picture attached
We let the slant height be x.
This would make the shorter base be x and the longer base be 1 1/2 x
In order to know the value of x we would need to perform the Pythagorean Theorem which is
The area of a trapezoid is: | 218 | 867 | {"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-2017-09 | latest | en | 0.920949 |
https://math.stackexchange.com/questions/3403759/x-n-n-1-infty-y-n-n-1-infty-subset-m-compact-then-exis | 1,701,474,509,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00156.warc.gz | 439,233,894 | 37,082 | # $\{x_n\}_{n=1}^\infty$, $\{y_n\}_{n=1}^\infty$ $\subset M$ (compact) then $\exists$ $\{x_{n_k}\}, \{y_{n_k}\}$ s.t. they both converge to same point.
Full question:
$$\left< M,\rho \right>$$ is a compact metric space with sequences $$\{x_n\}_{n=1}^\infty$$, $$\{y_n\}_{n=1}^\infty$$ and $$\displaystyle{\lim_{n \to \infty} \rho(x_n, y_n)=0}$$.
Prove $$\exists$$ subsequences $$\{x_{n_k}\}_{k=1}^\infty$$, $$\{y_{n_k}\}_{k=1}^\infty$$ and a point $$x \in M$$ s.t.
$$\displaystyle{\lim_{k \to \infty} \rho(x_{n_k}, x)=0}$$ and $$\displaystyle{\lim_{k \to \infty} \rho(y_{n_k}, x)=0}$$
Below is my attempt and I would like to know if there is any mistake or room for improvement.
Proof:
$$M$$ is compact so $$\{x_n\}_{n=1}^\infty$$ has a subsequence $$\{x_{n_k}\}_{k=1}^\infty$$ that converges to a point in $$M$$. Call this point $$x \in M$$.
So $$\displaystyle{\lim_{k \to \infty} x_{n_k}=x}$$ $$\implies$$ $$\displaystyle{\lim_{k \to \infty} \rho(x_{n_k}, x)=0}$$
So given an $$\epsilon \gt 0$$, $$\exists N_1 \in \mathbb{N}$$ s.t. if $$k \geq N_1$$ then $$\rho(x_{n_k}, x) \lt \frac{\epsilon}{2}$$
Also given $$\displaystyle{\lim_{n \to \infty} \rho(x_n,y_n)=0}$$, so $$\exists N_2 \in \mathbb{N}$$ s.t. if $$n \geq N_2$$ then $$\rho(x_n,y_n) \lt \frac{\epsilon}{2}$$
But $$n_k \geq n \ \ \forall n,k \in \mathbb{N}$$
So $$\rho(x_{n_k},y_{n_k}) \lt \frac{\epsilon}{2}$$ if $$k \geq max\{N_1,N_2\}$$
Now, let $$k \geq max\{N_1,N_2\}$$ then
$$\rho(y_{n_k}, x) \leq \rho(y_{n_k},x_{n_k}) + \rho(x_{n_k}, x)$$ $$\lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$
So for any given $$\epsilon \gt 0$$ we have shown there is an index $$N$$ s.t. if $$n \geq N$$ then $$\rho(y_{n_k}, x) \lt \epsilon$$ and $$\rho(x_{n_k}, x) \lt \epsilon$$ which means
$$\displaystyle{\lim_{n \to \infty} \rho(x_n,x)=0}$$ and $$\displaystyle{\lim_{n \to \infty} \rho(y_n,x)=0}$$ $$\ \ \ \ \ \ \Box$$
It is almost correct. The only problem that I see in it lies in the assertion “But $$n_k\geqslant n\ \forall n,k\in\mathbb N$$”, which is false. I suppose that you meant to write that $$n_k\geqslant k\ \forall k\in\mathbb N$$.
• That makes sense. I meant to write that the terms in a subsequence appear later in the sequence compared to the main sequence. Which is the same as writing what you wrote. Thanks for pointing that out.
– Sun
Oct 22, 2019 at 4:54
Rephrasing:
0) $$M$$ compact, there exist a convergent subsequence $$(x_{n_k}) \in M$$.
1) Let $$x \in M$$ be $$\lim_{k \rightarrow \infty} x_{n_k}=x$$,
i.e. for $$\epsilon >0$$ there is a $$K_0$$ s.t $$k \ge K_0$$,
implies $$d(x_{n_k},x) <\epsilon$$.
2)Given:
$$\lim_{n \rightarrow \infty} d(x_n,y_n)=0$$, i.e.
there is a $$N$$ s.t. $$n \ge N$$ implies $$d(x_n,y_n) < \epsilon$$.
3) Let $$K_1$$ be s.t for $$k \ge K_1$$,
$$n_k \ge n_{K_1} \ge N$$, implies
$$d(x_{n_k},y_{n _k}) <\epsilon$$.
4) Choose $$K_2=\max(K_0,K_1)$$.
5) Then for $$k \ge K_2$$
$$d(x,y_{n_k}) \le d(x,x_{n_k})+d(x_{n_k},y_{n_k})<2\epsilon$$. | 1,223 | 2,987 | {"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": 61, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-50 | latest | en | 0.520283 |
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## #26 2013-07-11 07:07:53
EbenezerSon
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### Re: Simplify the following:
#### bob bundy wrote:
At first, I did not spot that this is the difference of two squares. But then I noticed that (q^2 - 6qr + 9r^2) is a perfect square:
So try this:
put P = 3p and Q = (q - 3r) and use
Bob
Last but not the least could you help me with an illustration, how you got the (q - 3r)^2 ?. :-)
Last edited by EbenezerSon (2013-07-11 07:11:04)
## #27 2013-07-11 07:34:09
bob bundy
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### Re: Simplify the following:
OK. Here's an example of a perfect square. I work best with pictures so I've made a picture below for this example.
Could this be a perfect square ?
Well it has x^2. That's a good start. And it has 9y^2 = (3y)^2 so it's looking like it might be a perfect square.
Have a look at my picture. Click it to make it bigger.
With a perfect square there are four boxes.
Two are squares: x^2 and (3y)^2
The other two boxes have to be the same rectangle shape, once one way round, and then again turned 90 degrees.
So can the third term in the expression, 6xy, be made into two equal rectangles ... it must be 3xy and 3xy.
So now I can factorise the expression into two equal brackets:
q^2 tick
9r^2 = (3r)^2 tick
Split the middle term in two: -6qr = -3qr - 3qr tick
so
Now I shall have to log out so here's a hint for number (4)
Three of these terms will make a perfect square. You need to decide which three.
Then choose P and Q and use the difference of two squares.
I'll check tomorrow to see how you are getting on.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #28 2013-07-11 21:59:12
EbenezerSon
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### Re: Simplify the following:
4a^2 - 12ab - c^2 + 9b^2
= 4a^2 - 12ab - 9b^2 - c^2
= 4a^2 - 6ab - 6ab + 9b^2
= 2a(2a - 3b) -3b(2a - 3b)
= (2a - 3b)(2a - 3b) = (2a - 3b)^2.
------------------------------------------
-----------------------------------------
(2a - 3b)^2 - c^2 = (2a - 3b - c)(2a - 3b + c). Is the final answer correct?
Last edited by EbenezerSon (2013-07-12 02:21:20)
## #29 2013-07-11 22:03:28
EbenezerSon
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### Re: Simplify the following:
The book gave this as the final answer:
(2a - 3b - c)(2a - 3b '- C')
I don't understand how it got the '-C', I think it should be '+C' instead.
What do you say?
Last edited by EbenezerSon (2013-07-11 22:05:36)
## #30 2013-07-11 22:34:54
bob bundy
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### Re: Simplify the following:
Got to go out. I'll check again later.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #31 2013-07-12 02:31:11
bob bundy
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### Re: Simplify the following:
hi EbenezerSon
Just got back home. There's a way to check algebra when you want to know if you've done it correctly. Give me a moment and I'll post what you have to do.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #32 2013-07-12 02:34:21
EbenezerSon
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### Re: Simplify the following:
#### bob bundy wrote:
hi Ebenezerson
I think you'll need to multiply these brackets out like this
Then you can start to simplify.
Can you fill in the dots ?
Alternative way to do this:
difference of two squares:
Have you met this before ?
If so, you could put p = 2m + 2k and q = m - 2k
Bob
Sir this one, I tried resolving it my self with my understanding, but got different result, please I would be much obliged if you would take the pain to explain it to my understanding once more by manipulating each step. Please.
## #33 2013-07-12 02:42:55
bob bundy
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### Re: Simplify the following:
OK. here we go.
The rules of algebra are the same as the rules of arithmetic. So you can check your algebra by substituting some numbers.
Let's use Q4 as an example.
You started with 4a^2 -12ab - c^2 + 9b^2
You've got one answer. The book has another. Which is right?
Choose numbers for a, b, and c. It's best to choose different numbers and not 0 or 1. So I'll make
a = 2
b = 3
c = 5
All different with no common factors.
The original expression is 4a^2 -12ab - c^2 + 9b^2 = 4 x 2 x 2 - 12 x 2 x 3 - 5 x 5 + 9 x 3 x 3 = 16 - 72 - 25 + 81 = 97 - 97 = 0
Wow! I didn't plan that to be zero. It was just a lucky chance!
(2a - 3b - c)(2a - 3b + c) = (2 x 2 - 3 x 3 - 5) x ( 2 x 2 - 3 x 3 + 5) = (4 - 9 - 5) x (4 - 9 + 5) = -10 x 0 = 0
This is the same as what you started with.
(2a - 3b - c)(2a - 3b - C) = (2 x 2 - 3 x 3 - 5) x (2 x 2 - 3 x 3 - 5) = (4 - 9 - 5) x (4 - 9 - 5) = -10 x -10 = 100.
This isn't correct. The book answer must be wrong.
There's a small chance that you made two errors that 'cancelled out' but it's unlikely.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #34 2013-07-12 02:43:51
bob bundy
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### Re: Simplify the following:
Sir this one, I tried resolving it my self with my understanding, but got different result, please I would be much obliged if you would take the pain to explain it to my understanding once more by manipulating each step. Please.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #35 2013-07-12 05:08:23
EbenezerSon
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### Re: Simplify the following:
Errr sir, that's good, so always to check, should I pick the numbers randomly in my head to do the substitution? To see if will get (0) as result for both? Or the figure I will get should be the same in each case? And not necessarily (0)?
## #36 2013-07-12 05:12:33
EbenezerSon
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### Re: Simplify the following:
I got:
3m^2 - 8km + 13k^2, Which is impossible for me to split the middle term.
## #37 2013-07-12 07:06:47
bobbym
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### Re: Simplify the following:
Hi EbenezerSon;
Do you need to factor that?
Because there is always a factorization of an expression like that with a lot of work and help from a computer to check the expansions this is what I found:
This involves complex numbers and I am sure this is not what they want so have you copied the problem correctly?
Here are some easy to factor possibilities.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #38 2013-07-12 17:01:35
bob bundy
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### Re: Simplify the following:
#### Ebenezerson wrote:
I got:
3m^2 - 8km + 13k^2, Which is impossible for me to split the middle term.
I showed you this one back in post 16.
The -8km term should have been +12mk + 4mk so it looks like you've made two sign mistakes.
The 13k^2 should have been 9k^2 - 4k^2 so once again a sign mistake.
I recommend you do some practice on expanding brackets with some negative signs. You find this page and exercises helpful:
http://www.mathsisfun.com/algebra/expanding.html
Errr sir, that's good, so always to check, should I pick the numbers randomly in my head to do the substitution? To see if will get (0) as result for both? Or the figure I will get should be the same in each case? And not necessarily (0)?
Yes, you can pick random numbers. You may miss a mistake if you choose the same number for 'a' and 'b' so pick different numbers. It was a complete chance that the expression came out as 0. With a different choice for a, b and c it would have produced a completely different result. What you are looking for is the same result from both expressions.
Let's use your other question as an example.
(1) (2m + 3k)^2 - (m - 2k)^2.
Let's choose m = 2 and k = 3
(2m + 3k)^2 - (m - 2k)^2 = (2 x 2 + 3 x 3)^2 - (2 - 2 x 3)^2 = (4 + 9)^2 - (2 - 6)^2 = 13^2 - (-4)^2 = 169 - 16 = 153
So the correct answer should come to 153.
Your answer 3m^2 - 8km + 13k^2 = 3 x 2 x 2 - 8 x 2 x 3 + 13 x 3 x 3 = 12 - 48 + 117 = 129 - 48 = 81 so this must have an error.
My result 3m^2 + 5k^2 + 16mk = 3 x 2 x 2 + 5 x 3 x 3 + 16 x 2 x 3 = 12 + 45 + 96 = 153
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #39 2013-07-13 05:53:16
EbenezerSon
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### Re: Simplify the following:
I have understood #38 very well, but please, see how I came by that final answer, I have tried to understand how you solved it at #19 but still I haven't grasp it.
Last edited by EbenezerSon (2013-07-13 05:57:46)
## #40 2013-07-13 06:18:12
EbenezerSon
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### Re: Simplify the following:
(2m^2 + 3k)^2 - (m - 2k)^2 . Please see how I solved it, and correct where I have made the mistakes so that I could identify it. Please.
Here I go:
(2m + 3k)^2 - (m - 2k)^2 = (2m + 3k)(2m + 3k) - (m - 2k)(m - 2k) . by expansion.
= Multiplying the first expansion > (2m + 3k)(2m + 3k) = 4m^2 + 6mk + 6mk + 9k^2.
Multiplying the second expansion > (m - 2k)(m - 2k) = m^2 - 2mk - 2mk + 4k^2.
= (4m^2 + 6mk + 6mk + 9k^2) - (m^2 - 2mk - 2mk + 4k^2)
= grouping like terms > 4m^2 - m^2 + 6mk +6mk - 2mk - 2mk + 9k^2 + 4k^2
= adding them > 3mk^2 + 12mk - 4mk + 13k^2
= 3mk^2 + 8mk + 13k^2.
Please this is how I got my answer, I know I am wrong but I cant help.
## #41 2013-07-13 07:27:06
bob bundy
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### Re: Simplify the following:
(2m + 3k)^2 - (m - 2k)^2 = (2m + 3k)(2m + 3k) - (m - 2k)(m - 2k) . by expansion. Correct!
= Multiplying the first expansion > (2m + 3k)(2m + 3k) = 4m^2 + 6mk + 6mk + 9k^2. Correct!
Multiplying the second expansion > (m - 2k)(m - 2k) = m^2 - 2mk - 2mk + 4k^2. Correct!
= (4m^2 + 6mk + 6mk + 9k^2) - (m^2 - 2mk - 2mk + 4k^2) Correct!
= grouping like terms > 4m^2 - m^2 + 6mk +6mk - 2mk - 2mk + 9k^2 + 4k^2 This is where the mistake is.
-(m^2 - 2mk -2mk + 4k^2) = -m^2 + 2mk + 2mk - 4k^2
This is -1 times the bracket, so every term changes sign.
= adding them > 3mk^2 + 12mk - 4mk + 13k^2 Should be 3m^2 + + 12mk + 4mk + 9k^2 - 4k^2
= 3mk^2 + 8mk + 13k^2. Should be 3m^2 + 16mk - 5k^2
I recommend that you practise expanding brackets with negatives.
http://www.mathsisfun.com/algebra/expanding.html
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #42 2013-07-16 07:57:33
EbenezerSon
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### Re: Simplify the following:
#### bob bundy wrote:
= grouping like terms > 4m^2 - m^2 + 6mk +6mk - 2mk - 2mk + 9k^2 + 4k^2 This is where the mistake is.
-(m^2 - 2mk -2mk + 4k^2) = -m^2 + 2mk + 2mk - 4k^2[/b]
This is -1 times the bracket, so every term changes sign.
Thank you very much indeed Sir!
Errrr! That was where the mistakes stem from, candidly speaking I had thought, I one has to take off the bracket, after multiplication has taken place within the bracket.
I have now grown wary to multiply any sign outside the bracket after multiplcation has taken place within the bracket, especially negative signs.
Thanks Sir for the provision of the link, God bless.
## #43 2013-07-18 08:42:03
EbenezerSon
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### Re: Simplify the following:
Simplify the following:
Please help me solve this. (x^3/2 + x^1/2) (x^1/2 - x^-1/2)/(x^3/2 - x^1/2)^2. I got x^2 - x + x/x^3 - x^2 - x^2 as the final answer. The book has x + 1/x(x - 1) as its answer which I don't comprehend. The book did not show the procedures.
Last edited by EbenezerSon (2013-07-19 00:47:33)
## #44 2013-07-19 01:02:25
bobbym
Offline
### Re: Simplify the following:
Hi;
Is this the way the problem looks?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #45 2013-07-19 03:24:49
EbenezerSon
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### Re: Simplify the following:
This is how the top right bracket looks: (x^1/2 - x^-1/2) The last X raises to the power negative one over two(x^-1/2).
Thanks.
## #46 2013-07-19 04:35:58
bobbym
Offline
### Re: Simplify the following:
Hi;
Those are both the same. But to put it into the form you want then we have?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #47 2013-07-22 00:06:34
EbenezerSon
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### Re: Simplify the following:
Errr okay, I see!
But Sirs please solve for me.
## #48 2013-07-22 00:19:27
bobbym
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### Re: Simplify the following:
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #49 2013-07-22 00:59:13
EbenezerSon
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### Re: Simplify the following:
Thanks once more.
## #50 2013-07-22 01:00:54
bobbym
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### Re: Simplify the following:
I multiplied the numerator out and then the denominator. Just plain multiplication.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof. | 4,598 | 13,630 | {"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-2014-15 | longest | en | 0.848304 |
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perplexus dot info
A Reciprocal And Square Problem (Posted on 2007-07-10)
Find all real pairs (p, q) satisfying the following system of equations:
p - 1/p - q2 = 0
q/p + pq = 4
Submitted by K Sengupta Rating: 3.5000 (2 votes) Solution: (Hide) The given equations yield; p – 1/p = q^2 p+ 1/p = 4/q Thus, 16/(q^2) – q^4 = 4, giving: u^3 + 4u – 16 = 0, where u = q^2 Or, (u-2)(u^2 +2u +8) = 0 Or, u =2, ignoring the complex roots of u which are inadmissible. or, q = +/- √2) If q = √2, then p = (1/2)*(q^2 + 4/q) = 1+ √2 Similarly, q = - √2 gives p = 1 – √2 Thus (p, q) = (√2, 1 + √2), (-√2, 1- √2) are the only possible solutions.
Subject Author Date re: Analytical solution - corrected Ady TZIDON 2007-07-11 07:19:10 Sol Praneeth Yalavarthi 2007-07-10 11:44:22 Analytical solution Federico Kereki 2007-07-10 11:38:51 re(2): Solution Praneeth Yalavarthi 2007-07-10 11:30:14 computer aided solution Charlie 2007-07-10 11:16:04 re: Solution Charlie 2007-07-10 11:08:38 Solution Praneeth Yalavarthi 2007-07-10 10:15:27
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# If the perimeter of the triangle is 40 cm, find the area of the triangle.
Viewed 692918 times
1
The area of the triangle is, ? cm2.
2
• 102 cm²
Step-by-step explanation:
Perimeter is:
• P = 40 cm
• and
• P = 2x - 4 + 5x - 13 + x + 9 = 8x - 8
Comparing the value:
• 8x - 8 = 40
• 8x = 48
• x = 8 cm
Are formula:
• A = 1/2ah
Replacing values:
• A = 1/2(x + 9)(2x - 4)
• A = 1/2(8 + 9)(2*8 - 4) = 1/2*17*12 = 102 cm²
##### Mitchel Klein
15.5k 3 10 26
2
60cm²
Step-by-step explanation:
To find :-
• The Area of the triangle .
We know that Perimeter is the sum of sides of a triangle. Here the perimeter is 40cm .
→ 2x - 4 +5x - 13 + x + 9 = 40cm .
→ 8x - 8 = 40cm
→ 8x = 40+8 cm
→ 8x = 48 cm
→ x = 6cm .
We know that the area of the triangle is half the product of base and height .
• Base = x + 9 = 15cm
• Height = 2x -4 = 8cm .
→ A = 1/2 × b × h
→ A = 1/2 × 8cm × 15cm
→ A = 60cm²
15.5k 3 10 26 | 407 | 926 | {"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} | 4.5 | 4 | CC-MAIN-2022-49 | latest | en | 0.732534 |
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A060918 Expansion of e.g.f.: exp((-1)^k/k*LambertW(-x)^k)/(k-1)!, k=4. 4
1, 20, 360, 6860, 143570, 3321864, 84756000, 2372001720, 72384192540, 2394775746220, 85443353291296, 3271908306712500, 133893717061821080, 5832748749666611920, 269542701201588099840, 13172225935626444660144, 678788199609330554538000, 36790272488566573278647940 (list; graph; refs; listen; history; text; internal format)
OFFSET 4,2 COMMENTS a(n) = A243098(n,4)/6. - Alois P. Heinz, Aug 19 2014 LINKS Vincenzo Librandi, Table of n, a(n) for n = 4..200 FORMULA a(n) = (n-1)!/(k-1)!*Sum_{i=0..floor((n-k)/k)} 1/(i!*k^i)*n^(n-(i+1)*k)/(n-(i+1)*k)!, k=4. a(n) ~ 1/6*exp(1/4)*n^(n-1). - Vaclav Kotesovec, Nov 27 2012 MATHEMATICA CoefficientList[Series[E^(1/4*LambertW[-x]^4)/6, {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Nov 27 2012 *) PROG (PARI) x='x+O('x^30); Vec(serlaplace(exp(lambertw(-x)^4/4)/3! - 1/3!)) \\ G. C. Greubel, Feb 19 2018 CROSSREFS Cf. A057817, A060917, A243098. Sequence in context: A323961 A004292 A053508 * A115100 A049683 A228330 Adjacent sequences: A060915 A060916 A060917 * A060919 A060920 A060921 KEYWORD easy,nonn AUTHOR Vladeta Jovovic, Apr 10 2001 STATUS approved
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# Teaching Outlines
## Concept #
### Notes for teachers
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### Activity No #
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• Evaluation (Questions for assessment of the child)
• Question Corner
### Activity No #
• Estimated Time
• Materials/ Resources needed
• Prerequisites/Instructions, if any
• Multimedia resources
• Website interactives/ links/ Geogebra Applets
• Process (How to do the activity)
• Developmental Questions (What discussion questions)
• Evaluation (Questions for assessment of the child)
• Question Corner
## Concept #
### Notes for teachers
These are short notes that the teacher wants to share about the concept, any locally relevant information, specific instructions on what kind of methodology used and common misconceptions/mistakes.
### Activity No #
• Estimated Time
• Materials/ Resources needed
• Prerequisites/Instructions, if any
• Multimedia resources
• Website interactives/ links/ Geogebra Applets
• Process (How to do the activity)
• Developmental Questions (What discussion questions)
• Evaluation (Questions for assessment of the child)
• Question Corner
### Activity No #
• Estimated Time
• Materials/ Resources needed
• Prerequisites/Instructions, if any
• Multimedia resources
• Website interactives/ links/ Geogebra Applets
• Process (How to do the activity)
• Developmental Questions (What discussion questions)
• Evaluation (Questions for assessment of the child)
• Question Corner
# Math Fun
Usage
Create a new page and type {{subst:Math-Content}} to use this template
# TRIANGLES
Mark three non-collinear point P, Q and R on a paper. Join these pints in allpossible ways. The segments are PQ, QR and RP. A simple close curve formed by these three segments is called a triangle. It is named in one of the following ways.
Triangle PQR or Triangle PRQ or Triangle QRP or Triangle RPQ or Triangle RQP .
PQR
A triangle is one of the basic shapes of geometry: a polygon with three corners or vertices and three sides or edges which are line segments.
In fact, it is the polygon with the least number of sides.
A triangle PQR consists of all the points on the line segment PQ,QR and RP.
The three line segments, PQ, QR and RP that form the triangle PQ, are called the sides of the triangle PQR.
Angles:
A triangle has three angles. In figure, the three angles are ∠PQR ∠QRP and ∠RPQ
Parts of triangle:
A triangle has six parts, namely, three sides,PQ QRand RP.Three angles ∠PQR ∠QRP and ∠RPQ. These are also known as the elements of a triangle.
Vertices of a Triangle
The point of intersection of the sides of a triangle is known as its vertex. In figure, the three vertices are P, Q and R. In a triangle, an angle is formed at thevertex. Since it has three vertices, so three angles are formed. The word triangle =tri + angle ‘tri’ means three. So, triangle means closed figure of straight lines having three angles.
## Activities
### Activity 1: Identifying and Naming Triangles
#### Learning Objectives
Identify and name the triangles
#### Pre-requisites/Instructions
Identify and name the triangles in the following Figure.
#### Evaluation
1. Is it possible to construct a triangle with 3 collinear points?
2. Is it possible to construct a triangle whose sides are 3cm, 4cm and 9cm. Give reason.
# Classification of Triangles
Triangles can be classified in two groups:
Triangles differentiated on the basis of their sides.
## Equilateral Triangles:
A triangle with all sides equal to one another is called an equilateral triangle.
## Isosceles Triangle:
A triangle with a pair of equal sides is called an isosceles triangle.
## Scalene Triangle:
A triangle in which all the sides are of different lengths and no two sides are equal, the triangle is called a scalene triangle.
## Triangles differentiated on the basis of their angles.
Acute angled triangle.
A triangle whose all angles are acute is called an acute-angled triangle or simply an acute triangle.
Right Triangles :
A right triangle has one angle 90°
Right Isosceles Triangle :
Has a right angle (90°), and also two equal angles Can you guess what the equal angles are?
The Obtuse Triangle :
The Obtuse Triangle has an obtuse angle (an obtuse angle has more than 90°). In the picture on the left, the shaded angle is the obtuse angle that distinguishes this triangle
Since the total degrees in any triangle is 180°, an obtuse triangle can only have one angle that measures more than 90°.
## Interior angles of a triangle
The interior angles are those on the inside of the triangle.
## Exterior Angles of a triangle
An exterior angle is formed by extending any side of the triangle.
## Summary of triangle centres
There are many types of triangle centers. Below are four of the most common.
Incenter Located at intersection of the angle bisectors. See Triangle incenter definition Circumcenter Located at intersection of the perpendicular bisectors of the sides. See Triangle circumcenter definition Centroid Located at intersection of medians. See Centroid of a triangle Orthocenter Located at intersection of the altitudes of the triangle. See Orthocenter of a triangle
In the case of an equilateral triangle, all four of the above centers occur at the same point.
The Incenter of a triangle
Latin: in - "inside, within" centrum - "center"
The point where the three angle bisectors of a triangle meet. One of a triangle's points of concurrency.
Try this Drag the orange dots on each vertex to reshape the triangle. Note the way the three angle bisectors always meet at the incenter.
One of several centers the triangle can have, the incenter is the point where the angle bisectors intersect. The incenter is also the center of the triangle's incircle - the largest circle that will fit inside the triangle.
## Centroid of a Triangle
From Latin: centrum - "center", and Greek: -oid -"like" The point where the three medians of the triangle intersect. The 'center of gravity' of the triangle One of a triangle's points of concurrency. Try this Drag the orange dots at A,B or C and note where the centroid is for various triangle shapes.
Refer to the figure . Imagine you have a triangular metal plate, and try and balance it on a point - say a pencil tip. Once you have found the point at which it will balance, that is the centroid.
The centroid of a triangle is the point through which all the mass of a triangular plate seems to act. Also known as its 'center of gravity' , 'center of mass' , or barycenter.
A fascinating fact is that the centroid is the point where the triangle's medians intersect. See medians of a triangle for more information. In the diagram above, the medians of the triangle are shown as dotted blue lines.
Centroid facts
• The centroid is always inside the triangle
• Each median divides the triangle into two smaller triangles of equal area.
• The centroid is exactly two-thirds the way along each median.
Put another way, the centroid divides each median into two segments whose lengths are in the ratio 2:1, with the longest one nearest the vertex. These lengths are shown on the one of the medians in the figure at the top of the page so you can verify this property for yourself.
## Orthocenter of a Triangle
From Greek: orthos - "straight, true, correct, regular" The point where the three altitudes of a triangle intersect. One of a triangle's points of concurrency.
Try this Drag the orange dots on any vertex to reshape the triangle. Notice the location of the orthocenter.
The altitude of a triangle (in the sense it used here) is a line which passes through a vertex of the triangle and is perpendicular to the opposite side. There are therefore three altitudes possible, one from each vertex. See Altitude definition.
It turns out that all three altitudes always intersect at the same point - the so-called orthocenter of the triangle.
The orthocenter is not always inside the triangle. If the triangle is obtuse, it will be outside. To make this happen the altitude lines have to be extended so they cross. Adjust the figure above and create a triangle where the orthocenter is outside the triangle. Follow each line and convince yourself that the three altitudes, when extended the right way, do in fact intersect at the orthocenter.
### Activity 1 Types of Triangles
#### Learning Objectives
Be able to identify triangles.
#### Pre-requisites/Instructions
Identify the types of triangles.
#### Evaluation
1. Can a scalene triangle also be a right-angled triangle ? If yes can you draw one ?
### Activity 2 Similar Triangles
Learning Objective
To show similar planar figures, discuss congruence and properties of congruent/ similar triangles
#### Material and Resources Required
Blackboard
Geogebra files + projector
Calculator
• Planar figures and triangles
• Draw pairs of figures on the board [ both similar and dissimilar]; they can identify overlap of congruent figures
• Ask the children to identify
• If the children know the names of the theorem, ask them to explain- ask them what is SSS, AAA, ASA
• Show ratio and give the idea of proportionality
• Geogebra files. When I change the sides/ proportion, the triangles change in size. But the proportion remains the same, angle remains the same
• With calculator they verify proportion (this is very very useful for involving the whole class) they all can see the proportion remains constant though the size changes
• Show the arithmetic behind the proportion
#### Evaluation
[Activity evaluation - What should the teacher watch for when you do the activity; based on what they know change]
• Confusion between congruence and similarity
• When they give the theorem, if they cannot identify included side and angle
• When there is a wrong answer, identify what is the source of the confusion – sides, ratio and proportion
• Direct substitution
# Pythagorean Theorem
Pythagoras' Theorem was discovered by Pythagoras, a Greek mathematician and philosopher who lived between approximately 569 BC and 500 BC.
Pythagoras' Theorem states that:
In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. That is:
Pythagoras' Theorem in Three Dimensions
A three-dimensional object can be described by three measurements - length, width and height.
We can use Pythagoras' Theorem to find the length of the longest straw that will fit inside
the box or cylinder.
# Teachers Corner
Suchetha . S. S Asst. Teacher ( Mathematics ) GJC Thyamagondlu. Nelamangala Talluk Bangalore Rural District doing a lesson on similar triangles using GeoGebra in the classroom
## GeoGebra Contributions
1. The GeoGebra file below to understand Similar Triangles
1. Similar Triangles Part 1 http://www.karnatakaeducation.org.in/KOER/Maths/Similar_Triangles_1.html | 2,479 | 11,265 | {"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} | 3.71875 | 4 | CC-MAIN-2021-21 | latest | en | 0.703114 |
https://www.imlearningmath.com/how-can-you-give-someone-63-using-six-bills-without-using-any-one-dollar-bills/ | 1,603,998,342,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107905777.48/warc/CC-MAIN-20201029184716-20201029214716-00125.warc.gz | 755,662,449 | 24,203 | # How can you give someone $63 using six bills, without using any one dollar bills? Math Riddle: How can you give someone$63 using six bills, without using any one-dollar bills?
Answer: The correct answer is 1 – $50 bill, 1 –$5 bill , 4 – \$2 bills | 68 | 249 | {"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} | 3.5 | 4 | CC-MAIN-2020-45 | latest | en | 0.805261 |
http://www.sofsource.com/algebra-2-chapter-4-resource-book/relations/scale-ratio-formula.html | 1,480,955,579,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541773.2/warc/CC-MAIN-20161202170901-00175-ip-10-31-129-80.ec2.internal.warc.gz | 708,402,951 | 12,267 | Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
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scale ratio formula
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Author Message
Jliosomator
Registered: 05.06.2007
From: Texas
Posted: Friday 29th of Dec 09:12 Can someone help me with my assignment problems ? Most of them are based on scale ratio formula. I have read a some sample questions on monomials and logarithms but that didn’t really help me in finding solutions to the questions on my homework . I didn’t sleep last night since I have a deadline to meet . But the problem is no matter how much time I invest, I just don’t seem to be getting the hang of it. Every question poses a new problem , one which seems to be tougher than climbing Mt.Everest! I need some help as soon as possible. Somebody please guide me.
Jahm Xjardx
Registered: 07.08.2005
From: Odense, Denmark, EU
Posted: Sunday 31st of Dec 07:57 Algebrator is a useful program to solve scale ratio formula questions. It gives you step by step solutions along with explanations. I however would warn you not to just paste the answers from the software. It will not aid you in understanding the subject. Use it as a reference and solve the problems yourself as well.
Jrobhic
Registered: 09.08.2002
From: Chattanooga, TN
Posted: Monday 01st of Jan 10:26 I have tried out various software. I would without any doubt say that Algebrator has helped me to grapple with my difficulties on radical equations, multiplying fractions and logarithms. All I did was to just key in the problem. The reply showed up almost straight away showing all the steps to the result. It was quite easy to follow. I have relied on this for my math classes to figure out Algebra 2 and Remedial Algebra. I would highly advice you to try out Algebrator.
abusetemailatdrics
Registered: 24.03.2006
From: Northern Illinois
Posted: Monday 01st of Jan 17:11 Excellent . Just can’t believe it. Just the exact thing for me. Can you advise me where I can acquire this software?
Registered: 10.07.2002
From: NW AR, USA
Posted: Tuesday 02nd of Jan 17:11 I am a regular user of Algebrator. It not only helps me finish my homework faster, the detailed explanations provided makes understanding the concepts easier. I suggest using it to help improve problem solving skills.
erx
Registered: 26.10.2001
From: PL/DE/ES/GB/HU
Posted: Wednesday 03rd of Jan 07:15 http://www.sofsource.com/positive-integral-divisors.html. There you go. Hopefully you will not have to drop math. | 842 | 3,269 | {"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-2016-50 | longest | en | 0.921413 |
https://www.physicsforums.com/threads/complex-hermitian-conjugate-of-matrix.577558/ | 1,627,816,293,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154175.76/warc/CC-MAIN-20210801092716-20210801122716-00297.warc.gz | 963,051,594 | 13,788 | # Complex & Hermitian Conjugate of Matrix
## Homework Statement
1x2 Matrix A =
[(5) (-2i)]
What is the complex conjugate and Hermitian conjugate of this matrix?
## The Attempt at a Solution
D^T =
5
-2i
D^H =
5
+2i
What do you think of my answers?
Dick
Homework Helper
## Homework Statement
1x2 Matrix A =
[(5) (-2i)]
What is the complex conjugate and Hermitian conjugate of this matrix?
## The Attempt at a Solution
D^T =
5
-2i
D^H =
5
+2i
What do you think of my answers?
If the question were to write the transpose and the Hermitian conjugate then I'd agree with you. But the problem asks for the complex conjugate as well. | 189 | 643 | {"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} | 3.09375 | 3 | CC-MAIN-2021-31 | latest | en | 0.938338 |
http://www.colby.edu/chemistry/webmo/NBr3.html | 1,416,865,236,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400382386.21/warc/CC-MAIN-20141119123302-00212-ip-10-235-23-156.ec2.internal.warc.gz | 466,011,434 | 2,756 | ## NBr3
BR3 \ N1 - BR2 / BR4
Tell me about the atomic charges, dipole moment, bond lengths, angles, bond orders,
molecular orbital energies, or total energy.
Tell me about the best Lewis structure.
## Atomic Charges and Dipole Moment
N1 charge=-0.203
BR2 charge= 0.067
BR3 charge= 0.070
BR4 charge= 0.065
with a dipole moment of 0.62110 Debye
## Bond Lengths:
between N1 and BR2: distance=1.994 ang___ between N1 and BR3: distance=1.990 ang___
between N1 and BR4: distance=1.995 ang___
## Bond Angles:
for BR3-N1-BR2: angle=108.6 deg___ for BR4-N1-BR2: angle=107.9 deg___
## Bond Orders (Mulliken):
between N1 and BR2: order=0.935___ between N1 and BR3: order=0.936___
between N1 and BR4: order=0.935___
## Best Lewis Structure
The Lewis structure that is closest to your structure is determined. The hybridization of the atoms in this idealized Lewis structure is given in the table below.
### Hybridization in the Best Lewis Structure
1. A bonding orbital for N1-Br2 with 1.9928 electrons
__has 61.09% N 1 character in a s0.45 p3 hybrid
__has 38.91% Br 2 character in a s0.19 p3 hybrid
2. A bonding orbital for N1-Br3 with 1.9928 electrons
__has 61.17% N 1 character in a s0.45 p3 hybrid
__has 38.83% Br 3 character in a s0.20 p3 hybrid
3. A bonding orbital for N1-Br4 with 1.9928 electrons
__has 61.06% N 1 character in a s0.44 p3 hybrid
__has 38.94% Br 4 character in a s0.19 p3 hybrid
47. A lone pair orbital for N1 with 1.9957 electrons
48. A lone pair orbital for Br2 with 1.9994 electrons
49. A lone pair orbital for Br2 with 1.9847 electrons
50. A lone pair orbital for Br2 with 1.9836 electrons
__made from a s0.94 p3 hybrid
51. A lone pair orbital for Br3 with 1.9994 electrons
52. A lone pair orbital for Br3 with 1.9841 electrons
53. A lone pair orbital for Br3 with 1.9833 electrons
__made from a s0.94 p3 hybrid
54. A lone pair orbital for Br4 with 1.9994 electrons
55. A lone pair orbital for Br4 with 1.9849 electrons
56. A lone pair orbital for Br4 with 1.9838 electrons
__made from a s0.95 p3 hybrid
-With core pairs on: N 1 Br 2 Br 2 Br 2 Br 2 Br 2 Br 2 Br 2 Br 2 Br 2 Br 2 Br 2 Br 2 Br 2 Br 2 Br 3 Br 3 Br 3 Br 3 Br 3 Br 3 Br 3 Br 3 Br 3 Br 3 Br 3 Br 3 Br 3 Br 3 Br 4 Br 4 Br 4 Br 4 Br 4 Br 4 Br 4 Br 4 Br 4 Br 4 Br 4 Br 4 Br 4 Br 4 -
#### Donor Acceptor Interactions in the Best Lewis Structure
The localized orbitals in your best Lewis structure can interact strongly. A filled bonding or lone pair orbital can act as a donor and an empty or filled bonding, antibonding, or lone pair orbital can act as an acceptor. These interactions can strengthen and weaken bonds. For example, a lone pair donor->antibonding acceptor orbital interaction will weaken the bond associated with the antibonding orbital. Conversly, an interaction with a bonding pair as the acceptor will strengthen the bond. Strong electron delocalization in your best Lewis structure will also show up as donor-acceptor interactions.
Interactions greater than 20 kJ/mol for bonding and lone pair orbitals are listed below.
## Molecular Orbital Energies
The orbital energies are given in eV, where 1 eV=96.49 kJ/mol. Orbitals with very low energy are core 1s orbitals. More antibonding orbitals than you might expect are sometimes listed, because d orbitals are always included for heavy atoms and p orbitals are included for H atoms. Up spins are shown with a ^ and down spins are shown as v.
60 ----- 4.508
59 ----- -3.474
58 ----- -3.507
57 ----- -5.274
56 -^-v- -6.933
55 -^-v- -7.164
54 -^-v- -7.548
53 -^-v- -7.563
52 -^-v- -8.224
51 -^-v- -8.238
50 -^-v- -10.26
49 -^-v- -11.84
48 -^-v- -11.85
47 -^-v- -17.39
46 -^-v- -20.12
45 -^-v- -20.14
44 -^-v- -23.58
43 -^-v- -69.24 42 -^-v- -69.24 41 -^-v- -69.25 40 -^-v- -69.25
39 -^-v- -69.26 38 -^-v- -69.26
37 -^-v- -69.62 36 -^-v- -69.62 35 -^-v- -69.62 34 -^-v- -69.63 33 -^-v- -69.64 32 -^-v- -69.65
31 -^-v- -69.76 30 -^-v- -69.77
29 -^-v- -69.78
28 -^-v- -172.5 27 -^-v- -172.5 26 -^-v- -172.5 25 -^-v- -172.5 24 -^-v- -172.5
23 -^-v- -172.5
22 -^-v- -173.0 21 -^-v- -173.0
20 -^-v- -173.0
19 -^-v- -230.0 18 -^-v- -230.0
17 -^-v- -230.0
16 -^-v- -382.1
15 -^-v- -1518. 14 -^-v- -1518. 13 -^-v- -1518. 12 -^-v- -1518.
11 -^-v- -1518.
10 -^-v- -1518.
9 -^-v- -1518. 8 -^-v- -1518.
7 -^-v- -1518.
6 -^-v- -1681. 5 -^-v- -1681.
4 -^-v- -1681.
3 -^-v- -13068
2 -^-v- -13068
1 -^-v- -13068
## Total Electronic Energy
The total electronic energy is a very large number, so by convention the units are given in atomic units, that is Hartrees (H). One Hartree is 2625.5 kJ/mol. The energy reference is for totally dissociated atoms. In other words, the reference state is a gas consisting of nuclei and electrons all at infinite distance from each other. The electronic energy includes all electric interactions and the kinetic energy of the electrons. This energy does not include translation, rotation, or vibration of the the molecule.
Total electronic energy = -7776.6558733873 Hartrees | 1,762 | 4,986 | {"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} | 3.09375 | 3 | CC-MAIN-2014-49 | longest | en | 0.783672 |
http://sergeev.org/files/confs/pres25.htm | 1,532,099,393,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591683.78/warc/CC-MAIN-20180720135213-20180720155213-00473.warc.gz | 330,474,448 | 4,497 | LARGE ORDERS OF THE 1/n-EXPANSION IN ATOMIC PHYSICS
POPOV V.S., SERGEEV A.V.
Institute of Theoretical and Experimental Physics, Bol'shaya Cheriomushkinskaya 25, Moscow,
117259 Russian Federation,
S.I.Vavilov State Optical Institute,
Tuchkov per. 1, St.Petersburg, 199034 Russian Federation, e-mail sergeev@soi.spb.su
Introduction to the 1/n-expansion
An asymptotic expansion for the energy of atoms is developed in powers of 1/n, where n is the principal quantum number. We are considering the Rydberg states with large orbital momentum l corresponding at the limit . . to the circular orbits of an electron. So we assume that n=l+nr+1 while nr<<n. In an essence, such an approach is equivalent to recently developed well-known dimensional expansion, see for example: Goodson D.Z., Lopez-Cabrera M. et al, J.Chem.Phys., 97, 8481, 1992. This method is well-suited to the atoms in strong fields and nonseparable problems such as two-electron atoms.
The large n limit reduces to an exactly solvable classical electrostatic problem with an effective potential containing an additional centrifugal term. In this simplifying limit, the problem reduces to minization of the effective potential. For example, for a hydrogen atom the effective potential,
Veff(r) = -1/r + 1/2r2,
(we use atomic units, = 1) has a minimum at r=r0=1, =Veff(r0)=1/2:
At finite n an electron undergoes small oscillations about fixed position in an effective potential. The scaled energy = n2Enl is expanded in 1/n:
The 1/n-expansion is similar to the methods of molecular vibration analysis. It reduces to the Rayleigh - Schrodinger perturbation theory for anharmonic oscillator. The expansion coefficients can be calculated exactly and to high order, using recursive relations.
The divergence of large orders of 1/n- expansion
The results obtained by simply summing sequential terms in the expansion are quite poor. The reason is the divergence of the expansion, which renders convential summation methods ineffective beyond the lowest orders. As an illustrative example, we show the behaviour of ...... for the electronic energy of H2+ ion with fixed protons. R is the distance between two protons, and r=n-2R is a scaled distance.
The radius of convergence of the 1/n-expansion is zero, because tends to infinity.
It was confirmed by direct numerical calculation of expansion coefficients that the coefficients grow as factorials,
or
with
In this case, the energy can be accurately evaluated by means of Pade - Borel summation. The method consists of using Pade approximants for the Borel function F,
and than integrating it with the exponent,
The factorial increase of the coefficients leads to the singularity in the Borel function at z=a-1. So, by taking into account the divergent large-order behaviour of the expansion one can localize the nearest singularity to the origin in the Borel function and take full advantage of the Pade - Borel summation technique.
In an earlier paper (Popov V.S., Sergeev A.V., Phys.Lett.A, 172, 193, 1993) we examine parameter of the asymptotics a for the simpliest case of spherically-symmetric or separable problems, such as screened Coulomb potentials and Stark effect in a hydrogen atom.
We assume the effective potential to be of the form shown in the following figure.
All states are quasistationary because of the tunneling through the potential barrier between two turning points, r0 and r1. It leads to imaginary part of the energy, Im =-â/2, where â is the width of the level (from quasiclassical formula for decay rate),
is the classical action and is a constant. Supposing analiticity in the variable =1/n and using the dispersion relations in (which connect with the integral from the imaginary part of the energy), we obtain the parameters of the asymptotics: . and a=(2S)-1. For bound states, there are no more real turning points r1, so the large-order asymptotics contains only complex-conjugate terms.
Treating nonspherically-symmetric and two-electron atoms
Our aim is to extend the results for the behaviour of large orders of on multidimensional effective potentials.
As an example which has all essential features of general problem, we investigate in detail a hydrogen atom in parallel electric ( ) and magnetic ( ) fields. In this case, we deal with a two-dimensional quantum decay problem in an effective potential
where , z are cylindrical coordinates, F=n4 , B=n3 are scaled strengths of electric and magnetic fields, correspondingly. The 1/n-expansion is constructed around the classical circular orbit with the radius r0 which is the root of algebraic equation r(1-F2r4)2(1+B2r3/4)=1. The effective potential has a minimum only for small values of electric field, F<F*(B), where F*(B) is a "classical ionization threshold" at which a local minimum of Veff vanishes. Note, that if F>F*(B), the radius r0 and the coefficients become complex. This solution has no means in classical mechanics, but in quantum mechanics it describes both the position and the width of quasistationary state. As an illustration, we present here the effective potential for B=0.5 and F=0.2<F*(B) by means of isolevels.
The central problem (in order to calculate parameter a of the asymptotics) is the determination of the most probable escape path wich minimizes the classical action. We used two approaches. The first one is based on the method of characteristics (see: A.Schmid, Ann.Phys.(N.Y.) 170, 333, 1986). The classical trajectories in an inverted effective potential are calculated, a trajectory is chosen which terminates at a stopping point and which represents the most probable escape path. The parameter a equals to the reciprocal of the action along this trajectory. In the alternative approach, the action is expanded as a perturbation series around the minimum of the effective potential.
Our results of calculating the parameter a for magnetic fields B=0, 0.5 and 1.0 are presented in the figure (F*(B)=0.2081, 0.2532 and 0.3449, correspondingly). We found, that a while F F* by the typical law a (F-F*)-5/4.
For pure magnetic field (Zeeman effect) the parameter a becomes complex. The dependence of a on B is represented on the figure.
Also, we examine the parameter a for H2+ molecular ion and helium isoelectronic sequence. For example, for H2+ ion we found:
a=(z-Arth z)-1, (*)
where , t=r2(1-t)4, r=n- 2R<1.299, 0<t<1/3.
(3/2)r a-1
0.8 -0.47480062*
-0.474795L
1.0 -0.31384119*
-0.31384121L
1.2 -0.19751662152*
-0.19751661876L
* - Exact, formula (*)
L - results of M.Lopez-Cabrera et al, Phys.Rev.Lett., 68, 1992 (1992).
For two-electron atoms, we found the second Borel singularity = 1/a1 (the nearest to the origin Borel singularity =1/a remains to be calculated by our method):
Z
2 0.8849 **
0.3 3.7G
3 0.6327 **
0.3 3.5G
10 0.2343 **
-0.05 3.3G
** Exact results, obtained by method of characteristics
G D.Z.Goodson et al, J.Chem.Phys. 1992, 97, 8481
Back to Presentations on conferences.
Designed by A. Sergeev. | 1,767 | 7,007 | {"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-2018-30 | latest | en | 0.870392 |
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A256329 Number of partitions of 7n into exactly 4 parts. 3
0, 3, 23, 72, 169, 321, 551, 864, 1285, 1815, 2484, 3289, 4263, 5400, 6736, 8262, 10018, 11990, 14222, 16698, 19464, 22500, 25857, 29511, 33516, 37845, 42555, 47616, 53089, 58939, 65231, 71928, 79097, 86697, 94800, 103361, 112455, 122034, 132176, 142830 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS Colin Barker, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,-2,0,0,1,1,-1). FORMULA G.f.: x*(x^8+14*x^7+38*x^6+67*x^5+80*x^4+74*x^3+46*x^2+20*x+3) / ((x-1)^4*(x+1)^2*(x^2+1)*(x^2+x+1)). EXAMPLE For n=1 the 3 partitions of 7*1 = 7 are [1,1,1,4], [1,1,2,3] and [1,2,2,2]. PROG (PARI) concat(0, vector(40, n, k=0; forpart(p=7*n, k++, , [4, 4]); k)) (PARI) concat(0, Vec(x*(x^8+14*x^7+38*x^6+67*x^5+80*x^4+74*x^3+46*x^2+20*x+3) / ((x-1)^4*(x+1)^2*(x^2+1)*(x^2+x+1)) + O(x^100))) CROSSREFS Cf. A256327 (5n), A256328 (6n). Sequence in context: A163210 A163211 A126335 * A196649 A027701 A201482 Adjacent sequences: A256326 A256327 A256328 * A256330 A256331 A256332 KEYWORD nonn,easy AUTHOR Colin Barker, Mar 25 2015 STATUS approved
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Last modified January 17 10:30 EST 2019. Contains 319218 sequences. (Running on oeis4.) | 716 | 1,802 | {"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} | 3.578125 | 4 | CC-MAIN-2019-04 | latest | en | 0.592384 |
http://wiki.signature.net/index.php/IB_Statements/num | 1,701,894,737,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100603.33/warc/CC-MAIN-20231206194439-20231206224439-00869.warc.gz | 51,659,742 | 4,888 | # IB Statements/num
NUM function
Syntax: NUM(string-argument, numeric-error-argument)
Discussion: The NUM function converts a string-argument into its equivalent numeric value, only if the contents of the string-argument are numeric characters.
The string-argument may be a string constant, a single-element string variable, a string array element, a string expression, or a string function. The string-argument can contain only numeric digits (0 through 9), a decimal point, a leading or trailing sign (+ or -), and blank spaces.
Notes:
If both leading a trailing signs are present in the string-argument, the trailing sign determines the sign of the result. The sign cannot be imbedded within the string-argument. If the sign is omitted, the value is considered to be positive.
Only one decimal point is allowed in the string-argument. If no decimal point is present, it is assumed to follow the right-most digit.
A blank space cannot appear within the string-argument value, but may separate the value from the sign(s).
Dollar signs (\$) are not permitted in the string-argument. The numeric-error-argument is a numeric variable or numeric array element with a length and precision of at least 1.0 If the NUM function successfully converts the string-argument, the numeric-error-argument is set to 0. If the conversion is unsuccessful, it is set to 1.
Note: If the conversion is unsuccessful, the NUM function itself will return a value of zero.
The intermediate result of the NUM function is stored in an accumulator with a fixed precision of 16.8. Digits that are more than eight places on either side of the decimal point are truncated. If the result is stored in a receiving numeric variable, the precision is adjusted to the defined precision of that variable.
Example: VALUE = NUM(AMOUNT\$,ERR)
In the above example, the value contained in the string variable AMOUNT\$ is converted to numeric form. This result is then stored in the numeric variable VALUE. The numeric-error-argument ERR will be 0 if the conversion is successful or 1 if it is not successful.
For example, if AMOUNT\$ equals "5542.234-" (a string), then VALUE will equal 5542.234- (a number), and ERR will equal 0 (indicating a successful conversion).
Or, if AMOUNT\$ equals "555-1212" (a string), then VALUE will equal 0 and ERR will equal 1 (indicating an unsuccessful conversion due to the imbedded dash in the string-argument). | 535 | 2,422 | {"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} | 2.96875 | 3 | CC-MAIN-2023-50 | latest | en | 0.650395 |
https://www.scribd.com/document/60452808/15013335-soalan-Matematik-Pksr-1-Thn-3-Paper-2 | 1,571,367,195,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677412.35/warc/CC-MAIN-20191018005539-20191018033039-00246.warc.gz | 1,060,861,337 | 49,362 | You are on page 1of 4
# Instruction: Answer all the questions in standard written method.
2 Page 1 of 4
## Find the sum of 3376, 588 and 96
2 8 10 000 - 2 500 =
## 2 10 What number must be subtracted from 8400 to get 6368?
2 13 9 936 - 49 - 1 456 =
2 Page 2 of 4
11
## 2 14 ________ x 4 = 700 Fill in the blank.
2 15 Divide 54 by 6.
## 2 16 ______ 7 = 9 What is the missing number in the box?
3 Page 3 of 4
17
There are 1665 beads in box P. Box Q has 480 beads more than box P. How many beads are there in the two boxes altogether?
3 18 Mr Shahrul baked 1 200 butter buns and cheese buns. He sold both buns and had 156 buns left. How many buns did he sell?
3 19 Adam spends 15 minutes to answer a Mathematical question. How many minutes does he need to answer 8 questions?
3 20 Sofea has 240 doughnuts. She packs 4 doughnuts in each box. How many boxes does she need?
3
Prepared By; Pn Norizan Bt Awang@Rashid May 2009/PKSR 1
Page 4 of 4 | 326 | 956 | {"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} | 3.65625 | 4 | CC-MAIN-2019-43 | latest | en | 0.937333 |
https://dev.danilafe.com/Web-Projects/blog-static/commit/57aecc46be61e9791f33ce9f7ff6740e8043ae54 | 1,679,493,110,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943809.76/warc/CC-MAIN-20230322114226-20230322144226-00632.warc.gz | 238,974,515 | 14,205 | ### Update and publish catamorphism article
master
parent ded50480a8
commit 57aecc46be
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code/catamorphisms/Cata.hs
2. 39
`@ -99,3 +99,14 @@ invert :: BinaryTree a -> BinaryTree a` `invert = cata \$ \case` ` LeafF -> Leaf` ` NodeF a l r -> Node a r l` ``` ``` `data MaybeF a b = NothingF | JustF a deriving Functor` ``` ``` `instance Cata (Maybe a) (MaybeF a) where` ` out Nothing = NothingF` ` out (Just x) = JustF x` ``` ``` `getOrDefault :: a -> Maybe a -> a` `getOrDefault d = cata \$ \case` ` NothingF -> d` ` JustF a -> a`
`@ -1,10 +1,17 @@` `---` `title: "Generalizing Folds in Haskell"` `date: 2022-04-22T12:19:22-07:00` `draft: true` `tags: ["Haskell"]` `---` ``` ``` `Have you encountered Haskell's `foldr` function? Did you know that you can use it to express any ` `function on a list? What's more, there's a way to derive similar functions for` `{{< sidenote "right" "positive-note" "a large class of data types in Haskell." >}}` `Specifically, this is the class of inductive types.` `{{< /sidenote >}}` `This is precisely the focus of this post.` `Before we get into the details, it's good to review the underlying concepts in a more familiar setting: functions.` ``` ``` `### Recursive Functions` `Let's start off with a little bit of a warmup, and take a look at a simple recursive function:` ``length`. Here's a straightforward definition:` `@ -338,6 +345,36 @@ Given this, here's an implementation of that `invert` function we mentioned earl` ``` ``` `{{< codelines "Haskell" "catamorphisms/Cata.hs" 98 101 >}}` ``` ``` `#### Degenerate Cases` ``` ``` `Actually, the data types we consider don't have to be recursive. We can apply the same` `procedure of replacing recursive occurrences in a data type's definition with a new type parameter` `to `Maybe`; the only difference is that now the new parameter will not be used!` ``` ``` `{{< codelines "Haskell" "catamorphisms/Cata.hs" 103 107 >}}` ``` ``` `And then we can define a function on `Maybe` using `cata`:` ``` ``` `{{< codelines "Haskell" "catamorphisms/Cata.hs" 109 112 >}}` ``` ``` `This isn't _really_ useful, since we're still pattern matching on a type that looks` `identical to `Maybe` itself. There is one reason that I bring it up, though. Remember` `how `foldr` was equivalent to `cata` for `MyList`, because defining a function `MyListF a -> a` was` `the same as providing a base case `a` and a "combining function" `Int -> a -> a`? Well,` `defining a function `MaybeF x a -> a` is the same as providing a base case `a` (for `NothingF`)` `and a handler for the contained value, `x -> a`. So we might imagine the `foldr` function for `Maybe`` `to have type:` ``` ``` ````Haskell` `maybeFold :: a -> (x -> a) -> Maybe x -> a` ````` ``` ``` `This is exactly the function [`maybe` from `Data.Maybe`](https://hackage.haskell.org/package/base-4.16.1.0/docs/Data-Maybe.html#v:maybe)! Hopefully you can follow a similar process in your head to arrive at "fold"` `functions for `Either` and `Bool`. Indeed, there are functions that correspond to these data types` `in the Haskell standard library, named [`either`](https://hackage.haskell.org/package/base-4.16.1.0/docs/Data-Either.html#v:either) and [`bool`](https://hackage.haskell.org/package/base-4.16.1.0/docs/Data-Bool.html#v:bool).` `Much like `fold` can be used to represent any function on lists, `maybe`, `either`, and `bool` can be` `used to represent any function on their corresponding data types. I think that's neat.` ``` ``` `#### What About `Foldable`?` `If you've been around the Haskell ecosystem, you may know the `Foldable` type class.` `Isn't this exactly what we've been working towards here? No, not at all. Take` | 1,074 | 3,720 | {"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-2023-14 | latest | en | 0.736961 |
https://www.wiredfaculty.com/question/UTBKVFJVVk9UVUV4TVRBeE16QTROQT09 | 1,718,359,504,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00858.warc.gz | 1,014,602,298 | 11,889 | Permutations And Combinations
Question
# Find the sum to n terms of the sequence given by
Solution
The given sequence is defined by
Let denote the sum of n terms
∴
=
Since the series in the first bracket is a G.P. series with 1st term 2, common ratio 2 and the series in the second bracket is an A.P. with 1st term 3, common difference 3.
∴ | 97 | 352 | {"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-2024-26 | latest | en | 0.957788 |
https://rdrr.io/cran/soilphysics/man/llwr.html | 1,611,266,208,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527850.55/warc/CC-MAIN-20210121194330-20210121224330-00620.warc.gz | 556,820,248 | 11,520 | # llwr: Least Limiting Water Range (LLWR) In soilphysics: Soil Physical Analysis
## Description
Graphical solution for the Least Limiting Water Range and parameter estimation of the related water retention and penetration resistance curves. A summary containing standard errors and statistical significance of the parameters is also given.
## Usage
``` 1 2 3 4 5 6 7 8 9 10``` ```llwr(theta, h, Bd, Pr, particle.density, air, critical.PR, h.FC, h.WP, water.model = c("Silva", "Ross"), Pr.model = c("Busscher", "noBd"), pars.water = NULL, pars.Pr = NULL, graph = TRUE, graph2 = TRUE, xlab = expression(Bulk~Density~(Mg~m^{-3})), ylab = expression(theta~(m^{3}~m^{-3})), main = "Least Limiting Water Range", ...) ```
## Arguments
`theta` a numeric vector containing values of volumetric water content (m^3~m^{-3}). `h` a numeric vector containing values of matric head (cm, Psi, MPa, kPa, ...). `Bd` a numeric vector containing values of dry bulk density (Mg~m^{-3}). Note that `Bd` can also be a vector of length 1. See details. `Pr` a numeric vector containing values of penetration resistance (MPa, kPa, ...). `particle.density` the value of the soil particle density (Mg~m^{-3}). `air` the value of the limiting volumetric air content (m^3~m^{-3}). `critical.PR` the value of the critical soil penetration resistance. `h.FC` the value of matric head at the field capacity (cm, MPa, kPa, hPa, ...). `h.WP` the value of matric head at the wilting point (cm, MPa, kPa, hPa, ...). `water.model` a character; the model to be used for calculating the soil water content. It must be one of the two: `"Silva"` (default) or `"Ross"`. See details. `Pr.model` a character; the model to be used to predict soil penetration resistance. It must be one of the two: `"Busscher"` (default) or `"noBd"`. See details. `pars.water` optional; a numeric vector containing the estimates of the three parameters of the soil water retention model employed. If `NULL` (default), `llwr()` estimates them using a Newton-Raphson algorithm. See details. `pars.Pr` optional; a numeric vector containing estimates of the three parameters of the model proposed by Busscher (1990) for the functional relationship among `Pr`, `theta` and `Bd`. If NULL (default), `llwr()` estimates them using a Newton-Raphson algorithm. Moreover, if `Pr.model = "noBd"`, then the third value is considered to be null. `graph` logical; if TRUE (default) a graphical solution for the Least Limiting Water Range is plotted. `graph2` logical; if TRUE (default) a line of the Least Limiting Water Range as a function of bulk density is plotted. If `graph = FALSE`, then `llwr()` will automatically consider `graph2 = FALSE` too. `xlab` a title for the x axis; the default is Bulk~Density~(Mg~m^{-3}). `ylab` a title for the y axis; the default is θ~(m^{3}~m^{-3}). `main` a main title for the graphic; the default is "Least Limiting Water Range" `...` further graphical arguments.
## Details
The numeric vectors `theta`, `h`, `Bd` and `Pr` are supposed to have the same length, and their values should have appropriate unit of measurement. For fitting purposes, it is not advisable to use vectors with less than five values. It is possible to calculate the LLWR for a especific (unique) value of bulk density. In This case, `Bd` should be a vector of length 1 and, therfore, it is not possible to fit the models `"Silva"` and `"Busscher"`, for water content and penetration resistance, respectively.
The model employed by Silva et al. (1994) for the soil water content (θ) as a function of the soil bulk density (ρ) and the matric head (h) is:
θ = exp(a + b ρ)h^c
The model proposed by Ross et al. (1991) for the soil water content (θ) as a function of the matric head (h) is:
θ = a h^c
The penetration resistance model, as presented by Busscher (1990), is given by
Pr = b0 * (θ^{b1}) * (ρ^{b2})
If the agrument `Bd` receives a single value of bulk density, then `llwr()` fits the following simplified model (option `noBd`):
Pr = b0 * θ^{b1}
## Value
A list of
`limiting.theta ` a n x 4 matrix containing the limiting values of water content for each input value of bulk density at the volumetric air content (`thetaA`), penetration resistance (`thetaPR`), field capacity (`thetaFC`) and wilting point (`thetaWP`). `pars.water ` a "nls" object or a numeric vector containing estimates of the three parameters of the model employed by Silva et al. (1994) for the functional relationship among `theta`, `Bd` and `h`. `r.squared.water ` a "Rsq" object containing the pseudo and the adjusted R-squared for the water model. `pars.Pr ` a "nls" object or a numeric vector containing estimates of the three parameters of the penetration resistance model. `r.squared.Pr ` a "Rsq" object containing the pseudo and the adjusted R-squared for the penetration resistance model. `area ` numeric; the value of the shaded (LLWR) area. Calculated only when Bd is a vector of length > 1. `LLWR ` numeric; the value of LLWR (m^3~m^{-3}) corresponding to `Bd`. Calculated only when Bd is a single value.
## Author(s)
Anderson Rodrigo da Silva <anderson.agro@hotmail.com>
## References
Busscher, W. J. (1990). Adjustment of flat-tipped penetrometer resistance data to common water content. Transactions of the ASAE, 3:519-524.
Leao et al. (2005). An Algorithm for Calculating the Least Limiting Water Range of Soils. Agronomy Journal, 97:1210-1215.
Leao et al. (2006). Least limiting water range: A potential indicator of changes in near-surface soil physical quality after the conversion of Brazilian Savanna into pasture. Soil & Tillage Research, 88:279-285.
Ross et al. (1991). Equation for extending water-retention curves to dryness. Soil Science Society of America Journal, 55:923-927.
Silva et al. (1994). Characterization of the least limiting water range of soils. Soil Science Society of America Journal, 58:1775-1781.
`fitbusscher`
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26``` ```# Example 1 - part of the data set used by Leao et al. (2005) data(skp1994) ex1 <- with(skp1994, llwr(theta = W, h = h, Bd = BD, Pr = PR, particle.density = 2.65, air = 0.1, critical.PR = 2, h.FC = 100, h.WP = 15000)) ex1 # Example 2 - specifying the parameters (Leao et al., 2005) a <- c(-0.9175, -0.3027, -0.0835) # Silva et al. model of water content b <- c(0.0827, -1.6087, 3.0570) # Busscher's model ex2 <- with(skp1994, llwr(theta = W, h = h, Bd = BD, Pr = PR, particle.density = 2.65, air = 0.1, critical.PR = 2, h.FC = 0.1, h.WP = 1.5, pars.water = a, pars.Pr = b)) ex2 # Example 3 - specifying a single value for Bd ex3 <- with(skp1994, llwr(theta = W, h = h, Bd = 1.45, Pr = PR, particle.density = 2.65, air = 0.1, critical.PR = 2, h.FC = 100, h.WP = 15000)) ex3 # End (not run) ``` | 1,959 | 6,771 | {"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-2021-04 | latest | en | 0.576178 |
https://mail.python.org/pipermail/python-list/2006-May/368129.html | 1,566,532,146,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317817.76/warc/CC-MAIN-20190823020039-20190823042039-00477.warc.gz | 546,532,105 | 2,048 | # comparing values in two sets
Tim Chase python.list at tim.thechases.com
Mon May 15 02:19:33 CEST 2006
```> I'd like to compare the values in two different sets to
> test if any of the positions in either set share the same
> value (e.g., if the third element of each set is an 'a',
> then the test fails).
There's an inherant problem with this...sets by definition
are unordered, much like dictionaries. To compare them my
such means, you'd have to convert them to lists, sort the
lists by some ordering, and then compare the results.
Something like
s1 = set([1,3,5,7,9])
s2 = set([1,2,3])
list1 = list(s1)
list2 = list(s2)
list1.sort()
list2.sort()
if [(x,y) for x,y in zip(list1,list2) if x == y]:
print "There's an overlap"
else:
print "No matching elements"
Just to evidence matters, on my version of python (2.3.5 on
Debian), the following came back:
>>> set([1,3,5,7,9])
set([1,3,9,5,7])
That's not the original order, but the definition of a set
isn't hurt/changed by any ordering.
Thus, asking for the "position in a set" is an undefined
operation.
-tkc
PS: for the above was done in 2.3.5 using this line:
from sets import Set as set
``` | 348 | 1,162 | {"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-2019-35 | latest | en | 0.90244 |
https://www.coursehero.com/file/6235205/Lesson-51a-Oscillations/ | 1,524,814,033,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125949489.63/warc/CC-MAIN-20180427060505-20180427080505-00435.warc.gz | 742,605,281 | 221,040 | {[ promptMessage ]}
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Lesson 5.1a Oscillations
# Lesson 5.1a Oscillations - Oscillations If a loaded spring...
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Oscillations If a loaded spring is pulled down and released, the system moves up and own about its equilibrium position. This up and down motion of the load is called an oscillation . The rest position of the load is called its equilibrium position When the load is pulled down to a lower extreme , the work done in stretching the spring becomes potential energy of the spring which pulls the load back to the equilibrium position. 1
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The kinetic energy at the equilibrium position does work compressing the spring so that the load moves to the upper extreme , compressing the spring. equilibrium Lower extreme The compressed spring has potential energy and it does work pushing the load back to the equilibrium position giving it kinetic energy. The spring has potential energy when the load is at either extreme. The load is at rest at the extremes and its kinetic energy is zero. 2
At the equilibrium position, the potential energy of the spring is zero. The load has maximum kinetic energy at this position. The amount of stretch of the spring from its equilibrium position is called the displacement (y) of the load. The maximum displacement of the load from the equilibrium position is called the amplitude (A) of the oscillation. Equilibrium Upper Extreme Lower Extreme A U = 0, K max K = 0, U max K = 0, U max 3
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As the load oscillates between the two extremes, its displacement y changes reaching a maximum at the upper extreme ( y = A ) and a minimum at the lower extreme ( y = -A ). We will now obtain an equation for the displacement y as a function of time. Equilibrium Upper Extreme Lower Extreme http://www.upscale.utoronto.ca/PVB/Harrison/Flash/ClassMec A -A 4
A θ B C BC is a diameter of the circle. O A perpendicular drawn from Q meets BC at O . O is called the projection of Q onto BC If the center of the circle is the origin, when the object is at P , the projection O is at the origin. When the object is at Q , the projection O moves up. y , the distance it moves up is called the displacement of the projection. y P Q Consider an object moving around a circle of radius A . At t = 0, the object is at P. A little while later, the object is at the point Q so that the radius rotates through an angle θ . 5
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A θ B C O y P Q When the object is at B, the projection O also is at B.
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{[ snackBarMessage ]} | 661 | 2,920 | {"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-2018-17 | latest | en | 0.875734 |
http://forums.wolfram.com/mathgroup/archive/2009/Feb/msg00470.html | 1,579,376,703,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593937.27/warc/CC-MAIN-20200118193018-20200118221018-00393.warc.gz | 64,289,439 | 7,970 | Re: Scalar plot in 3D
• To: mathgroup at smc.vnet.net
• Subject: [mg96409] Re: Scalar plot in 3D
• From: dh <dh at metrohm.com>
• Date: Fri, 13 Feb 2009 03:41:22 -0500 (EST)
• References: <gn11bt\$7tg\$1@smc.vnet.net>
```
Hi Franco,
here is an example where the z-value is color coded:
ParametricPlot3D[{{4 + (3 + Cos[v]) Sin[u], 4 + (3 + Cos[v]) Cos[u],
4 + Sin[v]}}, {u, 0, 2 Pi}, {v, 0, 2 Pi},
ColorFunction -> Function[{x, y, z}, Hue[z]]]
her another one where the distance from the origin defines the color:
ParametricPlot3D[{{4 + (3 + Cos[v]) Sin[u], 4 + (3 + Cos[v]) Cos[u],
4 + Sin[v]}}, {u, 0, 2 Pi}, {v, 0, 2 Pi},
ColorFunction -> Function[{x, y, z}, Hue[Norm[{x, y, z}]]]]
hope this helps, Daniel
Franco Milicchio wrote:
> Hi everyone! I am still struggling with my "rendering hassles" :)
> I don't really know how to implement this in Mathematica, although the idea
> is quite simple. I have a list of points in 3D, their relative scalar value,
> and I'd like to have them nicely rendered on the domain on which they are
> defined. In other words, I have sampled a scalar field on a torus, and I'd
> like to plot its value (by hue or any other color means). Simple as it is, I
> cannot figure how to obtain this!
>
> Basically, I know the torus' radii (rMIN and rMAX), and I have my good
> vectors:
>
> coords = {{x1, y1, z1}, {x2, y2, z2}, ...}
> values = {v1, v2, ...}
>
> Is there anyone who can help me with this? I have seen ListVectorPlot3D, but
> it won't show in the plot the domain itself (so arrows laying on a torus),
> and ineed, it requires a vector field. Although I can fake a vector field,
> I'd like something more "visually appealing"! :)
>
> Thanks to anyone!
> Franco
```
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• Next by thread: Re: Re: Scalar plot in 3D | 632 | 1,981 | {"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-2020-05 | longest | en | 0.877463 |
https://www.medicalalgorithms.com/the-prst-score-of-evans-and-davies-for-depth-of-anesthesia | 1,716,834,732,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.25/warc/CC-MAIN-20240527175049-20240527205049-00319.warc.gz | 781,205,477 | 8,238 | ### Description
The PRST (pressure, rate, sweating, tears) score can be used to evaluate the depth of anesthesia. It uses physiologic responses to determine if sufficient anesthetic has been administered. The authors are from Oxford, England.
Limits of some classic measures for the depth of anesthesia:
(1) pupillary size may be affected by morphine, atropine or other medications
(2) muscle tone may be affected by use of muscle relaxants
Parameters:
(1) systolic arterial blood pressure
(2) heart rate
(3) sweating
(4) tears
Parameter
Finding
Points
systolic arterial blood pressure
less than control + 15 mm Hg
0
>= (control + 15) AND <= (control + 30 mm Hg)
1
greater than control + 30 mm Hg
2
heart rate
less than control + 15 beats per minute
0
>= (control + 15) AND <= (control + 30 beats per minute)
1
greater than control + 30 beats per minute
2
sweating
nil
0
skin moist to touch
1
2
tears
no excess of tears in open eye
0
excess of tears in open eye
1
tear overflow from closed eyes
2
where:
• The control value is presumed to be the baseline levels prior to anesthesia.
• In Table III of Evans et al (1987) the values for systolic blood pressure and heart rate do not cover values equal to 15 or 30. I adjusted the table above to cover these values.
PRST score =
= SUM(points for all 4 parameters)
Interpretation:
• minimum score: 0
• maximum score: 8
• A score > 3 indicates an insufficient depth of anesthesia.
• During anesthesia the score is maintained in the range of 0 to 3. | 399 | 1,542 | {"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-2024-22 | latest | en | 0.75619 |
https://docs.gpytorch.ai/en/v1.9.0/_modules/gpytorch/variational/batch_decoupled_variational_strategy.html | 1,675,610,024,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00094.warc.gz | 236,035,455 | 15,881 | # Source code for gpytorch.variational.batch_decoupled_variational_strategy
#!/usr/bin/env python3
import torch
from linear_operator.operators import MatmulLinearOperator, SumLinearOperator
from torch.distributions.kl import kl_divergence
from ..distributions import Delta, MultivariateNormal
from ..utils.errors import CachingError
from ..utils.memoize import pop_from_cache_ignore_args
from .delta_variational_distribution import DeltaVariationalDistribution
from .variational_strategy import VariationalStrategy
[docs]class BatchDecoupledVariationalStrategy(VariationalStrategy):
r"""
A VariationalStrategy that uses a different set of inducing points for the
variational mean and variational covar. It follows the "decoupled" model
proposed by Jankowiak et al. (2020)_ (which is roughly based on the strategies
proposed by Cheng et al. (2017)_.
Let :math:\mathbf Z_\mu and :math:\mathbf Z_\sigma be the mean/variance
inducing points. The variational distribution for an input :math:\mathbf
x is given by:
.. math::
\begin{align*}
\mathbb E[ f(\mathbf x) ] &= \mathbf k_{\mathbf Z_\mu \mathbf x}^\top
\mathbf K_{\mathbf Z_\mu \mathbf Z_\mu}^{-1} \mathbf m
\\
\text{Var}[ f(\mathbf x) ] &= k_{\mathbf x \mathbf x} - \mathbf k_{\mathbf Z_\sigma \mathbf x}^\top
\mathbf K_{\mathbf Z_\sigma \mathbf Z_\sigma}^{-1}
\left( \mathbf K_{\mathbf Z_\sigma} - \mathbf S \right)
\mathbf K_{\mathbf Z_\sigma \mathbf Z_\sigma}^{-1}
\mathbf k_{\mathbf Z_\sigma \mathbf x}
\end{align*}
where :math:\mathbf m and :math:\mathbf S are the variational parameters.
Unlike the original proposed implementation, :math:\mathbf Z_\mu and :math:\mathbf Z_\sigma
have **the same number of inducing points**, which allows us to perform batched operations.
Additionally, you can use a different set of kernel hyperparameters for the mean and the variance function.
We recommend using this feature only with the :obj:~gpytorch.mlls.PredictiveLogLikelihood objective function
as proposed in "Parametric Gaussian Process Regressors" (Jankowiak et al. (2020)_).
Use the mean_var_batch_dim to indicate which batch dimension corresponds to the different mean/var
kernels.
.. note::
We recommend using the "right-most" batch dimension (i.e. mean_var_batch_dim=-1) for the dimension
that corresponds to the different mean/variance kernel parameters.
Assuming you want b1 many independent GPs, the :obj:~gpytorch.variational._VariationalDistribution
objects should have a batch shape of b1, and the mean/covar modules
of the GP should have a batch shape of b1 x 2.
(The 2 corresponds to the mean/variance hyperparameters.)
.. seealso::
:obj:~gpytorch.variational.OrthogonallyDecoupledVariationalStrategy (a variant proposed by
Salimbeni et al. (2018)_ that uses orthogonal projections.)
:param ~gpytorch.models.ApproximateGP model: Model this strategy is applied to.
Typically passed in when the VariationalStrategy is created in the
__init__ method of the user defined model.
:param torch.Tensor inducing_points: Tensor containing a set of inducing
points to use for variational inference.
:param ~gpytorch.variational.VariationalDistribution variational_distribution: A
VariationalDistribution object that represents the form of the variational distribution :math:q(\mathbf u)
:param learn_inducing_locations: (Default True): Whether or not
the inducing point locations :math:\mathbf Z should be learned (i.e. are they
parameters of the model).
:type learn_inducing_locations: bool, optional
:type mean_var_batch_dim: int, optional
:param mean_var_batch_dim: (Default None):
Set this parameter (ideally to -1) to indicate which dimension corresponds to different
kernel hyperparameters for the mean/variance functions.
.. _Cheng et al. (2017):
https://arxiv.org/abs/1711.10127
.. _Salimbeni et al. (2018):
https://arxiv.org/abs/1809.08820
.. _Jankowiak et al. (2020):
https://arxiv.org/abs/1910.07123
Example (**different** hypers for mean/variance):
>>> class MeanFieldDecoupledModel(gpytorch.models.ApproximateGP):
>>> '''
>>> A batch of 3 independent MeanFieldDecoupled PPGPR models.
>>> '''
>>> def __init__(self, inducing_points):
>>> # The variational parameters have a batch_shape of [3]
>>> variational_distribution = gpytorch.variational.MeanFieldVariationalDistribution(
>>> inducing_points.size(-1), batch_shape=torch.Size([3]),
>>> )
>>> variational_strategy = gpytorch.variational.BatchDecoupledVariationalStrategy(
>>> self, inducing_points, variational_distribution, learn_inducing_locations=True,
>>> mean_var_batch_dim=-1
>>> )
>>>
>>> # The mean/covar modules have a batch_shape of [3, 2]
>>> # where the last batch dim corresponds to the mean & variance hyperparameters
>>> super().__init__(variational_strategy)
>>> self.mean_module = gpytorch.means.ConstantMean(batch_shape=torch.Size([3, 2]))
>>> self.covar_module = gpytorch.kernels.ScaleKernel(
>>> gpytorch.kernels.RBFKernel(batch_shape=torch.Size([3, 2])),
>>> batch_shape=torch.Size([3, 2]),
>>> )
Example (**shared** hypers for mean/variance):
>>> class MeanFieldDecoupledModel(gpytorch.models.ApproximateGP):
>>> '''
>>> A batch of 3 independent MeanFieldDecoupled PPGPR models.
>>> '''
>>> def __init__(self, inducing_points):
>>> # The variational parameters have a batch_shape of [3]
>>> variational_distribution = gpytorch.variational.MeanFieldVariationalDistribution(
>>> inducing_points.size(-1), batch_shape=torch.Size([3]),
>>> )
>>> variational_strategy = gpytorch.variational.BatchDecoupledVariationalStrategy(
>>> self, inducing_points, variational_distribution, learn_inducing_locations=True,
>>> )
>>>
>>> # The mean/covar modules have a batch_shape of [3, 1]
>>> # where the singleton dimension corresponds to the shared mean/variance hyperparameters
>>> super().__init__(variational_strategy)
>>> self.mean_module = gpytorch.means.ConstantMean(batch_shape=torch.Size([3, 1]))
>>> self.covar_module = gpytorch.kernels.ScaleKernel(
>>> gpytorch.kernels.RBFKernel(batch_shape=torch.Size([3, 1])),
>>> batch_shape=torch.Size([3, 1]),
>>> )
"""
def __init__(
self, model, inducing_points, variational_distribution, learn_inducing_locations=True, mean_var_batch_dim=None
):
if isinstance(variational_distribution, DeltaVariationalDistribution):
raise NotImplementedError(
"BatchDecoupledVariationalStrategy does not work with DeltaVariationalDistribution"
)
if mean_var_batch_dim is not None and mean_var_batch_dim >= 0:
raise ValueError(f"mean_var_batch_dim should be negative indexed, got {mean_var_batch_dim}")
self.mean_var_batch_dim = mean_var_batch_dim
# Maybe unsqueeze inducing points
if inducing_points.dim() == 1:
inducing_points = inducing_points.unsqueeze(-1)
# We're going to create two set of inducing points
# One set for computing the mean, one set for computing the variance
if self.mean_var_batch_dim is not None:
inducing_points = torch.stack([inducing_points, inducing_points], dim=(self.mean_var_batch_dim - 2))
else:
inducing_points = torch.stack([inducing_points, inducing_points], dim=-3)
super().__init__(model, inducing_points, variational_distribution, learn_inducing_locations)
def _expand_inputs(self, x, inducing_points):
# If we haven't explicitly marked a dimension as batch, add the corresponding batch dimension to the input
if self.mean_var_batch_dim is None:
x = x.unsqueeze(-3)
else:
x = x.unsqueeze(self.mean_var_batch_dim - 2)
return super()._expand_inputs(x, inducing_points)
def forward(self, x, inducing_points, inducing_values, variational_inducing_covar=None, **kwargs):
# We'll compute the covariance, and cross-covariance terms for both the
# pred-mean and pred-covar, using their different inducing points (and maybe kernel hypers)
mean_var_batch_dim = self.mean_var_batch_dim or -1
# Compute full prior distribution
full_inputs = torch.cat([inducing_points, x], dim=-2)
full_output = self.model.forward(full_inputs, **kwargs)
full_covar = full_output.lazy_covariance_matrix
# Covariance terms
num_induc = inducing_points.size(-2)
test_mean = full_output.mean[..., num_induc:]
induc_induc_covar = full_covar[..., :num_induc, :num_induc].add_jitter()
induc_data_covar = full_covar[..., :num_induc, num_induc:].to_dense()
data_data_covar = full_covar[..., num_induc:, num_induc:]
# Compute interpolation terms
# K_ZZ^{-1/2} K_ZX
# K_ZZ^{-1/2} \mu_Z
L = self._cholesky_factor(induc_induc_covar)
if L.shape != induc_induc_covar.shape:
# Aggressive caching can cause nasty shape incompatibilies when evaluating with different batch shapes
# TODO: Use a hook to make this cleaner
try:
pop_from_cache_ignore_args(self, "cholesky_factor")
except CachingError:
pass
L = self._cholesky_factor(induc_induc_covar)
interp_term = L.solve(induc_data_covar.double()).to(full_inputs.dtype)
mean_interp_term = interp_term.select(mean_var_batch_dim - 2, 0)
var_interp_term = interp_term.select(mean_var_batch_dim - 2, 1)
# Compute the mean of q(f)
# k_XZ K_ZZ^{-1/2} m + \mu_X
# Here we're using the terms that correspond to the mean's inducing points
torch.matmul(mean_interp_term.transpose(-1, -2), inducing_values.unsqueeze(-1)).squeeze(-1),
test_mean.select(mean_var_batch_dim - 1, 0),
)
# Compute the covariance of q(f)
# K_XX + k_XZ K_ZZ^{-1/2} (S - I) K_ZZ^{-1/2} k_ZX
middle_term = self.prior_distribution.lazy_covariance_matrix.mul(-1)
if variational_inducing_covar is not None:
middle_term = SumLinearOperator(variational_inducing_covar, middle_term)
predictive_covar = SumLinearOperator(
data_data_covar.add_jitter(1e-4).to_dense().select(mean_var_batch_dim - 2, 1),
MatmulLinearOperator(var_interp_term.transpose(-1, -2), middle_term @ var_interp_term),
)
return MultivariateNormal(predictive_mean, predictive_covar)
def kl_divergence(self):
variational_dist = self.variational_distribution
prior_dist = self.prior_distribution
mean_dist = Delta(variational_dist.mean)
covar_dist = MultivariateNormal(
torch.zeros_like(variational_dist.mean), variational_dist.lazy_covariance_matrix
)
return kl_divergence(mean_dist, prior_dist) + kl_divergence(covar_dist, prior_dist) | 2,617 | 10,312 | {"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} | 2.921875 | 3 | CC-MAIN-2023-06 | latest | en | 0.60474 |
http://www.truth-revelations.org/?page_id=1066 | 1,558,842,991,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258621.77/warc/CC-MAIN-20190526025014-20190526051014-00021.warc.gz | 358,148,158 | 6,312 | # 164. Music of the spheres
During the second world war the Nazi’s altered the musical standard to 440 Hertz while it used to be 432 Hertz. I will not go into the reasons for this and leave this to the conspiracy theorists.
Let’s have a look, if A is 432 Hertz then the octaves of it are 27 Hertz, 54, 108, 216, 864, 1728 etc. The D note will then be 576 Hertz with the octaves 9 hertz, 18, 36, 72, 144, 288, 1152. The E note will then be 324 and its octaves , 81 , 162 , 648 , 1296 etc. For those of you who have studied the articles will recognize all the familiar numbers of the star of Bethlehem again and also notice that there are mirror images again. But they will also have noticed that these numbers can be divided by 9 and that 4+3+2=9 also, yes it all is based on the very same structure.
Now let’s look at the numbers of the cycles given in the old scriptures. The Jews have 144.000 chosen ones. The Indian culture have this great cycles: Kali Yuga is 432.000 years, Satyug is 1728.000 year, Treta is 1.296.000 year and Dwapar is 864.000 years. Then the cycles mentioned by the Mayan’s : Tun is 18 uinal is 360 days, Katun is 20 tun is 7,200 days, Baktun is 20 katun is 144.000 days.
Not only are they related to the processional movement, but life itself, it influences the chakras, plants etc.
As you know from the articles this cycle of the zodiac takes 25920 years and is divided by 12 signs of the zodiac each lasting 2160 years. Our basis of geometry and the circle of 360 degrees(25920 / 72 =360) so it may come to no surprise that much of our math, science and spirituality were based on this knowledge, but just look at how far science and spirituality are moved away from each other, just like astronomy and astrology .
A little more, take this cycle of 25920 and divide it by 60 will give you the base note again of 432. While 25920/360 gives you the octave of D again 72.
You might recall the fact that there were 12 disciples and Christ making 13. That there are 13 chakras, 12 zodiacal signs and the snake or stick(recall the story of Moses in Egypt turning the stick or cane into a snake and back again). So let’s take the 13 times 2160 is 28080 (288) and divide it by the base note 432 is 65 (the 6 and 5 star in reverse). The zero between the eights means passing through the eye of the thorus.
Light is 2997… km divided by 45 (the ark of Noah) is 666.
You will have also heard that 144 is light,
144 x 56,25=8100
144 x 31,25=4500
51,84 / 144=36
The difference between 8100 and 4500 is 3600.
25920/1800 is 1440.
12 times 144 is 1728 and number 13 makes it 1872.
18 x 72 is 1296 a turning point (96) 12 x 96 is 1152,
which is twice 576 or the tree of good and evil.
This 96 is located where the 6 and 5 pointed star join.
The 1728 17 x 28 is 476 (100 short of 576).
11 times 476 is 5236. 33 times is 15708 which is 3 times 5236 and as you might recall the opening
from the 5 pointed star into the 6 pointed star is in the 15th division and reached through the 7th. Daath is the mystical 33 of the cabbala, the Commander of the interior of the temple of the free masons. Just look at the old measures 12 ounces troy is 5760 grains, and 12 ounces apothecaries is 96 drams apothecaries is 288 scruples is 5760 grains.
Now you might realize that the term master builder is related to this and that in the old days they wished to build according gods law or measure, even the songs!
Some of you say, I am not into numbers but you see numbers only as quantity, they were in fact the first language from which all other languages came. Even if you only look at them and recognize that there is only one structure upon which they all fit, then you might realize all major religions are built around this holy grail and start to realize that I reveal to you the light and the way.
As for you, O Daniel, make secret the words and seal up the book, until the time of the end (Daniel 12:4).
Gods will is that all sorts of men should be saved and come to accurate knowledge of truth Timothy 2:4. Do not delay. Take in the knowledge that means ever-lasting life (John 17:3).
30-03-2008
Moshiya van den Broek | 1,149 | 4,148 | {"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 | 3 | CC-MAIN-2019-22 | latest | en | 0.935829 |
http://learningyou.com.au/recommended-books/strand/number/strategy/look-for-a-pattern/ | 1,566,199,315,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314667.60/warc/CC-MAIN-20190819052133-20190819074133-00213.warc.gz | 117,038,555 | 18,449 | ## Recommended Books
### Filter books by:
Strand NumberAlgebraMeasurementSpaceStatistics and ProbabilityProfessional ReadingProblem Solving Year Year 5Year 6Year 7Year 2Year 3Year 8Year 4Foundation YearYear 1Year 9Year 10Maths AMaths B Numeracy Strategy Multiplication and Division TriangleAct it OutPart-Part-WholeLook for a patternAdditive to Multiplicative ThinkingBenchmarkingSee the Parts to match the wholeNumber LineTables and GraphsDraw a diagramVisualisationFind the PatternDraw what you can't seeMental StrategiesVarious
## Anno’s Mysterious Multiplying Jar
##### Australian Curriculum: Description
Select and apply efficient mental and written strategies and appropriate digital technologies to solve problems involving all four operations with whole numbers (ACMNA123)
##### Teaching ideas
At a year 6 level, factorial numbers can be used to help students understand and be able to explain the power of the operation of multiplication in contrast with addition. Specialist Mathematics in the Senior Secondary Mathematics Curriculum (ACARA) covers the concept of factorial notation (ACMSM003).
## Anno’s Magic Seeds
##### Australian Curriculum: Description
Solve word problems by using number sentences involving multiplication or division where there is no remainder (ACMNA082); Find unknown quantities in number sentences involving addition and subtraction and identify equivalent number sentences involving addition and subtraction (ACMNA083)
##### Teaching ideas
Explore other patterns such as square numbers, triangular numbers, Fibonacci sequence etc.
## Matherpieces: The Art of Problem-Solving
##### Australian Curriculum: Description
Subitise small collections of objects (ACMNA003); Represent and solve simple addition and subtraction problems using a range of strategies including counting on, partitioning and rearranging parts (ACMNA015); Describe, continue, and create number patterns resulting from performing addition or subtraction
##### Teaching ideas
Students investigate other situations involing subitising and counting on with and without distractors in routine and non-routine situations. Students could write the number sentences represented by art in this book (representational to abstract).
## Anno’s Counting Book
##### Australian Curriculum: Description
F. Y – Establish understanding of the language and processes of counting by naming numbers in sequences, initially to and from 20, moving from any starting point (ACMNA001); Year 2 – Describe patterns with numbers and identify missing elements (ACMNA035)
##### Teaching ideas
Explore other mediums with patterns e.g. wrapping paper and fabric. Show students how patterns can help to predict --> expect --> plan.
## When the King rides by
##### Australian Curriculum: Description
F. Y – Establish understanding of the language and processes of counting by naming numbers in sequences, initially to and from 20, moving from any starting point (ACMNA001); Year 2 – Describe patterns with numbers and identify missing elements (ACMNA035)
##### Teaching ideas
Explore other mediums with patterns e.g. wrapping paper and fabric. Show students how patterns can help to predict --> expect --> plan.
## How Many Snails?
##### Australian Curriculum: Description
F. Y – Establish understanding of the language and processes of counting by naming numbers in sequences, initially to and from 20, moving from any starting point (ACMNA001); Year 2 – Describe patterns with numbers and identify missing elements (ACMNA035)
##### Teaching ideas
Explore other mediums with patterns e.g. wrapping paper and fabric. Show students how patterns can help to predict --> expect --> plan. | 732 | 3,693 | {"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} | 3.765625 | 4 | CC-MAIN-2019-35 | latest | en | 0.773278 |
https://www.coursehero.com/file/16039947/bus-220-hw1/ | 1,604,066,535,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910815.89/warc/CC-MAIN-20201030122851-20201030152851-00120.warc.gz | 662,047,185 | 34,855 | bus 220 hw1 - Microsoft Excel 14.4 Answer Report Worksheet[bus 220 hw]Sheet1 Report Created 2:31:52 PM Result Solver found a solution All constraints
# bus 220 hw1 - Microsoft Excel 14.4 Answer Report...
This preview shows page 1 - 4 out of 11 pages.
Microsoft Excel 14.4 Answer ReportWorksheet: [bus 220 hw.xlsx]Sheet1Report Created: 2/18/2016 2:31:52 PMResult: Solver found a solution. All constraints and optimality conditions are satisfied.Solver Engine Engine: Simplex LP Solution Time: 0.705116 Seconds. Iterations: 0 Subproblems: 0Solver Options Max Time Unlimited , Iterations Unlimited , Precision 0.000001Max Subproblems Unlimited , Max Integer Sols Unlimited , Integer Tolerance 1%, SolveObjective Cell (Max)CellNameOriginal ValueFinal Value\$B\$13obje. Value(z)00Variable Cells CellNameOriginal ValueFinal ValueInteger \$C\$12Dec. var. x100 Contin \$D\$12Dec. var. x200 Contin ConstraintsCellNameCell ValueFormulaStatusSlack\$C\$15const1 LHS0 \$C\$15<=\$E\$15Not Binding8960\$C\$16const2 LHS0 \$C\$16<=\$E\$16Not Binding4160\$C\$17const3 LHS0 \$C\$17<=\$E\$17Not Binding3200
e Without Integer Constraints , Assume NonNegative
Microsoft Excel 14.4 Sensitivity ReportWorksheet: [bus 220 hw.xlsx]Sheet1Report Created: 2/18/2016 2:31:52 PMVariable Cells FinalReducedObjectiveAllowableAllowableCellNameValueCostCoefficientIncreaseDecrease\$C\$12Dec. var. x10000 1.00E+030\$D\$12Dec. var. x20000 1.00E+030ConstraintsFinalShadowConstraintAllowableAllowableCellName
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• Spring '08
• drexel
• LHS, Dec. Var | 461 | 1,590 | {"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-2020-45 | latest | en | 0.521228 |
https://es.mathworks.com/matlabcentral/answers/805386-how-to-take-average-of-lines-in-plot-plot-it-as-one-graph | 1,670,615,259,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711475.44/warc/CC-MAIN-20221209181231-20221209211231-00179.warc.gz | 275,630,460 | 25,972 | # How to take average of lines in Plot & Plot it as one graph?
140 views (last 30 days)
Nihar Jariwala on 18 Apr 2021
Hi Guys,
I have a graph with 140 lines, each of them having 200 points. I want to plot one line that has to be the average of 1 graph.
For Example: 1st Point = Averages of all the 1st Point of 140 lines.
2nd Point = Averages of all the 2nd Point of 140 lines.
Thanks.
Do you have access to the data of all those graphs as a matrix? If so you can do it in a very straight forward way:
data = randn(200,140); % Substitute by your data
dimensionContainingTheLines = 2;
averageResult = mean(data,dimensionContainingTheLines);
plot(data,'color',[0.9,0.9,0.9]); hold on
plot(averageResult,'linewidth',4);
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Translated by | 261 | 950 | {"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-2022-49 | latest | en | 0.910656 |
http://forums.wolfram.com/mathgroup/archive/2004/Sep/msg00509.html | 1,721,776,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00578.warc.gz | 11,539,830 | 7,637 | Re: Simple questions with Complex Numbers
• To: mathgroup at smc.vnet.net
• Subject: [mg50886] Re: [mg50876] Simple questions with Complex Numbers
• From: "David Park" <djmp at earthlink.net>
• Date: Sun, 26 Sep 2004 05:31:47 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
```Sunil,
When working with complex expressions, you will almost always want to and
need to use ComplexExpand. There is also a Conjugate command. So..
z = s + I t
ComplexExpand[z Conjugate[z]]
s^2 + t^2
David Park
From: Sunil Pinnamaneni [mailto:pinnama at cims.nyu.edu]
To: mathgroup at smc.vnet.net
I'm having difficulty getting mathematica to do simplifications with Complex
Numbers. For instance,
z = s+ I t;
zbar = s- I t;
FullSimplify[Re[1/(z zbar)]]
yields
Re[1/(s^2 +t^2)]
Is there any way to get Mathematica to yield the result 1/(s^2 +t^2) without
the Re? It seems like Mathematica should have been programmed to do easy
simplifications using Complex Numbers.
I'm interested in finding the Real part of expressions like 1/z^5, but
Mathematica doesn't seem to want to multiply the denominator by the
conjugate, and carry out the simpification.
Can Mathematica do such things, or does one have to specifically program it
to handle these simplifications.
Thanks,
Sunil
```
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# Geometry Puzzle
Teacher: Ms. Warren
Across
A triangle with 2 equal measures. One measure will be smaller or larger than the other 2.
A triangle with more than a 90 degree angle. Larger than right & acute triangles.
A triangle with 3 unequal sides. Each side has a different measure.
A statement that can be written in if-then form. If you give a mouse a cookie, then he'll want a glass of milk.
The "B" & "C" in proportions.
The two angles inside the triangle that don't share a vertex within the exterior angles.
The ratio of the length of the leg opposite angle (X)(opp) to the length of the leg adjacent to angle (X)(adj.)
Asserting that we know a mathematical concept is true. A chart showing A+B=C.
A line that intersects 2 or more lines in a plane at different points
Down
A triangle which all 3 sides are equal. Each side has the same measure.
A triangle that has all angles 90 degrees or less. Smaller than right & obtuse triangles.
Forming a logical chain of statements linking the given to what you are trying to prove. If you throw a ball up, then gravity will pull it down.
If part of an "if then" statement
Angle formed by the line of sight & the horizontal plane for an object above the horizontal.
Formed by negating both the hypothesis and the conclusion. If you don't give a mouse a cookie, then he wont want a glass of milk.
The "A" &"D" in proportions.
An equation stating that 2 ratios are equal. A/B=C/D.
A triangle which measures 90 degrees.
The line from the center of the circle measuring to the edge. Half the diameter. | 404 | 1,690 | {"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} | 3.578125 | 4 | CC-MAIN-2018-51 | longest | en | 0.913878 |
https://www.callutheran.edu/college-arts-sciences/mathematics/senior-capstone-projects.html | 1,726,251,132,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00203.warc.gz | 654,622,372 | 11,463 | College of Arts & Sciences
# Senior Capstone Projects
Graduating seniors majoring in Mathematics are required to complete the Capstone course. This is currently offered during the Fall semester. Students are strongly encouraged to that a Capstone Preparation course during the Spring of the Junior year to identify potential topics, investigate background material, and obtain practice with the proposal process.
As part of the Capstone course, each student proposes and carries out an individual research project under the direction of a faculty advisor. Examples of student projects are listed below.
#### Fall 2023
• Optimizing a Film Schedule to Minimize Production Costs Using PERT, CPM, and Stochastic Optimization
• Snakes and Ladders and Markov Chains
• Surveilling Disneyland
• Streaky Hitting in Professional Baseball and Softball
• Optimizing Substitutions in Professional Soccer
• Volleyball Statistics: Which One Should Be Used to Rank NCAA Teams
#### Fall 2022
• Modeling Point Differential in the NBA with Linear Regression
• An Exploration of the Cantor-Zassenhaus Algorithm
• Using Networks to Understand Police Misconduct
• A Re-creation of a Proof of the Abel-Ruffini Theorem Using Galois Theory
• Fermat’s Little Theorem: A Collection of Different Proof Approaches
• Conditions for a Euclidean Domain and an Extension of the Construction of Continued Fractions
• The Effects of Inverted Teaching on Student Learning in Secondary Math Classrooms and Applications for Students with Disabilities
#### Fall 2021
• Exploring different Voting Methods in the Heisman Trophy Voting
• Affine Invariants of a Curve: A Connection to High School Algebra
• Representations and Applications of Catalan Numbers.
• Affects of Switching Your Major on Years to Graduation at CLU.
• A Comparative Analysis of Performance Outcomes in Lesson Planning on Derivatives
• March Madness: What is the likelihood of a Perfect Bracket?
#### Fall 2020
• Fibonacci Sequence Mod m
• The Prisoner's Dilemma's Relation to Toilet Paper During a Pandemic
• How to End Ride Arguments at Disneyland
• Modeling Runs Scored in Baseball
• Optimization of Political Lying
• A Generalization of the Hardy Distribution for Gold Hole Scores
• Modeling Positive Goal-Scoring in Premier League
• A Markov Chain Analysis of XFL Overtime Rules
#### Fall 2019
• Algebraic Curves in Projective Space
• Curves Without Tritangent Planes
• Optimization of Parking at CLU Using Linear Programming
• Optimization of Cal Lutheran Parking Using Optimizations Research
• Analyzing CLU Voting by Shapley-Shubik Power Index
• On the Classification Theorem for Surfaces
• Picks, Bans, and Nash Equilibrium
• Graph-theoretic Theorems in Proofs from THE BOOK
• Analyzing Weighted Bloc Voting Models in the United Nations Security Council using Power Indices
• 9 Ways to Prove the Infinitude of Primes
• Cardinality Bounds on Topological Spaces
• Generalization of the Monty Hall Problem
• An Examination of Hyperbolic Geometry in HyperRogue
• Metrizability in Topology
#### Fall 2018
• Using Stochastic Optimization to Predict Sports Outcomes
• Magic Squares and Elliptic Curves
• Analyzing the Use of Stochastic Optimization through Variations of the News Vendor Problem
• Determining a Model for the California Lutheran University ART Committee Election
• Analyzing California Wildfire Trends Using Bayesian Data Analysis
• Tiling Rectangles and Deficient Rectangles with Polyominoes
• Bayesian Statistics: searching beneath the surface
• Surface Mapping
• RSA and Elliptic Curve Cryptography
• Probability of Playoff Formats
• Mathematical models for the island fox, feral pig, golden eagle system incorporating a strong demographic Allee effect in the island fox
#### Fall 2017
• Predator Prey models
• Optimizing a route at Disneyland
• Bayesian Networks
• Counting Lego stacks
• Analysis of Frieze Patterns
• Knot Invariants
#### Fall 2016
• Micro Economic Theory as a Motivation For Nonlinear Programming
• Bayesian Data Analysis Applied to the Women's SCIAC Swimming Championships
• Applications of Markov Chains in Baseball
• Representations of the Three-Dimensional Rotation Matrices
#### Fall 2015
• Analysis Basic Game Scheduling by TSP and Edge Labeling
• An Overview of Metrization Theorems
• Orbits: A Look at the Symmetry of 4 by 4 Sudoku Puzzles
• Limiting the Size of Topological Spaces Through Cardinal Invariants and Arhangel'skii's Theorem
• An Introduction to Translating Audio Signals into Wavelet Space
• The History of and a Complex Analytic Proof of the Fundamental Theorem of Algebra
• An Analysis fo 2015 Faculty Voting at California Lutheran University
• Finding the Minimum Starting Values of a Sudoku Puzzle that Produces a Unique Solution
• The Pythagorean Triple Tree and Its Many Branches
• Showing the Computational Complexity of Graph Coloring is NP-Complete
• A Mathematical Approach to Analyzing Voting Power with California as 6 States
• How to Find Patterns of Elliptic Solutions Using Modular Arithmetic
• Discovering Hyperbolic Functions Secret Identities through Trigonometric Functions
• Algorithms and Proofs of Three Graph Theoretic Problems
• Entries and Position Impact the Maximum Number of Solutions in a Sudoku Puzzle
• Topological Homogeneity and the Hilbert Cube
#### Fall 2014
• Holling Function Responses: A Look Into Predator-Prey Modeling
• An Array of Cryptosystems and Some Applications
• Influence of Matrix Symmetries on Eigenvalues and Dressing Actions
• Mathematical Modeling of the Hypothalamic-Pituitary-Adrenal Axis
• Duality of Invariants
• A Study of Pascal's Triangle, Square Pyramid, and Tetrahedron Fractal Dimensions and Patterns
• Game Theory and Dominion
#### Fall 2013
• Polynomial Equations and The Fundamental Theorem of Algebra
• Stability Analyses of Several Predator-Prey Models
• Lie Matrix Groups: An In-Depth Look at the Flip Transpose Group
• The Dynamics of Newton's Method for Complex Polynomials
• Patterns of Pascal's Triangle and Pyramid with Combinatorics
• Conics and Their Zero Sets
• Elliptic Curves Over Finite Fields
#### Fall 2012
• The Zombie Apocalypse
• The Sprague-Grundy Theorem
• Knot Theory: A Proof of Reidemeister's Theorem
• Lansey's Hypothesis
• Stochastic Optimization and the Newsboy Problem
• Steganography
#### Fall 2011
• Autonomous Robotic Movement Planning
• Burnside's Lemma and Polya's Theorem
• Monopoly and Markov Chains
• The Field of Cryptography Exposed
• p-adic Analysis, or, The Prime that Killed Calculus
• Evolutionary Game Theory
• Investigation into the Mathematical Framework of General Relativity
• Pinpoint: The Colors of Graph Theory and Adjacency Matrices
#### Fall 2010
• Wallpaper Groups- Euclidean and Hyperbolic
• Describing the Pursuit Curve of Missiles en route to Destroy Satellites
• Markov Chain Analysis of NFL Overtime
• Markov Applications to Soccer Formations
• The Hunt for the Classification Theorem for Surfaces
#### Spring 2010
• The Real Line is Uncountable: A Topological Proof
• Riemann Zeta-Functions and Quadratic Fields
• Mathematical Enigmas: Carroll's Guessing Game and a Strategy Puzzle
• Public Key Cryptosystems
#### Spring 2009
• An Investigation of Hyperbolic Triangles
• Being Transcendental: Searching for Transcendental Numbers
• Magic Square Rubik's Cubes
• Explorations of the Four Color Theorem
• A Statistical Analysis of NASCAR Scoring Systems
• Cellular Automata and Group Theory
• Logarithms in High School and Beyond
• Modeling and Predicting Earthquakes with Markov Chains
#### Spring 2008
• The Rank Number of Prism Graphs
• Cracking the Code
• Deal or No Deal: Making Smart Choices Toward Fortune
• Square Patterns in Rayleigh-Bénard Convection
• When Does a Product of Group Elements Equal Its Reverse?
• Shut the Box
• Voting Theory
• A Quantum Mechanical View of the Navier-Stokes Equations
#### Fall 2008
• Voting Theory: A Look at Arrow's Theorem
#### Spring 2007
• What are the real numbers and how do we teach them in the middle/high schools?
• The Fibonacci Golf Ball
• Teaching Logarithms
• Schwarzschild Geometry
• Permutation Statistics on Generalized Derangements and Desarrangements
#### Fall 2007
• Meta-Problems in Mathematics
• Magic Squares
#### Spring 2006
• Portfolio Management with the Lagrange Multiplier Technique
• Multi-Strain SIR Model Variations
• Julia Sets Over the Quaternions
• Solutions to Polynomial Equations: Their Role in High School and Beyond
• Potential Orderings of Polyhedra
#### Spring 2005
• Mathematics Behind Patience Sorting
• History and Role of Proofs in Secondary Mathematics Education: a Pedagogical Perspective
• Course Scheduling via Network Flows
• Human Knot
• Is God Rational?
• Analysis of the Current Math Placement Program at CLU
• Blackjack: A Beatable Game
• On the Symmetries of Pascal's Pyramid
#### Spring 2004
• Graph Pegging
• Applying the Hungarian Algorithm to NFL Scheduling
• Voting Blocs in Academic Divisions of CLU: A Mathematical Explanation of Faculty Power
• Mathematical Modeling: Stress Relaxation of Viscoelastic Materials
#### Fall 2004
• Game Theory and Military Scenarios
#### Spring 2003
• Databases for City Planning
• Graph Pebbling
• Number Theoretical Graph Pebbling
• Techniques for Teaching Proofs in High School Geometry
• Twin Primes: A Survey
• Fractals
• Program Scheduling
#### Spring 2002
• Scheduling a Scramble Golf Tournament
• Algebraic Coding Theory
• Comparing Craps Table Odds
#### Spring 2001
• What Energy Bar is Best to Fit Your Daily Calorie Needs?
• Earthquakes in California
• Traffic Scenarios
• Soccer Penalty Kicks: Are They Unfair to Goalkeepers?
• Batter Up: A Computer Simulated Look at Baseball
• The Perfect Toss: How to Survive the MCM
• Voting Schemes
• Pin Action!
#### Spring 2000
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• Salaries of Starting Professional Baseball Pitchers | 2,335 | 10,027 | {"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} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.832851 |
https://brainly.com/question/125579 | 1,484,571,458,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279176.20/warc/CC-MAIN-20170116095119-00305-ip-10-171-10-70.ec2.internal.warc.gz | 799,629,100 | 8,774 | 2014-09-22T13:24:21-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Ok so we need to subtract the area of the triangle from the area of the segment and this will equal 100.
We know that the area of the segment is:
And that the area of the triangle is:
Therefore:
We can simplify it through these steps:
Therefore r=22.04cm (4sf)
thank you very much !!!! | 146 | 623 | {"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} | 3.390625 | 3 | CC-MAIN-2017-04 | latest | en | 0.93609 |
http://www.mahendraguru.com/2017/12/quantitative-aptitude-questions-for18.html | 1,547,939,284,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583684033.26/warc/CC-MAIN-20190119221320-20190120003320-00586.warc.gz | 344,636,923 | 33,354 | Subscribe Now
# Quantitative Aptitude Questions For IBPS Clerk Mains : 18 - 12 - 17
mahendra Guru
Quantitative Aptitude quiz is basically to test your mathematical calculation and approach. It is to find out how fast you can solve a given question with the right methodology of solving the problem. If you know the formulas and short tricks of some important topic in Maths, you will definitely score good marks. So, it is important to know the basic concept of all the topic so you can apply the short tricks and solve the question with the new concept in lesser time while giving the quiz.
Quantitative Aptitude Quiz helps to evaluate your preparation for banking exam so you can improve your preparation level. Mahendra Guru provides you Quantitative Aptitude Quiz for Bank examination based on the latest pattern. So that you can practice on regular basis. It will definitely help you to score good marks in the exam. It is the most important section for all the govt exam like IBPS PO/ Clerk/SO/RRB, RBI, SBI, Insurance, SSC-MTS, CGL, CHSLState Level and other Competitive exams.
Q.1. What value will come in place of question mark (?) in the questions given below?
164.78 - 19.91 + 20.16 = ? + 110.35
(1) 56.25
(2) 54.68
(3) 52.96
(4) 53.32
(5) 59.32
Sol :
? = 165.03-110.35
= 54.68
Q.2. What value will come in place of question mark (?) in the questions given below?
28 + 46 × 1443 ÷ 39 = ? + 82.50
(1) 1634.8
(2) 1447.5
(3) 1643.2
(4) 1674.6
(5) 1647.5
Sol :
28+1702 = ? + 82.50
? = 1730 - 82.50
= 1647.50
Q.3. What value will come in place of question mark (?) in the questions given below?
144 × 18 - 80.5 × 8 = ? + 12.80
(1) 1955.2
(2) 1825.6
(3) 2018.4
(4) 1748.5
(5) 1935.2
Sol :
2592 - 644 = ? + 12.80
1948 = ? + 12.80
? = 1935.2
Q.4. What value will come in place of question mark (?) in the questions given below?
106% of 7725 - 39% of 7720= ?
(1) 5177.7
(2) 5228.6
(3) 5316.2
(4) 5422.4
(5) 5022.4
Sol :
?= 8188.5 - 3010.8
= 5177.7
Q.5. What value will come in place of question mark (?) in the questions given below?
-5871 + 5854.75 + 2362.35 = ? + (12)3
(1) 526.2
(2) 736.8
(3) 445.6
(4) 618.1
(5) 645.6
Sol:
? = 2346.1 - 1728
= 618.1
Q.6. What value will come in place of question mark (?) in the questions given below?
30% of 1960 + ?% of 1250 = 25% of 4534
(1) 42.18
(2) 39.26
(3) 44.38
(4) 43.64
(5) 45.64
Sol:
588 + 12.5 × ? = 1133.5
? = 43.64
Q.7. What value will come in place of question mark (?) in the questions given below?
(17 × 14) + (6 × 5 ×3) - (5 × 8)= ? × 4
(1) 69
(2) 74
(3) 72
(4) 75
(5) 78
Sol :
238 + 90 - 40 = ? × 4
? = 72
Q.8. What value will come in place of question mark (?) in the questions given below?
(73732 - 58607) ÷ 25 = ?
(1) 610
(2) 615
(3) 625
(4) 605
(5) 620
Sol :
? =
? = 605
Q.9. What value will come in place of question mark (?) in the questions given below?
(1)
(2) 590
(3) 576
(4) 24
(5)
Sol :
Q.10. What value will come in place of question mark (?) in the questions given below?
462.721 + 231.027 – 118.079 = 521.211 + ?
(1) 53.456
(2) 54.558
(3) 54.459
(4) 54.458
(5) 44.458
Sol :
? = 575.669 – 521.211 = 54.458
Q.1. (2)
Q.2. (5)
Q.3. (5)
Q.4. (1)
Q.5. (4)
Q.6. (4)
Q.7. (3)
Q.8. (4)
Q.9. (1)
Q.10 (2) | 1,327 | 3,325 | {"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} | 4.09375 | 4 | CC-MAIN-2019-04 | latest | en | 0.735371 |
https://winmundo.com/function-what-does-return-true-false-actually-do-python/ | 1,685,240,222,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643462.13/warc/CC-MAIN-20230528015553-20230528045553-00174.warc.gz | 675,861,490 | 15,567 | # function – What does return True/False actually do? (Python)
## function – What does return True/False actually do? (Python)
Because most of the time youll want to do something with the result other than print it. By using a return value, the function does one thing: do the calculation, and the caller can do whatever it wants with it: print it, write it to a database, use it as part of some larger calculation, whatever. The idea is called composability.
Also your example doesnt currently make much sense as it will always return `False` after evaluating `x > y` but doing nothing with the result. Perhaps you meant something like:
``````def is_greater(x, y):
if x > y:
return True
else:
return False
``````
Then `is_greater` is something that can easily be used in whatever context you want. You could do:
``````x = is_greater(a, b)
write_to_super_secret_database(x)
``````
(In real life you probably wouldnt use a function to do something so trivial, but hopefully the example makes sense.)
Functions which return True / False are used in further statements like IF:
Like this:
``````def is_bigger(x,y):
return x > y
if is_bigger(10,9):
do_something
elif:
print math dont work anymore
``````
you should take a look at variables and control structures:
http://www.tutorialspoint.com/python/python_if_else.htm | 315 | 1,327 | {"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} | 2.609375 | 3 | CC-MAIN-2023-23 | latest | en | 0.879669 |
http://cgm.cs.mcgill.ca/~athens/cs644/2002projects/vleves_probing/node_3_4.htm | 1,513,025,354,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948514051.18/warc/CC-MAIN-20171211203107-20171211223107-00477.warc.gz | 54,800,765 | 1,553 | Information Gained By Probing ::
### Colinear Contact Points Define Edges or Edge Segments
Consider three colinear contact points p1, p2, p3 as shown in the diagram below. As explained before, these three points are part of the bounding polygon. By definition of a convex polygon, these three point must be part of a single edge or edge segment of the bounding polygon. Moreover, the infinite line described by the three points automatically discards points on its exterior side into the outside set. As we will see in later sections this property allows us to verify our hypothesis about the shape of the polygon. This diagram illustrates this observation.
Probe Paths Define the Outside Set Summary | 140 | 704 | {"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} | 2.71875 | 3 | CC-MAIN-2017-51 | latest | en | 0.866336 |
https://community.fabric.microsoft.com/t5/Desktop/Creating-banding-based-on-measure/td-p/1614909 | 1,721,455,757,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515020.57/warc/CC-MAIN-20240720052626-20240720082626-00130.warc.gz | 146,764,466 | 52,523 | cancel
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Anonymous
Not applicable
Creating banding based on measure
Hi Experts
using live connection to cube.
I have a measure which give me Margin % values from 1% to 50%.
How could I group and or band these into the follwing bucket.
<10%
10 to 20%
21 to 40%
>40%
So I could add this a a filter over my data..
3 REPLIES 3
Community Support
Hi @Anonymous ,
Here are the steps you can follow：
1. Create measure.
``````Measure 2 =
SWITCH(
TRUE(),
'Table'[Measure]<=0.1,"<10%",
'Table'[Measure]>0.1&&'Table'[Measure]<=0.2,">10%&&<=20%",
'Table'[Measure]>0.2&&'Table'[Measure]<=0.4,">20%&&<40%",
'Table'[Measure]>0.4,">40%"
)``````
2. Result.
Best Regards,
Liu Yang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Super User
@Anonymous check this post for dynamic segmentation.
Check my latest blog post Compare Budgeted Scenarios vs. Actuals I would Kudos if my solution helped. 👉 If you can spend time posting the question, you can also make efforts to give Kudos to whoever helped to solve your problem. It is a token of appreciation!
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Anonymous
Not applicable
Hi perry2k
I cannot use that method as I am connecting to a cube live connection.. I have to some how utilise the measure..and or use an alternative method
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Top Solution Authors
Top Kudoed Authors | 615 | 2,475 | {"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-2024-30 | latest | en | 0.902243 |
https://www.askiitians.com/forums/Vectors/let-o-be-the-circumcentre-and-o-be-the-orthocentr_238192.htm | 1,722,825,291,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640427760.16/warc/CC-MAIN-20240805021234-20240805051234-00017.warc.gz | 519,134,363 | 42,683 | # Let O be the circumcentre and O' be the orthocentre of triangle ABC prove that OA + OB +OC = OO' in vector
Arun
25750 Points
5 years ago
Dear student
we know that O'G =2GO
where G is the centroid of triangle
let a point D between B and C
So, OD =(OB+OC )/2
now OA + OB + OC = OA + 2OD
we know that G devide the point A and mib point of apposite side(D) in ratiio 2:1
So, OG =(OA 2OD)/(2+1)
So, OA + OB + OC =3OG =2OG + OG =GO'+OG =OO' | 161 | 437 | {"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} | 3.75 | 4 | CC-MAIN-2024-33 | latest | en | 0.843761 |
https://web2.0calc.com/questions/hlp_7 | 1,586,040,103,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370525223.55/warc/CC-MAIN-20200404200523-20200404230523-00523.warc.gz | 781,096,328 | 5,827 | +0
HLP
+1
233
1
+91
A polynomial of degree \$13\$ is divided by \$d(x)\$ to give a quotient of degree \$7\$ and a remainder of \$3x^3+4x^2-x+12\$. What is \$\deg d\$?
Jul 14, 2019 | 75 | 183 | {"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-2020-16 | latest | en | 0.922765 |
https://www.airmilescalculator.com/distance/lft-to-shr/ | 1,603,563,846,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107884322.44/warc/CC-MAIN-20201024164841-20201024194841-00630.warc.gz | 614,910,866 | 52,293 | # Distance between Lafayette, LA (LFT) and Sheridan, WY (SHR)
Flight distance from Lafayette to Sheridan (Lafayette Regional Airport – Sheridan County Airport) is 1295 miles / 2083 kilometers / 1125 nautical miles. Estimated flight time is 2 hours 57 minutes.
Driving distance from Lafayette (LFT) to Sheridan (SHR) is 1606 miles / 2585 kilometers and travel time by car is about 27 hours 27 minutes.
## Map of flight path and driving directions from Lafayette to Sheridan.
Shortest flight path between Lafayette Regional Airport (LFT) and Sheridan County Airport (SHR).
## How far is Sheridan from Lafayette?
There are several ways to calculate distances between Lafayette and Sheridan. Here are two common methods:
Vincenty's formula (applied above)
• 1294.614 miles
• 2083.480 kilometers
• 1124.989 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1294.858 miles
• 2083.872 kilometers
• 1125.201 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).
## Airport information
A Lafayette Regional Airport
City: Lafayette, LA
Country: United States
IATA Code: LFT
ICAO Code: KLFT
Coordinates: 30°12′19″N, 91°59′15″W
B Sheridan County Airport
City: Sheridan, WY
Country: United States
IATA Code: SHR
ICAO Code: KSHR
Coordinates: 44°46′9″N, 106°58′48″W
## Time difference and current local times
The time difference between Lafayette and Sheridan is 1 hour. Sheridan is 1 hour behind Lafayette.
CDT
MDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 167 kg (367 pounds).
## Frequent Flyer Miles Calculator
Lafayette (LFT) → Sheridan (SHR).
Distance:
1295
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1295
Round trip? | 497 | 1,964 | {"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} | 2.84375 | 3 | CC-MAIN-2020-45 | latest | en | 0.824013 |
https://www.w3resource.com/excel/formulas/count/count-no-of-cells-containing-any-text.php | 1,716,387,911,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00632.warc.gz | 922,859,509 | 27,614 | # Excel Formulas - Count number of cells containing any text
## Count number of cells containing text (exclude only number, blank, or error)
Syntax of used function(s)
```COUNTIF(criteria_range, criteria)
```
The COUNTIF function is used to count the number of cells that meet a certain criteria.
Explanation
To count number of cells within a range of cells which contain any text except number, blank, or error the COUNTIF function can be used.
Formula
``````
=COUNTIF(Sample_Text,"*")
``````
How this formula works
In this formula the COUNTIF function search each cell in the range whether it contain any text except number, blank, or error and count if met the criteria.
## Count number of cells containing no text (excluding text )
Formula
``````
=COUNTIF(Sample_Text,"<>*")
``````
How this formula works
In this formula the COUNTIF function search each cell in the range and ignore counting if contain any text (except number, and error).
## Count number of cells containing text with no blank cell
Syntax of used function(s)
```COUNTIFS(criteria_range1, criteria1, [criteria_range2, criteria2]…)
```
The COUNTIFS function applies criteria to cells across multiple ranges and counts the number of times all criteria are met
Formula
``````
=COUNTIFS(Sample_Text,"*",Sample_Text,"<> ")
``````
How this formula works
In this formula the COUNTIFS function search each cell in the range and ignore counting if contain any blank cells.
## Count number of cells containing only text using SUMPRODUCT()
Syntax of used function(s)
```SUMPRODUCT(array1, [array2], [array3], ...)
ISTEXT(value)
```
The SUMPRODUCT function is used to multiplies the corresponding components in the given arrays, and returns the sum of those products. The ISTEXT funtion returns True if the value refers to a text.
Formula
``````
=SUMPRODUCT(--ISTEXT(Sample_Text))
``````
How this formula works
In the above formula the ISTEXT function search within the range Sample_Text and returns true is text found and creates an array like -
``````
{TRUE;TRUE;FALSE;TURE;FALSE;FALSE;TRUE;}
``````
Threfore the double negative (--) sign converts it like -
``````
{1;1;0;1;0;0;1;}
``````
and SUMPRODUCT then counts and return result.
| 521 | 2,239 | {"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-2024-22 | latest | en | 0.733428 |
https://www.nagwa.com/en/videos/794192952382/ | 1,582,767,281,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146643.49/warc/CC-MAIN-20200227002351-20200227032351-00282.warc.gz | 791,540,227 | 5,190 | Video: KS1-M18 • Paper 2 • Question 13
Amy made 25 using different shapes for tens and ones. Amy made a new number. What is Amy’s new number?
01:12
Video Transcript
Amy made 25 using different shapes for tens and ones. Amy made a new number. What is Amy’s new number?
Amy made 25 to begin with. She made her 25 from two triangles and five squares. This means that each triangle is worth 10. Two tens are 20. And five ones or five squares makes five. Two tens and five ones makes 25. So let’s work out what Amy’s new number is.
Three tens and four ones. Our three tens or three triangles makes 30 plus our four ones makes 34. Three tens and four ones is 34. Amy’s new number is number 34. | 180 | 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} | 3.875 | 4 | CC-MAIN-2020-10 | latest | en | 0.911872 |
http://www.einstein-online.info/dictionary/amplitude.html | 1,534,614,098,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213693.23/warc/CC-MAIN-20180818173743-20180818193743-00323.warc.gz | 507,501,136 | 7,841 | Sie sind hier: Startseite amplitude
# amplitude
For a physical quantity that changes periodically, the maximal value reached in the course of one period. The simplest example is a sine oscillation as shown here:
Over time (plotted from left to right), the sine curve oscillates between its minimum and its maximum values.
Depending on the nature of the oscillation or wave, the amplitude will have different meanings. For a pendulum swinging back and forth, the amplitude is the maximum angle between the vertical direction and the pendulum string. For an electromagnetic wave, the amplitude is the maximal value of the electric field or equivalently (since the two maxima are related) the maximum of the magnetic field. For a (weak) gravitational wave, the amplitude is a direct measure of the changes in distance caused by the wave - as a simple gravitational wave of amplitude A passes, there are two directions in which distances are alternately stretched by up to a factor (1+A/2) and compressed by a factor (1-A/2).
The amplitude can change over time. For instance, for an ordinary pendulum, air friction will slow the pendulum bob down, and for each period - for each time the pendulum bob travels back and forth - the amplitude will be less than for the previous period. For a wave, the amplitude will also in general vary with location. Typically, the amplitude of a wave will decrease with the distance from the wave's source. | 296 | 1,441 | {"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} | 3.078125 | 3 | CC-MAIN-2018-34 | longest | en | 0.8774 |
https://fellrnr.com/wiki/Special:MobileDiff/690 | 1,642,454,799,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300624.10/warc/CC-MAIN-20220117212242-20220118002242-00304.warc.gz | 304,935,881 | 5,751 | # Changes
## Relative Running Economy
, 19:05, 24 December 2010
m
no edit summary
==Introduction==There are two components to running ability; fitness and economy. Fitness is the ability of our bodies to generate energy for running and is the focus of a lot of our training. Economy is how far and fast you can run with a given amount of energy. Good economy is a critical part of running, and [Cadence] is one element I focus on. ==Measuring Efficiency==In an ideal world, we’d be able to easily measure our running economy and see if things are improving. If we could measure our breath, find out how much O<sub>2</sub> we consumed and how much CO<sub>2</sub> we produce, we’d know how much energy we burned (and from fat or carbohydrate). Sadly, this is not practical, so the best measure we have of energy consumption is our heart rate. This is far from perfect, as heart rate can vary for other reasons besides supplying O<sub>2</sub> for energy production. However, I believe it is a useful approximation. ==The Calculator==Assuming you know the distance you ran, your average heart rate and the time it took, you can calculate your efficiency. If you know your resting heart rate, enter that as well to optimize the calculation.
<html>
</table>
</td>
</tr>
<tr>
<td>Average Heart Rate</td>
<td><input maxlength="3" size="3" name="AverageHeartRate" value="150"></td>
</tr>
<tr>
<td>Resting Heart Rate</td>
<td><input maxlength="3" size="3" name="RestingHeartRate" value="40"></td>
</tr>
<tr>
<td>Distance (miles)</td>
<td><input maxlength="3" size="3" name="Distance" value="1"></td>
<button name="Calculate">Calculate</button><br>
</html>
==The Efficiency Calculation==
The formula that is used is<br>
<code>
Total Beats = (Average Heart Rate – Resting Heart Rate) * Time in Minutes
Work Per Mile = Total Beats / Distance in Miles
Efficiency = 1 / Work Per Mile * 100,000
</code>
8,036
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https://kmmiles.com/792807-km-in-miles | 1,685,991,723,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652161.52/warc/CC-MAIN-20230605185809-20230605215809-00755.warc.gz | 379,940,602 | 5,899 | kmmiles.com
Search
792807 km in miles
Result
792807 km equals 492333.147 miles
You can also convert 792807 km to mph.
Conversion formula
Multiply the amount of km by the conversion factor to get the result in miles:
792807 km × 0.621 = 492333.147 mi
How to convert 792807 km to miles?
The conversion factor from km to miles is 0.621, which means that 1 km is equal to 0.621 miles:
1 km = 0.621 mi
To convert 792807 km into miles we have to multiply 792807 by the conversion factor in order to get the amount from km to miles. We can also form a proportion to calculate the result:
1 km → 0.621 mi
792807 km → L(mi)
Solve the above proportion to obtain the length L in miles:
L(mi) = 792807 km × 0.621 mi
L(mi) = 492333.147 mi
The final result is:
792807 km → 492333.147 mi
We conclude that 792807 km is equivalent to 492333.147 miles:
792807 km = 492333.147 miles
Result approximation
For practical purposes we can round our final result to an approximate numerical value. In this case seven hundred ninety-two thousand eight hundred seven km is approximately four hundred ninety-two thousand three hundred thirty-three point one four seven miles:
792807 km ≅ 492333.147 miles
Conversion table
For quick reference purposes, below is the kilometers to miles conversion table:
kilometers (km) miles (mi)
792808 km 492333.768 miles
792809 km 492334.389 miles
792810 km 492335.01 miles
792811 km 492335.631 miles
792812 km 492336.252 miles
792813 km 492336.873 miles
792814 km 492337.494 miles
792815 km 492338.115 miles
792816 km 492338.736 miles
792817 km 492339.357 miles
Units definitions
The units involved in this conversion are kilometers and miles. This is how they are defined:
Kilometers
The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.
Miles
A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada. | 625 | 2,339 | {"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} | 4.0625 | 4 | CC-MAIN-2023-23 | latest | en | 0.793522 |
https://byjusexamprep.com/study-notes-on-venn-diagram-i | 1,669,729,355,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710698.62/warc/CC-MAIN-20221129132340-20221129162340-00240.warc.gz | 195,176,456 | 62,407 | # Study Notes on Venn Diagram
By Gaurav Mohanty|Updated : December 20th, 2021
This is a very important part of logical reasoning. These types of questions aim to test a candidate’s ability to classify and relate certain groups of items. This section is one of the most scoring sections in Reasoning for BBA & IPM Entrance Exams as all one has to do is just look at the image and data asked/given and answer the question straight away.
But many candidates fail to utilize the simplicity of these questions as they don’t know where to look for and what information.
What exactly is the Venn diagram?
• These diagrams were given by John Venn. To put it simply, these diagrams show all possible logical relations between a number of elements.
• In a typical Venn diagram, usually, there’s a use of geometrical figures like Circles, Triangles, Squares & Rectangles.
• A basic Venn diagram has data represented in ‘Circles’.
1. In a country three persons A, B and C live. They are three different persons. This information can be represented as:
Here, we can see that A, B and C are different elements so they’ve represented by different circles.
2. if we were to represent information in which two elements are intermingled while the third one is different we’ll do that a bit differently.
For example, Hindu, Indian, Parrot. Now, logically we know that Some Hindus are Indian (as some Hindu might be living abroad and be Australian or any other country’s citizen) and also, no Parrot is Hindu(as animals have no religion) also no Parrot is Indian (as no animal has ethnicity). This information of Hindu, Indian, Parrot can be represented as follows:
Here, the shaded area shows those Hindus who are Indians at the same time. Parrot is represented in a different circle.
3. Suppose, we need to convey this: Dog, Animal, Cow.
Now we know that all dogs are animals (clearly no dog is human) so the circle of ‘dog’ will have to be completely surrounded by a circle of the animal through the circle of animal can have some spare space aside from dog as the dog isn’t the only animal. Similarly, all ‘cows’ are animals so the circle of ‘cow’ will have to be completely surrounded by the circle of animal through the circle of animal can have some spare space aside from cow as the cow isn’t the only animal. This information can be represented as:
Here, we can see that ‘Animal’ has been represented by a big circle which encompasses the circles for both ‘cow’ and ‘dog. Notice, the circle for ‘animal’ has some spare space as this can contain other types of animals because ‘cow’ and ‘dog’ aren’t the only types of animal.
Types of questions asked in Competition Exam:
Let’s have a look at the type of questions asked specifically in SSC exam. There are basically two types of questions:
1) Finding relationship: To solve these kinds of questions, we need to have a strong grip on common relationships that exist in the world around us. Like to define the relationship between Catholics & Christian we need to know that Catholics are the type of Christians hence we can easily conclude that all Catholics are Christian but some Christians will not be Catholics as they will be the other type of Christians. This information can be represented as:
A typical question might look like this: Dean, Painter, Singer.
We live in a diverse world where people can be multi-talented also people possess just one talent so this info can be represented by 7 categories of people:
a) Who are only Dean
b) Who are only Painter
c) Who are only Singer
d) Who are both Dean & Painter
e) Who are both Painter & Singer
f) Who are both Singer & Dean
g) Who are all Dean, Painter & Singer?
This information can be represented by Venn-diagram as follow: (for reader’s convenience, the different regions have been labelled as named above but in exams, questions haven't marked this way)
2) Finding the exact region: These are the reverse version of the questions discussed above. Here, the diagram with labelled image is given and we’ve to identify the region specifically asked the question.
For example, an image like below will be given:
Circle S stands for households having a scooter, Circle T stands for households having a TV set,
Circle W stands for households having a Washing Machine, Circle C stands for households having a car.
Find household having both TV set, Car and Washing Machine but not scooter. (Question ends)
Now, if we look closely we can see that four distinct items have been, these 4 distinct items can be seen as:
Now, there are places where only ‘Circle S and Circle T’ meet, such place can be represented in the figure below with ‘orange’ colour, similarly, following colours have been used to represent different regions:
1) Green = Only (T, W and S) 2) Yellow = Only (T and W) 3) Purple = Only (W and C) 4) Blue = Only (S, W and C) 5) Baby Pink = Only (S and T) 6) White = Only (S, T and C) 7) Light brown = Only (T, W and C) 8) Red = All S,T, W and C.
Note: You don’t have to make such colorful representation in the exam. It’s been only colourized to help you visualized the different regions with specific labels.
Now, we have to find the household with TV, Washing machine, car but not scooter.
We know that TV = Circle T, Washing Machine = Circle W, Car = Circle C, Scooter = Circle S
So, we have to find where Circle T, W and C meet but not S.
We can clearly see that such region is represented by light brown region which was marked as ‘7’ in the original question figure.
The similar question can be given which represents different elements using different figures like the rectangle, triangle etc. as shown below:
Here, Circle represents college Professors, the triangle represents Surgeons and chemists are shown by the rectangle.
Find the area where Surgeons who are Chemists but not Professors are represented.
To find the area representing only Surgeons and Chemists, we need to look for where ONLY Triangle(=Surgeons) and Rectangle(=Chemist) meet and no sign of Circle(=professor).
Clearly, such area is shown by region marked as Z only (and not Y because that would include Circle also).
========================================== | 1,391 | 6,174 | {"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} | 4.625 | 5 | CC-MAIN-2022-49 | latest | en | 0.953149 |
https://number.academy/76103 | 1,659,913,641,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570730.59/warc/CC-MAIN-20220807211157-20220808001157-00701.warc.gz | 402,472,949 | 11,910 | # Number 76103
Number 76,103 spell 🔊, write in words: seventy-six thousand, one hundred and three . Ordinal number 76103th is said 🔊 and write: seventy-six thousand, one hundred and third. The meaning of number 76103 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 76103. What is 76103 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 76103.
## What is 76,103 in other units
The decimal (Arabic) number 76103 converted to a Roman number is (L)(X)(X)(V)MCIII. Roman and decimal number conversions.
#### Weight conversion
76103 kilograms (kg) = 167776.7 pounds (lbs)
76103 pounds (lbs) = 34520.1 kilograms (kg)
#### Length conversion
76103 kilometers (km) equals to 47289 miles (mi).
76103 miles (mi) equals to 122476 kilometers (km).
76103 meters (m) equals to 249679 feet (ft).
76103 feet (ft) equals 23197 meters (m).
76103 centimeters (cm) equals to 29961.8 inches (in).
76103 inches (in) equals to 193301.6 centimeters (cm).
#### Temperature conversion
76103° Fahrenheit (°F) equals to 42261.7° Celsius (°C)
76103° Celsius (°C) equals to 137017.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
76103 seconds equals to 21 hours, 8 minutes, 23 seconds
76103 minutes equals to 1 month, 3 weeks, 3 days, 20 hours, 23 minutes
### Codes and images of the number 76103
Number 76103 morse code: --... -.... .---- ----- ...--
Sign language for number 76103:
Number 76103 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 76103
### Multiplications
#### Multiplication table of 76103
76103 multiplied by two equals 152206 (76103 x 2 = 152206).
76103 multiplied by three equals 228309 (76103 x 3 = 228309).
76103 multiplied by four equals 304412 (76103 x 4 = 304412).
76103 multiplied by five equals 380515 (76103 x 5 = 380515).
76103 multiplied by six equals 456618 (76103 x 6 = 456618).
76103 multiplied by seven equals 532721 (76103 x 7 = 532721).
76103 multiplied by eight equals 608824 (76103 x 8 = 608824).
76103 multiplied by nine equals 684927 (76103 x 9 = 684927).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 76103
Half of 76103 is 38051,5 (76103 / 2 = 38051,5 = 38051 1/2).
One third of 76103 is 25367,6667 (76103 / 3 = 25367,6667 = 25367 2/3).
One quarter of 76103 is 19025,75 (76103 / 4 = 19025,75 = 19025 3/4).
One fifth of 76103 is 15220,6 (76103 / 5 = 15220,6 = 15220 3/5).
One sixth of 76103 is 12683,8333 (76103 / 6 = 12683,8333 = 12683 5/6).
One seventh of 76103 is 10871,8571 (76103 / 7 = 10871,8571 = 10871 6/7).
One eighth of 76103 is 9512,875 (76103 / 8 = 9512,875 = 9512 7/8).
One ninth of 76103 is 8455,8889 (76103 / 9 = 8455,8889 = 8455 8/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
76103
#### Is Prime?
The number 76103 is a prime number. The closest prime numbers are 76099, 76123.
#### Factorization and factors (dividers)
The prime factors of 76103
Prime numbers have no prime factors less than themselves.
The factors of 76103 are 1 , 76103
Total factors 2.
Sum of factors 76104 (1).
#### Prime factor tree
76103 is a prime number.
#### Powers
The second power of 761032 is 5.791.666.609.
The third power of 761033 is 440.763.203.944.727.
#### Roots
The square root √76103 is 275,867722.
The cube root of 376103 is 42,377363.
#### Logarithms
The natural logarithm of No. ln 76103 = loge 76103 = 11,239843.
The logarithm to base 10 of No. log10 76103 = 4,881402.
The Napierian logarithm of No. log1/e 76103 = -11,239843.
### Trigonometric functions
The cosine of 76103 is 0,489256.
The sine of 76103 is 0,87214.
The tangent of 76103 is 1,782583.
### Properties of the number 76103
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 76103 in Computer Science
Code typeCode value
76103 Number of bytes74.3KB
Unix timeUnix time 76103 is equal to Thursday Jan. 1, 1970, 9:08:23 p.m. GMT
IPv4, IPv6Number 76103 internet address in dotted format v4 0.1.41.71, v6 ::1:2947
76103 Decimal = 10010100101000111 Binary
76103 Decimal = 10212101122 Ternary
76103 Decimal = 224507 Octal
76103 Decimal = 12947 Hexadecimal (0x12947 hex)
76103 BASE64NzYxMDM=
76103 MD59b33d45b795a58e58ab585b10f9c49aa
76103 SHA1db309b344633738fc80379aa6f58455e9d54f0c2
76103 SHA22427a22b2e884a95c9f6b4e8c10f5e59d0436ac70702a2c46af0dace04
76103 SHA2562e3934f9db32802eb18803787dea497e8b930970a65271c257912aeac203b9ef
76103 SHA3840b2750783c34062526281d81ae47a4b7c95588011ce0ae06e289f5423ca48e59c101477790bdfa7ed8d4eb8797dfdbae
More SHA codes related to the number 76103 ...
If you know something interesting about the 76103 number that you did not find on this page, do not hesitate to write us here.
## Numerology 76103
### Character frequency in number 76103
Character (importance) frequency for numerology.
Character: Frequency: 7 1 6 1 1 1 0 1 3 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 76103, the numbers 7+6+1+0+3 = 1+7 = 8 are added and the meaning of the number 8 is sought.
## Interesting facts about the number 76103
### Asteroids
• (76103) 2000 DF103 is asteroid number 76103. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 2/29/2000.
## № 76,103 in other languages
How to say or write the number seventy-six thousand, one hundred and three in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 76.103) setenta y seis mil ciento tres German: 🔊 (Anzahl 76.103) sechsundsiebzigtausendeinhundertdrei French: 🔊 (nombre 76 103) soixante-seize mille cent trois Portuguese: 🔊 (número 76 103) setenta e seis mil, cento e três Chinese: 🔊 (数 76 103) 七万六千一百零三 Arabian: 🔊 (عدد 76,103) ستة و سبعون ألفاً و مائة و ثلاثة Czech: 🔊 (číslo 76 103) sedmdesát šest tisíc sto tři Korean: 🔊 (번호 76,103) 칠만 육천백삼 Danish: 🔊 (nummer 76 103) seksoghalvfjerdstusinde og ethundrede og tre Dutch: 🔊 (nummer 76 103) zesenzeventigduizendhonderddrie Japanese: 🔊 (数 76,103) 七万六千百三 Indonesian: 🔊 (jumlah 76.103) tujuh puluh enam ribu seratus tiga Italian: 🔊 (numero 76 103) settantaseimilacentotre Norwegian: 🔊 (nummer 76 103) sytti-seks tusen, en hundre og tre Polish: 🔊 (liczba 76 103) siedemdziesiąt sześć tysięcy sto trzy Russian: 🔊 (номер 76 103) семьдесят шесть тысяч сто три Turkish: 🔊 (numara 76,103) yetmişaltıbinyüzüç Thai: 🔊 (จำนวน 76 103) เจ็ดหมื่นหกพันหนึ่งร้อยสาม Ukrainian: 🔊 (номер 76 103) сiмдесят шiсть тисяч сто три Vietnamese: 🔊 (con số 76.103) bảy mươi sáu nghìn một trăm lẻ ba Other languages ...
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## Comment
If you know something interesting about the number 76103 or any natural number (positive integer) please write us here or on facebook. | 2,431 | 7,076 | {"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} | 2.953125 | 3 | CC-MAIN-2022-33 | latest | en | 0.624997 |
https://solvedlib.com/how-do-you-solve-log,296700 | 1,657,166,262,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683683.99/warc/CC-MAIN-20220707033101-20220707063101-00581.warc.gz | 575,998,617 | 17,418 | # How do you solve log_(1/7)x=-1?
###### Question:
How do you solve log_(1/7)x=-1?
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##### Drag and drop your Iiles Or click to browse03 9 points)Determine whether the following statements are true Or false If true, provide justification_ If false, provide counterexample_ (a) If lim f(r) exists and I = l is in the domain of f (c), then f(r) is continuous at I = 1 (b) If f(r) is continous on (0,1) and f (0) < 0 < f(1), then there exists number € in (0.1) such that f(c) = 0. Suppose f(r) and g(1) are continuous on [0.1] and that f(0) f(1) _ 1. 9(0) = 1 and 9(1) _ 0.Then the gra
Drag and drop your Iiles Or click to browse 03 9 points) Determine whether the following statements are true Or false If true, provide justification_ If false, provide counterexample_ (a) If lim f(r) exists and I = l is in the domain of f (c), then f(r) is continuous at I = 1 (b) If f(r) is contino... | 704 | 2,957 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 2, "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} | 2.921875 | 3 | CC-MAIN-2022-27 | latest | en | 0.941835 |
http://stackoverflow.com/questions/tagged/piecewise?sort=unanswered&pageSize=15 | 1,386,270,631,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163047212/warc/CC-MAIN-20131204131727-00029-ip-10-33-133-15.ec2.internal.warc.gz | 178,144,509 | 11,130 | # Tagged Questions
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### MATLAB: evaluation of a piecewise polynomial (pchip) with ppval
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### Solving System of Piecewise Equations in Matlab, keep getting “Function 'lt' is not implemented for MuPAD symbolic objects”
I'm trying to run this code syms x m eq3 = m-['(heaviside(x)-heaviside(x-1))*x^3/6 ']; %... % '(heaviside(x-1)-heaviside(x-2))*(1/6)*(-3*x) + ' ... % ...
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### slopes of piecewise regression
We want to make a piecewise regression and find the slopes for each piece of the regression. We have a dataset like this: df = data.frame ( ID = c(1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2), time = ... | 651 | 2,384 | {"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} | 3.03125 | 3 | CC-MAIN-2013-48 | latest | en | 0.84293 |
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Contributor II
## Line chart : Average value from start until value of point
I have a table with these fields Symbol,Code,Value:
Symbol Code Value S1 A1 110 S1 A2 100 S1 A5 105 S1 A6 108 S2 B3 120 S2 B5 150
I want to show a line chart with dim="Symbol" and exp="Value/AVG", that AVG is average of value before current value. for example for symbol=S1 and Code=A5 the AVG is 105=(110+100)/2 and for Symbol=S1 and Code=A6, AVG is 105.75=(110+100+105+108)/4.
How I can do this?
1 Solution
Accepted Solutions
MVP
Something like this?
3 Replies
MVP
Do you mean to have Code as your dimension?
MVP
Something like this?
Contributor II
Author
Thanks. You help me alot.
Community Browser | 220 | 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} | 3.03125 | 3 | CC-MAIN-2024-10 | latest | en | 0.804121 |
https://wiki.math.ucr.edu/index.php/009B_Sample_Midterm_3,_Problem_3_Detailed_Solution | 1,638,230,929,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358847.80/warc/CC-MAIN-20211129225145-20211130015145-00270.warc.gz | 679,676,124 | 7,699 | # 009B Sample Midterm 3, Problem 3 Detailed Solution
Find a curve ${\displaystyle y=f(x)}$ with the following properties:
(i) ${\displaystyle f''(x)=6x}$
(ii) Its graph passes through the point ${\displaystyle (0,1)}$ and has a horizontal tangent there.
Background Information:
1. If the graph of ${\displaystyle f(x)}$ passes through the point ${\displaystyle (a,b),}$ then ${\displaystyle f(a)=b.}$
2. If ${\displaystyle f(x)}$ has a horizontal tangent at the point ${\displaystyle (a,b),}$ then ${\displaystyle f'(a)=0.}$
Solution:
Step 1:
Since ${\displaystyle f(x)}$ passes through the point ${\displaystyle (0,1),}$
${\displaystyle f(0)=1.}$
Since ${\displaystyle f(x)}$ has a horizontal tangent at ${\displaystyle (0,1),}$
${\displaystyle f'(0)=0.}$
Step 2:
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\int f''(x)~dx}\\&&\\&=&\displaystyle {\int 6x~dx}\\&&\\&=&\displaystyle {3x^{2}+C.}\\\end{array}}}$
Since ${\displaystyle f'(0)=0,}$ we have
${\displaystyle {\begin{array}{rcl}\displaystyle {0}&=&\displaystyle {f'(0)}\\&&\\&=&\displaystyle {3(0)^{2}+C}\\&&\\&=&\displaystyle {C.}\\\end{array}}}$
Hence,
${\displaystyle f'(x)=3x^{2}.}$
Step 3:
We have
${\displaystyle {\begin{array}{rcl}\displaystyle {f(x)}&=&\displaystyle {\int f'(x)~dx}\\&&\\&=&\displaystyle {\int 3x^{2}~dx}\\&&\\&=&\displaystyle {x^{3}+D.}\\\end{array}}}$
Since ${\displaystyle f(0)=1,}$ we have
${\displaystyle {\begin{array}{rcl}\displaystyle {1}&=&\displaystyle {f(0)}\\&&\\&=&\displaystyle {(0)^{3}+D}\\&&\\&=&\displaystyle {D.}\\\end{array}}}$
Hence,
${\displaystyle f(x)=x^{3}+1.}$
${\displaystyle f(x)=x^{3}+1}$ | 642 | 1,687 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 24, "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} | 4.625 | 5 | CC-MAIN-2021-49 | latest | en | 0.565233 |
http://smlnj-gforge.cs.uchicago.edu/scm/viewvc.php/sml/trunk/src/MLRISC/library/union-find.sml?view=markup&revision=245&root=smlnj&sortdir=down&pathrev=1054 | 1,563,863,918,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528869.90/warc/CC-MAIN-20190723043719-20190723065719-00150.warc.gz | 138,997,380 | 4,763 | Home My Page Projects Code Snippets Project Openings SML/NJ
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View of /sml/trunk/src/MLRISC/library/union-find.sml
Sat Apr 17 18:47:12 1999 UTC (20 years, 3 months ago) by monnier
Original Path: sml/branches/SMLNJ/src/MLRISC/library/union-find.sml
File size: 765 byte(s)
```version 110.16
```
```signature UNION_FIND =
sig
type 'a union_find
val union_find : int * (int -> 'a) -> 'a union_find
val find : 'a union_find -> int -> 'a
val union' : 'a union_find -> int * int -> bool
val union : 'a union_find -> ('a * 'a -> 'a) -> int * int -> bool
val == : 'a union_find -> int * int -> bool
end
structure Unionfind :> UNION_FIND =
struct
structure A = Array
structure U = UnionFindRef
type 'a union_find = 'a U.uref A.array
fun union_find (n,f) = A.tabulate(n,(fn i => U.uref(f i)))
fun find U x = U.!!(A.sub(U,x))
fun union' U (x,y) = U.union' (A.sub(U,x),A.sub(U,y))
fun union U f (x,y) = U.union f (A.sub(U,x),A.sub(U,y))
fun == U (x,y) = U.==(A.sub(U,x),A.sub(U,y))
end
``` | 411 | 1,227 | {"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-2019-30 | latest | en | 0.414823 |
https://www.crumplab.com/statistics/10-MixedANOVA.html | 1,695,547,156,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506632.31/warc/CC-MAIN-20230924091344-20230924121344-00824.warc.gz | 798,915,264 | 13,520 | # 10More On Factorial Designs
Author
Matthew J. C. Crump
We are going to do a couple things in this chapter. The most important thing is more exposure to factorial designs. The second thing we do is show that you can mix it up with ANOVA. You already know that you can have more than one IV. And, you know that research designs can be between-subjects or within-subjects (repeated-measures). When you have more than one IV, they can all be between-subjects variables, they can all be within-subject repeated measures, or they can be a mix: say one between-subject variable and one within-subject variable. You can use ANOVA to analyze all of these kinds of designs. You always get one main effect for each IV, and a number of interactions, or just one, depending on the number of IVs.
## 10.1 Looking at main effects and interactions
Designs with multiple factors are very common. When you read a research article you will often see graphs that show the results from designs with multiple factors. It would be good for you if you were comfortable interpreting the meaning of those results. The skill here is to be able to look at a graph and see the pattern of main effects and interactions. This skill is important, because the patterns in the data can quickly become very complicated looking, especially when there are more than two independent variables, with more than two levels.
### 10.1.1 2x2 designs
Let’s take the case of 2x2 designs. There will always be the possibility of two main effects and one interaction. You will always be able to compare the means for each main effect and interaction. If the appropriate means are different then there is a main effect or interaction. Here’s the thing, there a bunch of ways all of this can turn out. Check out the ways, there are 8 of them:
1. no IV1 main effect, no IV2 main effect, no interaction
2. IV1 main effect, no IV2 main effect, no interaction
3. IV1 main effect, no IV2 main effect, interaction
4. IV1 main effect, IV2 main effect, no interaction
5. IV1 main effect, IV2 main effect, interaction
6. no IV1 main effect, IV2 main effect, no interaction
7. no IV1 main effect, IV2 main effect, interaction
8. no IV1 main effect, no IV2 main effect, interaction
OK, so if you run a 2x2, any of these 8 general patterns could occur in your data. That’s a lot to keep track of isn’t. As you develop your skills in examining graphs that plot means, you should be able to look at the graph and visually guesstimate if there is, or is not, a main effect or interaction. You will need you inferential statistics to tell you for sure, but it is worth knowing how to know see the patterns.
Figure 10.1 shows the possible patterns of main effects and interactions in bar graph form. Here is a legend for the labels in the panels.
• 1 = there was a main effect for IV1.
• ~1 = there was not a main effect for IV1
• 2 = there was a main effect for IV2
• ~2 = there was not a main effect of IV2
• 1x2 = there was an interaction
• ~1x2 = there was not an interaction
Figure 10.2 shows the same eight patterns in line graph form:
The line graphs accentuates the presence of interaction effects. Whenever the lines cross, or would cross if they kept going, you have a possibility of an interaction. Whenever the lines are parallel, there can’t be an interaction. When both of the points on the A side are higher or lower than both of the points on the B side, then you have a main effect for IV1 (A vs B). Whenever the green line is above or below the red line, then you have a main effect for IV2 (1 vs. 2). We know this is complicated. You should see what all the possibilities look like when we start adding more levels or more IVs. It gets nuts. Because of this nuttiness, it is often good practice to make your research designs simple (as few IVs and levels as possible to test your question). That way it will be easier to interpret your data. Whenever you see that someone ran a 4x3x7x2 design, your head should spin. It’s just too complicated.
## 10.2 Interpreting main effects and interactions
The interpretation of main effects and interactions can get tricky. Consider the concept of a main effect. This is the idea that a particular IV has a consistent effect. For example, drinking 5 cups of coffee makes you more awake compared to not drinking 5 cups of coffee. The main effect of drinking 5 cups of coffee vs not drinking coffee will generally be true across the levels of other IVs in our life. For example, let’s say you conducted an experiment testing whether the effect of drinking 5 cups of coffee vs not, changes depending on whether you are in your house or in a car. Perhaps the situation matters? No, probably not so much. You will probably still be more awake in your house, or your car, after having 5 cups of coffee, compared to if you hadn’t.
The coffee example is a reasonably good example of a consistent main effect. Another silly kind of example might be the main effect of shoes on your height. For example, if your IV was wearing shoes or not, and your DV was height, then we could expect to find a main effect of wearing shoes on your measurement of height. When you wear shoes, you will become taller compared to when you don’t wear shoes. Wearing shoes adds to your total height. In fact, it’s hard to imagine how the effect of wearing shoes on your total height would ever interact with other kinds of variables. You will be always be that extra bit taller wearing shoes. Indeed, if there was another manipulation that could cause an interaction that would truly be strange. For example, imagine if the effect of being inside a bodega or outside a bodega interacted with the effect of wearing shoes on your height. That could mean that shoes make you taller when you are outside a bodega, but when you step inside, your shoes make you shorter…but, obviously this is just totally ridiculous. That’s correct, it is often ridiculous to expect that one IV will have an influence on the effect of another, especially when there is no good reason.
The summary here is that it is convenient to think of main effects as a consistent influence of one manipulation. However, when an interaction is observed, this messes up the consistency of the main effect. That is the very definition of an interaction. It means that some main effect is not behaving consistently across different situations. Indeed, whenever we find an interaction, sometimes we can question whether or not there really is a general consistent effect of some manipulation, or instead whether that effect only happens in specific situations.
For this reason, you will often see that researchers report their findings this way:
“We found a main effect of X, BUT, this main effect was qualified by an interaction between X and Y”.
Notice the big BUT. Why is it there? The sentence points out that before they talk about the main effect, they need to first talk about the interaction, which is making the main effect behave inconsistently. In other words, the interpretation of the main effect depends on the interaction, the two things have to be thought of together to make sense of them.
Here are two examples to help you make sense of these issues:
### 10.2.1 A consistent main effect and an interaction
Figure 10.3 shows a main effect and interaction. There is a main effect of IV2: the level 1 means (red points and line) are both lower than the level 2 means (aqua points and line). There is also an interaction. The size of the difference between the red and aqua points in the A condition (left) is bigger than the size of the difference in the B condition.
How would we interpret this? We could say there WAS a main effect of IV2, BUT it was qualified by an IV1 x IV2 interaction.
What’s the qualification? The size of the IV2 effect changed as a function of the levels of IV1. It was big for level A, and small for level B of IV1.
What does the qualification mean for the main effect? Well, first it means the main effect can be changed by the other IV. That’s important to know. Does it also mean that the main effect is not a real main effect because there was an interaction? Not really, there is a generally consistent effect of IV2. The green points are above the red points in all cases. Whatever IV2 is doing, it seems to work in at least a couple situations, even if the other IV also causes some change to the influence.
### 10.2.2 An inconsistent main effect and an interaction
Figure 10.4 shows another 2x2 design. You should see an interaction here straight away. The difference between the aqua and red points in condition A (left two dots) is huge, and there is 0 difference between them in condition B. Is there an interaction? Yes!
Are there any main effects here? With data like this, sometimes an ANOVA will suggest that you do have significant main effects. For example, what is the mean difference between level 1 and 2 of IV2? That is the average of the green points ( (10+5)/2 = 15/2= 7.5 ) compared to the average of the red points (5). There will be a difference of 2.5 for the main effect (7.5 vs. 5).
Starting to see the issue here? From the perspective of the main effect (which collapses over everything and ignores the interaction), there is an overall effect of 2.5. In other words, level 2 adds 2.5 in general compared to level 1. However, we can see from the graph that IV2 does not do anything in general. It does not add 2.5s everywhere. It adds 5 in condition A, and nothing in condition B. It only does one thing in one condition.
What is happening here is that a “main effect” is produced by the process of averaging over a clear interaction.
How would we interpret this? We might have to say there was a main effect of IV2, BUT we would definitely say it was qualified by an IV1 x IV2 interaction.
What’s the qualification? The size of the IV2 effect completely changes as a function of the levels of IV1. It was big for level A, and nonexistent for level B of IV1.
What does the qualification mean for the main effect? In this case, we might doubt whether there is a main effect of IV2 at all. It could turn out that IV2 does not have a general influence over the DV all of the time, it may only do something in very specific circumstances, in combination with the presence of other factors.
## 10.3 Mixed Designs
Throughout this book we keep reminding you that research designs can take different forms. The manipulations can be between-subjects (different subjects in each group), or within-subjects (everybody contributes data in all conditions). If you have more than one manipulation, you can have a mixed design when one of your IVs is between-subjects and one of the other ones is within-subjects.
The only “trick” to these designs is to use the appropriate error terms to construct the F-values for each effect. Effects that have a within-subjects repeated measure (IV) use different error terms than effects that only have a between-subject IV. In principle, you could run an ANOVA with any number of IVs, and any of them good be between or within-subjects variables.
Because this is an introductory textbook, we leave out a full discussion on mixed designs. What we are leaving out are the formulas to construct ANOVA tables that show how to use the correct error terms for each effect. There are many good more advanced textbooks that discuss these issues in much more depth. And, these things can all be Googled. This is a bit of a cop-out on our part, and we may return to fill in this section at some point in the future (or perhaps someone else will add a chapter about this).
In the lab manual, you will learn how to conduct a mixed design ANOVA using software. Generally speaking, the software takes care of the problem of using the correct error terms to construct the ANOVA table.
## 10.4 More complicated designs
Up until now we have focused on the simplest case for factorial designs, the 2x2 design, with two IVs, each with 2 levels. It is worth spending some time looking at a few more complicated designs and how to interpret them.
### 10.4.1 2x3 design
In a 2x3 design there are two IVs. IV1 has two levels, and IV2 has three levels. Typically, there would be one DV. Let’s talk about the main effects and interaction for this design.
First, let’s make the design concrete. Let’s imagine we are running a memory experiment. We give people some words to remember, and then test them to see how many they can correctly remember. Our DV is proportion correct. We know that people forget things over time. Our first IV will be time of test, immediate vs. 1 week. The time of test IV will produce a forgetting effect. Generally, people will have a higher proportion correct on an immediate test of their memory for things they just saw, compared to testing a week later.
We might be interested in manipulations that reduce the amount of forgetting that happens over the week. The second IV could be many things. Let’s make it the number of time people got to study the items before the memory test, once, twice or three times. We call IV2 the repetition manipulation.
We might expect data like shown in Figure 10.5:
The figure shows some pretend means in all conditions. Let’s talk about the main effects and interaction.
First, the main effect of delay (time of test) is very obvious, the red line is way above the aqua line. Proportion correct on the memory test is always higher when the memory test is taken immediately compared to after one week.
Second, the main effect of repetition seems to be clearly present. The more times people saw the items in the memory test (once, twice, or three times), the more they remembered, as measured by increasingly higher proportion correct as a function of number of repetitions.
Is there an interaction? Yes, there is. Remember, an interaction occurs when the effect of one IV depends on the levels of an another. The delay IV measures the forgetting effect. Does the size of the forgetting effect change across the levels of the repetition variable? Yes it does. With one repetition the forgetting effect is .9-.6 =.4. With two repetitions, the forgetting effect is a little bit smaller, and with three, the repetition is even smaller still. So, the size of the forgetting effect changes as a function of the levels of the repetition IV. There is evidence in the means for an interaction. You would have to conduct an inferential test on the interaction term to see if these differences were likely or unlikely to be due to sampling error.
If there was no interaction and no main effect of repetition, we would see something like the pattern in Figure 10.6.
What would you say about the interaction if you saw the pattern in Figure 10.7?
The correct answer is that there is evidence in the means for an interaction. Remember, we are measuring the forgetting effect (effect of delay) three times. The forgetting effect is the same for repetition condition 1 and 2, but it is much smaller for repetition condition 3. The size of the forgetting effect depends on the levels of the repetition IV, so here again there is an interaction.
### 10.4.2 2x2x2 designs
Let’s take it up a notch and look at a 2x2x2 design. Here, there are three IVs with 2 levels each. There are three main effects, three two-way (2x2) interactions, and one 3-way (2x2x2) interaction.
We will use the same example as before but add an additional manipulation of the kind of material that is to be remembered. For example, we could present words during an encoding phase either visually or spoken (auditory) over headphones.
Figure 10.8 has two panels one for auditory and one for visual. You can think of the 2x2x2, as two 2x2s, one for auditory and one for visual. What’s the take home from this example data? We can see that the graphs for auditory and visual are the same. They both show a 2x2 interaction between delay and repetition. People forgot more things across the week when they studied the material once, compared to when they studied the material twice. There is a main effect of delay, there is a main effect of repetition, there is no main effect of modality, and there is no three-way interaction.
What is a three-way interaction anyway? That would occur if there was a difference between the 2x2 interactions. For example, consider the pattern of results in Figure 10.9.
We are looking at a 3-way interaction between modality, repetition and delay. What is going on here? These results would be very strange, here is an interpretation.
For auditory stimuli, we see that there is a small forgetting effect when people studied things once, but the forgetting effect gets bigger if they studies things twice. A pattern like this would generally be very strange, usually people would do better if they got to review the material twice.
The visual stimuli show a different pattern. Here, the forgetting effect is large when studying visual things once, and it get’s smaller when studying visual things twice.
We see that there is an interaction between delay (the forgetting effect) and repetition for the auditory stimuli; BUT, this interaction effect is different from the interaction effect we see for the visual stimuli. The 2x2 interaction for the auditory stimuli is different from the 2x2 interaction for the visual stimuli. In other words, there is an interaction between the two interactions, as a result there is a three-way interaction, called a 2x2x2 interaction.
We will note a general pattern here. Imagine you had a 2x2x2x2 design. That would have a 4-way interaction. What would that mean? It would mean that the pattern of the 2x2x2 interaction changes across the levels of the 4th IV. If two three-way interactions are different, then there is a four-way interaction. | 3,977 | 17,788 | {"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} | 3.59375 | 4 | CC-MAIN-2023-40 | latest | en | 0.945129 |
https://bsiegelwax.medium.com/comparing-entangled-states-f39681864ea5 | 1,618,136,721,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038061820.19/warc/CC-MAIN-20210411085610-20210411115610-00127.warc.gz | 268,815,845 | 31,753 | Comparing Entangled States
The SWAP Test is used to compare quantum states. It is constructed out of a Fredkin gate (controlled-SWAP) sandwiched between two Hadamard gates. The control qubit and Hadamard gates are placed on an ancilla qubit, and the SWAP gates are placed on the qubits that you want to compare. You measure only the ancilla qubit.
Two identical states measure 0 with a probability of 1, and two maximally-different states measure 0 with a probability of 50%. All other results either show how close the states are by having a probability of measuring 0 that is closer to 1, or how different they are by having a probability of measuring 0 that is closer to one-half.
The SWAP Test can also be used to compare entangled states. For purposes of this article, I’m just using the most basic GHZ states. If you have read the supplementary articles listed above, you’ll see that the approach is remarkably similar. You simply make a few adjustments to handle the additional qubits.
This SWAP Test compares the two most basic GHZ states: 000 and 111. The Hadamard gates and measurement are still on the ancilla qubit, but now there are three Fredkin gates. If you look at the lowercase “b” and “c,” you’ll see that the first Fredkin gate compares the top two qubits, the second compares the middle two qubits, and the last compares the bottom two qubits.
As should be expected from a SWAP Test, two maximally-different entangled states measure 0 with a probability of about 50%.
This is the same circuit as before, except that one GHZ state has a 1/3 pi rotation around the y axis and the other has a 2/3 pi rotation around the y axis.
The probability of measuring 0 noticeably increased as the entangled states were no longer maximally-different, but also not identical either.
This circuit keeps the GHZ state with the 1/3 pi rotation around the y axis, but substitutes a 1/4 pi rotation for the 2/3 pi rotation.
Because the two states are much closer together, the resultant histogram shows them as almost-but-not-quite identical.
This circuit keeps the GHZ states with the 1/3 pi and 1/4 pi rotations around the y axis, but adds a 2/3 pi rotation around the z axis for the latter.
Because of the rotation around the z axis, the result shows that the states have noticeably moved apart.
But, are the previous four circuits really comparing entangled states? All three qubits in each GHZ state should ideally measure the same, so what if I only measure the first qubit of each GHZ state, the ones to which I applied the rotations?
Comparing this histogram to the immediately previous one, we can see that there is indeed a difference between comparing the entire GHZ states and comparing only one qubit from each GHZ state.
But, what if we perform three SWAP Tests, separately comparing all three qubits in each GHZ state?
Any evidence of entanglement disappears. The SWAP Tests individually produce roughly similar results, but not in the synchronized way that you should expect from entangled states. Although this looks like a GHZ state that was run on a NISQ (keyword: noisy) device, it was actually run on the IBM Q simulator with no added noise model. Therefore, we are seeing an absence of entanglement, not the presence of noise.
Future Work
The circuits in this article only compare the most basic GHZ states. Future articles may show more practical comparisons, perhaps applicable to Quantum Machine Learning (QML) and Quantum Chemistry.
More from Brian N. Siegelwax
Independent Quantum Algorithm Designer https://www.linkedin.com/in/brian-siegelwax https://twitter.com/BSiegelwax?s=09 https://github.com/bsiegelwax
Convolutional Neural Networks for Multiclass Image Classification — A Beginners Guide to Understand
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# Mathematics – Class 8 – Chapter 13 – Direct and Inverse Proportion – Exercise 13.2 – NCERT Exercise Solution
1. Which of the following are in inverse proportion?
(i) The number of workers on a job and the time to complete the job.
Solution:
They are inversely proportional because if the number of workers increase the job will take less time to complete, also less workers will take more time to complete the job.
(ii) The time taken for a journey and the distance travelled in a uniform speed.
Solution:
They are not inversely proportional because more time require for more distance. Hence, time and distance covered in direct proportion.
(iii) Area of cultivated land and the crop harvested.
Solution:
They are not inversely proportional because more area of cultivated land will yield more crops. Hence, they are direct proportion.
(iv) The time taken for a fixed journey and the speed of the vehicle.
Solution:
They are inversely proportional because it will take less time to complete the journey if the speed is increase.
(v) The population of a country and the area of land per person.
Solution:
They are inversely proportional because If the population of a country increases, then the area of land per person will be decrease.
2. In a Television game show, the prize money of Rs.1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners:
Solution:
We observe that:
1 x 100,000 = 2 x 50,000
1,00,000 = 1,00,000
Here, number of winners and prize money are in inverse proportion because winners are increasing, prize money is decreasing.
Now,
Let the blank spaces be a, b, c, d and e respectively.
2 x 20,000 = 4 x a
a = (2 x 50,000)/4
= 25,000
2 x 50,000 = 5 x b
b = (2 x 50,000)/5
= 20,000
2 x 50,000 = 8 x c
c = (2 x 50,000)/8
= 12,500
2 x 50,000 = 10 x d
d = (2 x 50,000)/10
= 10,000
2 x 50,000 = 20 x e
e = (2 x 50,000)/20
= 5,000
3. Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table:
(i) Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40 degree?
Solution:
From the table we observe that the number of spokes are increasing and the angle between a pair of consecutive spokes is decreasing. So, it is inversely proportion and angle at the centre of a circle is 360o.
4 x 90o = 6 x 60o
360o = 360o
Let the blank space can say a, b and c
4 x 90o = 8 x a
a = 45o
4 x 90o = 10 x b
b = 36o
4 x 90o = 12 x c
c = 30o
required table is:
Now,
(i) Yes, they are in inversely proportion.
(ii) When the number of spokes is 15, then angle between a pair of consecutive spokes
4 x 90o = 15 x X
x = 24o.
(iii) The number of spokes would be needed:
If the angle between two consecutive spokes is 400, then
4 x 90o = y x 40o
Y = 9
Hence, the required number of spokes = 9
4. If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?
Solution:
According to question:
They are inversely proportional.
Let required number of sweets be b
24 x 5 = 20 x b
b = 24 x 5 /20
=6
Hence, the required number of sweets = 6
5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Solution:
According to question:
They are inversely proportional.
Let the required number of days be b.
Total number of animals = 20+10 = 30
Now,
20 x 6 = 30 x b
b = 4
Hence, the food will last for four days.
6. A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?
Solution:
According to question:
Here, it will take less number of days to complete the job when the number of persons will increase. Hence, they are inversely proportional.
Let time taken to complete the job be “b” days.
Now,
3 x 4 = 4 x b
b = 3
Hence, 4 persons will take 3 days to complete the job.
7. A batch of bottles was packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?
Solution:
According to question:
They are inversely proportional.
Let the required number of boxes be b.
Now,
25 x 12 = b x 20
b = 15
Hence, 15 boxes would be filled.
8. A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?
Solution:
According to question:
They are inversely proportional.
Let the required number of machines be b.
Now,
42 x 63 = b x 54
b = 49
Hence, 49 machines would be required to produce the article.
9. A car takes 2 hours to reach a destination by travelling at the speed of 60 km/hr. How long will it take when the car travels at the speed of 80 km/hr?
Solution:
According to question:
They are inversely proportional.
Let the required number of hours be b.
Now,
60 x 2 = 80 x b
b = (60 x 2)/80
= 3/2
= 1.5 h
Hence, required time = 1.5 hours
10. Two persons could fit new windows in a house in 3 days.
(i) One of the persons fell ill before the work started. How long would the job take now?
(ii) How many persons would be needed to fit the windows in one day?
Solution:
According to question:
(i)From the table, we observe that they are inversely proportional.
Let the required number of days be b.
2/1 = b/3
b = 6
Hence, the required number of days = 6
(ii)From the table, we observe that they are inversely proportional.
Let the required number of persons be p.
2/p = 1/3
p = 6
Hence, the required number of persons = 6
11. A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?
Solution:
According to question:
They are inversely proportional.
Let the required duration of each period be b.
8/9 = b/45
8×45 = 9b
b = 40
Hence, the required duration of each period would be 40 minutes.
👍👍👍
error: | 1,775 | 6,644 | {"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} | 4.6875 | 5 | CC-MAIN-2023-50 | latest | en | 0.927446 |
https://swoopfunding.com/uk/business-glossary/accounting-rate-of-return/ | 1,723,564,044,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641079807.82/warc/CC-MAIN-20240813141635-20240813171635-00224.warc.gz | 431,130,924 | 54,127 | # Accounting rate of return (ARR)
Page written by AI. Reviewed internally on July 11, 2024.
### Definition
The accounting rate of return (ARR) is a financial metric used to evaluate the profitability of an investment project or asset by comparing the average accounting profit generated by the investment to the initial investment cost.
### What is the accounting rate of return?
ARR is often expressed as a percentage and provides insight into the average annual return generated by the investment relative to its initial cost.
To calculate the accounting rate of return, the following formula is typically used:
ARR = (Average accounting profit / Initial Investment) x 100
ARR is relatively simple to calculate and understand compared to other investment appraisal methods, making it a popular metric for small businesses or projects where complex financial analysis may not be feasible.
Keep in mind that ARR does not explicitly consider the time value of money, as it does not discount future cash flows back to their present value. This can be a limitation, especially when comparing investment projects with different cash flow timing.
While ARR may not be suitable for comparing investment projects with different durations or cash flow profiles, it can be useful for comparing similar projects or investment alternatives within an organisation.
##### Accounting rate of return vs. required rate of return
The accounting rate of return (ARR) compares a project’s average accounting profit to its initial investment. It measures profitability based on accounting metrics but does not account for the time value of money or risk. In contrast, the required rate of return (RRR) is the minimum return an investor expects to earn from an investment to compensate for its risk. It considers factors like opportunity cost, inflation, and the project’s risk profile. RRR guides investment decisions by making sure projects meet or exceed the investor’s expectations for profitability relative to the associated risk.
##### What’s the difference between accounting rate of return and internal rate of return?
The accounting rate of return (ARR) and internal rate of return (IRR) are both metrics used in financial analysis but differ in their approach and insights. ARR calculates a project’s profitability based on accounting measures, specifically comparing average accounting profit to the initial investment. In contrast, IRR determines the discount rate at which the net present value (NPV) of cash flows from a project equals zero. It incorporates the time value of money, providing a more comprehensive assessment of a project’s potential return relative to its risk.
### Example of the accounting rate of return
Let’s say a company invests £50,000 in a new project. Over the next five years, the project generates the following annual accounting profits:
• Year 1: £10,000
• Year 2: £12,000
• Year 3: £14,000
• Year 4: £16,000
• Year 5: £18,000
Calculate the average accounting profit:
Average accounting profit = (£10,000 + £12,000 + £14,000 + £16,000 + £18,000) / 5 = £14,000
Use the formula to calculate ARR:
ARR = (£14,000 / £50,000) x 100 = 28%
So, the accounting rate of return (ARR) for this project is 28%. This means that, on average, the project generates a return of 28% per year relative to its initial investment of £50,000. | 734 | 3,399 | {"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} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.927048 |
https://www.johncanessa.com/2020/12/08/balance-brackets/ | 1,656,497,075,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00297.warc.gz | 908,235,418 | 14,205 | # Balance Brackets
It is pitch dark in the Twin Cities of Minneapolis and St. Paul. Before calling it a day I decided to try one more practice problem from the Facebook web site. Balance Brackets is a very common problem. In general the idea is that you are given a set of brackets and are sked to determine if the brackets are balanced.
I knew I have solved similar versions of the problem. When done with the code I move to a different computer and type in the contents for the post. I looked up in my web site for the string “balanced brackets” and found Balanced Brackets and Balanced Brackets – Possible Second Attempt. I generated the post in 2016 and 2017 respectively. I guess, we will have a third version.
I will solve the problem using the Java 8 programming language on a Windows 10 machine using the VSCode IDE. If you are planning on interviewing with Facebook I would suggest writing your code directly in the IDE presented by Facebook. Most (never generalize) IDEs have auto completion and when you decide to use a standard class the import is generated automatically. This is not the case with the Facebook IDE.
```A bracket is considered to be any one of the following characters: (, ), {, }, [, or ].
We consider two brackets to be matching if the first element is an open-bracket, e.g., (, {, or [,
and the second bracket is a close-bracket of the same type,
e.g., ( and ), [ and ], and { and } are the only pairs of matching brackets.
Furthermore, a sequence of brackets is said to be balanced if the following conditions are met:
o The sequence is empty, or
o The sequence is composed of two, non-empty, sequences both of which are balanced, or
o The first and last brackets of the sequence are matching,
and the portion of the sequence without the first and last elements is balanced.
You are given a string of brackets.
Your task is to determine whether each sequence of brackets is balanced.
If a string is balanced, return true, otherwise, return false.
```
Not much to say. The requirements seem to be simple and quite clear.
```boolean isBalanced(String s) {
}
```
This is the signature of the function we need to complete. We are presented with the input string and we need to return true if the brackets are balanced or false if not.
```{[()]}
main <<< s ==>{[()]}<==
main <<< isBalanced: true
{}()
main <<< s ==>{}()<==
main <<< isBalanced: true
{(})
main <<< s ==>{(})<==
main <<< isBalanced: false
)
main <<< s ==>)<==
main <<< isBalanced: false
(
main <<< s ==>(<==
main <<< isBalanced: false
```
The requirements provide the first 4 examples. I added the fifth.
Since I solved the problem in my computer I needed a test scaffolding to accept a string, pass it to the function and display the results. The test scaffolding I wrote IS NOT PART OF THE SOLUTION.
``` /**
* Test scaffolding
*
* @throws IOException
*/
public static void main(String[] args) throws IOException {
// **** open buffered reader ****
// **** close buffered reader ****
br.close();
// ???? ????
System.out.println("main <<< s ==>" + s + "<==");
// **** process string and display result ****
System.out.println("main <<< isBalanced: " + isBalanced(s));
}
```
The idea is simple. We read in a string and display it to make sure all is well so far. We then call the function we need to complete and display the returned value.
``` /**
* If a string is balanced, return true, otherwise, return false.
* Runtime O(n)
*/
static boolean isBalanced(String s) {
// **** sanity check(s) ****
if (s.length() == 0)
return true;
if (s.length() == 1)
return false;
// **** initialization ****
Stack<Character> stack = new Stack<>();
// **** process the string one character at a time ****
for (char ch : s.toCharArray()) {
// **** closing bracket (pop from stack)****
if (ch == ')' || ch == ']' || ch == '}') {
// **** check if stack is empty ****
if (stack.isEmpty())
return false;
// **** pop top from stack (if matching) ****
if (ch == ')' && stack.peek() == '(')
stack.pop();
else if (ch == ']' && stack.peek() == '[')
stack.pop();
else if (ch == '}' && stack.peek() == '{')
stack.pop();
else
return false;
}
// **** opening bracket (push to stack) ****
else if (ch == '(' || ch == '[' || ch == '{') {
stack.push(ch);
}
// **** not a bracket ****
else {
System.out.println("isBalanced <<< unexpected ch: " + ch);
return false;
}
}
// **** check if stack is not empty ****
if (!stack.isEmpty())
return false;
// **** brackets are balanced ****
return true;
}
```
I like to start with sanity checks. They tend to speed up processing if you get a lot of noise. In this case we have two test cases. Not sure if it made much sense.
We will use a stack to keep track of the opening and closing brackets.
We loop through all the characters in the input string. We push an opening bracket, we pop from the stack a matching closing bracket, or we just exit if we get some other character.
For an opening bracket we just push it into the stack. For a closing bracket we first check if the stack is empty. If not we match the character at the top of the stack to match the current character.
There is one more thing we need to do after we exit the loop. We need to make sure the stack is empty. We could have received more opening brackets that closing ones. For this reason we need to check if the stack is not empty. This is why I added the last test case.
Please note that since the software had two test cases, my solution code might have issues. If you find a problem please leave me a message so I can address it.
Hope you enjoyed solving this problem as much as I did. The entire code for this project can be found in my GitHub repository.
If you have comments or questions regarding this, or any other post in this blog, or if you would like for me to help out with any phase in the SDLC (Software Development Life Cycle) of a project associated with a product or service, please do not hesitate and leave me a note below. If you prefer, send me a private e-mail message. I will reply as soon as possible.
Keep on reading and experimenting. It is the best way to learn, become proficient, refresh your knowledge and enhance your developer toolset!
One last thing, many thanks to all 4,873 subscribers to this blog!!!
Keep safe during the COVID-19 pandemic and help restart the world economy.
Regards;
John
john.canessa@gmail.com
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,497 | 6,458 | {"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} | 3.125 | 3 | CC-MAIN-2022-27 | longest | en | 0.933601 |
https://alison.com/course/proyecto-matematicas-strand-1-probabilidad | 1,571,663,369,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987773711.75/warc/CC-MAIN-20191021120639-20191021144139-00328.warc.gz | 358,309,379 | 50,524 | en
# Leaving Certificate - Probability and Statistics Higher Level
## This course covers Probability and Statistics for the Irish Leaving Certificate maths curriculum.
Probability and Statistics
Free Course
This course covers Probability and Statistics for the Irish Leaving Certificate maths curriculum.
2-3 Hours
Yes
Yes
XSIQ
## Description
Probability and Statistics is one of two strands introduced in the first phase of the new Project Maths Course in the Irish curriculum. This topic covers up to half of the new Paper 2 in the Leaving Certificate Paper.
Statistics are used in real life to make sense of the information around us and how it affects us. Statistics looks at the data handling cycle and analysis of the data collected. This involves posing a question, collecting data on that question, presenting that data, analysing the data (using measures of spread and centre) and interpreting the results. In answering questions, it is essential that you can contextualise and justify your findings.
Probability is concerned with the likelihood of an event(s) happening. The information can be used to make informed decisions. The use of probability is commonly utilised in the world of finance, insurance and sport among others. Probability can also be used to infer the fairness of an event or series of events. It can be evaluated using a diagram or a rule-based approach. A combination of Probability and Statistics can be used to prove/disprove a given conjecture or statement (Hypothesis Testing (HL only)).
This Strand attempts to merge the mathematical aspects of Probability and Statistics with its real-life application. It is an interesting topic that is very accessible to all students.
Start Course Now
## Learning Outcomes
• - count the arrangements of n distinct objects (n!)
• - count the number of ways of arranging r objects from n distinct objects
• - count the number of ways of selecting r objects from n distinct objects
1.2 Concepts of probability
• - discuss basic rules of probability (AND/OR, mutually exclusive) through the use of Venn diagrams
• - calculate expected value and understand that this does not need to be one of the outcomes
• - recognise the role of expected value in decision making and explore the issue of fair games
• - extend their understanding of the basic rules of probability (AND/OR, mutually exclusive) through the use of formulae
• - use the Addition Rule, Multiplication Rule (Independent events), Multiplication Rule (General case)
• - solve problems involving conditional probability in a systematic way
1.3 Outcomes of random processes
• - find the probability that two independent events both occur
• - apply an understanding of Bernoulli trials
• - solve problems involving up to 3 Bernoulli trials
• - calculate the probability that the 1st success occurs on the nth Bernoulli trial where n is specified
• - solve problems involving calculating the probability of k successes in n repeated Bernoulli trials (normal approximation not required)
• - calculate the probability that the kth success occurs on the nth Bernoulli trial
• - use simulations to explore the variability of sample statistics from a known population and to construct sampling distributions
• - solve problems involving reading probabilities from the normal distribution tables
1.4 Statistical reasoning with an aim to becoming a statistically aware consumer
• - work with different types of bivariate data
1.5 Finding, collecting and organising data
• - discuss different types of studies: sample surveys, observational studies and designed experiments
• - design a plan and collect data on the basis of above knowledge
• - recognise the importance of randomisation and the role of the control group in studies
• - recognise biases, limitations and ethical issues of each type of study
• - select a sample (stratified, cluster, quota – no formulae required, just definitions of these)
• - design a plan and collect data on the basis of above knowledge
1.6 Representing data graphically and numerically1.6a Graphical
• - describe the sample (both univariate and bivariate data) by selecting appropriate graphical or numerical methods
• - explore the distribution of data, including concepts of symmetry and skewness
• - compare data sets using appropriate displays, including back-to-back stem and leaf plots
• - determine the relationship between variables using scatterplots
• - recognise that correlation is a value from -1 to +1 and that it measures the extent of the linear relationship between two variables
• - match correlation coefficient values to appropriate scatter plots
• - understand that correlation does not imply causality
• - analyse plots of the data to explain differences in measures of centre and spread
• - draw the line of best fit by eye
• - make predictions based on the line of best fit
• - calculate the correlation coefficient by calculator
1.6b Numerical
• - recognise standard deviation and interquartile range as measures of variability
• - use a calculator to calculate standard deviation
• - find quartiles and the inter-quartile range
• - use the interquartile range appropriately when analysing data
• - recognise the existence of outliers
• - recognise the effect of outliers
• - use percentiles to assign relative standing
1.7 Analysing, interpreting and drawing inferences from data
• - interpret a histogram in terms of distribution of data
• - make decisions based on the empirical rule
• - recognise the concept of a hypothesis test
• - calculate the margin of error for a population proportion
• - conduct a hypothesis test on a population proportion using the margin of error
1.8 Synthesis and problem-solving skills
• - explore patterns and formulate conjectures
• - explain findings
• - justify conclusions
• - communicate mathematics verbally and in written form
• - apply their knowledge and skills to solve problems in familiar and unfamiliar contexts
• - analyse information presented verbally and translate it into mathematical form
• - devise, select and use appropriate mathematical models, formulae or techniques to process information and to draw relevant conclusions
## Certification
All Alison courses are free to enrol, study and complete. To successfully complete this Certificate course and become an Alison Graduate, you need to achieve 80% or higher in each course assessment. Once you have completed this Certificate course, you have the option to acquire official Certification, which is a great way to share your achievement with the world. Your Alison Certification is:
Ideal for sharing with potential employers - include it in your CV, professional social media profiles and job applications
An indication of your commitment to continuously learn, upskill and achieve high results
An incentive for you to continue empowering yourself through lifelong learning
Alison offers 3 types of Certification for completed Certificate courses:
Digital Certificate - a downloadable Certificate in PDF format, immediately available to you when you complete your purchase
Certificate - a physical version of your officially branded and security-marked Certificate, posted to you with FREE shipping
Framed Certificate - a physical version of your officially branded and security-marked Certificate in a stylish frame, posted to you with FREE shipping | 1,435 | 7,381 | {"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} | 4.125 | 4 | CC-MAIN-2019-43 | latest | en | 0.897157 |
https://studylib.net/doc/8537234/chapter-12--tolerancing | 1,591,478,086,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348519531.94/warc/CC-MAIN-20200606190934-20200606220934-00001.warc.gz | 538,739,326 | 12,750 | # Chapter 12: Tolerancing
```Chapter 12
Tolerancing
1.
What is the purpose of Tolerancing?
2.
When does the maximum clearance occur?
3.
When does the minimum clearance occur?
4.
Define “Nominal Size”.
5.
What is the “Basic Hole System”?
6.
What is the “Basic Shaft System”?
7.
Why is the basic-hole system more common than the basic-shaft system?
8.
Which method is the preferred method of expressing tolerances?
9.
What do the two numbers of a limit dimension mean?
10. List five classes of fit.
11. What does the tem “Hole-Basis system of preferred fits” mean in reference to the metric system of
tolerances and fits?
12. What does the tem “Shaft-Basis system of preferred fits” mean in reference to the metric system of
tolerances and fits?
13. How is the tolerance grade specified in the metric system of tolerances and fits?
Tolerance Chart Assignment…Determine the limit dimensions for the hole and shaft for each of the
basic sizes identified. The values for .375, RC5 have been provided as an example.
Basic
Size
0.375
0.750
1.875
Name:
RC 5
Hole
Shaft
.3759
.3740
.3750
.3734
LC 1
Hole
LT4
Shaft
Hole
FN1
Shaft
Hole
Shaft
``` | 321 | 1,135 | {"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-2020-24 | latest | en | 0.851566 |
https://minneatairu.com/kjxnev75/478917-how-to-find-mass-number | 1,628,057,767,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154796.71/warc/CC-MAIN-20210804045226-20210804075226-00054.warc.gz | 376,890,327 | 12,464 | Categories
# how to find mass number
Sodium's number of neutrons is 12. We use the mass number in naming isotopes, like Carbon-12 or Oxygen-17. Contact the Department of Revenue of your state and ask them for specific information on this subject. The number of neutrons is not listed on the periodic table. If an element only has one isotope, relative atomic mass = relative mass of this isotope. One mole of oxygen contains 6.02 × 10 23 of oxygen molecules. Therefore, to find the number of neutrons of sodium, subtract sodium's number of protons, which is 11, from its atomic weight, which is 23. Now all numbers which stored as text are converted to numbers immediately. Mass Number is the number of protons and neutrons in an isotope. Add neutrons and protons together. This is a whole number. So in the nucleus there's only one proton and zero neutrons, so one plus zero gives us a mass number of one. The following diagram shows how to convert between Mass, Mole and Number of particles. A good example is fluorine. Choose from 500 different sets of term:mass atomic number = how to find the number of neutrons flashcards on Quizlet. Add the number of neutrons and protons together. After heating, the mass of the anhydrous compound is found to be 3.22 g. Determine the formula of the hydrate and then write out the name of the hydrate. On the other hand, the process of requesting the state tax ID number is different depending on the location. mass number = protons + neutrons. You can choose which way you'd like to search for your case. Select the connective cells or multiple ranges you need to convert to numbers, and then click Kutools > Content > Convert between Text and Number. To find the number of valence electrons of an element, we must only refer to the periodic table and seek the elementâs position within it. Answer: Together,the number of protons and the number of neutrons determine an elements isotopes have slightly different mass numbers, the atomic mass is calculated by obtaining the mean of the mass numbers for its isotopes. The chemical properties of an element are determined by its Atomic Number not its Mass Number which is why atomic numbers are shown on the Periodic table ⦠With nuclide symbols, of course! Solution: 1) Determine mass of water driven off: 4.31 â 3.22 = 1.09 g of water. The Mass Number describes the total number of protons and neutrons in the nucleus. Learn term:atomic mass number = how to find number of neutrons with free interactive flashcards. Answer:Together, the number of protons and the number of neutrons determine an element's mass number: mass number = protons + neutrons. Atomic mass is about weight of the atom. A = mass number = total number of nucleons. The periodic table is a neat arrangement of all the elements we have discovered to this point. Molar mass. Scroll down the page for more examples and solutions. mass number is the sum of protons and neutrons. Mass number, in nuclear physics, the sum of the numbers of protons and neutrons present in the nucleus of an atom.The mass number is commonly cited in distinguishing among the isotopes of an element, all of which have the same atomic number (number of protons) and are represented by the same literal symbol; for example, the two best known isotopes of uranium (those with mass numbers ⦠A firearm license PIN is required to purchase a firearm and to report a personal sale or transfer of a firearm through the Massachusetts Gun Transaction Portal. Chemists generally use the mole as the unit for the number of atoms or molecules of a material. Most criminal cases can only be searched by docket number. The sum of these two values is the mass number. If you want to calculate⦠mass number = total number of nucleons. Mass number definition, the integer nearest in value to the atomic weight of an atom and equal to the number of nucleons in the nucleus of the atom. 69 At first glance, it may seem like the problem didn't provide you with enough information. Because an electron has negligible mass relative to that of a proton or a neutron, the mass number is calculated by the sum of the number of protons and neutrons. The molar mass of an element is the mass in g of one mole of the element. This table has the atomic and mass number written on it. After you complete these fields, search tabs for name, case type, case number, and ticket/citation # will appear. The mass of one mole of a substance is called the molar mass. Thus, sodium's mass number is 23. read more However, if you have access to a periodic table, you can determine the mass number of the second isotope by examining its atomic mass. In other words, 1 amu ⦠The number of protons in an atom is known as the Atomic Number, which is how the periodic table is organized (they are listed in order of atomic number from left to right). Mass number has been given the symbol A so. It's equal to Avogadro's number (6.02 X 10 23) of atoms. 10 mol of carbon dioxide has a mass of 440 g. To the nearest whole number, isotopic mass = mass number for a specific isotope. Example: One mole of carbon contains 6.02 × 10 23 of carbon atoms. The mass number (symbol A, from the German word Atomgewicht (atomic weight)), also called atomic mass number or nucleon number, is the total number of protons and neutrons (together known as nucleons) in an atomic nucleus. First, find the atomic weight, listed in the bottom of the element's box. By definition, the mass of one mole of an element, or Avogadro's number (6.02 x 10 23) of atoms, is equal to its atomic mass in grams. Atomic mass is about weight of the atom. What is atomic mass number? Example #2: A hydrate of Na 2 CO 3 has a mass of 4.31 g before heating. 2) Determine moles of Na 2 CO 3 and water: Divide the number of atoms by 6.02 x 10 23, the number of atoms in one mole of the element â also known as Avogadro's number. A mole is a unit chemists use. How do we represent an atom, with all of its protons, neutrons, and electrons? Determine the molar mass by placing the units g/mol after the atomic weight of the element. The number is experimentally determined based on measuring the number of atoms in precisely 12 grams of the carbon-12 isotope, giving a value of approximately 6.022 x 10 23. Find out your numbers for BMI, belly fat, healthy weight, target heart rate, interval training, and more. So A is the mass number, which is equal to the number of protons, that's the atomic number which we symbolized by Z, plus the number of neutrons. Get tips, too. An atom in its typical state will not have a charge, as the positive particles (protons) balanced out the negative ones (electrons). By definition, the weight of one mole of an element (its molar mass) is ⦠So, the key to this problem is the fact that gallium has two isotopes. We can find atomic mass and mass number in chemical elements. Round this number to the nearest whole number to obtain the mass number. Symbol: A See more. Mass Number. In the Convert between Text and Number dialog box, select the Text to number option, and then click the OK button. This chemistry video tutorial explains how to calculate the number of protons, neutrons, and electrons in an atom or in an ion. This includes electrons as well as a slight change in mass due to binding energy. 2. To calculate the number of atoms in a sample, you need to find how many moles of the element the sample contains. Finally, subtract the atomic number from the mass number. You can use Avogadro's number in conjunction with atomic mass to convert a number of atoms or molecules into the number of grams. Mass Numbers - How to find Mass Numbers The mass number is established by rounding the atomic weight to the nearest whole number. If an atom has a mass number of 79 (A = 79), then the nucleus of this atom contains 79 nucleons. How to find the mass, moles, number of atoms, and elements? Atomic Mass is the mass of the entire atom of an isotope. The Periodic Table with Atomic Mass will give you the atomic weight, or atomic mass, of the elements. If an atom of an element is known to have 32 nucleons, then its mass number is 32 or A = 32. One mole (abbreviated mol) is equal to 6.022×10 23 molecular entities (Avogadroâs number), and each element has a different molar mass depending on the weight of 6.022×10 23 of its atoms (1 mole). Choose from 500 different sets of term:atomic mass number = how to find number of neutrons flashcards on Quizlet. Each proton and neutron's mass is approximately one atomic mass unit (AMU). If you're searching by docket number, make sure you enter the number exactly how it appears. It is simple to know the number of protons, electrons, and neutrons by using the periodic table. Step 2: Determine the molar mass of the element. How do you find the mass number of an element - 12805112 Liantic3645 Liantic3645 06/11/2019 Chemistry High School How do you find the mass number of an element 2 Learn term:mass atomic number = how to find the number of neutrons with free interactive flashcards. We can find atomic mass and mass number in chemical elements. Similarly, the number of neutrons in Na + = Mass Number â Atomic Number = 23 â 10 = 13. However, you can calculate the number of neutrons in an atom of an element. mass = number of moles × relative formula mass = 2 × 44 = 88 g Finding the relative formula mass Question. 1. All fluorine atoms have a mass of 19 (19 F), therefore its relative atomic mass ⦠The mass number is different for each different isotope of a chemical element. The mass of an atom is mostly localized to the nucleus. So A is equal to Z plus N. And for protium, let's look at protium here. Valence electrons and the Periodic Table. This means that the percent abundances of these two isotopes must amount to 100%. | 2,299 | 9,802 | {"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-2021-31 | latest | en | 0.891297 |
https://www.proofwiki.org/wiki/Definition:Vector_Sum/Component_Definition | 1,660,615,354,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00481.warc.gz | 830,003,497 | 12,287 | # Definition:Vector Sum/Component Definition
## Definition
Let $\mathbf u$ and $\mathbf v$ be vector quantities of the same physical property.
Let $\mathbf u$ and $\mathbf v$ be represented by their components considered to be embedded in a real $n$-space:
$\ds \mathbf u$ $=$ $\ds \tuple {u_1, u_2, \ldots, u_n}$ $\ds \mathbf v$ $=$ $\ds \tuple {v_1, v_2, \ldots, v_n}$
Then the (vector) sum of $\mathbf u$ and $\mathbf v$ is defined as:
$\mathbf u + \mathbf v := \tuple {u_1 + v_1, u_2 + v_2, \ldots, u_n + v_n}$
Note that the $+$ on the right hand side is conventional addition of numbers, while the $+$ on the left hand side takes on a different meaning.
The distinction is implied by which operands are involved.
## Examples
### Example $1$
Let:
$\ds \mathbf a$ $=$ $\ds 6 \mathbf i + 4 \mathbf j + 3 \mathbf k$ $\ds \mathbf b$ $=$ $\ds 2 \mathbf i - 3 \mathbf j - 3 \mathbf k$
Then:
$\mathbf a + \mathbf b = 8 \mathbf i + \mathbf j$ | 324 | 954 | {"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} | 3.875 | 4 | CC-MAIN-2022-33 | latest | en | 0.618979 |
https://52ostreet.com/slot-limited-dimension-drawing/ | 1,686,313,970,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00675.warc.gz | 99,826,666 | 13,109 | # Slot Limited Dimension Drawing7 min read
Sep 28, 2022 5 min
## Slot Limited Dimension Drawing7 min read
What is Slot Limited Dimension Drawing?
Slot limited dimension drawing is a method of representing an object within two dimensions by using a series of slots. The object is first represented by a number of points, and then each point is assigned a slot. The object is then drawn by connecting the slots.
This method can be used to create drawings of any object, no matter how complex it may be. It is particularly useful for creating technical drawings, as it allows the object to be accurately represented in a limited amount of space.
How is Slot Limited Dimension Drawing Used?
Slot limited dimension drawing is used to create drawings of objects that cannot be accurately represented in two dimensions. It is particularly useful for creating technical drawings, as it allows the object to be accurately represented in a limited amount of space.
Slot limited dimension drawing can be used to create drawings of any object, no matter how complex it may be. However, it is often used to create technical drawings, as these drawings require a high level of accuracy.
What are the Advantages of Slot Limited Dimension Drawing?
The main advantage of slot limited dimension drawing is that it allows the object to be accurately represented in a limited amount of space. This is particularly useful for creating technical drawings, as it allows the object to be clearly and concisely represented.
Slot limited dimension drawing is also very easy to learn and use. This means that anyone can learn how to use it to create accurate drawings of any object.
## How do you dimension a slot in engineering drawing?
A slot is a rectangular or square opening in a plate, bar, or other member that is less than the thickness of the member. A slot is usually used to receive a fastener, such as a bolt, screw, or rivet. Slots can also be used to receive other parts, such as gears, cams, and pulleys.
In engineering drawings, slots are dimensioned in two ways. The first way is by the size of the slot in terms of width, height, and depth. The second way is by the size of the opening in the member in which the slot is located.
The width of a slot is the distance from one side to the other side. The height of a slot is the distance from the top of the slot to the bottom of the slot. The depth of a slot is the distance from the front of the slot to the back of the slot.
The size of the opening in the member in which the slot is located is usually given in terms of the diameter of a circle that would fit into the opening. The diameter of a circle is the distance from one side of the circle to the other side of the circle.
## How do you indicate a slot on a drawing?
When indicating a slot on a drawing, you can use different types of symbols to show its location and size. In technical drawings, a slot is a rectangular opening that is used to receive a fastener, such as a screw or bolt. There are a few different ways to indicate a slot on a drawing, depending on the software you are using.
One way to indicate a slot is to use a symbol that looks like an open box. This symbol can be used to show the size and location of the slot. You can also use this symbol to indicate the size and placement of the hole that will be drilled to create the slot.
Another way to indicate a slot is to use a symbol that looks like a filled-in box. This symbol can be used to show the size and location of the slot. You can also use this symbol to indicate the size and placement of the hole that will be drilled to create the slot.
A third way to indicate a slot is to use a symbol that looks like a T-shaped diagram. This symbol can be used to show the size and location of the slot. You can also use this symbol to indicate the size and placement of the hole that will be drilled to create the slot.
If you are using a CAD program to create your drawings, you can also use the “slot” function to create a slot. This function will create a rectangular opening in your drawing that can be used to receive a fastener.
## What are the 3 types of drawing dimensions?
There are three types of drawing dimensions: linear, angular, and volume.
Linear dimensions are the most basic type of dimension and are used to measure the length or distance between two points. They are represented by a line with either arrowheads at each end or a tick mark on one end and an arrowhead on the other.
Angular dimensions are used to measure the angle between two lines or points. They are represented by a line with arrowheads at each end and are usually abbreviated as “degrees” (°).
Volume dimensions are used to measure the three-dimensional size of an object and are represented by a cube with either tick marks or arrowheads on each corner.
## What is a minimally dimensioned drawing?
A minimally dimensioned drawing is a type of engineering drawing that is used to communicate the simplest geometric details of an object. It is typically used when the object is small and has a simple shape.
A minimally dimensioned drawing is created by using basic geometric shapes, such as circles, squares, and triangles, to represent the object. These shapes are then connected to create the object’s outline. The dimensions of the object are not included on a minimally dimensioned drawing.
minimally dimensioned drawing is a type of engineering drawing that is used to communicate the simplest geometric details of an object.
## How is a slot dimensions?
A slot is an opening in a machine or tool that is specifically shaped to receive and hold a particular tool or part. The dimensions of a slot are determined by the size and shape of the tool or part that is to be inserted into it. A slot must be large enough to accommodate the tool or part, but not so large that it becomes difficult to handle or is susceptible to damage. The dimensions of a slot can vary depending on the application, but typically range from about 1/8 inch to 2 inches wide and from 1/8 inch to 1 inch deep.
## What is slot length?
What is slot length?
Slot length is the measure of how long a particular slot in a particular frame is. This length is determined by the number of bits in the slot and the number of bits in the frame.
Slot length is important because it helps to ensure that the bits in a frame are properly aligned. This alignment is necessary in order to ensure that the bits are properly interpreted.
There are a few different factors that can affect slot length. The first is the type of transmission. There are two main types of transmissions: synchronous and asynchronous. Synchronous transmissions use a clock signal to keep the bits in the frame aligned. Asynchronous transmissions do not use a clock signal and, as a result, the bits in the frame can be out of alignment.
The second factor that can affect slot length is the type of encoding. There are two main types of encoding: block encoding and character encoding. Block encoding splits the bits into blocks, which are then transmitted. Character encoding transmits the bits one character at a time.
The third factor that can affect slot length is the type of media. There are two main types of media: serial and parallel. Serial media transmits the bits one at a time. Parallel media transmits the bits multiple at a time.
The final factor that can affect slot length is the type of protocol. There are two main types of protocols: connection-oriented and connectionless. Connection-oriented protocols establish a connection between the sender and the receiver before transmitting the data. Connectionless protocols do not establish a connection between the sender and the receiver before transmitting the data.
## What are the types of dimensions?
Dimensions in geometry are the length, width, and height of a figure. There are other types of dimensions, too.
One type of dimension is time. Time is measured in seconds, minutes, hours, days, weeks, months, and years.
Another type of dimension is size. Size is measured in inches, feet, yards, and miles.
Another type of dimension is weight. Weight is measured in pounds and kilograms.
Another type of dimension is temperature. Temperature is measured in degrees Fahrenheit and degrees Celsius.
Another type of dimension is volume. Volume is measured in cubic inches, cubic feet, cubic yards, and cubic meters.
### Jim Miller
Jim Miller is an experienced graphic designer and writer who has been designing professionally since 2000. He has been writing for us since its inception in 2017, and his work has helped us become one of the most popular design resources on the web. When he's not working on new design projects, Jim enjoys spending time with his wife and kids. | 1,794 | 8,793 | {"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} | 2.96875 | 3 | CC-MAIN-2023-23 | latest | en | 0.953604 |
http://math.stackexchange.com/questions/228630/finding-a-conformal-map-from-unit-disk-to-half-plane/228638 | 1,467,283,578,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783398516.82/warc/CC-MAIN-20160624154958-00107-ip-10-164-35-72.ec2.internal.warc.gz | 193,555,955 | 17,514 | # Finding a conformal map from unit disk to half-plane
I'm trying to find a conformal map $f$ from the open unit disk to the set $\mathbb{C}-[-1/4,-\infty)$ (I think this means the half-plane Re$(w)>-1/4$ with the properties $f(0)=0$ and $f'(0)>0$. I know that the mapping $$f(z)=\frac{i+z}{i-z}$$ returns the right half-plane Re$(w)>0$ from the open unit disk, but subtracting 1/4 from it doesn't satisfy $f(0)=0$. I can't seem to find a lot of other examples. Are there any other conformal maps that I should try?
-
Here's an outline; I'll leave the details to you:
The map you have will send the unit disc to a half plane. To get from a half plane to all of $\mathbb{C}$ minus a ray, postcompose with $z\mapsto z^2$. Now, to get the missing ray where you want it, rotate and translate.
Lastly, look at the pre-image of $0$. You can precompose with an automorphism of the disk sending $0$ to that point. Then all that's left is to check that, when you compose all these maps, the derivative is a positive number.
-
So the image of the right half-plane under $z^2$ is $\mathbb{C}$ without the ray from $0$ to $-\infty$, and translating this by -1/4 gives a conformal map that maps from the domain to the range that we want. So $f(z)=((i+z)/(i-z))^2-1/4$, and $f(-i/3)=0$. How would I rotate $z=-i/3$ on the unit disk to $0$? – ff90 Nov 4 '12 at 4:45
I think that I would have to translate $-i/3$ to $0$, which after precomposing gives $z\mapsto (\frac{2i+3z}{4i+3z})^2-1/4$. But then if I'm not mistaken, $f'(0)$ is not $>0$.... – ff90 Nov 4 '12 at 6:14
Sorry, that's not a rotation. I'll adjust my answer above. – Brett Frankel Nov 4 '12 at 14:45
Thanks! Choosing a specific automorphism of the unit disk makes it work out and gives the simple-looking $f(z)=\frac{z}{(z-1)^2}$ as the answer. – ff90 Nov 5 '12 at 0:38
$$L_1(z)=\frac{i}{2} \frac{z+1}{1-z}$$
Then rotate -90 degrees to get the right half plane. Then $z^2$ etc.... | 630 | 1,936 | {"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} | 3.8125 | 4 | CC-MAIN-2016-26 | latest | en | 0.881552 |
https://oeis.org/A174225 | 1,618,659,581,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038119532.50/warc/CC-MAIN-20210417102129-20210417132129-00543.warc.gz | 522,722,110 | 4,001 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A174225 Array: row n consists of the positive integers x for which there are exactly n positive integers y such that x+y divides x*y. 0
2, 3, 4, 5, 9, 8, 7, 25, 27, 6, 11, 49, 125, 10, 32, 13, 121, 343, 14, 243, 64, 17, 169, 1331, 15, 3125, 729, 12, 19, 289, 2197, 16, 16807, 15625, 18, 256, 23, 361, 4913, 21, 161051, 117649, 20, 6561, 512, 29, 529, 6859, 22 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Many properties of the array follow easily from Comments at A063647: 1. Every positive integer except 1 occurs, exactly once. 2. Row 1 consists of the primes. 3. Row k includes p^k for all primes p, for k>=1. 4. Row 4 includes all products of two distinct primes. 5. Column 1 consists of even numbers. LINKS EXAMPLE Corner of the array: 2....3....5....7....11....13.... 4....9....25...49...121...169... 8....27...125..343..1331..2197.. 6....10...14...15...16....21.... 6 is in row 4 because there are 4 numbers y for which 6+y divides 6*y; they are 3,6,12,30. CROSSREFS Cf. A063647. Sequence in context: A335910 A266531 A284311 * A182945 A052270 A265335 Adjacent sequences: A174222 A174223 A174224 * A174226 A174227 A174228 KEYWORD nonn,tabl AUTHOR Clark Kimberling, Mar 12 2010 STATUS approved
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Last modified April 17 06:46 EDT 2021. Contains 343059 sequences. (Running on oeis4.) | 597 | 1,760 | {"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} | 2.890625 | 3 | CC-MAIN-2021-17 | latest | en | 0.761781 |
https://www.teacherspayteachers.com/Browse/Search:geometric%20solids%20worksheets | 1,566,056,022,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313428.28/warc/CC-MAIN-20190817143039-20190817165039-00278.warc.gz | 968,027,645 | 55,705 | # Results forgeometric solids worksheets
Geometric Solids: Worksheets - An extension lesson for children familiar with the following geometric solids: sphere, cube, cone, square based pyramid, ellipsoid, cylinder, ovoid, rectangular prism, triangular prism, triangular prism, and triangular based pyramid.Includes:black line masters for 12 g
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Geometric solids 3 part cards in solid blue, clip art images, AND high quality photographs of common objects are all included in this product! Download 9 of the 10 solid blue cards here. In addition to the cards, there are several follow up activities: Low-mess flip-flap booklet. Cut along dot
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36 Ratings
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Use these colorful clip art geometric solids (300 dpi) to make your own games, flashcards, manipulatives, and worksheets. Younger students can use them for patterning and counting, while older students may use them to identify faces, vertices, and edges, and to calculate volume. *Cubes, Cones, Rect
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29 Ratings
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Clip Art for geometric solids that should fit almost every need. There is a total of 385 png's. Please see preview or the description below for a detailed list and explanation of what's included. There are two types or categories of shapes included. I call them shaded and non shaded. The non-s
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23 Ratings
4.0
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Allow students to explore advanced three-dimensional solids with these five foldable nets. This packet features nets of a(n): Cuboid, Octahedron, Dodecahedron, Pentagonal Prism, and Hexagonal Pyramid Use this assortment as hands-on construction activities, as templates, or as worksheets for iden
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11 Ratings
4.0
PDF (81.29 KB)
This package contains the following activities: Step-by-step worksheets to construct 3-dimensional geometric solids (using the nets provided) and to calculate the surface area and volume of those solids. There are 4 worksheets: • Construct a cylinder • Construct a pyramid • Construct a rectangular
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UPDATED: Grade levels changed to reflect Common Core Shift. Common Core Standards tagged. This is a hands-on experience with 3-D shapes that your students will love. It would be great to use as an introduction to 3-dimensional figures and a helpful experience for students to use the math vocabulary
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8 Ratings
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I have used this organizer so many different ways. I have had students look at geometric solids as they fill it in, fill it it based on memory and even fill it is as they make some of the geomtric solids with toothpicks and marshmallows. Students can work independently or in small groups. I hav
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6 Ratings
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This is an excerpt from my popular line of Bossy Brocci Math & Big Science workbooks on Amazon. Printing should be: LANDSCAPE and DOUBLE-SIDED, with the Flip being along the 'SHORT' edge or side Want MORE Power for your Dollar? Give Brocci Bundles a Try before you buy! To get the Bundles &
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6 Ratings
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This is an excerpt from my popular line of Bossy Brocci Math & Big Science workbooks on Amazon. Printing should be: LANDSCAPE and DOUBLE-SIDED, with the Flip being along the 'SHORT' edge or side Want MORE Power for your Dollar? Give Brocci Bundles a Try before you buy! To get the Bundles &
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This is an excerpt from my popular line of Bossy Brocci Math & Big Science workbooks on Amazon. Printing should be: LANDSCAPE and DOUBLE-SIDED, with the Flip being along the 'SHORT' edge or side Want MORE Power for your Dollar? Give Brocci Bundles a Try before you buy! To get the Bundles &
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This is an excerpt from my popular line of Bossy Brocci Math & Big Science workbooks on Amazon. Printing should be: LANDSCAPE and DOUBLE-SIDED, with the Flip being along the 'SHORT' edge or side Want MORE Power for your Dollar? Give Brocci Bundles a Try before you buy! To get the Bundles &
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In this two page practice activity, students will identify common geometric solids by their shapes. We have included cubes, rectangular prisms, triangular prisms, cylinders, cones and spheres of differing sizes. Next, using crayons, markers or colored pencils, students will color the shapes the ap
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The 8 reproducible worksheets in this product constitute Unit 5 in the MathQuest 1 workbook. This unit focuses on manipulating concrete materials and using pictorial representations to help students learn the properties of various geometric solids. Students distinguish between different solids and r
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Spun-off from my popular series of Bossy Brocci Math & Big Science Workbooks on Amazon, this Surface Area & Volume of Common Geometric Solids mini-Bundle contains: 2 Lessons 10 Worksheets – and 69 Problems! Individual cost would be \$5.00, but the Bundle price is just \$3.50! 30 slides are p
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86 clip art images of different geometric solids. Images are both in color and black and white and are perfect to include on assessments or worksheets. Images include both right and oblique prisms and pyramids, cone, cylinder and sphere. Thumbnails show all black and white images (colored images a
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Week 13 of a series of weeks of daily math warm ups brought to you by Dino Might Duo! Week 13 has a mini lesson on Geometric Solids that is aligned with Common Core Standards. It can be used as warm ups, reviews, mini lessons, or however you see fit! Every day can be printed and used as a printab
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Geometric Solids Clip Art This product is brought to you by Sunny Boy Creations. This clip art can be used on worksheets, power point presentations and TpT products. This collection of clip art has 15 black & white images in png format. You can see all the images in the previews. Terms of Us
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Complete your unit lessons with this Three-Dimensional Geometry Lesson Bundle. Students begin with fantastic hands-on activities like cutting clay and wrapping cubes. With this foundation, they continue to surface area and volume using fillable solids and nets. This lesson bundle includes plenty of
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Self-Contained Learning Handout for Students Pages 1 and 2 of this documents are for printing a front-back student handout. Page 3 is the key for the Practice portion. Included in the printable student handout: *Student learning objective starts the lesson right *Basic notes for class di
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Interactive notes with colorful pictures, practice with counting faces, edges and vertices, a study guide, and 2 quizzes bundle packl
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Geometric Solids Properties
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Primary Montessori Geometry Bundle - This bundle contains 22 primary Montessori geometry materials. All of the materials included in this bundled unit download are sold separately on TpT. Includes:Angles - CardsAngles - BookDetective Adjective Command CardsDetective Adjective Game - Colored Detectiv
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Identifying the number of faces, vertices, and edges on geometric solids (3-D shapes).
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,049 | 8,108 | {"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} | 2.609375 | 3 | CC-MAIN-2019-35 | latest | en | 0.87734 |
https://www.tutoreye.com/homework-help/science/physics/digital-electronics | 1,679,787,029,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945376.29/warc/CC-MAIN-20230325222822-20230326012822-00356.warc.gz | 1,156,450,090 | 26,821 | Digital Electronics homework Help at TutorEye
# Best Homework Help For Digital Electronics
## Digital Electronics:
Digital electronics is a stream of electronics. It involves the study of devices related to digital signals. The electronic circuits study helps in various integrated circuits of digital electronic devices.
## Digital Electronics Sample Questions:
Question 1: The OR operation in boolean algebra is performed by which properties?
a) Commutative property b) Associative property
c) Distributive property d) All of these
Explanation: OR operation follows associative, commutative and distributive property. Therefore, OR operation is performed by all the above properties.
Question 2: _____________ is a process where letters or words are represented by the group of symbols.
a) Encoding b) Decoding
c) Both (a) and (b) d) None of these
Explanation: In order to secure the transmission of information, letters and words are denoted by different symbols. Encoding extracts codes from the computer to process the information.
Question 3: Which of the following represents absorption law?
a) A + BA = B b) A + AB = A
c) AB + A = A d) A + AA = A
Answer: A + BA = B
Explanation: The expression for absorption law is given by
A + AB = A(1+B) = A
A + AB = A (Since 1 + B = 1 as per 1’s property)
Question 4: Which of the following combinations is possible to construct exclusive-OR (XOR) logic gates?
a) OR b) AND, NOT
c) AND, OR, NOT d) OR, NOT
Explanation: The expression for XOR :
OR, AND, NOT are used to construct the above expression.
In total, 2 AND gates with two inputs and 1 OR are combined to construct XOR logic gates.
Question 5: How many truth tables are necessary for a three-input circuit?
a) 4 b) 8
c) 16 d) 25
Explanation: For 3 input : 23=8
8 truth tables are necessary for a three-input circuit.
Question 6: All logic operations can be obtained by means of ____________.
a) OR and NOR b) OR and NOT
c) NAND and NOR d) All of the above
Explanation: NOR and NAND are universal logic gates. They can be used to design all the three basic gates AND, OR and NOT. All the operations are obtained from these gates.
Question 7: Which of the following devices is a two stable device?
a) Latch b) Multivibrator
c) Monostable vibrator d) None of these
Explanation: Latch works on the principle of bistable multivibrators. This circuit is stable in either of two states. It changes from one state to another.
Question 8: ________ is used to facilitate error correction in digital communications.
a) Gray code b) Binary code
c) BCD code d) ASCII code
Explanation: In digital communication systems such as television and cable system, gray codes are used to facilitate error correction.
Question 9: There are __________ types of counters?
a) 2 b) 3
c) 4 d) 5
Explanation: There are three counters types : asynchronous/synchronous, single and multi-mode, modulus counters.
Question 10: Which of the following counters is also called an asynchronous counter?
a) Ripple counter b) SSI counter
c) Decade counter d) None of these
Explanation: In an asynchronous counter, flip flop is connected to an external clock while the rest of the flip flops have their preceding flip flop output as clock to them. | 823 | 3,542 | {"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-2023-14 | latest | en | 0.808185 |
http://karabincreative.com/6wpbj/the-diagonals-of-a-parallelogram-bisect-each-other-proof-844e9a | 1,627,103,442,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150129.50/warc/CC-MAIN-20210724032221-20210724062221-00469.warc.gz | 26,017,799 | 10,987 | now you can say perpendiculars are equal>> hence proved>>:-) This means that diagonals of a parallelogram bisect each other. She starts by assigning coordinates as given. For Study plan details. This Lesson (Proof: The diagonals of parallelogram bisect each other) was created by by chillaks(0) : View Source, Show About chillaks : am a freelancer In this lesson we will prove the basic property of parallelogram in which diagonals bisect each other. ABCD is a rectangle. To Prove: Quadrilateral ABCD is a square. Glven: ABCD … Get the answers you need, now! Geometry. First we join the diagonals and where they intersect is point E. Angle ECD and EBA are equal in measure because lines CD and AB are parallel and that makes them alternate angles. ABCD is a rectangle. This is a general property of any parallelogram. A C = B D ..... (1) and the diagonals bisect each other at right angles. Anmol proves that a quadrilateral is a parallelogram if and only if its diagonals bisect each other. The diagonals of rectangles bisect each other; Any two adjacent angles are supplementary (obviously, since they all measure 90°) The opposite angles are equal (again, obviously, since all interior angles measure 90°) But because the angles are all equal, there is an additional property of rectangles that we will now prove - that the diagonals of a rectangle are equal in length. So let me see. Geometry. That is, write a coordinate geometry proof that formally proves what this applet informally illustrates. ON OFF. Perpendicular from a line to an external point, Dividing a line into an equal amount of parts, Construct an Equilateral Triangle given one side, Construct an isosceles Triangle given the base and altitude, Construct an Isosceles Triangle given the leg and apex angle, Construct a Triangle 30°, 60°, 90° given the hypotenuse, Construct a Triangle given the base angles and the base length, Construct a Triangle give two sides and an angle, Construct a Equilateral Triangle with a given a perimeter, Construct a Triangle with a given a perimeter in the ratio 2:3:4, Prove that the angle in the same segment of a circle is equal, Calculate the angle at the centre of a circle, Construct an exterior tangent to the given circles, Construct an Interior tangent to the given circles, The sum of the interior angles in a Quadrilateral add up to 360°, Prove the diagonals of a parallelogram bisect each other, Proving the Diagonals of a Parallelogram bisect each other. can you fill in the bottom portion? ... a quadrilateral with diagonals that do not bisect each other is ____ a parallelogram. Given that, we want to prove that this is a parallelogram. For triangles AOB and COD, angle 1 is equal to angle 2, as they are vertical angles. both diagonals bisect at right angles=90degrees. never. Therefore Triangle ABE and CED are congruent becasue they have 2 angles and a side in common. Proof - Diagonals of a Parallelogram Bisect Each Other. 10:00 AM to 7:00 PM IST all days. Which of the following names can be appropriately applied to the diagram at the right? When studying geometry is one of the 2-column deductive proofs a student is expected to work out. To Prove: Diagonals of the rectangle bisect each other. Proof: 1. The diagonals AC and BD bisect each other as the diagonals of the parallelogram in accordance with the lesson Properties of diagonals of parallelograms (under the current topic Parallelograms of the section Geometry in this site). CPCTC can ____ be used in a proof before two triangles have been proven congruent. Parallelograms Properites, Shape, Diagonals, Area and Side Lengths plus interactive applet. And as a square is a special parallelogram, which has all the parallelogram's basic properties, this is true for a square as well. A rhombus is a parallelogram, so we will use what we already know about parallelograms - that the diagonals bisect each other. Angles EDC and EAB are equal in measure for the same reason. - 20538968 View Answer If the perimeter of a parallelogram is 1 4 0 m , the distance between a pair of opposite sides is 7 meter and its area is 2 1 0 s q . You can put this solution on YOUR website! The diagonals of a parallelogram bisect each other. 3. GIVEN: A parallelogram ABCD , Its diagonals, AC & BD intersect at O. Be sure to assign appropriate variable … sometimes. In a square, the diagonals bisect each other. The Diagonals of a Parallelogram Bisect Each Other In this lesson, we will prove that in a parallelogram, each diagonal bisects the other diagonal. prove them similar as parallelograms have equal and opposite sides and their diagonal will be same. Crystal is writing a coordinate proof to show that the diagonals of a parallelogram bisect each other. Published on Feb 27, 2018. a quadrilateral with one pair of opposite sides congruent and one pair of parallel sides is ____ a parallelogram. Glven: ABCD … Get the answers you need, now! 3 option is true, becuase if you find the coordinates of midpoints of both diagonals and these coordinates coincides, then these midpoints are placed in one point on the coordinate plane. So we're going to assume that the two diagonals are bisecting each other. can you fill in the bottom portion? 1800-212-7858 / 9372462318. So we've just proved-- so this is interesting. Subtitles; Subtitles info; Activity; Edit subtitles Follow. Theorem 8.7 If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. Complete the proof that the diagonals of a parallelogram bisect each other as a two-column proof. prove that the diagonals of a parallelogram bisect each other - Mathematics - TopperLearning.com | w62ig1q11. Crystal is writing a coordinate proof to show that the diagonals of a parallelogram bisect each other. For a rhombus, where all the sides are equal, we've shown that not only do they bisect each other but they're perpendicular bisectors of each other. A rhombus is a parallelogram, so we will use what we already know about parallelograms - that the diagonals bisect each other. A line that intersects another line segment and separates it into two equal parts is called a bisector. Theorem - diagonals of a parallelogram bisect each other – kaufen Sie diese Illustration und finden Sie ähnliche Illustrationen auf Adobe Stock Select all that apply. If you draw the figure, you'll see . Therefore, the straight segment CP, which is or own an. Theorem If ABCD is a parallelogram, then prove that the diagonals of ABCD bisect each other. Steps (a), (b), and (c) outline a proof of this theorem. Answer: Given : A rectangle A B CD. Further a rhombus is also a parallelgram and hence exhibits properties of a parallelogram and that diagonals of a parallelogram bisect each other. To Prove: Diagonals of the rectangle bisect each other. Ask Question Asked 3 years, 4 months ago. the diagonals of a parallelogram ____ bisect each other. Solution for Different Methods of Proof Two-Column Proofs (Continued) 3. The diagonals of a parallelogram bisect each other. Given above is Quadrilateral ABCD and we want to prove the diagonals bisects each other into equal lengths. Franchisee/Partner Enquiry (North) 8356912811. If the diagonals of a parallelogram are equal in length, then prove that the parallelogram is a rectangle. Use coordinate geometry to prove that the diagonals of a parallelogram bisect each other. Transform the two-column proof into a paragraph proof. The diagonals of a parallelogram bisect each other; therefore, they have the same midpoint. never. Hence line CE and EB are equal and AE and ED are equal due to congruent triangles. We have step-by-step solutions for … So we're assuming that that is equal to that and that that right over there is equal to that. 0:03 - 0:07 So, what we wanna prove is that it's diagonals bisect each other. Geometry, Parallelogram, Triangles Use coordinate geometry to prove that the diagonals of a parallelogram bisect each other. Contact. x*c - y*d = a (Eq 3) where: x = the scalar fraction along the diagonal 'c' y = the scalar fraction along the diagonal 'd' Hence line CE and EB are equal and AE and ED are equal due to congruent triangles. Need assistance? In addition, if we label the vertices P, Q, R, and S starting from the origin and going clockwise, then … the diagonals of a parallelogram ____ bisect each other. (See Exercise 25 for a particular instance of this… … AO = OC and BO = OD because it is given that diagonals bisect each other. c = a + b (Eq 1) d = b - a (Eq 2) Now, they intersect at point 'Q'. If we let a and b be the side lengths of the parallelogram and c as its altitude, then, the coordinates of the vertices can be easily determined as shown below.. the altitude of a triangle is ____ perpendicular to one side of a triangle. 0. In a rhombus all sides are equal and opposite sides are parallel. (See Exercise 25 for a particular instance of this… The diagonals of a parallelogram bisect each other. To prove that diagonals of a parallelogram bisect each other, Xavier first wants to establish that triangles APD and CPB are congruent. The diagonals of a parallelogram bisect each other. This is a general property of any parallelogram. A line that intersects another line segment and separates it into two equal parts is called a bisector . (See Exercise 25 for a particular instance of this… "The diagonals of a parallelogram bisect each other " …is a property of parallelogram. draw both the diagonals, take any two opposite triangles (not the adjacent ones). Geometry Theorem: The line joining the two end points of two equal and parallel line segment to … →AB ║CD→ Definition of a Parallelogram #AO=CO# - diagonals of a parallelogram bisect each other. Therefore the diagonals of a parallelogram do bisect each other into equal parts. 2 option is false, because it shows that the diagonals of parallelogram have different lengths. Related. 5.7 Proofs Using Coordinate Geometry. Line CD and AB are equal in length because opposite sides in a parallelogram are are equal. Theorem 8.7 If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. By accessing or using this website, you agree to abide by the Terms of Service and Privacy Policy. 0:01 - 0:03 So, we have a parallelogram right over here. 1 Answer Shwetank Mauria Mar 3, 2018 Please see below. The diagonals of a parallelogram bisect each other. always. Steps (a), (b), and (c) outline a proof of this theorem. Rectangle, trapezoid, quadrilateral. Hence Option C is the correct answer \ C) Prove that AC and BD have the same midpoint. And to do that, we just have to remind … If you're seeing this message, it means we're having trouble loading external resources on … Interactive of Proof. Problem. Answer: Given : A rectangle A B CD. Find an alternative way to prove that the diagonals of a parallelogram bisect each other. →AB ║CD→ Definition of a Parallelogram Which statement describes the properties of a rhombus select all that apply. Proof: 1. Hence in #DeltasABO# and #BCO#, we have. prove that the diagonals of a parallelogram bisect each other - Mathematics - TopperLearning.com | w62ig1q11 The Diagonals of a Parallelogram Bisect Each Other, intersects another line segment and separates it into two equal parts is called a, « Isosceles Triangles: the Median to the Base is Perpendicular to the Base, The Diagonals of Squares are Perpendicular to Each Other », the opposite sides of a parallelogram are equal in size, Opposite sides of a parallelogram are equal in size, if the diagonals of a quadrilateral bisect each other, then that quadrilateral is a parallelogram. We have already proven this property for any parallelogram. And as a square is a special parallelogram, which has all the parallelogram's basic properties, this is true for a square as well. Use vector methods to show that the diagonals of a parallelogram bisect each other. Given: The diagonals AC and BD of a quadrilateral ABCD are equal and bisect each other at right angles. Click hereto get an answer to your question ️ Show that if the diagonals of the quadrilateral are equal and bisect each other at right angles, then it is a square. Textbook solution for Elementary Geometry For College Students, 7e 7th Edition Alexander Chapter 4.1 Problem 33E. Consider parallelogram ORPQ with diagonals PR and OQ, the coordinates of the end points are O(0,0), R(a,0), P(b,c), Q(a+b,c). O A = O C; O B = O D ..... (2) ∠ A O B = ∠ B O C = ∠ C O D = ∠ A O D = 9 0 0 ..... (3) Proof: Consider A O B and C O B. O A = O C ....[from (2)] ∠ A O B = ∠ C O B. O B is the common side. a quadrilateral with diagonals that do not bisect each other is ____ a parallelogram. For triangles AOB and COD, angle 1 is equal to angle 2 as they are vertical angles. Steps (a), (b), and (c) outline a proof of this theorem. 3. let both diagonals bisect each other, then choose two alternate triangles. We have already proven this property for any parallelogram. Diagonals of parallelogram bisect each other , it means the diagonals cut each other into two equal halves. This video is suited for class-9 (Class-IX) or grade-9 kids. I am having such a hard time with Plane Geometry, please help me. always . Holt, Rinehart, and Winston . In this lesson, we will prove that in a parallelogram, each diagonal bisects the other diagonal. Crystal is writing a coordinate proof to show that the diagonals of a parallelogram bisect each other. AO = OC and BO = OD because it is given that diagonals bisect each other. I need to make a formal proof of the above and I'm just confused. Hence , they must have the same mid points. You can also proof this statement by doing constructions. Academic Partner. Theorem 4-16 If two sides of a quadrilateral are equal and parallel, then the quadrilateral is a parallelogram. Prove With Vectors That a Parallelogram's Diagonals Bisect. Create your own unique website with customizable templates. Steps (a), (b), and (c) outline a proof of this theorem. She starts… Then we go ahead and prove this theorem. Then the two diagonals are. let both diagonals bisect each other, then choose two alternate triangles. Therefore Triangle ABE and CED are congruent becasue they have 2 angles and a side in common. Clara writes the following proof for the theorem: If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram: Clara's proof 1. We will show that in a parallelogram, each diagonal bisects the other diagonal. (See Exercise 25 for a particular instance of this… A parallelogram, the diagonals bisect each other. Theorem 8.6 The diagonals of a parallelogram bisect each other Given : ABCD is a Parallelogram with AC and BD diagonals & O is the point of intersection of AC and BD To Prove : OA = OC & OB = OD Proof : Since, opposite sides of Parallelogram are parallel. Geometry. #AB=BC# - sides of a rhombus. both diagonals bisect at right angles=90degrees. Become our . Melissa writes the following proof for the theorem: If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram: Melissa's proof. That is, write a coordinate geometry proof that formally proves what this applet informally illustrates. Now to prove that OP and QR bisect each other, we need to show that the diagonals have the same midpoint. Therefore the diagonals of a parallelogram do bisect each other into equal parts. never. the altitude of a triangle is ____ perpendicular ... sometimes. The opposite angles are congruent, the diagonals bisect each other, the opposite sides are parallel, the diagonals bisect the angles . She starts by assigning coordinates as given. She starts by assigning coordinates as given. Hence Option C is the correct answer \ C) Prove that AC and BD have the same midpoint. Privacy policy. The proof can be simplified by placing a vertex of the parallelogram at the origin and one side coinciding with the x-axis. now you can say perpendiculars are equal>> hence proved>>:-) In AOD and BOC OAD = OCB AD = CB ODA = OBC AOD BOC So, OA = OC & OB = OD Hence Proved And you see the diagonals intersect at a 90-degree angle. Complete the proof that the diagonals of a parallelogram bisect each other as a two-column proof. If you have any questions while trying to complete this investigation, or suggestions that would be useful, especially for use at the high school level, please send e-mail to esiwdivad@yahoo.com . Be sure to assign appropriate variable coordinates to your parallelogram's vertices! Draw a parallelogram with two short parallel sides 'a' and two long parallel sides 'b'. Click hereto get an answer to your question ️ Prove by vector method that a quadrilateral is a rhombus if and only if diagonals are congruent and bisect each other at right angles. In a square, the diagonals bisect each other. In this video, we learn that the diagonals of a parallelogram bisect each other. How to prove the diagonals of a parallelogram bisect each other into equal length. Our diagonals intersect at point O, so we'd need to show the two linear angles formed at that intersection point are equal, and we can do that with triangle congruency. Proof that diagonals bisect each other To prove that the diagonals of a parallelogram bisect each other, we will use congruenttriangles: ∠ABE≅∠CDE{\displaystyle \angle ABE\cong \angle CDE}(alternate interior angles are equal in measure) In a quadrangle, the line connecting two opposite corners is called a diagonal. Proof: diagonals of a parallelogram bisect each other? 2. QU07 Proof that Diagonals of a Parallelogram Bisect Each Other Vector proof for midpoints of 2 sides and diagonal intersection. m , find the length of two adjacent sides of the parallelogram. The diagonals of a parallelogram bisect each other. 1 Answer Shwetank Mauria Mar 3, 2018 Please see below. Remember to mark all given… Diagonals are equal. Contact us on below numbers. Proof: Diagonals of a parallelogram bisect each other (Hindi) Anmol proves that a quadrilateral is a parallelogram if and only if its diagonals bisect each other. Diagonals of parallelogram bisect each other , it means the diagonals cut each other into two equal halves. prove them similar as parallelograms have equal and opposite sides and their diagonal will be same. Let's prove to ourselves that if we have two diagonals of a quadrilateral that are bisecting each other, that we are dealing with a parallelogram. Hence , they must have the same mid points. Find an answer to your question HELP PLEASE (: Crystal is writing a coordinate proof to show that the diagonals of a parallelogram bisect each other. Education Franchise × Contact Us. Complete the following proof by filling in each statement. Our diagonals intersect at point O, so we'd need to show the two linear angles formed at that intersection point are equal, and we can do that with triangle congruency. * Use the concepts of the coordinate proofs to solve problems on the coordinate plane. The diagonals of a parallelogram bisect each other; therefore, they have the same midpoint. 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We already know about parallelograms - that the two diagonals are bisecting each at... Is writing a coordinate proof to show that the diagonals bisect each.! Cut each other ; therefore, they have the same midpoint Get the answers you need now! And Privacy Policy line CE and EB are equal in length because opposite sides are equal opposite... | 6,804 | 28,705 | {"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} | 4.53125 | 5 | CC-MAIN-2021-31 | latest | en | 0.894616 |
https://www.bartleby.com/solution-answer/chapter-35-problem-15e-single-variable-calculus-concepts-and-contexts-enhanced-edition-4th-edition/9781337687805/190a1620-e0c0-4721-b085-be89bc89f332 | 1,632,538,271,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00620.warc.gz | 702,512,083 | 65,440 | # The derivative of the function e y cos x = 1 + sin ( x y ) by implicit differentiation.
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
#### Solutions
Chapter 3.5, Problem 15E
To determine
## To calculate: The derivative of the function eycosx=1+sin(xy) by implicit differentiation.
Expert Solution
The derivative of the function is y'=ycos(xy)+eysinxeycosxxcos(xy) .
### Explanation of Solution
Given information:
The function eycosx=1+sin(xy) .
Formula used:
Thechain rule for differentiation is if f is a function of gthen ddx(f(g(x)))=f'(g(x))g'(x) .
Power rule for differentiation is ddxxn=nxn1 .
Product rule for differentiation is ddx(fg)=f'(x)g(x)+f(x)g'(x) where f and g are functions of x .
Calculation:
Consider the function eycosx=1+sin(xy) .
Differentiate both sides with respect to x ,
ddx(eycosx)=ddx(1+sin(xy))
Recall that power rule for differentiation is ddxxn=nxn1 and chain rule for differentiation is if f is a function of gthen ddx(f(g(x)))=f'(g(x))g'(x) .
Also for the terms of the above expression, apply the product rule for differentiation.
Recall that product rule for differentiation is ddx(fg)=f'(x)g(x)+f(x)g'(x) where f and g are functions of x .
Apply it. Also observe that y is a function of x,
ddx(eycosx)=ddx(1+sin(xy))eycosxy'eysinx=cos(xy)[y+xy']eycosxy'eysinx=ycos(xy)+xcos(xy)y'
Isolate the value of y' on left hand side and simplify,
eycosxy'eysinx=ycos(xy)+xcos(xy)y'(eycosxxcos(xy))y'=ycos(xy)+eysinxy'=ycos(xy)+eysinxeycosxxcos(xy)
Thus, the derivative of the function is y'=ycos(xy)+eysinxeycosxxcos(xy) .
### Have a homework question?
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers! | 568 | 1,956 | {"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-2021-39 | latest | en | 0.757202 |
http://www.perlmonks.org/index.pl?node_id=465349 | 1,527,062,067,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865456.57/warc/CC-MAIN-20180523063435-20180523083435-00629.warc.gz | 438,095,665 | 10,526 | Do you know where your variables are? PerlMonks
### Loops loosing Performance
by PerlingTheUK (Hermit)
on Jun 09, 2005 at 23:07 UTC Need Help??
PerlingTheUK has asked for the wisdom of the Perl Monks concerning the following question:
Dear Monks, I have currently a set of values for weekdays, that is stored as integer. To clarify here are example binary values:
Mon = 0b0000001 = 1
Tue = 0b0000010 = 2
Mon to Fri = 0b0011111 = 31
Sat to Sun = 0b1100000 = 96
I need to know how many days of running two sets of data have in common. This is a simple task, however I got stuck in a vicious circle between simplicity and performance. I st up five test functions and compared their speed. It did not surprise me much, that the fastes way is a bitwise evaluation. But if I ran through all days with a loop, the function is a lot slower. Can anyone suggest a better way to approach this? Please find the current code below.
```sub test4 {
my ( \$iDays, \$iCmp ) = @_;
my \$iRes = 0;
for ( 1, 2, 4, 8, 16, 32, 64 ){
if ( \$iDays & \$iCmp & \$_ ){
\$iRes++;
}
}
return \$iRes;
}
sub test5 {
my ( \$iDays, \$iCmp ) = @_;
my \$iRes = 0;
\$iRes++ if ( \$iDays & \$iCmp & 1 );
\$iRes++ if ( \$iDays & \$iCmp & 2 );
\$iRes++ if ( \$iDays & \$iCmp & 4 );
\$iRes++ if ( \$iDays & \$iCmp & 8 );
\$iRes++ if ( \$iDays & \$iCmp & 16 );
\$iRes++ if ( \$iDays & \$iCmp & 32 );
\$iRes++ if ( \$iDays & \$iCmp & 64 );
return \$iRes;
}
cmpthese(\$count, {
'Test4' => sub { test4( int ( rand(127) ), int( rand(127) ) ) },
'Test5' => sub { test5( int ( rand(127) ), int( rand(127) ) ) }
});
__DATA__
Rate Test4 Test5
Test4 162690/s -- -46%
Test5 300000/s 84% --
The performance lost by the loop is quite steep, but the code nicer.
Cheers,
PerlingTheUK
Replies are listed 'Best First'.
Re: Loops loosing Performance
by kaif (Friar) on Jun 09, 2005 at 23:53 UTC
Are you a hacker? Are you not worried if your code looks like crazy old magic? Then you'll love this solution:
```sub test0 {
return (((\$_[0] & \$_[1]) * 2113665) & 17895697) % 15;
}
(Go ahead! Check that it's correct for all input values!) Benchmarks on my computer, against your code and hv's:
``` Rate Test4 Test5 Test7 Test6 Test0
Test4 64233/s -- -33% -33% -52% -72%
Test5 95523/s 49% -- -1% -29% -58%
Test7 96399/s 50% 1% -- -28% -58%
Test6 133602/s 108% 40% 39% -- -41%
Test0 227756/s 255% 138% 136% 70% --
Are you a hacker? Don't you know the C precedence table by heart? Then why do you need those crazy parentheses around the multiplication?
Isn't this enough:
```sub test1 {
((\$_[0] & \$_[1]) * 2113665 & 17895697) % 15;
}
Anyway, I like your method. Nice idea to search in HAKMEM. As you can see below, I've given a less concise solution.
Tried and tested it, but next year by this time, I still want to understand what the heck I did. Unless set a link in my comments that would be impossible.
Cheers,
PerlingTheUK
Kaif, I'd looove to understand what's going on there in test0. I'm not going to ask you for an explanation because I reckon that'd take some time, but would you mind pointing me in the right direction? What do I need to learn first in order to take that apart? Thanks.
Here's a hint: 0b1000000100000010000001 == 2113665 and 0b1000100010001000100010001 == 17895697. I'd gladly explain what's going on, but it seems that some people have already done it online, so I'll link to them instead. In short, multiplying by the first value "repeats" the input four times, while the second knocks out all but every fourth bit. Finally, we "cast out 15s".
Online resources: spurperl, below, linked to Bit Twiddling Hacks by Sean Eron Anderson of Stanford which contains this and many other bit hacks. Anderson, in turn, links to A Modulo Based Bit Counting Approach For 64 Bit Systems by Roshan James, an explanation of my solution above (more or less). Simply note the differences in integer size (they have 64 bits, while most of us only have 32) and requirements (our input is at most 7 bits, and they allow 16). I personally was inspired by a section of the HAKMEM, as I linked above (this also has a difference in size --- the PDP-1110 had 36-bit integers --- and requirements --- 9 bit input).
Re: Loops loosing Performance
by hv (Parson) on Jun 09, 2005 at 23:18 UTC
For speed, I'd suggest:
``` sub test6 {
my \$n = \$_[0] & \$_[1];
my \$count = 0;
++\$count, \$n &= \$n - 1 while \$n;
\$count;
}
For beauty I'd suggest hiding that behind a suitable function name - it costs an extra function call, but you can probably lose that in the way you call it when not benchmarking:
``` sub countbits {
my(\$n, \$count) = (\$_[0], 0);
++\$count, \$n &= \$n - 1 while \$n;
return \$count;
}
sub test7 {
my(\$iDays, \$iCmp) = @_;
return countbits(\$iDays & \$iCmp);
}
You might even want to add some comments. :)
Benchmarks:
``` Rate Test4 Test5 Test7 Test6
Test4 106192/s -- -38% -40% -57%
Test5 172463/s 62% -- -3% -31%
Test7 176987/s 67% 3% -- -29%
Test6 248686/s 134% 44% 41% --
Hugo
Thank you, test6 is what I was looking for. Quick but not a line for each bit of byte.
Cheers,
PerlingTheUK
Re: Loops loosing Performance
by spurperl (Priest) on Jun 10, 2005 at 05:02 UTC
What you have here is the classical "bit counting problem". By ANDing the two numbers all you need is find the amount of 1s in the result. The bit counting problem is a terrific example of the tradeoffs between space and time efficiency, and is one my my favorite interview questions.
You can find a lot of info about it online, for example here.
To make a long story short, the fastest technique is table lookup. Precompute the counts for all bytes (256 of them) in a table and do a simple lookup to find results later.
What you've been shown by kaif and others are the more arcane and enjoyable methods, but table lookup will be faster.
spurperl is right, table lookup is fastest. Using the notation op for original poster, lt for "lookup table", and the rest obvious, the overall benchmark looks something like this:
``` Rate op4 op5 hv7 hv6 roy4 kaif lt
op4 64233/s -- -33% -33% -52% -57% -72% -76%
op5 96098/s 50% -- -0% -28% -36% -58% -64%
hv7 96399/s 50% 0% -- -28% -35% -58% -64%
hv6 134032/s 109% 39% 39% -- -10% -41% -50%
roy4 149447/s 133% 56% 55% 12% -- -34% -45%
kaif 227019/s 253% 136% 135% 69% 52% -- -16%
lt 270514/s 321% 181% 181% 102% 81% 19% --
Just be sure to initialize your table outside of your function, that is, initialize only once.
Re: Loops losing Performance
by Roy Johnson (Monsignor) on Jun 10, 2005 at 01:47 UTC
Yet another way to do it. Somewhat slower and less arcane than kaif's, but maybe adaptable to other situations. And it saves a few keystrokes. :-)
```# Reused your sub name
sub test4 {
return unpack("%4b*", pack('C', \$_[0] & \$_[1]));
}
__END__
Rate Test5 Test4 Test0
Test5 88345/s -- -34% -47%
Test4 133240/s 51% -- -20%
Test0 167395/s 89% 26% --
Caution: Contents may have been coded under pressure.
Re: Loops loosing Performance
by ambrus (Abbot) on Jun 10, 2005 at 10:08 UTC
The problem is essentially to count the number of bits in an integer. There's a good solution for this, which I'll show now.
This subroutine from MMIXware is supposed to do exactly this. This is for a 32-bit word though, not just 7 bits.
Now if you understand how this works, you can probably convert it to efficent perl. I'm not sure I understand it in full, so the solution I'll write here may not be the same, thus may be slower.
```#!perl
use warnings;
use strict;
# the following subroutines couns the number of bits in an 8-bit integ
+er
sub sadd8_a { # my first solution -- without looking closely to your c
+ode
my(\$x) = @_;
my \$n = 0;
for (my \$k = 1; \$k < 0x100; \$k <<= 1) {
0 != (\$x & \$k) and \$n++;
}
\$n;
}
sub sadd8_b { # one of your solutions, modified a bit
my ( \$iDays ) = @_;
my \$iRes = 0;
for ( 1, 2, 4, 8, 16, 32, 64, 128 ){
if ( \$iDays & \$_ ){
\$iRes++;
}
}
return \$iRes;
}
my(\$x) = @_;
\$x = (\$x & 0b01010101) + (\$x >> 1 & 0b01010101);
\$x = (\$x & 0b00110011) + (\$x >> 2 & 0b00110011);
(\$x & 0b00001111) + (\$x >> 4);
}
# let's print an example
for my \$n (0 .. 31) {
print sadd8_a(\$n), " ";
}
print "\n";
# test
for my \$n (0 .. 255) {
my \$a = sadd8_a(\$n);
my \$b = sadd8_b(\$n);
my \$c = sadd8_c(\$n);
\$a == \$b && \$b == \$c or
die "wrong results: \$n \$a \$b \$c";
}
__END__
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Notices? | 2,916 | 8,953 | {"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} | 2.8125 | 3 | CC-MAIN-2018-22 | latest | en | 0.717639 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/756/1/z/a/ | 1,607,211,284,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141750841.83/warc/CC-MAIN-20201205211729-20201206001729-00413.warc.gz | 748,117,980 | 53,390 | # Properties
Label 756.1.z.a Level $756$ Weight $1$ Character orbit 756.z Analytic conductor $0.377$ Analytic rank $0$ Dimension $2$ Projective image $D_{6}$ CM discriminant -3 Inner twists $4$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$756 = 2^{2} \cdot 3^{3} \cdot 7$$ Weight: $$k$$ $$=$$ $$1$$ Character orbit: $$[\chi]$$ $$=$$ 756.z (of order $$6$$, degree $$2$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$0.377293149551$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\zeta_{6})$$ Defining polynomial: $$x^{2} - x + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{7}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Projective image $$D_{6}$$ Projective field Galois closure of 6.0.196036848.1
## $q$-expansion
The $$q$$-expansion and trace form are shown below.
$$f(q)$$ $$=$$ $$q -\zeta_{6}^{2} q^{7} +O(q^{10})$$ $$q -\zeta_{6}^{2} q^{7} + ( 1 - \zeta_{6}^{2} ) q^{19} + \zeta_{6}^{2} q^{25} + ( -1 - \zeta_{6} ) q^{31} + 2 \zeta_{6} q^{37} + q^{43} -\zeta_{6} q^{49} + ( -1 + \zeta_{6}^{2} ) q^{61} + 2 \zeta_{6}^{2} q^{67} + ( -1 - \zeta_{6} ) q^{73} -2 \zeta_{6} q^{79} + ( \zeta_{6} + \zeta_{6}^{2} ) q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q + q^{7} + O(q^{10})$$ $$2q + q^{7} + 3q^{19} - q^{25} - 3q^{31} + 2q^{37} + 2q^{43} - q^{49} - 3q^{61} - 2q^{67} - 3q^{73} - 2q^{79} + O(q^{100})$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/756\mathbb{Z}\right)^\times$$.
$$n$$ $$29$$ $$325$$ $$379$$ $$\chi(n)$$ $$1$$ $$\zeta_{6}$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
325.1
0.5 + 0.866025i 0.5 − 0.866025i
0 0 0 0 0 0.500000 0.866025i 0 0 0
649.1 0 0 0 0 0 0.500000 + 0.866025i 0 0 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
3.b odd 2 1 CM by $$\Q(\sqrt{-3})$$
7.d odd 6 1 inner
21.g even 6 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 756.1.z.a 2
3.b odd 2 1 CM 756.1.z.a 2
4.b odd 2 1 3024.1.cg.a 2
7.d odd 6 1 inner 756.1.z.a 2
9.c even 3 1 2268.1.p.a 2
9.c even 3 1 2268.1.bd.b 2
9.d odd 6 1 2268.1.p.a 2
9.d odd 6 1 2268.1.bd.b 2
12.b even 2 1 3024.1.cg.a 2
21.g even 6 1 inner 756.1.z.a 2
28.f even 6 1 3024.1.cg.a 2
63.i even 6 1 2268.1.p.a 2
63.k odd 6 1 2268.1.bd.b 2
63.s even 6 1 2268.1.bd.b 2
63.t odd 6 1 2268.1.p.a 2
84.j odd 6 1 3024.1.cg.a 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
756.1.z.a 2 1.a even 1 1 trivial
756.1.z.a 2 3.b odd 2 1 CM
756.1.z.a 2 7.d odd 6 1 inner
756.1.z.a 2 21.g even 6 1 inner
2268.1.p.a 2 9.c even 3 1
2268.1.p.a 2 9.d odd 6 1
2268.1.p.a 2 63.i even 6 1
2268.1.p.a 2 63.t odd 6 1
2268.1.bd.b 2 9.c even 3 1
2268.1.bd.b 2 9.d odd 6 1
2268.1.bd.b 2 63.k odd 6 1
2268.1.bd.b 2 63.s even 6 1
3024.1.cg.a 2 4.b odd 2 1
3024.1.cg.a 2 12.b even 2 1
3024.1.cg.a 2 28.f even 6 1
3024.1.cg.a 2 84.j odd 6 1
## Hecke kernels
This newform subspace is the entire newspace $$S_{1}^{\mathrm{new}}(756, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$T^{2}$$
$5$ $$T^{2}$$
$7$ $$1 - T + T^{2}$$
$11$ $$T^{2}$$
$13$ $$T^{2}$$
$17$ $$T^{2}$$
$19$ $$3 - 3 T + T^{2}$$
$23$ $$T^{2}$$
$29$ $$T^{2}$$
$31$ $$3 + 3 T + T^{2}$$
$37$ $$4 - 2 T + T^{2}$$
$41$ $$T^{2}$$
$43$ $$( -1 + T )^{2}$$
$47$ $$T^{2}$$
$53$ $$T^{2}$$
$59$ $$T^{2}$$
$61$ $$3 + 3 T + T^{2}$$
$67$ $$4 + 2 T + T^{2}$$
$71$ $$T^{2}$$
$73$ $$3 + 3 T + T^{2}$$
$79$ $$4 + 2 T + T^{2}$$
$83$ $$T^{2}$$
$89$ $$T^{2}$$
$97$ $$3 + T^{2}$$ | 1,904 | 3,927 | {"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} | 2.5625 | 3 | CC-MAIN-2020-50 | latest | en | 0.419424 |
http://time-price-research-astrofin.blogspot.com/2012/07/spx-vs-lunar-mystery-cycle.html | 1,596,792,889,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737172.50/warc/CC-MAIN-20200807083754-20200807113754-00361.warc.gz | 108,189,260 | 32,324 | ## Thursday, July 19, 2012
### SPX vs Lunar Mystery Cycle
This mysterious cycle is derived from numerology and beyond common reasoning:
(1) Divide the number of The Infinite (= 1) by The Soul number (= 7) = 0.1428571
(2) Float the decimals to the front by multiplying 0.1428571 by 100,000 = 14,285.71
(3) Assume this would be degrees of Lunar Motion, and divide them by 360 degrees = 39.68 Lunar Months of 29.53 Calendar Days (CD) = 39 Lunations + 20.15 Calendar Days = 1,171.79 CD or divided by the length of the mean solar year (365.24 CD) = 3.21 Years
Any pivot-date in e.g. the SPX plus approximately 1,172 CD will become another pivot usually even of the same polarity, e.g. May 8, 2009 High + 1,172 CD = Jul 23, 2012 High (see SPX chart below). Stunningly the results are exact within hours (see 2. chart). However, this cycle cannot forecast the magnitude and importance of the projected highs and lows.
An ancient approximation to π (pi) is 22 / 7 = 3.142857. This has been associated with Jewish mysticism. Certain Kabbalists know the 22 characters of the Hebrew alphabet represent a complete circumference that, when divided by 7 = the sacred number of cycles, produces the Kabbalistic π, also known as the perfect π
However, 1 / 7 = 0.142857142857142857142857142857... and if 142857 is multiplied by 2, 3, 4, 5, or 6, the answer will be a cyclic permutation of itself, and will correspond to the repeating digits of 2/7, 3/7, 4/7, 5/7, or 6/7, respectively:
1 × 142,857 = 142,857
2 × 142,857 = 285,714
3 × 142,857 = 428,571
4 × 142,857 = 571,428
5 × 142,857 = 714,285
6 × 142,857 = 857,142
7 × 142,857 = 999,999 = closest to The Inifinite
Multiplying 142857 by 2 one gets 285714, which is the same digits, rotated 4 positions, by 3 one gets 428571 (shifted 5 places), times 7 is 999999, the closest number to The Inifinite. There is a lot more number-plays with 142857 involved, e.g.
1 / 7 = 0 . 142857 142857 142857 14...
2 / 7 = 0 . 285714 285714 285714 28...
3 / 7 = 0 . 428571 428571 428571 42...
4 / 7 = 0 . 571428 571428 571428 57...
5 / 7 = 0 . 714285 714285 714285 71...
6 / 7 = 0 . 857142 857142 857142 85...
7 / 7 = 0 . 999999 999999 999999 99...
8 / 7 = 1 . 142857 142857 142857 14...
9 / 7 = 1 . 285714 285714 285714 28...
Ennead is greek and derived from ennea = the number 9 = 3 x 3, the first square of an odd number. A rare usage identifies the derived adjective, enneœteric, "a cycle of nine years", and Enneatical year, "every ninth year of life".
In viewing an Enneagram (this is the TPR - logo), one immediately notes the design consisting of a triangle 3-6-9 and a rather unique emblem formed by the numbers 142857. The numbers were listed in a counter-clockwise direction, given the fact most ancient cultures read circles that way. With this path of reading 142857 from the Enneagram, George Gurdjieff used to explain and visualize the dynamics of the interaction between the two great laws of the Universe, the Law of Three and the Law of Seven. The path of 142857 is also portrayed in Gurdjieff's Sacred Dances, known as the Movements.
And here is another nice one, recently presented by Christian Hagglof & JMH @ TPR I
Going on a Square of Nine from 64 to 100 is a 360°- round trip.
Square-root of 64 = 8
8 + 2 = 10
10 re-squared = 100
100 - 64 = 36
36 / 252 Trading Days (TD) in 1 Solar Year = 0.1428571
252 TD / 36 = 7
36 / 252 = 1 / 7 = 0.1428571
More clues on the sacred number 142857 can be found especially HERE & HERE as well as in Olney H. Richmond's "The Mystic Test Book" (p. 22, 30, 32, 34).
Well, and last not least 3.21 years is also a reproductive cycle in Yellowstone's grizzly bear population.
HERE | 1,181 | 3,661 | {"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} | 3.546875 | 4 | CC-MAIN-2020-34 | latest | en | 0.846405 |
http://www.show-my-homework.com/2012/05/thermal-expansion-law.html | 1,519,155,163,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813088.82/warc/CC-MAIN-20180220185145-20180220205145-00264.warc.gz | 516,352,671 | 9,566 | Thermal Expansion Law
1. An iron railroad rail is 2100 ft long when the temperature is 25°C. What is its length when the temperature is -15°C?
2. A copper vat is 12m long at room temperature (20°C). How much longer is it when it contains boiling water at 1atm pressure?
The law of thermal expansion for relative small intervals of temperatures is
$\Delta(L)/L0 = \alpha*\Delta(T)$
where delta is the variation, L the length, T the temperature and $\alpha$ the coefficient of thermal linear expansion.
$\alpha_{iron} = 11.1*10^{-6} (1/Celsius)$
$\alpha_{Cu} = 17*10^{-6} (1/Celsius)$
$\Delta(L) = L0*\alpha_{iron}*(25+15) =2100*11.1*10^{-6}*40 =0.9324 foot$
$L = L0+\Delta(L) =0.9324+2100 =2100.9324 feet = 636.35 m$
$\Delta(L) =L0*\alpha_{Cu}*(100-20) =12*17*10^{-6} *80 =0.01632 m$
$L = L0+\Delta(L) =12 +0.01632 =12.01632 m$ | 287 | 834 | {"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} | 3.953125 | 4 | CC-MAIN-2018-09 | latest | en | 0.72355 |
https://astronomy.stackexchange.com/questions/48453/why-does-the-gas-cloud-collapse-in-regions-of-high-density | 1,660,960,685,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573876.92/warc/CC-MAIN-20220820012448-20220820042448-00332.warc.gz | 123,020,735 | 65,283 | # Why does the gas cloud collapse in regions of high density?
Stars form when gas cloud collapse under gravity, becoming hot and subsequently initiating nuclear fusion. I have read that the collapse is triggered by density fluctuations, where regions of high density accumulate mass and then collapse? Why is that the case? When we say regions of high density, what does it mean? That they are regions where there are more gas particles? If that is the case, it should have more pressure and since particles move from high pressure to low pressure areas, accumulation of mass shouldn't happen right? Am I missing something here ?
The gas cloud will have a tendency to collapse because of gravity, but that collapse is counteracted by the pressure of the gas as you expect.
When something happens that causes local density fluctuations (e.g. a nearby supernova) the gas will be contracted locally. As this globule of cloud contracts, pressure will build up (increasing the total kinetic energy of the cloud) but so will gravity (increasing the potential energy). If the mass of the gas is below a specific threshold, the pressure will cause the gas to expand again as you expected, but if the mass of gas is above the threshold, the pressure increase will not be able to counteract the increase in gravity and the globule of gas will continue to contract into a protostar.
This contraction will continue until some other process (e.g. pressure caused by nuclear fusion or centrifugal force due to rotation) is able to counteract gravity.
Using math and physics:
In a stable gas cloud the virial theorem holds, which states that the total kinetic energy of the cloud is equal to half the potential energy.
$$K = \frac{1}{2}U$$ where $$K$$ is the kinetic energy and $$U$$ is the potential energy.
The kinetic energy is due to the motions of the particles in the gas cloud. This motion of particles and their interactions is what constitutes pressure. The kinetic energy is, therefore a measure of pressure. When $$N$$ is the number of particles, $$T$$ is the temperature and $$k$$ is Boltzmann's constant we have a kinetic energy equal to:
$$K = \frac{3}{2}NkT$$
The potential energy is due to gravity and is equal to:
$$U = -\frac{3}{5} \frac{GM^2}{R}$$
where $$G$$ is the gravitational constant, $$M$$ is the mass and $$R$$ is the size of the cloud.
The virial theorem then becomes:
$$NkT = \frac{1}{5}\frac{GM^2}{R}$$
And because $$N=M/m$$, where $$m$$ is the average mass of the particles, we can write: $$kT\frac{M}{m} = \frac{1}{5}\frac{GM^2}{R}$$ If this equality holds, the gas cloud is stable and is in virial equilibrium.
If, however, the right side of the equation is bigger, the cloud will collapse. The right side depends on the potential (gravitational) energy, so if gravity is stronger the cloud collapses:
$$kT\frac{M}{m} < \frac{1}{5}\frac{GM^2}{R}$$
If, on the other hand the kinetic energy wins out, the left side is bigger and the cloud will expand. This can be due to an increase in temperature which increases the average speed of the particles (and thereby the cloud's pressure).
To find the mass at which a cloud collapses given its size we can work some on the equation:
$$kT\frac{M}{m} < \frac{1}{5}\frac{GM^2}{R}$$
$$\frac{kT}{m} < \frac{1}{5}\frac{GM}{R}$$
$$M > \frac{5kT}{Gm}R$$
If we assume that the density $$\rho$$ is constant we can relate the radius $$R$$ to the mass $$M$$. The mass of the cloud is equal to the volume $$V = \frac{4}{3}\pi R^3$$ times the density, which gives us: $$M = \frac{4}{3}\pi \rho R^3$$ which gives us an expression for the radius $$R$$: $$R = \left[ \frac{3 M}{4 \pi \rho} \right]^{\frac{1}{3}}$$ Using this equation for $$R$$, we get: $$M > \frac{5kT}{Gm}\left[ \frac{3 M}{4 \pi \rho} \right]^{\frac{1}{3}}$$ $$M^{\frac{2}{3}} > \frac{5kT}{Gm}\left[ \frac{3}{4 \pi \rho} \right]^{\frac{1}{3}}$$
$$M > \left[ \frac{5kT}{Gm}\right]^{\frac{3}{2}}\left[ \frac{3}{4 \pi \rho} \right]^{\frac{1}{2}}$$
This value of $$M$$ is the threshold at which a globule of gas will start to contract and is called the Jean's mass $$M_J$$:
$$M_J = \left[ \frac{5kT}{Gm}\right]^{\frac{3}{2}}\left[ \frac{3}{4 \pi \rho} \right]^{\frac{1}{2}}$$
• If $$M > M_J$$, gravity is stronger --> Collapse
• If $$M < M_J$$, kinetic energy is stronger --> Expansion
References:
• I can't resist wondering how this applies to the famous what-if.xkcd.com/4 Feb 9 at 13:02
• Pressure and gravity have different dimensions/units. They can never be equal. Feb 10 at 0:45
• @ProfRob You are correct of course, pressure is the force (of gas molecules) per square metre and gravity is the force on a molecule or on a volume of space containing some mass. I will edit the answer accordingly. Feb 10 at 8:51
• I have removed the section as it was not really conducive to the argument. Feb 10 at 9:00
Whilst the answer given by @Dieudonné addresses one of the fundamental criteria for gravitational collapse here (the 'Jeans criterion') it contains some inaccuracies and omissions (which have been adopted from the cited references).
First of all, the equation assumed for the gravitational potential energy in the derivation implies a sphere of constant density. In hydrostatic equilibrium however (the assumption made in the reference this answer is based on) the density in an isothermal cloud must decrease like $$1/R^2$$ (as is not too difficult to show) which results in a gravitational potential energy
$$U=-\frac{GM^2}{R}$$
i.e. a factor $$5/3$$ larger than assumed here, with the Jeans Mass changing correspondingly.
For arbitrary density distributions, the potential energy is
$$U=-\frac{kGM^2}{R}$$
where $$\frac{1}{2}\le k\lt \infty$$ (see also this SE post)
More importantly, the impression is given here that star formation would proceed as long as the Jeans criterion (in whatever form) is fulfilled. This crucially ignores the fact that for a closed system the total energy must be constant, so the potential energy lost during collapse must be turned into kinetic energy i.e. the temperature must increase (given that the thermal collision times/distances for atoms are usually orders of magnitude shorter than the free fall times/ dimensions of the cloud). This would stop any collapse rather soon. From the equations
$$U=-\frac{GM^2}{R}$$ $$E=\frac{U}{2}$$
(where $$E$$ is the total energy as per the virial theorem) we can see that
$$R=-\frac{GM^2}{2E}$$
Since the total energy $$E$$ is constant, this means the radius $$R$$ must be (on average) constant as well. Even if we assume that initially the kinetic energy i.e. the temperature is zero (in which case we would have $$E=U$$ instead) the cloud could collapse only to half its original size.
This means that for star formation to occur, they gas cloud must continuously lose energy. This can only happen through inelastic collisions between the atoms and molecules, with the lost energy radiated away in the resulting atomic decay processes. A theoretical problem is here that the initial temperature of the cloud is much too low for any electronic states of atoms to be excited from the ground state. Hence it is generally assumed that the excitation of vibrational and rotational states of certain molecules are responsible for the energy losses. It is however also possible that highly excited states ('Rydberg states') of hydrogen atoms contribute to this. Although their density will be very small, the excitation cross section of these Rydberg states is extremely large, which could result in a significant effect (I have theoretically examined the significance of highly excited Rydberg states for the scattering of radio waves, which is a different context and applied to different physical conditions, but the general principles should be transferable).
Regions of higher density mean that there are indeed some extra atoms of gas in there relative to their surroundings, and the gravitational pull (if you will ignore GR here and go with the Newtonian models) would not balance for a particular atom situated in the center of two spheres of mismatching number of gas that exert unequal forces. By that force, all the regions with the slightest of a "fluctuation" relative to the other surrounding regions, will slowly but steadily accumulate more matter and will start rotating due to unbalanced forces on it too by its neighbors, and then it will spin and then collapse into itself.
(the heat and temperature will be enough to trigger nuclear fusion reactions and the gravitational pressure from outside is then balanced from the exothermic reaction which produces heat and causes an outward tendency to expand: forming a star when after many oscillations, the protostar settles to a definite volume)
To answer why indeed do density fluctuations occur in the first place,
1. God made the universe with a specific set of chosen (for unknown reasons) like this and let it evolve thereafter: we as a species developed in those conditions as a result-there might be other life developed in non-Earth conditions as well.
2)Our observable universe maybe just one of the many with innumerable conditions-it is the way it is because it is. +chaotic boundary conditions (either the universe is spatially infinite or there are infinite universes-ours is just one of them)
1. The weak anthropic principle
So, the conditions may arise due to divine inspiration or as predicted by the weak anthropic principle.
The fluctuations themselves arise because the universe functions according to quantum mechanics and relativity, and with the uncertainty principle we know that the early universe must have been uncertain with just the minimum fluctuations in the positions-(possibly causing clumps) and velocity. The small irregularities would then have been blown up by the rapid expansion of the universe as the inflationary models suggest.
All changes and people who point out errors are welcome.
• Hi Aveer, mentioning divine intervention before Jeans criterion on this (mainstream physics) site is probably going to frowned upon.
– pela
Feb 9 at 11:06 | 2,409 | 10,093 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 47, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-33 | longest | en | 0.941336 |
http://www.chestnut.com/en/eg/7/mathematics/2/7/examples/ | 1,477,333,565,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719677.59/warc/CC-MAIN-20161020183839-00285-ip-10-171-6-4.ec2.internal.warc.gz | 362,265,627 | 52,774 | # 7.2.7. Dividing Algebraic Expression by a Monomial
Find the quotient of .
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Find the quotient of .
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Show Solution
Find the quotient of .
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Find the quotient of .
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If , evaluate .
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If , find .
• A29
• B19
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Divide by , and then add the result to .
• A
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Find the multiplication result of by , then divide it by .
• A
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Complete .
• A,
• B,
• C,
• D,
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Determine the length of a rectangle having an area of and a width of .
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Find the length of a rectangle having an area of and a width of .
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Find the length of a rectangle having an area of and a width of .
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Find the height of a cuboid that has a volume of and a square-shaped base with a side length of .
• A
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Use the data on the figure below that shows two rectangles and to find the length of , given that the area of the coloured part is .
• Acm
• Bcm
• Ccm
• Dcm
Show Solution
If the area of a triangle is , and the length of its base is cm, find its height with respect to the given base.
• Acm
• Bcm
• Ccm
• Dcm
Show Solution
Find the result of without using the calculator.
Show Solution
Without using a calculator, find the result of .
Show Solution
Using mental math, find the result of .
Show Solution | 455 | 1,504 | {"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-2016-44 | latest | en | 0.737976 |
https://classroom.thenational.academy/lessons/multiply-2-digit-numbers-by-6-using-the-partitioning-method-ccvkar | 1,620,404,576,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988796.88/warc/CC-MAIN-20210507150814-20210507180814-00309.warc.gz | 211,817,337 | 41,248 | # Multiply 2-digit numbers by 6 using the partitioning method
In this lesson, we are going to build on our knowledge of the 6 times table and use partitioning to multiply larger numbers accurately.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
## Question 4
Q1.5 x 8 =
1/4
Q2.? x 8 = 96
2/4
Q3.What does this array represent? Tick 2
Select two (2) boxes
3/4
Q4.If I buy 7 books costing £8 each, how much do I spend in total?
4/4
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
## Question 4
Q1.5 x 8 =
1/4
Q2.? x 8 = 96
2/4
Q3.What does this array represent? Tick 2
Select two (2) boxes
3/4
Q4.If I buy 7 books costing £8 each, how much do I spend in total?
4/4
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
Quiz:
# Multiply 2-digit numbers by 6 using the partitioning method
Let's see what you have learnt about multiplying 2-digit numbers by 6
## Question 4
Q1.15 x 6 = ?
1/4
Q2.30 x 6 = ?
2/4
Q3.Which of these is a multiple of 6 (tick 2)
Select two (2) boxes
3/4
Q4.For 6 days in a row I spend £11 on my lunch. How much did I spent in total?
4/4
Quiz:
# Multiply 2-digit numbers by 6 using the partitioning method
Let's see what you have learnt about multiplying 2-digit numbers by 6
## Question 4
Q1.15 x 6 = ?
1/4
Q2.30 x 6 = ?
2/4
Q3.Which of these is a multiple of 6 (tick 2)
Select two (2) boxes
3/4
Q4.For 6 days in a row I spend £11 on my lunch. How much did I spent in total?
4/4
# Lesson summary: Multiply 2-digit numbers by 6 using the partitioning method
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... approximately equal to 1 in (10 1.5 ) 5,000,000 = 10 7,500,000 . So each time the monkey starts typing, there is a chance of p =1/10 7,500,000 of completing Shakespeare. This probability is rather small but it ... | 759 | 2,474 | {"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-2018-39 | longest | en | 0.904931 |
https://spreadsheetpoint.com/round-numbers-in-google-sheets/ | 1,679,512,776,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.74/warc/CC-MAIN-20230322180852-20230322210852-00197.warc.gz | 628,740,707 | 93,711 | # How to Round Numbers in Google Sheets (Easy Formulas)
Decimals in your data are a necessary evil. While it’s good to keep as many decimal places as you can (to ensure accuracy), it may make your spreadsheet difficult and sort of unappetizing to read.
Moreover, for the sake of uniformity and symmetry, it’s always a good idea to have all your data values rounded off to the same number of decimal places.
There are a few methods of how to round numbers in Google Sheets , and we are going to take a look at 5 of them in this tutorial.
## Different Rounding Functions in Google Sheets
There are a few different Google Sheets round function types, each of which serves a slightly different purpose. Depending on your requirement, you can choose anyone to round off your data values:
• ROUND
• ROUNDUP
• ROUNDDOWN
• MROUND
## How Does Rounding of Numbers Work (The Logic Behind It)
The ROUND function takes a numeric value and rounds it to a specified number of decimal places, according to standard rules.
The standard rules of rounding are as follows:
• If the digit to the right of the digit to be rounded is less than five, then it remains unchanged. In the other words, the number is ‘rounded down’ to the nearest digit
• If the digit to the right of the digit to be rounded is greater than or equal to 5, then it is incremented by 1. In other words, the number is ‘rounded up’ to the nearest digit.
For example, when rounding off the number 1.263 to the second decimal place, the digit to be rounded is 6. The digit to the right of 6 is 3 (which is less than 5). So the digit to be rounded remains the same, and the final result after rounding is 1.26.
If instead, you are trying to round the number 1.267 to the second decimal place, the digit to be rounded is 6. The digit to the right of 6 is 7 (which is greater than 5). So the digit to be rounded is increased by 1 and the final result after rounding is 1.27.
## How to ROUND Numbers in Google Sheets (using the ROUND function)
The syntax for the ROUND function is as follows:
`ROUND(value, [places])`
Here,
• value is the number that you want to round. This may be a numeric value or reference to a cell containing a numeric value.
• places is the number of digits or decimal places to which you want to round value. This parameter is optional. When it is not specified, its value is assumed to be 0 by default.
### An Example Use Google Sheets to Round to 2 Decimal Places ( Or Any Other Choice)
Let us take a look at some examples of the round function Google Sheets uses to round the given values of column A to the number of places specified in column B, using the ROUND function:
From the above image you will observe that Google Sheets limits decimal places like so:
• In row 2, we want to round off the value 213.146 to 1 decimal place. According to standard rounding rules, the rounding digit, 1 remains the same. So, the rounded number now becomes 213.1. We have essentially ‘rounded down’ the value 213.145.
• In row 3, we want to round off the value 213.146 to 2 decimal places. According to standard rounding rules, the rounding digit, 4 is increased by 1. The rounded number now becomes 213.15. We have essentially ‘rounded up the value 213.145.
• In row 5, there is no places parameter provided in the formula, so the default value 0 will be used. This means we want to round off the value 213.146 to 0 decimal places, in other words, to the nearest integer. According to standard rounding rules, the nearest integer to 213.146 is 213. Again, we essentially ‘rounded down’ the value 213.146.
• In row 10, there is again no places parameter. So we again need to round off the value to the nearest integer. The nearest integer to 213.642 is 214. Here, we essentially ‘rounded up’ the value 213.642.
### Limit Decimal Places With Negative Values in Google Sheets
The ROUND function can also be used with negative values for the places parameter. In such cases, the value is rounded to the left of the decimal point. So,
• if places is -1, then the ROUND function will round the value to the nearest tens.
• if places is -2, then the ROUND function will round the value to the nearest hundreds.
• if places is -3, then the ROUND function will round the value to the nearest thousands.
and so on.
Check also: Limitations of Google Sheets
Let us take a look at a few more examples to understand how the ROUND function works with negative values for the place parameter:
From the above image you will observe that:
• In row 2, we round off the value 213.146 to -1 places. The function removes all digits on the right of the decimal point. It then rounds the value to the left of the decimal point to the nearest tens. The nearest tens to the number 13 is 10. So, the function rounds down the value to 210.
• In row 3, we round off the value 213.146 to -2 places. The function rounds the integer portion of the value to the nearest hundreds. The nearest hundreds to the number 213 is 200. So, the function rounds down the value to 200.
• In row 6, we round off the value 266.142 to -1 places. The function rounds the integer portion of the value to the nearest tens. The nearest tens to the number 66 is 70. So, the function rounds up the value to 270.
• In row 9, we round off the value 656.142 to -3 places. The function rounds the integer portion of the value to the nearest thousands. The nearest thousands to the number 656 is 1000. So, the function rounds up the value to 1000.
It’s clear from the above examples that the ROUND function either rounds up or rounds down the given value, depending on the standard rounding rules. But what if you wanted to ensure that your value only gets rounded up and never down?
For such cases, you can make use of the ROUNDUP Google Sheets function.
## How to Round Up in Google Sheets – The ROUNDUP Function
The ROUNDUP function works the same way as the ROUND function, except that it always rounds the value upward. This functions allows Google Sheets to roundup to the nearest one or any other choice you may have.
### Syntax of the ROUNDUP Function
The syntax for the ROUNDUP function is the same as that of the ROUND function:
`ROUNDUP(value, [places])`
### Examples using the ROUNDUP Function to Limit Decimal Places
Let us take a look at some examples to see how the ROUNDUP function works:
From the above image, it is quite clear that the ROUNDUP function always rounds up the value to the given number of decimal places.
Similar to the ROUND function, the ROUNDUP function also supports negative values for the places parameter.
Here are some examples to help you understand how the ROUNDUP function works with negative values for the places parameters:
## How to Round Down Numbers in Google Sheets
The ROUNDDOWN function works the same way as the ROUND function, except that it always rounds the value downward.
### Syntax of the ROUNDDOWN Function
The syntax for the ROUNDDOWN function is the same as that of the ROUND function:
`ROUNDDOWN(value, [places])`
### Examples using the ROUNDDOWN Function
Let us take a look at some examples to see how the ROUNDDOWN function works:
From the above image, it is quite clear that the ROUNDDOWN function always rounds down the value to the given number of decimal places.
Similar to the ROUND function, the ROUNDDOWN function also supports negative values for the places parameter.
Here are some examples to help you understand how the ROUNDDOWN function works with negative values for the places parameters:
## How to Round in Google Sheets to the Nearest Integer Multiple (MROUND Function)
Using negative values with the above rounding functions can help convert your numbers to the nearest multiples of 10, 100, 1000, etc. However, what if you want to convert values to the nearest multiple of some other number, like 2, 3, 15, etc.?
Google Sheets has something for that too. It lets you use the MROUND function. This function works the same way as the ROUND function, except that it lets you round a value to the nearest integer multiple of another value.
### Syntax of the MROUND Function
The syntax for the MROUND function is similar to that of the ROUND function:
`MROUND(value, factor)`
Here,
• value is the number you want to round.
• factor is the number to whose multiple the value will be rounded.
Unlike ROUND, ROUNDDOWN, and ROUNDUP functions, you cannot use negative values in the second parameter of the MROUND function, unless the first parameter is also a negative number.
### Examples using the MROUND Function
Let us take a look at some examples to see how the MROUND function works:
From the above image you will observe that:
• In row 2, the MROUND function rounds the value 213.142 to the nearest multiple of 2. So, we get the result as 214.
• Similarly, in row 9, the MROUND function rounds the value 565.142 to the nearest multiple of 15. So, we get the result as 570.
In this tutorial, we showed you how to round numbers in Google Sheets using four different rounding functions in Google Sheets. These included the ROUND, ROUNDUP, ROUNDDOWN, and MROUND functions.
The ROUND function can help you round values according to standard rules, while the ROUNDUP function ensures that your values always get rounded upwards. Similarly, the ROUNDDOWN function ensures that your values always get rounded downwards.
The MROUND function, on the other hand, lets you round values to multiples of a certain integer.
## Using the CEILING Function in Google Sheets to Round Numbers
You can also use the CEILING function to round up numbers in Google Sheets to the nearest 1 or any other specified place.
The syntax for the CEILING function goes as follows:
`=CEILING(Number, Significance)`
In this syntax:
• Number is the number you want to round (usually a cell reference)
• Significance is the decimal place value that you wish the number to be rounded up to
Check out the below example
As you can see to show the result to one decimal place the significance is set to 0.1, which means the result will be shown in multiples of 0.1
A similar approach is taken to show results in multiples of 0.01, 0.001, and 1. You could also use the CEILING function to round up into multiples of 10s, 100s, 1000s etc.
## Which Google Sheets Round Function Should You Use?
Once you know what each rounding function does, applying them according to your needs becomes second nature. Just think about whether you need to round up, down, or to the nearest figure.
We hope this tutorial has enabled you to understand the differences between each of the rounding functions and now you know how to round numbers in Google Sheets.
Other Google Sheets tutorials you may like:
### How to Apply Formula to Entire Column in Google Sheets
#### Sumit
Sumit is a Google Sheets and Microsoft Excel Expert. He provides spreadsheet training to corporates and has been awarded the prestigious Excel MVP award by Microsoft for his contributions in sharing his Excel knowledge and helping people.
### 2 thoughts on “How to Round Numbers in Google Sheets (Easy Formulas)”
1. I think your explanation of rounding has the terms ’rounded up’ and ’rounded down’ reversed. You said:
The ROUND function takes a numeric value and rounds it to a specified number of decimal places, according to standard rules.
The standard rules of rounding are as follows:
If the digit to the right of the digit to be rounded is less than five, then it remains unchanged. In the other words, the number is ‘rounded up’ to the nearest digit
If the digit to the right of the digit to be rounded is greater than or equal to 5, then it is incremented by 1. In other words, the number is ‘rounded down’ to the nearest digit. | 2,692 | 11,762 | {"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-2023-14 | longest | en | 0.895617 |
http://forums.wolfram.com/mathgroup/archive/2005/Sep/msg00289.html | 1,527,302,499,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867277.64/warc/CC-MAIN-20180526014543-20180526034543-00330.warc.gz | 116,248,594 | 7,194 | Re: Set of strings reducing problem (Again!)
• To: mathgroup at smc.vnet.net
• Subject: [mg60348] Re: Set of strings reducing problem (Again!)
• From: "Valeri Astanoff" <astanoff at yahoo.fr>
• Date: Wed, 14 Sep 2005 03:27:15 -0400 (EDT)
• References: <dg69pb\$a96\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
Edson,
This is what I modified :
-1- Make Dispatch table order independant
-2- Insert z___List between p and q for not neighbor cases
In[1]:=
Unprotect[D];
U={"2","X"};
M={"1","2"};
D={"1","X"};
T={"1","2","X"};
L=Flatten[Outer[StringJoin,T,T,T,D]];
L = Select[L, Count[Characters[#], "1"] > 1 &];
cl = Characters /@ L;
r = Dispatch[{"1" + "X" -> "D",
"1" + "2" -> "M",
"1" + "U" -> "T",
"X" + "2" -> "U",
"X" + "M" -> "T",
"2" + "D" -> "T",
"X" + "1" -> "D",
"2" + "1" -> "M",
"U" + "1" -> "T",
"2" + "X" -> "U",
"M" + "X" -> "T",
"D" + "2" -> "T"}];
ncl = StringJoin @@@ (
cl //. {x___List,
{a___, p_String, c___},
z___List,
{a___, q_String, c___},
y___List} :>
{x, {a, p + q /. r, c}, z,
y} /; StringQ[p + q /.
r])
Out[10]=
{11TD,1U1D,1UU1,U11D,U1U1,UU11}
Seems to work...
v.a.
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• Next by thread: Re: Set of strings reducing problem (Again!) | 523 | 1,343 | {"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-2018-22 | latest | en | 0.797077 |
http://mathhelpforum.com/advanced-algebra/131378-prove-cubed-inverse-matrix-same-inverse-cubed-matrix-print.html | 1,495,483,524,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607046.17/warc/CC-MAIN-20170522190443-20170522210443-00312.warc.gz | 246,558,176 | 3,091 | # Prove a cubed inverse matrix is the same as an inverse cubed matrix
• Mar 1st 2010, 03:02 AM
Orbent
Prove a cubed inverse matrix is the same as an inverse cubed matrix
I have been asked to prove that the for a generic matrix A that, $(A^3)^{-1}$ is the same as $(A^{-1})^3$. I have no idea how to start this, any ideas would be much appreciated.
Pretty much the wording is prove that A cubed, then inversed is the same as A inversed then cubed.
• Mar 1st 2010, 03:30 AM
HallsofIvy
$\left(A^3\right)^{-1}$ is, by definition, the matrix, B, such that $B(A^3)= (A^3)B= I$, the identity matrix. Show that this is true for $(A^{-1})^3A^3$ and $A^3(A^{-1})^3$.
• Mar 1st 2010, 04:20 AM
Orbent
Done.
Thank you very much, i know what to do. I'll do it just in case anyone else looks at this thread.
First $(A^{-1})^3A^3$
$(A^{-1})^3$ becomes $A^{-1}A^{-1}A^{-1}$
$(A^3)$ becomes $AAA$
and $AA^{-1} = I$ and $AI=A$ and $A^{-1}I=A^{-1}$
Combining these facts makes:
$AAAA^{-1}A^{-1}A^{-1}$ this becomes,
$AAIA^{-1}A^{-1}$ which becomes,
$AAA^{-1}A^{-1}$, after we do that a few times we get,
$I$
This also works for,
$A^{-1}A^{-1}A^{-1}AAA$
Please let me know if i did that right :)
• Mar 1st 2010, 06:22 AM
HallsofIvy
Yes, that's exactly what you needed to do.
Now you use the fact that a matrix has at most one inverse to argue that $(A^{-1})^3= (A^3)^{-1}$. | 492 | 1,356 | {"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": 20, "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} | 4.28125 | 4 | CC-MAIN-2017-22 | longest | en | 0.906405 |
http://www.oxforddictionaries.com/search/english/?q=stokes&multi=1 | 1,429,400,476,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246636255.43/warc/CC-MAIN-20150417045716-00118-ip-10-235-10-82.ec2.internal.warc.gz | 718,321,571 | 9,859 | # Dictionary search results
Showing 1-11 of 11 results
## stokes British & World English
The cgs unit of kinematic viscosity, corresponding to a dynamic viscosity of 1 poise and a density of 1 gram per cubic centimetre, equivalent to 10−4 square metres per second
## stoke British & World English
Add coal or other solid fuel to (a fire, furnace, boiler, etc.)
## Stokes, Carl British & World English
(1927–96), US politician; full name Carl Burton Stokes. He was the first African-American mayor of a major US city (Cleveland 1968–72). After serving as mayor, he was a newscaster and a municipal court judge before being appointed as US ambassador to the Seychelles 1993–96
## Stokes in Cheyne–Stokes breathing British & World English
A cyclical pattern of breathing in which movement gradually decreases to a complete stop and then returns to normal. It occurs in various medical conditions, and at high altitudes
## Stokes in stokes British & World English
The cgs unit of kinematic viscosity, corresponding to a dynamic viscosity of 1 poise and a density of 1 gram per cubic centimetre, equivalent to 10−4 square metres per second
## Stokes in Stokes' law British & World English
A law stating that in fluorescence the wavelength of the emitted radiation is longer than that of the radiation causing it. This is not true in all cases
## Stokes in Stokes' theorem British & World English
A theorem proposing that the surface integral of the curl of a function over any surface bounded by a closed path is equal to the line integral of a particular vector function round that path
## Stokes' law British & World English
A law stating that in fluorescence the wavelength of the emitted radiation is longer than that of the radiation causing it. This is not true in all cases
## Stokes' theorem British & World English
A theorem proposing that the surface integral of the curl of a function over any surface bounded by a closed path is equal to the line integral of a particular vector function round that path
## Cheyne–Stokes breathing British & World English
A cyclical pattern of breathing in which movement gradually decreases to a complete stop and then returns to normal. It occurs in various medical conditions, and at high altitudes
## Carl Burton Stokes in Stokes, Carl British & World English
(1927–96), US politician; full name Carl Burton Stokes. He was the first African-American mayor of a major US city (Cleveland 1968–72). After serving as mayor, he was a newscaster and a municipal court judge before being appointed as US ambassador to the Seychelles 1993–96 | 570 | 2,600 | {"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-2015-18 | latest | en | 0.923157 |
https://emedia.rmit.edu.au/learninglab/content/a11-algebraic-operations | 1,696,295,303,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511023.76/warc/CC-MAIN-20231002232712-20231003022712-00003.warc.gz | 240,339,037 | 23,704 | ## A1.1 Algebraic operations
A page from Al-Khwarizmis “al-Kitab al-muhtasar fi hisab al-gabr wa-l-muqabala”, an early book on algebra. (Image from en.wikipedia.org )
This section introduces the basic skills for addition, subtraction, multiplication and division (+ - × ÷) of algebraic expressions.
Courses in science, engineering and other fields require you to have these skills.
#### Transcript
M: Hi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on operations with algebra. First of all make sure you can tell the difference between like terms and unlike terms. Like terms contain exactly the same pro-numerals, in other words letters or variables, so look at the like terms down the left hand column and compare them with the unlike terms in the right hand column, in particular look at the last term, seven E F and eight F E, they are in fact like terms. Only like terms can be added or subtracted. For instance, in this example, seven E plus 10 E, the terms are alike, therefore they can be added to give you 17 E. But look at this one, be careful because here we have two terms, the X squared term and the X term so we have to add them separately, so three X squared minus X squared minus four X squared gives us minus two X squared, and two X, which is different to the X squared term must be left alone so the answer is minus two X squared plus two X.
Similarly if you look at this example in red, there’s a U V term which is the same as V U and that is different to the U term therefore that gives us minus two U V plus three U, we keep them separate, we can’t add minus two U V to three U. And finally six R squared S minus two R S squared, they look similar but they are not the same, these are two unlike terms so we can’t simplify this expression.
Now before we move on to multiplying and dividing algebraic terms make sure you are familiar with the multiplication and dividing of positive and negative numbers on this slide. So study this carefully before moving onto the next slide.
But let’s multiply some pro-numerals. Here’s the first example, minus four times three B, multiply the numbers together, minus four and minus three gives you plus 12 and then just tag on the B afterwards so the answer is 12 B. Here’s another one, multiply out the numbers first, three times minus 15 gives you minus 15, now multiply the pro-numeral, E and E squared is E cubed, so the answer is minus 15 E cubed.
The next example in red multiply out, as before, the numbers, minus two times minus four is plus eight, then the pro-numerals, there’s a U squared term and there are Vs there, V times V is a V squared, so the answer is eight U squared V squared. Notice in these examples that because we’re multiplying the actual terms can be different, so make sure you know the difference between adding and subtracting where the pro‑numerals must be exactly the same, multiplying and dividing the pro-numerals can be different.
Finally minus three P Q times minus two Q times P, multiply out the numbers, that gives you plus six, P times P is P squared, Q times Q is Q squared, so the answer is six P squared Q squared. Now in this slide we’re dividing pro-numerals, so the first term minus 12 W Y Z divided by the second term three Y Z, express that as a fraction first and look for terms that will cancel top and bottom, in this case the Ys cancel top and bottom and the Zs cancel top and bottom, notice three will divide into 12 four times, so the answer is minus four W.
Here’s another one slightly more complicated. Now in this slide we’re dividing pro‑numerals, so the first term minus 12 W Y Z divided by the second term three Y Z, express that as a fraction first and look for terms that will cancel top and bottom, in this case the Ys cancel top and bottom and the Zs cancel top and bottom, notice three will divide into 12 four times so the answer is minus four W.
Here’s another one slightly more complicated. Writing it as a fraction, minus two M squared N over six M N squared, and then in the third term I’ve expanded the M squared so it’s M times N and then I’ve tacked the N on the end of that and I’ve done the same thing with the denominator with N squared N times N, so you can clearly see where the Ms and the Ns cancel, giving you an answer of minus M over three N. Notice minus on the top divided by plus on the bottom gives you an answer of minus.
And finally in blue there’s a multiplication and a division so you again express that as a fraction, expand the A squared and then cancel top and bottom terms giving you an answer of two A. Now we’ll move on to some slightly harder problems with algebra, but notice the rule that we use first of all, in other words if an expression has brackets in it that must be done first, secondly if there are indices you do that next, thirdly you do multiplying and dividing and lastly you do adding and subtraction.
So let’s do some examples. Three S T minus three S times four T, notice we have a negative sign and a multiplication sign, so the times is worked out first, so three S times four T is 12 S T, notice now we have subtraction, three S T minus 12 S T, notice the terms are alike therefore we can subtract them, three minus 12 is minus nine S T.
Let’s look at this one. We’ve got two terms in brackets and a division, so the brackets are worked out first, even though there’s a plus inside the brackets must be worked out first, which gives us six X Y, then we divide that by two X, express that as a fraction, cancel any terms, Xs cancel here, twos go into six three times so the answer is three Y.
And finally look at this one. We have brackets, multiplication, addition and division. So the brackets is worked out first, that gives as minus 10 A squared B squared in the second line, then I write out the rest, plus two A cubed B cubed divided by A B, what do we do next, well we have division which becomes ... comes before addition, therefore we divide the two A cubed B cubed by A B, writing that as a fraction, simplifying by cancelling like terms and then in the final line we’ve just got two terms which we can add together. Notice again that the terms are exactly the same, A squared B squared, therefore we can add them, minus 10 plus two gives us minus eight A squared B squared.
Now try some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short video.
### Like and Unlike Terms
Like terms contain exactly the same pro-numerals. 1 A pro-numeral is a combination of letters, symbols and numbers. For example, $$3x$$ is a pro-numeral. In algebra, when we write $$3x$$ we really mean $$3\times x$$. Another example of a pro-numeral is $$5ab$$. This means $$5\times a\times b$$. Note that $$5ab=5ba$$. The order of the letters and numbers is not important but usually we write numbers first and use alphabetical order for the letters. While $$5ba$$ is correct, it is more common to write $$5ab$$. The following table gives some examples of like and unlike terms.
Like Terms Unlike Terms
$$3x,5x$$ $$3x,5y$$
$$2a,-3a$$ $$3a,3$$
$$m^{2},7m^{2}$$ $$m^{2},7m$$
$$2ab,3ab$$ $$2a^{2}b,3ab^{2}$$
$$3xyz,5xyz$$ $$3xyz,3xy$$
$$3ef,3fe$$ $$3ef,\,3eg$$
$$6\alpha\beta$$, $$\beta\alpha$$ $$6\alpha\beta\gamma$$, $$6\alpha\beta$$
For some practice, please go to Exercise 1 below.
The essential rule is:
Only like terms may be added or subtracted.
Example 1. \begin{align*} 7e+10e & =7\times e+10\times e\\ & =\left(7+10\right)e\\ & =17e. \end{align*} In practice we would not put in the first line. It is included to reinforce what $$7e$$ and $$10e$$ mean. In practice we would probably go straight to the answer. That is $$7e+10e=17e$$.
Example 2. \begin{align*} 3x^{2}-x^{2}-4x^{2} & =\left(3-1-4\right)x^{2}\\ & =-2x^{2}. \end{align*}
Example 3. 2 This example shows how we can gather different like terms. In this case $$m$$ and $$n$$. \begin{align*} 3m-4n+6m+n & =\left(3+6\right)m+\left(-4+1\right)n\\ & =9m-3n. \end{align*}
Example 4. \begin{align*} 3a-b-5a+4ab-3b+ab & =\left(3-5\right)a+\left(-1-3\right)b+\left(4+1\right)ab\\ & =-2a-4b+5ab. \end{align*}
Example 5. \begin{align*} 3x-x^{2} & \quad\mathrm{there\textrm{ are no like terms so nothing can be done.}} \end{align*}
Example 6. \begin{align*} p+2p-3 & =3p-3. \end{align*}
Example 7. $8uv+3u-10vu=-2uv+3u.$
Example 8. $6r^{2}s-2rs^{2}\quad\mathrm{there\textrm{ are no like terms so nothing can be done.}}$
For some practice, please go to Exercise 2 below.
### Multiplication
Both like and unlike terms may be multiplied. When multiplying two or more terms consider: 1. The sign of the answer.3 In doing the multiplication, remember the basic rules for determining the sign of the product. \begin{align*} \left(+ve\right)\times\left(+ve\right) & =\left(+ve\right)\\ \left(+ve\right)\times\left(-ve\right) & =\left(-ve\right)\\ \left(-ve\right)\times\left(+ve\right) & =\left(-ve\right)\\ \left(-ve\right)\times\left(-ve\right) & =\left(+ve\right). \end{align*}
1. The product of the terms.
Here are some examples:
$\begin{array}{l} 1.\quad\left(-4\right)\times\left(-3b\right)=12b\\ 2.\quad-2\times6y=-12y\\ 3.\quad2e\times\left(-5e^{2}\right)=-10e^{3}\\ 4.\quad\left(-2u^{2}v\right)\times\left(-4v\right)=8u^{2}v^{2}\\ 5.\quad-3pq\times\left(-2q\right)\times p=6p^{2}q^{2} \end{array}$
### Division
In algebra, we often use fractions. For example: 4 In this case the top term, 2 is called the numerator, the bottom term 5, is called the denominator and the horizontal line between them is called the fraction bar, division bar or vinculum. $\frac{2}{5}=2\div5.$ Similarly for algebraic terms: $\frac{2x^{2}yz}{3xy^{2}}=2x^{2}yz\div3xy^{2}.$ When dividing algebraic terms:
1. Rewrite as a fraction if necessary
2. Expand any powers
3. Establish the sign of the answer5 In doing the division, remember the basic rules for determining the sign of the quotient. \begin{align*} \left(+ve\right)\div\left(+ve\right) & =\left(+ve\right)\\ \left(+ve\right)\div\left(-ve\right) & =\left(-ve\right)\\ \left(-ve\right)\div\left(+ve\right) & =\left(-ve\right)\\ \left(-ve\right)\div\left(-ve\right) & =\left(+ve\right). \end{align*}
4. Cancel any common factors
#### Examples
1. Divide $$-12wyz$$ by $$3yz$$.
Solution: \begin{align*} -12wyz\div3yz & =\frac{-12wyz}{3yz}\quad\textrm{write as a fraction}\\ & =-\frac{12wyz}{3yz}\quad\textrm{determine sign}\\ & =-4w\quad\textrm{cancel common factors}. \end{align*}
2. Divide $$2m^{2}n$$ by $$-6mn^{2}$$.
Solution: \begin{align*} 2m^{2}n\div6mn^{2} & =\frac{2m^{2}n}{-6mn^{2}}\\ & =-\frac{2m^{2}n}{6mn^{2}}\\ & =-\frac{m}{3n}. \end{align*}
3. Simplify $$6a^{2}\times4ab\div12ab.$$
Solution: \begin{align*} 6a^{2}\times4ab\div12ab & =\frac{6a^{2}\times4ab}{12ab}\quad\textrm{write as a fraction}\\ & =\frac{24a^{3}b}{12ab}\quad\textrm{establish sign and expand powers}\\ & =2a^{2}\quad\textrm{cancel common factors}. \end{align*}
4. Divide $$m+2$$ by $$4m$$ $\left(m+2\right)\div4m=\frac{m+2}{4m}.$ There are no common factors and so it is not possible to simplify further.
For practice examples see Exercise 3.
### Order of Operations
The basic operations of arithmetic are multiplication, division, addition and subtraction. Usually the order in which we perform the operations is important.
For example, what is the answer to the following arithmetic problem $4+3\times2+5\;?$ At first sight there are several possible answers.
Working from left to right we get: \begin{align*} 4+3\times2+5 & =7\times2+5\\ & =14+5\\ & =19. \end{align*} Working from right to left we get: \begin{align*} 4+3\times2+5 & =4+3\times7\\ & =4+21\\ & =25. \end{align*}
If we do the multiplication first we get another answer: \begin{align*} 4+3\times2+5 & =4+6+5\\ & =15. \end{align*} Yet another answer is obtained by doing the additions first: \begin{align*} 4+3\times2+5 & =7\times7\\ & =49. \end{align*} Mathematicians don’t like imprecision. In order to allow only one answer to the problem we perform operations in the following order:
1. Brackets
2. Indices
3. Multiplication and Division from left to right
Using the first letters (bold) we get the abbreviation BIMDAS.
Applying this rule the answer to our problem is \begin{align*} 4+3\times2+5 & =4+6+5\quad\textrm{multiplication first}\\ & =15\quad\textrm{then addition}. \end{align*}
#### Examples:
1. $$3\times2+4=6+4=10.$$
2. $$3+2\times4=3+8=11$$.
3. $$\left(3+2\right)\times4=5\times4=20.$$
4. $$3-2^{2}=3-2\times2=3-4=-1$$.
5. $$\left(3-2\right)^{2}=\left(3-2\right)\times\left(3-2\right)=1\times1=1$$.
6. $$3^{2}-2^{2}=3\times3-2\times2=9-4=5$$.
7. $$3st-3s\times4t=3st-12st=-9st$$.
For some practice, see Exercise 4.
### Exercise 1
Which of the five terms on the right is a like term with the term on the left?
$\begin{array}{llllllll} 1) & 3x & | & 3 & 2x & 3x^{2} & 2xy & 4x^{2}a\\ 2) & 2ab & | & 2a & 2b & 3x^{2} & 6abc & 12ab\\ 3) & 2x^{2} & | & x & 2x & 5x^{2} & 4x & 4x^{3}\\ 4) & 3xy^{2} & | & 3 & 3xy & 3x^{2}xy^{2} & 3y^{2} & y^2x\\ 5) & 2m^{2}n & | & n & mn^{2} & 8mn & 2m^{2} & 4nm^{2}\\ 6) & 4ab^{2}c & | & 4 & 2ab^{2}c & 4abc & 8b^{2}c & 4cba \end{array}$
$\begin{array}{lllllllllllllll} 1.\;2x & & 2.\;12ab & & 3.\;5x^{2} & & 4.\;y^2x & & 5.\;4nm^{2} & & 6.\;2ab^{2}c\end{array}$
### Exercise 2
Simplify each of the following:
$\begin{array}{lllllll} & & a)\;13x+4-3x-1 & & & & b)\;10mn+5m+12mn+6m\\ & & c)\;3xy^{2}+2xy+5xy^{2}+3xy & & & & d)\;5xy+6m-2xy-2m\\ & & e)\;x+y+2x-y & & & & f)\;3a+5b-a-6b\\ & & g)\;x+4y-x-2y & & & & h)\;7x-4m-5x-3m\\ & & i)\;4x-5x-3y+5x & & & & j)\;9mn-3m-n+4m^{2} \end{array}$
$\begin{array}{llllllllll} & a)\;10x+3 & & b)\;22mn+11m & & c)\;8xy^{2}+5xy & & d)\;3xy+4m & & e)\;3x\\ & f)\;2a-b & & g)\;2y & & h)\;2x-7m & & i)\;5x-3y & & j)\;9mn-3m-n+4m^{2} \end{array}$
### Exercise 3
Simplify the following algebraic expressions:
$\begin{array}{lllllll} 1. & & a)\;5\times2k & & & & b)\;4a\times3ab\\ & & c)\;y\times3y & & & & d)\;4m\times\left(-3mn\right)\\ & & e)\;m\times39\times5 & & & & f)\;2ab\times3bc\times\left(-4\right)\\ & & g)\;2ab^{2}\times3ac & & & & h)\;4m\times\left(-5kmp\right)\\ \\ \\ 2. & & a)\;18ef\div6f & & & & b)\;-100uvw\div100w\\ & & c)\;24gh^{2}\div8gh & & & & d)\;3m^{2}n\div12mn^{2}\\ & & e)\;rs\times2st\div2s & & & & f)\;3jk\times12km\div9jkm\\ & & g)\;10p\times3qp\div16pq & & & & h)\;4yz\times5w^{2}z\div10wy \end{array}$
$\begin{array}{llllllll} 1. & a)\;10k & & b)\;12a^{2}b & & c)\;3y^{2} & & d)\;-12m^{2}n\\ & e)\;15mp & & f)\;-24ab^{2}c & & g)\;6a^{2}b^{2}c & & h)\;-20km^{2}p\\ \\ 2. & a)\;3e & & b)\;-uv & & c)\;3h & & d)\;m/4n\\ & e)\;rst & & f)\;4k & & g)\;15p/8 & & h)\;2wz^{2} \end{array}$
### Exercise 4
Simplify the following expressions:
$\begin{array}{lllll} 1)\;18+\left(3\times\left(-5\right)\right) & & & & 2)\;3\times\left(-4\right)+\left(8\times2\right)\\ 3)\;10-5^{2}+3 & & & & 4)\;\left(10-5\right)^{2}+3\\ 5)\;10^{2}-5^{2} & & & & 6)\;\left(10-5\right)^{2}\\ 7)\;3m+2\times3m & & & & 8)\;6ab-3a\times4b\\ 9)\;16gh-4gh\times4 & & & & 10)\;9b-3b\times2k+2k\times b \end{array}$
$\begin{array}{lllllllll} 1.\;3 & & 2.\;4 & & 3.\;-12 & & 4.\;28 & & 5.\;75\\ 6.\;25 & & 7.\;9m & & 8.\;-6ab & & 9.\;0 & & 10.\;9b-4bk \end{array}$ | 5,140 | 15,234 | {"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} | 4.53125 | 5 | CC-MAIN-2023-40 | latest | en | 0.929303 |
http://classroom.synonym.com/school-project-idea-order-operations-12201494.html | 1,516,608,300,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891196.79/warc/CC-MAIN-20180122073932-20180122093932-00624.warc.gz | 78,744,417 | 11,268 | Using a project-based approach for teaching the order of operations to young math students allows them to practice and show their knowledge of math facts in an interesting way. As students create their projects, they use the procedures they have learned in class to demonstrate the order of operations in action or explain the process to an audience.
## Calendar Project
Assigning a project based on the current month’s calendar allows students to delve into using the order of operations coupled with their problem-solving skills to work backward and find equations to go with a provided answer. The goal of the project is for students to write equations using multiple numbers and operations to result in the number of each day in the month.
For example, for November 12, the student would create an equation that equals 12. Students may come up with (1 + 6) + 10 - 5 and write this equation along with the steps for solution in that calendar space. Provide criteria for the amount of numbers or steps you would like in each equation and limiting the use of numbers one and zero.
## Informational Brochure
By creating an informational brochure about the order of operations, students not only work through problems providing examples, but also explain the steps for using the order of operations. Assign a brochure project in which students fold a piece of paper into a tri-fold and create an appealing document that could help someone learn about the order of operations. You may want to assign criteria for what should be in each of the sections, such as a detailed explanation of each step, a way to remember the steps, examples of equations solved using the order of observations, and a statement of why using the order of operations is important.
## Mobile Operations
Creating a mobile is a fun way for students to highlight the step-by-step process of solving a multi-step equation using order of operations. Students begin at the top of the mobile with an equation using various operations and groupings. To create each subsequent layer, students use the order of operations to simplify the equation step by step on strips of paper. Students assemble them using a coat hanger and string with the original equation at the top, the steps to solve in the middle and the answer at the bottom.
For example, if the top layer was 3 x (5 + 8), the layer underneath would be 3 x 13. Hanging the mobiles up in the classroom reminds the students to approach these problems one step at a time and work their way down until they find the answer.
## Room Number Poster
As students take on a project to create posters labeling the classrooms in their school or hallway, they practice using the order of operations and writing equations to arrive at a specific answer. Give each student a different classroom number as the answer they must arrive at. Provide criteria for the number of steps that the equation should entail, and make it a requirement to use all of the operations. Students create a poster showing the steps of the order of operations they use to arrive at their assigned classroom number. When finished, students may enjoy displaying their work throughout the school. | 617 | 3,189 | {"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} | 4.28125 | 4 | CC-MAIN-2018-05 | latest | en | 0.938762 |
https://lampposthomeschool.com/how-to-make-a-math-memory/ | 1,722,995,866,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640667712.32/warc/CC-MAIN-20240807015121-20240807045121-00740.warc.gz | 278,838,743 | 32,464 | Home » Homeschool Math » How to Make a Math Memory
# How to Make a Math Memory
## Making math memories with your children is easier than you think.
One of my favorite memories of my grandmother is all about math, but not the way you think.
I spent a couple of summers with her in the beautiful piney woods of East Texas. The first summer my mother sent geography flashcards with me. My grandmother used them to teach me the countries of the world and their capitals. That was my happy geography memory from third grade.
Between 6th and 7th grade, I spent the summer again. She did not drive so trips to the grocery store were dependent on walking or having someone else drive her. Nana had very bad osteoporosis and walking several blocks with groceries was very difficult for her. Hence, it became the twice-weekly job of yours truly. It was the makings of my math memory with her.
Each week her small town newspaper had a flyer for the local grocery store. She read it with me and pointed out the best prices. Together we made a list with all the bargains and planned our meals around it. If there were any coupons—rare back then—she found them! It was my job to walk to the store and buy the groceries on our list. My reward for my work was a quarter to spend as I pleased.
With my hard-earned quarter, I usually bought an ice cream sandwich, which always melted before I made it home—unlike the ones you might have read about recently. Walking in the hot Texas sun, it’s hard to eat an ice cream sandwich and carry a bag of groceries at the same time. Occasionally, I splurged on one of those bigger comic books: Superman or Archie.
The homeschool style math lessons/memories served me well. Later after I learned to drive, I grocery shopped for my mom. Even now, I usually buy mostly the foods on sale. I mentally keep track of how much I am spending–probably to make sure I have enough to have my precious quarter left over. :-) It was my job to make sure I had 25 cents leftover from the five dollar bill my grandmother gave me for the groceries.
### It’s Easy to Make a Math Memory
Here are some quick and easy tips to help you create a math memory with your children.
• Choose a math activity you can do together with your children without a math textbook by:
• Checking your math teacher’s guides for ideas
• Looking for Hands-on Math Activity books at your local library
• Brainstorming ideas with your children to create something special for your family
• Searching the internet for family math activities
• Ask the children to mark your calendar with the time and place
• Before the event, be sure to gather the supplies you need
• Repeat if desired or start over and create another math memory
Are you making happy math memories with your children? I’d love to hear some ideas from you.
Blessings,
Harriet
Scroll to Top | 620 | 2,848 | {"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-2024-33 | latest | en | 0.984707 |
https://www.cbsesyllabus.in/vocational/data-science-class-10-syllabus | 1,660,838,701,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573242.55/warc/CC-MAIN-20220818154820-20220818184820-00658.warc.gz | 616,122,252 | 4,658 | # Data Science Class 10 Syllabus
The syllabus consists of five units: (i) Use of statistics in Data Science (ii) Distributions in Data Science (iii) Identifying Patterns (v) Data Merging (v) Ethics in Data Science.
### 1. Use of statistics in Data Science
1. Introduction
2. What are subsets?
3. Two-way frequency table
4. Interpreting two-way tables
5. Two-way relative frequency table
6. Meaning of mean
7. Median
8. Mean Absolute Deviation
9. What is Standard Deviation?
### 2. Distributions in Data Science
1. Introduction
2. What is distribution in data science?
3. What are different types of distributions?
4. Statistical Problem Solving Process
### 3. Identifying Patterns
1. What is partiality, preference and prejudice?
2. How to identify the partiality, preference and prejudice?
3. Probability for Statistics
4. The Central Limit Theorem
5. Why is the Central Limit Theorem important?
### 4. Data Merging
1. Overview of Data Merging
2. What is Z-Score?
3. How to calculate a Z-score?
4. How to interpret the Z-score?
5. Why is a Z-score so important?
6. Concept of Percentiles
7. Quartiles
8. Deciles
### 5. Ethics in Data Science
1. Note about data governance framework
2. Ethical guidelines around data analysis | 317 | 1,236 | {"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} | 2.765625 | 3 | CC-MAIN-2022-33 | latest | en | 0.77014 |
https://codedocs.org/what-is/recursion | 1,701,591,393,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100489.16/warc/CC-MAIN-20231203062445-20231203092445-00845.warc.gz | 207,023,966 | 20,848 | # Recursion
17:26
A visual form of recursion known as the Droste effect. The woman in this image holds an object that contains a smaller image of her holding an identical object, which in turn contains a smaller image of herself holding an identical object, and so forth. 1904 Droste cocoa tin, designed by Jan Misset
Recursion (adjective: recursive) occurs when a thing is defined in terms of itself or of its type. Recursion is used in a variety of disciplines ranging from linguistics to logic. The most common application of recursion is in mathematics and computer science, where a function being defined is applied within its own definition. While this apparently defines an infinite number of instances (function values), it is often done in such a way that no infinite loop or infinite chain of references can occur.
## Formal definitions
Ouroboros, an ancient symbol depicting a serpent or dragon eating its own tail.
In mathematics and computer science, a class of objects or methods exhibits recursive behavior when it can be defined by two properties:
• A simple base case (or cases) — a terminating scenario that does not use recursion to produce an answer
• A recursive step — a set of rules that reduces all successive cases toward the base case.
For example, the following is a recursive definition of a person's ancestor. One's ancestor is either:
• One's parent (base case), or
• One's parent's ancestor (recursive step).
The Fibonacci sequence is another classic example of recursion:
Fib(0) = 0 as base case 1,
Fib(1) = 1 as base case 2,
For all integers n > 1, Fib(n) = Fib(n − 1) + Fib(n − 2).
Many mathematical axioms are based upon recursive rules. For example, the formal definition of the natural numbers by the Peano axioms can be described as: "Zero is a natural number, and each natural number has a successor, which is also a natural number."[1] By this base case and recursive rule, one can generate the set of all natural numbers.
Other recursively defined mathematical objects include factorials, functions (e.g., recurrence relations), sets (e.g., Cantor ternary set), and fractals.[2]
There are various more tongue-in-cheek definitions of recursion; see recursive humor.
## Informal definition
Recently refreshed sourdough, bubbling through fermentation: the recipe calls for some sourdough left over from the last time the same recipe was made.
Recursion is the process a procedure goes through when one of the steps of the procedure involves invoking the procedure itself. A procedure that goes through recursion is said to be 'recursive'.[3]
To understand recursion, one must recognize the distinction between a procedure and the running of a procedure. A procedure is a set of steps based on a set of rules, while the running of a procedure involves actually following the rules and performing the steps.
Recursion is related to, but not the same as, a reference within the specification of a procedure to the execution of some other procedure.
When a procedure is defined as such, this immediately creates the possibility of an endless loop; recursion can only be properly used in a definition if the step in question is skipped in certain cases so that the procedure can complete.
But even if it is properly defined, a recursive procedure is not easy for humans to perform, as it requires distinguishing the new from the old, partially executed invocation of the procedure; this requires some administration as to how far various simultaneous instances of the procedures have progressed. For this reason, recursive definitions are very rare in everyday situations.
## In language
Linguist Noam Chomsky, among many others, has argued that the lack of an upper bound on the number of grammatical sentences in a language, and the lack of an upper bound on grammatical sentence length (beyond practical constraints such as the time available to utter one), can be explained as the consequence of recursion in natural language.[4][5]
This can be understood in terms of a recursive definition of a syntactic category, such as a sentence. A sentence can have a structure in which what follows the verb is another sentence: Dorothy thinks witches are dangerous, in which the sentence witches are dangerous occurs in the larger one. So a sentence can be defined recursively (very roughly) as something with a structure that includes a noun phrase, a verb, and optionally another sentence. This is really just a special case of the mathematical definition of recursion.
This provides a way of understanding the creativity of language—the unbounded number of grammatical sentences—because it immediately predicts that sentences can be of arbitrary length: Dorothy thinks that Toto suspects that Tin Man said that.... There are many structures apart from sentences that can be defined recursively, and therefore many ways in which a sentence can embed instances of one category inside another.[6] Over the years, languages in general have proved amenable to this kind of analysis.
Recently, however, the generally accepted idea that recursion is an essential property of human language has been challenged by Daniel Everett on the basis of his claims about the Pirahã language. Andrew Nevins, David Pesetsky and Cilene Rodrigues are among many who have argued against this.[7] Literary self-reference can in any case be argued to be different in kind from mathematical or logical recursion.[8]
Recursion plays a crucial role not only in syntax, but also in natural language semantics. The word and, for example, can be construed as a function that can apply to sentence meanings to create new sentences, and likewise for noun phrase meanings, verb phrase meanings, and others. It can also apply to intransitive verbs, transitive verbs, or ditransitive verbs. In order to provide a single denotation for it that is suitably flexible, and is typically defined so that it can take any of these different types of meanings as arguments. This can be done by defining it for a simple case in which it combines sentences, and then defining the other cases recursively in terms of the simple one.[9]
A recursive grammar is a formal grammar that contains recursive production rules.[10]
### Recursive humor
Recursion is sometimes used humorously in computer science, programming, philosophy, or mathematics textbooks, generally by giving a circular definition or self-reference, in which the putative recursive step does not get closer to a base case, but instead leads to an infinite regress. It is not unusual for such books to include a joke entry in their glossary along the lines of:
Recursion, see Recursion.[11]
A variation is found on page 269 in the index of some editions of Brian Kernighan and Dennis Ritchie's book The C Programming Language; the index entry recursively references itself ("recursion 86, 139, 141, 182, 202, 269"). Early versions of this joke can be found in Let's talk Lisp by Laurent Siklóssy (published by Prentice Hall PTR on December 1, 1975 with a copyright date of 1976) and in Software Tools by Kernighan and Plauger (published by Addison-Wesley Professional on January 11, 1976). The joke also appears in The UNIX Programming Environment by Kernighan and Pike. It did not appear in the first edition of The C Programming Language. The joke is part of the Functional programming folklore and was already widespread in the functional programming community before the publication of the aforementioned books.
Another joke is that "To understand recursion, you must understand recursion."[11] In the English-language version of the Google web search engine, when a search for "recursion" is made, the site suggests "Did you mean: recursion."[12] An alternative form is the following, from Andrew Plotkin: "If you already know what recursion is, just remember the answer. Otherwise, find someone who is standing closer to Douglas Hofstadter than you are; then ask him or her what recursion is."
Recursive acronyms are other examples of recursive humor. PHP, for example, stands for "PHP Hypertext Preprocessor", WINE stands for "WINE Is Not an Emulator" GNU stands for "GNU's not Unix", and SPARQL denotes the "SPARQL Protocol and RDF Query Language".
## In mathematics
The Sierpinski triangle—a confined recursion of triangles that form a fractal
### Recursively defined sets
#### Example: the natural numbers
The canonical example of a recursively defined set is given by the natural numbers:
0 is in ${\displaystyle \mathbb {N} }$
if n is in ${\displaystyle \mathbb {N} }$, then n + 1 is in ${\displaystyle \mathbb {N} }$
The set of natural numbers is the smallest set satisfying the previous two properties.
In mathematical logic, the Peano axioms (or Peano postulates or Dedekind–Peano axioms), are axioms for the natural numbers presented in the 19th century by the German mathematician Richard Dedekind and by the Italian mathematician Giuseppe Peano. The Peano Axioms define the natural numbers referring to a recursive successor function and addition and multiplication as recursive functions.
#### Example: Proof procedure
Another interesting example is the set of all "provable" propositions in an axiomatic system that are defined in terms of a proof procedure which is inductively (or recursively) defined as follows:
• If a proposition is an axiom, it is a provable proposition.
• If a proposition can be derived from true reachable propositions by means of inference rules, it is a provable proposition.
• The set of provable propositions is the smallest set of propositions satisfying these conditions.
### Finite subdivision rules
Finite subdivision rules are a geometric form of recursion, which can be used to create fractal-like images. A subdivision rule starts with a collection of polygons labelled by finitely many labels, and then each polygon is subdivided into smaller labelled polygons in a way that depends only on the labels of the original polygon. This process can be iterated. The standard middle thirds' technique for creating the Cantor set is a subdivision rule, as is barycentric subdivision.
### Functional recursion
A function may be recursively defined in terms of itself. A familiar example is the Fibonacci number sequence: F(n) = F(n − 1) + F(n − 2). For such a definition to be useful, it must be reducible to non-recursively defined values: in this case F(0) = 0 and F(1) = 1.
A famous recursive function is the Ackermann function, which, unlike the Fibonacci sequence, cannot be expressed without recursion.[citation needed]
### Proofs involving recursive definitions
Applying the standard technique of proof by cases to recursively defined sets or functions, as in the preceding sections, yields structural induction — a powerful generalization of mathematical induction widely used to derive proofs in mathematical logic and computer science.
### Recursive optimization
Dynamic programming is an approach to optimization that restates a multiperiod or multistep optimization problem in recursive form. The key result in dynamic programming is the Bellman equation, which writes the value of the optimization problem at an earlier time (or earlier step) in terms of its value at a later time (or later step).
### The recursion theorem
In set theory, this is a theorem guaranteeing that recursively defined functions exist. Given a set X, an element a of X and a function f: XX, the theorem states that there is a unique function ${\displaystyle F:\mathbb {N} \to X}$ (where ${\displaystyle \mathbb {N} }$ denotes the set of natural numbers including zero) such that
${\displaystyle F(0)=a}$
${\displaystyle F(n+1)=f(F(n))}$
for any natural number n.
#### Proof of uniqueness
Take two functions ${\displaystyle F:\mathbb {N} \to X}$ and ${\displaystyle G:\mathbb {N} \to X}$ such that:
${\displaystyle F(0)=a}$
${\displaystyle G(0)=a}$
${\displaystyle F(n+1)=f(F(n))}$
${\displaystyle G(n+1)=f(G(n))}$
where a is an element of X.
It can be proved by mathematical induction that F(n) = G(n) for all natural numbers n:
Base Case: F(0) = a = G(0) so the equality holds for n = 0.
Inductive Step: Suppose F(k) = G(k) for some ${\displaystyle k\in \mathbb {N} }$. Then F(k + 1) = f(F(k)) = f(G(k)) = G(k + 1).
Hence F(k) = G(k) implies F(k + 1) = G(k + 1).
By induction, F(n) = G(n) for all ${\displaystyle n\in \mathbb {N} }$.
## In computer science
A common method of simplification is to divide a problem into subproblems of the same type. As a computer programming technique, this is called divide and conquer and is key to the design of many important algorithms. Divide and conquer serves as a top-down approach to problem solving, where problems are solved by solving smaller and smaller instances. A contrary approach is dynamic programming. This approach serves as a bottom-up approach, where problems are solved by solving larger and larger instances, until the desired size is reached.
A classic example of recursion is the definition of the factorial function, given here in C code:
unsigned int factorial(unsigned int n) {
if (n == 0) {
return 1;
} else {
return n * factorial(n - 1);
}
}
The function calls itself recursively on a smaller version of the input (n - 1) and multiplies the result of the recursive call by n`, until reaching the base case, analogously to the mathematical definition of factorial.
Recursion in computer programming is exemplified when a function is defined in terms of simpler, often smaller versions of itself. The solution to the problem is then devised by combining the solutions obtained from the simpler versions of the problem. One example application of recursion is in parsers for programming languages. The great advantage of recursion is that an infinite set of possible sentences, designs or other data can be defined, parsed or produced by a finite computer program.
Recurrence relations are equations which define one or more sequences recursively. Some specific kinds of recurrence relation can be "solved" to obtain a non-recursive definition (e.g., a closed-form expression).
Use of recursion in an algorithm has both advantages and disadvantages. The main advantage is usually the simplicity of instructions. The main disadvantage is that the memory usage of recursive algorithms may grow very quickly, rendering them impractical for larger instances.
## In biology
Shapes that seem to have been created by recursive processes sometimes appear in plants and animals, such as in branching structures in which one large part branches out into two or more similar smaller parts. One example is Romanesco broccoli.[13]
## In art
Recursive dolls: the original set of Matryoshka dolls by Zvyozdochkin and Malyutin, 1892
Front face of Giotto's Stefaneschi Triptych, 1320, recursively contains an image of itself (held up by the kneeling figure in the central panel).
The Russian Doll or Matryoshka doll is a physical artistic example of the recursive concept.[14]
Recursion has been used in paintings since Giotto's Stefaneschi Triptych, made in 1320. Its central panel contains the kneeling figure of Cardinal Stefaneschi, holding up the triptych itself as an offering.[15][failed verification]
M. C. Escher's Print Gallery (1956) is a print which depicts a distorted city containing a gallery which recursively contains the picture, and so ad infinitum.[16]
• Corecursion
• Course-of-values recursion
• Digital infinity
• A Dream Within a Dream (poem)
• Droste effect
• False awakening
• Fixed point combinator
• Infinite compositions of analytic functions
• Infinite loop
• Infinite regress
• Infinitism
• Infinity mirror
• Iterated function
• Mathematical induction
• Mise en abyme
• Reentrant (subroutine)
• Self-reference
• Spiegel im Spiegel
• Strange loop
• Tail recursion
• Tupper's self-referential formula
• Turtles all the way down
## References
1. ^ "Peano axioms | mathematics". Encyclopedia Britannica. Retrieved 2019-10-24.
2. ^ "The Definitive Glossary of Higher Mathematical Jargon — Recursion". Math Vault. 2019-08-01. Retrieved 2019-10-24.
3. ^ "Definition of RECURSIVE". www.merriam-webster.com. Retrieved 2019-10-24.
4. ^ Pinker, Steven (1994). The Language Instinct. William Morrow.
5. ^ Pinker, Steven; Jackendoff, Ray (2005). "The faculty of language: What's so special about it?". Cognition. 95 (2): 201–236. CiteSeerX . doi:10.1016/j.cognition.2004.08.004. PMID 15694646. S2CID 1599505.
6. ^ Nordquist, Richard. "What Is Recursion in English Grammar?". ThoughtCo. Retrieved 2019-10-24.
7. ^ Nevins, Andrew; Pesetsky, David; Rodrigues, Cilene (2009). "Evidence and argumentation: A reply to Everett (2009)" (PDF). Language. 85 (3): 671–681. doi:10.1353/lan.0.0140. S2CID 16915455. Archived from the original (PDF) on 2012-01-06.
8. ^ Drucker, Thomas (4 January 2008). Perspectives on the History of Mathematical Logic. Springer Science & Business Media. p. 110. ISBN 978-0-8176-4768-1.
9. ^ Barbara Partee and Mats Rooth. 1983. In Rainer Bäuerle et al., Meaning, Use, and Interpretation of Language. Reprinted in Paul Portner and Barbara Partee, eds. 2002. Formal Semantics: The Essential Readings. Blackwell.
10. ^ Nederhof, Mark-Jan; Satta, Giorgio (2002), "Parsing Non-recursive Context-free Grammars", Proceedings of the 40th Annual Meeting on Association for Computational Linguistics (ACL '02), Stroudsburg, PA, USA: Association for Computational Linguistics, pp. 112–119, doi:.
11. ^ a b Hunter, David (2011). Essentials of Discrete Mathematics. Jones and Bartlett. p. 494. ISBN 9781449604424.
13. ^ "Picture of the Day: Fractal Cauliflower". Retrieved 19 April 2020.
14. ^ Tang, Daisy. "Recursion". Retrieved 24 September 2015. More examples of recursion: Russian Matryoshka dolls. Each doll is made of solid wood or is hollow and contains another Matryoshka doll inside it.
15. ^ "Giotto di Bondone and assistants: Stefaneschi triptych". The Vatican. Retrieved 16 September 2015.
16. ^ Cooper, Jonathan (5 September 2007). "Art and Mathematics". Retrieved 5 July 2020.
## Bibliography
• Dijkstra, Edsger W. (1960). "Recursive Programming". Numerische Mathematik. 2 (1): 312–318. doi:10.1007/BF01386232. S2CID 127891023.
• Johnsonbaugh, Richard (2004). Discrete Mathematics. Prentice Hall. ISBN 978-0-13-117686-7.
• Hofstadter, Douglas (1999). Gödel, Escher, Bach: an Eternal Golden Braid. Basic Books. ISBN 978-0-465-02656-2.
• Shoenfield, Joseph R. (2000). Recursion Theory. A K Peters Ltd. ISBN 978-1-56881-149-9.
• Causey, Robert L. (2001). Logic, Sets, and Recursion. Jones & Bartlett. ISBN 978-0-7637-1695-0.
• Cori, Rene; Lascar, Daniel; Pelletier, Donald H. (2001). Recursion Theory, Gödel's Theorems, Set Theory, Model Theory. Oxford University Press. ISBN 978-0-19-850050-6.
• Barwise, Jon; Moss, Lawrence S. (1996). Vicious Circles. Stanford Univ Center for the Study of Language and Information. ISBN 978-0-19-850050-6. - offers a treatment of corecursion.
• Rosen, Kenneth H. (2002). Discrete Mathematics and Its Applications. McGraw-Hill College. ISBN 978-0-07-293033-7.
• Cormen, Thomas H.; Leiserson, Charles E.; Rivest, Ronald L.; Stein, Clifford (2001). Introduction to Algorithms. Mit Pr. ISBN 978-0-262-03293-3.
• Kernighan, B.; Ritchie, D. (1988). The C programming Language. Prentice Hall. ISBN 978-0-13-110362-7.
• Stokey, Nancy; Robert Lucas; Edward Prescott (1989). Recursive Methods in Economic Dynamics. Harvard University Press. ISBN 978-0-674-75096-8.
• Hungerford (1980). Algebra. Springer. ISBN 978-0-387-90518-1., first chapter on set theory.
By: Wikipedia.org
Edited: 2021-06-18 18:02:21
Source: Wikipedia.org | 4,625 | 19,609 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 15, "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} | 4.0625 | 4 | CC-MAIN-2023-50 | latest | en | 0.91122 |
artificialinformer.com | 1,621,281,507,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992440.69/warc/CC-MAIN-20210517180757-20210517210757-00294.warc.gz | 127,746,422 | 8,245 | Artificial Informer - Issue One
## Glossary of Terminology
1. Dataset A collection of machine readable records, typically from a single source. A dataset can be a single file (Excel or CSV), a database table, or a collection of documents. In machine learning, a dataset is commonly called a corpus. When the dataset is being used to train[18] a machine learning model[12], it can be called a training dataset (a.k.a. a training set). Datasets need to be transformed into a matrix[11] before they can be used by a machine learning model.
Further reading: Training, Validation and Test Sets - Wikipedia
2. Distance Function, Distance Metric A method for quantifying how dissimilar, or far apart, two records are. Euclidean distance, the simplest distance metric used, is attributed to the Ancient Greek mathematician Euclid. This distance metric finds the length of a straight line between two points, as if using a ruler. Cosine distance is another popular metric which measures the angle between two points using trigonometry.
3. Document Object Model, DOM A representation of a HTML page using a hierarchical tree. This is the way that browsers "see" web pages. As an example, we have a simple HTML page and its corresponding DOM tree:
``` HTML DOM
---------------------------------------------------------------
<html> | html
<title>Example DOM</title> | ---------------
<style>*{margin: 0;}</style> | | |
<body> | | |
<h1>Example Page!</h1> | -------- ----------
<div> | | | | | |
<button>Save</button> | title style h1 div footer
</div> | |
<footer>A footer</footer> | button
</body> |
</html> |
```
4. Feature A column in a dataset representing a specific type of values. A feature is typically represented as a variable in a machine learning model. For example, in a campaign finance dataset, a feature might be "contribution amount" or "candidate name". The number of features in a dataset determines its dimensionality. In many machine learning algorithms, high dimensional data (data with lots of features) is notoriously difficult to work with.
5. Hash A short label or string of characters identifying a piece of data. Hashes are generated by a hash function. An example of this comes from the most common use case: password hashes. Instead of storing passwords in a database for anyone to read (and steal), password hashes are stored. For example, the password "Thing99" might get turned into something like `b3aca92c793ee0e9b1a9b0a5f5fc044e05140df3` by a hash function and saved in a database. When logging in, the website will hash the provided password and check it against the one in the database. A strong cryptographic hash function can't feasibly be reversed and uniquely identifies a record. In other usages, such as in_ LSH_, a hash may identify a group of similar records. Hashes are a fixed length, unlike the input data used to create them.
6. k-NN, k-Nearest Neighbors An algorithm for finding the k number of most similar records to a given record, or query point. k-NN can use a variety of distance metrics[2] to measure dissimilarity, or distance_, between points. In k-NN, when _k is equal to 1, the algorithm will return the single most similar record. When k is greater than 1, the algorithm will return multiple records. A common practice is to take the similar records, average them and make educated guesses about the query point.
7. Locality Sensitive Hashing, LSH A method, similar in application to k-NN[6], for identifying similar records given a query record. LSH uses some statistical tricks like hashing[5] and projection[9] to do this as a performance optimization. Due to this, it can be used on large amounts of data. The penalty for this is that it's possible for false records to turn up in the results and, inversely, for actual similar records to be missed.
8. Preprocessing A step in data analysis that happens before any actual analysis occurs to transform the data into a specific format or to clean it. A common preprocessing task is lowercasing and stripping symbols. Vectorization[20] is a common preprocessing step found in machine learning and statistics.
9. Projection A mathematical method for taking an input vector[20] and transforming it into another dimension. Typically, this is done by taking high dimensional data (data with a large number of columns) and converting it to a lower dimension. A simple example of this would be taking a 3D coordinate and turning it into a 2D point. This is one of the key concepts behind LSH[7].
Further reading: Random Projection - Wikipedia
10. Machine Learning, ML A field of statistics and computer science focused on building or using algorithms that can perform tasks, without being told specifically how to accomplish them, by learning from data. The type of data required by the machine learning algorithm, labeled or unlabeled, splits the field into two major groups: supervised[17] and unsupervised[19], respectively. Machine learning is a subfield of Artificial Intelligence.
11. Matrix Rectangularly arranged data made up of rows and columns. In the machine learning context, every cell in a matrix is a number. The numbers in a matrix may represent a letter, number or category.
``` Example m-by-n matrix (2x3)
n columns (3)
.---------------------------------.
m | 11.347271944951725 | 2203 | 2.0 | <- row vector
rows |--------------------+------+-----|
(2) | 9.411296351528783 | 1867 | 1.0 |
`---------------------------------'
\
This is element (2,1)
```
Each row, which represents a single record, is known as a vector[20]. The process of turning source data into a matrix is known as vectorization.
12. Model, Statistical Model A collection of assumptions that a machine learning algorithm has learned from a dataset. Fundamentally, a model consists of numbers, known as weights, that can be be plugged into a machine learning algorithm. We use models to get data out of machine learning algorithms.
13. Outlier Detection A method for identifying records that are out of place in the context of a dataset. These outlying data points can be thought of as strange, suspicious, fraudulent, rare, unique, etc. Outlier detection is a major subfield of machine learning with applications in fraud detection, quality assurance and alert systems.
14. Regression A statistical method for identifying relationships among the features[4] in a dataset.
15. Scraping, Web Scraping The process of loading a web page, extracting information and collecting it into a specific structure (a database, spreadsheet, etc). Typically web scraping is done automatically with a program, or tool, known as a web scraper.
16. String A piece of data, arranged sequentially, made up of letters, numbers or symbols. Technically speaking, computers represent everything as numbers, but they are converted to letters when needed. Numbers, words, sentences, paragraphs and even entire documents can be represented as strings.
17. Supervised Machine Learning, Supervised Learning A subfield of machine learning where algorithms learn to predict values or categories from human-labeled data. Examples of supervised machine learning problems: (1) predicting the temperature of a future day using a dataset of historical weather readings and (2) classifying emails by whether or not they are spam from a set of categorized emails. The goal of supervised machine learning is to learn from one dataset and then make accurate predictions on new data (this is known as generalization).
18. Training The process of feeding a statistical or machine learning algorithm data for the purpose of learning to predict, identifying structure, or extracting knowledge. As an example, consider a list of legitimate campaign contributions. Once an algorithm has been shown this data, it generates a model[12] representing how these contributions typically look. This model can be used to spot unusual contributions, since the model has learned what normal ones look like. There are many different methods for training models, but most of them are iterative, step-based procedures that slowly improve over time. A common analogy for how models are trained is hill climbing: knowing that a flat area (a good solution) is at the top of a hill, but only being able to see a short distance due to thick fog, the top can be found by following steep paths. Training is also known as model fitting.
19. Unsupervised Machine Learning, Unsupervised Learning A subfield of machine learning where algorithms learn to identify the structure or find patterns within a dataset. Unsupervised algorithms don't require human labeling or organization, and therefore can be used on a wide variety of datasets and in many situations. Examples of unsupervised use cases: (1) discovering natural groups of records in a dataset, (2) finding similar documents in a dataset and (3) identifying the way that events normally occur and using this to detect unusual events (a.k.a. outlier detection and anomaly detection[13]).
20. Vectorization, Vector The process of turning a raw source dataset into a numerical matrix[11]. Each record becomes a row of the matrix, known as a vector.
21. Weight A number that is used to either increase or decrease the importance of a feature[4]. Weights are used in supervised machine learning to quantify how well one variable predicts another; in unsupervised learning, weights are used to emphasize features that segment a dataset into groups.
22. XPath A description of the location of an element on a web page. From the browser's perspective, a web page is represented as a hierarchical tree known as the Document Object Model (DOM)[3]. An XPath selector describes a route through this tree that leads to a specific part of the page. | 2,091 | 10,059 | {"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-2021-21 | longest | en | 0.787054 |
https://www.esaral.com/q/construct-a-abc-in-which-base-ab-5-cm-a-30-and-ac-bc-2-5-cm-16407 | 1,723,584,041,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641085898.84/warc/CC-MAIN-20240813204036-20240813234036-00360.warc.gz | 569,679,109 | 11,901 | # Construct a ∆ABC in which base AB = 5 cm, ∠A = 30° and AC – BC = 2.5 cm.
Question:
Construct a ∆ABC in which base AB = 5 cm, ∠= 30° and AC – BC = 2.5 cm. Justify your construction.
Solution:
Steps of construction:
1. Draw a line segment AB = 5 cm.
2. Construct $\angle B A X=30^{\circ}$.
3. Set off AD = 2.5 cm.
4. Join DB.
5. Draw the right bisector of DB, meeting DB produced at C.
6. Join CB.
Thus, $\triangle A B C$ is the required triangle.
Justification:
Point C lies on the perpendicular bisector of DB.
⇒ CD = BC
Now
AD = AC − DC = AC − BC | 202 | 557 | {"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 | 4 | CC-MAIN-2024-33 | latest | en | 0.653986 |
https://discourse.julialang.org/t/help-with-using-the-plots-package/57036 | 1,656,565,902,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103661137.41/warc/CC-MAIN-20220630031950-20220630061950-00363.warc.gz | 257,906,844 | 8,034 | # Help with using the Plots package
Hello
How to plot a two-dimensional graph?
In this example I am trying to plot (x, y) and in response Julia returns a three-dimensional graph.
``````using Plots
x = 1:4
y = [0.16, 12, 8, 0.13]
y1 = [0.07, 13, 7]
y2 = y,y1
plot!(x,y2)
``````
Hm: you make `y2` into a tuple consisting of `y` and `y1`, and plot `x` vs. `(y, y1)`.
Could you post your output from a fresh REPL session? Using your code I get:
I.e. the second element of your `y2` tuple is ignored. Could you also explain what you expect to happen - why do you construct the tuple in the first place?
I just stopped and started Julia again, and he returned exactly what you put here.
I would like to plot two lines that would be y and y1
The problem is your `y1` has 3 elements, while `x` and `y` have four elements. If I add an element to `y1`, I get:
``````x = 1:4
y = [0.16, 12, 8, 0.13]
y1 = [0.07, 13, 7,2]
y2 = [y, y1]
plot(x,y2)
``````
4 Likes
… if you only have 3 elements for `y1`, do as follows:
``````y1 = [0.07, 13, 7, NaN]
``````
and then nothing will be plotted for the last point.
1 Like
Thanks!!!
You could also do it as follows:
``````plot(x,y)
plot!(x,y1)
``````
Others have answered your question pretty well, but just wanted to give my 2 cents.
I like this syntax when making plots:
``````p = plot() #You can insert all attributes you want here like "xlabel","size","title" etc.
plot!(x,y,label="first line")
plot!(x1,y1,label="second line")
``````
This allows me to get a plot element “p” and add plots to these as I please, while only having to specify most attirbutes once. If for example you someday want to make two different plots in a for loop you can do:
``````p1 = plot()
p2 = plot()
plot!(p1,x,y) #Add this plot only to p1
plot!(p2,x1,y1) #Add this plot only to p2
``````
Just wanted to share
Kind regards
4 Likes
I know that this example is not compilable, but the idea was to put this example inside a loop. I tried to follow the example above but I couldn’t.
``````x = 1:4
y1 = getvalue(r[1,:]+ o[1,:] + s[1,:])
y2 = getvalue(r[2,:]+ o[2,:] + s[2,:])
y3 = getvalue(r[3,:]+ o[3,:] + s[3,:])
y4 = getvalue(r[4,:]+ o[4,:] + s[4,:])
y=[y1,y2,y3,y4]
plot(x,y)
``````
``````using Plots
r=rand(4,4)
o=rand(4,4)
s=rand(4,4)
x=1:4
getvalue(u) = u
y=[]
for i=1:4
push!(y,getvalue(r[i,:]+ o[i,:] + s[i,:]))
end
plot(x,y)
``````
2 Likes
Minor alternative to Jeff’s solution:
``````using Plots
r=rand(4,4)
o=rand(4,4)
s=rand(4,4)
x=1:4
#
y = [r[i,:]+o[i,:]+s[i,:] for i in 1:size(r,1)]
plot()
for yi in y
plot!(x,yi)
end
plot!()
``````
I like the improved generality, but shouldn’t it be `size(r,1)` since i is used as the first index of r?
Ooops… of course. I’ll fix it.
@BLI, it seems enough to do in this case:
``````y = r + o + s
plot(x,y')
``````
1 Like
I included the trivial getvalue as a placeholder for OP’s example.
1 Like
True, and simpler.
However, if one saves the data in other data structures, e.g., a tuple – `z = Tuple(y)`, then
``````plot(x,z)
``````
doesn’t work, while iterating over `z` works:
``````plot()
for zi in z
plot!(x,zi)
end
plot!()
``````
Although this is elegant and noise-free, I think that OP’s coding skills are at a level that terse solutions like this won’t be much help to her.
1 Like
Then do:
``````z = Tuple(y)
plot(x, [z...])
``````
1 Like
Yet another case: if one has not decided whether to plot `r+o+s` row-wise or column-wise:
``````Y = r+o+s
#
plot()
for i in 1:size(Y,1)
plot!(x,Y[i,:])
end
plot!()
``````
vs.
``````plot()
for i in 1:size(Y,2)
plot!(x,Y[:,i])
end
plot!()
`````` | 1,261 | 3,616 | {"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} | 3.296875 | 3 | CC-MAIN-2022-27 | latest | en | 0.928169 |
http://mizar.uwb.edu.pl/version/current/html/proofs/transgeo/85 | 1,568,761,514,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-39/segments/1568514573124.40/warc/CC-MAIN-20190917223332-20190918005332-00205.warc.gz | 134,462,711 | 2,192 | let AFS be AffinSpace; :: thesis: for f being Permutation of the carrier of AFS st f is dilatation holds
for a, b, c, d being Element of AFS holds
( a,b // c,d iff f . a,f . b // f . c,f . d )
let f be Permutation of the carrier of AFS; :: thesis: ( f is dilatation implies for a, b, c, d being Element of AFS holds
( a,b // c,d iff f . a,f . b // f . c,f . d ) )
assume A1: f is dilatation ; :: thesis: for a, b, c, d being Element of AFS holds
( a,b // c,d iff f . a,f . b // f . c,f . d )
let a, b, c, d be Element of AFS; :: thesis: ( a,b // c,d iff f . a,f . b // f . c,f . d )
A2: c,d // f . c,f . d by ;
A3: a,b // f . a,f . b by ;
A4: now :: thesis: ( f . a,f . b // f . c,f . d implies a,b // c,d )
A5: now :: thesis: ( a,b // f . c,f . d implies a,b // c,d )
A6: now :: thesis: ( f . c = f . d implies a,b // c,d )
assume f . c = f . d ; :: thesis: a,b // c,d
then c = d by FUNCT_2:58;
hence a,b // c,d by AFF_1:3; :: thesis: verum
end;
assume a,b // f . c,f . d ; :: thesis: a,b // c,d
hence a,b // c,d by ; :: thesis: verum
end;
A7: now :: thesis: ( f . a = f . b implies a,b // c,d )
assume f . a = f . b ; :: thesis: a,b // c,d
then a = b by FUNCT_2:58;
hence a,b // c,d by AFF_1:3; :: thesis: verum
end;
assume f . a,f . b // f . c,f . d ; :: thesis: a,b // c,d
hence a,b // c,d by ; :: thesis: verum
end;
now :: thesis: ( a,b // c,d implies f . a,f . b // f . c,f . d )
A8: now :: thesis: ( f . a,f . b // c,d implies f . a,f . b // f . c,f . d )
A9: ( c = d implies f . a,f . b // f . c,f . d ) by AFF_1:3;
assume f . a,f . b // c,d ; :: thesis: f . a,f . b // f . c,f . d
hence f . a,f . b // f . c,f . d by ; :: thesis: verum
end;
assume a,b // c,d ; :: thesis: f . a,f . b // f . c,f . d
then ( f . a,f . b // c,d or a = b ) by ;
hence f . a,f . b // f . c,f . d by ; :: thesis: verum
end;
hence ( a,b // c,d iff f . a,f . b // f . c,f . d ) by A4; :: thesis: verum | 755 | 1,888 | {"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} | 3.421875 | 3 | CC-MAIN-2019-39 | latest | en | 0.460741 |
https://www.unitmeasurement.com/mass-conversion/carats.html | 1,623,837,815,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487623596.16/warc/CC-MAIN-20210616093937-20210616123937-00034.warc.gz | 858,432,926 | 4,355 | # Carats(ct) Conversion
## Convert Carats(ct) to other Mass units
### Carats Conversion Calculators
The Carats is a measurement unit of Mass, but it is a non-SI unit. The unit symbol of Carats is ct, 1 Carats is equal to 0.0002 Kilograms.
Carats to Kilograms Conversion.(ct to kg)
Carats to Grams Conversion.(ct to g)
Carats to Ounces Conversion.(ct to oz)
Carats to Pounds Conversion.(ct to lb)
Carats to Milligrams Conversion.(ct to mg)
Carats to Stones Conversion.(ct to st)
Carats to Decigrams Conversion.(ct to dg)
Carats to Centigrams Conversion.(ct to cg)
Carats to Micrograms Conversion.(ct to µg)
Carats to Nanograms Conversion.(ct to ng)
Carats to Picograms Conversion.(ct to pg)
Carats to Metric Tons Conversion.(ct to t)
Carats to Short Tons Conversion.(ct to ton)
Carats to Ounces Troy Conversion.(ct to oz t)
Carats to Slugs Conversion.(ct to slug)
### Common Mass Conversions
How many kilograms are in one grams?
1 Kilograms is equal to 1000 Grams.
How many grams are in one ounces?
1 Grams is equal to 0.03527 Ounces.
How many pounds are in one kilograms?
1 Pounds is equal to 0.45359 Kilograms.
How many grams are in one ounces?
1 Grams is equal to 0.03527 Ounces.
How many kilograms are in one pounds?
1 Kilograms is equal to 2.20462 Pounds.
How many ounces are in one grams?
1 Ounces is equal to 28.3495 Grams.
How many kilograms are in one ounces?
1 Kilograms is equal to 35.27399 Ounces.
### Mass Unit Conversion Table
Convert From KilogramsGramsOuncesPounds
1 Carats0.0002 kg0.2 g0.00705 oz0.00044 lb
2 Carats0.0004 kg0.4 g0.0141 oz0.00088 lb
3 Carats0.0006 kg0.6 g0.02115 oz0.00132 lb
4 Carats0.0008 kg0.8 g0.0282 oz0.00176 lb
5 Carats0.001 kg1 g0.03525 oz0.0022 lb
6 Carats0.0012 kg1.2 g0.0423 oz0.00264 lb
7 Carats0.0014 kg1.4 g0.04935 oz0.00308 lb
8 Carats0.0016 kg1.6 g0.0564 oz0.00352 lb
9 Carats0.0018 kg1.8 g0.06345 oz0.00396 lb
10 Carats0.002 kg2 g0.0705 oz0.0044 lb
11 Carats0.0022 kg2.2 g0.07755 oz0.00484 lb
12 Carats0.0024 kg2.4 g0.0846 oz0.00528 lb
13 Carats0.0026 kg2.6 g0.09165 oz0.00572 lb
14 Carats0.0028 kg2.8 g0.0987 oz0.00616 lb
15 Carats0.003 kg3 g0.10575 oz0.0066 lb
16 Carats0.0032 kg3.2 g0.1128 oz0.00704 lb
17 Carats0.0034 kg3.4 g0.11985 oz0.00748 lb
18 Carats0.0036 kg3.6 g0.1269 oz0.00792 lb
19 Carats0.0038 kg3.8 g0.13395 oz0.00836 lb
20 Carats0.004 kg4 g0.141 oz0.0088 lb | 866 | 2,358 | {"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-2021-25 | latest | en | 0.791255 |
https://www.joyk.com/dig/detail/1651050046511986 | 1,696,380,245,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511284.37/warc/CC-MAIN-20231003224357-20231004014357-00821.warc.gz | 900,184,362 | 6,234 | 计算旋转矩形相交面积
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计算旋转矩形相交面积
``````矩形2的4个点 [(bx1, by1), (bx2, by2), (bx3, by3), (bx4, by4)]
``````
1. 将4个点按照时钟序排列(我实现的是逆时针)
2. (PS:这一步不重要，我只是为了利用python里给定的函数求交点必须要转成这个数据结构)得到矩形的数据结构表示 ((x,y), (w,h), angle) #(x,y)表示矩形的中心点, (w,h)表示边长，w不一定大于h，而是w表示y值最小的点(如果相等就x最小)右侧的边长，同理h表示左侧的边长。angle表示为度数(360度一圈)，是w对应的边长和x轴正方向的夹角。
3. 求解出两个矩形相交的点以及某个矩形的顶点落在另一个矩形内的点集合S
4. 对S求个凸包 hull
5. hull是个凸多边形，根据向量的叉积求解面积即可(如找到一个点p，将每一条边都和这个点相连的三角形通过叉积求面积进行加和)
利用的函数
1. scipy.spatial.ConvexHull 求凸包
2. cv2.rotatedRectangleIntersection(rect1, rect2) 求两个矩形的交点(就是我第三步所说)
3. cv2.contourArea 求多边形的面积
python代码实现
import cv2
import math
import numpy as np
from scipy.spatial import ConvexHull
class RotatedRect:
def __init__(self, points):
self.points = sort_rotatedrect_points(points)
self.rect = rotate_rect(points) # ((centerX, centerY), (width,height), angle)
def intersection_area(self, rotated_rect):
return intersect_area(self.rect, rotated_rect.rect)
def union_area(self, rotated_rect):
return self.rect[1][0]*self.rect[1][1] + \
rotated_rect.rect[1][0]*rotated_rect.rect[1][1] - \
self.intersection_area(rotated_rect)
def iou(self, rotated_rect):
return self.intersection_area(rotated_rect) / self.union_area(rotated_rect)
def distance(a, b): # 计算点对之间的距离
return math.sqrt((a[0]-b[0])*(a[0]-b[0]) + (a[1]-b[1])*(a[1]-b[1]))
def sort_rotatedrect_points(points): # 将旋转矩形的四个点按照逆时针输出，且保证起始位置是y值最小的点
print('before sort: ', points)
points = sorted(points, key=lambda a: (a[1], a[0]))
print('after sort: ', points)
tmp1 = points[1].copy()
tmp2 = points[2].copy()
tmp3 = points[3].copy()
if points[1][0] < points[0][0]:
points[1] = tmp2
points[2] = tmp3
points[3] = tmp1
else:
points[2] = tmp3
points[3] = tmp2
print('after update sort: ', points)
return points
def rotate_rect(points):
:param points: points 表示旋转矩形的4个点 [(x1, y1), (x2, y2), (x3, y3), (x4, y4)]
:return: (centerx,centery), (width,height), angle
assert len(points) == 4, 'The points nums != 4'
points = sort_rotatedrect_points(points)
assert abs(points[0][0] + points[2][0] - points[1][0] - points[3][0]) < 1e-6, 'These points can\'t form a rectangle'
assert ( abs(points[0][1] + points[2][1] - points[1][1] - points[3][1]) < 1e-6 ), 'These points can\'t form a rectangle'
centerx = (points[0][0] + points[2][0]) / 2
centery = (points[0][1] + points[2][1]) / 2
center = (centerx, centery)
width = distance(points[0], points[1])
height = distance(points[0], points[3])
angle = math.atan2(points[1][1]-points[0][1], points[1][0]-points[0][0])
PI = math.acos(-1.0)
return center, (width, height), angle/PI*180
def convexhull(points):
:param unordered points: np.array([(x1, y1), (x2, y2), (x3, y3), ..., (xn, yn)])
:return: ordered points of convex
hull = ConvexHull(points)
arg_index = hull.vertices # 返回的是重新排序后的点坐标
hull = points[arg_index] # 重排后凸包后的顺序点
return hull
def intersect_area(rect1, rect2): # rect的数据格式要求((centerx,centery), (width,height), angle)(角度，以360作为一圈)
r1 = cv2.rotatedRectangleIntersection(rect1, rect2)
len_p = r1[1].shape[0]
for i in range(len_p):
p.append((r1[1][i][0][0], r1[1][i][0][1]))
p = np.array(p)
hull = convexhull(p)
r2 = cv2.contourArea(hull) # 根据凸包顺序点求面积
return r2
# 中心点 矩形的w h, 逆时针旋转的theta（角度，不是弧度）
rect1 = ((0,0),(1,1),30)
rect2 = ((1.5,0),(4,3),0)
r1 = cv2.rotatedRectangleIntersection(rect1, rect2)
print (r1)
points1 = [[1., 0], [-1., 0], [0, -1.], [0, 1.]]
rect1 = rotate_rect(points1)
points2 = [[-0.5, -0.5], [-0.5, 1.5], [1.5, -0.5], [1.5, 1.5]]
rect2 = rotate_rect(points2)
print(rect1)
print(rect2)
print(intersect_area(rect1, rect2)) | 1,534 | 3,728 | {"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} | 4.0625 | 4 | CC-MAIN-2023-40 | latest | en | 0.321555 |
https://cstheory.stackexchange.com/questions/53003/is-it-necessary-to-apply-alpha-conversion-in-this-term-to-perform-beta-reduction | 1,716,318,110,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058512.80/warc/CC-MAIN-20240521183800-20240521213800-00724.warc.gz | 158,247,375 | 39,813 | # Is it necessary to apply alpha conversion in this term to perform beta reduction in lambda calculus?
I am trying to prove that the expression ((λx.(λx.x))(ab)) does not require alpha conversion for beta reduction since there is no variable overlap, but how could I demonstrate this more formally?
• Proving that every $\alpha$-equivalent term ($\beta$) reduces to an $\alpha$-equivalent term should do the trick! It's a strange thing to prove though, unless this is homework.
– cody
Jun 28, 2023 at 15:13
The answer depends on the definition of the substitution you use.
With the standard definition:
$$(\lambda y.M)[N/x] := \lambda y.M[N/x]$$ provided $$x \neq y$$ and $$y \not\in \mathrm{FV}(N)$$
the reduction of your term does require α-conversion.
Similarly, if you choose the Barendregt variable convention (see 2.1.13 in the book), you aren't even allowed to write your term like you do — which seems a hygienic prohibition to me...
If you choose some alternative definition, like adding this rule:
$$(\lambda x.M)[N/x] := \lambda x.M$$
to the previous one, then the β-reduction normalizes in one step without α-conversion.
• Btw, the downvote would be more helpful if accompanied by a comment! Jun 28, 2023 at 21:24 | 322 | 1,236 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-22 | latest | en | 0.891586 |
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