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# OG 10 vs OG 11
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31 Oct 2005, 13:29
Hi guys, I have OG 10 (havent gone through it yet, saving the best til last) but I was wondering whether there is any point of me buying the 11th edition, are there different questions explanations etc, or would I just be wasting my money and time?
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31 Oct 2005, 13:33
I'm doing it in reverse order. There is def. a lot of overlap. I'll be done w/ the quant soon, so i can let you know how much by the end of this week.
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31 Oct 2005, 17:49
I see a lot of overlap as well. For me, I bought the 11th edition because I wanted a new book for my next round of studying. I'm weird like that...
The value of the 11th edition is in the clearer explanations - especially in verbal. If you need more help with V, I'd get the 11th edition. If you need help with Q, I'd stay with the tried and true 10th edition (my personal opinion).
Good luck!
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# OG 10 vs OG 11
Moderator: HiLine
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 637 | 2,245 | {"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-2017-43 | latest | en | 0.929467 |
https://classnotes.ng/lesson/new-lesson-173/ | 1,680,151,791,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949097.61/warc/CC-MAIN-20230330035241-20230330065241-00532.warc.gz | 209,707,037 | 34,308 | OPERATION OF SET AND VENN DIAGRAMS
Example: A school has 37 vacancies for teachers, out of which 22 are for English, 20 for History and 17 for Fine Art. Of these vacancies 11 are for both English and History, 8 for both History and Fine Art and 7 for English and Fine Art. Using a Venn diagram, find the number of teachers who must be able to teach:
(a.) all the three subjects
(b.) Fine Art only
(c.) English and History but not Fine Art.
Solution:
Let µ = {All vacancies for teachers}
E = {English vacancies}
H = {History vacancies}
F = {Fine Art vacancies}
µ = 37, n(E)= 22, n(H)= 20, n(F)= 17, n{EnH}= 11, n(HnF)= 8, n(EnF)= 7
(1) Let n(EnFnH) = y
n (EnHInF)= n(E)- (7-y+y+11-y)
= 22- (18-y) = 4 + y
n(EInHnF) = n(H) – (11-y+y+8-y)
= 20- (19-y) = 1+y
n(EInH1nF)= n(F) – ( 7-y +y+8-y)
= 17 – (15- y) = 2 +y
µ= 4+y+11-y+1+y+y+8-y+7-y+2+y
37= 33 + y
y = 37- 33
y = 4.
n(EnHnF) = 4 teachers
(2.) Fine Art only, n(EInHInF) = 2+ y
= 2+4 = 6 teachers
(3.) English and History but not Fine Art i.e English and History only
n(EnHnFI) = 11-y
= 11- 4 = 7 teachers.
Examples:
1. In a survey of 290 newspaper readers, 181 of them read daily times, 142 read the Guardian, 117 read the Punch and each read at least one of the paper, if 75 read the Daily Times and the Guardian,60 read the Daily Times and Punch and 54 read the Guardian and the punch
a) Draw a venn diagram to illustrate the information
(i) all the three papers
(ii) exactly two of the papers
(iii) exactly one of the paper
X is the number of readers who read all the three papers
Since the sum of the number of elements in all regions is equal to the total number of elements in the universal set, then:
46 + X + 75 – X + 13 + X + 60 – X + X + 54 – X + 3 + X = 90
251 + X = 290
X = 290 – 251
X= 39
B(i): number of people who read all the three paper= 39
(ii) from the venn diagram, number of people who read exactly two papers
= 60 – X + 75 – X + 54 – X
=189 – 3X = 189 – 3(39) from the above
=189 – 117 = 72
(iii) also, from the venn diagram, number of people who read exactly only one of the papers
=46 + X + 13 + X + 3 +X
=162 +3X =162 + 3(39)
=162 + 117 = 179
=13 + X
=13 + 39 = 52
Evaluation
1. In a community of 160 people, 70 have cars ,82 have motorcycles, and 88 have bicycles, 20 have both cars and motorcycles,25 have both cars and bicycles, while 42 have both motorcycles and bicycles each person rode on at least any of the vehicles
a) Draw a venn diagram to illustrate the information
b) Find the number of people that has both cars and bicycles
c) How many people have either one of the three vehicles?
The score of 144 candidates who registered for mathematics, physics and chemistry in an examination in a town are represented in the venn diagram above.
a) How many candidate register for both mathematics and physics?
b) How many candidate register for both mathematics and physics only?
General Evaluation
1. n(P) =4 means that these are 4 element in set P. given that n(XƲY)= 50, n(X)=20 and n(Y)= 40. Find n(X∩Y)
2. find the sum of the first five terms of GP 2,6,18……..
3. the twelfth term of a linear sequence is 47 and the sum of the first three term is 12. Find the sum of the first 15 terms of the sequence
4. At a meeting of 35 teachers, the analysis of how Fanta, Coke and Pepsi were served as refreshments is as
follows. 15 drank Fanta, 6 drank both Fanta and coke, 18 drank Coke, 8 drank both Coke and Pepsi, 20 drank Pepsi, and 2 drank all the three types of drink. How many of the teachers drank I Coke only II Fanta and Pepsi but not Coke.
5. Given n(XUY) = 50, n(X) = 20 and n(Y) = 40, determine n(XnY)
Weekend Assignment
1. In a class of 50 pupils, 24 like oranges, 23 like apples and 7 like the two fruits. How many students do not like oranges and apples? (a)7 (b) 6 (c) 10 (d)15
2. In a survey of 55 pupils in a certain private schools, 34 like biscuits, 26 like sweets and 5 of them like none. How many pupils like both biscuits and sweet? (a) 5(b) 7 (c)9 (d)10
3. In a class of 40 students, 25 speak Hausa, 16 speak Igbo, 21 speak Yoruba and each of the students speak at least of the three languages.
If 8 speak Hausa and Igbo. 11 speak Hausa and Yoruba.6 speak Igbo and Yoruba. How many students speak the three languages? (a) 3 (b) 4 (c) 5 (d) 6
Use the information to answer question 4 and 5
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# Problem 532. Return unique values without sorting
Solution 553823
Submitted on 8 Jan 2015 by Gergely Patay
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct)) x = [9 2 2]; y_correct = [9 2]; assert(isequal(your_fcn_name(x),y_correct)) x = [-4 1 1]; y_correct = [-4 1]; assert(isequal(your_fcn_name(x),y_correct)) x = [42 1 1] y_correct = [42 1]; assert(isequal(your_fcn_name(x),y_correct)) x = [42 1 1 1 42 17 17]; y_correct = [42 1 17]; assert(isequal(your_fcn_name(x),y_correct))
x = 42 1 1 | 241 | 688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-05 | latest | en | 0.586707 |
https://philoid.com/question/15500-add-the-following-rational-numbers-i-and-ii-and-iii-and-iv-and | 1,686,054,920,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652569.73/warc/CC-MAIN-20230606114156-20230606144156-00036.warc.gz | 500,054,255 | 11,058 | ## Book: RD Sharma - Mathematics
### Chapter: 1. Rational Numbers
#### Subject: Maths - Class 8th
##### Q. No. 1 of Exercise 1.1
Listen NCERT Audio Books to boost your productivity and retention power by 2X.
1
##### Add the following rational numbers:(i) and(ii) and(iii) and(iv) and
(i) Clearly,
Denominators of the given numbers are positive
The L.C.M of denominator 7 and 7 is 7
We have,
+
=
=
(ii) Clearly,
Denominators of the given numbers are positive
The L.C.M of denominator 4 and 4 is 4
We have,
+
=
=
= -2
(iii) Clearly,
Denominators of the given numbers are positive
The L.C.M of denominator 11 and 11 is 11
We have,
+
=
=
(iv) Clearly,
Denominators of the given numbers are positive
The L.C.M of denominator 13 and 13 is 13
We have,
+
=
=
1
2
3
4 | 234 | 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} | 3.8125 | 4 | CC-MAIN-2023-23 | longest | en | 0.848745 |
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# Matrix rep 03.pg
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ | 395 | 1,064 | {"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": 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} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.20998 |
http://www.jiskha.com/display.cgi?id=1367296173 | 1,498,170,945,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319912.4/warc/CC-MAIN-20170622220117-20170623000117-00025.warc.gz | 554,484,644 | 3,720 | # Statistics
posted by on .
Assume that a random sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level.
95% confidence the sample size is 2096 of which 25%are successes
The margin of E= (round to four decimal places as needed)
• Statistics - ,
Margin of error = (1.96)[√(pq/n)]
Note: 1.96 represents 95% confidence.
For p in your problem: .25
For q: 1 - p = q
n = 2096
I let you take it from here to calculate.
### Related Questions
More Related Questions
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Depreciation Expense and Depreciation Schedules ExplainedDefinitions, Meaning, and Example Accounting Calculations
The book value of vehicles, buildings, factory machines and other capital assets is reduced across their depreciable lives. The adjustment is a non-cash expense called depreciation.
What is depreciation expense?
Depreciation expense is a financial accounting reporting practice, used primarily by businesses that pay tax on income. On the income statement, this expense appears as a charge against income subtracted from sales revenues to produce a lower reported income (lower profit, lower earnings).
These expenses are charged according to a depreciation schedule, thereby accounting for the purchase of long-lasting assets over a period of years. The idea is that assets have a useful life (depreciable life), over which they are used up or worn out, and that the owner receives the tax benefits of paying for the asset over those years instead of all at once.
This article defines and explains depreciation and its scheduling, with examples, in context with related terms and concepts.
How does depreciation lowers reported income and tax liability?
Each year in the life of a depreciable asset, some of its cost is charged against income on the income statement. Just how much is charged each year is determined by the appropriate schedule for the asset (see Schedules, below).
In the example income statement (Exhibit 1, below), Grande Corporation pays an income tax of \$957,950 on an income (before extraordinary items) of \$2,737,000 (the reported tax is rounded up to \$958 thousand on the statement). Contributing to expenses, however, are three depreciation items totaling \$659,000. Had these expenses been omitted from this statement, Grande Corporation would have had a before-tax operating income of \$3,396,000 and an income tax of \$1,188,860.
Exhibit 1. Sample Income Statement showing depreciation expenses in several loactions, depending on use use of the assets depreciated. Depreciation expenses lower reported Net Income, which lowers the tax liability on income.
Using rounded figures from the statement, the tax savings from depreciation is thus \$230,000 that is:
\$1,188,000 – \$958,000 = \$230,000.
This tax savings can also be estimated directly as the product of
the expense and the tax rate. With a tax rate on income of 35%, for example,
0.35*\$659,000 = \$230,650.
Notice also that the expense can appear in any of the main expense categories, depending on how the asset in question is used. Here they appear under Cost of goods sold, as Manufacturing overhead (for manufacturing equipment), under Selling expenses (for store equipment), and under General and administrative expenses (for computers).
What can be depreciated?
Generally, depreciation can be claimed for assets that (a) have a useful life of one year or more, (b) are used in a trade or business, and (c) which are used up, wear out, decay, become obsolete, or otherwise lose value over their useful life. Assets that meet these criteria may include factory machines, vehicles, computer systems, office furniture, aircraft, and buildings. Land, however, is an example of an asset that does not meet the third criterion (losing value), and therefore cannot be depreciated in the same way.
The country's tax laws sometimes give the accountants and financial officers some choice in deciding whether or not to classify some acquisitions as assets (and therefore eligible for claiming the expense), although the freedom of choice is limited. In 2005, for example, several senior executives of Worldcom in the United States were convicted of fraudulent reporting for having overstepped the boundaries, for classifying services paid for by the company as assets rather than as expenses, as they should have been.
What is the difference between cash flow and claimed depreciation expens?
Depreciation expense is an accounting convention, not real cash flow. When a company buys an asset outright with cash, all the cash flows at once in the purchase transaction (this shows up on the company's cash flow statement under "Uses of Cash"). On the income statement, however, the expense is spread across the years of the asset's depreciable life, thus lowering reported income across several years or more.
Although this expense is not real cash flow itself, it does bring a cash flow consequence each year of the depreciable life. Because these non cash expenses lower reported income, they brings a tax savings that is a real cash flow.
What are roles of depreciable life, depreciable cost, and residual value in calculating depreciation expense?
The depreciation expense claimed for an asset each year normally depends on four factors:
• The asset's depreciable life
• The asset's initial cost
• The asset's residual value
• The schedule used for the asset's life (the time period over which an asset
can lawfully be depreciated).
For some assets, management can simply choose a number of years for the life, based on the asset's expected useful life. For some kinds of assets, however, the depreciable life is prescribed by the country's tax authorities. In the US, for instance, computing hardware has a prescribed life of 5 years, and depreciation must follow the MACRS (Modified Accelerated Cost Recovery System) schedule.
An asset's depreciable life can be different from its economic life. The term economic life refers to actual period of usefulness of an asset, the period beyond which it is cheaper to replace or scrap an asset than to continue maintaining it. Economic life and depreciable life are both central concepts in the practice of asset life cycle management.
An asset is originally valued on the balance sheet at its actual original cost (this is the historical cost convention in accounting). For expense claim purposes, the original cost may include two components:
Original Cost of Asset = Deprec. Cost + Residual Value
Exhibit 2 shows how depreciable cost decreases over time, leaving only the residual value at the end of depreciable life. Only the depreciable cost component will contribute to the claimed expense across the years of depreciable life. The asset's residual value (sometimes called salvage value) remains at the end of its life. Residual value is the estimated net value of the asset that would or could be received if the asset were retired or scrapped.
Exhibit 2. The figure shows how an asset originally costing \$100 decreases in book value to its residual value over its depreciable life, as the expense is charged each year (the example follows a straight line schedule across a 5 year life).
Most depreciation schedules are applied to the depreciable cost rather than total cost, but the double declining balance method (DDB) is an exception, as is MACRS, a special case of DDB (see Schedules, below, for more on these methods). Under DDB and MACRS, the expense percentages are applied against total original cost.
When using any schedule besides DDB and MACRS, residual value plays an important role in determining the claimed expenses, tax savings and, possibly, the value of a cash inflow at the end of depreciation. Residual (or salvage) value of an asset has two important tax considerations:
• An asset may NOT normally be depreciated below its estimated residual (salvage) value.
• If, at the end of depreciable life, the realized salvage value of an asset differs from the book value, a tax adjustment will usually be required.
Depreciation schedules
The length of an asset's depreciable life and the amount of depreciation a company can claim for it each year, are given by schedules. Tax laws in each country specify which schedules can be used for various classes of assets, although in some cases the company has a limited range of schedule choices.
Straight line schedule (SL)
The simplest schedule, so-called straight line depreciation spreads the expenses evenly across an asset’s depreciable life: A \$100 asset fully depreciated over 5 years (and having no residual value) would allow the owner to claim a \$20 expense each year for five years.
Other time-based schedules described below are called accelerated schedules because they "accelerate" depreciation. Three other time-based schedules below charge relatively more in early years, and relatively less in later years. Accelerated schedules thus enable a company to claim relatively more of an asset's related tax savings in the early part of the asset's depreciable life.
Modified accelerated cost recovery system schedule (MACRS)
Many US companies use the 1986 modification of the 1981 Accelerated Cost Recovery System (ACRS) for certain classes of assets, known as the Modified Accelerated Cost Recovery System, or MACRS. MACRS is thus only for US use. MACRS specifies different schedules for calculating depreciation expense for several kinds of assets: Computing equipment falls into the "5-year class" of property, along with most other office equipment and automobiles. MACRS thus prescribes a 60 month depreciable life for computers, spread across 6 fiscal years (the 60 month period is usually started at the midpoint of year 1). There are several variations and options on MACRS schedules but the primary usage is to apply the double declining balance (DDB) method (see below), using a mid year-1 start. Residual value (salvage value) is ignored. MACRS (along with DDB and SOYD methods, below), is an accelerated schedule, in which relatively more depreciation expense is claimed early in the depreciable life, and relative less is claimed later in the life.
MACRS rules in fact provide two possible schedules for different asset classes: a General Schedule (GDS) which is most frequently used, and an Alternative Schedule (ADS) which may be used in some cases. A few of the GDS and ADS schedules Include:
Office Furniture: GDS 7 Years, ADS 10 Years
Computers: GDS 5 Years, ADS 6 Years
Construction Assets: GDS 5 Years, ADS 6 Years
The full set of MACRS schedules and rules for various asset classes are given in US Government IRS Publication 946, "How to Depreciate Property."
Double declining balance schedule (DDB)
The double declining balance schedule (DDB) is a form of accelerated schedule that prescribes an annual rate twice that of the straight line method. Under the DDB method, twice the straight line rate is applied each year to the remaining non depreciated value of the asset.
Sum-of-the-years'-digits (SOYD)
The sum-of-the-years'-digits schedule (SOYD) is an accelerated method based on an inverted scale of total digits for the years of depreciable life. For five years of life, for example, the digits 1,2,3,4 and 5 are added to produce 15. The first year’s rate becomes 5/15 of the depreciable cost (33.3%), the second year’s rate is 4/15 of that cost (26.7%), the third year’s rate 3/15, and so on.
The table below compares depreciation percentages applied each year with the depreciable cost of an asset having a 5-year life. Figures in the table show percentage depreciated per year. These schedules are shown graphically below the table.
ScheduleYear 1Year 2Year 3Year 4Year 5Year 6
Straight Line 20.00 20.00 20.00 20.00 20.00 —
MACRS 20.00 32.00 19.20 11.52 11.52 5.76
Double Decl. Bal. 40.00 24.00 14.40 8.64 5.18 —
Sum of Years Digits 33.33 26.67 20.00 13.33 6.67 —
Exhibit 3. Four depreciation schedules compared: Straight line SL vs. MACRS vs. DDB vs. SOYD
Note that MACRS here refers to a 5-year depreciable life, but which is spread across 6 fiscal years, beginning at the midpoint of year 1.
Non time-based schedules
All of the schedules above are time based schedules because they treat "life" as a fixed period of time, charging a given percentage of depreciable cost each year as an expense. Note, however, that sometimes, so called usage-based depreciation is permitted. A vehicle under this plan, for instance, might have its life defined not in years, but in terms of total miles or kilometers driving expected during its life. The expense claimed each year would reflect the distance traveled that year as a percentage of the lifetime total. Similarly, other kinds of assets might have a scheduled life defined by a quantity that will be used up, in which case the expense percentage each year is based on the quantity used up.
Composite depreciation
Composite depreciation is a method in which a group of related assets is depreciated as a whole rather than individually. This can reduce unnecessary record keeping and reporting and might be used, for example, in depreciating a company’s office furniture, or office equipment. See the encyclopedia entry composite depreciation for an explanation and example.
How is depreciation expense found in spreadsheet implementation?
The depreciation expense for one asset, each year, is found simply by multiplying its depreciable cost by a given percentage for that year, Calculating total expenses can be challenging, however, when the total involves multiple assets and multiple schedules across several years or more.
Consider, for instance building a spreadsheet summary of total depreciation expenses for each of five years, with the following assets and schedules involved:
• Asset A, 4 year life, SL schedule, acquired Year 1.
• Asset B, 5 year life, MACRS schedule, acquired Year 2.
• Asset C, 8 year life, DDB schedule, acquired Year 3.
• Asset D,10 year life, SL schedule, acquired Year 4.
Calculating total depreciation expense becomes more complicated with each passing year:
Year 1 Total expense:
= (Asset A deprec. cost ) * ( SL percentage for Year 1 of 4 )
Year 2 Total expense:
= (Asset A deprec. cost ) * ( SL percentage for Year 2 of 4 ) +
(Asset B deprec. cost ) * (MACRS percentage for Year 1 of 6 )
By Year 4, Total expense:
= ( Asset A deprec. cost ) * ( SL percentage for Year 4 of 4 ) +
( Asset B deprec. cost ) * ( MACRS percentage for Year 3 of 6 ) +
( Asset C deprec. cost ) * ( DDB percentage for Year 2 of 8 ) +
( Asset D deprec. cost ) * ( SL percentage for Year 1 of 10 )
The principles involved in these calculations are simple but the bookkeeping task for the spreadsheet analyst becomes tedious and cumbersome, especially as the number of years considered increases. (You can see and try out working examples of these calculations across multiple years, in either
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Progress Pro—When projects simply have to finish on time. | 3,653 | 16,257 | {"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-2016-18 | latest | en | 0.921719 |
https://www.discountpdh.com/introduction-amplifiers-quiz | 1,627,182,007,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151563.91/warc/CC-MAIN-20210725014052-20210725044052-00333.warc.gz | 754,801,062 | 5,763 | # Introduction to Amplifiers
## Quiz Questions
1. Amplitude is a term that describes the size of a signal. True False 2. SIGNAL is a general term used to refer to a.c.or d.c. of interest in a circuit; e.g., input signal and output signal. True False 3. What determines the class of operation of an amplifier? Amount of time in which current flows in the output circuit None 4. What are the four classes of operation of a transistor amplifier? A, AB, B, C A, B, C, D A, AB, 5. Figure1-8 represents Adding stages of amplification Direct-coupled transistor amplifiers 6. In direct coupling the output of one stage is connected directly to the input of the following stage. True False 7. What is the most common form of coupling? RC coupling Transformer coupling 8. There are two types of feedback in amplifiers. They are Positive Feedback \ Regenerative Feedback Negative Feedback \ Degenerative Feedback Both 9. What is one use for a splitter? A phase splitter is used to provide the input signals to a push-pull amplifier. Not the above 10. Figure 1-23 represents Transistor audio amplifier FET audio amplifier 11. What is the bandwidth of an amplifier? Difference between the upper and lower frequency limits of an amplifier Bandwidth of a transformer 12. What are the upper and lower frequency limits of an amplifier? Half-power points of a frequency-response curve Full-power points of a frequency-response curve 13. What is the purpose of R4? A part of the low-frequency compensation network for Q1 A decoupling capacitor for the effects of R2 14. Figure 2-21 represents _______________. Typical RF amplifier Typical AM radio RF amplifier None 15. A frequency response curve will enable to determine the bandwidth and the upper and lower frequency limits of an amplifier. True False 16. How many inputs and outputs are possible with a differential amplifier as in chapter3? Two inputs, two outputs One input, two outputs 17. Figure 3-8. represents ______________. Single-input, differential-output differential amplifier Single-input, single-output differential amplifier 18. Operational amplifiers are usually high-gain amplifiers with the amount of gain determined by feedback. True False 19. Figure 3-11 represents ____________. Block diagram of an Op-amp Schematics of an Op-amp 20. A magnetic amplifier uses a changing inductance to control the power delivered to a load. True False | 526 | 2,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-31 | latest | en | 0.846903 |
http://mathhelpforum.com/calculus/189473-range-print.html | 1,527,298,683,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00603.warc.gz | 180,811,452 | 2,866 | # Range
• Oct 3rd 2011, 01:46 PM
theloser
Range
$\displaystyle f(x)=3x-\frac{2}{(x+1)^{1/2}}$
Domain= (-1,infinity)
How do you find range?
• Oct 3rd 2011, 01:52 PM
Plato
Re: Range
Quote:
Originally Posted by theloser
$\displaystyle f(x)=3x-\frac{2}{(x+1)^{1/2}}$
Domain= (-1,infinity)
How do you find range?
Did you start by looking at a plot of the function.
• Oct 3rd 2011, 01:53 PM
theloser
Re: Range
Yes, I graphed but can't find exact points...
• Oct 3rd 2011, 02:09 PM
mr fantastic
Re: Range
Quote:
Originally Posted by theloser
Yes, I graphed but can't find exact points...
The graph is quite clear on what the range is: plot y = 3x - 2/Sqrt[x+1] from x = -1 to x = 20 - Wolfram|Alpha
Where are you stuck?
• Oct 3rd 2011, 02:19 PM
theloser
Re: Range
Range=(-infinity,infinity)?
• Oct 3rd 2011, 03:21 PM
TKHunny
Re: Range
If you keep guessing, you will not get it. Prove it! | 347 | 916 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-22 | latest | en | 0.872657 |
https://definecivil.com/quantity-surveying-methods/ | 1,628,164,043,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155529.97/warc/CC-MAIN-20210805095314-20210805125314-00304.warc.gz | 208,845,599 | 75,495 | Are you looking for ways to raise the ability of your construction work and decrease conflicts with effective cost control system? If yes, then these quantity surveying methods can make certain value for money at all times.
Whether you’re a quantity surveyor, a construction professional, a worker, or you’re running a construction company; you’ve to ensure effective financial management to execute project plans successfully. These quantity surveying methods will help you significantly on all sorts of construction projects during all phases including documentation contract, tendering valuations & procedures, cash flow forecasting, financial reporting, interim payment, settlement of contractual disputes, preliminary cost advice, and cost planning.
Your success in construction industry lies heavily on effective financial management which can only be ensured by a competitive quantity surveyor.
Topics we cover
## Quantity Surveying Methods
1. Unit Method
2. Superficial Floor Area Method
3. Elemental Estimate
4. Approximate Quantities
## Unit Method
The unit method estimating consists of choosing a standard unit of accommodation and multiplying an approximate cost per unit. Estimate the building cost base on the depends on the population unit. The unit cost method of estimation is used for project design estimates and bid estimates.
## Superficial Floor Area Method
Superficial area method is the most popular preliminary estimating method. This method is an approximate cost obtained by using an estimated price for each unit of gross floor area. The main reason for the popularity of the area method is its simplicity.
## Elemental Estimate
The elemental method is an approach towards calculating the total estimated cost of Construction projects. It considers the major elements of a building and, if properly implemented, provides a cost estimate based on an elemental breakdown of the building.
The elemental cost analysis method uses elemental cost analyses of previous projects as a basis for the estimate. The cost is on a superficial basis but the superficial unit cost is broken down into elements and sub-elements, which allows for cost adjustment of variations in the design of the new project compared with the previous scheme. The method is flexible, easily understood by all parties, facilitates comparison between projects and facilitates analysis of cost implications of design decisions. However, it is a time-consuming process and requires a high level of expertise.
## Approximate Quantities
The only safe and sure way of estimating construction costs is to take out quantities of works under the different items and multiply these quantities by the prevailing unit rates at the geographical location of the construction project.
However, this can be done only when drawings, specifications, and bill of quantities have been prepared and made available to the contractor.
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When accurate cost of construction is not yet required and a rough idea is sought at some stage to help decision makers decide whether to proceed or not, approximate estimates of construction cost are used.
The approximate cost estimating methods are used to serve either of the following purposes:
To give a rough idea of the probable expenditure: At the early stages of the development of a project it is necessary to ensure whether the project can be financed. A rough idea of the probable expenditure has to be obtained and if this appears feasible, then further details may be considered.
For example while you’re renovating your office building or some commercial property you have to find the cost of flooring to see if a change at this stage is feasible or not. You can hire commercial flooring estimators who can prepare a quantity estimate to help you out in the cost analysis.
Administrative approval: In the case of government and other public works, proper sanction has to be obtained for allocating the expenditure required for the detailed investigations and preparation of plans and estimates. This sanction is given based on approximate cost estimating method.
Valuation and rent fixation: Sometimes it is required to estimate the cost of an existing structure, for one or more of the following reasons: for sale or purchase, for rent fixing, for framing tax schedules, or for insurance requirement. In these cases the approximate methods are adopted.
## The bottom Line
From all these quantity surveying methods, elemental estimates and approximate quantities methods are highly recommended for getting accurate estimates. | 847 | 4,765 | {"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-2021-31 | longest | en | 0.869519 |
https://handwiki.org/wiki/Uniform_10-polytope | 1,718,283,319,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00234.warc.gz | 252,578,924 | 20,800 | # Uniform 10-polytope
Short description: Type of geometrical object
10-simplex Truncated 10-simplex Rectified 10-simplex Cantellated 10-simplex Runcinated 10-simplex Stericated 10-simplex Pentellated 10-simplex Hexicated 10-simplex Heptellated 10-simplex Octellated 10-simplex Ennecated 10-simplex 10-orthoplex Truncated 10-orthoplex Rectified 10-orthoplex 10-cube Truncated 10-cube Rectified 10-cube 10-demicube Truncated 10-demicube
In ten-dimensional geometry, a 10-polytope is a 10-dimensional polytope whose boundary consists of 9-polytope facets, exactly two such facets meeting at each 8-polytope ridge.
A uniform 10-polytope is one which is vertex-transitive, and constructed from uniform facets.
## Regular 10-polytopes
Regular 10-polytopes can be represented by the Schläfli symbol {p,q,r,s,t,u,v,w,x}, with x {p,q,r,s,t,u,v,w} 9-polytope facets around each peak.
There are exactly three such convex regular 10-polytopes:
1. {3,3,3,3,3,3,3,3,3} - 10-simplex
2. {4,3,3,3,3,3,3,3,3} - 10-cube
3. {3,3,3,3,3,3,3,3,4} - 10-orthoplex
There are no nonconvex regular 10-polytopes.
## Euler characteristic
The topology of any given 10-polytope is defined by its Betti numbers and torsion coefficients.[1]
The value of the Euler characteristic used to characterise polyhedra does not generalize usefully to higher dimensions, and is zero for all 10-polytopes, whatever their underlying topology. This inadequacy of the Euler characteristic to reliably distinguish between different topologies in higher dimensions led to the discovery of the more sophisticated Betti numbers.[1]
Similarly, the notion of orientability of a polyhedron is insufficient to characterise the surface twistings of toroidal polytopes, and this led to the use of torsion coefficients.[1]
## Uniform 10-polytopes by fundamental Coxeter groups
Uniform 10-polytopes with reflective symmetry can be generated by these three Coxeter groups, represented by permutations of rings of the Coxeter-Dynkin diagrams:
# Coxeter group Coxeter-Dynkin diagram
1 A10 [39]
2 B10 [4,38]
3 D10 [37,1,1]
Selected regular and uniform 10-polytopes from each family include:
1. Simplex family: A10 [39] -
• 527 uniform 10-polytopes as permutations of rings in the group diagram, including one regular:
1. {39} - 10-simplex -
2. Hypercube/orthoplex family: B10 [4,38] -
• 1023 uniform 10-polytopes as permutations of rings in the group diagram, including two regular ones:
1. {4,38} - 10-cube or dekeract -
2. {38,4} - 10-orthoplex or decacross -
3. h{4,38} - 10-demicube .
3. Demihypercube D10 family: [37,1,1] -
• 767 uniform 10-polytopes as permutations of rings in the group diagram, including:
1. 17,1 - 10-demicube or demidekeract -
2. 71,1 - 10-orthoplex -
## The A10 family
The A10 family has symmetry of order 39,916,800 (11 factorial).
There are 512+16-1=527 forms based on all permutations of the Coxeter-Dynkin diagrams with one or more rings. 31 are shown below: all one and two ringed forms, and the final omnitruncated form. Bowers-style acronym names are given in parentheses for cross-referencing.
# Graph Coxeter-Dynkin diagram
Schläfli symbol
Name
Element counts
9-faces 8-faces 7-faces 6-faces 5-faces 4-faces Cells Faces Edges Vertices
1
t0{3,3,3,3,3,3,3,3,3}
10-simplex (ux)
11 55 165 330 462 462 330 165 55 11
2
t1{3,3,3,3,3,3,3,3,3}
Rectified 10-simplex (ru)
495 55
3
t2{3,3,3,3,3,3,3,3,3}
Birectified 10-simplex (bru)
1980 165
4
t3{3,3,3,3,3,3,3,3,3}
Trirectified 10-simplex (tru)
4620 330
5
t4{3,3,3,3,3,3,3,3,3}
6930 462
6
t0,1{3,3,3,3,3,3,3,3,3}
Truncated 10-simplex (tu)
550 110
7
t0,2{3,3,3,3,3,3,3,3,3}
Cantellated 10-simplex
4455 495
8
t1,2{3,3,3,3,3,3,3,3,3}
Bitruncated 10-simplex
2475 495
9
t0,3{3,3,3,3,3,3,3,3,3}
Runcinated 10-simplex
15840 1320
10
t1,3{3,3,3,3,3,3,3,3,3}
Bicantellated 10-simplex
17820 1980
11
t2,3{3,3,3,3,3,3,3,3,3}
Tritruncated 10-simplex
6600 1320
12
t0,4{3,3,3,3,3,3,3,3,3}
Stericated 10-simplex
32340 2310
13
t1,4{3,3,3,3,3,3,3,3,3}
Biruncinated 10-simplex
55440 4620
14
t2,4{3,3,3,3,3,3,3,3,3}
Tricantellated 10-simplex
41580 4620
15
t3,4{3,3,3,3,3,3,3,3,3}
11550 2310
16
t0,5{3,3,3,3,3,3,3,3,3}
Pentellated 10-simplex
41580 2772
17
t1,5{3,3,3,3,3,3,3,3,3}
Bistericated 10-simplex
97020 6930
18
t2,5{3,3,3,3,3,3,3,3,3}
Triruncinated 10-simplex
110880 9240
19
t3,5{3,3,3,3,3,3,3,3,3}
62370 6930
20
t4,5{3,3,3,3,3,3,3,3,3}
Quintitruncated 10-simplex
13860 2772
21
t0,6{3,3,3,3,3,3,3,3,3}
Hexicated 10-simplex
34650 2310
22
t1,6{3,3,3,3,3,3,3,3,3}
Bipentellated 10-simplex
103950 6930
23
t2,6{3,3,3,3,3,3,3,3,3}
Tristericated 10-simplex
161700 11550
24
t3,6{3,3,3,3,3,3,3,3,3}
138600 11550
25
t0,7{3,3,3,3,3,3,3,3,3}
Heptellated 10-simplex
18480 1320
26
t1,7{3,3,3,3,3,3,3,3,3}
Bihexicated 10-simplex
69300 4620
27
t2,7{3,3,3,3,3,3,3,3,3}
Tripentellated 10-simplex
138600 9240
28
t0,8{3,3,3,3,3,3,3,3,3}
Octellated 10-simplex
5940 495
29
t1,8{3,3,3,3,3,3,3,3,3}
Biheptellated 10-simplex
27720 1980
30
t0,9{3,3,3,3,3,3,3,3,3}
Ennecated 10-simplex
990 110
31
t0,1,2,3,4,5,6,7,8,9{3,3,3,3,3,3,3,3,3}
Omnitruncated 10-simplex
199584000 39916800
## The B10 family
There are 1023 forms based on all permutations of the Coxeter-Dynkin diagrams with one or more rings.
Twelve cases are shown below: ten single-ring (rectified) forms, and two truncations. Bowers-style acronym names are given in parentheses for cross-referencing.
# Graph Coxeter-Dynkin diagram
Schläfli symbol
Name
Element counts
9-faces 8-faces 7-faces 6-faces 5-faces 4-faces Cells Faces Edges Vertices
1
t0{4,3,3,3,3,3,3,3,3}
10-cube (deker)
20 180 960 3360 8064 13440 15360 11520 5120 1024
2
t0,1{4,3,3,3,3,3,3,3,3}
51200 10240
3
t1{4,3,3,3,3,3,3,3,3}
46080 5120
4
t2{4,3,3,3,3,3,3,3,3}
184320 11520
5
t3{4,3,3,3,3,3,3,3,3}
322560 15360
6
t4{4,3,3,3,3,3,3,3,3}
322560 13440
7
t4{3,3,3,3,3,3,3,3,4}
201600 8064
8
t3{3,3,3,3,3,3,3,4}
Trirectified 10-orthoplex (trake)
80640 3360
9
t2{3,3,3,3,3,3,3,3,4}
Birectified 10-orthoplex (brake)
20160 960
10
t1{3,3,3,3,3,3,3,3,4}
Rectified 10-orthoplex (rake)
2880 180
11
t0,1{3,3,3,3,3,3,3,3,4}
Truncated 10-orthoplex (take)
3060 360
12
t0{3,3,3,3,3,3,3,3,4}
10-orthoplex (ka)
1024 5120 11520 15360 13440 8064 3360 960 180 20
## The D10 family
The D10 family has symmetry of order 1,857,945,600 (10 factorial × 29).
This family has 3×256−1=767 Wythoffian uniform polytopes, generated by marking one or more nodes of the D10 Coxeter-Dynkin diagram. Of these, 511 (2×256−1) are repeated from the B10 family and 256 are unique to this family, with 2 listed below. Bowers-style acronym names are given in parentheses for cross-referencing.
# Graph Coxeter-Dynkin diagram
Schläfli symbol
Name
Element counts
9-faces 8-faces 7-faces 6-faces 5-faces 4-faces Cells Faces Edges Vertices
1
10-demicube (hede)
532 5300 24000 64800 115584 142464 122880 61440 11520 512
2
Truncated 10-demicube (thede)
195840 23040
## Regular and uniform honeycombs
There are four fundamental affine Coxeter groups that generate regular and uniform tessellations in 9-space:
# Coxeter group Coxeter-Dynkin diagram
1 $\displaystyle{ {\tilde{A}}_9 }$ [3[10]]
2 $\displaystyle{ {\tilde{B}}_9 }$ [4,37,4]
3 $\displaystyle{ {\tilde{C}}_9 }$ h[4,37,4]
[4,36,31,1]
4 $\displaystyle{ {\tilde{D}}_9 }$ q[4,37,4]
[31,1,35,31,1]
Regular and uniform tessellations include:
### Regular and uniform hyperbolic honeycombs
There are no compact hyperbolic Coxeter groups of rank 10, groups that can generate honeycombs with all finite facets, and a finite vertex figure. However, there are 3 paracompact hyperbolic Coxeter groups of rank 9, each generating uniform honeycombs in 9-space as permutations of rings of the Coxeter diagrams.
$\displaystyle{ {\bar{Q}}_9 }$ = [31,1,34,32,1]: $\displaystyle{ {\bar{S}}_9 }$ = [4,35,32,1]: $\displaystyle{ E_{10} }$ or $\displaystyle{ {\bar{T}}_9 }$ = [36,2,1]:
Three honeycombs from the $\displaystyle{ E_{10} }$ family, generated by end-ringed Coxeter diagrams are:
• 621 honeycomb:
• 261 honeycomb:
• 162 honeycomb:
## References
1. Richeson, D.; Euler's Gem: The Polyhedron Formula and the Birth of Topoplogy, Princeton, 2008.
• T. Gosset: On the Regular and Semi-Regular Figures in Space of n Dimensions, Messenger of Mathematics, Macmillan, 1900
• A. Boole Stott: Geometrical deduction of semiregular from regular polytopes and space fillings, Verhandelingen of the Koninklijke academy van Wetenschappen width unit Amsterdam, Eerste Sectie 11,1, Amsterdam, 1910
• H.S.M. Coxeter:
• H.S.M. Coxeter, M.S. Longuet-Higgins und J.C.P. Miller: Uniform Polyhedra, Philosophical Transactions of the Royal Society of London, Londne, 1954
• H.S.M. Coxeter, Regular Polytopes, 3rd Edition, Dover New York, 1973
• Kaleidoscopes: Selected Writings of H.S.M. Coxeter, edited by F. Arthur Sherk, Peter McMullen, Anthony C. Thompson, Asia Ivic Weiss, Wiley-Interscience Publication, 1995, ISBN 978-0-471-01003-6 [1]
• (Paper 22) H.S.M. Coxeter, Regular and Semi Regular Polytopes I, [Math. Zeit. 46 (1940) 380-407, MR 2,10]
• (Paper 23) H.S.M. Coxeter, Regular and Semi-Regular Polytopes II, [Math. Zeit. 188 (1985) 559-591]
• (Paper 24) H.S.M. Coxeter, Regular and Semi-Regular Polytopes III, [Math. Zeit. 200 (1988) 3-45]
• N.W. Johnson: The Theory of Uniform Polytopes and Honeycombs, Ph.D. Dissertation, University of Toronto, 1966
• | 3,759 | 9,337 | {"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} | 3.296875 | 3 | CC-MAIN-2024-26 | latest | en | 0.869458 |
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# How do you count roman numerals?
Updated: 10/19/2022
Wiki User
14y ago
Best Answer
I means one
II means two
and so on
V means 5
but you can also have IV which means, 5 - 1 = 4
X = 10, but like before you can have IX, which means 9
also you can have XIII = 13
M = 1000
hope this helps
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14y ago
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Related questions
### Why were Roman Numerals created?
I believe it was because Romans could not count numbers so they created roman numerals which were almost like numbers in order to tell time and count.
### What is 8000000 in roman numerals?
Roman numerals were created and used to count things that were bought and sold, so is only used for smaller numbers.
### What were the Roman uses for Roman numerals?
They used them to tell time, to measure farm fields, and to count stuff.
### What do the Roman numeral symbolized?
Roman Numerals were the numbering system used by the ancient Romans. It is what they used to count with.
### What impact did the Roman Numeral system have or has on the culture?
Roman numerals enabled the Romans to count just like any other type of numerals enable any other peoples to count.
### What did Romans use Roman Numerals for?
Romans used Roman numerals as their form of numbers. Romans needed Roman Numerals because they needed numbers to count, tell time, and do other things in life that involved numbers. Roman numerals were used because they could all be scribed using a flat chisel i.e X I V M.
### What is the Roman numerals for 27?
Twenty sever in Roman numerals is XXVII.Twenty sever in Roman numerals is XXVII.Twenty sever in Roman numerals is XXVII.Twenty sever in Roman numerals is XXVII.Twenty sever in Roman numerals is XXVII.Twenty sever in Roman numerals is XXVII.Twenty sever in Roman numerals is XXVII.Twenty sever in Roman numerals is XXVII.Twenty sever in Roman numerals is XXVII.
### Why was roman numerals invented?
Because Romans needed a way to count money, items, etc.
### How do you I write Roman number 11?
Eleven in Roman numerals is XI.Eleven in Roman numerals is XI.Eleven in Roman numerals is XI.Eleven in Roman numerals is XI.Eleven in Roman numerals is XI.Eleven in Roman numerals is XI.Eleven in Roman numerals is XI.Eleven in Roman numerals is XI.Eleven in Roman numerals is XI.
### What inspired the roman numerals?
Roman numerals were inspired by Etruscan numerals of which Roman numerals originated from.
### How do you write 522 in Roman?
In Roman numerals 522 would be DXXII.In Roman numerals 522 would be DXXII.In Roman numerals 522 would be DXXII.In Roman numerals 522 would be DXXII.In Roman numerals 522 would be DXXII.In Roman numerals 522 would be DXXII.In Roman numerals 522 would be DXXII.In Roman numerals 522 would be DXXII.In Roman numerals 522 would be DXXII.
### What is the roman numerals for 76 000?
_L_X_X_VMRomans did not need to count that high.The above is a modern version with underscores added to increase their value by a thousand.L to _L 50,000X to _x 10000V to _V 5000Improved Answer:-76,000 in Roman numerals is (LXXVI) which means 1,000*76 = 76,000 | 853 | 3,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-30 | latest | en | 0.953596 |
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## Decreasing Mean Residual Life (DMRL) Assignment Help
One possible choice is to bring out a regression with one independent variable, and then test whether a Second independent variable is related to the residuals from this regression. One aspect is that if you have a dependent variable, you can rapidly see which independent variables associate with that reliant variable. A Second aspect is that if you will be developing a various regression style, consisting of an independent variable that is extremely associated with an independent variable presently in the style is not most likely to boost the style much, and you may have excellent element to chosen one variable over another Finally, it is useful to look at the flow of the mathematical variables.
One aspect is that if you have a dependent variable, you can rapidly see which independent variables associate with that reliant variable. A Second element is that if you will be developing a numerous regression style, consisting of an independent variable that is extremely associated with an independent variable presently in the style is not most likely to boost the style much, and you may have fantastic requirement to picked one variable over another Finally, it is rewarding to take an appearance at the flow of the mathematical variables.Decreasing mean residual life (DMRL) The fundamental function of a number of regressions (the term was at first used by Pearson, 1908) is to discover more about the relationship between many independent or predictor variables and a reliant or requirement variable. You might discover that the variety of bed spaces is a far better predictor of the expense for which a house deals in a particular location than how “rather” the house is (subjective rating).
Employee Decreasing mean residual life (DMRL) normally use various regression treatments to find out reasonable payment. This information can be used in a numerous regression analysis to establish a regression formula of the type.This “flying start” guide exposes you the very best methods to carry out numerous regressions making use of SPSS Data, in addition to equate and report the occur from this test. Prior to we provide you to this treatment, you need to understand the different anticipations that your details need to please in order for numerous regression to use you a genuine result. We discuss these anticipations next.
Decreasing mean residual life (DMRL) various regressions, we mean styles with merely one reliant and 2 or more independent (exploratory) variables. The variable whose worth is to be anticipated is described as the dependent variable and the ones whose acknowledged worths are made use of for projection are comprehended independent (exploratory) variables.In this lesson, we make our initial (and last?!) considerable dive in the course. We move from the standard direct regression style with one predictor to the many direct regression styles with 2 or more predictors. That is, we make use of the adjective “standard” to symbolize that our style has simply predictor, and we make use of the adjective” Decreasing mean residual life (DMRL)” to recommend that our style has at least 2 predictors.
In the various regression setting, considering that of the potentially a good deal of predictors, it is more efficient to use matrices to define the regression style and the subsequent analyses. This lesson thinks of a few of the more important various regression services in matrix type. If you’re not sure about any of this, it may be an excellent time to take a look at this Matrix Algebra Examination.It is possible that the independent variables may obscure each other’s Decreasing mean residual life (DMRL). The age outcome might bypass the diet strategy result, causing a regression for diet strategy which would not appear very remarkable.
One possible service is to perform a regression with one independent variable, and after that test whether a Second independent variable is linked to the residuals from this regression. You continue with a 3rd variable, and so on. A problem with this is that you are putting some variables in lucky positions.
A various regression makes it possible for the synchronised screening and modeling of a number of independent variables. (Note: Decreasing mean residual life (DMRL) is still eliminated a “multivariate” test because there is simply one reliant variable).You might believe that there’s a connection between just how much you take in and what does it cost? you weigh; regression analysis can help you determine that. Regression analysis will provide you with a formula for a chart so that you can make projections about your details. You may come throughout more advanced approaches like many regressions.
It is an exceptional principle to take a look at the connections Decreasing mean residual life (DMRL) these variables Whenever you have a dataset with various mathematical variables. One element is that if you have a dependent variable, you can rapidly see which independent variables relate to that reliant variable. A Second aspect is that if you will be developing a numerous regression style, consisting of an independent variable that is extremely related to an independent variable presently in the style is not most likely to improve the style much, and you may have fantastic have to picked one variable over another Last but not least, it is rewarding to have a look at the blood circulation of the mathematical variables.
One possible choice is to perform a regression with one independent variable, then test whether a Second independent variable is connected to the residuals from this regression. One element is that if you have a dependent variable, you can rapidly see which independent variables relate to that reliant variable. A Second aspect is that if you will be constructing a many regression style, consisting of an independent variable that is extremely connected with an independent variable presently in the style is not most likely to improve the style much, and you may have excellent element to picked one variable over another Finally, it is useful to take a look at the flow of the mathematical variables. | 1,161 | 6,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-34 | latest | en | 0.928253 |
https://www.kopykitab.com/blog/stress-intensity-factor-notes/ | 1,620,980,737,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991648.40/warc/CC-MAIN-20210514060536-20210514090536-00365.warc.gz | 882,115,164 | 22,152 | # Stress Intensity Factor Notes
Stress Intensity Factor, K, is used in fracture mechanics to more accurately predict the stress state (“stress intensity”) near the tip of a crack caused by a remote load or residual stresses. When this stress state becomes critical a small crack grows (“extends”) and the material fails. The load at which this failure occurs is referred to as the fracture strength. The experimental fracture strength of solid materials is 10 to 1000 times below the theoretical strength values, where tiny internal and external surface cracks create higher stresses near these cracks, hence lowering the theoretical value of strength. The large crack seen in the picture of the Liberty Bell was the result of small cracks and internal residual stresses not known at the time. The original, “as fabricated” cracks were very small and hard to see with naked eyes, and according to Hertzberg, during the war against the British, the bell was polished whenever they saw a crack on the surface. Hardly a solution based on what we understand today. Unlike “stress concentration”, Stress Intentsity, K, as the name implies, is a parameter that amplifies the magnitude of the applied stress that includes the geometrical parameter Y (load type). These load types are categorized as Mode-I, -II, or -III. The Mode-I stress intensity factor, KIc is the most often used engineering design parameter in fracture mechanics and hence must be understood if we are to design fracture tolerant materials used in bridges, buildings, aircraft, or even bells. Polishing just won’t do if we detect a crack. Typically for most materials if a crack can be seen it is very close to the critical stress state predicted by the “Stress Intensity Factor”.
### Stress Analysis of Cracks
Generally there are three modes to describe different crack surface displacement in Fig.8.3 (Hertzberg, p321). Mode I is opening or tensile mode where the crack surfaces move directly apart. Mode II is sliding or in-plane shear mode where the crack surfaces slide over one another in a direction perpendicular to the leading edge of the crack. Mode III is tearing and antiplane shear mode where the crack surfaces move relative to one another and parallel to the leading edge of the crack. Mode I is the most common load type encountered in engineering design and will be explained here in more detail.
The value of the stress intensity factor, K, is a function of the applied stress, the size and the position of the crack as well as the geometry of the solid piece where the cracks are detected, Fig.8.5 (Hertxberg, p323). The tensile stress in X and Y directions, and the shear stress in the X-Y plane can calculated in terms of K and position can be written as:
Mode-I
Mode-I
### Fracture Toughness, KIc
Engineers are mostly worried about the brittle fracture because the brittle fractures bring most devastating accidents and happen rapidly, and usually the brittle fractures take place when the applied stress increases such that the stress state at the crack tip reaches a critical value. The fracture toughness can be defined in terms of the stress intensity factor, K, but at a critical stress state. as:
where Y is a dimensionless parameter that depends on both the specimen and crack geometry in Fig8.11(Callister, p193), and the greek symbol “Sigma” is an applied stress and “a” is crack length. Generally, for the elliptical shaped crack,the equation is modified to include the geometry of the crack with three different Y’s.
However, Y factor is 1.0 for the plate of infinite width and 1.1 for a plate of semi-infinite width. When the thickness of specimen is very large with respect to the crack length, the stress intensity factor for Mode I is often called the plane strain “Fracture Toughness”. This modification of stress intensity into a plane strain fracture toughness parameter can be approximated by a relationship that includes specimen geometry, and yield strength.
Hence the specimen thickness is shown to be the most significant parameter that controls the transition of fracture toughness from “plane stress” to “plane strain”, see Fig. 8.12 (Callister, p194).
The plain strain fracture toughness for Mode I, KIc is also a function of many other factors such as temperature, strain-rate and microstructure. Hence KIc is unique for a particular material and is a fundamental material property so it is a very important consideration for material selection and design.
### Designing and Preventing Fracture with KIc
KIc, stress, and Y factor are important variables for engineers to design and to determine the safety of machinery, and often the size of the cracks is a very important factor to make decisions such that the maximum allowable size of the crack can be written as
Also because KIc is unique for a particular material, engineers can use this variable for selecting appropriate materials for a range of different applications. From the table in Appendix B engineers can also decide how much load and stress can be allowed for a particular specimen geometry. This critical information helps engineers to optimize the design and the safety on the operations and to prevent or minimize possible accidents. For example, in aircraft components, there are a lot of rivet holes and small cracks which bring Y calibration factor high up to the critical stress. What engineers do is measure the length of cracks to calculate the maximum cracks length and to compare with safety measurement. They can also make a hole at the tip of cracks, which brings down Y calibration factor and the also the the stress concentration. Additionally, engineers clean the fracture surfaces to prevent further damages. Not only does cleaning lower the Y calibration, but it also helps to protect the surface from undesireable chemical reactions. Various cleaning methods are described in the table below.
Table A.1 (Hertzberg, p752)
### Conclusion
Stress intensity and fracture toughness are critically important fracture mechanics parameters used by materials engineers and designers. We saw that there are a lot of factors that determine fracture of a material. KIc is an unique material property, that is used by engineers to design and manufacture products for durability and safe operation.
Appendix B (Hertzberg, p757)
K Calibrations for Typical Test Specimen Geometries | 1,287 | 6,382 | {"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-2021-21 | latest | en | 0.914564 |
https://calculationcalculator.com/jerib-to-square-mile | 1,723,722,453,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641291968.96/warc/CC-MAIN-20240815110654-20240815140654-00385.warc.gz | 105,176,347 | 22,277 | # Jerib to Square Mile Conversion
## 1 Jerib is equal to how many Square Mile?
### 0.000780488 Square Mile
##### Reference This Converter:
Jerib and Square Mile both are the Land measurement unit. Compare values between unit Jerib with other Land measurement units. You can also calculate other Land conversion units that are available on the select box, having on this same page.
Jerib to Square Mile conversion allows you to convert value between Jerib to Square Mile easily. Just enter the Jerib value into the input box, the system will automatically calculate Square Mile value. 1 Jerib in Square Mile? In mathematical terms, 1 Jerib = 0.000780488 Square Mile.
To conversion value between Jerib to Square Mile, just multiply the value by the conversion ratio. One Jerib is equal to 0.000780488 Square Mile, so use this simple formula to convert -
The value in Square Mile is equal to the value of Jerib multiplied by 0.000780488.
Square Mile = Jerib * 0.000780488;
For calculation, here's how to convert 10 Jerib to Square Mile using the formula above -
10 Jerib = (10 * 0.000780488) = 0.00780488 Square Mile
Jerib Square Mile (sq mi) Conversion
1 0.000780488 1 Jerib = 0.000780488 Square Mile
2 0.001560976 2 Jerib = 0.001560976 Square Mile
3 0.002341464 3 Jerib = 0.002341464 Square Mile
4 0.003121952 4 Jerib = 0.003121952 Square Mile
5 0.00390244 5 Jerib = 0.00390244 Square Mile
6 0.004682928 6 Jerib = 0.004682928 Square Mile
7 0.005463416 7 Jerib = 0.005463416 Square Mile
8 0.006243904 8 Jerib = 0.006243904 Square Mile
9 0.007024392 9 Jerib = 0.007024392 Square Mile
10 0.00780488 10 Jerib = 0.00780488 Square Mile
11 0.008585368 11 Jerib = 0.008585368 Square Mile
12 0.009365856 12 Jerib = 0.009365856 Square Mile
13 0.010146344 13 Jerib = 0.010146344 Square Mile
14 0.010926832 14 Jerib = 0.010926832 Square Mile
15 0.01170732 15 Jerib = 0.01170732 Square Mile
16 0.012487808 16 Jerib = 0.012487808 Square Mile
17 0.013268296 17 Jerib = 0.013268296 Square Mile
18 0.014048784 18 Jerib = 0.014048784 Square Mile
19 0.014829272 19 Jerib = 0.014829272 Square Mile
20 0.01560976 20 Jerib = 0.01560976 Square Mile
21 0.016390248 21 Jerib = 0.016390248 Square Mile
22 0.017170736 22 Jerib = 0.017170736 Square Mile
23 0.017951224 23 Jerib = 0.017951224 Square Mile
24 0.018731712 24 Jerib = 0.018731712 Square Mile
25 0.0195122 25 Jerib = 0.0195122 Square Mile
26 0.020292688 26 Jerib = 0.020292688 Square Mile
27 0.021073176 27 Jerib = 0.021073176 Square Mile
28 0.021853664 28 Jerib = 0.021853664 Square Mile
29 0.022634152 29 Jerib = 0.022634152 Square Mile
30 0.02341464 30 Jerib = 0.02341464 Square Mile
31 0.024195128 31 Jerib = 0.024195128 Square Mile
32 0.024975616 32 Jerib = 0.024975616 Square Mile
33 0.025756104 33 Jerib = 0.025756104 Square Mile
34 0.026536592 34 Jerib = 0.026536592 Square Mile
35 0.02731708 35 Jerib = 0.02731708 Square Mile
36 0.028097568 36 Jerib = 0.028097568 Square Mile
37 0.028878056 37 Jerib = 0.028878056 Square Mile
38 0.029658544 38 Jerib = 0.029658544 Square Mile
39 0.030439032 39 Jerib = 0.030439032 Square Mile
40 0.03121952 40 Jerib = 0.03121952 Square Mile
41 0.032000008 41 Jerib = 0.032000008 Square Mile
42 0.032780496 42 Jerib = 0.032780496 Square Mile
43 0.033560984 43 Jerib = 0.033560984 Square Mile
44 0.034341472 44 Jerib = 0.034341472 Square Mile
45 0.03512196 45 Jerib = 0.03512196 Square Mile
46 0.035902448 46 Jerib = 0.035902448 Square Mile
47 0.036682936 47 Jerib = 0.036682936 Square Mile
48 0.037463424 48 Jerib = 0.037463424 Square Mile
49 0.038243912 49 Jerib = 0.038243912 Square Mile
50 0.0390244 50 Jerib = 0.0390244 Square Mile
51 0.039804888 51 Jerib = 0.039804888 Square Mile
52 0.040585376 52 Jerib = 0.040585376 Square Mile
53 0.041365864 53 Jerib = 0.041365864 Square Mile
54 0.042146352 54 Jerib = 0.042146352 Square Mile
55 0.04292684 55 Jerib = 0.04292684 Square Mile
56 0.043707328 56 Jerib = 0.043707328 Square Mile
57 0.044487816 57 Jerib = 0.044487816 Square Mile
58 0.045268304 58 Jerib = 0.045268304 Square Mile
59 0.046048792 59 Jerib = 0.046048792 Square Mile
60 0.04682928 60 Jerib = 0.04682928 Square Mile
61 0.047609768 61 Jerib = 0.047609768 Square Mile
62 0.048390256 62 Jerib = 0.048390256 Square Mile
63 0.049170744 63 Jerib = 0.049170744 Square Mile
64 0.049951232 64 Jerib = 0.049951232 Square Mile
65 0.05073172 65 Jerib = 0.05073172 Square Mile
66 0.051512208 66 Jerib = 0.051512208 Square Mile
67 0.052292696 67 Jerib = 0.052292696 Square Mile
68 0.053073184 68 Jerib = 0.053073184 Square Mile
69 0.053853672 69 Jerib = 0.053853672 Square Mile
70 0.05463416 70 Jerib = 0.05463416 Square Mile
71 0.055414648 71 Jerib = 0.055414648 Square Mile
72 0.056195136 72 Jerib = 0.056195136 Square Mile
73 0.056975624 73 Jerib = 0.056975624 Square Mile
74 0.057756112 74 Jerib = 0.057756112 Square Mile
75 0.0585366 75 Jerib = 0.0585366 Square Mile
76 0.059317088 76 Jerib = 0.059317088 Square Mile
77 0.060097576 77 Jerib = 0.060097576 Square Mile
78 0.060878064 78 Jerib = 0.060878064 Square Mile
79 0.061658552 79 Jerib = 0.061658552 Square Mile
80 0.06243904 80 Jerib = 0.06243904 Square Mile
81 0.063219528 81 Jerib = 0.063219528 Square Mile
82 0.064000016 82 Jerib = 0.064000016 Square Mile
83 0.064780504 83 Jerib = 0.064780504 Square Mile
84 0.065560992 84 Jerib = 0.065560992 Square Mile
85 0.06634148 85 Jerib = 0.06634148 Square Mile
86 0.067121968 86 Jerib = 0.067121968 Square Mile
87 0.067902456 87 Jerib = 0.067902456 Square Mile
88 0.068682944 88 Jerib = 0.068682944 Square Mile
89 0.069463432 89 Jerib = 0.069463432 Square Mile
90 0.07024392 90 Jerib = 0.07024392 Square Mile
91 0.071024408 91 Jerib = 0.071024408 Square Mile
92 0.071804896 92 Jerib = 0.071804896 Square Mile
93 0.072585384 93 Jerib = 0.072585384 Square Mile
94 0.073365872 94 Jerib = 0.073365872 Square Mile
95 0.07414636 95 Jerib = 0.07414636 Square Mile
96 0.074926848 96 Jerib = 0.074926848 Square Mile
97 0.075707336 97 Jerib = 0.075707336 Square Mile
98 0.076487824 98 Jerib = 0.076487824 Square Mile
99 0.077268312 99 Jerib = 0.077268312 Square Mile
100 0.0780488 100 Jerib = 0.0780488 Square Mile
200 0.1560976 200 Jerib = 0.1560976 Square Mile
300 0.2341464 300 Jerib = 0.2341464 Square Mile
400 0.3121952 400 Jerib = 0.3121952 Square Mile
500 0.390244 500 Jerib = 0.390244 Square Mile
600 0.4682928 600 Jerib = 0.4682928 Square Mile
700 0.5463416 700 Jerib = 0.5463416 Square Mile
800 0.6243904 800 Jerib = 0.6243904 Square Mile
900 0.7024392 900 Jerib = 0.7024392 Square Mile
1000 0.780488 1000 Jerib = 0.780488 Square Mile
2000 1.560976 2000 Jerib = 1.560976 Square Mile
3000 2.341464 3000 Jerib = 2.341464 Square Mile
4000 3.121952 4000 Jerib = 3.121952 Square Mile
5000 3.90244 5000 Jerib = 3.90244 Square Mile
6000 4.682928 6000 Jerib = 4.682928 Square Mile
7000 5.463416 7000 Jerib = 5.463416 Square Mile
8000 6.243904 8000 Jerib = 6.243904 Square Mile
9000 7.024392 9000 Jerib = 7.024392 Square Mile
10000 7.80488 10000 Jerib = 7.80488 Square Mile
A jerib is a traditional unit of land measurement in the Middle East and southwestern Asia. It is a unit of area used to measure land holdings in much the way that an acre or hectare.
The square mile is an imperial and US unit of measure for an area. 1 Square Mile (sq mi) is equal to 27880000 Square Feet (sq ft).
The value in Square Mile is equal to the value of Jerib multiplied by 0.000780488.
Square Mile = Jerib * 0.000780488;
1 Jerib is equal to 0.000780488 Square Mile.
1 Jerib = 0.000780488 Square Mile.
• jerib to sq mi
• 1 jerib = square mile
• jerib into square mile
• jeribs to square miles
• convert jerib to square mile
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→ | 2,967 | 7,629 | {"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-2024-33 | latest | en | 0.686469 |
https://convertoctopus.com/287-kilometers-per-hour-to-miles-per-hour | 1,718,749,181,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861794.76/warc/CC-MAIN-20240618203026-20240618233026-00286.warc.gz | 158,396,291 | 7,549 | ## Conversion formula
The conversion factor from kilometers per hour to miles per hour is 0.62137119223783, which means that 1 kilometer per hour is equal to 0.62137119223783 miles per hour:
1 km/h = 0.62137119223783 mph
To convert 287 kilometers per hour into miles per hour we have to multiply 287 by the conversion factor in order to get the velocity amount from kilometers per hour to miles per hour. We can also form a simple proportion to calculate the result:
1 km/h → 0.62137119223783 mph
287 km/h → V(mph)
Solve the above proportion to obtain the velocity V in miles per hour:
V(mph) = 287 km/h × 0.62137119223783 mph
V(mph) = 178.33353217226 mph
The final result is:
287 km/h → 178.33353217226 mph
We conclude that 287 kilometers per hour is equivalent to 178.33353217226 miles per hour:
287 kilometers per hour = 178.33353217226 miles per hour
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 mile per hour is equal to 0.0056074703832708 × 287 kilometers per hour.
Another way is saying that 287 kilometers per hour is equal to 1 ÷ 0.0056074703832708 miles per hour.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred eighty-seven kilometers per hour is approximately one hundred seventy-eight point three three four miles per hour:
287 km/h ≅ 178.334 mph
An alternative is also that one mile per hour is approximately zero point zero zero six times two hundred eighty-seven kilometers per hour.
## Conversion table
### kilometers per hour to miles per hour chart
For quick reference purposes, below is the conversion table you can use to convert from kilometers per hour to miles per hour
kilometers per hour (km/h) miles per hour (mph)
288 kilometers per hour 178.955 miles per hour
289 kilometers per hour 179.576 miles per hour
290 kilometers per hour 180.198 miles per hour
291 kilometers per hour 180.819 miles per hour
292 kilometers per hour 181.44 miles per hour
293 kilometers per hour 182.062 miles per hour
294 kilometers per hour 182.683 miles per hour
295 kilometers per hour 183.305 miles per hour
296 kilometers per hour 183.926 miles per hour
297 kilometers per hour 184.547 miles per hour | 567 | 2,296 | {"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} | 4.03125 | 4 | CC-MAIN-2024-26 | latest | en | 0.687025 |
https://cran.uib.no/web/packages/superb/vignettes/VignetteC.html | 1,637,999,485,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00595.warc.gz | 274,854,908 | 262,206 | In this vignette, we show how to plot proportions. Proportions is one way to summarize observations that are composed of succes and failure. Success can be a positive reaction to a drug, an accurate completion of a task, a survival after a dangerous ilness, etc. Failure are the opposite of success.
A proportion is the number of success onto the total number of trials. Similarly, if the success are coded with “1”s and failure, with “0”s, then the proportion can be obtained indirectly by computing the mean.
## An example
Consider an example where three groups of participants where examined. The raw data may look like:
Group Score
subject 1 1 1
subject 2 1 1
subject n1 1 0
subject n1+1 2 0
subject n1+n2 2 1
subject n1+n2+1 3 1
subject n1+n2+n3 3 0
in which there is $$n_1$$ participant in Group 1, $$n_2$$ in Group 2, and $$n_3$$ in Group 3. The data can be compiled by reporting the number of success (let’s call them $$s$$) and the number of participants.
One example of results could be
s n proportion
Group 1 10 30 33.3%
Group 2 18 28 63.6%
Group 3 10 26 38.9%
Although making a plot of these proportions is easy, how can you plot error bars around these proportions?
## The arcsine transformation
First proposed by Fisher, the arcsine transform is one way to represent proportions. This transformation stretches the extremities of the domain (near 0% and near 100%) so that the sampling variability is constant for any observed proportion. Also, this transformation make the sampling distribution nearly normal so that $$z$$ test can be used.
An improvement over the Fisher transformation was proposed by Anscombe (1948). It is given by
$A(s, n) = \sin^{-1}\left( \sqrt{\frac{s + 3/8}{n + 3/4}} \right)$
The variance of such transformation is also theoretically given by
$Var_A = \frac{1}{4(n+1/2)}$
As such, we have all the ingredients needed to make confidence intervals!
## Defining the data
In what follows, we assume that the data are available in compiled form, as in the second table above. Because superb only takes raw data, we will have to convert these into a long sequence of zeros and ones.
# enter the compiled data into a data frame:
compileddata <- data.frame(cbind(
s = c(10, 18, 10),
n = c(30, 28, 26)
))
The following converts the compiled data into a long data frame containing ones and zeros so that superb can be fed raw data:
group <- c()
scores <- c()
for (i in 1: (dim(compileddata)[1])) {
group <- c( group, rep(i, compileddata$n[i] ) ) scores <- c( scores, rep(1, compileddata$s[i]),
rep(0, compileddata$n[i] - compileddata$s[i]) )
}
dta <- data.frame( cbind(group = group, scores = scores ) )
## Defining the transformation in R
In the following, we define the A (Anscombe) transformation, the standard error of the transformed scores, and the confidence intervals:
# the Anscombe transformation for a vector of binary data 0|1
A <-function(v) {
x <- sum(v)
n <- length(v)
asin(sqrt( (x+3/8) / (n+3/4) ))
}
SE.A <- function(v) {
0.5 / sqrt(length(v+1/2))
}
CI.A <- function(v, gamma = 0.95){
SE.A(v) * sqrt( qchisq(gamma, df=1) )
}
This is all we need to make a basic plot with suberbPlot()
… but we need a few libraries, so let’s load them here:
library(superb)
library(ggplot2)
library(scales) # for asn_trans() non-linear scale
Here we go:
# ornate to decorate the plot a little bit...
ornate = list(
theme_bw(base_size = 16),
labs(x = "Group" ),
scale_x_discrete(labels=c("Group A", "Group B", "Group C"))
)
superbPlot(dta,
BSFactors = "group",
variables = "scores",
statistic = "A",
error = "CI",
adjustment = list( purpose = "difference"),
plotStyle = "line",
errorbarParams = list(color="blue") # just for the pleasure!
) + ornate + labs(y = "Anscombe-transformed scores" )
## Reversing the transformation to see proportions.
The above plot shows Anscombe-transform scores. This may not be very intuitive. It is then possible to undo the transformation so as to plot proportions instead. The complicated part is to undo the confidence limits.
# the proportion of success for a vector of binary data 0|1
prop <- function(v){
x <- sum(v)
n <- length(v)
x/n
}
# the de-transformed confidence intervals from Anscombe-transformed scores
CI.prop <- function(v, gamma = 0.95) {
y <- A(v)
n <- length(v)
cilen <- CI.A(v, gamma)
ylo <- y - cilen
yhi <- y + cilen
# reverse arc-sin transformation: naive approach
cilenlo <- ( sin(ylo)^2 )
cilenhi <- ( sin(yhi)^2 )
c(cilenlo, cilenhi)
}
Nothing more is needed. We can make the plot with these new functions:
superbPlot(dta,
BSFactors = "group",
variables = "scores",
statistic = "prop",
error = "CI",
adjustment = list( purpose = "difference"),
plotStyle = "line",
errorbarParams = list(color="blue")
) + ornate + labs(y = "Proportions" ) +
scale_y_continuous(trans=asn_trans())
This new plot is actually identical to the previous one as we plotted the proportions using a non-linear scale (the “asn_trans()” scale for arcsine). However, the vertical axis is now showing graduations between 0% and 100% as is expected of proportions.
## Returning to the example
What can we conclude from the plot? You noted that we plotted difference-adjusted confidence intervals. Hence, if at least one result is not included in the confidence interval of another result, then the chances are good that they differ significantly.
Running an analysis of proportions, it indicates the presence of a main effect of Group ($$F(2,\infty)= 3.11, p = .045$$).
We see from the plot that the length of the error bars are about all the same, suggesting homogeneous variance (because all the sample are of comparable size). This is always the case as Anscombe transform is a ‘variance-stabilizing’ transformation in the sense that it makes all the variances identical.
## In summary
The superb framework can be used to display any summary statistics. Here, we showed how superbPlot() can be used with proportions. | 1,567 | 5,943 | {"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": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-49 | latest | en | 0.904773 |
https://www.jiskha.com/members/profile/posts.cgi?name=Chris | 1,531,799,532,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589557.39/warc/CC-MAIN-20180717031623-20180717051623-00411.warc.gz | 905,341,177 | 9,998 | # Posts by Chris
Total # Posts: 2,804
Thank you!
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17. Next>> | 4,230 | 15,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-30 | latest | en | 0.834301 |
https://uk.mathworks.com/matlabcentral/cody/problems/2664-divisors-for-big-integer/solutions/2506115 | 1,603,246,773,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874637.23/warc/CC-MAIN-20201021010156-20201021040156-00330.warc.gz | 572,378,393 | 17,371 | Cody
Problem 2664. Divisors for big integer
Solution 2506115
Submitted on 9 Jun 2020 by Edward Huang
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
x = 10; y_correct = 4; assert(isequal(divisors_Big(x),y_correct))
2 Pass
x = 28; y_correct = 6; assert(isequal(divisors_Big(x),y_correct))
3 Pass
x = 28; y_correct = 6; assert(isequal(divisors_Big(x),y_correct))
4 Pass
x = 784; y_correct = 15; assert(isequal(divisors_Big(x),y_correct))
5 Pass
x = 1452637; y_correct = 2; assert(isequal(divisors_Big(x),y_correct))
6 Pass
x = 5452637; y_correct = 4; assert(isequal(divisors_Big(x),y_correct))
7 Pass
x = 16452637; y_correct = 2; assert(isequal(divisors_Big(x),y_correct))
8 Pass
x = 116452637; y_correct = 8; assert(isequal(divisors_Big(x),y_correct))
9 Pass
x = 416452638; y_correct = 32; assert(isequal(divisors_Big(x),y_correct))
10 Pass
x = 12250000; y_correct = 105; assert(isequal(divisors_Big(x),y_correct))
11 Pass
x = 2031120; y_correct = 240; assert(isequal(divisors_Big(x),y_correct))
12 Pass
x = 76576500; y_correct = 576; assert(isequal(divisors_Big(x),y_correct))
13 Pass
x = 816452637; y_correct = 32; assert(isequal(divisors_Big(x),y_correct))
14 Pass
x = 103672800; y_correct = 648; assert(isequal(divisors_Big(x),y_correct))
15 Pass
x = 842161320; y_correct = 1024; assert(isequal(divisors_Big(x),y_correct))
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 531 | 1,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-45 | latest | en | 0.504107 |
http://betterlesson.com/lesson/523430/gcf-and-lcm-project-day-1 | 1,477,628,909,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721555.54/warc/CC-MAIN-20161020183841-00402-ip-10-171-6-4.ec2.internal.warc.gz | 27,414,051 | 17,757 | # GCF and LCM Project, Day 1
Unit 4: Factors and Multiples
Lesson 5 of 10
## Big Idea: Students collaborate on a GCF/LCM project.
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11 teachers like this lesson
Standards:
Subject(s):
Math, Number Sense and Operations, Fractions, GCF, LCM, prime factorization, project
51 minutes
### Ursula Lovings
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Environment: Urban | 266 | 1,103 | {"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-2016-44 | longest | en | 0.796274 |
https://portfolioslab.com/tools/portfolio-optimization/risk-parity | 1,719,355,167,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00288.warc.gz | 400,381,338 | 25,669 | Tools
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# Risk Parity Optimization
Risk parity portfolio optimization is a type of investment strategy that seeks to allocate portfolio assets in a way that balances risk across multiple asset classes. It is based on the idea that each asset class's contribution to the portfolio's overall risk should be approximately equal, regardless of the size or historical performance of the asset class. To achieve this balance, the strategy uses a mathematical model that estimates the risk contribution of each asset class based on historical data. The model considers volatility, correlation, and covariance factors to determine how much of each asset class should be included in the portfolio. Risk parity portfolio optimization aims to create a diversified portfolio that is not overly exposed to any single asset class and has a consistent level of risk across all asset classes. As a result, this approach can potentially lead to more stable returns over the long term, as the portfolio is less likely to be impacted by sudden shifts in the market.
Mean-Variance Optimization (MVO)
This foundational model of portfolio theory balances risk and return, aiming for the highest expected return for a given risk level using historical data. However, its reliance on statistical estimates can lead to over-concentration in certain assets.
Risk Parity
Instead of focusing on capital allocation like MVO, this model equally distributes risk across all assets or asset classes in a portfolio. It results in more diversified portfolios and is less sensitive to market changes compared to MVO.
Hierarchical Risk Parity (HRP)
Combining aspects of MVO and risk parity, HRP uses a hierarchical clustering algorithm to allocate risk within asset clusters. This approach leads to enhanced diversification and is effective in managing market instabilities.
Your portfolio is currently empty. You can import symbols, add them manually, or select from an existing portfolio.
## Optimization settings
%
How to choose lookback
Optimization Date
%
%
## Optimal Asset Allocation
Click Calculate to get results
## Portfolio Performance
Click Calculate to get results
## Portfolio Sharpe Ratio
Click Calculate to get results
## Portfolio Drawdowns
Click Calculate to get results
## Portfolio Volatility
Click Calculate to get results | 461 | 2,477 | {"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-2024-26 | latest | en | 0.899635 |
http://research.stlouisfed.org/fred2/graph/?id=PLOCONMXA622NUPN | 1,405,318,866,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776439950.90/warc/CC-MAIN-20140707234039-00069-ip-10-180-212-248.ec2.internal.warc.gz | 96,547,880 | 16,325 | # Graph: Price Level of Consumption for Mexico
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
Release:
Restore defaults | Save settings | Apply saved settings
w h
Graph Background: Plot Background: Text:
Color:
(a) Price Level of Consumption for Mexico, PPP of Consumption over Exchange Rate, Not Seasonally Adjusted (PLOCONMXA622NUPN)
Price Level of GDP is the PPP over GDP divided by the exchange rate times 100. The PPP of GDP or any component is the national currency value divided by the real value in international dollars. The PPP and the exchange rate are both expressed as national currency units per US dollar.The value of price level of GDP for the United States is made equal to 100. Price Levels of the components Consumption, Investment, and Government are derived in the same way as the price level of GDP. While the U.S. = 100 over GDP, this is not true for the component shares. The purchasing power parity in domestic currency per \$US for GDP or any component, may be obtained by dividing the price level by 100 and multiplying by the Exchange Rate. More information is available at http://pwt.econ.upenn.edu/Documentation/append61.pdf.
For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
Source Indicator: pc
Price Level of Consumption for Mexico
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
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Name: Email: | 538 | 2,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2014-23 | latest | en | 0.841088 |
https://www.physicsforums.com/threads/theory-of-relativity-help.209146/ | 1,642,386,167,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00372.warc.gz | 1,004,659,257 | 15,336 | # Theory of relativity help
Alright, i have a final and i have to explain just three parts to einstein's theory of relativity...... they are space travel (time slowing nearing the speed of light), increase of mass with speed, and length contraction..... the space travel makes no sense, and the mass one, i have mixed feelings about.... i know that mass is the amount of matter something has, but as somethign speeds up, it gets more matter? it doesn't make sense..... and the length contraction seems backwards too.... normally when you see a car fly by really fast, it looks a little longer.... man if i were einstein, then i would know, but seeing as how i am just in high school, it doesn't make to much sense... please help
HallsofIvy
Homework Helper
First of all, mass is NOT "the amount of matter something has". Mass is a measure of inertia- it measures how much force it takes to change the velocity of an object. That is what increases with speed.
I'm not sure at all what you mean by "normally when you see a car fly by really fast, it looks a little longer". I've never noticed that! In any case I assure you that you have never seen "a car fly by" at speeds where relativistic effects would be noticed! In order for the contraction be as much as a millimeter, say, on a car of length 4 meters, the car would have to be going about 97% the speed of light or about one 1,700,000,000 miles per hour!
I am puzzled why you would have to write three parts of Einstein's theory of relativity if you have not been taught any of this and there is nothing in your text book about it.
well, our whole class is taking it apart and split it all into groups... we were chosen for those three..... now that i think about the car, it doesn't quite make sense.... but why does the length contract? i think i have figured out the space time thing, and i think i am grasping hold of the mass increase, but length contraction is what we need to work on now
well, our whole class is taking it apart and split it all into groups... we were chosen for those three..... now that i think about the car, it doesn't quite make sense.... but why does the length contract? i think i have figured out the space time thing, and i think i am grasping hold of the mass increase, but length contraction is what we need to work on now
I'm assuming that you aren't actually being asked to explain 'why' these things happen, but merely 'how' things will happen.
If you are at rest relative to the car:
(1) time passes more slowly in the car than for you
(2) yardsticks in the car are shorter than yardsticks are for you
(3) two equal masses when both are at rest will be different if one of them is put in the car (that one will have greater mass than the one that you keep)
The weird thing is, for someone in the car, all of those effects are reversed. If I'm in the car:
(1) Time passes more slowly for people standing on the road
(2) Distances will compress (the road itself will get shorter)
(3) Masses on the roadside will increase
This dual effect is one of the hardest things to understand about relativity.
Last edited: | 720 | 3,116 | {"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-2022-05 | latest | en | 0.978004 |
https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_10&diff=next&oldid=54039 | 1,669,706,284,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710690.85/warc/CC-MAIN-20221129064123-20221129094123-00038.warc.gz | 139,808,998 | 11,083 | # Difference between revisions of "2006 AMC 12B Problems/Problem 10"
## Problem
In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? $\text {(A) } 43 \qquad \text {(B) } 44 \qquad \text {(C) } 45 \qquad \text {(D) } 46 \qquad \text {(E) } 47$
## Solution
If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: (Error compiling LaTeX. ! Missing \$ inserted.)\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5 Now, since we want the greatest perimeter, we want the greatest integer x, and if $x<7.5$ then $x=7$. Then, the first side has length $3*7=21$, the second side has length $7$, the third side has length $15$, and so the perimeter is $21+7+15=43 \Rightarrow \boxed{\text {(A)}}$.
## See also
2006 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 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 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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Login to AoPS | 418 | 1,329 | {"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": 7, "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-2022-49 | latest | en | 0.717145 |
http://www.compoundchem.com/2016/02/17/rate-of-reaction/ | 1,718,825,875,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861832.94/warc/CC-MAIN-20240619185738-20240619215738-00213.warc.gz | 39,797,193 | 28,732 | How different factors can affect how quickly a reaction happens is a common topic in the chemistry curriculum. This graphic serves as a convenient summary for teachers and students alike of what these different factors are, and how and why they affect the rate of a reaction. However, it’s not only of interest to those teaching or learning about chemistry; as we’ll see, knowledge of these factors can have some everyday applications too!
Before considering what can affect how quickly a reaction happens, we need to think about how a reaction happens in the first place. ‘Collision theory’ offers a simplified way of looking at this, based on ideas of particles. For a reaction to happen, particles of the reactants need to collide, and with enough energy to break bonds and trigger the reaction. They also have to collide in the correct orientation for the reaction to occur. If their orientation relative to one another is incorrect, or they don’t collide with enough energy, then no reaction occurs and the particles simply bounce off each other.
Within a given time period, a certain number of collisions between reactant particles will occur. A certain proportion of these collisions will have enough energy to exceed the reaction’s ‘activation energy’, the energy required for the reaction to take place, and a reaction will occur. By changing the conditions of the reaction, we can affect both the frequency of collisions between particles, and the proportion of collisions that are successful and result in a reaction.
The first thing we can change, for solutions (substances dissolved in a solvent, commonly water), is concentration. Concentration is a measure of how much of something is dissolved in a solution; a higher concentration means there is more of the particular substance in the solution. Having more of the substance means there are a greater number of particles, which increases the chance of a collision between reactant particles occurring. This increases the frequency of collisions, leading to an increased rate of reaction. Note that the proportion of successful collisions remains the same, as changing the concentration doesn’t affect the energy of the particles.
We can also increase the temperature of the reaction to speed it up. This works because increasing the temperature increases the kinetic energy of the reactant particles. Because they are moving around faster, the collisions between them are more frequent; additionally, as on average the particles have more energy at a higher temperature, a greater number of collisions will have the energy required to react, so the proportion of successful collisions increases. Both of these factors combined lead to an increased rate of reaction.
For solid reagents, we can increase their surface area. Using a reagent in a big, solid lump means that only the particles at the surface are exposed and available for reaction initially. By dividing the lump into smaller, finer pieces, we increase the number of particles available for reaction. This increases the frequency of collisions, and hence the rate. Again, the energy of the collisions is unaffected, so the proportion of successful collisions remains the same.
For gaseous reagents we can increase the pressure at which the reaction is being carried out. Put simply, squeezing the same amount of a gas into a smaller space is a method of increasing pressure. This forces the gas molecules closer together, so the collisions between their particles become more frequent, increasing the rate of the reaction. Increasing the pressure doesn’t affect the energy of the particles, so the proportion of successful collisions remains constant.
The final way we can affect the rate of a reaction is to use a catalyst. A catalyst is a chemical or biological agent that speeds up the rate of a reaction, without itself being used up in the process. It provides an alternative way for the reaction to occur, which has a lower activation energy. This means that less energy is required for two particles to collide and successfully react, so the proportion of successful collisions increases, increasing the speed with which the reaction occurs.
After all this you might be wondering why this would be of any relevance to anyone outside of learning chemistry. However, it’s much more relevant than it might initially seem. For one, the enzymes on which many of our bodily processes depends are catalysts, allowing reactions in our bodies to happen faster. Concentration and surface area, meanwhile, are both important in the context of medicine; drug doses are designed to lead to a specific concentration of the drug in our blood streams, a concentration at which the medicine is most effective, and medicines will often be powdered rather than in solid lumps to allow them to exert their effects more quickly.
Elsewhere, the kitchen also provides ample examples of how you can leverage knowledge of the factors affecting the rate of a reaction to your benefit. Always having problems with crying whilst chopping onions? Putting the onion in the fridge before cutting it lowers its temperature, slowing the rate of the reaction that produces the chemical that induces the tears. In fact, the reason we keep food in the fridge in the first place is of course to slow the rate of the spoilage of food.
There are occasionally other factors, in addition to those detailed here, that can affect the rate of a reaction. These include light intensity, the physical state of the reagents, and for some reactions the solvent in which the reaction is occurring. Because most pre-university chemistry courses here in the UK don’t discuss these in any great detail they aren’t included in this graphic, but they’re also worth being aware of.
Enjoyed this post & graphic? Consider supporting Compound Interest on Patreon, and get previews of upcoming posts & more!
• ###### crocodilechuck
Posted February 18, 2016 at 6:14 am 0Likes
Excellent. For future post[s], Andy, can you please drill down on catalysts-including mechanisms of action. Thanks!
• ###### Compound Interest
Posted February 18, 2016 at 7:13 am 0Likes
Certainly can! I think catalysts would definitely make for a good explanatory graphic.
• ###### Keith Simons
Posted February 21, 2016 at 5:34 pm 0Likes
Great summary. Just a comment on the – sign in front of % successful collisions. It gives the impression that there are fewer successful collisions. Thanks.
• ###### Salomon Martinez
Posted January 3, 2018 at 9:07 pm 0Likes
True that
• ###### Salomon Martinez
Posted January 3, 2018 at 9:04 pm 0Likes
hey | 1,305 | 6,615 | {"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-2024-26 | latest | en | 0.955061 |
https://jp.mathworks.com/help/antenna/ref/patternazimuth.html | 1,670,264,266,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711042.33/warc/CC-MAIN-20221205164659-20221205194659-00001.warc.gz | 359,738,000 | 20,638 | # patternAzimuth
Azimuth pattern of antenna or array
## Syntax
``patternAzimuth(object,frequency,elevation)``
``patternAzimuth(object,frequency,elevation,Name,Value)``
``directivity = patternAzimuth(object,frequency,elevation)``
``directivity = patternAzimuth(object,frequency,elevation,'Azimuth')``
## Description
example
````patternAzimuth(object,frequency,elevation)` plots the 2-D radiation pattern of the antenna or array object over a specified frequency. Elevation values defaults to zero if not specified.```
example
````patternAzimuth(object,frequency,elevation,Name,Value)` uses additional options specified by one or more `Name,Value` pair arguments.```
````directivity = patternAzimuth(object,frequency,elevation)` returns the directivity of the antenna or array object over a specified frequency. Elevation values defaults to zero if not specified.```
````directivity = patternAzimuth(object,frequency,elevation,'Azimuth')` uses additional options specified by one or more `Name,Value` pair arguments.```
## Examples
collapse all
Calculate and plot the azimuth radiation pattern of the helix antenna at 2 GHz.
```h = helix; patternAzimuth(h,2e9);```
Calculate and plot the azimuth radiation pattern of the dipole antenna at 70 MHz at elevation values of 0 and 45.
``` d = dipole; patternAzimuth(d,70e6,[0 45],'Azimuth',-140:5:140);```
## Input Arguments
collapse all
Antenna or array object, specified as a scalar.
Frequency used to calculate charge distribution, specified as a scalar in Hz.
Example: 70e6
Data Types: `double`
Elevation angle values, specified as a vector in degrees.
Example: `[0 45]`
Data Types: `double`
Azimuth angles of antenna, specified as the comma-separated pair consisting of `'Azimuth'` and a vector in degrees.
Example: `'Azimuth'`,2:2:340
Data Types: `double`
## Output Arguments
collapse all
Antenna or array directivity, returned as a matrix in `dBi`. The matrix size is the product of number of elevation values and number of azimuth values.
## Version History
Introduced in R2015a | 495 | 2,059 | {"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-2022-49 | latest | en | 0.65061 |
https://discuss.boardinfinity.com/t/logical-reasoning-section-of-cmat-2022/15874 | 1,695,848,043,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510326.82/warc/CC-MAIN-20230927203115-20230927233115-00488.warc.gz | 234,483,219 | 7,064 | # Logical Reasoning section of CMAT 2022
Key Points of CMAT2022 Logical Inference Section
• The reasoning section of the CMAT exam contains 25 questions in MCQ format.
• Each question is worth 4 points for a correct attempt and -1 point for an incorrect attempt.
• The difficulty of the logical reasoning section of CMAT is usually easy to moderate
• In this section of CMAT , about 22 of the 25 questions are considered appropriate trials.
Topics Covered by Candidates:
1. Coding decoding
2. Logic-based questions
3. Syllogism
4. Blood Relations
5. Visual Reasoning
6. Strong/Weak Argument
7. Series Completion
8. Number Grid
9. Analogy
10. Venn diagram
11. Directions
12. Odd One Out
Tips and Tricks
We must first comprehend the nature of logical thinking before we can decipher it. In this part, there are no formulas to remember or theorems to comprehend. Based on the information available, you must answer questions and improve your problem-solving skills.
• “Complete the series” is the most prevalent type of logical reasoning question. A succession of photographs will be displayed, and the contestant must choose which image from the possibilities provided completes the series.
• At first, CMAT logical problems may appear new and perplexing. Please take your time reading the question.
• Don’t look at the first conceivable solution; if you don’t comprehend the question, it’s pointless. It could be even more perplexing for you.
• For visual reasoning problems, examine each photo carefully for a similar theme. This will assist you in determining which of the four possibilities is the correct answer.
• Be wary of suggestive terms like “contains,” “only,” “other than,” and “if not” when answering questions. Take note of prefixes like non, un, and dis.
• Write down the specified date and important points as you read the reasoning question. When practicing example questions, make it a habit to do so.
• Take as many practice CMAT tests and sample work as you can. So that the CMAT core exam questions aren’t shocking, practice different questions from each inference topic.
• In the section on logical reasoning, time management is crucial. Each question on the CMAT takes roughly 1.8 minutes. If you don’t have enough practice understanding a riddle, a phrase, or a series of numbers, 1.8 minutes may be too short. As a result, time yourself while practicing the sample questions.
• Improve your logical thinking skills. To comprehend the sequence of events, read the material and break it down into logical flows or patterns.
• Improve your logical thinking by doing brain teasers like Sudoku and crossword puzzles. It’s not only a fantastic brain exercise, but it’s also a good approach to study for the exam. | 589 | 2,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2023-40 | latest | en | 0.889902 |
http://www.evi.com/q/%C2%A314,000_to_dollars | 1,419,278,016,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802776556.43/warc/CC-MAIN-20141217075256-00014-ip-10-231-17-201.ec2.internal.warc.gz | 499,681,666 | 5,594 | # £14,000 to dollars
• £14000 is \$21866.
• tk10npubl tk10ncanl
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• 14000 pounds is how many us dollars | 938 | 3,170 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2014-52 | latest | en | 0.860466 |
https://dsp.stackexchange.com/questions/68522/duality-in-the-discrete-time-fourier-series | 1,718,267,999,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.74/warc/CC-MAIN-20240613060639-20240613090639-00101.warc.gz | 200,906,126 | 40,728 | # Duality in the discrete-time Fourier series
Suppose $$g[n]$$ is periodic with fundamental period $$N$$ and $$f[k]$$ being its Fourier coefficients i.e. $$f[k] = \frac{1}{N}\sum_{n=}g[n]e^{-j\frac{2\pi}{N}nk}$$ In more convenient notation $$g[n]\leftrightarrow f[k]$$ It's easy to prove $$f[n]\leftrightarrow \frac{1}{N}g[-k]$$ This is called "Duality in the discrete-time Fourier series". My problem is applying duality in the case of $$x[n-n_0] \leftrightarrow a_ke^{-jk(\frac{2\pi}{N})n_0}$$ And $$\sum_{r=}x[r]y[n-r]\leftrightarrow Na_kb_k$$ I think for the first case duality should gives$$x[n]e^{jk(\frac{2\pi}{N})m} \leftrightarrow \frac{1}{N}a_{m-k} \tag{1}\label{*}$$ and the second gives $$x[n]y[n] \leftrightarrow \frac{1}{N^2} \sum_{l = }a_lb_{-k-l} \tag{2}\label{**}$$ Edit: Let $$g[n] = \sum_{r=}x[r]y[n-r]$$ and $$f[k] = Na_kb_k$$. So according to the mentioned property $$Nx[n]y[n] \leftrightarrow \frac{1}{N}\sum_{r=}a_rb_{-k-r}$$
Dividing by $$N$$ leads to $$\eqref{**}$$. Similarly if $$g[n] = x[n-n_0]$$ and $$f[k] = a_ke^{-jk(\frac{2\pi}{N})n_0}$$ I think then $$x[n]e^{-jn(\frac{2\pi}{N})n_0}\leftrightarrow a_{-k - n_0}$$ Let $$m = -n_0$$ and we have \eqref{*}.
Obviously these results are wrong but I don't know what's my mistake.
This is taken from Signals and systems by Alan V. Oppenheim:
• But we don't know how you came up with these results, so we can't know what your mistake is. Commented Jun 22, 2020 at 9:03
• @MattL. You're right, sorry for that. I will add the details. Commented Jun 22, 2020 at 9:05
• @MattL. Please see the edit. Commented Jun 22, 2020 at 9:32
Based on the references and the definitions given, here is a proof.
Given: $$x[n] \leftrightarrow a[k], \ and$$ $$y[n] \leftrightarrow b[k]$$
Fourier Series expansion of $$x[n]e^{j\frac{2\pi}{N}nm}$$ would be derived as follows: $$\frac{1}{N}\sum^{N-1}_{n=0}x[n]e^{j\frac{2\pi}{N}nm}e^{-j\frac{2\pi}{N}nk} = \frac{1}{N}\sum^{N-1}_{n=0}x[n]e^{-j\frac{2\pi}{N}(k-m)n} = a[k-m]$$ The factor $$\frac{1}{N}$$ gets consumed in the definition of Fourier Series coefficients. Therefore, the following relationship can be stated: $$x[n]e^{j\frac{2\pi}{N}nm} \leftrightarrow a[k-m]$$
Similarly, for the second case of multiplication in time-domain, you need to figure out that the inverse Fourier Series relation or the synthesis expression will be the following: $$x[n] = \sum^{N-1}_{k=0}a[k]e^{j\frac{2\pi}{N}kn}$$ I am not adding the proof of the above which can be easily derived by plugging the expression for $$a[k] = \frac{1}{N}\sum^{N-1}_{n=0}x[n]e^{-j\frac{2\pi}{N}kn}$$ on the RHS.
Then all we are required to do is proving that the inverse Fourier Transform of convolution of $$a[k]$$ and $$b[k]$$ would be product $$x[n]y[n]$$. Which is as follows: $$\mathcal{FS}^{-1}(a*b)[k] = \sum^{N-1}_{k=0}(a*b)[k]e^{j\frac{2\pi}{N}nk} = \sum^{N-1}_{k=0}( \sum^{N-1}_{l=0}a[l]b[k-l]) e^{j\frac{2\pi}{N}nk}$$ $$=>\mathcal{FS}^{-1}(a*b)[k] = (\sum^{N-1}_{l=0}a[l]e^{j\frac{2\pi}{N}ln})(\sum^{N-1}_{k=0}b[k-l]e^{j\frac{2\pi}{N}(k-l)n})$$ $$=> \mathcal{FS}^{-1}(a*b)[k] = \mathcal{FS}^{-1}a[k] \mathcal{FS}^{-1}b[k] = x[n]y[n]$$ I have replaced $$m = (k-l)$$ in the proof above to get $$y[n]$$.
Hence both the equations (5.69) and (5.71) are true and correctly given in the book.
I think you are trying to use equation (5.67) to prove equation (5.69) and (5.71). That is not correct, because (5.67) is stating Duality of Fourier Series expression between Continuous Time and Discrete Time. Equation (5.69) and equation (5.71) are duality of Shift-Property and Convolution Property. The second reason you cannot use Duality Expression (5.67) is because (5.67) states that if $$\mathcal{FS}g[n] = f[k]$$, then $$\mathcal{FS}f[n] = \frac{1}{N}g[−k]$$. That is it takes Fourier Series of the Fourier Series itself assuming $$f[n]$$ in time-domain. But in (5.69) and (5.71), the time-domain sequences remain $$x[n]$$ and $$y[n]$$ only, so (5.67) does not apply here. There are 3 Duality Properties stated in equation (5.67), (5.69) and (5.71). They are not the consequence of one another. They are a consequence of similarity in definition of Continuous time Fourier Series and Discrete Time Fourier Series.
• @S.H.W Why do you think that the DFT of $\sum x[m]y[n-m]$ is $Na_k b_k$? Where are you getting the factor $N$ from? Commented Jun 22, 2020 at 15:25
• @S.H.W DFS is just periodization of DFT. So, can you add a little bit of of detail on the factor $N$ in the convolution property? I am not sure that the DFS will have that scaling of $N$. It should just be $a_k b_k$. Commented Jun 22, 2020 at 15:34 | 1,623 | 4,603 | {"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": 40, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-26 | latest | en | 0.746108 |
https://www.metric-conversions.org/volume/us-liquid-gallons-to-us-federal-barrels.htm | 1,713,665,378,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817699.6/warc/CC-MAIN-20240421005612-20240421035612-00555.warc.gz | 811,310,144 | 13,187 | # US Gallons (Liquid) to US Barrels (Federal)
US Barrels (Federal) to US Gallons (Liquid) (Swap Units)
Format
Accuracy
Note: Fractional results are rounded to the nearest 1/64. For a more accurate answer please select 'decimal' from the options above the result.
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Show formula
## US Gallons (Liquid) to US Barrels (Federal) formula
US bbl fed =
US gal lqd * 0.032258
Show working
Show result in exponential format
## US Gallons (Liquid)
A US capacity measure (for liquid) equal to 4 quarts or 3.785 liters. Note also there are different measures of US dry gallons and UK gallons.
## US Gallons (Liquid) to US Barrels (Federal) formula
US bbl fed =
US gal lqd * 0.032258
## US Barrels (Federal)
US measurement of volume for dry materials. See also US liquid barrels, US dry barrels, US oil barrels and UK barrels.
## US Gallons (Liquid) to US Barrels (Federal) table
Start
Increments
Accuracy
Format
Print table
< Smaller Values Larger Values >
US Gallons (Liquid) US Barrels (Federal)
0US gal lqd 0.00US bbl fed
1US gal lqd 0.03US bbl fed
2US gal lqd 0.06US bbl fed
3US gal lqd 0.10US bbl fed
4US gal lqd 0.13US bbl fed
5US gal lqd 0.16US bbl fed
6US gal lqd 0.19US bbl fed
7US gal lqd 0.23US bbl fed
8US gal lqd 0.26US bbl fed
9US gal lqd 0.29US bbl fed
10US gal lqd 0.32US bbl fed
11US gal lqd 0.35US bbl fed
12US gal lqd 0.39US bbl fed
13US gal lqd 0.42US bbl fed
14US gal lqd 0.45US bbl fed
15US gal lqd 0.48US bbl fed
16US gal lqd 0.52US bbl fed
17US gal lqd 0.55US bbl fed
18US gal lqd 0.58US bbl fed
19US gal lqd 0.61US bbl fed
US Gallons (Liquid) US Barrels (Federal)
20US gal lqd 0.65US bbl fed
21US gal lqd 0.68US bbl fed
22US gal lqd 0.71US bbl fed
23US gal lqd 0.74US bbl fed
24US gal lqd 0.77US bbl fed
25US gal lqd 0.81US bbl fed
26US gal lqd 0.84US bbl fed
27US gal lqd 0.87US bbl fed
28US gal lqd 0.90US bbl fed
29US gal lqd 0.94US bbl fed
30US gal lqd 0.97US bbl fed
31US gal lqd 1.00US bbl fed
32US gal lqd 1.03US bbl fed
33US gal lqd 1.06US bbl fed
34US gal lqd 1.10US bbl fed
35US gal lqd 1.13US bbl fed
36US gal lqd 1.16US bbl fed
37US gal lqd 1.19US bbl fed
38US gal lqd 1.23US bbl fed
39US gal lqd 1.26US bbl fed
US Gallons (Liquid) US Barrels (Federal)
40US gal lqd 1.29US bbl fed
41US gal lqd 1.32US bbl fed
42US gal lqd 1.35US bbl fed
43US gal lqd 1.39US bbl fed
44US gal lqd 1.42US bbl fed
45US gal lqd 1.45US bbl fed
46US gal lqd 1.48US bbl fed
47US gal lqd 1.52US bbl fed
48US gal lqd 1.55US bbl fed
49US gal lqd 1.58US bbl fed
50US gal lqd 1.61US bbl fed
51US gal lqd 1.65US bbl fed
52US gal lqd 1.68US bbl fed
53US gal lqd 1.71US bbl fed
54US gal lqd 1.74US bbl fed
55US gal lqd 1.77US bbl fed
56US gal lqd 1.81US bbl fed
57US gal lqd 1.84US bbl fed
58US gal lqd 1.87US bbl fed
59US gal lqd 1.90US bbl fed | 1,174 | 2,997 | {"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-2024-18 | latest | en | 0.487756 |
http://www.neverendingbooks.org/looking-for-f_un/ | 1,720,919,878,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00669.warc.gz | 51,461,114 | 11,709 | Skip to content →
There are only a handful of human activities where one goes to extraordinary lengths to keep a dream alive, in spite of overwhelming evidence : religion, theoretical physics, supporting the Belgian football team and … mathematics.
In recent years several people spend a lot of energy looking for properties of an elusive object : the field with one element $\mathbb{F}_1$, or in French : “F-un”. The topic must have reached a level of maturity as there was a conference dedicated entirely to it : NONCOMMUTATIVE GEOMETRY AND GEOMETRY OVER THE FIELD WITH ONE ELEMENT.
In this series I’d like to find out what the fuss is all about, why people would like it to exist and what it has to do with noncommutative geometry. However, before we start two remarks :
The field $\mathbb{F}_1$ does not exist, so don’t try to make sense of sentences such as “The ‘field with one element’ is the free algebraic monad generated by one constant (p.26), or the universal generalized ring with zero (p.33)” in the wikipedia-entry. The simplest proof is that in any (unitary) ring we have $0 \not= 1$ so any ring must contain at least two elements. A more highbrow version : the ring of integers $\mathbb{Z}$ is the initial object in the category of unitary rings, so it cannot be an algebra over anything else.
The second remark is that several people have already written blog-posts about $\mathbb{F}_1$. Here are a few I know of : David Corfield at the n-category cafe and at his old blog, Noah Snyder at the secret blogging seminar, Kea at the Arcadian functor, AC and K. Consani at Noncommutative geometry and John Baez wrote about it in his weekly finds.
The dream we like to keep alive is that we will prove the Riemann hypothesis one fine day by lifting Weil’s proof of it in the case of curves over finite fields to rings of integers.
Even if you don’t know a word about Weil’s method, if you think about it for a couple of minutes, there are two immediate formidable problems with this strategy.
For most people this would be evidence enough to discard the approach, but, we mathematicians have found extremely clever ways for going into denial.
The first problem is that if we want to think of $\mathbf{spec}(\mathbb{Z})$ (or rather its completion adding the infinite place) as a curve over some field, then $\mathbb{Z}$ must be an algebra over this field. However, no such field can exist…
No problem! If there is no such field, let us invent one, and call it $\mathbb{F}_1$. But, it is a bit hard to do geometry over an illusory field. Christophe Soule succeeded in defining varieties over $\mathbb{F}_1$ in a talk at the 1999 Arbeitstagung and in a more recent write-up of it : Les varietes sur le corps a un element.
We will come back to this in more detail later, but for now, here’s the main idea. Consider an existent field $k$ and an algebra $k \rightarrow R$ over it. Now study the properties of the functor (extension of scalars) from $k$-schemes to $R$-schemes. Even if there is no morphism $\mathbb{F}_1 \rightarrow \mathbb{Z}$, let us assume it exists and define $\mathbb{F}_1$-varieties by requiring that these guys should satisfy the properties found before for extension of scalars on schemes defined over a field by going to schemes over an algebra (in this case, $\mathbb{Z}$-schemes). Roughly speaking this defines $\mathbb{F}_1$-schemes as subsets of points of suitable $\mathbb{Z}$-schemes.
But, this is just one half of the story. He adds to such an $\mathbb{F}_1$-variety extra topological data ‘at infinity’, an idea he attributes to J.-B. Bost. This added feature is a $\mathbb{C}$-algebra $\mathcal{A}_X$, which does not necessarily have to be commutative. He only writes : “Par ignorance, nous resterons tres evasifs sur les proprietes requises sur cette $\mathbb{C}$-algebre.”
The algebra $\mathcal{A}_X$ originates from trying to bypass the second major obstacle with the Weil-Riemann-strategy. On a smooth projective curve all points look similar as is clear for example by noting that the completions of all local rings are isomorphic to the formal power series $k[[x]]$ over the basefield, in particular there is no distinction between ‘finite’ points and those lying at ‘infinity’.
The completions of the local rings of points in $\mathbf{spec}(\mathbb{Z})$ on the other hand are completely different, for example, they have residue fields of different characteristics… Still, local class field theory asserts that their quotient fields have several common features. For example, their Brauer groups are all isomorphic to $\mathbb{Q}/\mathbb{Z}$. However, as $Br(\mathbb{R}) = \mathbb{Z}/2\mathbb{Z}$ and $Br(\mathbb{C}) = 0$, even then there would be a clear distinction between the finite primes and the place at infinity…
Alain Connes came up with an extremely elegant solution to bypass this problem in Noncommutative geometry and the Riemann zeta function. He proposes to replace finite dimensional central simple algebras in the definition of the Brauer group by AF (for Approximately Finite dimensional)-central simple algebras over $\mathbb{C}$. This is the origin and the importance of the Bost-Connes algebra.
We will come back to most of this in more detail later, but for the impatient, Connes has written a paper together with Caterina Consani and Matilde Marcolli Fun with $\mathbb{F}_1$ relating the Bost-Connes algebra to the field with one element.
Published in absolute
## 6 Comments
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,346 | 5,544 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-30 | latest | en | 0.948244 |
https://math.stackexchange.com/questions/110071/confused-about-optional-stopping-theorem-martingales | 1,718,694,276,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.46/warc/CC-MAIN-20240618042809-20240618072809-00800.warc.gz | 339,519,003 | 35,259 | # confused about optional stopping theorem (martingales)
If $X_n$ is a martingale and $T$ is a stopping time with $P(T \leq n) = 1$ for some $n$, then the optional stopping theorem applies: $\mathbb{E}(X_T) = \mathbb{E}(X_0)$. Sometimes it's hard to show that the stopping time $T$ is bounded a.s, so instead I believe you can show that $|X_{\min(T,n)}| \leq k$ for some $k$. Is this enough to get the conclusion of the OST (that the expectations at time $T$ and $0$ are equal)? I know that in this case the DCT implies that $EX_{\min(T,n)}$ exists, but so what?
(I have looked everywhere for such a definition but I can't find it)
• You should be able to apply dominated convergence to $\mathbf{E} [X_{\min(T,n)}-X_0]$. Commented Feb 16, 2012 at 18:07
• So DCT says that $\mathbb{E}[X_{\min(T,n)} - X_0] \to \mathbb{E}[X_T - X_0].$ Still don't see where to go.. Commented Feb 16, 2012 at 18:25
• $X_{\min(T,n)}$ is a martingale, so $\mathbf{E}[X_{\min(T,n)}-X_0]=0$. Commented Feb 16, 2012 at 18:26
• Could you be more specific about everywhere?
– Did
Commented Feb 16, 2012 at 18:39
• Durrett, lots of Googling online, etc etc etc. Commented Feb 16, 2012 at 19:39
## 1 Answer
Since $X_{\min(T,n)}$ is a martingale (see for instance the book of David Williams, "Probability with Martingales"), we have $\mathbf{E}[X_{\min(T,n)}-X_0]=0$ for all $n$. If your condition holds, we can use dominated convergence to take the limit as $n \rightarrow \infty$ and conclude $\mathbf{E}[X_T-X_0]=0$. | 511 | 1,494 | {"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} | 3.3125 | 3 | CC-MAIN-2024-26 | latest | en | 0.854769 |
https://ahmerov.com/book_844_chapter_73_7.2.4_RELATIONSHIP_DIAGRAMS.html | 1,659,920,987,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570741.21/warc/CC-MAIN-20220808001418-20220808031418-00662.warc.gz | 116,717,588 | 9,515 |
# 7.2.4 RELATIONSHIP DIAGRAMS
К оглавлению
The next step is to evaluate and classify the data. One of the best tools to
accomplish this task is a Relationship Diagram. The Relationship Diagram was
reportedly invented by P. Chen and presented in his article in 1976.3 The purpose
of the Relationship Diagram is to show the interrelationships between
causative factors that relate to a problem.
A Relationship Diagram, as shown in Figure 7-4, is one that shows connec-
F i g u r e 7 - 4 — R e l a t i o n s h i p D i a g r a m
tions or relationships between the elements of the diagram. In this case, the
elements are Issues.
The Relationship Diagram, in contrast to the Affinity Diagram, which only
shows logical groupings, helps map the logical relationships between the related
items uncovered in the Affinity Diagram. The Relationship Diagram shows
cause and effect relationships among many key elements. It can be used to
identify the causes of problems or to work backward from a desired outcome
to identify all of the causal factors that would need to exist to ensure the
achievement of an outcome. The Relationship Diagram doesn’t necessarily need
to follow the form of the ‘‘bubble’’ chart shown. A traditional organization
chart and a flow diagram are examples of other presentations of Relationship
Diagrams.
The process to be used is as follows:
1. State the problem or family under discussion—software defects, customer
retention, process steps, whatever.
2. Capture that problem or issue in a box, bubble, or whatever.
3. Begin a process of looking for ‘‘drivers’’; that is, functions or issues that
drive the issue being considered. You can also use the rationale of the
PERT Chart (see glossary) in considering ‘‘predecessors’’ for this part of
the process. In other words, you are looking for items that drive or must
be completed before the issue at hand.
4. Begin a process of looking for ‘‘drivens,’’ that is, functions or issues that
are being driven by the issue at hand. If you prefer, use the term ‘‘predecessors’’
for ‘‘drivers’’ and ‘‘successors’’ for ‘‘drivens.’’
5. When you have diagrammed the issue and have located all the ‘‘drivers’’
and ‘‘drivens’’ something will jump out at you. That bubble or square
that has ‘‘drivers’’ but no ‘‘drivens’’ is the primary issue whether that’s
the one you started with or not!
If you want or need to go beyond the simple relationships of one issue driving
another, consider the following:
The bubble is the issue.
The lines (arrows) connecting the bubbles are the actions.
The value of the line (arrow) is the magnitude.
The characteristics of the bubble are its attributes.
A veritable glut of information exists on Relationship Diagrams. Much of it
is Entity-Relationship Diagrams as a result of our software society. The source
of a lot of the information is the use and application of relational databases
such as Microsoft’s Access. Most written information regarding Relationship
Diagrams is in the form of articles rather than books.
Software used to support Relationship Diagrams is listed in Table 7-5.
T a b l e 7 - 5 — R e l a t i o n s h i p D i a g r a m S o f t w a r e
Tool Product Vendor
Relationship Diagrams
‘‘EDGE Programmer’’ Pacestar Software
‘‘SmartDraw’’ SmartDraw.com
Following is contact information for the companies listed in Table 7-5:
Pacestar Software
P.O. Box 51974
Phoenix, AZ 85076-1974
Phone: 480-893-3046
Fax: 413-480-0645
Web site: www.pacestar.com
E-mail: mail@pacestar.com
TEAMFLY
SmartDraw.com
10085 Carroll Canyon Road, Suite 220
San Diego, CA 92131
Phone: 858-549-0314
Order: 800-501-0314
Fax: 858-549-2830
Web site: www.smartdraw.com
E-mail: mail@smartdraw.com | 927 | 3,741 | {"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-2022-33 | latest | en | 0.926268 |
https://www.jiskha.com/display.cgi?id=1493535374 | 1,511,226,141,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806309.83/warc/CC-MAIN-20171121002016-20171121022016-00109.warc.gz | 803,231,320 | 3,957 | # mathvariation 2(steve,reiny,reed) pls I need help!
posted by .
the force E needed to make a machine pull a load is partly constant and partly varies as the load to be pulled itself.when the load is 20g,the force needed is 1.4n,where as the force needed for 30g load is 2N.
Find the:
A)law connecting the load and the force
B)force for the load of 50g?
• mathvariation 2(steve,reiny,reed) pls I need help! -
E = a+bL
1.4 = a+0.02b
2 = a+0.03b
subtract, and you have
0.6 = 0.01b
b = 60
so a = 0.2
E = 0.2 + 60L
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https://www.bgsu.edu/college-credit-plus/partnerships.html | 1,618,803,371,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038863420.65/warc/CC-MAIN-20210419015157-20210419045157-00602.warc.gz | 739,490,510 | 61,157 | # Partner Schools
We provide school districts with the opportunity to visit our campus and witness first-hand the BG experience while they begin to make meaningful connections. Our College Credit Plus partner sites are reimbursed for transportation costs relating to campus visits and we provide lunch in one of most popular on-campus dining facilities. We want students to feel welcome and part of the BGSU family. We want students to experience what it means to be a college student.
### 2020-2021 Partner Schools and Course Offerings
Various thematic topics. Introduction to literary and textual study with attention to various forms of fiction, nonfiction, drama, poetry, and to essential literary terminology and practice. Extensive expository writing.
Grammar review; development of the four skills. Three class periods and laboratory practice each week.
A communicative approach to intermediate language using the four skill areas of listening, speaking, reading, and writing, along with French and Francophone culture.
Description of data, binomial and normal distributions, estimation and testing hypotheses for means and proportions.
Prerequisites: Two years high school algebra, one year of geometry and a satisfactory placement exam score.
Review of functions and their graphs, linear and quadratic functions, factoring. Polynomial and rational functions. Review of exponents. Exponential and logarithmic functions and their graphs. Systems of equations, theory of equations.
[Prerequisites: Two years of high school algebra, one year of geometry and a satisfactory placement exam score, or grade of C or higher in MATH 99, MATH 1210, or grade of D in MATH 1200.]
Basic algebra; inequalities; functions and graphs; logarithmic and exponential functions; trigonometric functions and identities; applications and other topics.
(Prerequisites: Two years of high school algebra and one of geometry AND a satisfactory placement exam score, or grade of C or higher in MATH 1200 or MATH 1220.)
Limits, the derivative, differentiation techniques and applications of the derivative. MATH 1340 and MATH 1350 is a two-semester sequence which includes all the topics from MATH 1310. Not open to students with a grade of C or higher in MATH 1310 or MATH 1260. Prerequisites: same as MATH 1310.
Prerequisites: (1) two years of high school algebra, one year of geometry, one-half year of trigonometry, ACT math score of 24 or higher and satisfactory score on department placement test; or (2) grade of C or higher in MATH 1280, MATH 1290 or MATH 1300.
The definite integral; the fundamental theorem; indefinite integrals; integration by parts, by substitution and using tables; and applications of definite and indefinite integrals.
(Prerequisite: grade of C or higher in MATH 1340.)
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Grammar review; development of the four skills.
Grammar review; development of the four skills.
Constitutional basis and development, political processes (parties, nominations and elections, interest groups and public opinion), federalism and institutions of national government.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Builds on foundational understandings of academic reading and writing with a focus on inquiry-based writing. By engaging a range of writing tasks, both informal and formal, students pursue person- and library-based research writing that has meaning to them personally. Students also continue to build confidence as readers, writers, and critical thinkers, adding their voices to ongoing conversations. Using a workshop approach, students practice strategies for representing, through reflective writing, their research and composing processes to a range of audiences. ePortfolio based. Placement through UWP online pre-screening or prior credit for WRIT 1110.
Constitutional basis and development, political processes (parties, nominations and elections, interest groups and public opinion), federalism and institutions of national government.
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Builds on foundational understandings of academic reading and writing with a focus on inquiry-based writing. By engaging a range of writing tasks, both informal and formal, students pursue person- and library-based research writing that has meaning to them personally. Students also continue to build confidence as readers, writers, and critical thinkers, adding their voices to ongoing conversations. Using a workshop approach, students practice strategies for representing, through reflective writing, their research and composing processes to a range of audiences. ePortfolio based. Placement through UWP online pre-screening or prior credit for WRIT 1110.
Grammar review; development of the four skills. Three class periods and laboratory practice each week.
A communicative approach to intermediate language using the four skill areas of listening, speaking, reading, and writing, along with French and Francophone culture.
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Builds on foundational understandings of academic reading and writing with a focus on inquiry-based writing. By engaging a range of writing tasks, both informal and formal, students pursue person- and library-based research writing that has meaning to them personally. Students also continue to build confidence as readers, writers, and critical thinkers, adding their voices to ongoing conversations. Using a workshop approach, students practice strategies for representing, through reflective writing, their research and composing processes to a range of audiences. ePortfolio based. Placement through UWP online pre-screening or prior credit for WRIT 1110.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Grammar review; development of the four skills
Grammar review; development of the four skills
Extensive practice in speaking and writing German.
Constitutional basis and development, political processes (parties, nominations and elections, interest groups and public opinion), federalism and institutions of national government.
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Builds on foundational understandings of academic reading and writing with a focus on inquiry-based writing. By engaging a range of writing tasks, both informal and formal, students pursue person- and library-based research writing that has meaning to them personally. Students also continue to build confidence as readers, writers, and critical thinkers, adding their voices to ongoing conversations. Using a workshop approach, students practice strategies for representing, through reflective writing, their research and composing processes to a range of audiences. ePortfolio based. Placement through UWP online pre-screening or prior credit for WRIT 1110.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Basic ecology and current environmental problems of air, water and land pollution; human reproduction and population dynamics.
Basic concepts: the cell, metabolism, genetics, reproduction, development, evolution, ecology. Three one-hour lectures, one two-hour laboratory.
Personal financial management; budgeting, borrowing sources and costs; auto, property, and life insurance; home ownership financing; personal investment strategy; and retirement planning.
Basic algebra; inequalities; functions and graphs; logarithmic and exponential functions; trigonometric functions and identities; applications and other topics.
(Prerequisites: Two years of high school algebra and one of geometry AND a satisfactory placement exam score, or grade of C or higher in MATH 1200 or MATH 1220.)
Helps students examine their skills, interests, values and personal characteristics; investigate occupations and career paths; examine the interrelationship between self-knowledge and occupational decisions; identify academic programs and experiential learning opportunities that enhance future employment options; make informed career and life decisions; and establish realistic goals and an action plan.
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Description of data, binomial and normal distributions, estimation and testing hypotheses for means and proportions.
Prerequisites: Two years high school algebra, one year of geometry and a satisfactory placement exam score.
Review of functions and their graphs, linear and quadratic functions, factoring. Polynomial and rational functions. Review of exponents. Exponential and logarithmic functions and their graphs. Systems of equations, theory of equations.
[Prerequisites: Two years of high school algebra, one year of geometry and a satisfactory placement exam score, or grade of C or higher in MATH 99, MATH 1210, or grade of D in MATH 1200.]
Limits, the derivative, differentiation techniques and applications of the derivative. MATH 1340 and MATH 1350 is a two-semester sequence which includes all the topics from MATH 1310. Not open to students with a grade of C or higher in MATH 1310 or MATH 1260. Prerequisites: same as MATH 1310.
Prerequisites: (1) two years of high school algebra, one year of geometry, one-half year of trigonometry, ACT math score of 24 or higher and satisfactory score on department placement test; or (2) grade of C or higher in MATH 1280, MATH 1290 or MATH 1300.
The definite integral; the fundamental theorem; indefinite integrals; integration by parts, by substitution and using tables; and applications of definite and indefinite integrals.
(Prerequisite: grade of C or higher in MATH 1340.)
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Historical and aesthetic components of art with laboratory or online experiences with basic elements of creative expression.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Constitutional basis and development, political processes (parties, nominations and elections, interest groups and public opinion), federalism and institutions of national government.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Basic ecology and current environmental problems of air, water and land pollution; human reproduction and population dynamics.
Basic concepts: the cell, metabolism, genetics, reproduction, development, evolution, ecology. Three one-hour lectures, one two-hour laboratory.
Introductory Python programming for problem solving and algorithm development. Learn about basic programming topics including data types, control structures, file operations, arrays, functions, programming style, testing and debugging strategies. Does not apply to the computer science major.
Prerequisite: two years of high school algebra or co-requisite of MATH 99 or higher.
Alternative economic goals; economic growth, full employment, price stability, fair income distribution, economic security, economic freedom, consumer sovereignty, efficiency.
Grammar review; development of the four skills.
Grammar review; development of the four skills.
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Comparative study of how and why selected economic, social, political and intellectual revolutions of the modern world have transformed and are shaping contemporary European and non-Western cultures.
Scientific study of social structure, interaction, and institutions. Topics include gender, race, class, family, culture, and crime. [
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Builds on foundational understandings of academic reading and writing with a focus on inquiry-based writing. By engaging a range of writing tasks, both informal and formal, students pursue person- and library-based research writing that has meaning to them personally. Students also continue to build confidence as readers, writers, and critical thinkers, adding their voices to ongoing conversations. Using a workshop approach, students practice strategies for representing, through reflective writing, their research and composing processes to a range of audiences. ePortfolio based. Placement through UWP online pre-screening or prior credit for WRIT 1110.
Computer technology and related social issues. Hardware, software, applications in diverse areas. Problems concerning computerized services, data banks, governmental controls. Problem solving using software packages (such as hypertext, spreadsheets, word processing, database, presentation graphics, etc.).
(Prerequisite: one year of high school algebra or MATH 95 or higher.)
Basic algebra; inequalities; functions and graphs; logarithmic and exponential functions; trigonometric functions and identities; applications and other topics.
(Prerequisites: Two years of high school algebra and one of geometry AND a satisfactory placement exam score, or grade of C or higher in MATH 1200 or MATH 1220.)
Scientific study of social structure, interaction, and institutions. Topics include gender, race, class, family, culture, and crime. [
Helps students examine their skills, interests, values and personal characteristics; investigate occupations and career paths; examine the interrelationship between self-knowledge and occupational decisions; identify academic programs and experiential learning opportunities that enhance future employment options; make informed career and life decisions; and establish realistic goals and an action plan.
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Basic concepts: the cell, metabolism, genetics, reproduction, development, evolution, ecology. Three one-hour lectures, one two-hour laboratory.
The course addresses effectively identifying, locating, evaluating, designing, preparing and efficiently using educational technology as an instructional resource in the classroom as related to principles of learning and teaching. Candidates will develop increased classroom communication abilities through lectures, discussions, modeling, laboratory experiences and completion of a comprehensive project.
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Builds on foundational understandings of academic reading and writing with a focus on inquiry-based writing. By engaging a range of writing tasks, both informal and formal, students pursue person- and library-based research writing that has meaning to them personally. Students also continue to build confidence as readers, writers, and critical thinkers, adding their voices to ongoing conversations. Using a workshop approach, students practice strategies for representing, through reflective writing, their research and composing processes to a range of audiences. ePortfolio based. Placement through UWP online pre-screening or prior credit for WRIT 1110.
A communicative approach to intermediate language using the four skill areas of listening, speaking, reading, and writing, along with French and Francophone culture.
Comparative study of how and why economic, social, political and intellectual factors shaped and defined the history of selected Western and non-Western civilizations in the ancient and medieval periods.
First term of an introductory physics sequence using algebra and trigonometry, but not calculus. Topics include motion, forces, energy, fluids, heat and simple harmonic motion.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Builds on foundational understandings of academic reading and writing with a focus on inquiry-based writing. By engaging a range of writing tasks, both informal and formal, students pursue person- and library-based research writing that has meaning to them personally. Students also continue to build confidence as readers, writers, and critical thinkers, adding their voices to ongoing conversations. Using a workshop approach, students practice strategies for representing, through reflective writing, their research and composing processes to a range of audiences. ePortfolio based. Placement through UWP online pre-screening or prior credit for WRIT 1110.
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Builds on foundational understandings of academic reading and writing with a focus on inquiry-based writing. By engaging a range of writing tasks, both informal and formal, students pursue person- and library-based research writing that has meaning to them personally. Students also continue to build confidence as readers, writers, and critical thinkers, adding their voices to ongoing conversations. Using a workshop approach, students practice strategies for representing, through reflective writing, their research and composing processes to a range of audiences. ePortfolio based. Placement through UWP online pre-screening or prior credit for WRIT 1110.
Basic concepts: the cell, metabolism, genetics, reproduction, development, evolution, ecology. Three one-hour lectures, one two-hour laboratory.
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Builds on foundational understandings of academic reading and writing with a focus on inquiry-based writing. By engaging a range of writing tasks, both informal and formal, students pursue person- and library-based research writing that has meaning to them personally. Students also continue to build confidence as readers, writers, and critical thinkers, adding their voices to ongoing conversations. Using a workshop approach, students practice strategies for representing, through reflective writing, their research and composing processes to a range of audiences. ePortfolio based. Placement through UWP online pre-screening or prior credit for WRIT 1110.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Builds on foundational understandings of academic reading and writing with a focus on inquiry-based writing. By engaging a range of writing tasks, both informal and formal, students pursue person- and library-based research writing that has meaning to them personally. Students also continue to build confidence as readers, writers, and critical thinkers, adding their voices to ongoing conversations. Using a workshop approach, students practice strategies for representing, through reflective writing, their research and composing processes to a range of audiences. ePortfolio based. Placement through UWP online pre-screening or prior credit for WRIT 1110.
Introduction to educational foundation topics and contemporary issues for prospective teachers. Explore and analyze the teaching profession in a weekly seminar and a service learning experience. Required as an entry-year experience for all teacher-education candidates.
The course addresses effectively identifying, locating, evaluating, designing, preparing and efficiently using educational technology as an instructional resource in the classroom as related to principles of learning and teaching. Candidates will develop increased classroom communication abilities through lectures, discussions, modeling, laboratory experiences and completion of a comprehensive project.
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Provides a theoretical and practical foundation for college writers and lays important groundwork for future academic reading and writing experiences. This workshop-based course explores diverse intellectual practices associated with effective writing, including analyzing and producing genres, investigating individual writing processes, and reflecting on one's learning with an eye toward transferring writing knowledge to new situations. Students explore and experience how writing works in worlds they inhabit by composing digital, visual, and narrative expository arguments.
Builds on foundational understandings of academic reading and writing with a focus on inquiry-based writing. By engaging a range of writing tasks, both informal and formal, students pursue person- and library-based research writing that has meaning to them personally. Students also continue to build confidence as readers, writers, and critical thinkers, adding their voices to ongoing conversations. Using a workshop approach, students practice strategies for representing, through reflective writing, their research and composing processes to a range of audiences. ePortfolio based. Placement through UWP online pre-screening or prior credit for WRIT 1110.
### 2018-2019 Partner Schools and Course Offerings
Historical and aesthetic components of art with laboratory or online experiences with basic elements of creative expression.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Grammar review; development of the four skills. Three class periods and laboratory practice each week.
A communicative approach to intermediate language using the four skill areas of listening, speaking, reading, and writing, along with French and Francophone culture.
Grammar review; development of the four skills.
Grammar review; development of the four skills.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Constitutional basis and development, political processes (parties, nominations and elections, interest groups and public opinion), federalism and institutions of national government.
Basic principles of public speaking. Focuses on informative and persuasive speaking in both extemporaneous and impromptu styles. Emphasizes adapting to diverse audiences, reducing communication apprehension, presenting in varied contexts, and using technology effectively.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Grammar review; development of the four skills
Grammar review; development of the four skills
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Constitutional basis and development, political processes (parties, nominations and elections, interest groups and public opinion), federalism and institutions of national government.
Grammatical review and reading of ancient Latin texts.
Grammatical review and reading of ancient Latin texts.
Basic algebra; inequalities; functions and graphs; logarithmic and exponential functions; trigonometric functions and identities; applications and other topics.
(Prerequisites: Two years of high school algebra and one of geometry AND a satisfactory placement exam score, or grade of C or higher in MATH 1200 or MATH 1220.)
Limits, the derivative, differentiation techniques and applications of the derivative. MATH 1340 and MATH 1350 is a two-semester sequence which includes all the topics from MATH 1310. Not open to students with a grade of C or higher in MATH 1310 or MATH 1260. Prerequisites: same as MATH 1310.
The definite integral; the fundamental theorem; indefinite integrals; integration by parts, by substitution and using tables; and applications of definite and indefinite integrals.
(Prerequisite: grade of C or higher in MATH 1340.)
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Description of data, binomial and normal distributions, estimation and testing hypotheses for means and proportions.
(Placement: ACT Match sub-score of 22 or Math Placement required).
Review of functions and their graphs, linear and quadratic functions, factoring. Polynomial and rational functions. Review of exponents. Exponential and logarithmic functions and their graphs. Systems of equations, theory of equations.
[Prerequisites: Two years of high school algebra, one year of geometry and a satisfactory placement exam score, or grade of C or higher in MATH 99, MATH 1210, or grade of D in MATH 1200.]
Limits, the derivative, differentiation techniques and applications of the derivative. MATH 1340 and MATH 1350 is a two-semester sequence which includes all the topics from MATH 1310. Not open to students with a grade of C or higher in MATH 1310 or MATH 1260. Prerequisites: same as MATH 1310.
The definite integral; the fundamental theorem; indefinite integrals; integration by parts, by substitution and using tables; and applications of definite and indefinite integrals.
(Prerequisite: grade of C or higher in MATH 1340.)
Laboratory course for non-science majors. Emphasis on scientific data analysis and the meaning of scientific knowledge. Not acceptable toward physics major or minor. Two lectures and one two-hour laboratory.
Scientific approaches to the study of behavior of organisms. Application to personal and social behavior.
Scientific study of social structure, interaction, and institutions. Topics include gender, race, class, family, culture, and crime.
Historical and aesthetic components of art with laboratory or online experiences with basic elements of creative expression.
Scientific approaches to the study of behavior of organisms. Application to personal and social behavior.
Constitutional basis and development, political processes (parties, nominations and elections, interest groups and public opinion), federalism and institutions of national government.
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Basic ecology and current environmental problems of air, water and land pollution; human reproduction and population dynamics.
Scientific study of social structure, interaction, and institutions. Topics include gender, race, class, family, culture, and crime. [
Comparative study of how and why selected economic, social, political and intellectual revolutions of the modern world have transformed and are shaping contemporary European and non-Western cultures.
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Grammar review; development of the four skills.
Constitutional basis and development, political processes (parties, nominations and elections, interest groups and public opinion), federalism and institutions of national government.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Computer technology and related social issues. Hardware, software, applications in diverse areas. Problems concerning computerized services, data banks, governmental controls. Problem solving using software packages (such as hypertext, spreadsheets, word processing, database, presentation graphics, etc.).
(Prerequisite: one year of high school algebra or MATH 95 or higher.)
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Basic algebra; inequalities; functions and graphs; logarithmic and exponential functions; trigonometric functions and identities; applications and other topics.
(Prerequisites: Two years of high school algebra and one of geometry AND a satisfactory placement exam score, or grade of C or higher in MATH 1200 or MATH 1220.)
Helps students examine their skills, interests, values and personal characteristics; investigate occupations and career paths; examine the interrelationship between self-knowledge and occupational decisions; identify academic programs and experiential learning opportunities that enhance future employment options; make informed career and life decisions; and establish realistic goals and an action plan.
Limits, the derivative, differentiation techniques and applications of the derivative. MATH 1340 and MATH 1350 is a two-semester sequence which includes all the topics from MATH 1310. Not open to students with a grade of C or higher in MATH 1310 or MATH 1260. Prerequisites: same as MATH 1310.
The definite integral; the fundamental theorem; indefinite integrals; integration by parts, by substitution and using tables; and applications of definite and indefinite integrals.
(Prerequisite: grade of C or higher in MATH 1340.)
Introduction to functional areas and environments of business administration. Knowledge and skill development in accounting, finance, management and marketing. Overview of legal, economic, ethical and social/cultural aspects of business in domestic and global markets.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Description of data, binomial and normal distributions, estimation and testing hypotheses for means and proportions.
(Placement: ACT Match sub-score of 22 or Math Placement required).
Review of functions and their graphs, linear and quadratic functions, factoring. Polynomial and rational functions. Review of exponents. Exponential and logarithmic functions and their graphs. Systems of equations, theory of equations.
[Prerequisites: Two years of high school algebra, one year of geometry and a satisfactory placement exam score, or grade of C or higher in MATH 99, MATH 1210, or grade of D in MATH 1200.]
Introduction to educational foundation topics and contemporary issues for prospective teachers. Explore and analyze the teaching profession in a weekly seminar and a service learning experience. Required as an entry-year experience for all teacher-education candidates.
Basic concepts: the cell, metabolism, genetics, reproduction, development, evolution, ecology. Three one-hour lectures, one two-hour laboratory.
Basic ecology and current environmental problems of air, water and land pollution; human reproduction and population dynamics.
Basic concepts: the cell, metabolism, genetics, reproduction, development, evolution, ecology. Three one-hour lectures, one two-hour laboratory.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Grammar review; development of the four skills. Three class periods and laboratory practice each week
A communicative approach to intermediate language using the four skill areas of listening, speaking, reading, and writing, along with French and Francophone culture.
Comparative study of how and why economic, social, political and intellectual factors shaped and defined the history of selected Western and non-Western civilizations in the ancient and medieval periods. [
First term of an introductory physics sequence using algebra and trigonometry, but not calculus. Topics include motion, forces, energy, fluids, heat and simple harmonic motion.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Introduction to Spanish language and to Hispanic cultures. Communicative approach to teach beginning language use in the four skill areas: listening, speaking, reading, writing.
SPAN 1010 continued. Prerequisite: SPAN 1010 or a satisfactory placement exam score.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Basic ecology and current environmental problems of air, water and land pollution; human reproduction and population dynamics.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Comparative study of how and why economic, social, political and intellectual factors shaped and defined the history of selected Western and non-Western civilizations in the ancient and medieval periods. [
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Provides an overview of the development and characteristics of students with exceptional learning needs; historical, philosophical, and legal issues in special education; instructional strategies and adaptations of learning environments; and consultation and collaboration to meet the needs of all students.
(Prerequisite: EDTL 2010)
Focuses on the broad continuum of cognitive, social, emotional and physical development of children emphasizing conception through grade 3. Examines children growing up in diverse families, communities, and cultural contexts through various observational techniques, application of developmental theory, and instruction in research methodology.
Introduction to educational foundation topics and contemporary issues for prospective teachers. Explore and analyze the teaching profession in a weekly seminar and a service learning experience. Required as an entry-year experience for all teacher-education candidates.
The course addresses effectively identifying, locating, evaluating, designing, preparing and efficiently using educational technology as an instructional resource in the classroom as related to principles of learning and teaching. Candidates will develop increased classroom communication abilities through lectures, discussions, modeling, laboratory experiences and completion of a comprehensive project.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
### 2018-2019 Partner Schools and Course Offerings
Historical and aesthetic components of art with laboratory or online experiences with basic elements of creative expression.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Grammar review; development of the four skills. Three class periods and laboratory practice each week.
A communicative approach to intermediate language using the four skill areas of listening, speaking, reading, and writing, along with French and Francophone culture.
Grammar review; development of the four skills.
Grammar review; development of the four skills.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Constitutional basis and development, political processes (parties, nominations and elections, interest groups and public opinion), federalism and institutions of national government.
Basic principles of public speaking. Focuses on informative and persuasive speaking in both extemporaneous and impromptu styles. Emphasizes adapting to diverse audiences, reducing communication apprehension, presenting in varied contexts, and using technology effectively.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Grammar review; development of the four skills
Grammar review; development of the four skills
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Constitutional basis and development, political processes (parties, nominations and elections, interest groups and public opinion), federalism and institutions of national government.
Grammatical review and reading of ancient Latin texts.
Grammatical review and reading of ancient Latin texts.
Basic algebra; inequalities; functions and graphs; logarithmic and exponential functions; trigonometric functions and identities; applications and other topics.
(Prerequisites: Two years of high school algebra and one of geometry AND a satisfactory placement exam score, or grade of C or higher in MATH 1200 or MATH 1220.)
Limits, the derivative, differentiation techniques and applications of the derivative. MATH 1340 and MATH 1350 is a two-semester sequence which includes all the topics from MATH 1310. Not open to students with a grade of C or higher in MATH 1310 or MATH 1260. Prerequisites: same as MATH 1310.
The definite integral; the fundamental theorem; indefinite integrals; integration by parts, by substitution and using tables; and applications of definite and indefinite integrals.
(Prerequisite: grade of C or higher in MATH 1340.)
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Description of data, binomial and normal distributions, estimation and testing hypotheses for means and proportions.
(Placement: ACT Match sub-score of 22 or Math Placement required).
Review of functions and their graphs, linear and quadratic functions, factoring. Polynomial and rational functions. Review of exponents. Exponential and logarithmic functions and their graphs. Systems of equations, theory of equations.
[Prerequisites: Two years of high school algebra, one year of geometry and a satisfactory placement exam score, or grade of C or higher in MATH 99, MATH 1210, or grade of D in MATH 1200.]
Limits, the derivative, differentiation techniques and applications of the derivative. MATH 1340 and MATH 1350 is a two-semester sequence which includes all the topics from MATH 1310. Not open to students with a grade of C or higher in MATH 1310 or MATH 1260. Prerequisites: same as MATH 1310.
The definite integral; the fundamental theorem; indefinite integrals; integration by parts, by substitution and using tables; and applications of definite and indefinite integrals.
(Prerequisite: grade of C or higher in MATH 1340.)
Laboratory course for non-science majors. Emphasis on scientific data analysis and the meaning of scientific knowledge. Not acceptable toward physics major or minor. Two lectures and one two-hour laboratory.
Scientific approaches to the study of behavior of organisms. Application to personal and social behavior.
Scientific study of social structure, interaction, and institutions. Topics include gender, race, class, family, culture, and crime.
Historical and aesthetic components of art with laboratory or online experiences with basic elements of creative expression.
Scientific approaches to the study of behavior of organisms. Application to personal and social behavior.
Constitutional basis and development, political processes (parties, nominations and elections, interest groups and public opinion), federalism and institutions of national government.
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Basic ecology and current environmental problems of air, water and land pollution; human reproduction and population dynamics.
Scientific study of social structure, interaction, and institutions. Topics include gender, race, class, family, culture, and crime. [
Comparative study of how and why selected economic, social, political and intellectual revolutions of the modern world have transformed and are shaping contemporary European and non-Western cultures.
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Grammar review; development of the four skills.
Constitutional basis and development, political processes (parties, nominations and elections, interest groups and public opinion), federalism and institutions of national government.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Computer technology and related social issues. Hardware, software, applications in diverse areas. Problems concerning computerized services, data banks, governmental controls. Problem solving using software packages (such as hypertext, spreadsheets, word processing, database, presentation graphics, etc.).
(Prerequisite: one year of high school algebra or MATH 95 or higher.)
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Basic algebra; inequalities; functions and graphs; logarithmic and exponential functions; trigonometric functions and identities; applications and other topics.
(Prerequisites: Two years of high school algebra and one of geometry AND a satisfactory placement exam score, or grade of C or higher in MATH 1200 or MATH 1220.)
Helps students examine their skills, interests, values and personal characteristics; investigate occupations and career paths; examine the interrelationship between self-knowledge and occupational decisions; identify academic programs and experiential learning opportunities that enhance future employment options; make informed career and life decisions; and establish realistic goals and an action plan.
Limits, the derivative, differentiation techniques and applications of the derivative. MATH 1340 and MATH 1350 is a two-semester sequence which includes all the topics from MATH 1310. Not open to students with a grade of C or higher in MATH 1310 or MATH 1260. Prerequisites: same as MATH 1310.
The definite integral; the fundamental theorem; indefinite integrals; integration by parts, by substitution and using tables; and applications of definite and indefinite integrals.
(Prerequisite: grade of C or higher in MATH 1340.)
Introduction to functional areas and environments of business administration. Knowledge and skill development in accounting, finance, management and marketing. Overview of legal, economic, ethical and social/cultural aspects of business in domestic and global markets.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Description of data, binomial and normal distributions, estimation and testing hypotheses for means and proportions.
(Placement: ACT Match sub-score of 22 or Math Placement required).
Review of functions and their graphs, linear and quadratic functions, factoring. Polynomial and rational functions. Review of exponents. Exponential and logarithmic functions and their graphs. Systems of equations, theory of equations.
[Prerequisites: Two years of high school algebra, one year of geometry and a satisfactory placement exam score, or grade of C or higher in MATH 99, MATH 1210, or grade of D in MATH 1200.]
Introduction to educational foundation topics and contemporary issues for prospective teachers. Explore and analyze the teaching profession in a weekly seminar and a service learning experience. Required as an entry-year experience for all teacher-education candidates.
Basic concepts: the cell, metabolism, genetics, reproduction, development, evolution, ecology. Three one-hour lectures, one two-hour laboratory.
Basic ecology and current environmental problems of air, water and land pollution; human reproduction and population dynamics.
Basic concepts: the cell, metabolism, genetics, reproduction, development, evolution, ecology. Three one-hour lectures, one two-hour laboratory.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Grammar review; development of the four skills. Three class periods and laboratory practice each week
A communicative approach to intermediate language using the four skill areas of listening, speaking, reading, and writing, along with French and Francophone culture.
Comparative study of how and why economic, social, political and intellectual factors shaped and defined the history of selected Western and non-Western civilizations in the ancient and medieval periods. [
First term of an introductory physics sequence using algebra and trigonometry, but not calculus. Topics include motion, forces, energy, fluids, heat and simple harmonic motion.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Introduction to Spanish language and to Hispanic cultures. Communicative approach to teach beginning language use in the four skill areas: listening, speaking, reading, writing.
SPAN 1010 continued. Prerequisite: SPAN 1010 or a satisfactory placement exam score.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Basic ecology and current environmental problems of air, water and land pollution; human reproduction and population dynamics.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Comparative study of how and why economic, social, political and intellectual factors shaped and defined the history of selected Western and non-Western civilizations in the ancient and medieval periods. [
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Provides an overview of the development and characteristics of students with exceptional learning needs; historical, philosophical, and legal issues in special education; instructional strategies and adaptations of learning environments; and consultation and collaboration to meet the needs of all students.
(Prerequisite: EDTL 2010)
Focuses on the broad continuum of cognitive, social, emotional and physical development of children emphasizing conception through grade 3. Examines children growing up in diverse families, communities, and cultural contexts through various observational techniques, application of developmental theory, and instruction in research methodology.
Introduction to educational foundation topics and contemporary issues for prospective teachers. Explore and analyze the teaching profession in a weekly seminar and a service learning experience. Required as an entry-year experience for all teacher-education candidates.
The course addresses effectively identifying, locating, evaluating, designing, preparing and efficiently using educational technology as an instructional resource in the classroom as related to principles of learning and teaching. Candidates will develop increased classroom communication abilities through lectures, discussions, modeling, laboratory experiences and completion of a comprehensive project.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
### 2017-2018 Partner Schools and Course Offerings
Historical and aesthetic components of art with laboratory or online experiences with basic elements of creative expression.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Grammar review; development of the four skills. Three class periods and laboratory practice each week.
A communicative approach to intermediate language using the four skill areas of listening, speaking, reading, and writing, along with French and Francophone culture.
Grammar review; development of the four skills.
Grammar review; development of the four skills.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Basic principles of public speaking. Focuses on informative and persuasive speaking in both extemporaneous and impromptu styles. Emphasizes adapting to diverse audiences, reducing communication apprehension, presenting in varied contexts, and using technology effectively.
Grammar review; development of the four skills. Three class periods and laboratory practice each week
A communicative approach to intermediate language using the four skill areas of listening, speaking, reading, and writing, along with French and Francophone culture.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Basic concepts: the cell, metabolism, genetics, reproduction, development, evolution, ecology. Three one-hour lectures, one two-hour laboratory.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Personal financial management; budgeting, borrowing sources and costs; auto, property, and life insurance; home ownership financing; personal investment strategy; and retirement planning.
Comparative study of how and why economic, social, political and intellectual factors shaped and defined the history of selected Western and non-Western civilizations in the ancient and medieval periods. [
Comparative study of how and why selected economic, social, political and intellectual revolutions of the modern world have transformed and are shaping contemporary European and non-Western cultures.
First term of an introductory physics sequence using algebra and trigonometry, but not calculus. Topics include motion, forces, energy, fluids, heat and simple harmonic motion.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Grammar review; development of the four skills
Grammar review; development of the four skills
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Constitutional basis and development, political processes (parties, nominations and elections, interest groups and public opinion), federalism and institutions of national government.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Grammatical review and reading of ancient Latin texts.
Grammatical review and reading of ancient Latin texts.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Description of data, binomial and normal distributions, estimation and testing hypotheses for means and proportions.
(Placement: ACT Match sub-score of 22 or Math Placement required).
The definite integral; the fundamental theorem; indefinite integrals; integration by parts, by substitution and using tables; and applications of definite and indefinite integrals.
(Prerequisite: grade of C or higher in MATH 1340.)
Laboratory course for non-science majors. Emphasis on scientific data analysis and the meaning of scientific knowledge. Not acceptable toward physics major or minor. Two lectures and one two-hour laboratory.
Scientific approaches to the study of behavior of organisms. Application to personal and social behavior.
Scientific study of social structure, interaction, and institutions. Topics include gender, race, class, family, culture, and crime.
First term of an introductory physics sequence using algebra and trigonometry, but not calculus. Topics include motion, forces, energy, fluids, heat and simple harmonic motion.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Grammar review; development of the four skills.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Scientific study of social structure, interaction, and institutions. Topics include gender, race, class, family, culture, and crime. [
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Introduction to functional areas and environments of business administration. Knowledge and skill development in accounting, finance, management and marketing. Overview of legal, economic, ethical and social/cultural aspects of business in domestic and global markets.
Computer technology and related social issues. Hardware, software, applications in diverse areas. Problems concerning computerized services, data banks, governmental controls. Problem solving using software packages (such as hypertext, spreadsheets, word processing, database, presentation graphics, etc.).
(Prerequisite: one year of high school algebra or MATH 95 or higher.)
Basic algebra; inequalities; functions and graphs; logarithmic and exponential functions; trigonometric functions and identities; applications and other topics.
(Prerequisites: Two years of high school algebra and one of geometry AND a satisfactory placement exam score, or grade of C or higher in MATH 1200 or MATH 1220.)
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Scientific study of social structure, interaction, and institutions. Topics include gender, race, class, family, culture, and crime. [
Helps students examine their skills, interests, values and personal characteristics; investigate occupations and career paths; examine the interrelationship between self-knowledge and occupational decisions; identify academic programs and experiential learning opportunities that enhance future employment options; make informed career and life decisions; and establish realistic goals and an action plan.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Laboratory course for non-science majors. Emphasis on scientific data analysis and the meaning of scientific knowledge. Not acceptable toward physics major or minor. Two lectures and one two-hour laboratory.
Basic ecology and current environmental problems of air, water and land pollution; human reproduction and population dynamics.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Review of functions and their graphs, linear and quadratic functions, factoring. Polynomial and rational functions. Review of exponents. Exponential and logarithmic functions and their graphs. Systems of equations, theory of equations.
[Prerequisites: Two years of high school algebra, one year of geometry and a satisfactory placement exam score, or grade of C or higher in MATH 99, MATH 1210, or grade of D in MATH 1200.]
Description of data, binomial and normal distributions, estimation and testing hypotheses for means and proportions.
(Placement: ACT Match sub-score of 22 or Math Placement required).
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Basic concepts: the cell, metabolism, genetics, reproduction, development, evolution, ecology. Three one-hour lectures, one two-hour laboratory.
Basic ecology and current environmental problems of air, water and land pollution; human reproduction and population dynamics.
Shore and ocean environments, variety and adaptations of marine life. Observations of marine organisms in marine laboratory.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Comparative study of how and why selected economic, social, political and intellectual revolutions of the modern world have transformed and are shaping contemporary European and non-Western cultures.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Grammar review; development of the four skills. Three class periods and laboratory practice each week
A communicative approach to intermediate language using the four skill areas of listening, speaking, reading, and writing, along with French and Francophone culture.
Comparative study of how and why economic, social, political and intellectual factors shaped and defined the history of selected Western and non-Western civilizations in the ancient and medieval periods. [
First term of an introductory physics sequence using algebra and trigonometry, but not calculus. Topics include motion, forces, energy, fluids, heat and simple harmonic motion.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Communicative approach to teach intermediate language use in the four skill areas: listening, speaking, reading, writing (emphasis on composition). Reading and discussion in Spanish of cultural readings.
Basic ecology and current environmental problems of air, water and land pollution; human reproduction and population dynamics.
Comparative study of how and why economic, social, political and intellectual factors shaped and defined the history of selected Western and non-Western civilizations in the ancient and medieval periods. [
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Provides an overview of the development and characteristics of students with exceptional learning needs; historical, philosophical, and legal issues in special education; instructional strategies and adaptations of learning environments; and consultation and collaboration to meet the needs of all students.
(Prerequisite: EDTL 2010)
Focuses on the broad continuum of cognitive, social, emotional and physical development of children emphasizing conception through grade 3. Examines children growing up in diverse families, communities, and cultural contexts through various observational techniques, application of developmental theory, and instruction in research methodology.
Introduction to educational foundation topics and contemporary issues for prospective teachers. Explore and analyze the teaching profession in a weekly seminar and a service learning experience. Required as an entry-year experience for all teacher-education candidates.
The course addresses effectively identifying, locating, evaluating, designing, preparing and efficiently using educational technology as an instructional resource in the classroom as related to principles of learning and teaching. Candidates will develop increased classroom communication abilities through lectures, discussions, modeling, laboratory experiences and completion of a comprehensive project.
Basic expository writing; emphasis on organizing and developing coherent essays of at least 800 words for college-educated audiences.
Expository writing including research paper; emphasis on analytical writing based on critical reading. Placement through pretesting or successful completion of GSW 1110. Students must complete course and program portfolio assessment successfully to receive a passing grade.
Selected constitutional, intellectual, political and social developments that defined and shaped America between its first European settlement and the end of Reconstruction.
How and why selected economic, intellectual, political and social developments transformed post-Civil War America and shaped 20th-century American society.
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http://www.cirro.dk/39wcdgou/75078e-orbitals-s-p-d-f | 1,679,570,799,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945144.17/warc/CC-MAIN-20230323100829-20230323130829-00377.warc.gz | 66,620,302 | 10,257 | Counting the 4s, 4p, and 4d orbitals, this makes a total of 16 orbitals in the fourth level. It . If you go on to learn more about chemistry, you ⦠Multimedia: interviste, video, animazioni, GiovediScienza 2020 – 35esima edizione (on line). At any one energy level, we have three absolutely equivalent p orbitals pointing mutually at right angles to each other. Fortunately, you will probably not have to memorize the shapes of the f orbitals. Not all electrons inhabit s orbitals. All we can do is draw a shape that will include the electron most of the time, say 95% of the time. How many #3d_(z^2)# orbitals have #n = 3# and #l = 2#? #color(white)(.....)"s" color(white)(..............) 0 color(white)(............) 2(2(0) + 1) = 2# How many d orbitals can there be in one energy level? b) 4f A given set of p orbitals consists of how many orbitals? Electrons Share. The orbital shapes are: s, p, d, and f. Summarize Aufbau’s rule for filling orbitals. The letters and words refer to the visual impression left by the fine structure of the spectral lines which occurs due to the first relativistic corrections, especially the spin-orbital interaction. Helmenstine, Ph.D. Anne Marie. How many orbitals are found in the d sublevel? Since an electron can theoretically occupy all space, it is impossible to draw an orbital. How many p-orbitals are occupied in a K atom? How many atoms does each element have? These orbitals look like two dumbbells in the same plane. The valence shell of the element X contains 2 electrons in a 5s subshell. Why can higher energy levels accommodate more electrons? Sublevels are designated with lower-case letters. ... P orbitals have 3 different rotations along the x y and z axes. They are named s,p,d,f .The s, p, d, and f stand for sharp, principal, diffuse and fundamental, respectively. Orientasi dan Bentuk Orbital s p d f, Elektron, Bilangan Kuantum, Atom - Bentuk orbital ditentukan oleh subkulit dari elektron atau ditentukan bilangan kuantum azimutnya. -s orbitals have the largest screening effect for a given n value since s electrons are closer to the nucleus.-p orbital's have the next highest screening effect and then comes d and then f orbital's. What is the structural difference between a 2P and a 3P orbital? In which main energy level does the 's' sublevel first appear? s, p, d, and f orbitals are available at all higher energy levels as well. For 1-First shell-, S-S orbital. Which atomic orbitals of which subshells have a dumbbell shape? Shapes of Orbitals and Electron Density Patterns . d and f orbitals. What is the maximum number of orbitals in the p sublevel? Shape of d-orbitals . Elements in the long form of periodic table have been divided into four blocks i.e. How many electrons can occupy the p orbitals at each energy level? The five 3d orbitals are called This is because one p orbital is composed of three sub orbitals named as p x, p y and p z. What is the maximum electron capacity of the "s" orbital of an atom? When orbitals are not filled, energy of all orbitals are same while when the are filled energy of orbitals are s p d f. Does the 3rd electron shell have a capacity for 8e- or 18e-? An S orbital has the shape of a sphere. The letters s, p, d, and f come from the descriptions of alkali metal spectroscopy lines as appearing sharp, principal, diffuse, or fundamental. This is simply for convenience, because what you might think of as the x, y or z direction changes constantly as the atom tumbles in space. The letter refers to the shape of the orbital. Now, you'll also hear the term, subshell, subshell, or sometimes people will say sublevels and that's where they're talking about s or p or d and eventually f so if I circle this, I'm talking about that first shell. Choose from 500 different sets of term:orbitals = s, p, d, f flashcards on Quizlet. Orbitals are the regions of space in which electrons are most likely to be found. D â Orbitals. How many p orbitals are there in a neon atom? An orbital wants to fill it's self. The letters s, p, d, and f were assigned for historical reasons that need not concern us. Classification of Elements based on s,p,d and f Orbitals the long form of the periodic table divides the elements into four major blocks known as s, p, d, and f. this division represents the name of the orbital and received from the last electron of the shell. Jadi, apabila suatu elektron memiliki bilangan kuantum azimut sama, maka bentuk orbitalnya juga sama, sehingga yang membedakan hanyalah tingkat energinya. In the Schrödinger equation for this system of one negative and one positive particle, the atomic orbitals are the eigenstates of the Hamiltonian operatorfor the energy. Learn term:orbitals = s, p, d, f with free interactive flashcards. There are similar orbitals at subsequent levels: 3px, 3py, 3pz, 4px, 4py, 4pz and so on. Click the images to see the various 3d orbitals There are a total of five d orbitals and each orbital can hold two electrons. d and f orbitals. However, although there is only one s orbital in the s subshell, there are 3 p orbitals in the p subshell, 5 d orbitals in the d subshell, and 7 f orbitals in the 5 subshell. Which sublevel is filled after the 5s sub level? 2) Orbitals are combined when bonds form between atoms in a molecule. Q:- Come è possibile notare, quattro dei cinque orbitali di tipo d sono tetralobati con la funzione d'onda Ψ che assume segno alternativamente opposto. Orbitals in the 2p sublevel are degenerate orbitals â Which means that the 2px, 2py, and 2pz orbitals have the exact same energy, as illustrated in the diagram provided below. Explain? Alright, so why would the atom want to have 8 electrons in it's outer most shell, good question. For example, the electron in a hydrogen (H) atom would have the values n=1 and l=0. The f orbital is more complex, but follows the same rules based on proton alignment as the p and d orbitals. Now consider sin 2 x, the square of the original function.In quantum chemistry Ψ 2 provides us with the electron density - it defines the size and shapes of the familiar orbitals s, p, d, f, etc. Sublevels are designated with lower-case letters. How would you find how many orbitals a sublevel has? Perché gli orbitali s, p, d, f si chiamano così? How many orbitals are in each of the sublevels (s,p,d,f)? What is the maximum number of electrons that can occupy the 3d orbitals? Video \(\PageIndex{1}\): Energy levels, sublevels and orbitals. An s orbital is spherically symmetric around the nucleus of the atom, like a hollow ball made of rather fluffy material with the nucleus at its centre. The letters, s, p, d, and f designate the shape of the orbital. Now, letâs look at a cross-section of these orbitals. s ,p ,d and f. This division is based upon the name of the orbitals which receives the last electron. The f orbital has 15 protons to complete a fifth level of a tetrahedral structure. 3dyz Non so dirti, però, se l’introduzione di questa nomenclatura sia da attribuire ad uno spettroscopista in particolare. Where does the maximum electron density occur for 2s and 2p orbitals in hydrogen atom? s- sharp for L=0 p- principal for L=1 d- diffuse for L=2 f- fundamental for L=3 source: Electron configuration - Wikipedia. Because it's an s orbital and it's the first shell it's labelled 1S. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Oxygen reacts with fluorine to form only #OF_2#, but sulphur which is in the same group 16 as oxygen, reacts with fluorine to form #SF_2#, #SF_4# and #SF_6#. If #â# is the angular quantum number of subshell then maximum electrons it can hold is #2(2 â + 1)#, #underline(bb("Sub-shell" color(white)(.....) â color(white)(.....) "Maximum electrons"))# When filling similar orbitals, distribute one electron Each sublevel has differing numbers of orbitals. Each p-orbital … How many electrons can the 2nd shell accommodate? Just remember that there seven f orbitals in each level from level 4 and onwards. c) 3s Thus in simple terms the screening effect decreases in order :s orbitals > p orbitals> d orbitals> f orbitals. At the first energy level, the only orbital available to electrons is the 1s orbital. As shown, each element’s electron configuration is unique to its position on the periodic table. How many atomic orbitals are there in the 4p sublevel? What is the maximum number of f orbitals in any single energy level in an atom? Secondary Quantum Number/Orbital Shape Quantum number: represents the shape of the orbital- s, p, f, d. l is a range of n-1. However, although there is only one s orbital in the s subshell, there are 3 p orbitals in the p subshell, 5 d orbitals in the d subshell, and 7 f orbitals in the 5 subshell. It is sort of like a hollow tennis ball. S block elements 1)Elements in which the last electron enters the s orbital of ⦠Why does an electron found in a 2s orbital have a lower energy than an electron found in a 2p orbital in multielectron systems? This you could view as the in and out of the page so you could view that as the z-dimension. To make sense of the names, we need to look at them in two groups. How does the 3s orbital differ from the 2s orbital? Libretexts. Just like the s â orbitals, with an increase in size and energy of p orbitals quantum number ( 4p > 3p > 2p ), the size and energy of p orbitals also increase. C’è chi si pente e rimuoverlo potrebbe essere difficile: oggi sappiamo perché. What is the maximum number of d orbitals in a principal energy level? Element X also has a partially filled 4d subshell. Each sublevel has differing numbers of orbitals. There are four types of orbitals that you should be familiar with s, p, d and f (sharp, principle, diffuse and fundamental). s, p, d and f orbitals are then available at all higher energy levels as well. How can we know what orbitals are higher in energy? f-14. Can someone compare s, p, d, and f orbitals in terms of size, shape, and energy? These orbitals are designated as P x, P y & P z orbitals. First off, the terminology is obsolete and historical. As the energy levels increase, the electrons are located further from the nucleus, so the orbitals get bigger. Each orbital is denoted by a number and a letter. They can be obtained analytically, meaning that the resulting orbitals are products of a polyno… For example, the 1 shell has an S orbital. How many total orbitals are within the 2s and 2p sublevels of the second energy level? 2. An atom of any other element ionized down to a single electron is very similar to hydrogen, and the orbitals take the same form. The main difference between s orbital and p orbital is that s orbitals are spherical shaped whereas p orbitals are dumbbell shaped. Atom there are seven f orbitals are there in an individual d orbital is spherical its... Spettroscopista in particolare ( degenerate state ) but differ in their orientations at subsequent:..., 4px, 4py, 4pz and so on # 2d #, # 1d,... F orbital is denoted by a number and a sample electron configuration explained... Equivalent p orbitals pointing mutually at right angles to each letter, 4px, 4py, 4pz and on! 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Atomic xenon equal in energy differ from a 2py orbital are s, p and d. an illustration of #... Difficile: oggi sappiamo perché are a total of 16 orbitals in an 3p subshell in the fourth higher. D- diffuse for L=2 f- fundamental for L=3 source: electron configuration is unique its. The name of the s, p, d, and f orbitals are in each level from 4. Are available at all higher energy levels, there are seven f in. The s/p/d sum p z orbitals ( z^2 ) # atomic orbital in fourth. 4F c ) 3s d ) 2d e ) 2p like spheres but # p # orbitals that the... Space ) and energies ( e.g the second energy level in an metal... # '' be '' ^- # s # orbitals have # n = 3 # and 3f. Hold the electrons begin to pair up to make sense of the time, say %. D orbital, a p sublevel terms of size is 1s < 2s < 3s â¦... The third main energy level of the shape of the lobes is pointing between two of the s orbitals then... Consequence of the axes, not along them 's labelled 1s called 2px, 2py and.... Are five values of m namely -2, -1, 0, 1, 2 orbitals which! Available here so on are the orbital shapes are: s, p d..., 0, +1 electrons are contained in the orbital ( s, p, d, and orbitals! 0.1 m in NaOH the x-z plane, and # l = 1, 2 3 2015... D sublevel spherical shaped whereas p orbitals at each energy level of the orbitals! Sharp for L=0 p- principal for L=1 d- diffuse for L=2 f- fundamental for L=3 source: electron -! Balls where there is a surface between the two balls where there is an incorrect for... Size, shape, and more with flashcards, games, and f. Summarize Aufbau ’ s rule filling. 2, there are some combinations of orbitals in an atom there are three,! More complex, but follows the same plane go before the # 3d_ ( z^2 #! The 3rd electron shell have a dumbbell shape # 2d #, and orbitals! The 3rd electron shell have a dumbbell shape fourth shell, good question and! A 2px orbital differ from a 2p orbital be found combined when form! Of nine orbitals altogether different sets of term: orbitals = s p! Like a hollow tennis ball is composed of three sub orbitals named as p x p. Three different directions along x, p and d. an illustration of the orbital shapes:! Electrons allowed in an atom first group contains the 3dxy, 3dxz and 3dyz orbitals four!, j, etc sublevel has d- diffuse for L=2 f- fundamental for L=3:... From top to bottom as shown, each element ’ s electron configuration, explained later ) ( )... | 4,966 | 19,174 | {"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} | 2.609375 | 3 | CC-MAIN-2023-14 | latest | en | 0.822432 |
https://proxieslive.com/name-of-this-2d-automata/ | 1,603,709,999,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891203.69/warc/CC-MAIN-20201026090458-20201026120458-00443.warc.gz | 498,635,591 | 7,065 | # Name of this 2D automata
Today I had a weird dream about a kind of automata. Each automaton has a finite 2D grid wrapped around like a torus, formatted as a table. Execution starts in one cell. Each step it executes whatever is in the current cell, and in each cell there’s a symbol and a numeric operator. In one variant the numeric part is said to be optional. The symbol can be “n” or “a”, or some other letters that I don’t remember. They are not arbitrary but have meanings.
If the symbol is “n”, the numeric operator will rotate the current row or column by a number of cells like in a Rubik’s cube(without changing the “execution pointer”). It’s written like this: +1 for rotating a row to the right, -1 to the left (presumably other numbers are allowed too but I didn’t see examples in my dream), $$+\frac{1}{-}$$ for rotating a column upwards, $$-\frac{1}{-}$$ for rotating downwards. There’s no separate state apart from what’s on the grid. I’m not sure what “a” means, but it may have meant rotating the whole grid in the given direction while keeping the execution cell position fixed. I saw an example in my dream with just one cell with “n” in it, described the the smallest automaton that doesn’t do anything.
I didn’t see any examples of explicit machine input or output. Maybe the input is encoded in the grid, or maybe it’s designed to be interactive and the animation created by the moving cells is the intended output. I’m not sure if there’s any way to halt/accept/reject in it.
Does this kind of automata happen to have a name? Is there any related research? | 374 | 1,586 | {"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-45 | latest | en | 0.908967 |
https://id.scribd.com/document/425392020/Cambridge-Ext2-Ch1-Complex-Numbers-IWEB | 1,576,218,121,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540548544.83/warc/CC-MAIN-20191213043650-20191213071650-00553.warc.gz | 406,892,190 | 130,290 | Anda di halaman 1dari 62
# CHAPTER ONE
Complex Numbers I
## Chapter Overview: One of the significant properties of the real numbers is
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that any of the four arithmetic operations of addition, subtraction, multiplication
and division can be applied to any pair of real numbers, with the exception that
division by zero is undefined. As a result, every linear equation
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ax + b = 0 where a 6= 0
can be solved.
D E
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The situation is not so satisfactory when quadratic equations are considered.
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There are some quadratic equations that can be solved, but others, like
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x2 + 2x + 3 = 0 ,
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have no real solution. This apparent inconsistency, that some quadratics have a
solution whilst others do not, can be resolved by the introduction of a new type
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## of number, the complex number.
But there is more to complex numbers than just solving quadratic equations. In
this chapter the reader is shown an application to geometry and how they can be
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applied to the study of polynomials. These new numbers have many applications
beyond this course, such as in evaluating certain integrals and in solving problems
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## in electrical engineering. Complex numbers also provide links between seemingly
unrelated quantities and areas of mathematics. Here is a stunning example. The
four most significant real numbers encountered so far are 0, 1, e and π. As will be
shown in a later chapter on complex numbers, these four numbers are connected
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eiπ + 1 = 0 .
C
N
## Introducing A New Type of Number: The investigation is begun by examining
the roots of various quadratic equations. For convenience in presenting the new
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## work, the method of completing the square is used exclusively.
First consider only those quadratic equations with rational solutions, such as the
equation x2 − 4x − 12 = 0 . Completing the square:
(x − 2)2 = 16
so x − 2 = 4 or − 4
which leads to the two roots
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
2 CHAPTER 1: Complex Numbers
α = 6 and β = −2 .
Note that α + β = 4 and αβ = −12 .
Repeating this process for a number of quadratics with rational solutions, it soon
becomes evident that if ax2 + bx + c = 0 has solutions α and β then
b c
α+β = − and αβ = .
a a
Further investigation reveals that there are some quadratic equations which do
not have rational solutions, such as x2 − 4x − 1 = 0 . Completing the square:
(x − 2)2 = 5 .
Herein lies a problem since there is no rational number which when squared
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equals 5 . This problem is overcome by introducing a new type of number, in
√ √ 2
this case the irrational number 5 which has the property that 5 = 5.
Assuming that the normal rules of algebra apply to this new number, it follows
√ 2 √ 2 √ √
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that − 5 = 5 = 5 , so that 5 has two square roots, namely 5 and − 5 .
D E
If the introduction of this new type of number is valid then the solution may
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proceed. Thus
G √
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x − 2 = 5 or − 5
which leads to the two roots
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√ √
α = 2 + 5 and β = 2 − 5 .
Note that α + β = 4 and αβ = −1 .
EC BR
## Repeating this process for a number of quadratics with irrational solutions, it
soon becomes evident that if ax2 + bx + c = 0 has irrational roots α and β then
b c
α+β = − and αβ = .
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a a
Since this is consistent with the quadratic equations with rational solutions, it
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seems that the introduction of surds into the number system is valid. Indeed
surds have been used since Year 8 and students will be proficient in their use.
Yet further investigation reveals that there are some quadratic equations which
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## have neither rational nor irrational solutions, such as x2 −4x+5 = 0 . Completing
the square yields:
(x − 2)2 = −1 .
Again there is a problem since there is no known number which when squared
equals −1 . Just as before, this problem is overcome by introducing a new type
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of number. In this case the so called imaginary number i is introduced which has
the property that i2 = −1. Assuming that the normal rules of algebra apply to
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this new number, it follows that (−i)2 = i2 = −1 , so that −1 has two square
roots, namely i and −i . If the introduction of this new type of number is valid
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## then the solution may proceed. Thus
x − 2 = i or − i
which leads to the two roots
α = 2 + i and β = 2 − i .
Note that α + β = 4 and
αβ = (2 − i)(2 + i)
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1A The Arithmetic of Complex Numbers 3
## = 22 − i2 (difference of two squares)
=4+1
= 5.
Repeating this process for a number of quadratics with solutions which involve
the imaginary number i, it soon becomes evident that if ax2 + bx + c = 0 has
solutions α and β then
b c
α+β = − and αβ = .
a a
Since this is consistent with all previously encountered quadratic equations, it
seems reasonable to include the imaginary number i in the number system.
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A New Number in Arithmetic: The imaginary number i is now formally included
into the system of numbers. It has the special property that i2 = −1. This new
number i will be treated as if it were an algebraic pronumeral when it is combined
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with real numbers using the four arithmetic operations of addition, subtraction,
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multiplication and division.
O
G
A NEW NUMBER: The new number i has the special property that
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i2 = −1 .
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1
It may be used like a pronumeral with real numbers in addition, subtraction,
multiplication and division.
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It is instructive to write out the first four positive powers of i. They are:
i1 = i i2 = −1 i3 = i2 × i i4 = i3 × i
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(by definition) = −1 × i = −i × i
= −i =1
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Writing out the next four powers of i, it is found that this sequence repeats.
2
i5 = i4 × i i6 = i4 × i2 i7 = i4 × i3 i8 = i4
= 1×i = 1 × (−1) = 1 × (−i) =1
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=i = −1 = −i
It should be clear from these calculations that the sequence continues to cycle.
In general, the result can be determined from the remainder when the index is
divided by 4.
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POWERS OF THE IMAGINARY NUMBER: A power of i may take only one of four possible
values. If k is an integer, then these values are:
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2
i4k = 1 , i4k+1 = i , i4k+2 = −1 , i4k+3 = −i .
U
## WORKED EXAMPLE 1: Simplify: (a) i23 (b) i7 + i9
SOLUTION:
(a) i23 = i4×5+3 (b) i7 + i9 = −i + i
= −i =0
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
4 CHAPTER 1: Complex Numbers
Complex Numbers: Since i has been included in the number system and since it
is to be treated as a pronumeral, the number system must now include the real
numbers plus new quantities like
2i , −7i , 5 + 4i and 6 − 3i .
The set which includes all such quantities as well as the real numbers is given
the
√ symbol C. Each quantity in C is called a complex number. Thus 5, 2i and
6 − 3i are all examples of complex numbers. In the first case, 5 is also a real
number, and the real numbers form a special subset of the complex numbers. The
number 2i is an example of another special subset of the complex numbers. This
set consists of all the real multiples of i , which are called imaginary numbers.
Thus −7i is another example of an imaginary number.
FS
TWO NEW TYPES OF NUMBERS: Let a and b be real numbers.
COMPLEX NUMBERS: Numbers of the form a + ib are called complex numbers.
3 The set of all complex numbers is given the symbol C.
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IMAGINARY NUMBERS: Numbers of the form ib, that is the complex numbers for
which a = 0, are called imaginary numbers.
D E
O
Again noting that i is treated as a pronumeral, the addition, subtraction and
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multiplication of complex numbers presents no problem.
PR
(2 − 3i) + (5 + 7i) = 7 + 4i , (7 + 2i) − (5 − 3i) = 2 + 5i ,
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√ √ √
3(−5 + 7i) = −15 + 21i , 3(2 + i 3) = 2 3 + 3i .
The following worked examples of multiplication involve binomial expansions and
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## the property that i2 = −1.
WORKED EXAMPLE 2: Simplify:
(a) (2 − 3i)(5 + 7i) (b) (3 − 2i)2 (c) (4 + 3i)2 (d) (2 + 5i)(2 − 5i)
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SOLUTION:
(a) (2 − 3i)(5 + 7i) = 10 − i − 21i2 (c) (4 + 3i)2 = 16 + 24i + 9i2
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= 10 − i + 21 = 16 + 24i − 9
= 31 − i = 7 + 24i
(b) (3 − 2i)2 = 9 − 12i + 4i2 (d) (2 + 5i)(2 − 5i) = 4 − 25i2
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= 9 − 12i − 4 = 4 + 25
= 5 − 12i = 29
The last three examples above demonstrate the expansions of (x + iy)2 , (x − iy)2
and (x + iy)(x − iy) for real values of x and y . Note that in the final example,
the result is the sum of two squares and is a real number. This will always be
C
N
## THE SUM OF TWO SQUARES: Let x and y be real numbers, then
4 (x + iy)(x − iy) = x2 + y 2
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## which is always a real number.
Complex Conjugates: The last result is significant and will be used frequently.
Clearly the pair of numbers x + iy and x − iy have a special relationship. They
are called complex conjugates. Thus the complex conjugate of 3 + 2i is 3 − 2i.
Similarly the conjugate of 7 − 5i is 7 + 5i.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1A The Arithmetic of Complex Numbers 5
When the conjugate is required, the complex number is written with a bar above
it. Thus:
2+i = 2−i −3i = 3i
−1 + 4i = −1 − 4i −3 − 5i = −3 + 5i
COMPLEX CONJUGATES: Let x and y be real numbers, then the two complex numbers
x + iy and x − iy are called complex conjugates.
5
A: The conjugate of x + iy is x + iy = x − iy .
B: The conjugate of x − iy is x − iy = x + iy .
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Division: Just like real numbers, division by zero is undefined. Dividing a complex
number by any other real number presents no problem. As with rational numbers,
fractions should be simplified wherever possible by cancelling out common factors.
6 + 8i −2 − 6i
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= 3 + 4i = − 23 − 2i
√ 2 3
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√ −12 + 21i −4 + 7i
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2 − 2i = or − 45 + 57 i
√ =1−i 2 15 5
2
G
PR
There is a potential problem if one complex number is divided by another, such
2+i
as in 3−i . As it stands, it is not clear that this sort of quantity is even allowed
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in the new number system, since it is not in the standard form x + iy.
The problem is resolved by taking a similar approach to that used to deal with
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surds in the denominator. The process here is called realising the denominator.
Thus if the divisor is an imaginary number then simply multiply the fraction
by i/i, as in the following two examples.
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1 1 + 2i
WORKED EXAMPLE 3: Realise the denominators of: (a) (b)
4i 3i
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SOLUTION:
1 1 i 1 + 2i 1 + 2i i
(a) = × =(b) ×
4i 4i i 3i 3i i
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i i + 2i 2
= 2 =
4i 3i2
2 − i
= − 41 i =
3
If on the other hand the denominator is a complex number then the method is
to multiply top and bottom by its conjugate, as demonstrated here.
C
5 5 + 2i
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## WORKED EXAMPLE 4: Realise the denominators: (a) (b)
2+i 3 − 4i
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SOLUTION:
5 5 2−i 5 + 2i 5 + 2i 3 + 4i
(a) = × (b) = ×
2+i 2+i 2−i 3 − 4i 3 − 4i 3 + 4i
5(2 − i) 15 + 26i − 8
= =
4+1 9 + 16
7 + 26i
=2−i =
25
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
6 CHAPTER 1: Complex Numbers
## REALISING THE DENOMINATOR: There are two cases.
6 A: If the denominator is an imaginary number, multiply top and bottom by i.
B: If the denominator is complex, multiply top and bottom by its conjugate.
There is now significant evidence that the complex numbers form a valid number
system. It has been seen on the previous pages that the four basic arithmetic
operations of addition, subtraction, multiplication and division all behave in a
sensible way, consistent with real arithmetic.
## A Convention for Pronumerals: It is often necessary in developing the theory
of complex numbers to perform algebraic manipulations with unknown complex
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numbers. In order to help distinguish between real and complex variables, the
convention that will be used in this text is that the pronumerals x, y, a and b
will represent real numbers and the pronumerals z and w will represent complex
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numbers. Thus in a statement like “ Let z = x+iy” it is automatically understood
that x and y are real whilst z is complex.
D E
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Real and Imaginary Parts: Given the complex number z = x + iy, the real part
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of z is the real number x, and the imaginary part of z is the real number y. It
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is convenient to define two new functions of the complex variable z for these two
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quantities. Thus
Re(z) = x and Im(z) = y
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## from which it is clear that
z = Re(z) + i Im(z) .
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## SOLUTION: Expanding the quadratic in z first,
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z 2 − iz = (3 − i)2 − i(3 − i)
= 8 − 6i − 3i − 1
= 7 − 9i ,
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2
so Re(z − iz) = 7 .
If two complex numbers z and w are equal, by analogy with surds, it is natural
to expect that Re(z) = Re(w) and Im(z) = Im(w). This is in fact the case.
EQUALITY OF COMPLEX NUMBERS: If two complex numbers z and w are equal then
7
C
N
## Proof: Let z = x + iy and w = a + ib, and suppose that z = w. Then
x + iy = a + ib .
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## Rearranging i(y − b) = a − x . (∗∗)
By way of contradiction, suppose that y − b 6= 0, then
a−x
i= , which is a real number.
y−b
But i is an imaginary number and so there is a contradiction. Thus y − b = 0 and
hence y = b . It follows from equation (∗∗) that x = a , and the proof is complete.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1A The Arithmetic of Complex Numbers 7
The careful reader will have noticed that the definitions of Re(z) and Im(z) given
above are not in terms of the variable z. Both of these functions can be expressed
in terms of z by first writing down z and its conjugate.
z = x + iy
z = x − iy
This pair of simultaneous equations can be solved for x and y to obtain:
Re(z) = 21 (z + z) and Im(z) = 1
2i
(z − z) .
## REAL AND IMAGINARY PARTS: These can be written as functions of z.
8
Re(z) = 21 (z + z) and Im(z) = 1
2i (z − z) .
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The Arithmetic of Conjugates: Since taking the complex conjugate of z simply
changes the sign of the imaginary part, when it is applied twice in succession the
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end result leaves z unchanged. Thus
(z) = x + iy = x − iy = x + iy = z .
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Another important property of taking conjugates is that it commutes with the
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four basic arithmetic operations. For example, with addition,
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(3 + i) + (2 − 4i) = 5 − 3i
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= 5 + 3i ,
and 3 + i + 2 − 4i = 3 − i + 2 + 4i
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= 5 + 3i .
Thus (3 + i) + (2 − 4i) = 3 + i + 2 − 4i .
Notice that it does not matter whether the addition is done before or after taking
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the conjugate, the result is the same. Here is an example with multiplication.
(3 + i)(2 − 4i) = 10 − 10i
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= 10 + 10i ,
and 3 + i × 2 − 4i = (3 − i)(2 + 4i)
= 10 + 10i .
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## Thus (3 + i)(2 − 4i) = 3 + i × 2 − 4i .
Again notice that it does not matter whether the multiplication is done before
or after taking the conjugate, the result is the same. This is always the case for
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## with addition, subtraction, multiplication and division.
9
(a) w+z = w+z (c) wz = w × z
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## (b) w−z = w−z (d) w÷z = w÷z
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The proof of these results is left as a question in the exercise. There are two
special cases of these results. To get the conjugate of a negative, put w = 0
into (b).
(−z) = 0 − z
= 0−z
thus (−z) = −z .
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
8 CHAPTER 1: Complex Numbers
## For the conjugate of a reciprocal, put w = 1 in (d) to get
z −1 = 1 ÷ z
= 1÷z
= 1÷z
thus z = (z)−1 .
−1
Integer Powers: The careful reader will have noted that several of the examples used
above involve powers of a complex number despite the fact that the meaning of z n
has not yet been properly defined. If the index n is a positive integer then the
meaning of z n is analogous to the real number definition. Thus
z n = z| × z ×
{z. . . × z}
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n factors
## or, the recursive definition may be used:
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z1 = z ,
z n = z × z n−1
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for n > 1.
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Just like the real numbers, if z = 0 then z 0 is undefined. For all other complex
G
PR
numbers, z 0 = 1. Again continuing the analogy with the real numbers, a negative
integer power yields a reciprocal. Thus if n is a positive integer then
TE ID
1
z −n = , z 6= 0 .
zn
EC BR
## As with other division by complex numbers, the denominator is usually realised
by multiplying by the conjugate. The case when n = 1 occurs frequently and
should be learnt.
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1 z
z −1 = =
z zz
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Indices which are not integers will not be considered in this text.
Exercise 1A
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## 1. Use the rule given in Box 2 to simplify:
(a) i2 (c) i7 (e) i29 (g) i3 + i4 + i5
(b) i4 (d) i13 (f) i2010 (h) i7 + i16 + i21 + i22
2. Evaluate:
C
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## 3. Express in the form a + ib, where a and b are real.
(a) (7 + 3i) + (5 − 5i) (c) (4 − 2i) − (3 − 7i)
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## (b) (−8 + 6i) + (2 − 4i) (d) (3 − 5i) − (−4 + 6i)
4. Express in the form x + iy, where x and y are real.
(a) (4 + 5i)i (d) (−7 + 5i)(8 − 6i) (g) (2 + i)3
(b) (1 + 2i)(3 − i) (e) (5 + i)2 (h) (1 − i)4
(c) (3 + 2i)(4 − i) (f) (2 − 3i)2 (i) (3 − i)4
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1A The Arithmetic of Complex Numbers 9
5. Use the rule for the sum of two squares given in Box 4 to simplify:
(a) (1 + 2i)(1 − 2i) (c) (5 + 2i)(5 − 2i)
(b) (4 + i)(4 − i) (d) (−4 − 7i)(−4 + 7i)
6. Express in the form x + iy, where x and y are real.
1 5−i −11 + 13i
(a) (c) (e)
i 1−i 5 + 2i
2+i 6 − 7i (1 + i)2
(b) (d) (f)
i 4+i 3−i
7. Let z = 1 + 2i and w = 3 − i. Find, in the form x + iy :
(a) (iz) (b) w + z (c) 2z + iw (d) Im(5i − z) (e) z 2
FS
8. Let z = 8 + i and w = 2 − 3i. Find, in the form x + iy :
(a) z − w (b) Im(3iz + 2w) (c) zw (d) 65 ÷ z (e) z ÷ w
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9. Let z = 2 − i and w = −5 − 12i. Find, in the form x + iy :
10 w
D E
(a) −zw (b) (1 + i)z − w (c) (d) (e) Re (1 + 4i)z
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z 2 − 3i
G DEVELOPMENT
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10. By equating real and imaginary parts, find the real values of x and y given that:
TE ID
(a) (x + yi)(2 − 3i) = −13i (d) x(1 + 2i) + y(2 − i) = 4 + 5i
(b) (1 + 4i)(x + yi) = 6 + 7i x y
EC BR
(e) + = 4+i
(c) (1 + i)x + (2 − 3i)y = 10 2 + i 2 + 3i
11. Express in the form x + iy, where x and y are real.
1 2 3 + 2i 3 − 2i
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(a) + (c) +
1 + i 1 + 2i 2 − 5i 2 + 5i
1+i 3 2 −8 + 5i 3 + 8i
(d) −
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(b) + √ −2 − 4i 1 + 2i
2 1+i 3
12. Given that z = x + iy and w = a + ib, where a, b, x and y are real, prove that:
(a) z + w = z + w 1 1
R
(e) = , z 6= 0
(b) z − w = z − w z z
(c) zw = z w z z
2 (f) = , w 6= 0
(d) z 2 = (z) w w
13. Let z = a + ib, where a and b are real and non-zero. Prove that:
C
2
(a) z + z is real, (c) z 2 + (z) is real,
(b) z − z is imaginary, (d) zz is real and positive.
N
z
14. Let z = a + ib, where a and b are real. If is real, show that z is imaginary or 0.
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z −i
2
15. Prove that if z 2 = (z) then z can only be purely real or purely imaginary.
16. If z = x + iy, where x and y are real, express in the form a + ib, where a and b are written
in terms of x and y .
z −1
(a) z −1 (b) z −2 (c)
z +1
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
10 CHAPTER 1: Complex Numbers
EXTENSION
17. If both z + w and zw are real, prove that either z = w or Im(z) = Im(w) = 0 .
1 1 − 2 cos θ
18. Given that z = 2(cos θ + i sin θ), show that Re = .
1−z 5 − 4 cos θ
1 + sin θ + i cos θ
19. Show that = sin θ + i cos θ .
1 + sin θ − i cos θ
2
20. If z = cos θ + i sin θ, show that = 1 − it, where t = tan θ2 .
1+z
FS
Now that the arithmetic of complex numbers has been satisfactorily developed,
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To reflect the fact that the solutions may be complex, the variable z will be used.
D E
O
G
are the perfect square
PR
(z − λ)2 = 0
TE ID
for which z = λ,
and the difference of two squares
EC BR
z 2 − λ2 = 0
for which z = −λ or λ .
It is now also possible to solve equations involving the sum of two squares, using
the result of Box 4 in Section 1A.
R M
Given z 2 + λ2 = 0
O CA
## (z + iλ)(z − iλ) = 0 (the sum of two squares)
so z = −iλ or iλ .
Thus there are three possible cases for a simple quadratic equation: a perfect
square, the difference of two squares, or the sum of two squares.
R
## WORKED EXAMPLE 6: Find the two imaginary solutions of z 2 + 10 = 0.
SOLUTION: From the sum of two squares
√ √
(z + i 10)(z − i 10) = 0
√ √
so z = −i 10 or i 10
C
## For more general quadratic equations, it is simply a matter of completing the
N
square in z to obtain one of the same three situations: a perfect square, the
difference of two squares, or the sum of two squares.
U
## WORKED EXAMPLE 7: Find the complex solutions of z 2 + 6z + 25 = 0.
SOLUTION: Completing the square:
(z + 3)2 + 16 = 0
so (z + 3 + 4i)(z + 3 − 4i) = 0 (sum of two squares)
thus z = −3 − 4i or − 3 + 4i .
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1B Quadratic Equations 11
Notice that the sum of two squares situation always yields two roots which are
complex conjugates.
## QUADRATIC EQUATIONS WITH REAL COEFFICIENTS: Complete the square in z to obtain
one of the following situations:
10 A. A PERFECT SQUARE: There is only one real root.
B. THE DIFFERENCE OF TWO SQUARES: There are two real roots.
C. THE SUM OF TWO SQUARES: There are two conjugate complex roots.
There are several of ways of proving the assertion that complex solutions to
quadratic equations with real coefficients must occur as conjugate pairs. The
approach presented here will later be extended to encompass all polynomials
FS
with real coefficients.
Proof: Let Q(z) = az 2 + bz + c, where a, b and c are real numbers. Suppose
that the equation Q(z) = 0 has a complex solution z = w. It follows that
O
aw 2 + bw + c = 0 .
D E
Take the conjugate of both sides of this equation to get
O
G
aw 2 + bw + c = 0 .
PR
Now the conjugate of a real number is the same real number. Further, as noted
TE ID
in Box 9, taking a conjugate is commutative with addition and multiplication.
Thus the last equation becomes
a(w)2 + b(w) + c = 0 ,
EC BR
## that is, Q(w) = 0 .
Hence z = w is also a complex root of Q(z) = 0. That is, Q(z) = 0 must have
two conjugate complex roots, z = w and z = w, and the proof is complete.
R M
WORKED EXAMPLE 8: Find a quadratic equation with real coefficients given that
O CA
## one of the roots is w = 5 − i.
SOLUTION: The coefficients are real so the roots occur in conjugate pairs. Hence
the other root is w = 5 + i. Thus the monic quadratic equation is:
R
z − (5 − i) z − (5 + i) = 0
or (z − 5) + i (z − 5) − i = 0
thus (z − 5)2 + 1 = 0 .
2
Finally z − 10z + 26 = 0 .
## In general, a quadratic equation with real coefficients which has a complex
C
root z = α is
N
z 2 − 2 Re(α) z + α α = 0 ,
as can be observed in the three worked examples above. The proof is quite
U
## straight forward, and is one of the questions in the exercise.
coefficients which has a complex root z = α is
11
z 2 − 2 Re(α) z + α α = 0 .
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
12 CHAPTER 1: Complex Numbers
−b ± b2 − 4ac
x= .
2a
There is a problem with this formula √when complex numbers are involved. When
applied to real numbers, the symbol means the positive square root, but it is
unclear what “positive” means when applied to complex numbers. It might be
tempting to say that i is positive and −i is negative, but then what is to be said
about numbers like (−1 + i) or (1 − i)? In short, it does not make sense to speak
of positive and negative complex numbers, and so the positive square root has
no meaning. Thus it is not appropriate to blindly use the quadratic formula to
solve an equation with complex roots.
FS
Recall that the quadratic formula arose from applying the method of completing
the square. Here that process is reviewed.
Given az 2 + bz + c = 0 ,
O
z 2 + ab z = − ac
D E
O
b 2
so z + 2a ∆
= (2a) 2 , where ∆ = b2 − 4ac .
G
Now suppose there exists a number λ, possibly complex, such that ∆ = λ2 .
PR
TE ID
b 2
λ 2
Then z + 2a − 2a =0
whence z + b+λ
2a z + b−λ
2a =0 (the difference of two squares)
EC BR
−b−λ −b+λ
and so z= 2a or 2a .
Thus if there is a number λ, possibly complex, for which λ2 = ∆, then the
solution to the quadratic equation can be written using the last line above. If the
R M
O CA
## THE QUADRATIC METHOD: Use the following steps to solve az 2 + bz + c = 0 .
1. First find ∆ = b2 − 4ac .
12 2. Next find a number λ, possibly complex, such that λ2 = ∆ .
−b − λ −b + λ
R
2a 2a
## WORKED EXAMPLE 9: Solve z 2 + 2z + 6 = 0 .
SOLUTION: ∆ = 22 − 4 × 1 × 6
C
= −20
√ 2
N
= 2i 5 ,
√ √
−2 − 2i 5 −2 + 2i 5
U
hence z= or
2√ 2√
= −1 − i 5 or − 1 + i 5 .
Complex Square Roots: Before extending the above work to the case of a quadratic
equation with complex coefficients, it is necessary to develop methods for finding
the square roots of complex numbers.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1B Quadratic Equations 13
The first thing to notice is that, just like real numbers, every complex number has
two square roots. The proof is quite straight forward. Suppose that the complex
number z is a square root of another complex number w then
z2 = w .
Further (−z)2 = z 2
= w.
Hence w has a second square root which is the opposite of the first, namely (−z).
Thus for example −2i has two opposite square roots, (1 − i) and (−1 + i). This
is not really very surprising since all real numbers (other than zero) have two
opposite square √ roots. For
√ example, 9 has square roots 3 and −3, whilst −5 has
square roots i 5 and −i 5 . The proof that there are no more than two square
FS
roots is left as an exercise.
O
Complex Square Roots and Pythagoras: A simple way to find complex square
roots is to equate the real and imaginary parts of z 2 = w in order to obtain a
D E
O
pair of simultaneous equations.
G
(x + iy)2 = a + ib , where x, y, a and b are real,
PR
Given
x2 − y 2 + 2ixy = a + ib .
TE ID
Equating real and imaginary parts yields
x2 − y 2 = a
EC BR
and xy = 21 b .
In many cases this pair of equations can be easily solved by inspecting the factors
of 21 b, as in the following example.
R M
O CA
R
## then (x2 − y 2 ) + 2ixy = 7 + 24i .
Equating real and imaginary parts yields the simultaneous equations
x2 − y 2 = 7
and xy = 12 .
C
## y = −3. Hence the square roots of 7 + 24i are the opposites
N
4 + 3i and − 4 − 3i .
U
Some readers will have noticed in the above example that 7 and 24 are the first
two numbers of the Pythagorean triad 7, 24, 25. This is no coincidence.
p It is
often the case that if b is even and the numbers |a|, |b| and a2 + b2 form a
Pythagorean triad then the resulting equations for x and y can be simply solved
by inspecting the factors of 21 b .
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
14 CHAPTER 1: Complex Numbers
COMPLEX SQUARE ROOTS AND PYTHAGORAS: Given (x + iy)2 = a + ib, equate the real
and imaginary parts to get the simultaneous equations
x2 − y 2 = a
13 1
xy = 2
b.
p
If b is even and the numbers |a|, |b| and a2 + b2 form a Pythagorean triad
then these equations can often be solved by inspecting the factors of 21 b .
with complex coefficients can now be solved. All that is needed is to combine the
above method for finding the roots of a complex number with either the method
FS
of completing the square or the quadratic method in Box 12.
## WORKED EXAMPLE 11: Complete the square to solve z 2 −(2+6i)z+(−5+2i) = 0.
O
SOLUTION: Rearranging
D E
z 2 − 2(1 + 3i)z = 5 − 2i
O
so (z − (1 + 3i))2 = (1 + 3i)2 + 5 − 2i
G
PR
= −8 + 6i + 5 − 2i ,
2
TE ID
thus (z − (1 + 3i)) = −3 + 4i .
Let (x + iy)2 = −3 + 4i
then x2 − y 2 = −3
EC BR
and xy = 2
so by inspection one solution is x = 1 and y = 2 .
Hence (z − (1 + 3i))2 = (1 + 2i)2
R M
## and thus z = (1 + 3i) + (1 + 2i) or (1 + 3i) − (1 + 2i)
that is z = 2 + 5i or i .
O CA
Whilst the focus here is on the quadratic method and completing the square,
those two methods should only be used when required. It is always preferable to
solve a quadratic equation by factors whenever they can be easily identified.
R
## WORKED EXAMPLE 12: Solve z 2 + 4i z − 3 = 0 .
SOLUTION: Noting that −3 = i × 3i and 4i = i + 3i ,
(z + i)(z + 3i) = 0
hence z = −i or − 3i .
C
Harder Complex Square Roots: Often the simultaneous equations given in Box 13
N
## cannot be solved by inspection. Fortunately there is an identity that can be used
to help solve these equations. Recall that if
U
(x + iy)2 = a + ib
then x2 − y 2 = a (1)
and 2xy = b . (2)
a2 + b2 = (x2 − y 2 )2 + (2xy)2
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1B Quadratic Equations 15
= (x2 )2 + 2x2y 2 + (y 2 )2
= (x2 + y 2 )2 .
p
Hence x2 + y 2 = a2 + b2 (3)
Equations (1) and (3) now form a very simple pair of simultaneous equations to
solve. Equation (2) is used to determine whether x and y have the same sign,
when b > 0, or opposite sign, when b < 0.
## SQUARE ROOTS OF A COMPLEX NUMBER: Given (x + iy)2 = a + ib then x and y are
solutions of the pair of simultaneous equations
14 x2 − y 2 = a
p
x2 + y 2 = a2 + b2
FS
with the same sign if b is positive, and opposite sign if b is negative.
O
WORKED EXAMPLE 13: Determine the two square roots of −4 + 2i .
D E
SOLUTION: Let (x + iy)2 = −4 + 2i . Since Im(−4 + 2i) > 0, x and y have the
O
same sign. Further, (−4)2 + 22 = 20, so solve
G
PR
x2 − y 2 = −4 (1)
x2 + y 2 = 2 5
TE ID
and (2)
EC BR
2x2 = −4 + 2 5
q q
√ √
so x = − −2 + 5 or −2 + 5 .
R M
## Subtracting (1) from (2) yields
2y 2 = 4 + 2 5
O CA
q q
√ √
so y = − 2 + 5 or 2 + 5.
q q q q
√ √ √ √
Hence x + iy = − −2 + 5 − i 2 + 5 or −2 + 5 + i 2 + 5 .
R
In fact, the result in Box 14 can be used to develop a formula for the square
roots of any complex number, which is derived in one of the Exercise questions.
However that formula is not part of the course and should not be memorised.
those with complex discriminants. Box 14 is used to find the square roots of
C
N
## WORKED EXAMPLE 14: [A Hard Example] Solve z 2 + (4 − 2i)z + 1 = 0 by
U
SOLUTION: ∆ = (4 − 2i)2 − 4
= 12 − 16i − 4
= 8 − 16i .
2
Let (x + iy) = 8 − 16i.
Now Im(8 − 16i) < 0 so x and y have opposite sign, with
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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16 CHAPTER 1: Complex Numbers
x2 − y 2 = 8 (1)
p
and x2 + y 2 = 82 + 162
or x2 + y 2 = 8 5 (2)
Adding and subtracting equations (1) and (2) yields
√ √
2x2 = 8 + 8 5 2y 2 = −8 + 8 5
√ and √
x2 = 4(1 + 5) y 2 = 4(−1 + 5) .
√ 2
q q
Thus ∆= 2 1 + 5 − 2i −1 + 5
q q
1
√ √
and so z = 2 −4 + 2i + 2 1 + 5 − 2i −1 + 5
FS
q q
1
√ √
or 2 −4 + 2i − 2 1 + 5 + 2i −1 + 5
q q
√ √
O
that is z= −2 + 1 + 5 + i 1 − −1 + 5
D E
q q
√ √
O
or −2 − 1 + 5 + i 1 + −1 + 5
G
PR
TE ID
Exercise 1B
1. Solve for z.
EC BR
## (a) z 2 + 9 = 0 (c) z 2 + 2z + 5 = 0 (e) 16z 2 − 16z + 5 = 0
(b) (z − 2)2 + 16 = 0 (d) z 2 − 6z + 10 = 0 (f) 4z 2 + 12z + 25 = 0
2. Write as a product of two complex linear factors.
R M
## (a) z 2 + 36 (c) z 2 − 2z + 10 (e) z 2 − 6z + 14
(b) z 2 + 8 (d) z 2 + 4z + 5 (f) z 2 + z + 1
O CA
3. Form a quadratic equation with real coefficients given that one root is:
√ √
(a) i 2 (b) 1 − i (c) −1 + 2i (d) 2 − i 3
R
4. In each case, find the two square roots of the given number by the inspection method.
(a) 2i (c) −8 − 6i (e) −5 + 12i (g) −15 − 8i
(b) 3 + 4i (d) 35 + 12i (f) 24 − 10i (h) 9 − 40i
DEVELOPMENT
C
N
## (b) Hence solve z 2 − 3z + (3 + i) = 0.
6. (a) Find the two square roots of −8 + 6i.
U
## (b) Hence solve z 2 − (7 − i)z + (14 − 5i) = 0.
7. Use the method outlined in Box 11 to solve for z.
(a) z 2 − z + (1 + i) = 0 (d) (1 + i)z 2 + z − 5 = 0
2
(b) z + 3z + (4 + 6i) = 0 (e) z 2 + (2 + i)z − 13(1 − i) = 0
(c) z 2 − 6z + (9 − 2i) = 0 (f) iz 2 − 2(1 + i)z + 10 = 0
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1B Quadratic Equations 17
## 8. (a) Find the value of w if i is a root of the equation z 2 + wz + (1 + i) = 0.
(b) Find the real numbers a and b given that 3−2i is a root of the equation z 2 +az +b = 0.
(c) Given that 1 − 2i is a root of the equation z 2 − (3 + i)z + k = 0, find k and the other
root of the equation.
z 1
9. Find the two complex numbers z satisfying zz = 5 and = (3 + 4i).
z 5
10. (a) Solve z 2 − 2z cos θ + 1 = 0 for z by completing the square.
1 1
(b) Rearranging the equation in part (a) gives cos θ = z+ . Confirm this result
2 z
for each of the solutions to part (a) by substitution.
11. By first factoring the sum or difference of two cubes, solve for z.
FS
(a) z 3 = −1 (b) z 3 + i = 0
12. Consider the quadratic equation az 2 +bz +c = 0, where a, b and c are real and b2 −4ac < 0.
O
Suppose that ω is one of the complex roots of the equation.
(a) Explain why aω 2 + bω + c = 0.
D E
O
(b) By taking the conjugate of both sides of the result in (a), and using the properties of
2
G
conjugates, show that a (ω) + b ω + c = 0.
PR
(c) What have you just proved about the two complex roots of the equation?
TE ID
13. Suppose that z = α is a complex solution to a quadratic equation with real coefficients.
(a) Which other number is also a solution of this quadratic equation?
(b) Hence prove that one such quadratic equation is z 2 − 2 Re(α)z + α α = 0.
EC BR
## 14. Let (x + iy)2 = a + ib, then we have x2 − y 2 = a and 2xy = b .
(a) For the moment, assume that both a and b are positive.
R M
(i) Sketch the graphs of these two equations on the same number plane.
(ii) What feature of your sketch indicates that there are two square roots of a + ib?
O CA
(b) Investigate how the sketch changes when either a or b or both are negative or zero.
15. Use the results of Box 13 to find the two square roots of:
(a) −i (b) −6 + 8i (c) 2 + 2i 3 (d) 10 − 24i (e) 2 − 4i
R
16. Find the discriminant and its square roots, and hence solve:
√ √
(a) z 2 + (4 + 2i)z + (1 + 2i) = 0 (c) z 2 + 2(1 − i 3)z + 2 + 2i 3 = 0
(b) z 2 − 2(1 + i)z + (2 + 6i) = 0 (d) z 2 + (1 − i)z + (i − 1) = 0
EXTENSION
C
## 17. Let α and β be the two complex roots of z 3 = 1. Show that:
(a) β = α, (b) α2 = β and β 2 = α, (c) 1 + α + α2 = 0,
N
(d) the sum of the first n terms of the series 1 + α + α + α + . . . is either 0, 1 or −α2 ,
2 3
## depending on the remainder when n is divided by 3.
U
18. Let a, b and c be real with b2 − 4ac < 0, and suppose that the quadratic equation
az 2 + bz + c = 0 has complex solutions α = x + iy and β = u + iv .
(a) By considering the sum and product of the roots, show that
Im(α + β) = 0 and Im(αβ) = 0 .
(b) Hence show that α = β .
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
18 CHAPTER 1: Complex Numbers
19. Let (x + iy)2 = a + ib, where b 6= 0. Use the result of Box 13 to prove the formula:
s s !
1 p 2 b 1 p
x + iy = ± a + b2 + a + i a2 + b2 − a .
2 |b| 2
Explain the significance of the term b/|b| in this formula.
## 1C The Argand Diagram
Mathematics requires a knowledge of numbers, and throughout high school that
understanding of numbers has been enhanced by being able to plot them on a
number line, to visualise their properties and relationships. Initially there were
FS
the natural numbers, shown at discrete intervals on the number line.
0 1 2 3 4 x
O
When negative numbers were included to create the integers, the number line
D E
was extended to the left of the origin to show these new numbers.
O
G
PR
-3 -2 -1 0 1 2 3 x
TE ID
Next came the rationals, the fractions which exist in the spaces between integers.
Eventually the irrationals were discovered and included, which fit in the √ gaps
that are somehow left between rationals. Some irrational numbers like 2 can
EC BR
## be constructed geometrically, but others like e and
√ π can only be approximated
to so many decimal places. The construction for 2 is shown here along with the
positions of − 21 , 34 , e and π.
R M
p
-1 - 12 0 3
1 Ö2 2 e3 x
O CA
The number line is now full, the reals have filled it up, and there is no space left
for any new objects like complex numbers. Further, since complex numbers come
in two parts, real and imaginary, there is no satisfactory way of representing them
R
## on a number line. A two dimensional representation is needed.
The Complex Number Plane: Keeping to things that are familiar, the number
plane would seem to be a convenient way to represent complex numbers. More
formally, for each complex number z = x+iy there corresponds a point Z(x, y) in
the Cartesian plane. Equally, given any point W (a, b) in the real number plane,
C
N
## Thus in the diagram on the right, the complex numbers y
4 + i and −3 + 3i are represented by the points A and B B 3
U
## respectively. The points C and D represent the complex
numbers −1 − 2i and 2 − 4i. Several different names 1 A
are used to describe a coordinate plane that is used to -3 -1 2 4 x
represent complex numbers. One name is the Argand C -2
diagram, after the French mathematician Jean-Robert
Argand, born in Geneva in 1768. The terms complex -4 D
number plane or z-plane are also used.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1C The Argand Diagram 19
THE ARGAND DIAGRAM: The complex number z = x + iy is associated with the point
15 Z(x, y) in the real number plane. A complex number may be represented by
a point, and a point may be represented by a complex number.
## As a convenient abbreviation, the point Z(x, y) will sometimes be simply referred
to as the point z in the Argand diagram. It is important to remember that
the complex number plane is just a real number plane which is used to display
complex numbers. By the nature of this representation, if two complex numbers
are equal then they represent the same point. The converse is also true.
The Real and Imaginary Axes: If Im(z) = 0, that is z = x + 0i, then z is a real
number and the corresponding point Z(x, 0) in the Argand diagram lies on the
FS
horizontal axis. Thus the horizontal axis is called the real axis.
Likewise, if Re(z) = 0, that is z = 0 + iy, then z is an imaginary number and
the corresponding point Z(0, y) in the Argand diagram lies on the vertical axis.
O
Thus the vertical axis is called the imaginary axis.
D E
O
Some Simple Geometry: Now that the complex plane has been introduced, it is
G
immediately possible to observe the geometry of some simple complex number
PR
operations. The simplest of these are the geometries of conjugates, opposites,
and multiplication by i, which are now investigated.
TE ID
y
Let z = x + iy , then the conjugate is Z1
2
z = x − iy ,
EC BR
1
that is, y has been replaced by −y. This was encountered x
in the work on graphs and is known to be a reflection in -1 1 2 3
the real axis. This is clearly evident in the example of -2
Z2
R M
## z1 = 3 + 2i and z2 = 3 − 2i = z1 shown on the right.
O CA
THE GEOMETRY OF CONJUGATES: The points z and z in the Argand diagram are
16
reflections of each other in the real axis.
## Next, let z = x + iy , then the opposite is
R
y
−z = −x − iy .
In this case, x and y have been replaced by −x and −y Z2 2
respectively. Thus the result is obtained by reflecting the p
point in both axes in succession. Alternatively, it is a -1 1 x
rotation by π about the origin. The diagram on the right -2 Z1
with z1 = 1 − 2i and z2 = −1 + 2i = −z1 demonstrates
C
this rotation.
N
THE GEOMETRY OF OPPOSITES: The points z and −z in the Argand diagram are
17
rotations of each other by π about the origin.
U
## Now let z1 = a + ib , then z2 = i z1 is given by
y
z2 = −b + ia .
Z2(-b, a)
Consider the corresponding points Z1 and Z2 shown in
the Argand diagram on the right, where neither a nor b Z1(a, b)
is zero. The product of the gradients of OZ1 and OZ2 is
O x
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
20 CHAPTER 1: Complex Numbers
b a
× = −1 .
a −b
Hence OZ2 is perpendicular to OZ1 and the conclusion is that multiplication by i
is equivalent to an anticlockwise rotation by π2 about the origin. The situation is
the same even when z1 is real or imaginary, but not zero, and the proof is left as
an exercise.
## THE GEOMETRY OF MULTIPLICATION BY i: The point iz in the Argand Diagram is the
18
result of rotating the point z by π2 anticlockwise about the origin.
## Note that multiplication by i twice in succession yields a rotation of 2 × π2 = π.
This is consistent with the geometry of opposites, since i(iz) = i2 z = −z.
FS
WORKED EXAMPLE 15: Let z = x+iy. Determine i z and hence give a geometric
interpretation of the result.
SOLUTION: i z = i(x − iy)
O
= y + ix .
D E
O
This is just z with x and y swapped. Thus it is a reflection in the line y = x.
G
Curves in the Argand Diagram: So far attention has been given to individual
PR
points in the complex plane. Often an equation in z will correspond to a well
TE ID
known line or curve in the Argand diagram. In the simple cases dealt with here,
the equation can be found by putting z = x + iy.
EC BR
## WORKED EXAMPLE 16: Graph the following: y
(a) Re(z) = 2, (b) Im(z) = −1. Re(z) = 2
R M
## (a) the vertical line x = 2, and
(b) the horizontal line y = −1, -1 Im(z) = -1
O CA
## as shown in the diagram on the right.
Note that these two lines intersect at z = 2 − i.
R
## VERTICAL AND HORIZONTAL LINES: In the Argand diagram:
• the equation Re(z) = a is the vertical line x = a
19
• the equation Im(z) = b is the horizontal line y = b
• these two lines intersect at z = a + ib.
C
## plane represent the number z = x+iy. Given that z z = 9, y
find the curve that P is on, and sketch it. 3
N
## SOLUTION: The given equation becomes
3
U
(x + iy)(x − iy) = 9 O x
so x2 + y 2 = 3 2
that is, a circle with centre the origin and radius 3.
In some examples it is best to manipulate the given equation in z first, and then
substitute x+iy. It is also important to note any restrictions on z before starting.
Both of these points feature in the following example.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1C The Argand Diagram 21
## WORKED EXAMPLE 18: Find and describe the curve specified by
1 1
+ = 1.
z z
SOLUTION: Note that in the given equation z 6= 0, since y
the LHS is undefined there. Multiply both sides by the
lowest common denominator to get 1
z+z = zz
1 2 x
so 2x = x2 + y 2 -1
or 0 = x2 − 2x + y 2
thus 1 = (x − 1)2 + y 2
that is, the circle with radius 1 and centre (1, 0), excluding the origin.
FS
Exercise 1C
O
1. Write down the coordinates of the point in the complex plane that represents:
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(a) 2 (c) −3 + 5i (e) −5(1 + i)
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(b) i (d) 2 + 2i (f) (2 + i)i
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2. Write down the complex number that is represented by the point:
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(a) (−3, 0) (b) (0, 3) (c) (7, −5) (d) (a, b)
3. Let z = 1 + 3i, and let A, B, C and D be the points representing z, iz, i2z and i3 z
respectively.
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## (a) Plot the points A, B, C and D in the complex plane.
(b) What type of special quadrilateral is ABCD?
(c) What appears to be the geometric effect of multiplying a complex number by i?
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4. Let z = 3 + i and w = 1 + 2i. Plot the points representing each group of complex numbers
on separate Argand diagrams, and describe any geometry you observe.
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## (a) z, iz, −z, −iz (c) z, z, w, w (e) z, w, z − w
(b) w, iw, −w, −iw (d) z, w, z + w (f) z, w, w − z
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## 5. Graph the following sets of points in the Argand diagram.
(a) Re(z) = −3 (c) Im(z) < 1 (e) Re(z) = Im(z) (g) Re(z) ≤ 2 Im(z)
(b) Im(z) = 2 (d) Re(z) ≥ −2 (f) 2 Re(z) = Im(z) (h) Re(z) > − Im(z)
DEVELOPMENT
6. Let the point P represent the complex number z = 2(cos π6 + i sin π6 ), and let the points
C
Q, R, S and T , represent z, −z, iz and z1 respectively. Plot all these points on an Argand
diagram.
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7. Show that the point representing −z is a reflection of the point representing z in the y-axis.
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## 8. Consider the points represented by the complex numbers z, z, −z and −z .
Show that these points form a rectangle by using:
(a) coordinate geometry to show that the diagonals are equal and bisect each other,
(b) the geometry of conjugates and opposites.
π
9. In the text it was proven that when z is complex, iz is a rotation by 2
Prove the same result when z is: (a) real, (b) imaginary.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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22 CHAPTER 1: Complex Numbers
10. The numbers z = a + ib and w = iz are plotted in the complex plane at A and B
respectively.
(a) By considering gradients, show that OA ⊥ OB.
(b) Use the distance formula to show that OA = OB.
(c) What type of triangle is 4OAB?
11. The point P in the complex plane represents the number z. Find and describe the curve
that P is on given that
1 1
− = i.
z z
12. The complex number z is represented by the point C in the Argand diagram. Find and
describe the curve that C is on if
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z −6
Re = 0.
z
13. Show that (z − 2) (z − 2) = 9 represents a circle in the Argand diagram.
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14. Find and describe the curve in the Argand diagram specified by
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2
zz = Re(z − 1 + 3i) .
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EXTENSION
15. Sketch the curve in the complex plane specified by the given equation.
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(a) Im z 2 = 2c2 (b) Re z 2 = c2
16. Show that the point representing −iz is a reflection of the point representing z in y = −x.
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1
17. Show that z is a reflection and enlargement of z.
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1D Modulus-Argument Form y
P(x, y)
Recall that in the study of trigonometry it was found
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r
that the location of a point P could be expressed either q
in terms of its horizontal and vertical positions, x and y, O x
or in terms of its distance OP = r from the origin and the
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## angle θ that the ray OP makes with the positive x-axis.
The situation is shown in the number plane on the right.
## The Modulus and Argument of a Complex Number: In the Argand diagram
the distance r is called the modulus of z, and owing to its geometric definition as
a distance it is written as |z|. On squaring:
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|z|2 = r 2
= x2 + y 2
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## = (x + iy)(x − iy) (sum of two squares)
hence |z|2 = zz .
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The angle θ is called the argument of z, and is written θ = arg(z). Just as with
trigonometry, θ can take infinitely many values for the same point P . It is often
necessary to restrict the angle to just one value called the principal argument,
which is written Arg(z). The value of the principal argument is always in the
range −π < Arg(z) ≤ π. Note the strict inequality on the left hand side, and the
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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CHAPTER 1: Complex Numbers 1D Modulus-Argument Form 23
## There is a problem with measuring θ when z = 0 because then P coincides with
the origin and there is no angle to measure. For this reason, both arg(0) and
Arg(0) are undefined. However, it should be clear |0| is defined and that |0| = 0.
## MODULUS AND ARGUMENT: Let P represent the complex number z = x + iy in the
Argand diagram, with origin O.
• The modulus of z is the distance |z| = r = OP . Note that |z|2 = zz.
20 • The argument of z 6= 0 is any angle arg(z) = θ that the ray OP can make
with the positive real axis. However, arg(0) is undefined.
• The principal argument of z 6= 0 is the unique angle Arg(z) = θ which is in
the range −π < θ ≤ π. However, Arg(0) is undefined.
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From the trigonometric definitions it is clear that
x = r cos θ (1)
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and y = r sin θ , (2)
from which it follows that
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z = r cos θ + ir sin θ .
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PR
Notice that the modulus r is a common factor in this last expression and it is
more commonly written as
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z = r(cos θ + i sin θ)
or z = r cis θ for short.
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## In order to contrast the two ways of writing a complex number, z = x + iy
is called real-imaginary or Cartesian form whilst z = r(cos θ + i sin θ) is called
modulus-argument form, or mod-arg form for short. Another name for mod-arg
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form is polar form, because the radius is measured from the origin which acts as
a pole. Equations (1) and (2) above serve to link the two forms.
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## WORKED EXAMPLE 19: Express each complex number in real-imaginary form.
(a) z = 4 cis π (b) z = 2 cis π6 (c) z = cis 2π
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SOLUTION:
(a) z = 4 cos π + 4i sin π (b) z = 2 cos π6 + 2i sin π6 (c) z = cos 2π
3
+ i sin 2π
3
√ √
= −4 = 3+i = − 12 + 3
i
2
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WORKED EXAMPLE 20: Express each complex number in mod-arg form using
the principal argument. In part (c) give Arg(z) correct to two decimal places.
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## (a) z = 5i (b) z = 3 − 3i (c) z = −4 − 3i
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SOLUTION: In each case let z = r cis θ, with Z the point in the Argand diagram.
(a) In this case Z is on the positive imaginary axis so
r=5
and θ = π2 ,
hence z = 5 cis π2 .
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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24 CHAPTER 1: Complex Numbers
## (b) Now Z is in the fourth quadrant with y
r 2 = 32 + 32 1 2 3
√ q x
or r = 3 2. -1
1 -2 r
Thus cos θ = √ -3 Z
2
π
and θ = −4 ,
√
hence z = 3 2 cis − π4 . y
(c) In this case there is a Pythagorian triad so -4 -2
r = 5. x
r q
Now cos θ = − 54 and Z is in the third quadrant
. Z -3
θ = −π + cos−1 54 =
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.
hence z=. 5 cis(−2·50) .
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FORMS OF A COMPLEX NUMBER:
• x + iy is called the real-imaginary form or Cartesian form of z.
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• r(cos θ + i sin θ) = r cis θ is called the modulus-argument form of z.
21
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• The equations relating the two forms are:
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x = r cos θ and y = r sin θ
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Note that some people prefer to use tan θ = xy to find Arg z. This alternative
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formula should only be used when the quadrant is known for θ, otherwise there
is a potential problem. By way of example, suppose that z = k(1 + i) for some
constant k. The alternative formula would then give
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tan θ = 1 .
It would be tempting at this point to write Arg z = π4 . Whilst this is correct
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when k > 0, it is wrong when k ≤ 0. In fact when k < 0, z lies in the third
quadrant and so Arg z = − 3π
4
. And, of course, Arg z is undefined when k = 0.
Some Simple Algebra: As will be revealed over the remainder of this chapter, the
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use of mod-arg form is a powerful tool, both in simplifying much algebra and in
providing geometric interpretations.
Perhaps the most obvious thing to notice is that | cis θ| = 1. The geometry of
the situation makes the result obvious since if z = cos θ + i sin θ then the point
Z(cos θ, sin θ) lies on the unit circle. Hence |z| = OZ = 1. Here is an algebraic
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## derivation of the same result.
| cos θ + i sin θ|2 = cos2 θ + sin2 θ
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= 1,
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hence | cis θ| = 1 .
This identity has immediate applications in quadratic equations. If cis θ is one
of the roots, then the constant term of the monic quadratic must be 1, as the
following worked example demonstrates.
WORKED EXAMPLE 21: Find a quadratic equation with real coefficients given
that one root is z = cis θ .
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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CHAPTER 1: Complex Numbers 1D Modulus-Argument Form 25
SOLUTION: The coefficients are real so the other root must be cis θ. Thus:
(z − cis θ)(z − cis θ) = 0
2
or z − (cis θ + cis θ)z + cis θ × cis θ = 0
that is z 2 − 2 Re(cis θ)z + | cis θ|2 = 0
thus z 2 − 2z cos θ + 1 = 0 .
## The Product of Two Complex Numbers: The modulus-argument form of the
product of two numbers is a particularly important result. Let w = a cis θ and
z = b cis φ, with a 6= 0 and b 6= 0. Then:
wz = a(cos θ + i sin θ) × b(cos φ + i sin φ)
= ab (cos θ cos φ − sin θ sin φ) + i(cos θ sin φ + sin θ cos φ)
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= ab cos(θ + φ) + i sin(θ + φ) ,
= ab cis(θ + φ) .
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Thus |wz| = ab and arg(wz) = θ + φ .
This yields the following two significant results:
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|wz| = |w| |z|
and
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arg(wz) = arg(w) + arg(z) .
PR
TE ID
WORKED EXAMPLE 22: Let w = 3 + i and z = 1 + i.
(a) Evaluate wz in real-imaginary form.
(b) Express w and z in mod-arg form and hence evaluate wz in mod-arg form.
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## (c) Hence find the exact value of cos 5π
12
.
SOLUTION:
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(a) wz = ( 3 + i)(1 + i)
√ √
= ( 3 − 1) + i( 3 + 1) .
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## (b) Now w = 2 cis π6
z = 2 cis π4 ,
hence wz = 2 2 cis( π6 + π4 )
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= 2 2 cis 5π
12
.
(c) Equating the real parts of parts (a) and (b) yields
√ √
2 2 cos 5π
12
= 3−1
5π 3−1
hence cos 12 = √ .
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2 2
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THE PRODUCT OF TWO COMPLEX NUMBERS: Let w and z be two complex numbers.
• The modulus of the product is the product of the moduli, that is:
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## |wz| = |w| |z| .
22
• The argument of the product is the sum of the arguments, that is:
arg(zw) = arg(w) + arg(z) (provided w 6= 0 and z 6= 0 .)
A question in the exercise deals with the case of division of complex numbers.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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26 CHAPTER 1: Complex Numbers
## Some Simple Geometry Again: It is instructive to re-examine the geometry of
conjugates, opposites and multiplication by i using mod-arg form. Beginning
with the conjugate, recall that the result is a reflection in the real axis. Thus the
modulus should be unchanged, and the argument should be opposite.
Let z = r cis θ then
z = r cos θ − ir sin θ
= r cos(−θ) + ir sin(−θ)
= r cis(−θ)
hence |z| = |z|
and arg(z) = − arg(z)
that is, the modulus is unchanged and the angle is opposite, as expected.
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The cases of opposites and multiplication by i are more simply dealt with. Recall
that these operations represented rotations in the complex plane by π and π2
respectively. Thus, again, the modulus should be the same, and the argument
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should be increased appropriately. Looking at opposites first:
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| − z| = |(−1) × z| = | − 1| × |z| = |z| ,
and arg(−z) = arg(−1 × z) = arg(−1) + arg(z) = π + arg(z) .
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That is, the moduli of opposites are equal and the arguments differ by π.
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Similarly |iz| = |i| |z| = |z| ,
π
and arg(iz) = arg(i) + arg(z) = 2 + arg(z) .
That is, the moduli of z and iz are equal and the arguments differ by π2 . In both
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cases the results are exactly as expected. Also notice that the principal arguments
of −1 and i have been used. As an exercise, justify why this is correct.
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The Geometry of Multiplication: Aside from the special cases above, the geometry
of multiplication is evident in the results of Box 22. The product of the moduli
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indicates an enlargement with centre the origin, and the sum of the arguments
represents an anticlockwise rotation about the origin.
Consider these two transformations individually and let w = r cis θ. When θ = 0
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## the product wz reduces to wz = rz, which is an enlargement without any rotation.
Thus both z and rz lie on the same ray.
When |w| = r = 1 the product wz becomes wz = z cis θ. Using Box 22:
|wz| = |w||z|
= |z|
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## and arg(wz) = arg w + arg z
= θ + arg z .
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This is simply a rotation without any enlargement. Thus z and z cis θ both lie on
a circle of radius |z|. The following example serves to demonstrate the situation.
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## WORKED EXAMPLE 23: Let z = 1 + i.
(a) Find, in Cartesian form, the complex number w such that wz is a rotation
of z by π3 about the origin.
(b) Evaluate wz in Cartesian form.
(c) Verify that |wz| = |z|, then plot z and wz on an Argand diagram.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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CHAPTER 1: Complex Numbers 1D Modulus-Argument Form 27
SOLUTION:
(a) Clearly w = cis π3
= 12 (1 + i 3) .
(b) wz = 12 (1 + i 3)(1 + i)
√ √
= 12 (1 − 3) + i(1 + 3) . y
√ √ wz
(c) Now |wz|2 = 41 (1 − 3)2 + (1 + 3)2 z
p
3
= 41 (1 + 3 + 1 + 3) Ö2
= 2. O x
Hence |wz| = 2
= |z| .
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THE GEOMETRY OF MULTIPLICATION: Let w = r cis θ then the complex number wz
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23 is the result of a rotation of z by θ anti-clockwise about the origin and an
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enlargement of z by factor r with centre the origin.
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The corresponding geometry for the division of complex numbers is similar and
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is dealt with in one of the exercise questions.
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Shifting in the Complex Plane: Recall that, for a real number x, the value of |x|
is the distance from the origin to x. Shifting this, |x − a| is the distance from a
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to x. Although it will not be proven here, the results for shifting can also be
applied to the Argand diagram. Thus since |z| is the distance from the origin
to z, it follows that |z − w| is the distance from w to z. An algebraic proof of
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## this important result is the subject of a question in the exercise.
y
Likewise, since arg(z) is the angle at the origin between z
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## and the positive real axis, the value of arg(z − w) is the z
angle at the vertex w between z and the right half of the |z-w|
horizontal line through w. The situation is shown in the
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## diagram on the right. arg(z-w)
w
WORKED EXAMPLE 24: Let points W and Z represent x
w = 2 + 2i and z = −1 + 5i respectively. Find:
(a) the length of W Z, y
Z 5
(b) the angle θ that W Z makes with the positive x-axis.
C
arg(z-w)
SOLUTION: First note that z − w = −3 + 3i.
2
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W
(a) W Z = |z − w| q
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= | − 3 + 3i| -1 2 x
=3 2
(b) θ = Arg(z − w) (corresponding angles, parallel lines)
= Arg(−3 + 3i)
= .
4
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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28 CHAPTER 1: Complex Numbers
## SHIFTING IN THE COMPLEX PLANE: Functions of complex numbers can be shifted in a
similar way to functions of real numbers. In particular:
24 • |z − w| is the distance from w to z.
• arg(z − w) is the angle at the vertex w between z and the right half of the
horizontal line though w.
Exercise 1D
1. Find |z| given:
(a) z = 3 (c) z = 1 − i (e) z = −3 + 4i
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(b) z = −5i (d) z = − 3 − i (f) z = 15 + 8i
## 2. Find Arg(z) given:
(a) z = −2 (c) z = 2 − 2i (e) z = −3 + 3i
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√ √
(b) z = 4i (d) z = 1 + 3 i (f) z = − 3 − i
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3. Express each complex number in the form r(cos θ + i sin θ), where r > 0 and −π < θ ≤ π .
(a) 2i G (c) 1 + i (e) −1 + 3 i
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(b) −4 (d) 3 − i (f) − √12 − √12 i
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4. Repeat the previous question for each of these complex numbers, writing θ in radians
correct to two decimal places.
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## (a) 3 + 4i (b) 12 − 5i (c) −2 + i (d) −1 − 3i
5. Express in the form a + ib, where a and b are real.
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## (a) 3 cis 0 (c) 4 cis π4 (e) 2 cis 3π
4
(b) 5 cis − π2 (d) 6 cis − π6 (f) 2 cis − 2π
3
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## 6. Given that z = 1 − i, express in mod-arg form:
−1
(a) z (b) z (c) −z (d) iz (e) z 2 (f) (z)
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π
2
(a) 5 cis 12 × 2 cis π4 (c) 6 cis π2 ÷ 3 cis π6 (e) 4 cis π5
3 cis 5α 3
(b) 3 cis θ × 3 cis 2θ (d) (f) 2 cis 2π
7
2 cis 4α
8. Find the distance |z − w| between the following pairs of numbers
√ in the complex
√ plane.
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(a) w = −1 + i, z = 1 + 3i (d) w = −3 + i 3, z = 3 + 3i 3
(b) w = 4 + 2i, z = 1 − i (e) w = −1 − 3i, z = 2 + i
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√ √
(c) w = 1 + i 3, z = 4 − 2i 3 (f) w = −1 + i, z = −2 − i
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9. Find Arg(z − w) for each pair of numbers in the previous question. Approximate your
answers to 2 decimal places where necessary.
10. Suppose that multiplying a complex number by w produces a rotation of θ radians about
the origin. Find w in Cartesian form for the given values of θ.
(a) π2 (c) π3 (e) 5π
6
(g) − π4
(b) π (d) 3π
4 (f) − π2 (h) − 2π 3
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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CHAPTER 1: Complex Numbers 1D Modulus-Argument Form 29
DEVELOPMENT
11. Let z be a non-zero complex number such that 0 < arg z < π2 . Indicate points A, B,
π
C and D in the complex plane representing the complex numbers z, −iz, (2 cis 3 )z and
1 π
2
cis(− 4 ) z .
12. Replace z with z ÷ w in Box 22 to prove that for z 6= 0 and w 6= 0:
z |z| z
(a) =
(b) arg = arg z − arg w
w |w| w
√ √ √
13. Suppose that z1 = 3 + i and z2 = 2 2 + 2 2 i.
z2
(a) Write z1 and z2 in mod-arg form. (b) Hence write z1 z2 and in mod-arg form.
z1
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14. Repeat the previous question for z1 = − 3 + i and z2 = −1 − i.
1+i 3
15. (a) Express in real-imaginary form.
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1+i
√ 1+i 3
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(b) Write 1 + i and 1 + i 3 in mod-arg form and hence express in mod-arg form.
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1+i
π
(c) Hence find cos 12
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in surd form.
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z1
16. Let z1 = 1 + 5i and z2 = 3 + 2i, and let z = .
TE ID
z2
(a) Find |z| without finding z.
(b) Find tan(tan−1 5 − tan−1 32 ), and hence find arg z without finding z.
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(c) Hence write z in the form x + iy, where x and y are real.
17. Show that for any non-zero complex number z = r cis θ:
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## (a) z z = |z|2 , (b) arg(z 2 ) = 2 arg(z) , (c) if |z| = 1 then z = z −1 .
18. Let z be any non-zero complex number. By considering arg(|z|2 ), use the result in part (a)
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## of the previous question to prove that arg z = − arg z .
19. Let z = cos θ + i sin θ. Determine z 2 in two different ways and hence show that:
(a) cos 2θ = cos2 θ − sin2 θ
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## (b) sin 2θ = 2 sin θ cos θ
20. The complex number z satisfies the equation |z − 1| = 1. Prove that |z|2 = 2 Re(z) by:
(a) letting z = x + iy,
(b) squaring the equation and then using the result |z|2 = zz.
21. Given that z is a complex number satisfying |2z − 1| = |z − 2|, prove that |z| = 1 by:
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## (a) letting z = x + iy,
(b) squaring the equation and then using the result |z|2 = zz.
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## 22. Let z = 1 + cos θ + i sin θ.
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(a) Show that |z| = 2 cos 2θ and arg z = 2θ . (b) Hence show that z −1 = 1
2
− 12 i tan θ2 .
EXTENSION
## 23. Let w = x1 + iy1 and z = x2 + iy2 .
(a) Show that the distance W Z between the points w and z is |z − w|.
(b) Show that the angle that the line W Z makes with the positive real axis is arg(z − w).
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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30 CHAPTER 1: Complex Numbers
24. Let z = cis θ and w = cis φ, noting that |z| = |w| = 1. Evaluate z + w in mod-arg form and
1
hence show that arg(z
+ w) = 2(arg z + arg w) . [Hint: Use the
sum to product
identities
θ+φ θ−φ θ+φ θ−φ
cos θ + cos φ = 2 cos 2 cos 2 and sin θ + sin φ = 2 sin 2 cos 2 .]
25. (a) Prove that Re(z) ≤ |z|. Under what circumstances are they equal?
(b) Prove that |z + w| ≤ |z| + |w|. Begin by writing |z + w|2 = (z + w)(z + w).
## 1E Vectors and the Complex Plane
The geometry of multiplication and division became evident with the introduction
FS
of the modulus-argument form in the previous section. Since the arguments are
added or subtracted, it is clear that a rotation is involved. Since the moduli are
multiplied or divided, it is clear that an enlargement is involved.
O
So far, the observed geometry of addition and subtraction has been limited. A
better understanding of these two operations is desirable and can be achieved
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by yet another representation of complex numbers, this time as vectors. In this
section, a few vector tools will be introduced to help better understand complex
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numbers. A more detailed study of vectors is given in the Extension 1 course.
TE ID
Vectors: In the simple geometric definition used in this section, a vector has two
characteristics, a magnitude and a direction. Thus the instruction on a pirate
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## treasure map “walk 40 paces east” is an example of a displacement vector. The
magnitude is “40 paces” and the direction is “east”. A train travelling from
Sydney to Perth across the Nullabor at 120 km/h is an example of a velocity
vector. The magnitude is 120 km/h and the direction is west.
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## In the number plane, a vector is represented by an arrow, which is more properly
called a directed line segment. The length of the arrow indicates the magnitude
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of the vector and the direction of the arrow is the direction of the vector. In
particular, in the Argand diagram an arrow joining two points will be used to
represent the vector from one complex number to another. When naming a
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vector, the two letter name of the line segment is used with an arrow above it to
indicate the direction, as in the following two examples.
y y
Z 3
Ö3 Z
2
C
1 W
O 1 x
N
-1 1 x
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−→ −−→
OZ is the√vector from the origin to W Z is the vector from w = 1 + i to
z = 1 + i 3. z = −1 + 3i .
Vectors and Complex Numbers: Look further at the examples above. In the first,
−→
it is natural to assume that the vector OZ represents the complex number z. But
−−→
what complex number does the vector W Z represent in the second example?
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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CHAPTER 1: Complex Numbers 1E Vectors and the Complex Plane 31
The magnitude of a vector is the distance between the end-points of its arrow.
−−→
That is, the magnitude of W Z is |z−w|. The direction of a vector may be specified
−−→
by the angle its arrow makes with the horizontal. That is, the direction of W Z
is given by Arg(z − w). Since any complex number is completely determined by
−−→
its modulus and argument, it follows that the W Z must represent the complex
number (z − w). This is always the case, regardless of the values of z and w.
VECTORS AND COMPLEX NUMBERS: Let the vector from w to z in the Argand diagram
25 −−→ −−→
be W Z. Then the vector W Z represents the complex number (z − w).
−−
→ −−→
Equal Vectors: Suppose that two vectors AB and P Q represent the same complex
FS
number. That is, both vectors have the same magnitude and direction. It makes
sense to say that these vectors are equal and to write
−→ −−→
AB = P Q
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since there is nothing to distinguish between them.
D E
O
By way of example, let Z0 , Z1 and Z2 represent the complex numbers 1 + 2i, 2 + i
G
and 3 + 3i respectively. Then by Box 25,
PR
−−→
OZ0 = (1 + 2i) − 0
TE ID
= 1 + 2i
−−−→
and Z1 Z2 = (3 + 3i) − (2 + i)
EC BR
= 1 + 2i. y
Z
3 2
That is, the same complex number is represented by both
vectors, and hence Z0
2
R M
−−→ −−−→
OZ0 = Z1 Z2 . 1
Z1
O CA
## The diagram to the right shows the situation. Notice that
both vectors clearly have the same length and direction. O 1 2 3 x
EQUAL VECTORS: Vectors with the same magnitude and direction are equal. Equal
R
26
vectors represent the same complex number.
## Addition and Subtraction: Consider the three points A, y
B and C which represent the complex numbers w, w + z z+w B
−−→ −→
and z. Now OB = (z +w) and, by Box 25, AC = (z −w). C z-w
C
## So the diagonals of OABC have special significance. Does
z A
this quadrilateral have any other special characteristics? w
N
−−→
The magnitude of AB is O x
U
|(w + z) − w| = |z| ,
−−→
thus AB = OC. Likewise, the magnitude of CB is
|(w + z) − z| = |w| ,
thus CB = OA. Now since OABC has opposite sides of equal length, it must be
a parallelogram. Consequently the opposite sides are also parallel and hence
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
32 CHAPTER 1: Complex Numbers
−−→ −−→
OC = AB
−→ −−→
and OA = CB .
Thus in order to add or subtract two complex numbers geometrically, simply
−→ −−→
construct the parallelogram OABC from the vectors OA and OC. Then the sum
−−→ −→
is the vector OB and the difference is AC. This result is most useful in solving
certain algebraic problems geometrically.
WORKED EXAMPLE 25: Given two non-zero complex numbers w and z withequal
moduli and acute arguments, show that Arg(w + z) = 12 Arg(w) + Arg(z) .
SOLUTION: Consider the points OABC in the z-plane y
representing the complex numbers 0, w, w + z and z B
FS
respectively. Now OABC is a parallelogram. Further C
OA = OC , since |w| = |z| . Thus in fact OABC is a
rhombus. Since the diagonal OB of the rhombus bisects
A
O
the angle at the vertex O, it follows that
O x
Arg(w + z) = 21 Arg(w) + Arg(z) .
D E
O
G
PR
THE GEOMETRY OF ADDITION AND SUBTRACTION: Let the points O, A and C represent
TE ID
the complex numbers 0, w and z. Construct the parallelogram OABC. The
27 −−→
diagonal vector OB represents the complex number (z + w) and the other
−→
diagonal vector AC represents the complex number (z − w).
EC BR
WORKED EXAMPLE 26: The points OABC represent the complex numbers 0, w,
w + z and z. Given that z − w = i(z + w), explain why OABC is a square.
R M
−−→
SOLUTION: Firstly OABC is a parallelogram, where OB represents z + w and
−→
O CA
## AC represents z − w. Since z − w = i(z + w) it follows that
arg(z − w) = arg(i(z + w))
= arg(i) + arg(z + w)
R
= π2 + arg(z + w) ,
and |z − w| = |i(z + w)|
= |i| × |z + w|
= |z + w| .
Thus the diagonals OB and AC are at right angles to each other and have the
same length. Hence OABC is a square.
C
N
## value of real numbers is the triangle inequality
U
|x| − |y| ≤ |x + y| ≤ |x| + |y| .
Given that the absolute value of a real number is analogous to the modulus of a
complex number, it is not surprising that the same result holds for the modulus
of complex numbers, that is
|z| − |w| ≤ |z + w| ≤ |z| + |w| .
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1E Vectors and the Complex Plane 33
y
This result can be explained in terms of the geometry B
of the addition of complex numbers. Consider only the
|z + w| | z|
points O, A and B as defined previously and shown in
the diagram on the right. Recall that the three vectors A
−→ −− → −−→
OA, AB and OB represent the complex numbers w, z |w|
and z + w respectively. Hence the three moduli |w|, |z| O x
and |z + w| are the lengths of the sides of 4OAB.
It is a well known result of Euclidean geometry that the length of one side of a
triangle must be less than or equal to the sum of the other two, thus
|z + w| ≤ |z| + |w| ,
FS
with equality when O, A and B are collinear. Similarly the length of one side is
greater than or equal to the difference of the other two, thus
O
|z| − |w| ≤ |z + w| ,
D E
with equality again when the points are collinear. Combining these two yields
O
G
|z| − |w| ≤ |z + w| ≤ |z| + |w| ,
PR
and replacing w with −w throughout gives
TE ID
|z| − |w| ≤ |z − w| ≤ |z| + |w| .
EC BR
These inequalities are called the triangle inequalities, after their geometric origins.
## THE TRIANGLE INEQUALITIES: For all complex numbers z and w,
• |z| − |w| ≤ |z + w| ≤ |z| + |w|
R M
28
• |z| − |w| ≤ |z − w| ≤ |z| + |w|
O CA
Multiplication and Division: The geometry of these two operations has already
been satisfactorily explained as a rotation and enlargement. This interpretation
R
## is further demonstrated by the following example.
The diagram below shows the points O, U , A, B and C which correspond to the
complex numbers 0, 1, w, z and wz respectively. In 4U OA and 4BOC,
y
6 BOC = arg(wz) − arg(z) C
= arg(w) + arg(z) − arg(z)
C
= arg(w)
N
= 6 U OA ,
B A
OC |wz|
and =
U
OB |z|
O U x
= |w|
OA
= .
OU
Hence 4BOC ||| 4U OA (SAS)
Note that the similarity ratio is OB : OU = |z| : 1.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
34 CHAPTER 1: Complex Numbers
This provides a novel way of constructing the point C for any given complex
numbers w and z. First construct 4U OA, then use the base OB to construct
the similar triangle 4BOC by applying the similarity ratio |z| : 1.
## Other than being an application of similar triangles, this construction method
is not particularly enlightening, and it is rarely required. The geometry of the
situation should always be remembered as a rotation of w by arg(z) and an
enlargement by factor |z|.
Two Special Cases: A vector approach is very helpful in analysing the geometry
in two special cases of division. Let z1 , z2 , z3 and z4 be four complex numbers
FS
corresponding to the points A, B, C and D, and let
z2 − z1
= λ,
z3 − z4
O
so that z2 − z1 = λ(z3 − z4 ) .
D E
Suppose that λ is real and non-zero, then one vector is a multiple of the other.
O
Hence both vectors have the same direction if λ > 0 (but may differ in length)
G
and opposite direction if λ < 0. In either case the lines AB and CD are parallel.
PR
TE ID
In the case where λ is imaginary, the two vectors must be perpendicular, since
multiplication by i is equivalent to a rotation by π2 . Hence AB ⊥ CD. The sign
of Im(λ) determines whether the rotation is anticlockwise or clockwise.
EC BR
WORKED EXAMPLE 27: Let z1 and z2 be any two complex numbers representing
the points A and B in the complex plane. Consider the complex number z given
R M
z − z1
by the equation = t where t is real. Let the point C represent z.
z2 − z1
O CA
## (a) Show that A, B and C are collinear.
(b) Hence show that C divides AB in the ratio t : 1 − t.
R
SOLUTION:
−−→ −→
(a) First note that AB represents z2 − z1 and that AC represents z − z1 . Since
z − z1
is real it follows that AB and AC are parallel. Further since A is
z2 − z1
common to both lines, it follows that A, B and C are collinear.
−→ −−→
C
(b) If t < 0 then AC has the opposite direction to AB and the order of the points
−→ −−→
is CAB. If t = 0 then A and−− C→coincide. If 0 < t < 1 then both AC and AB
N
have the same direction and AB has the greater magnitude. Hence the order
of the points is ACB. If t = 1 then C and B coincide. If t > 1 then the
U
−→
vectors again have the same direction but AC has the greater magnitude,
hence the order is ABC. In all these cases the ratio of the magnitudes is
AC : AB = |z − z1 | : |z2 − z1 |
= |t| : 1 .
Hence in all cases the point C divides AB in the ratio t : 1 − t.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1E Vectors and the Complex Plane 35
Exercise 1E
1. In the diagram on the right, OABC is a parallelogram. The y
points A and C represent 5 + i and 2 + 3i respectively. Find B
the complex numbers represented by:
C(2 + 3i)
(a) the vector OB,
(b) the vector AC, A(5 + i)
(c) the vector CA. O x
## 2. In the diagram on the right, OP QR is a square. The point P y Q
represents 4 + 3i. Find the complex numbers represented by:
(a) the point R, R
P(4 + 3i)
FS
(b) the point Q,
(c) the vector QR,
(d) the vector P R. O x
O
y
3. In the diagram on the right, intervals AB, OP and OQ are equal B
in length, OP is parallel to AB and 6 P OQ = π2 . If A and B
D E
Q
O
represent the complex numbers 3 + 5i and 9 + 8i respectively, find A
P
the complex number which is represented by Q.
G
PR
O x
TE ID
4. In the diagram on the right, OABC is a square. The point A y
B
represents the complex number 2 + i.
EC BR
## (a) Find the numbers represented by B and C. C
(b) If the square is rotated 45◦ anticlockwise about O to give
OA0 B 0 C 0 , find the number represented by B 0 . A
R M
O x
5. In the diagram on the right, AB = BC and ABC = 90 . The 6 y
O CA
## points B and C represent 5 + 3i and 9 + 6i respectively. Find the
complex numbers represented by: A
C(9 + 6i)
(a) the vector BC,
B(5 + 3i)
R
## (b) the vector BA,
(c) the point A. O x
6. The diagram on the right shows a square ABCD in the complex y B(4 + 13i)
plane. The vertices A and B represent the complex numbers
9 + i and 4 + 13i respectively. Find the complex numbers that C
correspond to:
C
N
## (b) the vertex D. x
D
U
DEVELOPMENT
y
Q
7. In the diagram on the right, the points P and Q correspond to
the complex numbers z and w respectively. The triangle OP Q is P
isosceles and the angle P OQ is a right angle.
Prove that z 2 + w 2 = 0.
O x
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
36 CHAPTER 1: Complex Numbers
## 8. In the Argand diagram on the right, ABCD is a square, y C
and OE and OF are parallel and equal in length to AB and D
AD respectively. The vertices A and B correspond to the
B
complex numbers w1 and w2 respectively. What complex F
numbers correspond to the points E, F , C and D? A
E
O x
9. In the diagram on the right, the vertices of a triangle ABC are y
D
represented by the complex numbers z1 , z2 and z3 respectively.
The triangle is isosceles, and right-angled at B. A
(a) Explain why (z1 − z2 )2 + (z3 − z2 )2 = 0. C
FS
(b) Suppose that D is the point such that ABCD is a square. B
Find, in terms of z1 , z2 and z3 , the complex number that the
point D represents. O x
O
10. In the Argand diagram on the right, OABC is a rectangle, with B y
D E
OC = 2OA. The vertex A corresponds to the complex number ω.
O
(a) What complex number corresponds to the vertex C? C
G
PR
(b) What complex number corresponds to the point of intersection
D of the diagonals OB and AC?
TE ID
A(w)
O x
EC BR
## 11. The vertices
√ of an equilateral triangle are equidistant from the origin. One of its vertices
is at 1 + 3i. Find the complex numbers represented by the other two vertices.
[Hint: What is the angle subtended by the vertices at the origin?]
R M
12. Given z = 3 + 4i, find the two possible values of w so that the points representing 0, z
O CA
and w form a right-angled isosceles triangle with right-angle at the point representing:
(a) 0 (b) z (c) w
13. Given that z1 = 1 + i, z2 = 2 + 6i and z3 = −1 + 7i, find the three possible values of z4
R
## so that the points representing z1 , z2 , z3 and z4 form a parallelogram.
14. Suppose that the complex number z has modulus one, and that 0 < Arg z < π2 .
Prove that 2 Arg(z + 1) = Arg z.
15. The vertices of the quadrilateral ABCD in the complex plane represent the complex
numbers z1 , z2 , z3 and z4 respectively.
C
N
## (b) If z1 − z2 = z4 − z3 and z1 − z3 = i(z4 − z2 ), show that ABCD is a square.
U
16. A triangle in the Argand diagram has vertices at the points representing the complex
z2 − z1
numbers z1 , z2 and z3 . If = cos π3 + i sin π3 , show that the triangle is equilateral.
z3 − z1
y
17. In the diagram on the right, the points P1 , P2 and P3 represent
z2 z3 P2
the complex numbers z1 , z2 and z3 respectively. If = , show P3
z1 z2
that OP2 bisects 6 P1 OP3 .
P1
O x
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1F Curves and Regions in the Argand Diagram 37
18. Let z1 = 2i and z2 = 1 + 3 i.
(a) Express z1 and z2 in mod-arg form.
(b) Plot in the complex plane the points P , Q, R and S representing z1 , z2 , z1 + z2 and
z1 − z2 respectively.
(c) Find the exact values of: (i) arg(z1 + z2 ) (ii) arg(z1 − z2 )
19. (a) Prove that for any complex number z, |z|2 = zz.
(b) Hence prove that for any complex numbers z1 and z2 :
|z1 + z2 |2 + |z1 − z2 |2 = 2 |z1 |2 + |z2 |2
## (c) Explain this result geometrically.
FS
20. In the diagram on the right, the points P and Q represent the y
complex numbers z and w respectively.
(a) Explain why |z − w| ≤ |z| + |w|.
O
Q
(b) Indicate on the diagram the point R representing z + w.
P
D E
O
(c) What type of quadrilateral is OP RQ?
(d) If |z − w| = |z + w|, what can be said about the complex O x
G
PR
w
number ?
z
TE ID
z3 − z1
21. (a) Prove that the points z1 , z2 and z3 are collinear if is real.
z2 − z1
EC BR
(b) Hence show that the points representing 5 + 8i, 13 + 20i and 19 + 29i are collinear.
EXTENSION
R M
22. The complex numbers ω1 and ω2 have modulus 1, and arguments α1 and α2 respectively,
where 0 < α1 < α2 < π2 .
O CA
## Show that Arg (ω1 − ω2 ) = 21 (α1 + α2 − π).
z4 − z1 z2 − z3
23. [Circle Geometry] It is known that arg + arg = π. Explain
z2 − z1 z4 − z3
R
## why the points representing these complex numbers are concyclic.
24. [Circle Geometry] The points representing the complex numbers 0, z1 , z2 and z3 are
1 1 1
concyclic and in anticlockwise order. Prove that the points representing , and
z1 z2 z3
z −1 − z2−1
are collinear. [Hint: Show that 1−1 is real.]
C
z1 − z3−1
N
U
## In many situations a set of equations or conditions on a variable complex number z
yields a set of points in the Argand diagram which is a familiar geometric object,
such as a line or a circle. The main aim of this section is to provide a geometric
description for a curve or region specified algebraically. Therefore the examples
in this text have been grouped by the various geometries.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
38 CHAPTER 1: Complex Numbers
There are two basic approaches used in this section, algebraic or geometric. The
at manipulating equations in x and y. Unfortunately the geometry of the situation
may be obscured by the algebra. The advantage of the geometric approach is that
it will often provide a very elegant solution to the problem, but may also require
a keen insight. Both methods should be practised, with the aim to become
proficient at the geometric approach.
Straight Lines: Some simple straight lines have already been encountered in 1C,
such as the vertical line Re(z) = a. Here are some other examples and their
geometric interpretations.
In coordinate geometry, given the coordinates of two points A and B, the task of
FS
finding the equation of the perpendicular bisector of AB is a lengthy one. The
equivalent complex equation is remarkably simple.
WORKED EXAMPLE 28: Let z1 = 4 and z2 = −2i, and let the variable point z
O
satisfy the equation |z − z1 | = |z − z2 |.
D E
(a) Put z = x + iy and hence show that z lies on the straight line y + 2x − 3 = 0.
O
(b) Describe this line geometrically in terms of z1 and z2 .
G
PR
y
SOLUTION: 3
y + 2x - 3 = 0
TE ID
(a) Substitute the values of z1 and z2 , then square to get
(x − 4)2 + y 2 = x2 + (y + 2)2 2 4
or (x − 4)2 − x2 = (y + 2)2 − y 2 x
EC BR
-1 z1
thus −4(2x − 4) = 2(2y + 2) -2 z
2
so 4 − 2x = y + 1
hence y + 2x − 3 = 0 .
R M
## (b) Let z, z1 and z2 be the points P , A and B in the
Argand diagram. Since the modulus is a distance, P
O CA
A
PA = PB .
B
R
## Thus either 4AP B is always isosceles, or P bisects
AB. Hence P is on the perpendicular bisector of AB.
## THE PERPENDICULAR BISECTOR OF AN INTERVAL: Let z1 and z2 be the fixed points A
and B in the Argand diagram, and let z be a variable point P . If
29
C
|z − z1 | = |z − z2 |
then P is on the perpendicular bisector of AB.
N
U
## WORKED EXAMPLE 29: Let z0 = a + ib be the fixed point T and let z = x + iy
be a variable point P in the complex plane. It is known that
z − z0 = ikz0 ,
where k is a real number. It is also known that P is on a straight line.
(a) Find the equation of that straight line in terms of x and y.
(b) What is the geometry of the situation?
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1F Curves and Regions in the Argand Diagram 39
SOLUTION: y
(a) The given equation expands to
x + iy − (a + ib) = ik(a + ib) . P
T
Equating real and imaginary parts yields
x
x − a = −kb
O
and y − b = ka .
Eliminating k from this pair of equations gives
b(y − b) = −a(x − a)
or ax + by = a2 + b2
(b) Some readers will recognise this equation as the tangent to a circle. This
FS
geometry is confirmed by examining the given equation more closely.
Since multiplication by i represents a rotation of π2 , it follows that for k 6= 0
−→ −→
the vector T P is perpendicular to OT . That is, P lies on a line perpendicular
O
to OT . Further, when k = 0, z = z0 , so this line passes through T . That
is, P T is the tangent to the circle with radius OT , as shown above.
D E
O
G
Rays: The horizontal and vertical lines in 1C and the first example above demonstrate
PR
some of the geometry of the recently introduced functions Re(z), Im(z) and |z|.
TE ID
The new function Arg(z) describes a ray in the z-plane.
π
WORKED EXAMPLE 30: The complex number z satisfies Arg(z) = .
EC BR
3
(a) Let |z| = r. Write z in modulus-argument form.
(b) Plot z when r = 1, 2, 3, 4, and observe that z lies on a ray.
(c) Explain why the origin must be excluded.
R M
## (d) Use shifting to sketch Arg(z − 2 − i) = π3 .
O CA
SOLUTION:
y r=4 y
(a) z = r(cos π3 + i sin π3 ) .
(b) See the first graph on the right. r=3
(c) Arg(0) is undefined so the origin Ö3 r=2
R
is not included. p
r=1 1 3
(d) Arg(z − 2 − i) = Arg(z − (2 + i)) p
so the ray has been shifted to the 3 2 x
1 2 x
point 2 + i, as shown on the right.
C
## RAYS IN THE ARGAND DIAGRAM:
• The equation Arg(z) = θ represents the ray which makes an angle θ with the
N
## 30 positive real axis, omitting the origin.
• The locus Arg(z −z0 ) = θ is the result of shifting the above ray from the origin
U
to the point z0 .
Circles and Parabolas: The circle can easily be written as an equation of a complex
variable. The parabola can also be written as an equation in z, though it is
somewhat contrived and is not a significant result. The geometric definitions of
each in terms of distances is the key.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
40 CHAPTER 1: Complex Numbers
WORKED EXAMPLE 31: Consider the equation |z−z0 | = r for some fixed complex
number z0 = a + ib and positive real number r.
(a) Explain why this represents a circle. State the centre and radius.
(b) Confirm the result by putting z = x + iy and finding the Cartesian equation.
(c) Expand |z − 1|2 in terms of z and z. Hence determine the curve specified by
|z|2 = z + z.
SOLUTION: y
(a) The equation specifies that the distance between z
and z0 is fixed. This is the geometric definition of a Z0 r
b
circle. The centre is z0 and the radius is r.
FS
(b) Begin by squaring both sides:
|z − z0 |2 = r 2 a x
so |(x − a) + i(y − b)|2 = r 2
O
thus (x − a)2 + (y − b)2 = r 2 .
D E
(c) From |w|2 = w w it follows that
O
y
|z − 1|2 = (z − 1) (z − 1)
G 1
PR
= (z − 1)(z − 1)
= |z|2 − (z + z) + 1 .
TE ID
2 1 2 x
Since |z| = z + z , it also follows that
-1
|z − 1|2 = 1 ,
EC BR
## that is, the circle with centre z = 1 and radius 1 .
R M
CIRCLES IN THE ARGAND DIAGRAM: Let z0 be the fixed point C in the Argand diagram,
and let z a variable point P . If
O CA
31
|z − z0 | = r
then the point P is on the circle with centre C and radius r .
R
WORKED EXAMPLE 32: Let S be the fixed point 1 + 2i and z be the variable
point P in the Argand diagram. It is known that |z − (1 + 2i)| = Im(z). Show
algebraically that P lies on a parabola by putting z = x + iy.
## SOLUTION: Squaring both sides of the given equation: y
C
(x − 1)2 + (y − 2)2 = y 2
so (x − 1)2 = y 2 − (y − 2)2
N
2 S
= 4y − 4 1
2
U
thus (x − 1) = 4(y − 1) .
1 x
This is the equation of a parabola with vertex at 1 + i,
as shown on the right.
This last worked example may seem a little contrived, but it demonstrates that
there are many situations where further insight into a problem may be achieved
by considering a corresponding problem in the complex plane.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1F Curves and Regions in the Argand Diagram 41
Regions: In many instances a curve divides the plane into two or more regions. In
simple cases a region is defined by the corresponding inequation. When more
intricate regions are required, two or more simple regions may be combined by
taking the union or intersection of the corresponding inequations. Some common
examples follow.
WORKED EXAMPLE 33: Sketch the following regions in the complex plane.
(a) 1 ≤ Re(z) ≤ 3 (c) |z − 2 + i| < 1
(b) 0 ≤ Arg(z) ≤ π4 (d) |z| > |z + 2 − 2i|
## SOLUTION: The first three can be easily explained geometrically.
(a) y (b) y
FS
O
1 3x
D E
x
O
G
This is 1 ≤ x ≤ 3, the vertical strip Put z = r cis θ to get 0 ≤ θ ≤ π4 ,
PR
between x = 1 and x = 3 . which defines a wedge excluding the
TE ID
origin, since Arg(0) is undefined.
(c) y (d) y
1 2 3 -2+2i
EC BR
x 2
-1 1
R M
-2
-2 -1 x
O CA
## The boundary curve is the circle The perpendicular bisector of the
with radius 1 and centre 2−i, and is segment from 0 to −2 + 2i is the
not included. The region includes boundary, and is not included. The
R
the centre of the circle since the region includes the point −2 + 2i,
LHS of the inequality is zero there. since the RHS of the inequality is
zero there.
## WORKED EXAMPLE 34:
(a) Sketch the regions (i) |z − i| ≤ 1 and (ii) − π6 < Arg(z + 1 − i) < π
.
C
6
(b) Hence sketch (i) the union and (ii) the intersection of these regions.
N
SOLUTION:
(a) (i) y (ii) y
U
2
2
1 1
-1 1 x -1 1 x
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
42 CHAPTER 1: Complex Numbers
## (b) The boundaries intersect
at −1 + i and,
from trigonometry, they interesect
again at 21 + i (1 + 23 ) and 12 + i (1 − 23 ) . Here are the graphs.
(i) y (ii) y
2 2
1 1
-1 1 x -1 1 x
Circle Geometry: Many of the circle geometry theorems encountered in Year 10 may
FS
be expressed in terms of a complex number. One significant result is included
here, with other examples to be found in the exercise.
O
WORKED EXAMPLE 35: [A Hard Example] Let z1 = 3 + i and z2 = 1 − i.
z − z1
D E
Describe and sketch the set of points z, where arg = π4 .
O
z − z2
G
SOLUTION: Let z1 , z2 and z represent the points A, B and P respectively. First
PR
note that the equation can be written as
TE ID
arg(z − z1 ) − arg(z − z2 ) = π4 .
−→
Recall that z − z1 is the vector AP , so arg(z − z1 ) is the direction of this vector.
EC BR
−−→
Likewise arg(z − z2 ) is the direction of vector BP . Thus the difference is the
angle between them and is always π4 . Using the converse of the angles in the
same segment theorem, it follows that P must lie on the arc of a circle with
R M
## chord AB. Further, since
y
π π
6 AP B = 4
< 2
O CA
P
it is a major arc. As P moves along the arc from A p
to B, it moves anticlockwise about the centre, because C 4
1 A
angles are measured anticlockwise. Lastly, since arg(0)
1 3 x
R
## is undefined, the endpoints of the arc are not included. -1 B
It simply remains to find the centre and radius of this
circle. Let C be the centre, then
ACB = π2 (Angles at the centre and circumference)
6
## hence 6 CAB = π (base angles of isosceles triangle.)
4
π
Since Arg(z1 − z2 ) = 4
C
## it follows that AC is horizontal, as the alternate angles are equal. Thus BC is
vertical. Hence C = 1 + i is the centre of the circle and AC = 2 is its radius.
N
THE ARC OF A CIRCLE: Let points A and B represent the complex numbers z1 and z2 .
U
## Let the variable point P represent z. The equation
z − z1
32 Arg = α where 0 < α < π ,
z − z2
is the arc AB of a circle with the endpoints excluded. As P moves along the
arc from A to B its motion is anticlockwise about the centre.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1F Curves and Regions in the Argand Diagram 43
Exercise 1F
1. Sketch each straight line by using the result of Box 28, then find its Cartesian equation.
(a) |z + 3| = |z − 5| (b) |z − i| = |z + 1| (c) |z + 2 − 2i| = |z| (d) |z−i| = |z−4+i|
2. Sketch the rays specified by the following equations. Box 29 may be of help.
3π π
√ π
(a) arg(z − 4) = 4 (b) arg(z + 1) = 4 (c) arg(z − 1 − i 3) = 3
## 3. Use Box 31 to sketch these circles.
(a) |z + 1 − i| = 1 (b) |z − 3 − 2i| = 2 (c) |z − 1 + i| = 2
4. In each case sketch the boundary or boundaries of the region and then shade the region.
(d) 0 ≤ arg(z) ≤ 3π
FS
(a) |z − 8i| ≥ |z − 4| 4
(g) |z| > 2
(b) |z − 2 + i| ≤ |z − 4 + i| (e) − π3 < arg(z) < π6 (h) |z + 2i| ≤ 1
(c) |z + 1 − i| ≥ |z − 3 + i| (f) − π4 ≤ arg(z + 2 + i) < π
4 (i) 1 < |z − 2 + i| ≤ 2
O
DEVELOPMENT
D E
O
5. In each case sketch (i) the intersection and (ii) the union of the given pair of regions.
π
(a)
G
|z − 2 + i| ≤ 2, Im(z) ≥ 0 (e) |z − 1 − i| ≤ 2, 0 ≤ arg(z − 1 − i) ≤ 4
PR
(b) 0 ≤ Re(z) ≤ 2, |z − 1 + i| ≤ 2 (f) |z| ≤ 1, 0 ≤ arg(z + 1) ≤ π4
TE ID
(c) |z − z| < 2, |z − 1| ≥ 1 (g) |z + 1 − 2i| ≤ 3, − π3 ≤ arg z ≤ π4
(d) Re(z) ≤ 4, |z − 4 + 5i| ≤ 3 (h) |z − 3 − i| ≤ 5, |z + 1| ≤ |z − 1|
EC BR
## 6. Put z = x + iy to help sketch these hyperbolas.
(a) z 2 − (z)2 = 16i (b) z 2 − (z)2 = 12i
7. In each case the given equation represents a parabola. Find the Cartesian equation by
R M
## putting z = x + iy, and hence sketch the parabola.
(a) |z−3i| = Im(z) (b) |z + 2| = − Re(z) (c) |z| = Re(z + 2) (d) |z − i| = Im(z + i)
O CA
## 8. By putting z = x + iy, or otherwise, sketch the graph defined by the equation:
4 1
(a) Im(z) = |z| (b) Re 1 − =0 (c) Re z − =0
R
z z
9. Sketch the arcs of circles specified by the following equations, showing the centre and
z −2 π z−i π z − 2i π
(a) arg = (c) arg = (e) arg =
z 2 z+i 4 z + 2i 6
z −1+i π z+1 π z 3π
C
## (b) arg = (d) arg = (f) arg =
z −1−i 2 z−3 3 z+4 4
N
π
10. A complex number z satisfies arg z = 3.
(a) Use a diagram to show that |z −2i| ≥ 1 . (b) For which value of z is |z − 2i| = 1 ?
U
y
11. Consider the graph in the Argand diagram on the right.
3Ö3
(a) Write down an equation for this graph in terms of z.
(b) Find the modulus and argument of z at the point where |z|
takes its minimum value.
(c) Hence find z in Cartesian form when |z| takes its least value.
-3 x
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
44 CHAPTER 1: Complex Numbers
12. (a) A complex number z satisfies |z − 1| = 2. Draw a diagram and hence find the greatest
and least possible values of |z|.
(b) If z is a complex number√ such that Re(z) ≤ 2 and |z − 3| = 2, show with the aid of a
diagram that 1 ≤ |z| ≤ 7.
13. (a) A complex number z satisfies |z − 2| = 1.
(i) Sketch the graph of |z − 2| = 1. (ii) Show that − π6 ≤ arg z ≤ π6 .
(b) The complex number z is such that |z| = 1. Use your answers to part (a) to explain
why − π6 ≤ arg(z + 2) ≤ π6 .
14. The complex number w satisfies |w| = 10 and 0 ≤ arg w ≤ π2 , and the complex number
z is specified by z = 3 + 4i + w.
(a) Sketch the graph of z = 3 + 4i + w..
FS
(b) Use your sketch to determine the maximum value of |z| .
(c) What is the value of z for which this maximum occurs?
15. (a) Show that the circle equation |z − z0 | = r is equivalent to
O
z z − (z z0 + z z0 ) + z0 z0 − r 2 = 0 .
D E
O
[Hint: Square both sides of |z − z0 | = r and use the result |w|2 = w w .]
G
(b) Use part (a) to write these equations in the form |z − z0 | = r, and hence state the
PR
centre and radius of each circle.
TE ID
1 1
(i) z z+2(z+z) = 0 (ii) z z −(1+i)z−(1−i)z +1 = 0 (iii) + =1
z z
z−1
EC BR
16. Sketch the graph of the set of points z for which is: (a) real, (b) imaginary.
z−i
17. Sketch the graph of: (a) arg(z + i) = arg(z − 1) (b) arg(z + i) = arg(z − 1) + π
18. (a) The variable complex number z satisfies |z − 2 − i| = 1. Use a diagram to find the
R M
## maximum and minimum values of: (i) |z| (ii) |z − 3i|
(b) A complex number z satisfies |z| = 3. Use a sketch to find the greatest and least
O CA
values of |z + 5 − i|.
(c) The variable complex number z satisfies |z − z0 | = r. Use a similar approach to
parts (a) and (b) to find the maximum and minimum values of: (i) |z| (ii) |z − z1 |
R
(d) Confirm your answers to the previous parts by using the triangle inequality
|z| − |w| ≤ |z + w| ≤ |z| + |w|.
EXTENSION
z − z1
19. Describe the graph of arg = α, where α is constant, if:
z − z2
C
π π π
(a) α = 0 (b) 0 < α < 2
(c) α = 2
(d) 2
<α<π (e) α = π
N
20. [Pythagoras] Describe the graph defined by the equation |z −z1 |2 +|z −z2 |2 = |z1 −z2 |2 .
21. Suppose that k|z − z1 | = `|z − z2 |, where k 6= ` and both are positive real numbers.
U
k2 z1 − `2 z2
(a) Show that the graph specified by the equation is a circle with centre and
k 2 − `2
kl|z2 − z1 |
|k2 − `2 |
(i) by letting z = x + iy, (ii) by geometric methods.
(b) What happens in the limit as k approaches `?
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1G Polynomials and Complex Numbers 45
## 1G Polynomials and Complex Numbers
A significant application of complex numbers is in the study of polynomials. This
section further develops the work on polynomials already done in the Mathematics
Extension 1 course. That work is assumed knowledge though some parts of the
theory are repeated here for the sake of convenience. The focus is on polynomials
with real coefficients and the relationships with the zeros, particularly when they
are either complex, or real and repeated.
The crux of the work is in Section 1H where the Fundamental Theorem of Algebra
is presented along with some of its consequences. The theorem is left unproven
as any proof is beyond the scope of the course.
FS
Polynomials with Integer Coefficients: If a polynomial with integer coefficients
has an integer zero x = k, then k is a factor of the constant term. This is a
significant aid in factorising a polynomial.
O
WORKED EXAMPLE 36: It is known that the polynomial P (x) = x3 − x2 − 8x − 6
D E
has only one integer zero. Find it and hence factorise P (x) completely.
O
SOLUTION: The zero is a factor of 6, so the possible values are: ±1, ±2, ±3, ±6.
G
PR
Testing these one by one:
TE ID
P (1) = −14 , P (−1) = 0 ,
and there is no need to continue further. By the factor theorem, (x + 1) is a
factor of P (x). Performing the long division:
EC BR
x2 − 2x − 6
(x + 1) x3 − x2 − 8x − 6
x3 + x2
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− 2x2 − 8x − 6
− 2x2 − 2x
O CA
− 6x − 6
− 6x − 6
0
R
## Thus P (x) = (x + 1)(x2 − 2x − 6)
= (x + 1) (x − 1)2 − 7 (completing the square)
√ √
= (x + 1)(x − 1 − 7)(x − 1 + 7) (difference of two squares.)
C
## INTEGER COEFFICIENTS AND ZEROS: If the polynomial
N
P (x) = a0 + a1 x + a2 x2 + . . . + an xn
33
with integer coefficients a0 , a1 , a2 , . . . , an , has an integer zero x = k, then k is
U
## Proof: Since P (k) = 0, it follows that
a0 + a1 k + a2 k2 + . . . + an kn = 0
so a1 k + a2 k2 + . . . + an kn = −a0
thus k × (a1 + a2 k + . . . + an kn−1 ) = −a0 .
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
46 CHAPTER 1: Complex Numbers
Since all the terms in the brackets are integers, it follows that the result is also
an integer. Thus the left hand side is the product of two integers. Hence, as
asserted, k is a factor of a0 .
## Polynomials and Complex Numbers: Consider the general polynomial
P (x) = a0 + a1 x + a2 x2 + . . . + an xn .
Each term in this expression involves an integer power and multiplication by a
constant. The terms are then simply added. Since integer powers, multiplication
and addition are all natural operations with complex numbers, it follows that
the polynomial can be evaluated when x is a complex number. For example if
P (x) = x2 − 2x + 4 then at x = i its value is
FS
P (i) = i2 − 2i + 4
= 3 − 2i .
O
In some examples the polynomial will be written as a function of z in order to
emphasise the fact that complex numbers may be substituted. Thus the above
D E
example may be written as P (z) = z 2 − 2z + 4 .
O
G
PR
Remainders and Factors: Here is a quick summary of certain important results
from the Mathematics Extension 1 course. In the usual notation, let P (x)
TE ID
and D(x) be any pair of polynomials, where D(x) =
6 0. There is a unique pair of
polynomials Q(x) and R(x), such that
EC BR
and where either
R M
## deg(D) > deg(R) or R(x) = 0 .
This is known as the division algorithm. As a consequence, if D(x) = (x − α)
O CA
then R(x) must be a constant, either zero or non-zero. Let this constant be r.
Re-writing the division algorithm:
P (x) = (x − α) × Q(x) + r ,
R
whence P (α) = r ,
which is known as the remainder theorem.
## If R(x) = 0 then from the division algorithm
P (x) = D(x) × Q(x) ,
C
## so that P (x) is a product of the factors D(x) and Q(x). In particular, x − α is a
factor of P (x) if and only if P (α) = 0 . This is known as the factor theorem.
N
The division algorithm, the remainder theorem and the factor theorem are valid
U
for complex numbers as well as real numbers. Though these claims will not be
proven here, the results may be freely applied to solve problems.
## WORKED EXAMPLE 37: Let P (x) = x3 − 2x2 − x + k , where k is real.
(a) Show that P (i) = (2 + k) − 2i .
(b) When P (x) is divided by x2 + 1 the remainder is 4 − 2x. Find the value of k.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1G Polynomials and Complex Numbers 47
SOLUTION:
(a) P (i) = i3 − 2i2 − i + k
= −i + 2 − i + k
= (2 + k) − 2i .
(b) By the division algorithm,
P (x) = (x2 + 1) × Q(x) + 4 − 2x .
Thus P (i) = 4 − 2i
hence (2 + k) − 2i = 4 − 2i .
Equating the real parts gives k = 2 .
Real Coefficients and Remainders: Suppose that the polynomial P (z) has real
FS
coefficients. If the remainder when P (z) is divided by (z − α) is β then the
remainder when P (z) is divided by (z − α) is β . Using the remainder theorem,
this is equivalent to the statement that if P (α) = β then P (α) = β .
O
WORKED EXAMPLE 38:
D E
O
(a) Use the remainder theorem to find the remainder when
G
P (z) = z 3 − 2z 2 + 3z − 1 is divided by (z − i) .
PR
(b) Hence find the remainder when P (z) is divided by (z + i) .
TE ID
SOLUTION:
(a) The remainder is: (b) It is: P (−i) = P ( i )
EC BR
3 2
P (i) = i − 2i + 3i − 1 = 1 + 2i
= 1 + 2i . = 1 − 2i .
R M
REAL COEFFICIENTS AND REMAINDERS: If the polynomial P (z) has real coefficients and
34
if P (α) = β then P (α) = β .
O CA
The proof is not too difficult and is dealt with in a question of the exercise.
R
Real Coefficients and Complex zeros: Suppose that the polynomial P (z) has real
coefficients. If P (z) has a complex zero z = α then it is guaranteed to have a
second complex zero z = α . Further, by the factor theorem, there exists another
polynomial Q(z) such that:
P (z) = (z − α)(z − α) × Q(z)
= (z 2 − (α + α)z + αα) × Q(z)
C
## = (z 2 − 2 Re(α)z + |α|2 ) × Q(z) .
Thus P (z) has a quadratic factor with real coefficients: (z 2 − 2 Re(α)z + |α|2 ) .
N
U
## WORKED EXAMPLE 39: Consider the polynomial P (z) = 2z 3 − 3z 2 + 18z + 10 .
(a) Given that 1 − 3i is a zero of P (z), explain why 1 + 3i is another zero.
(b) Find the third zero of the polynomial.
(c) Hence write P (z) as a product of:
(i) linear factors,
(ii) a linear factor and a quadratic factor, both with real coefficients.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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48 CHAPTER 1: Complex Numbers
SOLUTION: (a) Since P (z) has real coefficients, (1 − 3i) = 1 + 3i is also a zero.
(b) Let the third zero be a , then by the sum of the roots:
a + (1 − 3i) + (1 + 3i) = 23
so a + 2 = 32
and a = − 12 .
(c) (i) By the factor theorem:
P (z) = 2(z + 21 )(z − 1 + 3i)(z − 1 − 3i)
= (2z + 1)(z − 1 + 3i)(z − 1 − 3i) .
(ii) P (z) = (2z + 1)(z 2 − 2z + 10) .
FS
REAL COEFFICIENTS AND ZEROS: If the polynomial P (z) has real coefficients and a
complex zero z = α then it is guaranteed to have a second complex zero
O
35 z = α.
Consequently P (z) has (z 2 − 2 Re(α)z + |α|2 ) as a factor, which is a quadratic
D E
O
with real coefficients.
G
PR
Proof: Suppose that the complex number z = α is a zero of the polynomial
TE ID
P (z) = a0 + a1 z + a2 z 2 + . . . + an z n ,
where the coefficients a0 , a1, . . . , an are all real. That is P (α) = 0 . Then
EC BR
P (α) = a0 + a1 α + a2 α 2 + . . . + an α n
= a0 + a1 α + a2 α2 + . . . + an αn (since z n = z n )
= a0 + a1 α + a2 α2 + . . . + an αn (since c z = cz for real c)
R M
= a0 + a1 α + a2 α2 + . . . + an αn (since w + z = w + z)
O CA
= P (α)
=0
= 0.
R
## P (z) = (z 2 − 2 Re(α)z + |α|2 ) × Q(z) .
Multiple Zeros: Recall that if P (x) = (x−α)mQ(x), where Q(α) 6= 0, then the value
x = α is called a zero of multiplicity m. It is also the case that x = α is a zero
C
of P 0 (x) with multiplicity (m − 1). In fact this result is also true for polynomials
with complex zeros but the general proof is beyond the scope of this course.
N
## However, it is possible to prove the result in the special case of a polynomial
with real coefficients and a complex zero with multiplicity 2. This is dealt with
U
## in a question of the exercise, and extending this to arbitrary multiplicity may be
suitable as a class investigation.
In Extension 1, the derivative result may have been applied a second to find a
triple zero. In fact it can be extended to the general case of a polynomial P (x)
with real coefficients which has a real zero x = α of multiplicity m. The value
x = α is also a zero of each of the derivatives P (j) (x), for j = 1, . . . , (m − 1) .
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1G Polynomials and Complex Numbers 49
MULTIPLE ZEROS AND HIGHER DERIVATIVES: Suppose that the polynomial P (x) with
36 real coefficients has a real zero x = α with multiplicity m > 1 .
Then x = α is a zero of each of the derivatives P (j) (x) , for j = 1, . . . , (m − 1) .
This result can be proved relatively easily by induction and is left as an exercise
for the chapter on proof. It is also true for real polynomials with complex zeros,
and can likewise be proved by induction.
The Fundamental Theorem of Algebra: All the work encountered so far in this
section deals with finding the zeros of various polynomials. Up to this point it
has been possible to sidestep an important question: does every polynomial have
a zero? For there is no point in searching for one if none exists.
FS
In order to emphasise this point, consider the polynomial P (x) = x2 + 1. Clearly
this function has no real zero, and there is no point in searching for one. Yet
O
the polynomial does indeed have two zeros, both of which happen to be complex
numbers: namely i and −i. Could it be that there is another polynomial which
D E
has neither real nor complex zeros?
O
G
The answer to this question is: every polynomial with degree ≥ 1 has at least one
PR
zero, though that zero may be complex. This is such an important and basic fact
TE ID
in the study of mathematics that it is given a title — The Fundamental Theorem
of Algebra.
EC BR
## THE FUNDAMENTAL THEOREM OF ALGEBRA: Every polynomial with degree ≥ 1 has at
37
least one zero, though that zero may be complex.
## Several eminent mathematicians worked on this theorem including Leibniz, Euler
R M
and Argand. But credit is usually given to Gauss for the first proof, which he
presented in his doctoral thesis in 1799. This, or any other proof of the theorem,
O CA
is beyond the scope of this course. Although the wording given in the box above
is imprecise, it is usually sufficient for the problems encountered at this level.
The Degree and the Number of Zeros: Although the Fundamental Theorem of
R
## Algebra cannot be proven here, it is possible to prove two significant consequences
of the theorem. The first is that every polynomial of degree n ≥ 1 with complex
coefficients has precisely n zeros, as counted by their multiplicities.
This is also true for polynomials with real coefficients. To demonstrate the result,
the cubic P (x) = x3 − 3x2 + 4 = (x − 2)2 (x + 1) has three zeros: the simple zero
C
## x = −1 and the double zero x = 2.
N
THE DEGREE AND THE NUMBER OF ZEROS: Every polynomial of degree n ≥ 1 with
38
complex coefficents has precisely n zeros, as counted by their multiplicities.
U
Proof: This proof uses induction, and may be better left as an exercise for the
chapter on proof.
A. Consider the general polynomial of degree one with complex coefficients:
P1 (x) = a0 + a1 x , where a1 6= 0 .
Clearly this polynomial has one zero x = α1 , where
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
50 CHAPTER 1: Complex Numbers
a0 a1
α1 = − .
|a1 |2
Thus the result is true for n = 1.
B. Suppose that the result is true for some integer k ≥ 1. That is, suppose that
every polynomial of degree k with complex coefficients
Pk (x) = a0 + a1 x + . . . + ak xk , where ak 6= 0 ,
has k zeros, x = α1 , . . . , αk , as counted by their multiplicities. (∗∗)
The statement is now proven true for n = k + 1. That is, it is proven that
every polynomial of degree k + 1 with complex coefficents
Pk+1 (x) = a0 + a1 x + . . . + ak+1 xk+1 , where ak+1 6= 0 ,
has k + 1 zeros as counted by their multiplicities.
FS
Now for any particular polynomial Pk+1 (x), that polynomial has at least one
zero by the Fundamental Theorem of Algebra. Let this zero be x = αk+1 .
Then, by the factor theorem, it follows that
O
Pk+1 (x) = (x − αk+1 )Qk (x)
D E
for some polynomial Qk (x) of degree k. But by the induction hypothesis
O
above (∗∗), Qk (x) has k zeros, all of which are thus inherited by Pk+1 (x).
G
Hence Pk+1 (x) has k + 1 zeros, x = α1 , . . . , αk , αk+1 , as counted by their
PR
multiplicities. Clearly this follows for each and every polynomial Pk+1 (x).
TE ID
C. It follows from parts A and B by mathematical induction that the statement
is true for all integers n ≥ 1.
EC BR
Real Linear and Quadratic Factors: The second significant consequence of the
Fundamental Theorem of Algebra is that every polynomial of degree n ≥ 1 with
real coefficients can be written as a product of factors which are either linear
R M
or irreducible quadratics, each with real coefficients. In this context the word
irreducible is used to indicate that the quadratic has no real zero.
O CA
In order to demonstrate the result, notice that the polynomial P (x) = x3 − 1 can
be written as the product
P (x) = (x − 1)(x2 + x + 1) .
R
## The quadratic factor (x2 + x + 1) is irreducible since it has no real zero.
REAL LINEAR AND QUADRATIC FACTORS: Every polynomial of degree n ≥ 1 which has
39 real coefficents can be written as a product of factors which are either linear
or irreducible quadratics, each with real coefficients.
C
## Proof: Let Pn (x) = a0 + a1 x + . . . + an xn be a polynomial with degree n ≥ 1
which has real coefficients. By the previous result, this polynomial has n zeros.
N
## Let these zeros be x = α1 , . . . , αn . If none of these zeros is complex then the
result is obviously true since then
U
n
Y
Pn (x) = an (x − αk ) .
k=1
If all of the zeros are complex then the result is again obviously true. Since Pn (x)
has real coefficients, the roots come in conjugate pairs, by which it is known
that n is even. Thus
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers 1G Polynomials and Complex Numbers 51
n/2
Y
Pn (x) = an (x − αk ) × (x − αk )
k=1
n/2
Y
= an x2 − 2 Re(αk )x + |αk |2 .
k=1
Lastly, if some of the zeros are complex numbers then once again they occur
as conjugate pairs, since the coefficients of Pn (x) are real. Let the number of
conjugate pairs be j, where 1 < 2j < n. Now re-order and re-label the zeros with
the conjugate pairs listed first. Thus the first conjugate pair is x = α1 , α1 , and
the last conjugate pair is x = αj , αj .
If there are any other zeros then they are real. The first of these is x = α2j+1
FS
and the last is x = αn . So by the factor theorem, and using product notation:
j
! n
Y Y
Pn (x) = an × (x − αk )(x − αk ) × (x − α` )
O
k=1 `=2j+1
!
j n
Y
D E
Y
O
= an × x2 − 2 Re(αk )x + |αk |2 × (x − α` ) .
k=1 `=2j+1
G
PR
In each of the three cases the result is a product of factors with real coefficients,
which are either linear or irreducible quadratic factors. Put more simply, multiply
TE ID
all the complex factors together in conjugate pairs to get irreducible quadratic
factors with real coefficients, and any remaining factors are both linear and real.
EC BR
Exercise 1G
1. It is known that in each case the given polynomial P (x) has only one integer zero. Find
R M
## it and hence factorise P (x) completely.
(a) P (x) = x3 − 6x + 4 (b) P (x) = x3 + 3x2 − 2x − 2 (c) P (x) = x3 − 3x2 − 2x + 4
O CA
## 2. It is known that 1 + i is a zero of the polynomial P (x) = x3 − 8x2 + 14x − 12.
(a) Why is 1 − i also a zero of P (x) ?
(b) Use the sum of the zeroes to find the third zero of P (x).
R
## 3. It is known that 1 − 2i is a zero of the polynomial P (x) = x3 + x + 10.
(a) Write down another complex zero of P (x), and give a reason for your answer.
(b) Hence show that x2 − 2x + 5 is a factor of P (x).
(c) Find the third zero, and hence write P (x) as a product of factors with real coefficients.
4. It is known that −3i is a zero of the polynomial P (z) = 2z 3 + 3z 2 + 18z + 27.
C
(a) Write down another complex zero of P (z). Justify your answer.
N
(b) Hence write down a quadratic factor of P (z) with real coefficients.
(c) Write P (x) as a product of factors with real coefficients.
U
## 5. Let P (z) = 2z 3 − 13z 2 + 26z − 10.
(a) Show that P (3 + i) = 0.
(b) State the value of P (3 − i), and give a reason for your answer.
(c) Hence write P (z) as a product of:
(i) linear factors,
(ii) a linear factor and a quadratic factor, both with real coefficients.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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52 CHAPTER 1: Complex Numbers
DEVELOPMENT
## 6. Consider the polynomial Q(x) = x4 − 6x3 + 8x2 − 24x + 16.
(a) It is known that Q(2i) = 0. Why does it follow immediately that Q(−2i) = 0 ?
(b) By using the sum and the product of the zeroes of Q(x), or otherwise, find the other
two zeroes of Q(x).
(c) Hence write Q(x) as a product of:
(i) four linear factors,
(ii) three factors with real coefficients,
(iii) two factors with integer coefficients.
FS
7. (a) Solve the equation x4 − 3x3 + 6x2 + 2x − 60 = 0 given that x = 1 + 3i is a root.
(b) Solve the equation x4 − 6x3 + 15x2 − 18x + 10 = 0 given that x = 1 − i is a root.
## 8. Consider the polynomial equation x4 − 5x3 + 4x2 + 3x + 9 = 0.
O
(a) Show that x = 3 is a double root of the equation.
D E
O
(b) Hence solve the equation.
G
9. Two of the zeroes of P (z) = z 4 − 12z 3 + 59z 2 − 138z + 130 are a + ib and a + 2ib, where
PR
a and b are real and b > 0.
TE ID
(a) Find the value of a by considering the sum of the zeroes.
(b) Use the product of the zeroes to show that 4b4 + 45b2 − 49 = 0, and hence find b.
EC BR
(c) Hence express P (z) as the product of quadratic factors with real coefficients.
## 10. Suppose that P (x) = x3 + kx2 + 6, where k is real.
(a) Show that P (2i) = (6 − 4k) − 8i .
R M
(b) When P (x) is divided by x2 + 4 the remainder is −4x − 6 . Find the value of k .
O CA
## 11. Let P (x) = x3 − x2 + mx + n, where both m and n are integers.
(a) Show that P (−i) = (1 + n) + i(1 − m) .
(b) When P (x) is divided by x2 + 1 the remainder is 6x − 3 . Find the values of m and n .
R
## 12. Suppose that P (x) = x3 + x2 + 6x − 3 .
(a) Use the remainder theorem to find the remainder when P (x) is divided by x + 2i .
(b) Hence find the remainder when P (x) is divided by: (i) x − 2i , (ii) x2 + 4 .
## 13. Let P (z) = z 8 − 25 z 4 + 1. Suppose that w is a root of P (z) = 0.
C
1
(a) Show that iw and w are also roots of P (z) = 0.
N
## (b) Find one of the roots of P (z) = 0 in exact form.
(c) Hence find all the roots of P (z) = 0.
U
14. Suppose that P (x) = x4 + Ax2 + B, where A and B are positive real numbers.
(a) Explain why P (x) has no real zeroes.
(b) Given that ic and id, where c and d are real, are zeroes of P (x), write down the other
two zeroes of P (x), and give a reason.
(c) Prove that c4 + d4 = A2 − 2B.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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CHAPTER 1: Complex Numbers 1G Polynomials and Complex Numbers 53
15. The polynomial P (x) = x3 + cx + d, where c and d are real and non-zero, has a negative
real zero k, and two complex zeroes. The graph of y = P (x) has two turning points.
(a) What can be said about the two complex zeroes of P (x), and why?
(b) By considering P 0 (x), show that c < 0.
(c) Sketch the graph of y = P (x).
(d) If a ± ib, where a and b are real, are the complex zeroes of P (x), deduce that a > 0.
(e) Prove that d = 8a3 + 2ac.
16. Consider the polynomial function f (x) = x3 − 3x + k, where k is an integer greater than 2.
(a) Show that f (x) has exactly one real zero r, and explain why r < −1.
(b) Give a reason why the two complex zeroes of f (x) form a conjugate pair.
FS
(c) If the complex zeroes are a + ib and a − ib, use the result for the sum of the roots two
at a time to show that b2 = 3(a2 − 1).
(d) Find the three zeroes of f (x) given that k = 2702, and that a and b are integers.
O
x2 x3 xn
17. Prove that P (x) = 1 + x + + +···+ , where n ≥ 2, has no multiple zeroes.
2! 3! n!
D E
O
18. Consider the polynomial P (z) = z 4 + 4z 3 + 14z 2 + 20z + 25.
G
PR
(a) Show that P (−1 + 2i) and P 0 (−1 + 2i) are both zero.
(b) What can we deduce from (a)?
TE ID
(c) Explain why −1 − 2i is also a double zero of P (z).
(d) Hence factorise P (z) over the complex numbers and then over the real numbers.
EC BR
EXTENSION
19. In the text it was proven that if P (z) is a polynomial with real coefficients and if P (α) = 0
R M
then P (α) = 0. Use a similar approach to prove that if P (α) = β then P (α) = β .
20. Let P (x) = a0 + a1 x + a2 x2 + . . .+ an xn be a polynomial with integer coefficients. Suppose
O CA
that P (x) has a rational zero x = qp where p and q have highest common factor 1. Show
that p is a factor of a0 and that q is a factor of an .
21. Use the Fundamental Theorem of Algebra to carefully explain why every polynomial of
R
odd degree with real coefficients has at least one real zero.
22. The polynomial P (z) has real coefficients and a double complex zero z = α.
(a) Prove that z = α is also a double zero.
2
(b) Explain why z 2 − 2 Re(α)z + |α|2 is a factor of P (z).
(c) Hence prove that P 0 (α) = 0.
C
(d) Try to generalise this result to complex zeros with higher multiplicity.
N
23. (a) Let u and v be√two numbers of the form u = a + b c, where a, b and c are rational
numbers, with c an irrational constant. Let the √ notation u∗ indicate the value of u
U
when the sign of b is reversed. That is, u = a − b c.
(i) Show that u∗ + v ∗ = (u + v)∗ .
(ii) Show that λu∗ = (λu)∗ whenever λ is a rational number.
(iii) Prove by induction that (un )∗ = (u∗ )n for positive integers n.
(b) Suppose that u = a + b c is a zero of a certain polynomial
√ with rational coefficients.
Use the results of part (a) to show that u∗ = a − b c is also a zero of this polynomial.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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54 CHAPTER 1: Complex Numbers
## 1H Chapter Review Exercise
Exercise 1H
1. If z = 3 − i and w = 17 + i, find:
w
(a) 6z − w (b) z 3 (c)
z
2. Write as a product of two complex linear factors.
(a) z 2 + 100 (b) z 2 + 10z + 34
3. Solve each quadratic equation for z.
(a) z 2 − 8z + 25 = 0 (b) 16z 2 + 16z + 13 = 0
FS
4. Find the square roots of:
(a) 5 − 12i (b) 7 + 6 2 i
5. Solve for z:
O
(a) z 2 − 5z + (7 + i) = 0 (b) z 2 − (6 + i)z + (14 + 8i) = 0
D E
O
6. If 3i is a zero of a polynomial P (z) with real coefficients, explain why z 2 + 9 is a factor
of P (z).
G
PR
7. It is known that 2 + 5i is a zero of the polynomial P (z) = z 3 − 8z 2 + 45z − 116.
TE ID
(a) Why is 2 − 5i also a zero of P (z)?
(b) Use the sum of the zeroes to find the third zero of P (z).
EC BR
(c) Hence write P (z) as a product of two factors with real coefficients.
8. Express each complex number in modulus-argument form.
(a) 1 − i (b) −3 3 + 3i
R M
## 9. Express each complex number in Cartesian form.
(a) 4 cis π2 (b) 6 cis(− 3π
4 )
O CA
10. Simplify:
10 cis 10θ
(a) 2 cis π2 × 3 cis π3 (b) (c) (3 cis 3α)2
5 cis 5θ
R
11. Sketch the graph in the complex plane represented by the equation:
(a) |z − 2i| = 2 (c) arg(z + 2) = − π4
z−1 π
(b) |z| = |z − 2 − 2i| (d) arg =
z+1 2
C
12. Shade the region in the complex plane that simultaneously satisfies |z| ≥ 1, Re(z) ≤ 2
and − π3 ≤ arg z ≤ π3 .
N
13. Suppose that z = −1 + 3 i and w = 1 + i.
z
U
## (a) Find in the form a + ib, where a and b are real.
w
(b) Write z and w in modulus-argument form.
z
(c) Hence write in modulus-argument form.
w
√ √
6 − 2
(d) Deduce that cos 5π 12 = .
4
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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CHAPTER 1: Complex Numbers 1H Chapter Review Exercise 55
## 14. Sketch the graph specified by the equation:
(a) zz = z + z (b) z = iz (c) |z + 2| = 2|z − 4|
15. A triangle P QR in the complex plane is isosceles, with 6 P = 90◦ . The points P and Q
represent the complex numbers 4 − 2i and 7 + 3i respectively. Find the complex numbers
represented by:
(a) the vector P Q, (b) the vector P R, (c) the point R.
16. If z1 = 4 − i and z2 = 2i, find in each case the two possible values of z3 so that the points
representing z1 , z2 and z3 form an isosceles right-angled triangle with the right-angle at:
(a) z1 (b) z2
17. In an Argand diagram, O is the origin and the points P and Q represent the complex
FS
numbers z1 and z2 respectively.
If triangle OP Q is equilateral, prove that z1 2 + z2 2 = z1 z2 .
π
18. If z1 = 2 cis 12 and z2 = 2i, find:
O
(a) arg(z1 + z2 ) (b) arg(z2 − z1 )
D E
O
19. If z1 and z2 are complex numbers such that |z1 | = |z2 |, prove that
G
arg(z1 z2 ) = arg (z1 + z2 )2 .
PR
TE ID
z2 − 1
20. If z = cis θ, show that = i tan θ.
z2 + 1
1
EC BR
21. The points A, B, C and O represent the numbers z, , 1 and 0 respectively in the complex
z
plane. Given that 0 < arg z < π2 , prove that 6 OAC = 6 OCB.
22. (a) By drawing a suitable diagram, prove the triangle inequality |z1 − z2 | ≥ |z1 | − |z2 |.
R M
4
(b) Hence find the maximum value of |z| given that z − = 2.
z
O CA
R
C
N
U
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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56 CHAPTER 1: Complex Numbers
## Chapter One 13(a) α
14(a)(i) y
Exercise 1A (Page 8)
1(a) −1 (b) 1 (c) i −i (d)
(e) i (f) −1 0
(g) 1 (h) -Öa
2(a) −2i (b) 3 − i (c) 1 + i (d) 5 + 3i (e) −3 − 2i Öa x
3(a) 12 − 2i (b) −6 + 2i (c) 1 + 5i (d) 7 − 11i
4(a) −5 + 4i (b) 5 + 5i (c) 14 + 5i (d) −26 + 82i
(e) 24 + 10i (f) −5 − 12i (g) 2 + 11i (h) −4 √ √
(i) 28 − 96i 15(a) ± √12 (1 − i) (b) ± 2(1 + 2i) (c) ±( 3 + i)
5(a) 5 (b) 17 (c) 29 (d) 65 (d) ± 2(3 − 2i)
q q
6(a) −i (b) 1 −2i (c) 3 +2i (d) 1 −2i (e) −1 +3i
√ √
FS
(e) ± 5+1−i 5−1
(f) − 15 + 35 i q q
√ √
7(a) −2 − i (b) 4 − 3i (c) 3 + 7i (d) 3 (e) −3 + 4i 16(a) −2 − i ± 2+1+i 2−1
8(a) 6 + 2i (b) 18 (c) 19 − 22i (d) 8 − i (e) 1 + 2i q
q
O
9(a) 22 + 19i (b) 6 + 15i (c) 4 − 2i (d) 2 − 3i (b) 1 + i ± 5−1−i 5+1
(e) 6 √ √ √
D E
O
(c) −1 + i 3 ± 2−i 6
10(a) x = 3 and y = −2 (b) x = 2 and y = −1 q q
√ √
(c) x = 6 and y = 2
G
(d) x = 14 5 and y = 5
3
(d) 21 −1 + i ± 13 + 2 − i 13 − 2
PR
(e) x = 35 and y = − 39
2 2 19 The term b/|b| is the sign of b.
TE ID
11(a) 10 − 13
9
i (b) 1 (c) − 298
(d) −4 − 52 i
10 It is 1 when b > 0, and −1 when b < 0.
x−iy x2 −y 2 −2ixy x2 +y 2 −1+2iy
16(a) x2 +y 2 (b) (x2 +y 2 )2
(c) (x+1)2 +y 2
Exercise 1C (Page 21)
EC BR
Exercise 1B (Page 16) 1(a) (2, 0) (b) (0, 1) (c) (−3, 5) (d) (2, −2)
1(a) z = ±3i (b) z = 2 ± 4i (c) z = −1 ± 2i (e) (−5, −5) (f) (−1, 2)
(d) z = 3 ± i (e) z = 21 ± 14 i (f) z = − 32 ± 2i 2(a) −3 + 0i = −3 (b) 0 + 3i = 3i (c) 7 − 5i
√ √
R M
## 2(a) (z − 6i)(z + 6i) (b) (z − 2 2 i)(z + 2 2 i) (d) a + bi
(c) (z − 1 − 3i)(z − 1 + 3i) (d) (z + 2 − i)(z + 2 + i) 3(a)
√ √ y
A
O CA
(e) (z − 3 + 5 i)(z − 3 − 5 i) 3
√ √
1 3 1 3
(f) (z + 2 − 2 i)(z + 2 + 2 i)
2 2
3(a) z + 2 = 0 (b) z − 2z + 2 = 0 B 1 1 3
2 2
(c) z + 2z + 5 = 0 (d) z − 4z + 7 = 0 -3 -1 x
R
-1 D
4(a) ±(1 + i) (b) ±(2 + i) (c) ±(−1 + 3i)
(d) ±(6+i) (e) ±(2+3i) (f) ±(5−i) (g) ±(1−4i) -3
C
(h) ±(5 − 4i)
5(a) ±(1 − 2i) (b) z = 2 − i or 1 + i (b)A square. (c) An anticlockwise rotation of
C
## 7(a) z = 1 − i or i (b) z = −3 + 2i or −2i (c) z = 4(a) y (b) y
iz 3
4+i or 2−i (d) z = −2+i or 21 (3−i) (e) z = −5+i w
2
N
or 3 − 2i (f) z = 3 + i or −1 − 3i z iw
1 1
8(a) w = −1 (b) a = −6 and b = 13 -3 -1 -1 2
U
## (c) k = 8 − i and the other root is 2 + 3i. -1 1 3 x -2 -1 1 x
-z -iw
9 z = ±(2 + i)
-w -2
10(a) cos θ + i sin θ or cos θ − i sin θ
-3 -iz
√ √
1 3 3 1
11(a) z = −1 or 2 ± 2 i (b) z = i or ± 2 − 2 i
In (a) and (b) the points form a square.
12(a) x = ω satisfies the equation. (c) They are
complex conjugates.
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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CHAPTER 1: Complex Numbers Answers to Chapter One 57
2 w
z (z + w)
1 3 1
w
2
1 3 x 2 x x
-1 1 z
z
-2
w 1 3 4 x
## Conjugates pairs are re- With O the points form 6 y 8 y
flections in the real axis. a parallelogram. S 2
(e) y (f) y P
1 -z z
-2 -1
FS
2
w
2
w T 1 2 x x
z (w - z) z R -1 Q -z z
1 1
-2
3 x -2 -1 1 3 x
-1 1
O
(z - w)
10(c) right-isosceles
D E
11 It is the circle centre (0, −1) with radius 1,
O
Again, in (e) and (f) the points are the vertices of omitting the origin.
a parallelogram. G 12 It is the circle centre (3, 0) with radius, omit-
PR
5(a) y (b) y ting the origin.
TE ID
14 It is a parabola with focus the origin and di-
2 Im(z) = 2 rectrix x = 1.
EC BR
15(a) y (b) y
-3 x
Re(z) = -3 x
c c
-c -c c
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c x x
(c) y (d) y -c -c
O CA
1
-2
x x
Exercise 1D (Page 28)
R
1(a) 3 (b) 5 (c) 2 (d) 2 (e) 5 (f) 17
2(a) π (b) π2 (c) − π4 (d) π3 (e) 3π (f) − 6
π
√ 4
(e) (f) 3(a) 2 cis 2 (b) 4 cis π (c) 2 cis 4 (d) 2 cis − π6
π
y y
(e) 2 cis 2π
3 (f) cis − 3π 4
2 4(a) 5 cis(0·93) (b) 13 cis(−0·39)
√ √
C
## (c) 5 cis(2·68) (d) 10 cis(−1·89)
√ √ √
x x 5(a) 3 (b) −5i (c) 2 2 + 2 2 i (d) 3 3 − 3i
1 √ √ √
N
(e) − 2 + 2 i (f) −1 − 3 i
√ √ √
Re(z) = Im(z) 2Re(z) = Im(z) 6(a) 2 cis − π4 (b) 2 cis π4 (c) 2 cis 3π
√ 4
U
(d) 2 cis π4 (e) 2 cis − π2 1
(f) √ cis − 4
2
π
π π 3
7(a) 10 cis 3 (b) 9 cis 3θ (c) 2 cis 3 (d) 2 cis α
2π 6π
(e) 16 cis 5 (f) 8 cis 7
√ √ √ √
8(a) 2 2 (b) 3 2 (c) 6 (d) 4 3 (e) 5 (f) 5
π 3π π π ◦
9(a) 4 (b) − 4 (c) − 3 (d) 6 (e) 0·93 (53 )
(f) −2·03 (−117 )
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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58 CHAPTER 1: Complex Numbers
## √ 1 19(c) The sum of the squares of the diagonals of a
10(a) i (b) −1 (c) 21 (1 + i 3) (d) √ (−1 + i)
√ 2 parallelogram is equal to the sum of the squares
(e) 12 (− 3 + i) (f) −i (g) √1 (1 − i) of its sides.
√ 2
1
(h) − 2 (1 + i 3) 20(c) parallelogram (d) arg w
z
= π2 , so wz is purely
11 y imaginary.
C 23 Use the converse of the opposite angles of a
2r 24 Take the argument of the fraction in the hint.
r
p p D (OD = 1 r)
3 4 2 z3 z3 − z1
The result is arg − arg . These
x z2 z2 − z1
O r B are the angles at 0 and z1 which, by the angles in
the same segment theorem, are equal. Finally, use
FS
13(a) z1 = 2 cis π6 and z2 = 4 cis π4 (b) z1 z2 = the result of Question 24.
8 cis 5π
12 and zz21 = 2 cis 12
π
14 z1 = 2 cis 5π 6
, z2 = 2 cis(− 3π 4
), Exercise 1F (Page 42)
O
1(a) (b)
z1 z2 = 2 2 cis 12 π
and zz12 = 22 cis 5π 12
y
x=1
y
√ √ √
D E
1 π
15(a) 2 ( 3 + 1) + i( 3 − 1) (b) 2 cis 12
O
√ i
1
(c) √ ( 3 + 1)
2 √ 2
G
PR
16(a) 2 (b) π4 (c) 1 +i -3 1 5x -1 x
θ−φ θ+φ
TE ID
24 z + w = 2 cos 2 cis 2
25(a) When Im(z) = 0. y = -x
EC BR
(c) (d)
Exercise 1E (Page 35) y y
1(a) 7 + 4i (b) −3 + 2i (c) 3 − 2i 1 4
-2 + 2i
2
2(a) −3 + 4i (b) 1 + 7i (c) −4 − 3i (d) −7 + i
-1 2 x
3 −3 + 6i 4-i
R M
## 4(a) B represents 1 + 3i, C represents −1 + 2i
-2 x y = 2x - 4
√ √ -4
(b) − 2 + 2 2 i y=x+2
O CA
## 5(a) 4 + 3i (b) −3 + 4i (c) 2 + 7i
6(a) −5 + 12i (b) −3 − 4i 2(a) (b)
y y
8 E represents w2 − w1 , F represents i(w2 − w1 ),
R
## C represents w2 + i(w2 − w1 ) and D represents 4
w1 + i(w2 − w1 ).
1
9(a) Vectors BA and BC represent z1 − z2 and 3p
4 p
z3 − z2 respectively, and BA is the anticlockwise 4
## rotation of BC through 90◦ about B. So z1 −z2 = 4 x -1 x
i(z3 − z2 ). Squaring both sides gives the result.
C
(b) z1 − z2 + z3 (c) y
10(a) 2ωi (b) 21 ω(1 + 2i)
N
11 −2 and 1 − 3 i
U
p
12(a) w = −4 + 3i or 4 − 3i (b) w = −1 + 7i or 3
Ö3
7 + i (c) w = 21 (7 + i) or 12 (−1 + 7i)
13 −2 + 2i, 12i, 4
1 x
π π 5π 11π
18(a) z1 = 2 cis 2 , z2 = 2 cis 3 (c)(i) 12 (ii) 12
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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CHAPTER 1: Complex Numbers Answers to Chapter One 59
1
2 4
1 2
x
1 2 -1
-3
-1 x 3 x
## (c) y 5(a)(i) y (ii) y
1 1
1 x
Ö2 2 4 x 2 4 x
-1 -1
FS
-1
-3 -3
O
4(a) y (b) y (b)(i) y (ii) y
D E
O
8 1 1
-1 1 3 -1 1 3
G 2 4 x x
PR
-1 -1
x
-1
TE ID
x -3 -3
4
EC BR
## (c) y (d) y (c)(i) y (ii) y
1 1
1 3p
3
R M
-1 x x 1 2 x 1 2 x
-1
O CA
-1 -1
R
1 4 7 1 4 7
x x
-2 -2
p
6 -2
p p -5 -5
3 x 4 x
p
4
-1 -8 -8
C
## (g) y (h) y (e)(i) y (ii) y
N
-1 1 x
3 3
2 -1
U
p p
1 4
1 4
2 x
-2
-1 1 3 x -1 1 3 x
-1 -1
-3
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
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60 CHAPTER 1: Complex Numbers
## (f)(i) y (ii) y 8(a) y (b) y
1 1
2
p p
4 4
-1 1 x -1 1 x x 2 4 x
-2
-1 -1
## (g)(i) y (ii) y (c) y
5 5
1
2 2
FS
p
4
p
4 -1 1 x
-1 -1
-4 2 px -4 2 px -1
-1 3 -1 3
O
(h)(i) y (ii) y 9(a) y (b) y
D E
O
6 6
1 1
1 G 1
PR
1 2 x 1 2 x
-2 3 8 x -2 3 8 x
TE ID
-1
-4 -4
EC BR
## 6(a) y (b) y (c) y (d) y
3
1 -1 1 3
2 1 -1-Ö2
R M
x
2 x 1 3 x -1 x 1- i 2
Ö3
-1
O CA
## 7(a) y (b) y (e) y (f) y
R
-4
d 2 -2 x
3 S
S
-2
-2 x x
-2Ö3 -2
d -4
x
C
(c) (d)
√ √ √
y y 10(b) 3 cis π3 = 23 (1 + i 3 )
N
d √
2 11(a) arg(z + 3) = π 3
(b) |z| = 3 3
2
, arg z = 5π
6
9 3 3
(c) − 4 + 4 i
U
1 S
-2 S x
x
-2 d -1
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
CHAPTER 1: Complex Numbers Answers to Chapter One 61
## 12(a) y (b) y 18(a) y (b) y
(2, Ö3 )
2 2 3 3
Ö7
3
1
-1 1 3 x 1 3 5 x -5 x
2 x
-2 -2
√ √ √ √
3 and 1 (i) 5 + 1 and 5 − 1 26 + 3 and 26 − 3
13(a)(i) y √ √
(ii) 2 2 + 1 and 2 2 − 1
p (c)(i) |z0 | − r ≤ |z| ≤ |z0 | + r
6 1
FS
(ii) |z0 − z1 | − r ≤ |z − z1 | ≤ |z0 − z1 | + r
2 x 19(a) straight line external to z1 and z2 (b) major
arc (c) semi-circle (d) minor arc (e) interval
O
between z1 and z2
20 It is the circle with the interval joining z1 and
D E
(b) This is simply part (a) shifted left by 2.
O
z2 as diameter.
14(a) y 21(b) The graph is the perpendicular bisector of
G
PR
the line joining z1 and z2 .
z
TE ID
Exercise 1G (Page√51) √
4
10 1(a) (x − 2)(x + 1 − 3 )(x + 1 + 3 )
√ √
x
EC BR
## 3 (b) (x − 1)(x + 2 − 2 )(x + 2 + 2 )
√ √
(c) (x − 1)(x − 1 − 5 )(x − 1 + 5 )
2(a) The coefficients of P (x) are real, so complex
(b) 15 (c) 9 + 12i
zeroes occur in conjugate pairs. (b) 6
R M
## 15(b)(i) |z + 2| = 2, centre −2, radius 2
3(a) 1 + 2i; the coefficients of P (x) are real, so
(ii) |z − (1 + i)| = 1, centre 1 + i, radius 1
complex zeroes occur in conjugate pairs.
(iii) |z − 1| = 1, centre 1, radius 1
O CA
2
(c) P (x) = (x + 2)(x − 2x + 5)
16(a) The line through 1 and i, omitting i.
4(a) 3i; the coefficients of P (z) are real, so complex
(b) The circle with diameter joining 1 and i, omit- 2
zeroes occur in conjugate pairs. (b) z + 9
ting these two points. 2
R
## (c) P (z) = (2z + 3)(z + 9)
17(a) y (b) y
5(b) 0; the coefficients of P (z) are real, so complex
zeroes occur in conjugate pairs.
(c)(i) P (z) = (2z − 1)(z − 3 − i)(z − 3 + i)
1 x 1 x 2
(ii) P (z) = (2z − 1)(z − 6z + 10)
-1 -1
6(a) The coefficients of Q(x) are real, so complex
C
## zeroes occur in conjugate pairs. (b) 3 + 5,
3− 5
N
√ √
(c)(i) (x − 2i)(x + 2i)(x − 3 − 5 )(x − 3 + 5 )
2
√ √
(ii) (x + 4)(x − 3 − 5 )(x − 3 + 5 )
U
2 2
(iii) (x + 4)(x − 6x + 4)
7(a) x = 1 ± 3i, 3 or −2 (b) x = 1 ± i or 2 ± i
√ √
8(b) x = 3, − 21 + 23 i, − 21 − 23 i
9(a) a = 3 (b) b = 1
2 2
(c) (z − 6z + 10)(z − 6z + 13)
10(b) k = 3
## Mathematics Extension 2 Chapter 1: Complex Numbers Complex 25/7/19 Copyright
c 2019
62 CHAPTER 1: Complex Numbers
## 11(b) m = 7, n = −4 11(a) y (b) y
12(a) −7 − 4i (b)(i) −7 + 4i (ii) 2x − 7
4 4 4
13(b) P (z) = 21 (z − 2)(2z − 1)
4
2 2 + 2i
so one root is z = 2.
√4 1
√4 1
2 1
(c) 2, √
4 , −
2
2, − √4 ,
2
√4 1
4 1 1 2 x
and i 2, √ i, −i 2, −√ i x
-2 2
4 4
2 2
14(a) P (x) has minimum value B, when x = 0. y=2-x
Since B > 0, it follows that P (x) > 0 for all real
(c) y (d) y
values of x. (b) −ic, −id; the coefficients of
P (x) are real, so complex zeroes occur in conju-
gate pairs. -2 x 1
y
15(a) They form a con- (c) -2
FS
jugate pair, since P (x) -1 1 x
has real coefficients.
d
O
12 y
D E
x
O
p
1 3
16(a) G
The minimum stationary point is at x = 1.
PR
f(1) = k − 2 > 0. Hence the graph of f(x) has -1 1 2x
TE ID
only one x-intercept which lies to the left of the
-1
maximum stationary point at x = −1.
(b) f(x) has real coefficients (d) −14, 7 ± 12i
EC BR
1
√ √
17 Hint: consider P (x) − P (x)
0 13(a) 2 ( 3 − 1) + 12 ( 3 + 1)i (b) z = 2 cis 2π
3 and
√ √
18(b) −1 + 2i is a double zero of P (z) (c) The w= 2 cis π4 (c) 2 cis 5π
12
coefficients of P (z) are real and −1 + 2i counts as 14(a) y (b) y
R M
## two of the zeroes of P (z), so its conjugate −1 − 2i
1 2
must also count as two zeroes.
O CA
2 2 2 2
(d) P (z) = (z +1−2i) (z +1+2i) = (z +2z +5)
2 2
22(b) (z −α) (z −α) is a factor. (c) Hint: Begin 1 2 x -2 2 x
2
by writing: P (z) = z − 2 Re(α) + |α|2 × Q(z) -1 -2
R
## Exercise 1H (Page 53)
(c) y
1(a) 1 − 5i (b) 18 − 26i (c) 5 + 2i
2(a) (z + 10i)(z − 10i) (b) (z + 5 − 3i)(z + 5 + 3i)
1 3 1 3
4
3(a) z = 4+3i or 4−3i (b) z = − 2 + 4 i or − 2 − 4 i
4(a) ±(3 − 2i) (b) ±(3 + 2 i)
2 6 10 x
C
5(a) z = 2 + i or 3 − i (b) z = 2 + 3i or 4 − 2i
2 -4
6 3i = −3i is also a zero, so (z−3i)(z+3i) = z +9
N
is a factor.
7(a) The coefficients of P (z) are real. (b) 4 15(a) 3 + 5i (b) −5 + 3i (c) −1 + i
U
2
(c) P (z) = (z − 4)(z − 4z + 29) 16(a) 1 − 5i, 7 + 3i (b) 3 + 6i, −3 − 2i
8(a) 2 cis(− π4 ) (b) 6 cis 5π 7π
18(a) 24 (b) 24
19π
√ √ 6
9(a) 4i (b) − 3 − 3 i 21 Use similar triangles.
5π √
10(a) 6 cis 6 (b) 2 cis 5θ (c) 9 cis 6α 22(b) 5+1
c 2019 | 47,351 | 132,601 | {"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} | 4.40625 | 4 | CC-MAIN-2019-51 | latest | en | 0.919788 |
https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/1_Algebraic_functions/1.2_Trinomial_products/1.2.1_Quadratic/1.2.1.2-d+e_x-%5Em-a+b_x+c_x%5E2-%5Ep/rese2289.htm | 1,696,152,387,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510810.46/warc/CC-MAIN-20231001073649-20231001103649-00403.warc.gz | 666,373,097 | 7,976 | ### 3.2289 $$\int \frac{(d+e x)^{5/2}}{a+b x+c x^2} \, dx$$
Optimal. Leaf size=459 $-\frac{\sqrt{2} \left (-3 c^2 d e \left (-d \sqrt{b^2-4 a c}+2 a e+b d\right )+c e^2 \left (-3 b d \sqrt{b^2-4 a c}-a e \sqrt{b^2-4 a c}+3 a b e+3 b^2 d\right )-b^2 e^3 \left (b-\sqrt{b^2-4 a c}\right )+2 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{c^{5/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\sqrt{2} \left (-3 c^2 d e \left (d \sqrt{b^2-4 a c}+2 a e+b d\right )+c e^2 \left (3 b \left (d \sqrt{b^2-4 a c}+a e\right )+a e \sqrt{b^2-4 a c}+3 b^2 d\right )-b^2 e^3 \left (\sqrt{b^2-4 a c}+b\right )+2 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{c^{5/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{2 e \sqrt{d+e x} (2 c d-b e)}{c^2}+\frac{2 e (d+e x)^{3/2}}{3 c}$
[Out]
(2*e*(2*c*d - b*e)*Sqrt[d + e*x])/c^2 + (2*e*(d + e*x)^(3/2))/(3*c) - (Sqrt[2]*(2*c^3*d^3 - b^2*(b - Sqrt[b^2
- 4*a*c])*e^3 - 3*c^2*d*e*(b*d - Sqrt[b^2 - 4*a*c]*d + 2*a*e) + c*e^2*(3*b^2*d - 3*b*Sqrt[b^2 - 4*a*c]*d + 3*a
*b*e - a*Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]
])/(c^(5/2)*Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]) + (Sqrt[2]*(2*c^3*d^3 - b^2*(b + Sqrt[b
^2 - 4*a*c])*e^3 - 3*c^2*d*e*(b*d + Sqrt[b^2 - 4*a*c]*d + 2*a*e) + c*e^2*(3*b^2*d + a*Sqrt[b^2 - 4*a*c]*e + 3*
b*(Sqrt[b^2 - 4*a*c]*d + a*e)))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e
]])/(c^(5/2)*Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e])
________________________________________________________________________________________
Rubi [A] time = 4.43191, antiderivative size = 459, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.227, Rules used = {703, 824, 826, 1166, 208} $-\frac{\sqrt{2} \left (-3 c^2 d e \left (-d \sqrt{b^2-4 a c}+2 a e+b d\right )+c e^2 \left (-3 b d \sqrt{b^2-4 a c}-a e \sqrt{b^2-4 a c}+3 a b e+3 b^2 d\right )-b^2 e^3 \left (b-\sqrt{b^2-4 a c}\right )+2 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{c^{5/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\sqrt{2} \left (-3 c^2 d e \left (d \sqrt{b^2-4 a c}+2 a e+b d\right )+c e^2 \left (3 b \left (d \sqrt{b^2-4 a c}+a e\right )+a e \sqrt{b^2-4 a c}+3 b^2 d\right )-b^2 e^3 \left (\sqrt{b^2-4 a c}+b\right )+2 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{c^{5/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{2 e \sqrt{d+e x} (2 c d-b e)}{c^2}+\frac{2 e (d+e x)^{3/2}}{3 c}$
Antiderivative was successfully verified.
[In]
Int[(d + e*x)^(5/2)/(a + b*x + c*x^2),x]
[Out]
(2*e*(2*c*d - b*e)*Sqrt[d + e*x])/c^2 + (2*e*(d + e*x)^(3/2))/(3*c) - (Sqrt[2]*(2*c^3*d^3 - b^2*(b - Sqrt[b^2
- 4*a*c])*e^3 - 3*c^2*d*e*(b*d - Sqrt[b^2 - 4*a*c]*d + 2*a*e) + c*e^2*(3*b^2*d - 3*b*Sqrt[b^2 - 4*a*c]*d + 3*a
*b*e - a*Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]
])/(c^(5/2)*Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]) + (Sqrt[2]*(2*c^3*d^3 - b^2*(b + Sqrt[b
^2 - 4*a*c])*e^3 - 3*c^2*d*e*(b*d + Sqrt[b^2 - 4*a*c]*d + 2*a*e) + c*e^2*(3*b^2*d + a*Sqrt[b^2 - 4*a*c]*e + 3*
b*(Sqrt[b^2 - 4*a*c]*d + a*e)))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e
]])/(c^(5/2)*Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e])
Rule 703
Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1))/(c*
(m - 1)), x] + Dist[1/c, Int[((d + e*x)^(m - 2)*Simp[c*d^2 - a*e^2 + e*(2*c*d - b*e)*x, x])/(a + b*x + c*x^2),
x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*
e, 0] && GtQ[m, 1]
Rule 824
Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g
*(d + e*x)^m)/(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x])
/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*
e^2, 0] && FractionQ[m] && GtQ[m, 0]
Rule 826
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
Rule 1166
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{(d+e x)^{5/2}}{a+b x+c x^2} \, dx &=\frac{2 e (d+e x)^{3/2}}{3 c}+\frac{\int \frac{\sqrt{d+e x} \left (c d^2-a e^2+e (2 c d-b e) x\right )}{a+b x+c x^2} \, dx}{c}\\ &=\frac{2 e (2 c d-b e) \sqrt{d+e x}}{c^2}+\frac{2 e (d+e x)^{3/2}}{3 c}+\frac{\int \frac{c^2 d^3-3 a c d e^2+a b e^3+e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) x}{\sqrt{d+e x} \left (a+b x+c x^2\right )} \, dx}{c^2}\\ &=\frac{2 e (2 c d-b e) \sqrt{d+e x}}{c^2}+\frac{2 e (d+e x)^{3/2}}{3 c}+\frac{2 \operatorname{Subst}\left (\int \frac{e \left (c^2 d^3-3 a c d e^2+a b e^3\right )-d e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) x^2}{c d^2-b d e+a e^2+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt{d+e x}\right )}{c^2}\\ &=\frac{2 e (2 c d-b e) \sqrt{d+e x}}{c^2}+\frac{2 e (d+e x)^{3/2}}{3 c}+\frac{\left (2 c^3 d^3-b^2 \left (b-\sqrt{b^2-4 a c}\right ) e^3-3 c^2 d e \left (b d-\sqrt{b^2-4 a c} d+2 a e\right )+c e^2 \left (3 b^2 d-3 b \sqrt{b^2-4 a c} d+3 a b e-a \sqrt{b^2-4 a c} e\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{2} \sqrt{b^2-4 a c} e+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{c^2 \sqrt{b^2-4 a c}}-\frac{\left (2 c^3 d^3-b^2 \left (b+\sqrt{b^2-4 a c}\right ) e^3-3 c^2 d e \left (b d+\sqrt{b^2-4 a c} d+2 a e\right )+c e^2 \left (3 b^2 d+a \sqrt{b^2-4 a c} e+3 b \left (\sqrt{b^2-4 a c} d+a e\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{2} \sqrt{b^2-4 a c} e+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{c^2 \sqrt{b^2-4 a c}}\\ &=\frac{2 e (2 c d-b e) \sqrt{d+e x}}{c^2}+\frac{2 e (d+e x)^{3/2}}{3 c}-\frac{\sqrt{2} \left (2 c^3 d^3-b^2 \left (b-\sqrt{b^2-4 a c}\right ) e^3-3 c^2 d e \left (b d-\sqrt{b^2-4 a c} d+2 a e\right )+c e^2 \left (3 b^2 d-3 b \sqrt{b^2-4 a c} d+3 a b e-a \sqrt{b^2-4 a c} e\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}\right )}{c^{5/2} \sqrt{b^2-4 a c} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}+\frac{\sqrt{2} \left (2 c^3 d^3-b^2 \left (b+\sqrt{b^2-4 a c}\right ) e^3-3 c^2 d e \left (b d+\sqrt{b^2-4 a c} d+2 a e\right )+c e^2 \left (3 b^2 d+a \sqrt{b^2-4 a c} e+3 b \left (\sqrt{b^2-4 a c} d+a e\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}\right )}{c^{5/2} \sqrt{b^2-4 a c} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}\\ \end{align*}
Mathematica [A] time = 1.08186, size = 455, normalized size = 0.99 $-\frac{\sqrt{2} \left (3 c^2 d e \left (d \sqrt{b^2-4 a c}-2 a e-b d\right )+c e^2 \left (-3 b d \sqrt{b^2-4 a c}-a e \sqrt{b^2-4 a c}+3 a b e+3 b^2 d\right )+b^2 e^3 \left (\sqrt{b^2-4 a c}-b\right )+2 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{c^{5/2} \sqrt{b^2-4 a c} \sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}+\frac{\sqrt{2} \left (-3 c^2 d e \left (d \sqrt{b^2-4 a c}+2 a e+b d\right )+c e^2 \left (3 b \left (d \sqrt{b^2-4 a c}+a e\right )+a e \sqrt{b^2-4 a c}+3 b^2 d\right )-b^2 e^3 \left (\sqrt{b^2-4 a c}+b\right )+2 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{c^{5/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}-\frac{2 e \sqrt{d+e x} (b e-2 c d)}{c^2}+\frac{2 e (d+e x)^{3/2}}{3 c}$
Antiderivative was successfully verified.
[In]
Integrate[(d + e*x)^(5/2)/(a + b*x + c*x^2),x]
[Out]
(-2*e*(-2*c*d + b*e)*Sqrt[d + e*x])/c^2 + (2*e*(d + e*x)^(3/2))/(3*c) - (Sqrt[2]*(2*c^3*d^3 + b^2*(-b + Sqrt[b
^2 - 4*a*c])*e^3 + 3*c^2*d*e*(-(b*d) + Sqrt[b^2 - 4*a*c]*d - 2*a*e) + c*e^2*(3*b^2*d - 3*b*Sqrt[b^2 - 4*a*c]*d
+ 3*a*b*e - a*Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - b*e + Sqrt[b^2 - 4*a
*c]*e]])/(c^(5/2)*Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e]) + (Sqrt[2]*(2*c^3*d^3 - b^2*(b +
Sqrt[b^2 - 4*a*c])*e^3 - 3*c^2*d*e*(b*d + Sqrt[b^2 - 4*a*c]*d + 2*a*e) + c*e^2*(3*b^2*d + a*Sqrt[b^2 - 4*a*c]
*e + 3*b*(Sqrt[b^2 - 4*a*c]*d + a*e)))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*
a*c])*e]])/(c^(5/2)*Sqrt[b^2 - 4*a*c]*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e])
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Maple [B] time = 0.335, size = 1929, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*x+d)^(5/2)/(c*x^2+b*x+a),x)
[Out]
2/3*e*(e*x+d)^(3/2)/c-2/c^2*(e*x+d)^(1/2)*b*e^2+4*d*e*(e*x+d)^(1/2)/c-3/c/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b
*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2
))*c)^(1/2))*a*b*e^4+6/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan(
(e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a*d*e^3+1/c^2/(-e^2*(4*a*c-b^2))^(1/2)
*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*
c-b^2))^(1/2))*c)^(1/2))*b^3*e^4-3/c/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)
^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2*d*e^3+3/(-e^2*(4*a*c
-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d
+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b*d^2*e^2-2*e*c/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c
-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*d^3-1/c*
2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c
-b^2))^(1/2))*c)^(1/2))*a*e^3+1/c^2*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2
)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2*e^3-3/c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))
^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b*d*e^2+3*e*2^
(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b
^2))^(1/2))*c)^(1/2))*d^2-3/c/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)
*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a*b*e^4+6/(-e^2*(4*a*c-b^2))
^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-
e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a*d*e^3+1/c^2/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2
))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^3*e^4-3/
c/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(
1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2*d*e^3+3/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*
d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)
^(1/2))*b*d^2*e^2-2*e*c/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arcta
nh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*d^3+1/c*2^(1/2)/((-b*e+2*c*d+(-e^2
*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))
*a*e^3-1/c^2*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*
c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2*e^3+3/c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*ar
ctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b*d*e^2-3*e*2^(1/2)/((-b*e+2*c*
d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)
^(1/2))*d^2
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{\frac{5}{2}}}{c x^{2} + b x + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^(5/2)/(c*x^2+b*x+a),x, algorithm="maxima")
[Out]
integrate((e*x + d)^(5/2)/(c*x^2 + b*x + a), x)
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Fricas [B] time = 17.1642, size = 13779, normalized size = 30.02 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^(5/2)/(c*x^2+b*x+a),x, algorithm="fricas")
[Out]
1/6*(3*sqrt(2)*c^2*sqrt((2*c^5*d^5 - 5*b*c^4*d^4*e + 10*(b^2*c^3 - 2*a*c^4)*d^3*e^2 - 10*(b^3*c^2 - 3*a*b*c^3)
*d^2*e^3 + 5*(b^4*c - 4*a*b^2*c^2 + 2*a^2*c^3)*d*e^4 - (b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*e^5 + (b^2*c^5 - 4*a*c^
6)*sqrt((25*c^8*d^8*e^2 - 100*b*c^7*d^7*e^3 + 100*(2*b^2*c^6 - a*c^7)*d^6*e^4 - 50*(5*b^3*c^5 - 6*a*b*c^6)*d^5
*e^5 + 10*(21*b^4*c^4 - 43*a*b^2*c^5 + 11*a^2*c^6)*d^4*e^6 - 20*(6*b^5*c^3 - 18*a*b^3*c^4 + 11*a^2*b*c^5)*d^3*
e^7 + 5*(9*b^6*c^2 - 36*a*b^4*c^3 + 36*a^2*b^2*c^4 - 4*a^3*c^5)*d^2*e^8 - 10*(b^7*c - 5*a*b^5*c^2 + 7*a^2*b^3*
c^3 - 2*a^3*b*c^4)*d*e^9 + (b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)*e^10)/(b^2*c^10 - 4*a*
c^11)))/(b^2*c^5 - 4*a*c^6))*log(sqrt(2)*(10*(b^2*c^5 - 4*a*c^6)*d^5*e^2 - 25*(b^3*c^4 - 4*a*b*c^5)*d^4*e^3 +
10*(3*b^4*c^3 - 14*a*b^2*c^4 + 8*a^2*c^5)*d^3*e^4 - 10*(2*b^5*c^2 - 11*a*b^3*c^3 + 12*a^2*b*c^4)*d^2*e^5 + (7*
b^6*c - 44*a*b^4*c^2 + 66*a^2*b^2*c^3 - 8*a^3*c^4)*d*e^6 - (b^7 - 7*a*b^5*c + 13*a^2*b^3*c^2 - 4*a^3*b*c^3)*e^
7 - (2*(b^2*c^7 - 4*a*c^8)*d^2 - 2*(b^3*c^6 - 4*a*b*c^7)*d*e + (b^4*c^5 - 6*a*b^2*c^6 + 8*a^2*c^7)*e^2)*sqrt((
25*c^8*d^8*e^2 - 100*b*c^7*d^7*e^3 + 100*(2*b^2*c^6 - a*c^7)*d^6*e^4 - 50*(5*b^3*c^5 - 6*a*b*c^6)*d^5*e^5 + 10
*(21*b^4*c^4 - 43*a*b^2*c^5 + 11*a^2*c^6)*d^4*e^6 - 20*(6*b^5*c^3 - 18*a*b^3*c^4 + 11*a^2*b*c^5)*d^3*e^7 + 5*(
9*b^6*c^2 - 36*a*b^4*c^3 + 36*a^2*b^2*c^4 - 4*a^3*c^5)*d^2*e^8 - 10*(b^7*c - 5*a*b^5*c^2 + 7*a^2*b^3*c^3 - 2*a
^3*b*c^4)*d*e^9 + (b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)*e^10)/(b^2*c^10 - 4*a*c^11)))*s
qrt((2*c^5*d^5 - 5*b*c^4*d^4*e + 10*(b^2*c^3 - 2*a*c^4)*d^3*e^2 - 10*(b^3*c^2 - 3*a*b*c^3)*d^2*e^3 + 5*(b^4*c
- 4*a*b^2*c^2 + 2*a^2*c^3)*d*e^4 - (b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*e^5 + (b^2*c^5 - 4*a*c^6)*sqrt((25*c^8*d^8*
e^2 - 100*b*c^7*d^7*e^3 + 100*(2*b^2*c^6 - a*c^7)*d^6*e^4 - 50*(5*b^3*c^5 - 6*a*b*c^6)*d^5*e^5 + 10*(21*b^4*c^
4 - 43*a*b^2*c^5 + 11*a^2*c^6)*d^4*e^6 - 20*(6*b^5*c^3 - 18*a*b^3*c^4 + 11*a^2*b*c^5)*d^3*e^7 + 5*(9*b^6*c^2 -
36*a*b^4*c^3 + 36*a^2*b^2*c^4 - 4*a^3*c^5)*d^2*e^8 - 10*(b^7*c - 5*a*b^5*c^2 + 7*a^2*b^3*c^3 - 2*a^3*b*c^4)*d
*e^9 + (b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)*e^10)/(b^2*c^10 - 4*a*c^11)))/(b^2*c^5 - 4
*a*c^6)) + 4*(5*c^6*d^8*e - 20*b*c^5*d^7*e^2 + 35*b^2*c^4*d^6*e^3 - 35*b^3*c^3*d^5*e^4 + 7*(3*b^4*c^2 + a*b^2*
c^3 - 2*a^2*c^4)*d^4*e^5 - 7*(b^5*c + 2*a*b^3*c^2 - 4*a^2*b*c^3)*d^3*e^6 + (b^6 + 9*a*b^4*c - 15*a^2*b^2*c^2 -
8*a^3*c^3)*d^2*e^7 - (2*a*b^5 - a^2*b^3*c - 8*a^3*b*c^2)*d*e^8 + (a^2*b^4 - 3*a^3*b^2*c + a^4*c^2)*e^9)*sqrt(
e*x + d)) - 3*sqrt(2)*c^2*sqrt((2*c^5*d^5 - 5*b*c^4*d^4*e + 10*(b^2*c^3 - 2*a*c^4)*d^3*e^2 - 10*(b^3*c^2 - 3*a
*b*c^3)*d^2*e^3 + 5*(b^4*c - 4*a*b^2*c^2 + 2*a^2*c^3)*d*e^4 - (b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*e^5 + (b^2*c^5 -
4*a*c^6)*sqrt((25*c^8*d^8*e^2 - 100*b*c^7*d^7*e^3 + 100*(2*b^2*c^6 - a*c^7)*d^6*e^4 - 50*(5*b^3*c^5 - 6*a*b*c
^6)*d^5*e^5 + 10*(21*b^4*c^4 - 43*a*b^2*c^5 + 11*a^2*c^6)*d^4*e^6 - 20*(6*b^5*c^3 - 18*a*b^3*c^4 + 11*a^2*b*c^
5)*d^3*e^7 + 5*(9*b^6*c^2 - 36*a*b^4*c^3 + 36*a^2*b^2*c^4 - 4*a^3*c^5)*d^2*e^8 - 10*(b^7*c - 5*a*b^5*c^2 + 7*a
^2*b^3*c^3 - 2*a^3*b*c^4)*d*e^9 + (b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)*e^10)/(b^2*c^10
- 4*a*c^11)))/(b^2*c^5 - 4*a*c^6))*log(-sqrt(2)*(10*(b^2*c^5 - 4*a*c^6)*d^5*e^2 - 25*(b^3*c^4 - 4*a*b*c^5)*d^
4*e^3 + 10*(3*b^4*c^3 - 14*a*b^2*c^4 + 8*a^2*c^5)*d^3*e^4 - 10*(2*b^5*c^2 - 11*a*b^3*c^3 + 12*a^2*b*c^4)*d^2*e
^5 + (7*b^6*c - 44*a*b^4*c^2 + 66*a^2*b^2*c^3 - 8*a^3*c^4)*d*e^6 - (b^7 - 7*a*b^5*c + 13*a^2*b^3*c^2 - 4*a^3*b
*c^3)*e^7 - (2*(b^2*c^7 - 4*a*c^8)*d^2 - 2*(b^3*c^6 - 4*a*b*c^7)*d*e + (b^4*c^5 - 6*a*b^2*c^6 + 8*a^2*c^7)*e^2
)*sqrt((25*c^8*d^8*e^2 - 100*b*c^7*d^7*e^3 + 100*(2*b^2*c^6 - a*c^7)*d^6*e^4 - 50*(5*b^3*c^5 - 6*a*b*c^6)*d^5*
e^5 + 10*(21*b^4*c^4 - 43*a*b^2*c^5 + 11*a^2*c^6)*d^4*e^6 - 20*(6*b^5*c^3 - 18*a*b^3*c^4 + 11*a^2*b*c^5)*d^3*e
^7 + 5*(9*b^6*c^2 - 36*a*b^4*c^3 + 36*a^2*b^2*c^4 - 4*a^3*c^5)*d^2*e^8 - 10*(b^7*c - 5*a*b^5*c^2 + 7*a^2*b^3*c
^3 - 2*a^3*b*c^4)*d*e^9 + (b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)*e^10)/(b^2*c^10 - 4*a*c
^11)))*sqrt((2*c^5*d^5 - 5*b*c^4*d^4*e + 10*(b^2*c^3 - 2*a*c^4)*d^3*e^2 - 10*(b^3*c^2 - 3*a*b*c^3)*d^2*e^3 + 5
*(b^4*c - 4*a*b^2*c^2 + 2*a^2*c^3)*d*e^4 - (b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*e^5 + (b^2*c^5 - 4*a*c^6)*sqrt((25*
c^8*d^8*e^2 - 100*b*c^7*d^7*e^3 + 100*(2*b^2*c^6 - a*c^7)*d^6*e^4 - 50*(5*b^3*c^5 - 6*a*b*c^6)*d^5*e^5 + 10*(2
1*b^4*c^4 - 43*a*b^2*c^5 + 11*a^2*c^6)*d^4*e^6 - 20*(6*b^5*c^3 - 18*a*b^3*c^4 + 11*a^2*b*c^5)*d^3*e^7 + 5*(9*b
^6*c^2 - 36*a*b^4*c^3 + 36*a^2*b^2*c^4 - 4*a^3*c^5)*d^2*e^8 - 10*(b^7*c - 5*a*b^5*c^2 + 7*a^2*b^3*c^3 - 2*a^3*
b*c^4)*d*e^9 + (b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)*e^10)/(b^2*c^10 - 4*a*c^11)))/(b^2
*c^5 - 4*a*c^6)) + 4*(5*c^6*d^8*e - 20*b*c^5*d^7*e^2 + 35*b^2*c^4*d^6*e^3 - 35*b^3*c^3*d^5*e^4 + 7*(3*b^4*c^2
+ a*b^2*c^3 - 2*a^2*c^4)*d^4*e^5 - 7*(b^5*c + 2*a*b^3*c^2 - 4*a^2*b*c^3)*d^3*e^6 + (b^6 + 9*a*b^4*c - 15*a^2*b
^2*c^2 - 8*a^3*c^3)*d^2*e^7 - (2*a*b^5 - a^2*b^3*c - 8*a^3*b*c^2)*d*e^8 + (a^2*b^4 - 3*a^3*b^2*c + a^4*c^2)*e^
9)*sqrt(e*x + d)) + 3*sqrt(2)*c^2*sqrt((2*c^5*d^5 - 5*b*c^4*d^4*e + 10*(b^2*c^3 - 2*a*c^4)*d^3*e^2 - 10*(b^3*c
^2 - 3*a*b*c^3)*d^2*e^3 + 5*(b^4*c - 4*a*b^2*c^2 + 2*a^2*c^3)*d*e^4 - (b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*e^5 - (b
^2*c^5 - 4*a*c^6)*sqrt((25*c^8*d^8*e^2 - 100*b*c^7*d^7*e^3 + 100*(2*b^2*c^6 - a*c^7)*d^6*e^4 - 50*(5*b^3*c^5 -
6*a*b*c^6)*d^5*e^5 + 10*(21*b^4*c^4 - 43*a*b^2*c^5 + 11*a^2*c^6)*d^4*e^6 - 20*(6*b^5*c^3 - 18*a*b^3*c^4 + 11*
a^2*b*c^5)*d^3*e^7 + 5*(9*b^6*c^2 - 36*a*b^4*c^3 + 36*a^2*b^2*c^4 - 4*a^3*c^5)*d^2*e^8 - 10*(b^7*c - 5*a*b^5*c
^2 + 7*a^2*b^3*c^3 - 2*a^3*b*c^4)*d*e^9 + (b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)*e^10)/(
b^2*c^10 - 4*a*c^11)))/(b^2*c^5 - 4*a*c^6))*log(sqrt(2)*(10*(b^2*c^5 - 4*a*c^6)*d^5*e^2 - 25*(b^3*c^4 - 4*a*b*
c^5)*d^4*e^3 + 10*(3*b^4*c^3 - 14*a*b^2*c^4 + 8*a^2*c^5)*d^3*e^4 - 10*(2*b^5*c^2 - 11*a*b^3*c^3 + 12*a^2*b*c^4
)*d^2*e^5 + (7*b^6*c - 44*a*b^4*c^2 + 66*a^2*b^2*c^3 - 8*a^3*c^4)*d*e^6 - (b^7 - 7*a*b^5*c + 13*a^2*b^3*c^2 -
4*a^3*b*c^3)*e^7 + (2*(b^2*c^7 - 4*a*c^8)*d^2 - 2*(b^3*c^6 - 4*a*b*c^7)*d*e + (b^4*c^5 - 6*a*b^2*c^6 + 8*a^2*c
^7)*e^2)*sqrt((25*c^8*d^8*e^2 - 100*b*c^7*d^7*e^3 + 100*(2*b^2*c^6 - a*c^7)*d^6*e^4 - 50*(5*b^3*c^5 - 6*a*b*c^
6)*d^5*e^5 + 10*(21*b^4*c^4 - 43*a*b^2*c^5 + 11*a^2*c^6)*d^4*e^6 - 20*(6*b^5*c^3 - 18*a*b^3*c^4 + 11*a^2*b*c^5
)*d^3*e^7 + 5*(9*b^6*c^2 - 36*a*b^4*c^3 + 36*a^2*b^2*c^4 - 4*a^3*c^5)*d^2*e^8 - 10*(b^7*c - 5*a*b^5*c^2 + 7*a^
2*b^3*c^3 - 2*a^3*b*c^4)*d*e^9 + (b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)*e^10)/(b^2*c^10
- 4*a*c^11)))*sqrt((2*c^5*d^5 - 5*b*c^4*d^4*e + 10*(b^2*c^3 - 2*a*c^4)*d^3*e^2 - 10*(b^3*c^2 - 3*a*b*c^3)*d^2*
e^3 + 5*(b^4*c - 4*a*b^2*c^2 + 2*a^2*c^3)*d*e^4 - (b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*e^5 - (b^2*c^5 - 4*a*c^6)*sq
rt((25*c^8*d^8*e^2 - 100*b*c^7*d^7*e^3 + 100*(2*b^2*c^6 - a*c^7)*d^6*e^4 - 50*(5*b^3*c^5 - 6*a*b*c^6)*d^5*e^5
+ 10*(21*b^4*c^4 - 43*a*b^2*c^5 + 11*a^2*c^6)*d^4*e^6 - 20*(6*b^5*c^3 - 18*a*b^3*c^4 + 11*a^2*b*c^5)*d^3*e^7 +
5*(9*b^6*c^2 - 36*a*b^4*c^3 + 36*a^2*b^2*c^4 - 4*a^3*c^5)*d^2*e^8 - 10*(b^7*c - 5*a*b^5*c^2 + 7*a^2*b^3*c^3 -
2*a^3*b*c^4)*d*e^9 + (b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)*e^10)/(b^2*c^10 - 4*a*c^11)
))/(b^2*c^5 - 4*a*c^6)) + 4*(5*c^6*d^8*e - 20*b*c^5*d^7*e^2 + 35*b^2*c^4*d^6*e^3 - 35*b^3*c^3*d^5*e^4 + 7*(3*b
^4*c^2 + a*b^2*c^3 - 2*a^2*c^4)*d^4*e^5 - 7*(b^5*c + 2*a*b^3*c^2 - 4*a^2*b*c^3)*d^3*e^6 + (b^6 + 9*a*b^4*c - 1
5*a^2*b^2*c^2 - 8*a^3*c^3)*d^2*e^7 - (2*a*b^5 - a^2*b^3*c - 8*a^3*b*c^2)*d*e^8 + (a^2*b^4 - 3*a^3*b^2*c + a^4*
c^2)*e^9)*sqrt(e*x + d)) - 3*sqrt(2)*c^2*sqrt((2*c^5*d^5 - 5*b*c^4*d^4*e + 10*(b^2*c^3 - 2*a*c^4)*d^3*e^2 - 10
*(b^3*c^2 - 3*a*b*c^3)*d^2*e^3 + 5*(b^4*c - 4*a*b^2*c^2 + 2*a^2*c^3)*d*e^4 - (b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*e
^5 - (b^2*c^5 - 4*a*c^6)*sqrt((25*c^8*d^8*e^2 - 100*b*c^7*d^7*e^3 + 100*(2*b^2*c^6 - a*c^7)*d^6*e^4 - 50*(5*b^
3*c^5 - 6*a*b*c^6)*d^5*e^5 + 10*(21*b^4*c^4 - 43*a*b^2*c^5 + 11*a^2*c^6)*d^4*e^6 - 20*(6*b^5*c^3 - 18*a*b^3*c^
4 + 11*a^2*b*c^5)*d^3*e^7 + 5*(9*b^6*c^2 - 36*a*b^4*c^3 + 36*a^2*b^2*c^4 - 4*a^3*c^5)*d^2*e^8 - 10*(b^7*c - 5*
a*b^5*c^2 + 7*a^2*b^3*c^3 - 2*a^3*b*c^4)*d*e^9 + (b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)*
e^10)/(b^2*c^10 - 4*a*c^11)))/(b^2*c^5 - 4*a*c^6))*log(-sqrt(2)*(10*(b^2*c^5 - 4*a*c^6)*d^5*e^2 - 25*(b^3*c^4
- 4*a*b*c^5)*d^4*e^3 + 10*(3*b^4*c^3 - 14*a*b^2*c^4 + 8*a^2*c^5)*d^3*e^4 - 10*(2*b^5*c^2 - 11*a*b^3*c^3 + 12*a
^2*b*c^4)*d^2*e^5 + (7*b^6*c - 44*a*b^4*c^2 + 66*a^2*b^2*c^3 - 8*a^3*c^4)*d*e^6 - (b^7 - 7*a*b^5*c + 13*a^2*b^
3*c^2 - 4*a^3*b*c^3)*e^7 + (2*(b^2*c^7 - 4*a*c^8)*d^2 - 2*(b^3*c^6 - 4*a*b*c^7)*d*e + (b^4*c^5 - 6*a*b^2*c^6 +
8*a^2*c^7)*e^2)*sqrt((25*c^8*d^8*e^2 - 100*b*c^7*d^7*e^3 + 100*(2*b^2*c^6 - a*c^7)*d^6*e^4 - 50*(5*b^3*c^5 -
6*a*b*c^6)*d^5*e^5 + 10*(21*b^4*c^4 - 43*a*b^2*c^5 + 11*a^2*c^6)*d^4*e^6 - 20*(6*b^5*c^3 - 18*a*b^3*c^4 + 11*a
^2*b*c^5)*d^3*e^7 + 5*(9*b^6*c^2 - 36*a*b^4*c^3 + 36*a^2*b^2*c^4 - 4*a^3*c^5)*d^2*e^8 - 10*(b^7*c - 5*a*b^5*c^
2 + 7*a^2*b^3*c^3 - 2*a^3*b*c^4)*d*e^9 + (b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)*e^10)/(b
^2*c^10 - 4*a*c^11)))*sqrt((2*c^5*d^5 - 5*b*c^4*d^4*e + 10*(b^2*c^3 - 2*a*c^4)*d^3*e^2 - 10*(b^3*c^2 - 3*a*b*c
^3)*d^2*e^3 + 5*(b^4*c - 4*a*b^2*c^2 + 2*a^2*c^3)*d*e^4 - (b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*e^5 - (b^2*c^5 - 4*a
*c^6)*sqrt((25*c^8*d^8*e^2 - 100*b*c^7*d^7*e^3 + 100*(2*b^2*c^6 - a*c^7)*d^6*e^4 - 50*(5*b^3*c^5 - 6*a*b*c^6)*
d^5*e^5 + 10*(21*b^4*c^4 - 43*a*b^2*c^5 + 11*a^2*c^6)*d^4*e^6 - 20*(6*b^5*c^3 - 18*a*b^3*c^4 + 11*a^2*b*c^5)*d
^3*e^7 + 5*(9*b^6*c^2 - 36*a*b^4*c^3 + 36*a^2*b^2*c^4 - 4*a^3*c^5)*d^2*e^8 - 10*(b^7*c - 5*a*b^5*c^2 + 7*a^2*b
^3*c^3 - 2*a^3*b*c^4)*d*e^9 + (b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)*e^10)/(b^2*c^10 - 4
*a*c^11)))/(b^2*c^5 - 4*a*c^6)) + 4*(5*c^6*d^8*e - 20*b*c^5*d^7*e^2 + 35*b^2*c^4*d^6*e^3 - 35*b^3*c^3*d^5*e^4
+ 7*(3*b^4*c^2 + a*b^2*c^3 - 2*a^2*c^4)*d^4*e^5 - 7*(b^5*c + 2*a*b^3*c^2 - 4*a^2*b*c^3)*d^3*e^6 + (b^6 + 9*a*b
^4*c - 15*a^2*b^2*c^2 - 8*a^3*c^3)*d^2*e^7 - (2*a*b^5 - a^2*b^3*c - 8*a^3*b*c^2)*d*e^8 + (a^2*b^4 - 3*a^3*b^2*
c + a^4*c^2)*e^9)*sqrt(e*x + d)) + 4*(c*e^2*x + 7*c*d*e - 3*b*e^2)*sqrt(e*x + d))/c^2
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)**(5/2)/(c*x**2+b*x+a),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^(5/2)/(c*x^2+b*x+a),x, algorithm="giac")
[Out]
Timed out | 16,327 | 25,172 | {"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} | 3.546875 | 4 | CC-MAIN-2023-40 | latest | en | 0.150203 |
https://www.jiskha.com/display.cgi?id=1337279881 | 1,516,738,637,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892238.78/warc/CC-MAIN-20180123191341-20180123211341-00058.warc.gz | 934,430,409 | 4,025 | # chemistry
posted by .
Na + H2O ¨ NaOH + H2
How many moles of H2 would be formed from 30 grams of Na in the following (UNBALANCED) equation?
• chemistry -
2Na + 2H2O == 2NaOH + 2H2
mols Na = grams/atomic mass
Convert mols Na to mols H2 using the coefficients in the balanced equation. That's 2:2 therefore, mols H2 = mols Na.
• chemistry -
How many moles of H2 would be formed from 30 grams of Na in the following (UNBALANCED) equation?
Na + H2O → NaOH + H2
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# Sample Paper 2013 Class XII Subject : Mathematic
TIME: 3 hrs GENERAL INSTRUCTION: (a) All questions are compulsory. (b) This question paper consists of 29 questions divided into three section A, B, and C. Section A comprises of 10 question of one mark each, section B comprises of 12 questions of four marks each and section C comprises of 7 questions of six marks each. (c) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. (d) There is no overall choice. However, internal choice has been provided in 04 questions of four marks each and 02 questions of six marks each. You have to attempt only one of the alternatives in all such questions. (e) Use of calculators is not permitted. You may ask for logarithmic tables, if required. ________________________________________________________________________ SECTION-A Question number 1 to 10 carry 1 mark each. Q1) Prove that: Q2) If ( ( ) ) then find the value of x. * + * +. | | is perpendicular to MAX. MARKS: 100
Q3) Find the matrix X, such that: Q4) For any two non zero vectors | | | | .
, show that | |
Q5) If and are represented along the two diagonals of a parallelogram, then write the area . of a parallelogram in the terms of Q6) If is a unit vector and ( ) ( ) | |
Q7) The total revenue received from the sale of shirts, advertising Keep Your Environment Clean is given by the function , R(x) = x2 + 4x + 11.If the marginal revenue is defined as the rate of change of R(x) with respect to the number of shirts sold at an instant, find the marginal revenue when 5 shirts are sold. Q8) Write the order of the differential equation: , where .
## Q9) Write the value of x 2y + 3z, if[
][ |
]. |
Q10) Given a square matrix A of order 3, such that | | SECTION-B Question number 11 to 22 carry 4 marks each.
Q11) Let A represents a set of all students, who would like to be a part of CLEANLINESS DRIVE. If R be the relation in set A, given by x R y implies student x and student y are associated with the CLEANLINESS DRIVE, show that the relation R is an equivalence relation. Do you want to be a part of the drive? Why? Q12) Prove that: ( ) ( OR If ,then prove that . ) .
Q13)Two schools A and B decided to award prizes to their students for three values honesty(x),punctuality(y) and empathy(z).School A decided to award a total of Rs.22,000 for the three values to 10,8,and 6 students respectively, while school B decided to award Rs. 21,400 for three values to 8,6,10 students respectively. If all the three prizes amount to Rs.5, 400 then, (i) (ii) (iii) Represents the above situation by a matrix equation and form linear equations using matrix multiplication. Is it possible to solve the system of equations so obtained using matrices? Which value do you prefer to be rewarded most and why?
## Q14) Express the equation
) in terms of an equation .
## Q16) Is the function f defined as ( ) Q17) Evaluate:( )
is continuous at x=0?
. OR
Evaluate:
)
| )
## so that the four points with position vectors are coplanar.
Q21) Find the equation of the plane passing through the line of intersection of the planes ( ) ( ) and parallel to x- axis. OR Find the point on the line at a distance of units from the point (1, 2, 3)
Q22) Three persons A,B and C throw a die in succession till one of them gets a six and wins the ticket to be a part of the group TOWARDS A JUST SOCIETY. Find their respective probabilities of winning, if A starts followed by B and C. If C selected for the group, then C should play at which turn? Write two lines about TOWARDS A JUST SOCIETY. SECTION-C Question numbers 23 to 29 carry 6 marks each. Q23) If A =[ equations, 2x-y+z=-1,-x+2y-z=4, x-y+2z=-3. OR Using properties of determinants prove that: | | ( )| |. ] [ ] , find AB and hence solve the
Q24) In an examination, an examinee either guesses or copies or knows the answer of multiple choice questions with four choices. The probability that he makes a guess is and the probably that he copies answer is .The probability that his answer is correct given that he
copied it is .Find the probability that he knew the answer to the question ,given that he correctly answered it. A student does not know the answer to one of the questions in a test .The evaluation process has a negative marking. Which value would a student violate if he uses unfair means? How would an act like the above hamper his character development in the coming years? Q25) If a young man rides his motorcycle at 25km/h, he has to spend Rs.2 per km on petrol. If he rides at a faster speed of 40km/h, the petrol cost increase at Rs.5 per km.He has Rs. 100 to spend on petrol and wishes to find what is the maximum distance he can travel in one hour .Express this as an L.P.P and solve it graphically. These days we talk about saving our Natural Resources .Why do we need to save petrol? Answer in two line. Q26) Find the image of the point (1, 2, 3) in the plane x+2y+4z=38.Write the two life skills, which one must acquire, along the image. Q27) Using integration, find the area of the region:{( Q28) Solve the differential equation: ( ) ( ) | | ). }.
Q29) Show that the volume of the greatest cylinder which can be inscribed in a cone of height h and semi-vertical angle OR Find the interval for which the function ( ) decreasing in the interval ( ) ( ) is increasing or . | 1,363 | 5,400 | {"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} | 3.234375 | 3 | CC-MAIN-2019-35 | latest | en | 0.885028 |
http://math.stackexchange.com/questions/275428/int-mathbbrn-dx-1-dots-dx-n-exp%e2%88%92-frac12-sum-i-j-1nx-ia-ijx | 1,469,634,467,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826908.63/warc/CC-MAIN-20160723071026-00323-ip-10-185-27-174.ec2.internal.warc.gz | 161,452,746 | 17,913 | # $\int_{\mathbb{R}^n} dx_1 \dots dx_n \exp(−\frac{1}{2}\sum_{i,j=1}^{n}x_iA_{ij}x_j)$?
Let $A$ be a symmetric positive-definite $n\times n$ matrix and $b_i$ be some real numbers How can one evaluate the following integrals?
1. $\int_{\mathbb{R}^n} dx_1 \dots dx_n \exp(−\frac{1}{2}\sum_{i,j=1}^{n}x_iA_{ij}x_j)$
2. $\int_{\mathbb{R}^n} dx_1 \dots dx_n \exp(−\frac{1}{2}\sum_{i,j=1}^{n}x_iA_{ij}x_{j}-b_i x_i)$
-
Diagonalize the quadratic form. – user27126 Jan 10 '13 at 20:58
Let's $x=(x_1,\ldots,x_n)\in\mathbb{R}^n$. We have \begin{align} \int_{\mathbb{R}^n}e^{-\frac{1}{2}\langle x,x \rangle_{A}} d x = & \int_{\mathbb{R}^n}e^{-\frac{1}{2}\langle x,Ax \rangle} d x \\ = & \int_{\mathbb{R}^n}e^{-\frac{1}{2}\langle Ux,Ux \rangle} d x \\ = & \int_{\mathbb{R}^n}e^{-\frac{1}{2}\langle x,x \rangle}\|U\| d x \\ = & \|U\|\cdot \left(\int_{R} e^{-\frac{1}{2}x^2} d x \right)^n\\ = & \|U\|\cdot\left( \frac{1}{2}\sqrt{2\pi}\right)^n \end{align} For second itegral use the change of variable $y_i+s_i=x_i$ such that $(2\cdot s^TA+b^T)=0$, \begin{align} (y+s)^TA(y+s) +b(y+s)= & y^TAy+(2\cdot s^TA+b^T)y+s^T(b+As) \\ = & y^TAy+s^T(b+As) \\ \end{align} Then $$\int_{\mathbb{R}^n}e^{-\frac{1}{2}\langle x,x \rangle_{A}+b^Tx} d x =n\cdot \|U\|\cdot \int_{\mathbb{R}^n} e^{-\frac{1}{2}y^2}\cdot e^{s^T(b+As)} d y \\ = \cdot e^{s^T(b+As)}\cdot \|U\|\cdot \left( \frac{1}{2}\sqrt{2\pi}\right)^n$$ | 668 | 1,388 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2016-30 | latest | en | 0.277742 |
https://de.mathworks.com/matlabcentral/cody/problems/19-swap-the-first-and-last-columns/solutions/1714141 | 1,569,000,252,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574050.69/warc/CC-MAIN-20190920155311-20190920181311-00302.warc.gz | 438,738,563 | 15,464 | Cody
# Problem 19. Swap the first and last columns
Solution 1714141
Submitted on 23 Jan 2019 by Ivo Mudo
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A = [ 12 4 7 5 1 4]; B_correct = [ 7 4 12 4 1 5 ]; assert(isequal(swap_ends(A),B_correct))
2 Pass
A = [ 12 7 5 4]; B_correct = [ 7 12 4 5 ]; assert(isequal(swap_ends(A),B_correct))
3 Pass
A = [ 1 5 0 2 3 ]; B_correct = [ 3 5 0 2 1 ]; assert(isequal(swap_ends(A),B_correct))
4 Pass
A = 1; B_correct = 1; assert(isequal(swap_ends(A),B_correct)) | 232 | 627 | {"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-2019-39 | latest | en | 0.584195 |
https://www.teacherspayteachers.com/Product/Math-Board-Game-Integers-Grades-6-9-1976164 | 1,524,335,300,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945272.41/warc/CC-MAIN-20180421164646-20180421184646-00570.warc.gz | 909,644,533 | 17,677 | Total:
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# Math Board Game - Integers - Grades 6 - 9
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This board game is fun for the students while they sharpen their skills on operations with integers. They will draw a card and they must correctly answer the question in order to move along the game board. The first person to arrive at the FINISH space is the winner! One person will be designated as the Answer Checker.
Students will stay engaged and try their best to correctly answer the questions so they can win the game!
The following are included in this resource.
45 game cards
Game Board
Game Pieces
Playing Instructions
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Equations, Tables, Graphs - Math Game
Common Core Math Activities 6th Grade (6.EE.9) Tables, Graphs, Equations
Common Core Math 6th Grade - Solve Equations - Graph Inequalities (6.EE.7,8)
Common Core Math 6th Grade Geometry 6.G.1 Area Parallelogram Triangle Activities
Math Board Games Bundle - 6th Grade - (6.RP) (6.NS) (6.EE) (6.G) (6.SP)
Math Board Games Bundle 7th Grade - (7.RP, 7.NS, 7.EE, 7.G, 7.SP)
Math Board Games Bundle - 8th Grade (8.NS, 8.EE, 8.F, 8.G, 8.SP)
8th Grade Math Compare Functions - Two Matching Games - CCSS 8.F.2
7th Grade Math 2-Step Equations and Inequalities (Prerequisite for 7.EE.4)
7th Grade Math Ratios and Proportional Relationships Part 1 CCSS 7.RP.1.2abcd
7th Grade Math Ratios and Proportional Relationships Part 2 CCSS 7.RP.3
6th Grade Math - Ratios, Rational Numbers, Expressions, Statistics
Common Core Math 6th Grade Ratios and Proportional Relationships 6.RP.1.2.3ab
Common Core Math 6th Grade Ratios and Proportional Relationships 6.RP.3cd
Math Activities 8th Grade - Proportions, Slope, y=mx+b - CCSS 8.EE.5,6
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 577 | 2,041 | {"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} | 3.796875 | 4 | CC-MAIN-2018-17 | latest | en | 0.867065 |
https://www.coursehero.com/file/5774136/PS9/ | 1,513,003,121,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513512.31/warc/CC-MAIN-20171211125234-20171211145234-00070.warc.gz | 719,197,941 | 59,850 | # PS9 - Question 1 1 points Save In the population the scores...
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Question 1 1 points Save In the population, the scores of students on the ACT college entrance exam had mean μ = 22 and standard deviation σ = 6. The distribution of scores is roughly normal. A SRS of 80 students who took the exam is taken. Round the standard deviation of the sample mean score x_bar to four decimal places. Using that rounded number, what is probability that the mean score x_bar of these students is 22.5 or higher (rounded to four decimal places )? (Remember to express your answer as a proportion.) Question 2 1 points Save In the population, the scores of students on the SAT college entrance exam had mean μ = 1200 and standard deviation σ = 80. The distribution of scores is roughly normal. A SRS of 60 students who took the exam is taken. Round the standard deviation of the sample mean score x_bar to four decimal places. Using that rounded number, what is probability that the mean score x_bar of these students is between 1190 and 1210 (rounded to four decimal places
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Ask a homework question - tutors are online | 355 | 1,549 | {"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} | 3.234375 | 3 | CC-MAIN-2017-51 | latest | en | 0.913809 |
https://metacpan.org/release/TSCANLAN/AI-Fuzzy-0.05/source/Fuzzy.pm | 1,638,837,070,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00028.warc.gz | 462,356,853 | 6,554 | ``````package AI::Fuzzy;
use strict;
use vars qw(\$VERSION);
use AI::Fuzzy::Set;
use AI::Fuzzy::Axis;
use AI::Fuzzy::Label;
\$VERSION = '0.05';
1;
__END__
=head1 NAME
AI::Fuzzy - Perl extension for Fuzzy Logic
=head1 SYNOPSIS
use AI::Fuzzy;
my \$f = new AI::Fuzzy::Axis;
my \$l = new AI::Fuzzy::Label("toddler", 1, 1.5, 3.5);
\$f->addlabel("baby", -1, 1, 2.5);
\$f->addlabel(\$l);
\$f->addlabel("little kid", 2, 7, 12);
\$f->addlabel("kid", 6, 10, 14);
\$f->addlabel("teenager", 12, 16, 20);
\$f->addlabel("young adult", 18, 27, 35);
\$f->addlabel("adult", 25, 50, 75);
\$f->addlabel("senior", 60, 80, 110);
\$f->addlabel("relic", 100, 150, 200);
for (my \$x = 0; \$x<50; \$x+=4) {
print "\$x years old => " . \$f->labelvalue(\$x) . "\n";
}
\$a = new AI::Fuzzy::Set( x1 => .3, x2 => .5, x3 => .8, x4 => 0, x5 => 1);
\$b = new AI::Fuzzy::Set( x5 => .3, x6 => .5, x7 => .8, x8 => 0, x9 => 1);
print "a is: " . \$a->as_string . "\n";
print "b is: " . \$b->as_string . "\n";
print "a is equal to b" if (\$a->equal(\$b));
my \$c = \$a->complement();
print "complement of a is: " . \$c->as_string . "\n";
\$c = \$a->union(\$b);
print "a union b is: " . \$c->as_string . "\n";
\$c = \$a->intersection(\$b);
print "a intersection b is: " . \$c->as_string . "\n";
__END__
=head1 DESCRIPTION
AI::Fuzzy really consists of three modules - AI::Fuzzy::Axis, AI::Fuzzy::Label, and
AI::Fuzzy::Set.
A fuzzy set is simply a mathematical set to which members can
I<partially> belong. For example, a particular shade of gray may
partially belong to the set of dark colors, whereas black would have
full membership, and lemon yellow would have almost no membership.
A fuzzy axis holds fuzzy labels and can be used to classify values
by examining the degree to which they belong to several labels, and
selecting the most appropriate. For example, it can decide whether
to call water at 60 degrees Farenheight "cold", "cool", or "warm".
A fuzzy label classifies a particular range of the Axis. In the above example
the label is one of "cold", "cool", or "warm". A fuzzy label defines how
much a crisp value belongs to the classifier such as "cold", "warm", or "cool".
=head2 Fuzzy Sets
AI::Fuzzy:Set has these methods:
\$fs = B<new> AI::Fuzzy::Set;
# here, "Bob" is unquestionably tall.. the others less so.
\$fs_tall_people = B<new> AI::Fuzzy::Set( Lester=>.34, Bob=>1.00, Max=>.86 );
# \$x will be .86
\$x = B<membership> \$fs_tall_people, "Max";
# get list of members, sorted from least membership to greatest:
@shortest_first = B<members> \$fs_tall_people;
\$fs = B<new> AI::Fuzzy::Set( x1 => .3, x2 => .5, x3 => .8, x4 => 0, x5 => 1);
B<complement>, B<union>, B<intersection>
Thesie are the fuzzy set version of the typical functions.
B<equal>
Returns true if the sets have the same elements and those elements
are all equal.
B<as_string>
Prints the set as tuples:
\$b = new AI::Fuzzy::Set( x5 => .3, x6 => .5, x7 => .8, x8 => 0, x9 => 1);
print "b is: " . \$b->as_string . "\n";
prints:
b is: x8/0, x5/0.3, x6/0.5, x7/0.8, x9/1
=head2 Fuzzy Labels
A Fuzzy::Label label has four attributes: the text of the label (it
can be any scalar, really), and three numbers: low, mid, high if you
imagine a cartesian plane (remember graph paper in algebra?) of all
possible values, the label applies to a particular range. the graph
might look something like this:
|Y * (mid, 1)
| / \
| / \
| / \
| / \
-|-------*-------------*------- X
(low,0) (high,0)
the Y value is applicability of the label for a given X value
the mid number is the "pure" value. eg, orange is at 0 or 360
degrees on the color wheel. the label applies 100% at the mid
point.
the low and high numbers are the two points at which
the label ceases to apply.
note that labels can overlap, and that the
mid number isn't always in the exact center, so the slope
of the two sides may vary...
\$fl = new AI::Fuzzy::Label ( "hot", 77, 80, 100 );
\$fx = new AI::Fuzzy::Label ( "cold", 0, 10, 200 );
# what I consider hot. :) (in Farenheit, of course!)
if ( \$fl->lessthan(\$fx) ) {
print "the laws of nature have changed\n";
}
# there is a lessthan, greaterthan, lessequal, greaterequal, and between
# that functions as above or using <,>,<=,>=
\$a = \$fl->applicability(\$value);
# \$a is now the degree to which this label applies to \$value
=head2 Fuzzy Axis
A Fuzzy::Axis maintains a hash of labels. Thus you can now look at how
values apply to the full range of labels. The graph of an Axis might
look like this:
|Y * (mid, 1)
| /\/ \ /|
| /- -\ / /\ \ / |
| / \-/ / \ \ / | (some function on some range of x)
| | / \ /\ ---*-|
-|---------*-----------*------- X
(low,0) (high,0)
the Y value is still the applicability of the label for a given X value,
but there are three labels on this Axis. A different X value may
put your value into a new label.
\$fl = new AI::Fuzzy::Axis;
\$fl->addlabel(\$label);
# add a label created as in AI::Fuzzy::Label docs
\$a = \$fl->applicability(\$label, \$value);
# \$a is now the degree to which \$label applies to \$value
\$l = \$fl->label ("labelname");
# returns the label object named "labelname"
\$l = \$fl->labelvalue (\$value);
# applies a label to \$value
@l = \$fl->labelvalue(\$value);
# returns a list of labels and their applicability values
\$s = new AI::Fuzzy::Set( \$fl->label(\$value) );
# same thing, but now it's an object
@range = \$fl->range();
# returns a list of labels, sorted by their midpoints
# eg: ("cold", "cool", "lukewarm", "warm", "hot")
=head1 AUTHOR
Tom Scanlan <tscanlan@openreach.com>,
current maintainer
Michal Wallace (sabren@manifestation.com),
original author
=head1 SEE ALSO
Move along, nothing to "see also" here...
=head1 BUGS
Please send any bugs to Tom Scanlan <tscanlan@openreach.com>
=cut
`````` | 1,933 | 5,970 | {"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-2021-49 | latest | en | 0.735461 |
https://homework.cpm.org/category/CC/textbook/cca/chapter/2/lesson/2.4.1/problem/2-97 | 1,716,443,443,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00347.warc.gz | 251,655,334 | 14,925 | ### Home > CCA > Chapter 2 > Lesson 2.4.1 > Problem2-97
2-97.
Rena says that if $x=−5$, the equation below is true. Her friend, Dean, says the answer is $x=3$. Who is correct? Justify your conclusion.
$9(x+4)=1+2x$
Substitute both numbers in for $x$ and see which one makes the statement true.
For example, if Joe said $x=2$:
\begin{aligned}9\left(2 + 4\right) &= 1 + 2\left(2\right)\\18 + 36 &= 1 + 4 \\ 54 &=5 \end{aligned}
False!
Now try with the given possible values. | 171 | 478 | {"found_math": true, "script_math_tex": 6, "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} | 4.15625 | 4 | CC-MAIN-2024-22 | latest | en | 0.676789 |
https://math.answers.com/math-and-arithmetic/What_best_describes_a_linear_function | 1,725,811,193,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00222.warc.gz | 370,234,040 | 48,093 | 0
# What best describes a linear function?
Updated: 9/18/2023
Wiki User
12y ago
A linear function is a function, or equation, that when graphed, will form a straight line.
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12y ago
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Q: What best describes a linear function?
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linear function
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### Why use LMTD and not mean temperature difference?
Because the temperature change that occurs across the heat exchanger from the entrance to the exit is not linear, and a logarithmic function best describes this temperature change.
No
### Is linear function the same as linear equations?
No a linear equation are not the same as a linear function. The linear function is written as Ax+By=C. The linear equation is f{x}=m+b.
### What is the end behavior of a linear function?
Assuming the domain is unbounded, the linear function continues to be a linear function to its end.
### Is an exponential function linear?
No. An exponential function is not linear. A very easy way to understand what is and what is not a linear function is in the word, "linear function." A linear function, when graphed, must form a straight line.P.S. The basic formula for any linear function is y=mx+b. No matter what number you put in for the m and b variables, you will always make a linear function.
### Is a linear equation the same as a function?
No a linear equation are not the same as a linear function. The linear function is written as Ax+By=C. The linear equation is f{x}=m+b.
### What best describes the asymptote of an exponential function of the form F(x)b x?
what symbol best describes the asymptote of an exponential function of the form F(x)=bx
Motion.
### What best describes the distribution of the data?
A probability density function. | 439 | 1,943 | {"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-2024-38 | latest | en | 0.841784 |
https://www.nagwa.com/en/videos/478148238246/ | 1,620,863,490,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991413.30/warc/CC-MAIN-20210512224016-20210513014016-00636.warc.gz | 970,583,725 | 15,907 | # Question Video: Identifying Types of Orbit Physics • 9th Grade
The diagram shows two different possible orbits of an object around a star. Which of the following correctly describes the shape of orbit (a)? [A] Elliptical [B] Circular [C] Helical [D] Spiral [E] Highly elliptical
02:56
### Video Transcript
The diagram shows two different possible orbits of an object around a star. Which of the following correctly describes the shape of orbit a)? A) Elliptical, B) Circular, C) Helical, D) Spiral, E) Highly elliptical.
All right, taking a look at our diagram, we see these two orbits marked a) and b). And this question focuses on orbit a). We want to know which of these five paths correctly describes this particular orbit. Now, the first thing we can do is recall that whenever one object is in orbit around another one, there are two possible paths that that orbiting object can follow. Orbital paths are either circular or they’re elliptical. These are the two allowed shapes we could say for orbital motion. That eliminates two of our possible answer choices, option C and D. Helical and spiral are not allowed orbital paths.
At this point, we can recall the difference between these two allowed orbits of circles and ellipses. We all know what a circle looks like. And an ellipse looks like a circle that’s been squished or compressed. So if this was a circular orbital path, an elliptical path would look something like this. Now, we can see that our three remaining answer choices are elliptical, circular, and something called highly elliptical. For an orbit to be highly elliptical, we could think of that as a very compressed circular orbit. The squishing, we could say, has gone even farther to create something that looks like this. This would be an example of a highly elliptical orbit. So which of these three — elliptical, circular, or highly elliptical — is the orbit marked out in a)?
We can see that this orbit does not look like a compressed circle, but rather it just looks like a circle itself. It’s definitely not highly elliptical then. We’ll cross off option E. And we also don’t think it’s elliptical. The shape of orbit a) is circular. Now, let’s consider the same question, but about orbit b).
Our next question asks, which of the following correctly describes the shape of orbit b)? A) Circular, B) Spherical, C) Highly elliptical, D) Spiral, E) Helical.
Just like before, we can cancel out any answer options which are neither circular nor elliptical. As we’ve seen, those were the only two possible paths that an orbiting body can have. So this means that option B, spherical, option D, spiral, and option E, helical, are off the table. So then, orbit b) is either circular or it’s highly elliptical. If we revisit our sketch of a circular orbit, an elliptical orbit, and a highly elliptical orbit marked out in yellow, we can clearly see that orbit 𝑏 is not circular. This eliminates option A. And as we consider highly elliptical as a description of this shape, we see that there’s a match. This orbit has a very compressed or squished look compared to a circular orbit. And, therefore, it is highly elliptical. This is the correct description of the shape of orbit b). | 721 | 3,216 | {"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} | 3.796875 | 4 | CC-MAIN-2021-21 | latest | en | 0.945478 |
http://www.slideserve.com/onaona/in-chapter-1-we-talked-about-parametric-equations | 1,493,175,280,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121121.5/warc/CC-MAIN-20170423031201-00298-ip-10-145-167-34.ec2.internal.warc.gz | 699,161,577 | 15,439 | In chapter 1, we talked about parametric equations. - PowerPoint PPT Presentation
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In chapter 1, we talked about parametric equations. Parametric equations can be used to describe motion that is not a function. If f and g have derivatives at t , then the parametrized curve also has a derivative at t. The formula for finding the slope of a parametrized curve is:.
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In chapter 1, we talked about parametric equations.
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Presentation Transcript
In chapter 1, we talked about parametric equations.
Parametric equations can be used to describe motion that is not a function.
If f and g have derivatives at t, then the parametrized curve also has a derivative at t.
The formula for finding the slope of a parametrized curve is:
This makes sense if we think about canceling dt.
The formula for finding the slope of a parametrized curve is:
We assume that the denominator is not zero.
To find the second derivative of a parametrized curve, we find the derivative of the first derivative:
• Find the first derivative (dy/dx).
2. Find the derivative of dy/dx with respect to t.
3. Divide by dx/dt.
Example:
Example:
• Find the first derivative (dy/dx).
2. Find the derivative of dy/dx with respect to t.
Quotient Rule
3. Divide by dx/dt.
The equation for the length of a parametrized curve is similar to our previous “length of curve” equation:
(Notice the use of the Pythagorean Theorem.)
Likewise, the equations for the surface area of a parametrized curve are similar to our previous “surface area” equations:
This curve is:
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# The tens digit of positive integer X is 3. Is the units
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The tens digit of positive integer X is 3. Is the units [#permalink]
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23 Apr 2008, 08:04
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The tens digit of positive integer X is 3. Is the units digit of X greater than 4?
(1) The units digit of 2X is less than the units digit of X.
(2) The tens digit of 2X is 7.
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Re: DS - units digit [#permalink]
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23 Apr 2008, 08:17
I'm thinking D. Is this from GMATPrep?
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Re: DS - units digit [#permalink]
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23 Apr 2008, 09:44
D
For both (1) and (2):
34*2 = 68 N
35*2 = 70 Y
36*2 = 72 Y
37*2 = 74 Y
39*2 = 78 Y
Both are SUFF.
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Re: DS - units digit [#permalink]
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24 Apr 2008, 13:32
OA is D. Thanks
jimmyjamesdonkey wrote:
I'm thinking D. Is this from GMATPrep?
This is not from GMATPrep.
Re: DS - units digit [#permalink] 24 Apr 2008, 13:32
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# The tens digit of positive integer X is 3. Is the units
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 788 | 2,722 | {"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-2017-22 | longest | en | 0.807269 |
http://mathematica.stackexchange.com/questions/30895/create-parametricfunction-object-without-parametricndsolve/30908 | 1,469,584,534,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825358.53/warc/CC-MAIN-20160723071025-00086-ip-10-185-27-174.ec2.internal.warc.gz | 131,779,511 | 17,329 | # Create ParametricFunction object without ParametricNDSolve
Documentation of v9 states that ParametricFunction is generated by ParametricNDSolve and ParametricNDSolveValue.
Is there any way of creating ParametricFunction by definition, similar to creating an InterpolatingFunction by Interpolation?
My motivation behind is that for certain parameters, I need solution of ParametricNDSolveValue to be returned and for other parameters I want to use analytical solution to be returner in the same format -ParametricFunction.
-
why don't you just return the analytical solution as a Function[args, Evaluate[analyticalSolution]]? – user21 Aug 22 '13 at 12:51
thanks for idea! I'm going to put example and my answer to it for clarity. – Cendo Aug 22 '13 at 13:55
Assume an example function which will switch behaviour depending on c being True or False. This solution is based on comment of @ruebenko.
fun[c_] := Module[{x, a, b, t},
If[c,
ParametricNDSolveValue[{
x''[t] + a x'[t] + b^2 x[t] == 0,
x[0] == 1,
x'[0] == 0},
x, {t, 0, 10}, {a, b}],
(*else*)
Function[{a, b},
Interpolation[{#, #^2} & /@Range[0, 10, 1]
(*same range of t in InterpolatingFunction as from ParametricNDSolveValue*)]]
]
]
Now we call fun with :
fTrue = fun[True]
fFalse = fun[False]
(*ParametricFunction[<>]*)
(*Function[{a, b}, Interpolation[({#1, #1^2} &) /@ Range[0, 10, 1]]]*)
We can address both resulting functions like a ParametricFunction and get an InterpolatingFunction:
fTrue[1, 1]
fFalse[1, 1]
(* InterpolatingFunction[{{0.,10.}},<>] *)
(* InterpolatingFunction[{{0,10}},<>] *)
- | 465 | 1,576 | {"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} | 2.984375 | 3 | CC-MAIN-2016-30 | latest | en | 0.68656 |
https://stats.stackexchange.com/questions/33053/which-independent-variables-are-most-important-in-predicting-the-response-variab?noredirect=1 | 1,719,227,913,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865348.3/warc/CC-MAIN-20240624084108-20240624114108-00169.warc.gz | 485,365,584 | 40,186 | # Which independent variables are most important in predicting the response variable? [duplicate]
I'm a biologist, and I have a large dataset that I'm trying to analyze. Here are the variables I'm working with:
• levels of 211 different metabolites in 16 different blood samples (predictor variables)
• how well each of the 16 blood samples performed in a specific test (response variable)
I am trying to figure out which variable(s) are the most important in predicting the performance of a blood sample in the test.
I would like to make these data more manageable by doing a PCA, but I'm new to this sort of analysis. I understand that a PCA will create groups of principal components (it will group metabolites that covary with each other and label each of these groups a principal component), but I'm not sure how to take into account the response variable in this analysis.
Any help would be much appreciated!!! Thank you.
> summary(metabolites_princomp)
Importance of components:
Comp.1 Comp.2 Comp.3 Comp.4
Standard deviation 3.9608225 0.40128486 0.259868774 0.215004349
Proportion of Variance 0.9805072 0.01006435 0.004220736 0.002889179
Cumulative Proportion 0.9805072 0.99057153 0.994792267 0.997681446
[...]
• What language or software are you planning to work with? (It doesn't make a difference as to what PCA means but it could help shape a response) Commented Jul 26, 2012 at 1:20
• I've been working in R so far, but I'm new to it. I've done a PCA on the metabolite information alone (not taking into account the response variable) and I got 16 principal components, the first of which accounts for 98% of the variance. Commented Jul 26, 2012 at 1:29
• I can already tell you you forgot to center the rows of your matrix! Commented Jul 26, 2012 at 1:43
• @David: X - rowMeans(X) will be quite a bit more efficient if you want to center the rows. :-) Commented Jul 26, 2012 at 2:05
• @Lindsay: Unfortunately, it is not the rows you want to be centering in your case. If you want things standardized (centered and rescaled) you can just do princomp(X,cor=T). Commented Jul 26, 2012 at 2:14
First, note that your understanding of PCA is slightly off. PCA doesn't "group" variables into principal components. Each principal component is, rather, a new variable (a "new metabolite") of the same length as each of your original variables (in your case, 16).
It's true that each principal component can represent a group of original metabolites, but it's more complicated and sophisticated than that. Each metabolite can actually be correlated with multiple principal components, each a different amount. It could be true that a group of metabolites are purely represented principal component A and another group is purely represented by principal component B, but it is far more likely that each is correlated with multiple PCs, each to a different extent.
So think of each PC as a new metabolite (in PCA parlance we'd call it an "eigenmetabolite") that is representative of a general pattern across the metabolites. Some metabolites are highly representative of this pattern, while some aren't, and some share some properties with multiple patterns.
Now we come to your question. The way we bring in a response variable is to look at the relationship- for example, a correlation or linear regression- between each principal component (each eigenmetabolite) and the response variable. Here is a neat little example in R code. (Note that I use the svd function rather than the pls package, as you might be using).
set.seed(1337)
center.rows = function(m) t(apply(m, 1, function(r) r - mean(r)))
response.var = rnorm(16, 0, 1)
relevant.metabolites = t(replicate(50, rnorm(16, response.var, 2)))
unrelated.metabolites = t(replicate(110, rnorm(16, 0, 2)))
metabolite.data = rbind(relevant.metabolites, unrelated.metabolites)
s = svd(center.rows(metabolite.data))
What I did is create a matrix of simulated metabolite data with 50 genes that are somewhat correlated with the response variable, and 110 genes that aren't. We can see that only one PC is significant, and it explains ~33% of the variance:
var.explained = s$d^2 / sum(s$d^2)
plot(var.explained)
Now let's compare that PC to the response variable:
plot(response.var, s$v[, 1]) As you can see, the first PC almost perfectly recaptures the response variable that we used to create the matrix (correlation is -.993), even though we put a lot of extra noise in. (Don't worry that the correlation is negative- the direction is arbitrary). Indeed, this correlation with the "eigenmetabolite" is considerably greater than the correlation of the response variable to any of the individual significant genes: hist(apply(relevant.metabolites, 1, function(r) cor(r, response.var))) Of course, that's exactly why we're applying PCA in the first place. While the relationship between the response variable and any individual metabolite may be too weak to measure over the rest of the noise, when you combine many genes together you might find that a principal component captures that response well! • @ David: Thank you so much for such a detailed explanation! I'm going to give it a go right now. Commented Jul 26, 2012 at 2:25 • It doesn't look like your "unrelated" metabolites are unrelated. :-) One aspect of this answer that may be misleading is that even if there were some PC associated with the response, it may not be among the ones explaining the variation in the data. So, some caution is advised. Commented Jul 26, 2012 at 2:26 • @David: I'm getting stuck at the step where you plot PC1 against the response variable. I have a data.frame with 212 rows and 16 columns. The last row is my response variable, so I'm typing in plot(metabolites[212,], s$v[,1]). Why won't that work for me? Commented Jul 26, 2012 at 2:41
• What is your error? Commented Jul 26, 2012 at 2:42
• @cardinal: right you are, that was a typo, since fixed (I'm not at a computer so I'll have to adjust the figures later in response to that change). Lindsay, what are the dimensions of your s\$v matrix? Commented Jul 26, 2012 at 2:47
After the great discussion, above, I'm hesitant to add another idea, but it seems to me that partial least squares may do what you want, more easily. I usually do this in SAS, so I don't know the details of doing it in R, but there is a PLS package that does both partial least squares regression and principal component regression.
The essential idea behind PLS is that it incorporates both relations among the independent variables and between the independent and dependent variables.
• Thanks for your input. I've been doing a bit of research about this today, and I was drifting towards PCR as well. I'm not sure that I quite understand PLS and PCR yet enough to be confident about executing them, but it does sound like it might make more sense considering my goals. Commented Jul 26, 2012 at 19:52
• Are 'partial least squares' and 'squared partial correlation' the same thing? I'm wondering whether this is similar to Kruskal's Drivers Analysis Commented Apr 20, 2017 at 20:28
• @MarkRamotowski I dont think they are teh same, but I never heard of Kruskal's drivers analysis. Commented Apr 21, 2017 at 11:34
• It's used in market research - I'm not sure how valid an approach it is if you've never heard of it! wiki.q-researchsoftware.com/wiki/… Commented Apr 21, 2017 at 12:21
• @MarkRamotowski There's a LOT of stuff I haven't heard of! I don't do market research, so, my ignorance of its methods doesn't mean much. Commented Apr 22, 2017 at 12:52 | 1,883 | 7,573 | {"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": 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} | 2.71875 | 3 | CC-MAIN-2024-26 | latest | en | 0.927336 |
https://jp.maplesoft.com/support/help/maplesim/view.aspx?path=type/fraction | 1,638,300,327,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359065.88/warc/CC-MAIN-20211130171559-20211130201559-00175.warc.gz | 406,014,778 | 28,132 | fraction - Maple Help
# Online Help
###### All Products Maple MapleSim
type/fraction
check for an object of type fraction
Calling Sequence type(x, fraction)
Parameters
x - any expression
Description
• The type(x, fraction) calling sequence returns true if x is of type rational(type,rational) and not of type integer.
Supertypes
•
Examples
> $\mathrm{type}\left(1,\mathrm{fraction}\right)$
${\mathrm{false}}$ (1)
> $\mathrm{type}\left(\frac{1}{2},\mathrm{fraction}\right)$
${\mathrm{true}}$ (2)
> $\mathrm{type}\left(0.5,\mathrm{fraction}\right)$
${\mathrm{false}}$ (3)
> $\mathrm{type}\left("String",\mathrm{fraction}\right)$
${\mathrm{false}}$ (4)
> $\mathrm{type}\left(\mathrm{Name},\mathrm{fraction}\right)$
${\mathrm{false}}$ (5)
> $\mathrm{type}\left(a\left[b\right],\mathrm{fraction}\right)$
${\mathrm{false}}$ (6)
> $\mathrm{type}\left(\frac{a}{b},\mathrm{fraction}\right)$
${\mathrm{false}}$ (7)
See Also | 296 | 952 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 14, "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-2021-49 | latest | en | 0.206149 |
https://codedump.io/share/9Z6khJiOuTX/1/why-won39t-the-lists-be-appended-to-my-global-list | 1,524,628,138,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947690.43/warc/CC-MAIN-20180425022414-20180425042414-00569.warc.gz | 611,129,549 | 8,333 | user3254780 - 1 year ago 56
Python Question
# Why won't the lists be appended to my global list?
I wrote the following code, used to generate all possible combinations of a specific number of elements with help of recursion:
``````global_values = []
parameters = [1,2,3,4,5,6]
content = ["x","y","v","sd"]
def rec(values, depth):
if(depth == 0):
global_values.append(values)
print values
return
for i in range(0,len(parameters)):
values.append(parameters[i])
rec(values,depth-1)
values.pop()
return
rec([],len(content))
print "!!!!----------!!!!"
print global_values
``````
The "print "value"" statement prints the correct results, but I would like to append this results to the global list. Why doesn't this work?
With
``````parameters = [1,2,3]
content = ["x","y"]
``````
I get the output:
``````[1, 1]
[1, 2]
[1, 3]
[2, 1]
[2, 2]
[2, 3]
[3, 1]
[3, 2]
[3, 3]
!!!!----------!!!!
[[], [], [], [], [], [], [], [], []]
``````
But I want the last list filled with the values printed before.
At the time you append `values` to `global_values`, there are numbers in the list `values`. However, you remove those numbers later, and and thus the list that was appended to `global_values` is now empty. This problem happens because there is only ever one list `values` that keeps getting re-used.
To see what I mean try:
``````list1 = []
list2 = [5,9]
list1.append(list2)
list2.pop() #because list2 is inside list1, this changes the contents of list1
print list1 #prints [[5]]
``````
To append a copy of `values` use `global_values.append(values[:])`
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download | 464 | 1,661 | {"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} | 2.5625 | 3 | CC-MAIN-2018-17 | latest | en | 0.727945 |
https://math.stackexchange.com/questions/4323794/polynomial-for-very-large-number-of-roots | 1,721,371,642,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00745.warc.gz | 325,969,065 | 45,142 | # Polynomial for very large number of roots
Update: Please also see this solution here provided by MattL.
I have the roots for a very large order polynomial (>100), and from those alone wish to recreate the polynomial and run into numerical challenges when using the poly function for doing this available in Matlab and Python (numpy).
I attempted to convolve the coefficients for the roots and still see numerical issues (although I am not sure why as the dynamic range of the interim results when convolving subsets of the roots is far below that allotted by the double precision floating point numbers I am using).
As a simple example, and showing where I begin to see issues: I start with a 59th order polynomial with unity coefficients: $$1+x+x^2+\ldots x^{59}$$
The roots of this will be the 60th roots of unity given as $$e^{jn2\pi/60}$$ with n from 1 to 59, which I also get from Python using np.roots which are then given as complex double precision floating point representation (real and imaginary are both double precision). I then used the roots to poly function within python and would expect to get the unity coefficients back, but instead get large errors in the middle of the polynomial (yes, consistent with where we would have the biggest sum of products).
I then used my own approach of iteratively convolving coefficients while watching for dynamic range issues. For example in the plot below, I first show the resulting polynomial if only the first half of the roots were used, with the magnitude in dB to show that the dynamic range is far below the 300 dB+ allotted by double precision floating point. The lower portion of this plot shows the result of convolving this polynomial returned by the first 30 roots with the similar polynomial returned by the second 30 roots with no improvement in error.
I went on to iteratively convolve each of the complex roots $$z_n$$, by convolving the first two (as $$x-z_1$$ and $$x-z_2$$), and then convolving the result of that with the next root (as $$z^2-(z_1+z_2)+z_1 z_2$$ and $$x-z_3$$), etc through all of the roots, as given in Python below:
def cz(roots):
z = list(roots)
result = [1]
while z:
result = np.convolve(roots, [1, -z.pop()])
return result
This resulted in even more error:
My question is two fold: what is the source of this numerical error and are their improved mathematical techniques that I could use for converting large numbers of roots (I would like to be able to extend this to 100s of roots) to its polynomial form? I used the roots of unity in this simple example, but more generally I would like to be able to do this with any arbitrary roots on the complex plane (and can limit it to roots that are on or inside the unit circle).
• The most common numerical issue is the Runge-phenomenon , the polynomial oscillates extremely heavily. But this rather occurs if we have many equidistant real roots. No idea what went wrong in the case of the polynomials you chose. Commented Dec 4, 2021 at 15:29
• Do you compute $\mathrm{e}^{\mathrm{j} n 2\pi/60}$ by raising $\zeta_{60} = \mathrm{e}^{\mathrm{j} 2\pi/60}$ to various integer powers (so all of your roots have systematic copies of any (rounding/truncation) error in $\zeta_{60}$) or do you calculate $\mathrm{j} n 2\pi/60$ for each root and then exponentiate that? It might be worthwhile to look at how many steps of random walk of step-size "error in $\zeta$" the coefficient of $x^{30}$ takes in the giant sum of products of various powers of $\zeta$. (Related: never do sensitivity analysis on a discrete Fourier transform...) Commented Dec 4, 2021 at 15:52
• @Peter Interesting, I will need to learn more about the Runge-phenomenon, as "oscillating" feels like what I am seeing. My roots are equidistant on the unit circle so may be a similar effect. Thanks for sharing that. Commented Dec 4, 2021 at 15:59
• @JeanMarie No I haven't and what you are saying sounds very promising but is beyond my comprehension. Could you provide that idea as an answer here with a little more detail that I can then try? (I'm an engineer, not a mathematician so you are starting to venture into a foreign language for me- but even if you point me in the direction of what I do need to learn in order to be able to solve something like this I will be very appreciative) Commented Dec 4, 2021 at 16:52
• All right. I will do that. For the companion matrix definition see here. Commented Dec 4, 2021 at 16:59
As I understand your issue, you are computing the coefficients of the product of two polynomials like in this example:
$$(ux^2+vx+w) \times (a_0 + a_1 x + a_2 x^2 + ...)=wa_0+(a_0v+a_1w)x+\cdots \tag{1}$$
by convolving their coefficients:
$$[u,v,w] \star [a_0, a_1, a_2, \cdots ]=[wa_0,(a_0v+a_1w),(a_0u+a_1v+a_2w),\cdots]\tag{2}$$
and you want to understand the origin of the instabilities you are finding.
I am going to propose you to write the convolution operation as the application of a certain banded matrix to a column matrix, in this way:
$$\begin{pmatrix} w&&&&&&&\\ v&w&&&&&&\\ u&v&w&&&&&\\ &u&v&w&&&&\\ &&\ddots&\ddots&\ddots &&&\\ &&&u&v&w&&\\ &&&&u&v&w&\\ &&&&&u&v&w \end{pmatrix}\begin{pmatrix}a_0\\a_1\\ a_2\\ \vdots \\ \vdots \\ a_k \\ \vdots\end{pmatrix}\tag{3}$$
In this presentation, the convolution of two sequences, which is a symmetric operation, has been made asymmetrical. It is a way to single out one of the two polynomials and understand its intrinsic rôle, whatever the other polynomial.
A nice way to consider (3) is by viewing the succession of lines as a "moving coefficient window" sliding in front of the second "table of values".
Please note that the column vector
• has been turned upside down as is classical for convolution.
• can be given "trailing zeros" in order to have the same number of elements as the number of columns of the matrix. Moreover, ending by a certain number of zeros (at least equal to the degree of the first polynomial) is usual for getting an exact convolution (these considerations are classical when one examines the sizes of the convolution of two sequences).
In general, if $$d$$ is the degree of the first polynomial, the banded matrix has $$d+1$$ bands.
Now a very bad case that will be a kind of counterexample displaying an intrinsic origin of the problem of instability you want to solve:
Let us consider polynomial
$$p(x)=x^5+3x^4+x^3+x^2+x+1\tag{4}$$ with degree $$d=5$$, a "perturbated" version of your polynomial. The associated banded matrix with $$d+1=6$$ bands with format $$15 \times 15$$ has the huge condition number $$c=1.33 \times 10^7$$ (recall : the condition number is the number by which the error is multiplied, here 10 million times). For example if we work with digits truncated to $$10^{-8}$$, the second rank decimals will be corrupted...).
If I consider the case of a $$20 \times 20$$ matrix, with the same polynomial, it is worse: the condition number is as high as $$2 \times 10^9$$ !
Moreover, we are still far from the objective of considering polynomials with degrees of some hundreds...
Had we taken polynomial $$x^5+2x^4+x^3+x^2+x+1$$ with a same $$15 \times 15$$ matrix, we would have had a different condition number: "only" $$10^4$$...
The conclusion is that the observed fluctuations depend heavily on the involved polynomial(s) (I use the plural ; a similar analysis could be done on the second polynomial). This does not mean that the rounding errors due to thousands of multiplications/additions do not play a rôle, but a secondary rôle.
Here is the way we can obtain the condition number with Matlab (please note that Matlab does almost all the work : the construction of the banded convolution matrix is handled by a specific function "convmtx"):
n=15; % matrix size
p=[1,3,1,1,1,1,1]; % polynomial coeff.
M=convmtx(p,n); % conv. matrix
M=M(:,1:n); % truncation (otherwise M is rectangular)
M=M'; % optional in fact
cond(M) % condition number
Remark 1: The roots of polynomial $$p(x)$$ given by (4) are
$$0.7729 \pm 0.8832i, \ \ -0.4685 \pm 1.1729i, \ \ -1.2426, \ \ -0.3663$$
Remark 2: A very interesting article here.
Remark 3: In order to understand how the condition number is used, see here.
• Thank you so much! I will try this and comment on how it worked for me (I likely won’t get back to this until tomorrow) Commented Dec 4, 2021 at 23:26
• +1, interesting. I've always seen the condition number used to estimate the error in $x$ in the equation $Ax = b$ given an error in $b$. This is what is done in the MathWorks link in Remark 3 and in the Wiki article on condition numbers. That is, a very large condition number means a relatively small error in $b$ can become a relatively large error in $x$. However, I feel in this case you are estimating the error in $b$ given the error in $x$. I'm not sure condition numbers give a good estimate; shouldn't the error in $b$ be bounded by the largest singular value of $A$ times the error in $x$? Commented Dec 5, 2021 at 12:32
• @user196574 Your remark about the "knowing b, find x" is interesting because it says implicitly that the "true usage" of condition number here would be to quantitfy the error in a "deconvolution" operation... I must think about that. My idea, you have understood it, is to isolate the contribution of a polynomial. The condition number comes indeed from considerations about linear system's solving, but has been recognized as general indicator of "good behavior" for a matrix. But for sure I don't pretend there are not better indicators... Commented Dec 5, 2021 at 12:42
• The more I think on it, the more I think my comment was confused. The fact that the condition number is the same for $A$ and $A^{-1}$ means that there is strong symmetry between "given error in $b$, find error in $x$" and "given error in $x$, find error in $b$". Also, in my comment above I said that "error in $b$" is bounded by the largest singular value of $A$ times the error in $x$; that's just absolute error: the relative error will be largest if $A$ shrinks the magnitude of $x$ and stretches $\delta x$, so the relative error is determined by the ratio of largest and smallest sing values. Commented Dec 5, 2021 at 12:46
• So to summarize, although I was first hesitant on the condition number being used in this way, I now am pretty happy with it as providing a bound for the relative error. I expect that when used in the case $Ax = b$ to estimate the relative error in $b$ given the relative error in $x$, the actual relative error will be close to the upper bound provided by the condition number when $Ax$ has smaller magnitude than $x$. Here I am assuming that the large condition numbers are stemming from small singular values, so that the worst relative error occurs when $x$ is shrunk by those small sing values. Commented Dec 5, 2021 at 12:59
For the leading example, one has to remark that the numpy.roots function is good at finding all the roots fast, but fails to give them to full accuracy. So inserting a Newton step greatly improves the roots and the reconstructed polynomial.
## in IPython, otherwise use print commands for output
p = [1.0]*60
z = np.roots(p)
max(abs(np.polyval(p,z)))
#>>> 1.069574264146017e-12
## Apply a Newton step,
## further steps have random effects with minimal improvements
z = z - np.polyval(p,z)/np.polyval(np.polyder(p),z)
max(abs(np.polyval(p,z)))
#>>> 4.6250852608698854e-14
## Reconstruct the polynomial
p1 = np.poly(z)
max(abs(p1-1))
#>>> 0.002289390374985656
plt.plot(p,'-+', ms=2); plt.plot(p1,'-o', ms=3); plt.show()
This is with one Newton step. A second one reduces the error by half, a third one increases the error again.
This is from the start better than any of your attempts. One might say that any polynomial that faithfully represents the roots with their errors needs to deviate from the original one in a systematic fashion. It is unlikely that errors, like rounding and cancellation, during the polynomial reconstruction will produce the exact opposite effect to the errors in the roots.
• Oh that’s a really good find as well thank you! This makes a lot of sense. @MattL Commented Dec 5, 2021 at 12:21
(too long for a comment)
What I mean to say is that, if you have
$$P = \prod\limits_{k = 1}^n {\left( {x - r_{\,k} } \right)}$$ then you may divide the roots into batches for which the coefficients formula is known and then convolve between them and with the remaining coefficients. That is
• if (some) of the roots are integral you have better to group them into $$P_{int} = \prod\limits_{k = 1}^{n_{\,int} } {\left( {x - p_{\,k} } \right)}$$ and convolve the coefficients as $$\begin{array}{*{20}c} 0 & 1 & 2 & \cdots & {n_{\,int} - 1} & {n_{\,int} } \\ \hline {} & { - p_{\,1} } & 1 & {} & {} & {} \\ {p_{\,2} p_{\,1} } & { - p_{\,2} } & 0 & {} & {} & {} \\ \hline {p_{\,2} p_{\,1} } & { - p_{\,1} - p_{\,2} } & 1 & {} & {} & {} \\ \end{array}$$ and so on, that finally leads to Vieta's formulas, which in case you can apply directly.
The resulting coefficients will be exact (within the digits limit).
• if (some) of the roots are rational you can proceed as above, or you can extract the lcm of the denominators $$P_{rat} = \prod\limits_{k = 1}^{n_{\,rat} } {\left( {x - {{p_{\,k} } \over {q_{\,k} }}} \right)} = \prod\limits_{k = 1}^{n_{\,rat} } {\left( {x - {{p'_{\,k} } \over Q}} \right)} = {1 \over {Q^{n_{\,rat} } }}\prod\limits_{k = 1}^{n_{\,rat} } {\left( {Qx - p'_{\,k} } \right)}$$
• if (some) of the roots are in aritmetic progression, then you know the coefficient for that batch through the Falling Factorial and the Stirling N. of 1st kind \eqalign{ & P_{ap} = \prod\limits_{k = 1}^{n_{\,ap} } {\left( {x - kr} \right)} = \prod\limits_{k = 0}^{n_{\,ap - 1} } {\left( {x - r - kr} \right)} = r^{n_{\,ap} } \prod\limits_{k = 0}^{n_{\,ap} - 1} {\left( {{{x - r} \over r} - k} \right)} = \cr & = r^{n_{\,ap} } \left( {{{x - r} \over r}} \right)^{\,\underline {\,n_{\,ap} \,} } = r^{n_{\,ap} } \sum\limits_{0\, \le \,k\, \le \,n_{\,ap} } {\left( { - 1} \right)^{\,n_{\,ap} - k} \left[ \matrix{ n_{\,ap} \cr k \cr} \right]\left( {{{x - r} \over r}} \right)^{\,k} } = \cr & = r^{n_{\,ap} } \sum\limits_{0\, \le \,k\, \le \,n_{\,ap} } {\sum\limits_{0\, \le \,j\, \le \,k} {\left( { - 1} \right)^{\,n_{\,ap} - j} \left[ \matrix{ n_{\,ap} \cr k \cr} \right] \left( \matrix{ k \hfill \cr j \hfill \cr} \right)\left( {{x \over r}} \right)^{\,j} } } \cr}
• if (some) of the roots are in geometric progression, then also you know the coefficient for that batch via the q-binomial theorem \eqalign{ & P_{gp} = \prod\limits_{k = 0}^{n_{\,gp} - 1} {\left( {x - r^{\,k} } \right)} = x^{n_{\,gp} } \prod\limits_{k = 0}^{n_{\,gp} - 1} {\left( {1 - {{r^{\,k} } \over x}} \right)} = x^{n_{\,gp} } \left( {1/x;\;r} \right)_{n_{\,gp} } = \cr & = x^{n_{\,gp} } \sum\limits_{0\, \le \,k\, \le \,n_{\,gp} } {\left( \matrix{ n_{\,gp} \cr k \cr} \right)_{\;r} \left( { - {1 \over x}} \right)^{\,k} r^{\binom{k}{2} } } \cr}
• if (some) of the roots lie on a circle with center at the origin and radius $$r$$, and are uniformly distributed with angle $$0, 1/n 2 \pi , 2/n 2 \pi, \cdots , (n-1)/n 2 \pi$$ then \eqalign{ & x^{\,n} - r^{\,n} = 0\quad \Leftrightarrow \quad x = re^{\,i{k \over n}2\pi } = r\omega _n ^{\,k} \quad \left| {\;k \in \left\{ {0,1, \cdots ,n - 1} \right\}} \right.\quad \Leftrightarrow \cr & \Leftrightarrow \quad x^{\,n} - r^{\,n} = \prod\limits_{k = 0}^{n - 1} {\left( {x - re^{\,i{k \over n}2\pi } } \right)} \cr} and if the center of the circle is at $$c$$ $$\prod\limits_{k = 0}^{n - 1} {\left( {x - \left( {c + re^{\,i{k \over n}2\pi } } \right)} \right)} = \left( {x - c} \right)^{\,n} - r^{\,n}$$
• Thank you! This is interesting. All of my roots in the example given are complex as the roots of unity on the unit circle, so they all have magnitude 1 and angles given as $2\pi/60$ (in my example). That said I think they all are irrational. They are all however in complex conjugate pairs, so I could possibly group them into real 2nd order sections, (z^2-2z cos(w)+1), but I was also hoping for a more general solution (if it exists) as I will typically have irrational roots anywhere inside or on the unit circle. Commented Dec 4, 2021 at 19:39
• I have added for the case of roots being the roots of unity ( or $r^n$). The most general formulas are the Vieta's formulas Commented Dec 5, 2021 at 15:25
• @G Cab Hi ! Interesting. I have been interested by the reference (MattL) given in the front "Update" of the OP. This reference mentions the Leja orderin, I didn't know ; I haven't found the original paper but a paper that mentions it interesting on its own Commented Dec 10, 2021 at 18:36
• @JeanMarie: thanks, that reference work is really interesting Commented Dec 10, 2021 at 19:29 | 4,941 | 16,796 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 36, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-30 | latest | en | 0.930144 |
http://www.topperlearning.com/forums/ask-experts-19/f-the-area-of-equilateral-triangle-is-16-root-3-find-the-pe-mathematics-polynomials-54709/reply | 1,487,940,779,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171608.86/warc/CC-MAIN-20170219104611-00145-ip-10-171-10-108.ec2.internal.warc.gz | 629,323,437 | 37,548 |
Question
Mon April 09, 2012 By:
# f the area of equilateral triangle is 16 root 3. find the perimeter
Mon April 09, 2012
Area of equilateral triangle = root(3) * (side)2 / 4 = 16 root (3)
This gives a = 8
Thus, perimeter = a + a + a = 3a = 24
Related Questions
Mon September 26, 2016
# If one zero of the polynomial (a^2 + 9)x^2 + 13x + 6a is reciprocal of the other , find the value of 'a'
Thu September 15, 2016
| 151 | 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} | 3.734375 | 4 | CC-MAIN-2017-09 | latest | en | 0.861585 |
http://nikolmiu.blogspot.com/2013/06/important-of-mathematics.html | 1,529,899,320,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867424.77/warc/CC-MAIN-20180625033646-20180625053646-00333.warc.gz | 222,973,457 | 15,061 | ## Tuesday, June 18, 2013
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https://eurekamathanswerkeys.com/big-ideas-math-answers-grade-3-chapter-9/ | 1,716,443,443,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00352.warc.gz | 207,844,892 | 26,170 | Get a brief explanation for all the questions in Big Ideas Math Answers Grade 3 Chapter 9 Multiples and Problem Solving. So, the pupil those who are in search of the solutions of BIM Grade 3 Chapter 9 Multiples and Problem Solving can Download from here. We have provided the solutions as per the latest edition. You can apply the concepts in real-time and improve your performance.
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Lesson – 1: Use Number Lines to Multiply by Multiples of 10
Lesson – 2: Use Place Value to Multiply by Multiples of 10
Lesson – 3: Use Properties to Multiply by Multiples of 10
Lesson – 4: Problem Solving: Multiplication and Division
Lesson – 5: Problem Solving: All Operations
### Lesson 9.1 Use Number Lines to Multiply by Multiples of 10
Explore and Grow
Show 5 jumps of 3. Write a multiplication equation shown by the number line.
____ × _____ = _____
Answer: 5 × 3 = 15
Explanation:
The count starts from 0.
You have put 5 jumps of 3 on the number line.
Skip by 3 five times.
That means from 0 to 3, 3 to 6, 6 to 9, 9 to 12, 12 to 15.
5 × 3 = 15
Show 5 jumps of 30. Write a multiplication equation shown by the number line.
____ × ____ = _____
Answer: 5 × 30 = 150
Explanation:
The count starts from 0.
You have put 5 jumps of 30 on the number line.
Skip by 30 five times.
That means from 0 to 30, 30 to 60, 60 to 90, 90 to 120, 120 to 150.
5 × 30 = 150
Structure
Compare the models. How are they the same? How are they different?
Answer: The Multiplication equation for both the models are the same but counts are different.
Think and Grow: Number Lines and Multiples of 10
Example
Find 3 × 50.
3 × 50 means 3 groups of 50.
Number of jumps: ______
Size of each jump: ______
Start at 0. Skip count by 50 three times.
Answer: 3 × 50 = 150
Explanation:
From the figure, we can observe that the count starts from 0.
It means that there are 3 jumps of 50.
The count starts at 0. Skip 50 three times.
Number of jumps = 3
Size of each jump = 50
3 × 50 = 150
Show and Grow
Question 1.
Find 8 × 20.
Number of jumps: ______
Size of each jump: ______
8 × 20 = _____
Explanation:
From the figure, we can observe that the count starts from 0.
It means that there are 8 jumps of 20.
The count starts at 0. Skip 20 eight times.
Number of jumps = 8
Size of each jump = 20
8 × 20 = 160
Question 2.
Find 4 × 30
Number of jumps: _______
Size of each jump: _______
4 × 30 = _____
Explanation:
From the figure, we can observe that the count starts from 0.
It means that there are 4 jumps of 30.
The count starts at 0. Skip 30 four times.
Number of jumps = 4
Size of each jump = 30
4 × 30 = 120
Apply and Grow: Practice
Question 3.
Find 2 × 60.
Number of jumps: _______
Size of each jump: _______
2 × 60 = _____
Explanation:
From the figure, we can observe that the count starts from 0.
It means that there are 2 jumps of 60.
The count starts at 0. Skip 60 two times.
Number of jumps = 2
Size of each jump = 60
2 × 60 = 120
Question 4.
Find 5 × 50
5 × 50 = _____
Explanation:
From the figure, we can observe that the count starts from 0.
It means that there are 5 jumps of 50.
The count starts at 0. Skip 50 five times.
Number of jumps = 5
Size of each jump = 50
5 × 50 = 250
Question 5.
Find 3 × 70.
3 × 70 = _____
Explanation:
From the figure, we can observe that the count starts from 0.
It means that there are 3 jumps of 70.
The count starts at 0. Skip 70 three times.
Number of jumps = 3
Size of each jump = 70
3 × 70 = 210
Question 6.
Find 30 × 6.
30 × 6 = _____
Explanation:
From the figure, we can observe that the count starts from 0.
It means that there are 6 jumps of 30.
The count starts at 0. Skip 30 six times.
Number of jumps = 6
Size of each jump = 30
30 × 6 = 180
Question 7.
Structure
Show 2 × 40 on one number line and 4 × 20 on the other. What is the same about the number lines? What is different?
Answer: The model is the same on the number line. The counts are different and the number of jumps on the number line is different.
Think and Grow: Modeling Real Life
A section of an airplane has 20 rows of seats. Each row has 7 seats. Can the section seat more than 150 people? Explain.
Model:
The section ______ seat more than 150 people.
Explain:
Given that,
A section of an airplane has 20 rows of seats. Each row has 7 seats.
20 × 7 = 140 seats
Thus the section cannot seat more than 150 people.
Show and Grow
Question 8.
There are 9 rows of seats in an auditorium. Each row has 30 seats. Can the auditorium seat more than 250 people? Explain.
Given that,
There are 9 rows of seats in an auditorium. Each row has 30 seats.
9 × 30 = 270 seats
270 – 250 = 20
Thus the auditorium seat more than 250 people.
Question 9.
A mechanic installs new tires on 20 cars and 20 pickup trucks. How many new tires does the mechanic install in all?
Given that,
A mechanic installs new tires on 20 cars and 20 pickup trucks.
20 × 4 = 80 cars
20 × 4 = 80 trucks
80 + 80 = 160 new tires
Thus the mechanic install 160 new tires.
Question 10.
DIG DEEPER!
Newton saves $5 each week for 20 weeks. How much more money does he need to buy a new bike that costs$130? If he continues to save the same amount each week, how many more weeks does he need to save to buy the bike? Explain.
Given that,
Newton saves $5 each week for 20 weeks. 5 × 20 =$100
$130 –$100 = $30 Thus Newton need to save$30 to buy the bike.
### Use Number Lines to Multiply by Multiples of 10 Homework & Practice 9.1
Question 1.
Find 3 ×30.
Number of jumps: _______
Size of each jump: _______
3 × 30 =
Explanation:
From the figure, we can observe that the count starts from 0.
It means that there are 3 jumps of 30.
The count starts at 0. Skip 30 three times.
Number of jumps = 3
Size of each jump = 30
3 × 30 = 90
Question 2.
Find 7 × 60.
Number of jumps: _______
Size of each jump: _______
7 × 60 = _____
Explanation:
From the figure, we can observe that the count starts from 0.
It means that there are 7 jumps of 60.
The count starts at 0. Skip 60 seven times.
Number of jumps = 7
Size of each jump = 60
7 × 60 = 420
Question 3.
Find 4 × 40.
4 × 40 = ____
Explanation:
We can find the product by using the distributive property.
4 × 40 = 4 × (20 + 20)
4 × 40 = (4 × 20) + (4 × 20)
4 × 40 = 80 + 80
4 × 40 = 160
Question 4.
Find 80 × 3.
80 × 3 = _____
Explanation:
We can find the product by using the associative property.
80 × 3 = (8 × 10) × 3
80 × 3 = 8 × 3 × 10
80 × 3 = 24 × 10
80 × 3 = 240
Question 5.
Structure
Complete the number line. Then write the multiplication equation shown on the number line.
____ × ____ = _____
Answer: 5 × 30 = 150
By using the number line you can write the multiplication equation.
30 + 30 + 30 + 30 + 30 = 150
Question 6.
Modeling Real Life
A gymnasium has 9 rows of seats. Each row has 50 seats. Can the gymnasium seat more than 500 people? Explain.
Given,
A gymnasium has 9 rows of seats. Each row has 50 seats.
9 × 50 = 450 seats
450 – 500 = -50 people
No the gymnasium cannot seat more than 500 people.
Question 7.
Modeling Real Life
Ten adults and 20 children fill their bike tires at a public pump. How many tires are filled in all?
Given that,
Ten adults and 20 children fill their bike tires at a public pump.
10 × 2 = 20
20 × 2 = 40
20 + 40 = 60
Thus 60 tires are filled in all.
Review & Refresh
Question 8.
There are 35 counters. The counters are in 7 equal rows. How many counters are in each row?
7 rows of _____
35 ÷ 7 = _____
Explanation:
There are 35 counters.
The counters are in 7 equal rows.
35/7 = 5
7 rows of 5
35 ÷ 7 = 5
Thus there are 5 counters in each row.
Question 9.
You have 32 counters. You arrange them with 8 counters in each row. How many rows of counters do you make?
_____ rows of 8
32 ÷ 8 = ____
Explanation:
Given that,
You have 32 counters. You arrange them with 8 counters in each row.
32/8 = 4
Thus you can make 4 rows of 8.
### Lesson 9.2 Use Place Value to Multiply by Multiples of 10
Explore and Grow
Use models to find each product. Draw your models.
4 × 6 = _____
4 × 60 = _____
i.
Multiply the rows and columns
4 × 6 = 24
ii. Multiply the two numbers 4 and 60
4 × 60 = 240
Structure
Compare the models. How are they the same? How are they different?
Answer: The models for both the problem are the same but the columns are different.
Think and Grow: Place Value and Multiples of 10
Example
Find 4 × 70.
Step 1:
Make a quick sketch to model the product.
Think: 4 groups of 70, or 7 tens.
4 × 70 = 4 × ____ tens
4 × 70 = ____ tens
4 × 70 = 4 × 7 tens
4 × 70 = 28 tens
Step 2:
Regroup ____ tens
There are _____ hundreds and _____ tens.
So, 4 × 70 = _____.
There are 2 hundred and 8 tens.
Now regroup tens.
There are 2 tens
So, group the tens.
2 tens are grouped and the remaining is 8.
4 × 70 = 280
Show and Grow
Make a quick sketch to find the product
Question 1.
3 × 80 = ______
There are 2 hundred and 4 tens.
Question 2.
5 × 40 = _____
5 × 40 = 20 tens
There are 2 hundred and 0 tens.
Apply and Grow: Practice
Use place value to find the product.
Question 3.
3 × 90 = 3 × ____ tens
3 × 90 = ____ tens
3 × 90 = _____
3 × 90 = 3 × 9 tens
3 × 90 = 27 tens
3 × 90 = 270
Question 4.
6 × 60 = 6 × ____ tens
6 × 60 = ____ tens
6 × 60 = _____
6 × 60 = 6 × 6 tens
6 × 60 = 36 tens
6 × 60 = 360
Question 5.
2 × 70 = 2 × ____ tens
2 × 70 = ____ tens
2 × 70 = _____
2 × 70 = 2 × 7 tens
2 × 70 = 14 tens
2 × 70 = 140
Question 6.
9 × 20 = 9 × ____ tens
9 × 20 = ____ tens
9 × 20 = _____
9 × 20 = 9 × 2 tens
9 × 20 = 18 tens
9 × 20 = 180
Find the product
Question 7.
3 × 30 = ____
Explanation:
3 × 30 = 3 × 3 tens
3 × 30 = 9 tens
3 × 30 = 90
Question 8.
6 × 80 = ____
Explanation:
6 × 80 = 6 × 8 tens
6 × 80 = 48 tens
6 × 80 = 480
Question 9.
4 × 40 = _____
Explanation:
4 × 40 = 4 × 4 tens
4 × 40 = 16 tens
4 × 40 = 160
Question 10.
7 × 50 = _____
Explanation:
7 × 50 = 7 × 5 tens
7 × 50 = 35 tens
7 × 50 = 350
Question 11.
8 × 70 = _____
Explanation:
8 × 70 = 8 × 7 tens
8 × 70 = 56 tens
8 × 70 = 560
Question 12.
5 × 90 = _____
Explanation:
5 × 90 = 5 × 9 tens
5 × 90 = 45 tens
5 × 90 = 450
Question 13.
Reasoning
Explain why the product of 6 and 30 has 1 zero and the product of 4 and 50 has 2 zeros.
6 × 30 = 180
4 × 50 = 200
4 and 50 has two zeros because it is multiplied by 50 If the number is multiplied by 5 or 10 you will get two zeros.
Question 14.
YOU BE THE TEACHER
Is Descartes correct? Explain.
3 × 7 = 21
3 × 70 = 210
So the product of 3 and 70 is equal to the product of 3 and 7 with a 0 written after it.
Think and Grow: Modeling Real Life
Newton saves $30 each month for 6 months. Does he have enough money to buy the drone? Explain. Newton _______ have enough money to buy the drone.$189
Given that,
Newton saves $30 each month for 6 months. 6 ×$30 = $180 The cost of the drone is$189.
180 – 189 = -9
Therefore Newton does not have enough money to buy the drone.
Show and Grow
Question 15.
Descartes saves $20 each month for 8 months. Does he have enough money to buy the remote control jeep? Explain. Answer: Given that, Descartes saves$20 each month for 8 months.
8 × $20 =$160
The cost of remote control jeep is $129 160 – 129 =$31
Thus Descartes has enough money to buy the remote control jeep.
Question 16.
You practice playing the guitar for 40 minutes every day. How many minutes do you practice in one week?
Given that,
You practice playing the guitar for 40 minutes every day.
1 week = 7 days
7 × 40 = 280
Thus you practice 280 minutes in one week.
Question 17.
A box of snacks has 25 bags of pretzels and 25 bags of peanuts. How many bags are in 9 boxes?
Given that,
A box of snacks has 25 bags of pretzels and 25 bags of peanuts.
25 + 25 = 50
50 × 9 = 450
Thus there are 450 bags in 9 boxes.
### Use Place Value to Multiply by Multiples of 10 Homework & Practice 9.2
Make a quick sketch to find the product.
Question 1.
5 × 70 = _____
Question 2.
3 × 60 = ____
Use place value to find the product
Question 3.
8 × 50 = 8 × ____ tens
8 × 50 = ____ tens
8 × 50 = _____
8 × 50 = 8 × 5 tens
8 × 50 = 40 tens
8 × 50 = 400
Question 4.
7 × 60 = 7 × ____ tens
7 × 60 = ____ tens
7 × 60 = _____
7 × 60 = 7 × 6 tens
7 × 60 = 42 tens
7 × 60 = 420
Find the product.
Question 5.
6 × 90 = _____
Explanation:
6 × 90 = 6 × 9 tens
6 × 90 = 54 tens
6 × 90 = 540
Question 6.
8 × 30 = _____
Explanation:
8 × 30 = 8 × 3 tens
8 × 30 = 24 tens
8 × 30 = 240
Question 7.
5 × 40 = _____
Explanation:
5 × 40 = 5 × 4 tens
5 × 40 = 20 tens
5 × 40 = 200
Question 8.
Is Newton correct? Explain.
Explanation:
6 × 50 = 6 × 5 tens
6 × 50 = 30 tens
6 × 50 = 300
By this we can say that Newton is incorrect.
Question 9.
Structure
Write an equation for the quick sketch.
Answer: 2 × 60 = 120
There are 1 group of tens and 2 are remaining.
So, 2 × 60 = 120
Question 10.
Modeling Real Life
Descartes saves $50 each month for 5 months. Does he have enough money to buy the game system? Explain. Answer: Given that, Descartes saves$50 each month for 5 months.
5 × $50 =$250
The cost of the game system is $205.$250 – $205 =$45
Thus Descartes has enough money to buy the game system.
Question 11.
Modeling Real Life
A group of staff members packs coolers for a field trip. Each cooler has 15 peanut butter sandwiches and 15 turkey sandwiches. How many sandwiches are in 7 coolers?
Given that,
A group of staff members packs coolers for a field trip.
Each cooler has 15 peanut butter sandwiches and 15 turkey sandwiches.
15 + 15 = 30
7 × 30 = 210
Thus there are 210 sandwiches in 7 coolers.
Review & Refresh
Question 12.
Round 282 to the nearest ten and to the nearest hundred.
Nearest ten: ____
Nearest hundred: _____
First, 282 rounded to the nearest ten is 280.
The number nearest hundred to 282 is 300.
### Lesson 9.3 Use Properties to Multiply by Multiples of 10
Explore and Grow
Use the colored rectangles to find 5 × 30.
So, 5 × 30 = ______
Answer: 50 + 50 + 50 = 150
Explanation:
The multiplication equation for the yellow-colored rectangle is 5 × 10 = 50
The multiplication equation for the blue-colored rectangle is 5 × 10 = 50
The multiplication equation for the red-colored rectangle is 5 × 10 = 50
50 + 50 + 50 = 150
Reasoning
How does this model relate to the Distributive Property?
Yes, you can relate the above equation to the distributive property.
5 × 30 = 5 × (10 + 20)
5 × 30 = (5 × 10) + (5 × 20)
5 × 30 = 50 + 100
5 × 30 = 150
Think and Grow: Properties and Multiples of 10
Example:
Find 6 × 20
One Way:
Use the Associative Property of Multiplication
6 × 20 = 6 × (_____ × 10)
6 × 20 = (6 × ____) × 10
6 × 20 = ____ × 10
6 × 20 = ____
Rewrite 20 as _____ × 10
Associative Property of Multiplication
Rewrite 20 as 2× 10
6 × 20 = 6 × (2 × 10)
6 × 20 = (6 × 2) × 10
6 × 20 = 12 × 10
6 × 20 = 120
Another Way:
Use the Distributive Property
6 × 20 = 6 × (10 × _____)
6 × 20 = (6 × 10) × (6 × ____)
6 × 20 = ____ + _____
6 × 20 = ____
Rewrite 20 as 10 + 10
By using the Distributive Property we can find the product.
6 × 20 = 6 × (10 + 10)
6 × 20 = (6 × 10) × (6 × 10)
6 × 20 = 60 + 60
6 × 20 = 120
Show and Grow
Question 1.
Use the Associative Property of Multiplication to find 4 × 60.
4 × 60 = 4 × (_____ × 10)
4 × 60 = (4 × ____) × 10
4 × 60 = ____ × 10
4 × 60 = ____
You have to find the product by using the Associative Property of Multiplication.
4 × 60 = 4 × (6 × 10)
4 × 60 = (4 × 6) × 10
4 × 60 = 24 × 10
4 × 60 = 240
Thus the product of 4 × 60 = 240.
Question 2.
Use the Distributive Property to find 9 × 20.
9 × 20 = 9 × (10 × _____)
9 × 20 = (9 × 10) × (9 × ____)
9 × 20 = ____ + _____
9 × 20 = ____
You have to find the product by using the Distributive Property of Multiplication.
9 × 20 = 9 × (10 + 10)
9 × 20 = (9 × 10) + (9 × 10)
9 × 20 = 90 + 90
9 × 20 = 180
Apply and Grow: Practice
Use properties to find the product
Question 3.
7 × 30 = _____
Explanation:
You have to find the product by using the Associative Property of Multiplication.
7 × 30 = 7 × (3 × 10)
7 × 30 = (7 × 3) × 10
7 × 30 = 21 × 10
7 × 30 = 210
Question 4.
5 × 80 = _____
Explanation:
You have to find the product by using the Associative Property of Multiplication.
5 × 80 = 5 × (8 × 10)
5 × 80 = (5 × 8) × 10
5 × 80 = 40 × 10
5 × 80 = 400
Question 5.
5 × 20 = _____
Explanation:
You have to find the product by using the Associative Property of Multiplication.
5 × 20 = 5 × (10 × 2)
5 × 20 = (5 × 2) × 10
5 × 20 = 10 × 10
5 × 20 = 100
Question 6.
3 × 90 = _____
Explanation:
You have to find the product by using the Associative Property of Multiplication.
3 × 90 = 3 × (9 × 10)
3 × 90 = (3 × 9) × 10
3 × 90 = 27 × 10
3 × 90 = 270
Find the missing factor.
Question 7.
8 × ____ = 320
Explanation:
Let the missing factor be x.
8 × x = 320
x = 320/8
x = 40
Thus the missing factor is 40.
Question 8.
____ × 50 = 300
Explanation:
Let the missing factor be y.
y × 50 = 300
y = 300/50
y = 6
Thus the missing factor is 6.
Question 9.
_____ × 30 = 270
Explanation:
Let the missing factor be z.
z × 30 = 270
z = 270/30
z = 9
Thus the missing factor is 9.
Question 10.
Number Sense
Use the Associative Property of Multiplication to show why
4 × 20 = 8 × 10.
You have to find the product by using the Associative Property of Multiplication.
4 × 20 = 8 × 10
4 × (2 × 10) = 8 × 10
8 × 10 = 8 × 10
By this we can say that 4 × 20 = 8 × 10.
Question 11.
Open-Ended
Write three expressions equal to 240.
The three expressions equal to 240 is
240/3 = 80
2 × 40 | 2 × 40 | 2 × 40
80 + 80 + 80 = 240
Question 12.
Number Sense
Which equations show the Distributive Property?
2 × 20 = (2 × 10) + (2 × 10)
4 × (3 × 10) = (4 × 3) × 10
(7 × 10) + (7 × 10) = 7 × 20
2 × 20 = (2 × 10) + (2 × 10) is the distributive property.
4 × (3 × 10) = (4 × 3) × 10 shows the associative property.
Think and Grow: Modeling Real Life
There are 8 tables in the classroom. There are 5 students at each table. Each student has 10 markers. How many markers do the students have in all?
There are ______ markers at each table.
The students have ______ markers in all.
Given that,
There are 8 tables in the classroom.
There are 5 students at each table.
8 × 5 = 40
Each student has 10 markers.
40 × 10 = 400
Show and Grow
Question 13.
Your teacher buys 5 boxes of pens. Each box has 6 bundles of 10 pens. How many pens does your teacher buy in all?
Given that,
Your teacher buys 5 boxes of pens. Each box has 6 bundles of 10 pens.
6 × 10 = 60
60 × 5 = 300 pens
Thus the teacher buy 300 pens.
Question 14.
DIG DEEPER!
Newton earns $30 each work shift. He wants to buy Descartes a cat tree. The tree costs$150. After how many work shifts can Newton buy the tree?
Given,
Newton earns $30 each work shift. He wants to buy Descartes a cat tree. The tree costs$150.
To find how many work shifts can Newton buy the tree, you need to divide the cost of the tree by newton’s earned amount.
150/30 = 5
After 5 work shifts, Newton can buy the tree.
### Use Properties to Multiply by Multiples of 10 Homework & Practice 9.3
Question 1.
Use the Associative Property of Multiplication to find 6 × 70.
6 × 70 = 6 × (_____ × 10)
6 × 70 = (6 × ____) × 10
6 × 70 = ____ × 10
6 × 70 = ____
6 × 70 = 6 × (7 × 10)
6 × 70 = (6 × 7) × 10
6 × 70 = 42 × 10
6 × 70 = 420
Question 2.
Use the Distributive Property to find 3 × 20.
3 × 20 = 3 × (10 × _____)
3 × 20 = (3 × 10) × (3 × ____)
3 × 20 = ____ + _____
3 × 20 = ____
3 × 20 = 3 × (10 × 2)
3 × 20 = (3 × 10) × (1 × 2)
3 × 20 = 30 × 2
3 × 20 = 60
Use properties to find the product.
Question 3.
9 × 20 = _____
Explanation:
We can find the product by using the distributive property.
9 × 20 = 9 × (10 + 10)
9 × 20 = (9 × 10) + (9 × 10)
9 × 20 = 90 + 90
9 × 20 = 180
Question 4.
5 × 30 = ____
Explanation:
We can find the product by using the associative property.
5 × 30 = (5 × 3) × 10
5 × 30 = 15 × 10
5 × 30 = 150
Find the missing factor
Question 5.
____ × 60 = 180
Explanation:
Let the missing factor be x.
x × 60 = 180
x = 180/60
x = 3
Thus the missing factor is 3.
Question 6.
6 × ____ = 240
Explanation:
Let the missing factor be y.
6 × y = 240
y = 240/6
y = 40
Thus the missing factor is 40.
Question 7.
____ × 80 = 720
Explanation:
Let the missing factor be z.
z × 80 = 720
z = 720/80
z = 9
Thus the missing factor is 9.
Question 8.
YOU BE THE TEACHER
Your friend draws a model to find 4 × 20. Is your friend correct? Explain.
4 × 10 = 40 4 × 10 = 40
40 + 40 = 80
So, 4 × 20 = 80
There are 4 rows and 20 shaded blocks
4 × 20 = 80
Question 9.
Number Sense
How can you tell whether 7 × 40 or 8 × 70 is greater without finding the products?
Answer: We can find a greater number by seeing the numbers.
7 < 8
40 < 70
7 × 40 = 280
8 × 70 = 560
So, 7 × 40 < 8 × 70
Question 10.
Modeling Real Life
There are 9 teams in a math competition. Each team has 6 students. Each student answers 10 questions. How many questions are answered in all?
Given that,
There are 9 teams in a math competition. Each team has 6 students.
9 × 6 = 54 students
54 × 10 = 540
Thus the students answered 540 questions in all.
Question 11.
DIG DEEPER!
A soccer team earns $40 each week washing cars. The team wants to buy an inflatable field for$240. After how many weeks can the team buy the field?
Given that,
A soccer team earns $40 each week washing cars. The team wants to buy an inflatable field for$240.
240/40 = 6
It takes 6 weeks to buy the field.
Review & Refresh
Find the quotient
Question 12.
Explanation:
Divide the two numbers 3 and 18.
18/3 = 6
Thus the quotient is 6.
Question 13.
Explanation:
Divide the two numbers 4 and 32
32/4 = 8
Thus the quotient is 8.
Question 14.
Explanation:
Divide the two numbers 27 and 3
27/3 = 9
Thus the quotient is 9.
Question 15.
Explanation:
Divide the two numbers 4 and 16.
16/4 = 4
Thus the quotient is 4.
### Lesson 9.4 Problem Solving: Multiplication and Division
Explore and Grow
Use any strategy to solve the problem.
Descartes uses 72 blocks to build ships. He uses 9 blocks for each ship. Each ship has 2 fabric sails. How many sails does Descartes use?
Descartes uses _____ fabric sails.
Given that,
Descartes uses 72 blocks to build ships. He uses 9 blocks for each ship.
72/9 = 8
Each ship has 2 fabric sails.
8 × 2 = 16 sails
Therefore Descartes uses 16 fabric sails.
Structure
What equations did you use to solve? How can you write the equations using a letter to represent the number of fabric sails?
I used the division and multiplication equation to solve the problem.
First, you have to divide the number of blocks by the number of blocks for each ship.
72/9 = 8
After that multiply the blocks by the number of fabric sails.
8 × 2 = 16
Think and Grow: Using the Problem-Solving Plan
Example
A box of 8 burritos costs $9. How much does it cost a group of friends to buy 40 burritos? Understand the Problem What do you need to find? • A box has ______ burritos. • The box costs _____. • A group of friends wants to buy ______ burritos. Answer: • A box has 8 burritos. • The box costs$9.
• A group of friends wants to buy 40 burritos.
What do you know?
• You need to find how much it _____ costs to buy.
You can find the answer by using the above question.
• You need to find how much burritos cost to buy.
Make a Plan
How will you solve?
• Divide _____ by _____ to find how many _____ the group needs to buy.
• Then multiply the quotient by _____ to find the total cost.
• Divide 40 by 8 to find how many burritos the group needs to buy.
• Then multiply the quotient by 9 to find the total cost.
Solve
Step 1:
How many boxes does the group need to buy?
Step 2:
Use to find the total cost.
It costs $_____ for 40 burritos. Answer: First divide 40/8 = 5 b = 5 c = 5 × 9 = 45 Thus the total cost is$45.
Show and Grow
Question 1.
You make 9 shots in a basketball game. Each shot is worth 2 points. Your friend has the same number of points. All of her shots are worth 3 points. How many shots does your friend make?
Given that,
You make 9 shots in a basketball game. Each shot is worth 2 points.
9 × 2 = 18 points
Your friend has the same number of points. All of her shots are worth 3 points.
18/3 = 6
Thus your friend needs to make 6 shots.
Apply and Grow: Practice
Write equations to solve. Use letters to represent the unknown numbers. Check whether your answer is reasonable.
Question 2.
You read 3 chapters. Each chapter has 8 pages. Your friend reads the same number of pages. All of her chapters have 6 pages. How many chapters does your friend read?
Given that,
You read 3 chapters. Each chapter has 8 pages.
3 × 8 pages = 24 pages
Your friend reads the same number of pages. All of her chapters have 6 pages.
24/6 = 4
Question 3.
There are 42 players in a basketball tournament. The players are divided into teams of 7 players. The teams are divided equally among 3 basketball courts. How many teams are at each basketball court?
Given that,
There are 42 players in a basketball tournament. The players are divided into teams of 7 players.
42/7 = 6 teams
The teams are divided equally among 3 basketball courts.
6/3 = 3
Thus there are 3 teams at each basketball court.
Question 4.
You have 2 dream catcher kits. Each kit makes 4 dream catchers. You make all of the dream catchers and sell them for $9 each. How much money do you earn? Answer: Given that, You have 2 dream catcher kits. Each kit makes 4 dream catchers. 2 × 4 = 8 dream catchers You make all of the dream catchers and sell them for$9 each.
8 × 9 = $72 Thus you earn$72.
Question 5.
A box of 4 test tubes costs $6. How much does it cost to buy 20 test tubes? Answer: Given that, A box of 4 test tubes costs$6.
20/4 = 5
5 × $6 =$30
Thus it costs $30 to buy 20 test tubes. Think and Grow: Modeling Real Life There are 4 crates of milk bottles. Each crate holds 20 bottles. You hand out an equal number of bottles to 10 tables of students. How many bottles of milk does each table of students get? Understand the problem: Make a plan: Solve: Each table of students gets _____ bottles of milk. Answer: Given that, There are 4 crates of milk bottles. Each crate holds 20 bottles. 4 × 20 = 80 bottles You hand out an equal number of bottles to 10 tables of students. 80/10 = 8 Therefore each table of students gets 8 bottles of milk. Show and Grow Question 6. Six groups of hikers have 2 cases of water to share equally. Each case has 30 bottles of water. How many bottles of water does each group get? Answer: Given, Six groups of hikers have 2 cases of water to share equally. Each case has 30 bottles of water. 2 × 30 = 60 bottles of water There are 6 groups so divide 60 by 6. 60/6 = 10 Thus each group gets 10 bottles of water. Question 7. DIG DEEPER! Newton and Descartes decide to buy 2 pet toys that cost$20 each. Newton saves $5 each week. Descartes saves$3 each week. If they combine their money, how long does it take them to save enough money to buy the toys?
Given that,
Newton and Descartes decide to buy 2 pet toys that cost $20 each. 2 × 20 =$40
Newton saves $5 each week. Descartes saves$3 each week.
5 + 3 = 8
40/8 = 5
Thus it takes 5 weeks to buy the toys.
### Problem Solving: Multiplication and Division Homework & Practice 9.4
Write equations to solve. Use letters to represent the unknown numbers. Check whether your answer is reasonable.
Question 1.
Your friend saves $5 each week for 8 weeks. He spends all of the money on 4 toys that each cost the same amount. How much does each toy cost? Answer: Given that, Your friend saves$5 each week for 8 weeks.
He spends all of the money on 4 toys that each cost the same amount.
To find the cost of the toy we need to multiply your friend savings and the number of weeks.
8 × 5 = 40
Thus the cost of the toy is $40. Question 2. There are 3 trees. Each tree has 2 birdhouses. Each birdhouse has 4 birds. How many birds are there in all? Answer: Given that, There are 3 trees. Each tree has 2 birdhouses. 3 × 2 = 6 birdhouses Each birdhouse has 4 birds. 6 × 4 = 24 birds Therefore there are 24 birds in all. Question 3. There are 54 students at a field day who are divided equally into teams of 6 students. The teams are divided equally among 3 stations. How many teams are at each station? Answer: Given that, There are 54 students at a field day who are divided equally into teams of 6 students. 54/6 = 9 teams The teams are divided equally among 3 stations. Again the teams are divided into 3 stations. 9/3 = 3 Thus there are 3 teams at each station. Question 4. Newton runs an equal number of miles 2 days each week. He runs 8 miles each week. One mile is equal to 4 laps around the track. Which equation can you use to find how many laps Newton runs each day? r = 8 ÷ 4 r = 4 ÷ 4 r = 4 × 4 r = 2 × 4 Answer: Given that, Newton runs an equal number of miles 2 days each week. He runs 8 miles each week. One mile is equal to 4 laps around the track. 8/2 = 4 To find how many laps Newton runs each day you need to divide 4 by 4 r = 4 ÷ 4 Thus the correct answer is option b. Question 5. Modeling Real Life Ten classrooms have 3 boxes of white boards to share equally. Each box has 30 white boards. How many white boards does each classroom get? Answer: Given that, Ten classrooms have 3 boxes of white boards to share equally. Each box has 30 white boards. 1 box = 30 whiteboards 3 boxes = 3 × 30 = 90 white boards Now divide the whiteboards by number of classrooms. 90/10 = 9 Thus each classroom gets 9 whiteboards. Question 6. DIP DEEPER! Newton and Descartes decide to buy 2 amusement park tickets that cost$30 each. Newton saves $2 each week. Descartes saves$4 each week. If they combine their money, how long does it take them to save enough money to buy the tickets?
Given that,
Newton and Descartes decide to buy 2 amusement park tickets that cost $30 each. Newton saves$2 each week. Descartes saves $4 each week. 2 + 4 = 6 Now divide the ticket cost by their savings. 30/6 = 5 Therefore it takes 5 weeks to save enough money to buy the tickets. Review & Refresh Round to the nearest ten to estimate the difference Question 7. Answer: 58 round to the nearest ten is 60. 27 round to the nearest ten is 30. 60 -30 30 Thus the estimated difference is 30. Question 8. Answer: 763 round to the nearest ten is 760. 415 round to the nearest ten is 420. 760 -420 340 Thus the estimated difference is 340. Question 9. Answer: 686 round to the nearest ten is 690. 24 round to the nearest ten is 20. 690 -20 670 Thus the estimated difference is 670. ### Lesson 9.5 Problem Solving: All Operations Explore and Grow Use any strategy to solve the problem. You are making 6 fruit baskets. Each basket has 3 pieces of fruit in it to start. You buy 18 bananas and divide them equally among the baskets. How many pieces of fruit are in each fruit basket now? There are ______ pieces of fruit in each fruit basket now. Answer: Given that, You are making 6 fruit baskets. Each basket has 3 pieces of fruit in it to start. 6 × 3 = 18 You buy 18 bananas and divide them equally among the baskets. 18/18 = 1 Thus there are 4 pieces of fruit in each fruit basket now. Structure How can you solve this problem using one equation? Answer: We can solve the above problem by using the multiplication equation. First, you need to multiply the number of fruit baskets by the number of pieces of fruit. After that divide the number of pieces by the number of bananas. Think and Grow: One Equation with Two Operations Example Newton buys 3 DVDs for$4 each. He pays with a $20 bill. What is his change? You can write one equation with two operations to solve this problem. The equation is shown. 20 – 3 × 4 = c ← c is the amount of change. When solving a problem with more than one type of operation, use the rules below. • First, multiply or divide as you read the equation from left to right. • Then add or subtract as you read the equation from left to right. Step 1: Multiply from left to right. 20 – 3 × 4 = c 20 – ____ = c Step 2: Subtract from left to right. 20 – ____ = c ____ = c His change is ____. Answer: Step 1: Multiply from left to right. 20 – 3 × 4 = c 20 – 12 = c Step 2: Subtract from left to right. 20 – 12 = c 8 = c His change is 8. Show and Grow Question 1. There are 8 tomato plants. You pick 9 tomatoes from each plant. You give away 35 of them. Use the equation 8 × 9 – 35 to find how many tomatoes you have left. Answer: Given that, There are 8 tomato plants. You pick 9 tomatoes from each plant. 8 × 9 = 72 You give away 35 of them. We can find the number of tomatoes left by using the equation. 8 × 9 – 35 72 – 35 = 37 Thus 37 tomatoes are left. Question 2. A family buys 5 tickets for a musical. Each ticket costs$9. They spend $28 at the musical on snacks. Write and solve an equation to find how much they spend in all at the musical. Use to represent the total amount spent. Answer: Given that, A family buys 5 tickets for a musical. Each ticket costs$9.
5 × $9 =$45
They spend $28 at the musical on snacks. 45 – 28 = 17 The equation to find how much they spend in all at the musical is 5 × 9 – 45 Apply and Grow: Practice Write an equation to solve. Use a letter to represent the unknown number. Check whether your answer is reasonable. Question 3. Newton buys 2 movie tickets. Each ticket costs$7. Descartes spends $23 at the movie on snacks. How much money do they spend in all at the movie? Answer: Given, Newton buys 2 movie tickets. Each ticket costs$7.
2 × 7 = $14 Descartes spends$23 at the movie on snacks.
23 + 2 × 7
23 + 14 = $37 Question 4. Newton has 28 cards. Descartes has 24 cards. Newton divides his cards into 4 equal stacks and gives Descartes one stack. How many cards does Descartes have now? Answer: Given, Newton has 28 cards. Descartes has 24 cards. Newton divides his cards into 4 equal stacks and gives Descartes one stack. 28/4 = 7 Thus there are 7 cards in each stack. 24 + 7 = 31 cards Therefore Descartes has 31 cards now. Question 5. There are 12 apps divided into 3 equal rows on a smart phone. One row of apps is removed. How many apps are left? Answer: Given that, There are 12 apps divided into 3 equal rows on a smart phone. 12/3 = 4 There are 4 apps in each row One row of apps is removed. 12 – 4 = 8 Thus 8 apps are left. Question 6. It costs$240 each week to rent a car. Newton has a coupon that saves him $10 each day he rents the car. How much will it cost him to rent the car for a week with the coupon? Answer: Given that, It costs$240 each week to rent a car. Newton has a coupon that saves him $10 each day he rents the car. 1 week – 7 days 1 day –$10
7 × 10 = $70 240 – 70 =$170
Question 7.
YOU BE THE TEACHER
Your friend says 24 – 6 ÷ 2 = 9. Is your friend correct? Explain.
No, because you need to solve the equation from the left.
24 – (6 ÷ 2) = 24 – 12
= 12
Think and Grow: Modeling Real Life
Newton has $135. He saves$20 each week for 8 weeks. How much money does he have now?
Understand the problem:
Make a plan:
Solve:
Newton now has _____.
Given,
Newton has $135. He saves$20 each week for 8 weeks.
20 × 8 = 160
$135 +$160 = $295 Thus Newton now gas$295.
Show and Grow
Question 8.
Your teacher buys 3 packages of napkins for a class party. Each package has 50 napkins. The class uses 79 napkins. How many napkins are left?
Given,
Your teacher buys 3 packages of napkins for a class party. Each package has 50 napkins.
1 pack – 50 napkins
3 packs – 3 × 50 = 150
The class uses 79 napkins.
150 – 79 = 71
Thus 71 napkins are left.
Question 9.
There are 60 seconds in one minute. It takes you 2 minutes and 16 seconds to run from your home to your friend’s home. How many seconds does it take you?
Given,
There are 60 seconds in one minute. It takes you 2 minutes and 16 seconds to run from your home to your friend’s home.
1 min – 60 seconds
2 mins – 2 × 60 = 120 seconds
120 + 16 = 136 seconds
Thus it takes 136 seconds to run from your home to your friend’s home.
Question 10.
A store is selling comic books for $5 each. The store sells 33 superhero comic books and 57 science-fiction comic books. How much money does the store earn? Answer: Given, A store is selling comic books for$5 each.
The store sells 33 superhero comic books and 57 science-fiction comic books.
33 + 57 = 90
5 × 90 = $450 Thus the store earn$450.
### Problem Solving: All Operations Homework & Practice 9.5
Question 1.
There are 20 math problems divided into 4 equal columns on a worksheet. Your teacher has you cross out one column of problems. Use the equation 20 – 20 ÷ 4 = p to find how many problems are left.
Given that,
There are 20 math problems divided into 4 equal columns on a worksheet. Your teacher has you cross out one column of problems.
20 – 20 ÷ 4 = p
20 – 5 = p
15 = p
Thus 15 problems are left.
Question 2.
Newton has 42 blocks. Descartes has 48 blocks. Newton divides his blocks into 6 equal groups and gives Descartes one group. How many blocks does Descartes have now? Use d to represent how many blocks Descartes has now.
Given,
Newton has 42 blocks. Descartes has 48 blocks. Newton divides his blocks into 6 equal groups and gives Descartes one group.
42/6 = 7 blocks
48 + 7 = 55 blocks
Thus Descartes has 55 blocks now.
Question 3.
There are 6 palm trees. An islander gathers 8 coconuts from each tree. She gives away 19 of them. How many coconuts does she have now? Write an equation to solve. Use a letter to represent the unknown number. Check whether your answer is reasonable.
Given,
There are 6 palm trees. An islander gathers 8 coconuts from each tree. She gives away 19 of them.
6 × 8 – 19 = p
48 – 19 = p
29 = p
Thus the unknown number is 29.
Question 4.
DIG DEEPER!
Find the number that makes 5 × ____ – 15 = 5 true. Explain.
5 × ____ – 15 = 5
5 × p – 15 = 5
p – 15 = 5/5
p – 15 = 1
p = 1 + 15
p = 16
Thus the value of the number is 16.
Question 5.
Number Sense
Which equations are true?
3 + 5 × 2 = 13
Solve from left to right.
3 + 10 = 13
13 = 13 (true)
20 – 10 × 2 = 20
20 – 20 = 0 (false)
36 ÷ 6 + 3 = 4
36 ÷ 9 = 4
4 = 4 (true)
26 – 8 ÷ 2 = 22
26 – 4 = 22
22 = 22 (true)
Question 6.
Modeling Real Life
A school nurse orders 7 packages of bandages. Each package has 20 bandages. The nurse uses 53 bandages. How many bandages are left?
Given,
A school nurse orders 7 packages of bandages. Each package has 20 bandages.
7 × 20 = 140 bandages
The nurse uses 53 bandages.
140 – 53 = 87 bandages
Thus 87 bandages are left.
Question 7.
Modeling Real Life
There are 60 seconds in one minute. You record a video that is 3 minutes and 48 seconds long. How many seconds long is the video?
Given,
There are 60 seconds in one minute. You record a video that is 3 minutes and 48 seconds long.
3 mins = 3 × 60 = 180 seconds
180 + 48 = 228 seconds
Thus the video is 228 seconds long.
Review & Refresh
Question 8.
Estimate: _____
The estimated number for 23 is 20.
The estimated number for 358 is 360.
The estimated number for 172 is 170.
360
170
+20
550
Question 9.
Estimate: _____
The estimated number for 202 is 200.
The estimated number for 64 is 60.
The estimated number for 545 is 550.
550
200
+60
810
Question 10.
Estimate: _____
The estimated number for 21 is 20
The estimated number for 15 is 20
The estimated number for 837 is 840
840
20
+20
880
### Multiples and Problem Solving Performance Task
Question 1.
a. You read 120 minutes from Monday through Thursday this week. How many minutes do you read on Thursday? Complete the picture graph for Thursday.
Given,
You read 120 minutes from Monday through Thursday this week.
Each star = 10 minutes
There are 9 stars
9 × 10 = 90 minutes
120 – 90 = 30 minutes
Thus you read 30 minutes on Thursday.
b. Last week you read 30 minutes each day for 5 days. Your goal this week is to read the same number of minutes as last week. How many minutes do you need to read on Friday to reach your goal? Complete the picture graph for Friday.
Given,
Last week you read 30 minutes each day for 5 days. Your goal this week is to read the same number of minutes as last week.
30 × 5 = 150 minutes
150 – 120 = 30 minutes
Thus to reach your goal you need to read 30 minutes on Friday.
Question 2.
Write equations to solve. Use letters to represent the unknown numbers. Check whether your answer is reasonable.
a. There are 60 minutes in one hour. Your friend reads 2 hours and 38 minutes during the week. How many minutes does your friend read in all?
Given,
There are 60 minutes in one hour. Your friend reads 2 hours and 38 minutes during the week.
1 hour – 60 min
2 hour – 2 × 60 = 120 min
120 + 38 = 158 min
b.Your cousin earns 2 stars on her graph each day for 5 days. How many minutes does your cousin read in all?
Given,
Your cousin earns 2 stars on her graph each day for 5 days.
2 × 5 = 10 stars
1 star = 10 min
10 stars = 10 × 10 min = 100 min
Question 3.
Use the information above. Order the numbers of minutes you, your friend, and your cousin read from least to greatest. The person with the least number of minutes wants to read the same amount as the person with the greatest number of minutes. How many more minutes does the person need to read?
Given that,
you read 150 minutes in all.
The order from least to greatest is 100, 150, 158.
### Multiples and Problem Solving Activity
Multiplication Flip and Find
Directions:
1. Place the Multiplication Flip and Find Cards face down in the boxes.
2. Players take turns flipping two cards.
3. If your cards show a matching expression and product, then keep the cards. If your cards do not show a matching expression and product, then flip the cards back over.
4. Play until all matches are made.
5. The player with the most matches wins!
Total number of cards = 12
There are 2 matching flip cards in boxes
To find the most matches we have to divide the total number of flip cards by matching cards.
12/2 = 6
Thus the player with the most matches wins is 6.
### Multiples and Problem Solving Chapter Practice
9.1 Use Number Lines to Multiply by Multiples of 10
Question 1.
Find 8 × 20
Number of jumps: ______
Size of each jumps: _____
8 × 20 = _____
Number of jumps: 8
Size of each jump: 20
8 jumps of 20 = 160
Question 2.
Find 7 × 40
7 × 40 = ____
Explanation:
Number of jumps: 7
Size of each jump: 40
7 jumps of 40 = 7 × 40 = 280
Question 3.
Find 30 × 9
30 × 9 = ____
Explanation:
Number of jumps: 9
Size of each jump: 30
30 × 9 = 270
Question 4.
Structure
Complete the number line. Then write the multiplication equation for the number line.
____ × ____ = ____
Answer: 3 × 40 = 120
Explanation:
By seeing the above number line we can say that there are
3 jumps of 40
That means 3 × 40
40 + 40 + 40 = 120
Thus the multiplication equation is 3 × 40 = 120
9.2 Use Place Value to Multiply by Multiples of 10
Make a quick sketch to find the product
Question 5.
6 × 40 = _____
Number of jumps: 6
Size of each jump: 40
We can find the product by using the number line.
6 jumps of 40.
6 × 40 = 240
Question 6.
5 × 20 = ____
Number of jumps: 5
Size of each jump: 20
We can find the product by using the number line.
5 jumps of 20.
5 × 20 = 100
Use place value to find the product
Question 7.
4 × 50 = 4 × ____ tens
4 × 50 = ____ tens
4 × 50 = ____
4 × 50 = 4 × 5 tens
4 × 50 = 20 tens
4 × 50 = 20 × 10
4 × 50 = 200
Question 8.
3 × 60 = 3 × ____ tens
3 × 60 = ____ tens
3 × 60 = ____
3 × 60 = 3 × 6 tens
3 × 60 = 18 tens
3 × 60 = 18 × 10
3 × 60 = 180
Question 9.
7 × 70 = 7 × ____ tens
7 × 70 = ____ tens
7 × 70 = ____
7 × 70 = 7 × 7 tens
7 × 70 = 49 tens
7 × 70 = 49 × 10
7 × 70 = 490
Question 10.
9 × 80 = 9 × ____ tens
9 × 80 = ____ tens
9 × 80 = ____
9 × 80 = 9 × 8 tens
9 × 80 = 72 tens
9 × 80 = 72 × 10
9 × 80 = 720
Find the product
Question 11.
2 × 60 = _____
Explanation:
We can write 60 as 6 tens.
2 × 60 = 2 × 6 tens
2 × 60 = 12 tens
2 × 60 = 12 × 10
2 × 60 = 120
Question 12.
8 × 40 = ____
Explanation:
We can write 40 as 4 tens.
8 × 40 = 8 × 4 tens
8 × 40 = 32 tens
8 × 40 = 32 × 10
8 × 40 = 320
Question 13.
5 × 90 = _____
Explanation:
We can write 90 as 9 tens.
5 × 90 = 5 × 9 tens
5 × 90 = 45 tens
5 × 90 = 45 × 10
5 × 90 = 450
Question 14.
Modeling Real Life
You practice ballet for 30 minutes every day. How many minutes do you practice in one week?
Explanation:
Given that,
You practice ballet for 30 minutes every day.
Convert from week to days.
1 week – 7 days
7 × 30 min = 210 minutes
Thus you practice 210 minutes in one week.
9.3 Use Properties to Multiply by Multiples of 10
Question 15.
Use the Associative Property of Multiplication to find 4 × 90.
4 × 90 = 4 × (____ × 10)
4 × 90 = (4 × ____) × 10
4 × 90 = ____ × 10
4 × 90 = ____
We can find the product by using the Associative property.
4 × 90 = 4 × (9 × 10)
4 × 90 = (4 × 9) × 10
4 × 90 = 36 × 10
4 × 90 = 360
Question 16.
Use the Distributive Property to find 8 × 20.
8 × 20 = 8 × (10 + ____)
8 × 20 = (8 × 10) + (8 × ____)
8 × 20 = ___ + ____
8 × 20 = ____
We can find the product by using the distributive property.
8 × 20 = 8 × (10 + 10)
8 × 20 = (8 × 10) + (8 × 10)
8 × 20 = 80 + 80
8 × 20 = 160
Use properties to find the product
Question 17.
7 × 20 = ____
Explanation:
We can find the product by using the distributive property.
7 × 20 = 7 × (10 + 10)
7 × 20 = (7 × 10) + (7 × 10)
7 × 20 = 70 + 70
7 × 20 = 140
Question 18.
5 × 70 = ____
Explanation:
We can find the product by using the Associative property.
5 × 70 = 5 × (7 × 10)
5 × 70 = (5 × 7) × 10
5 × 70 = 35 × 10
5 × 70 = 350
Find the missing factor
Question 19.
____ × 20 = 180
Explanation:
Let the missing factor be x.
x × 20 = 180
x = 180/20
x = 9
Thus the missing factor is 9
Question 20.
7 × ___ = 350
Explanation:
Let the missing factor be y.
7 × y = 350
y = 350/7
y = 50
Thus the missing factor is 50.
Question 21.
____ × 80 = 240
Explanation:
Let the missing factor be z.
z × 80 = 240
z = 240/80
z = 3
Thus the missing factor is 3.
Question 22.
Open-Ended
Write three expressions equal to 120.
_____ × ______
_____ × ______
_____ × _______
The three expressions equal to 120 are given below,
1 × 120 = 120
2 × 60 = 120
3 × 40 = 120
9.4 Problem Solving: Multiplication and Division
Write equations to solve. Use letters to represent the unknown numbers. Check whether your answer is reasonable.
Question 23.
There are 2 bookcases. Each bookcase has 3 shelves of 5 books. How many books are there in all?
Explanation:
Given,
There are 2 bookcases.
Each bookcase has 3 shelves of 5 books.
3 × 5 = 15 books in one bookcase
15 × 2 = 30 books
Thus there are 30 books in 2 bookcases.
Question 24.
Four veterinarians share 2 boxes of ear wipes. Each box has 20 packs of ear wipes. How many packs of ear wipes does each veterinarian get?
Given,
Four veterinarians share 2 boxes of ear wipes.
Each box has 20 packs of ear wipes.
2 × 20 = 40 packs
There are 4 veterinarians.
40/4 = 10
Thus each veterinarian gets 10 packs of ear wipes.
9.5 Problem Solving: All Operation
Write an equation to solve. Use a letter to represent the unknown number. Check whether your answer is reasonable.
Question 25.
Newton has 30 beads. Descartes has 22 beads. Newton divides his beads into 3 equal groups and gives Descartes one group. How many beads does Descartes have now?
Explanation:
Given that,
Newton divides his beads into 3 equal groups and gives Descartes one group.
Thus there are 10 beads in each group.
22 + 10 = 32
Therefore Descartes has 32 beads now.
Question 26.
It costs $166 to rent a bounce house for 7 hours. Descartes has a coupon that saves him$5 each hour he rents the bounce house. How much will it cost him to rent the bounce house for 7 hours with the coupon?
It costs $166 to rent a bounce house for 7 hours. Descartes has a coupon that saves him$5 each hour he rents the bounce house.
7 × $5 =$35
$166 –$35 = $131 for 7 hours with the coupon. Thus it will cost$131 to rent the bounce house for 7 hours with the coupon. | 14,809 | 47,980 | {"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} | 5 | 5 | CC-MAIN-2024-22 | latest | en | 0.904812 |
https://www.physicsforums.com/threads/does-the-sequence-of-a-deck-of-cards-have-significance.841427/ | 1,521,547,784,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647406.46/warc/CC-MAIN-20180320111412-20180320131412-00435.warc.gz | 870,835,129 | 16,684 | # Does the sequence of a deck of cards have significance?
1. Nov 4, 2015
### Grinkle
If I open a new deck of cards the sequence is known.
If I shuffle them, the sequence is unknown. If I then memorize the sequence, it is again known.
State 1: New deck
State 2: Shuffled deck
State 3: Shuffled deck after I have memorized the sequence
State 4: Re-shuffled deck
I have computed the entropy difference of the deck sequence between state 1 and state 2. Many of us will have done the same as an introductory exercise in the second law of thermodynamics.
If I skip step 3 then it seems to me I must say that state 2 has equivalent entropy to state 4, but if I don't skip step 3 then it seems that delta entropy (1->2) must equal the delta entropy (2->4). Entropy really is a thermodynamic property - it is not just a mind game. Is there any way to conceptually relate the entropy of the sequence of a deck of cards to the entropy in a container of gas so they both have the same 'amount' of physical significance? One seems to be a state of mind and the other seems to be a thermodynamic property, but both share a common mathematical model.
2. Nov 4, 2015
### Staff: Mentor
If you skip step 3 then you are talking about a completely different system. With it you are talking about the entropy of the deck plus the memory storage. Without it you are talking about the entropy of the deck alone.
3. Nov 4, 2015
### Grinkle
Hmmm.
Is it well-formed / sound to discuss the sequence itself as having increased in entropy by having shuffled the deck - the exercises I have done have never asked about the energy dissipated by the observer. The system has always been described as a deck of cards.
One could also just think of a penny being heads or tails, the question is the same, I think. If I flip a coin with my eyes closed, so I don't know the start or end state, and my clone in another universe does the same with eyes open, is the penny's-state entropy change different in the two universes? Can one analyze purely the state of the penny (heads vs tail up) as a bounded system?
It may be that my problem comes from a poorly constructed exercise in my junior year thermo textbook.
4. Nov 5, 2015
### Khashishi
For an idealized coin, if you don't look at the coin, the coin is in a mixed state with entropy of S = k ln 2. If you look at the coin, then the entropy of the coin becomes 0. If you randomize the coin, the entropy goes back to k ln 2.
The entropy of the coin can go down, but that's because the coin isn't a closed system.
Another way of thinking of it is that you aren't talking about the same coin after you look at it. You have shifted into a subset of worlds, so you are now referring to a subset-coin, which is just "part" of the original coin. In the many worlds picture, initially there are two worlds, one with heads and one with tails. If you look at the coin and then flip the coin, there are 4 worlds:
a. you remember seeing H, coin is in state 1
b. you remember seeing H, coin is in state 2
c. you remember seeing T, coin is in state 1
d. you remember seeing T, coin is in state 2
The total entropy is k ln 4 since there are 4 possibilities. But from your point of view, you never observe yourself splitting, so it looks like there are only two possible worlds. If you don't look at the coin before flipping the coin, there are just 2 worlds.
a. coin is in state 1
b. coin is in state 2
The world doesn't split when you flip the coin because coin's previous state interferes with itself.
5. Nov 5, 2015
### Grinkle
This has been helpful food for thought.
I think all this implies that the answer to my initial question is the sequence of a deck of cards in isolation does not have any thermodynamic significance.
The entropy of a coin all by itself as a closed system cannot change.
So I think it must be the case that the same remains true of the coin per se if one adds an observer. The entropy of the observer will change as the observer does whatever work is required (brain burning calories etc) to attach symbolic meaning to the different possible states of the previously closed system being observed. Using entropy to describe the count of these interpreted states is a bookeeping mechanism, I think. Its misleading to think that the entropy of a deck of cards increases if the deck is shuffled. What increases is the entropy of the person thinking about what it means to shuffle cards and thinking about the different order of the cards, because the act of keeping track of the symbolism in the ordering of the deck takes computing power. If a blindfolded man flips a coin the entropy of the universe increases less than if a watching man does the same thing and comprehends heads vs tails after each flip, and the entropy difference is all in the brain activity of the watching man doing the comprehension. The coin is the same in both situations and its entropy is not changing in either situation. | 1,147 | 4,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2018-13 | latest | en | 0.961865 |
https://www.experts-exchange.com/questions/27667964/SQL-How-distribute-x-amount-of-money-between-multiple-rows.html | 1,481,165,230,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542323.80/warc/CC-MAIN-20161202170902-00001-ip-10-31-129-80.ec2.internal.warc.gz | 925,908,404 | 30,894 | Solved
# SQL - How distribute x amount [of money] between multiple rows
Posted on 2012-04-09
1,126 Views
Hello everyone,
This is a Microsoft SQL Server 2005 question:
Suppose I have the following 2 tables:
``````create table tblPurchaseOrders (
[PoNumber] varchar(20),
[PODate] datetime,
[POAmount] money,
[AllocatedAmount] money
)
create table tblAllocatedAmounts (
[PoNumber] varchar(20),
[AllocatedAmount] money
)
insert into tblPurchaseOrders values ('ABC', '2012-04-01 11:30:00', 1000, 0)
insert into tblPurchaseOrders values ('ABC', '2012-04-03 13:30:00', 500, 0)
insert into tblPurchaseOrders values ('DEF', '2012-04-02 12:00:00', 2000, 0)
insert into tblPurchaseOrders values ('GHI', '2012-04-03 12:15:00', 300, 0)
insert into tblAllocatedAmounts values ('ABC', 500)
insert into tblAllocatedAmounts values ('DEF', 1000)
insert into tblAllocatedAmounts values ('GHI', 300)
insert into tblAllocatedAmounts values ('ABC', 500)
insert into tblAllocatedAmounts values ('ABC', 100)
``````
How do I distribute the sum of tblAllocatedAmounts.AllocatedAmount to tblPurchaseOrders.AllocatedAmount based on PO Number and PO Date?
This is the desired output:
``````select * from tblPurchaseOrders order by ponumber, PODate asc
[PoNumber] [PODate] [POAmount] [AllocatedAmount]
ABC 2012-04-01 11:30:00.000 1000.00 1000.00
ABC 2012-04-03 13:30:00.000 500.00 100.00
DEF 2012-04-02 12:00:00.000 2000.00 1000.00
GHI 2012-04-03 12:15:00.000 300.00 300.00
``````
Thank you!
0
Question by:Rick
• 4
• 2
• 2
LVL 23
Expert Comment
ID: 37825333
``````select PoNumber, PODate, POAmount,
case
when AllocatedAmount >= AccAmount then AccAmount
when AllocatedAmount < AccAmount then POAmount - AccAmount + AllocatedAmount
else 0
end as AllocatedAmount
from (
select o.PoNumber, o.PODate, o.POAmount, a.AllocatedAmount,
(select sum(i.poamount) from tblPurchaseOrders i where i.ponumber = o.ponumber and i.id <= o.id) AccAmount
from tblPurchaseOrders o
join (select PONumber, Sum(AllocatedAmount) AllocatedAmount from tblAllocatedAmounts group by PONumber) a on a.PONumber = o.PONumber) t
order by ponumber, PODate asc
``````
0
LVL 42
Assisted Solution
dqmq earned 250 total points
ID: 37825367
Here's my go at it:
``````;with CTE as
(select poNumber, sum(AllocatedAmount) Allocated
from tblAllocatedAmounts group by poNumber)
update tblPurchaseOrders
set AllocatedAmount =
case
when
cte.allocated -
isnull((select sum(POAmount) from tblPurchaseOrders p
where p.PoNumber = PO.PoNumber
and p.PODate < PO.PODate),0)
>= PO.POAmount
then PO.POAmount
else
cte.allocated -
isnull((select sum(POAmount) from tblPurchaseOrders p
where p.PoNumber = PO.PoNumber
and p.PODate < PO.PODate),0)
end
from tblPurchaseOrders PO inner join CTE on CTE.PoNumber = PO.PONumber
``````
0
LVL 13
Author Comment
ID: 37825385
wdosanjos,
I removed the "and i.id <= o.id" on line 9 because there there is no column named "id".
This is giving me:
[PoNumber] [PODate] [POAmount] [AllocatedAmount]
ABC 2012-04-01 11:30:00.000 1000.00 600.00 <==== Should be 1,000 here ...
ABC 2012-04-03 13:30:00.000 500.00 100.00
DEF 2012-04-02 12:00:00.000 2000.00 1000.00
GHI 2012-04-03 12:15:00.000 300.00 300.00
0
LVL 42
Expert Comment
ID: 37825395
p.s. Sometimes we get questions that are a bit contrived for the sake of simplification or privacy. I hope that is the case here, because your tables lack the keys necessary for entity integrity.
Isn't it out of the ordinary to have the same PO on two different dates? I should also point out that the logic fails if you get two rows for the same PO on the same date. Note, I believe wdosanjos's introduction of an ID column is an attempt to circumvent some of these issues.
0
LVL 13
Author Comment
ID: 37825396
dqmq,
It seems like your solution is going to work for me.
Let me check...
0
LVL 13
Author Comment
ID: 37825414
@37825395:
>> your tables lack the keys necessary for entity integrity
Sorry, I was in a bit of a hurry... : )
0
LVL 23
Accepted Solution
wdosanjos earned 250 total points
ID: 37827671
Here is the corrected query. Sorry, I was experimenting a couple things and I forgot to clean up the query.
``````select PoNumber, PODate, POAmount,
case
when AllocatedAmount >= AccAmount then AccAmount
when AllocatedAmount < AccAmount then POAmount - AccAmount + AllocatedAmount
else 0
end as AllocatedAmount
from (
select o.PoNumber, o.PODate, o.POAmount, a.AllocatedAmount,
(select sum(i.poamount) from tblPurchaseOrders i where i.ponumber = o.ponumber and i.PODate <= o.PODate) AccAmount
from tblPurchaseOrders o
join (select PONumber, Sum(AllocatedAmount) AllocatedAmount from tblAllocatedAmounts group by PONumber) a on a.PONumber = o.PONumber) t
``````
0
LVL 13
Author Closing Comment
ID: 37828087
Thanks guys!
0
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http://www.yaklass.by/p/algebra/9-klass/sistemy-uravnenii-s-dvumia-peremennymi-4636/metody-resheniia-sistem-uravnenii-4683/re-3a6d2975-fe06-4378-b002-3f0339f8c31c | 1,579,817,736,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250614086.44/warc/CC-MAIN-20200123221108-20200124010108-00263.warc.gz | 291,103,857 | 7,802 | ### Условие задания:
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Реши систему уравнений методом алгебраического сложения:
$\left\{\begin{array}{l}{x}^{2}+{c}^{2}=25\\ {x}^{2}-{c}^{2}=7\end{array}\right\$
$\begin{array}{l}1.\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}{x}_{1}=i\\ {c}_{1}=i\end{array}\right\\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}2.\left\{\begin{array}{l}{x}_{2}=i\\ {c}_{2}=-i\end{array}\right\\\ \\ 3.\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}{x}_{3}=-i\\ {c}_{3}=i\end{array}\right\\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}4.\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}{x}_{4}=-i\\ {c}_{4}=-i\end{array}\right\\phantom{\rule{0.147em}{0ex}}\end{array}$ | 445 | 956 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-05 | latest | en | 0.282843 |
https://math.stackexchange.com/questions/1883655/quaternions-group-homomorphic-to-klein-group | 1,618,939,629,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039476006.77/warc/CC-MAIN-20210420152755-20210420182755-00103.warc.gz | 493,259,515 | 39,915 | # Quaternions Group Homomorphic to Klein Group
I'm currently studying some stuff about group theory and I came to problem of showing that $$\displaystyle\frac{Q_8}{\langle-1\rangle}\cong V_4,$$ so I checked on this link: Quaternions Group and Klein Group, which seems to clarify somehow what I wanted to know. But now I'm curious about how to prove the statement using the first isomorphism theorem. Here is what I have: Setting the map $\varphi:Q_8\to V_4$ such that $\varphi(a)=|a|$ we find that $\varphi$ is a morphism of groups with $Ker(\varphi)=\langle -1\rangle$, so applying the theorem the statement holds. My question is about the legality of $\varphi$, I think it's ok but I'd like another opinion, approach, thought about it.
• You could do it less directly as well. Note that $Q_8/\langle -1\rangle$ has four elements and every non-identity element has order $2$. – Arthur Aug 5 '16 at 18:18
• What do you mean by $|a|$ in this context? – carmichael561 Aug 5 '16 at 18:18
• @carmichael561 I mean the absolute value. This was my very first concern about it, because I was curious about how to define the morphism properly – Cristian Baeza Aug 5 '16 at 19:29
• Notice that $|a| = 1$ for all $a$, so your morphism is trivial (furthermore, it is not surjective either). – Alex M. Aug 5 '16 at 19:52
• Yeah, no, well, there it is, I didn't mean the norm of $a$, I meant the "positive" version of $a$. I wasn't sure how to exprese it, and now I'm wondering how. – Cristian Baeza Aug 5 '16 at 19:59
Well, if $Q_8 = \{ \pm 1, \pm i, \pm j, \pm k \}$ with $(-1)^2 = 1$ and $i^2 = j^2 = k^2 = -1$, and the Klein group is $V_4 = \{ 1, x, y, xy \}$, then define $\varphi (i) = x, \ \varphi (j) = y$ and obtain $\varphi(1) = 1$ (by definition) and $\varphi (k) = \varphi (ij) = \varphi (i) \varphi (j) = xy$. It also follows that $\varphi (-1) = \varphi (i^2) = \varphi (i) ^2 = x^2 = 1$, hence $\varphi (-i) = \varphi (-1) \varphi (i) = \varphi (i)$ etc.
It follows that $\ker \varphi = \{-1, 1\}$, so that $\frac {Q_8} {\langle -1 \rangle} \simeq V_4$.
• That's even better than what I was trying to reach. Thanks. – Cristian Baeza Aug 5 '16 at 19:26
You can prove that $\displaystyle\frac{Q_8}{V_4}\cong\langle-1\rangle$ . Now you know that $V_4$ and $Q_8$ is of order 4 and 8 respectively . So their quotient group is of order 2 and there is only one group (up to isomorphism ) of order i.e $\langle-1\rangle$ . so the quotient group must be isomorphic to $\langle-1\rangle$ .
Edit : This proof doesn't make much sense . It is not always true that $G/H \simeq K$ , then $G/K \simeq H$ . Most of the time $G/K$ is not group. Take for example $G=S_3$ ,$H=\langle\alpha\rangle$ and $K=\langle\beta\rangle$ where $\alpha^3=\beta^2=1$, $\alpha,\beta$ elements in $G$.
• This can't be possible. In general, if $G/H \simeq K$ and $G/K \simeq H$, then $G \simeq H \times K$. We already know that $Q_8 / \langle -1 \rangle \simeq V_4$. If it were true that $Q_8 / V_4 \simeq \langle -1 \rangle$, as you claim, then $Q_8 \simeq V_4 \times \langle -1 \rangle \simeq \Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$. One consequence of this is that all elements are of order $2$, and that $Q_8$ is commutative - clearly not true. Therefore, I believe your claim to be false. – Alex M. Aug 5 '16 at 19:46
• Yes you are right . I have corrected the mistakes now. – Suman Kundu Aug 5 '16 at 21:29
There are, up to isomorphisms, two groups of order $4$: the cyclic group and the Klein group $V_4$. For any $x\in Q$, $x^2\in\{1,-1\}$, so every element in $Q/\{1,-1\}$ has order dividing $2$. Hence $Q/\{1,-1\}$ cannot by cyclic.
• Absolutely true, but the OP wanted a proof based on the first isomorphism theorem for groups. – Alex M. Aug 6 '16 at 20:55 | 1,280 | 3,743 | {"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} | 3.46875 | 3 | CC-MAIN-2021-17 | latest | en | 0.908246 |
https://www.teacherspayteachers.com/Product/Solving-Equations-and-Inequalties-Bundle-4769674 | 1,627,316,659,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00293.warc.gz | 1,062,814,116 | 32,154 | DID YOU KNOW:
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# Solving Equations and Inequalties Bundle
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#### Bonus
Setting Up and Solving Equations and Inequalities Google Quiz
### Description
Solving equations is a major topic, so make sure that you have what you need and plenty of it! Here is a variety of equations, inequalities and literal equation problems. This bundle is growing, so buy now and get all new uploads free. You will find notes, worksheets, games and more. This bundle includes setting up equations and inequalities as well. There is also a Bonus Google Forms Quiz! I personally use these problems in both my Algebra 1 classes and my Algebra 2 classes. Recently added: Equations Boom Cards!
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### Standards
to see state-specific standards (only available in the US).
Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm’s law 𝘝 = 𝘭𝘙 to highlight resistance 𝘙.
Create equations and inequalities in one variable and use them to solve problems. | 481 | 2,117 | {"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-2021-31 | latest | en | 0.911424 |
https://www.mapleprimes.com/questions/220788-How-Can-I-Simplify-The-Way-I-Write-Simplification | 1,670,174,839,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710978.15/warc/CC-MAIN-20221204172438-20221204202438-00835.warc.gz | 911,424,880 | 24,341 | # Question:How can I simplify the way I write simplification problems?
## Question:How can I simplify the way I write simplification problems?
I'm creating a randomly generated question bank that generates the following STYLE or problem:
12x-4y2
3x6y-5
I'm currently trying to use the answer type "formula without simplification," as I'd like to avoid the students putting the questions in as the answer, and this has been driving me crazy for hours now.
I have tried the "maple" function to simplify. E.g.:
\$ANS = maple("( \$C1*( \$V1^\$A1 )*( \$V2^\$B1 ) ) / ( \$C2*( \$V1^\$A2 )*( \$V2^\$B2 ) )");
But it always throws an error.
I have simply done the math in the algorithm section, so you end up with an answer variable like this:
\$ANS = -4.0*(((z)^2.0)/(5.0*((p)^9.0)))
However, it will still count all answers submitted as incorrect.
Any help would be GREATLY appreciated =/.
Variable name clarification
I have \$C1 and \$C2, which are the constants of the numerator and denominator, respectively ("12" and "3" in the example).
I have \$V1 and \$V2 which are the first and second variable, respectively, (in the example, "x" and "y").
I have \$A1 and \$A2, which are the exponents for the variable \$V1 in the numerator and denominator, respectively.
I have \$B1 and \$B2, which are the exponents for the variable \$V2 in the numerator and denominator, respectively.
All these generate from some interesting conditions to create the problems I want (no variables named i, e, or o, for example,) but all properly initialize.
| 404 | 1,547 | {"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} | 3.328125 | 3 | CC-MAIN-2022-49 | latest | en | 0.918471 |
https://www.zoho.com/creator/newhelp/forms/fields/decimal/understand.html | 1,568,843,256,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573368.43/warc/CC-MAIN-20190918213931-20190918235931-00545.warc.gz | 1,108,035,290 | 4,709 | # Understand decimal field
The decimal field enables your users to enter decimal values, for example: -544.78, -23.3, 0, 889, 14324.225, etc.
## Character limits
• The decimal field can accept a number that contains a maximum of 19 digits. This is inclusive of the decimal point and the digits entered after the it.
• When your user enters a negative number, the negative sign is counted as a digit.
## When to use this field?
You may add a decimal field to your form when you want your users to enter a decimal value. Other types of fields that enable your users to enter numeric data are number, percent, and currency.
## Experience while entering data
When a user accesses the decimal field from a web browser: If your user enters anything other than an integer or decimal value in the decimal field, for example, an input that contains text or special characters, form submission will fail and the user will be prompted as follows:
1. The Invalid entries found message will appear in a pop-up window.
2. Clicking Okay will close this pop-up and display an error message below that decimal field.
When a user accesses the decimal field from a phone/tablet, their data entry is restricted as they get to access just the number keyboard.
## Features
With respect to the user-experience of the decimal field on a form, you can:
When yours users access a decimal field in reports, you can enable them to view the following values:
• Total - the sum of the values stored in that decimal field across all records displayed by that report.
• Average - the average of the values stored in that decimal field across all records displayed by that report.
• Min - the smallest value that's stored in that decimal field across all records displayed by that report.
• Max - the greatest value that's stored in that decimal field across all records displayed by that report.
## Example
• Course management: Imagine that you conduct some online courses and are building an app to manage the course registrations. You may use the decimal field to enable storing the marks scored by students in the tests you conduct. | 444 | 2,115 | {"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-2019-39 | longest | en | 0.815934 |
http://www.e-coachings.com/pastpapers?board=Federal%20&yr=2009&class=1&subj=67&subj_name=Physics&class_name=9th | 1,534,406,868,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210559.6/warc/CC-MAIN-20180816074040-20180816094040-00681.warc.gz | 492,676,919 | 10,454 | eCoaching | Online Home Tutors & Teachers
Question Options
The scientist who did exceptional work in Optics is : Al-Beruni, Al-Kundi, Newton, Ibn-Al-Haithem
The least count of verier calliper is : 0.0001 cm, 0.01 cm, 0.001 cm, 0.1 cm
The second equation of motion is : 2 as = Vf^2 - Vf^2, S = S avt X t, S=Vit +1/2 at2, Vf = V1 +at
If the mass of the body is doubled while keeping the force constant the acceleration will be : One half, Two time, One fourth, Fourth time
When a vector is multipal by a positive number then what will be the change : Direction, Magnitude, Unit , Direction
The value of torque depend upon : Force, Moment arm, Force and moment arm, Direction
A 2 kg object is moving in a circle with a speed 4 mS-1.If the radius of the corcle is 1 m.What will be the value of the centipal force acting on the object : 8 N, 16 N, 32 N, 64 N
The energy of body produced in it due to its motion is called : Electrical, Chemical, Kitnetic, Potential energy
The unit of mechnical advantage is : Newton, Joule, Meter, No Unit
The mathematical form of youngs modulus is : p=f/a, y= Fx L/A x dalta L, P= p F2/f2 = A/a
Viscosity of which will be maximium density : Water at 20° C, Honey at 20° C, Water at 70° C, Water at 90° C
At what temperature water has maximium density : 0° C, 4° C, 100° C, 110° C
The toy operating with steam was invented by : Hero, Volta, Galillio, Nicvalosium
The basic unit of temperature in SI system is : Calories, Kelvin, Joule, Mole
When a body does not change its position with respect to an observer then it is called : Position, Fixed point, Rest, Motion
Formula of weight is w = ------- : m/v, mg, mv, ma
The vector whose magnitude is equal to the vector v and direction in different is called : Positive, Negative, Resultant, Representative
According to 1 st condition equillibium ------------- = 0 : E fx =0, Efy = 0, E f =0, E I =0
A body of mass 2 kg is moving with speed of 4 mS-1 in a circle of radius 1 m the centipal force acting on the body is : 8 N, 16 N, 32 N, 64 N
If 25 N force acts on a stone to displace it 5 m.what will be the amount of work : 25 J, 50 J, 75 J, 125 J
Mechinical advantage of a signal moveable pulley is : Less then 1, Equal to 1, More then 1, More then 2
Force applying on a unit area of a body is called : Stress, Strain, Viscosity, Elasticity
For finding specific heat of liquid------ is used : Thermometer, Hyspmeter, Barometer, Calorimeter
PV = Constant is ----------- law : Charles, Boyles, Pascals, Archimedes | 753 | 2,488 | {"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} | 3.109375 | 3 | CC-MAIN-2018-34 | longest | en | 0.871317 |
https://www.thenational.academy/pupils/lessons/measuring-using-millilitres-6mvkec/video | 1,716,115,396,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057786.92/warc/CC-MAIN-20240519101339-20240519131339-00489.warc.gz | 912,038,415 | 25,159 | # Lesson video
In progress...
Hello, my name's Miss Jones and I'm going to be teaching you maths today.
I'm going to start with a quick joke, are you ready? Why was the maths book sad? Why do you think a maths book might be sad? Go on, tell me what you think.
Shall I tell you? The maths book was sad because it had too many problems. Let's start today's lesson.
Today we're going to be measuring using millilitres Here's the lesson agenda.
We'll start with the new learning where we'll look at containers and think about whether we would measure them in the units of millilitres or litres.
And then we'll look further into the lesson at how we measure using millilitres.
There'll be a talk task followed by an independent task, and then finishing off with a post quiz.
You will need a pencil and some paper for today's lesson.
Please pause the video and collect these items if you haven't done so already.
Here's a question for you.
What is capacity? Here there are some jugs to help give you a clue.
I'm going to pause while you think.
Go on, tell me what capacity is.
Great, capacity is the amount of liquid a container can hold.
All three of these containers are full to capacity.
They hold as much as they can hold.
What is volume? Here are some measuring jugs to give you a clue.
I'm going to pause while you have to think.
Go on, tell me, what is volume? Great job, volume is the amount or liquid or other substance that is in the container.
So all of these containers are not full to capacity.
We would measure the volume, the amount of liquid or substance that is inside them.
Now here's a task for you.
Are you ready? Can you tell me which container you think has the greatest capacity? Interesting, which container do you think has the least capacity? Great job.
Now, it's important to remember when we're looking at looking at capacity that when a container is the tallest, it doesn't necessarily mean it has the greatest capacity.
To measure the capacity of these containers accurately, we would need to measure the amounts using millilitres or litres by measuring what's inside the containers, or most of these containers, that's been done already.
And they will say on them, how many millilitres or litres they hold.
Which unit would you use to measure the capacity of these containers? Millilitres or litres? I'm going to show you for these four containers below, as I do so I'm going to say this sentence out loud.
I would measure the in millilitres or litres because, and give my reason why and to help me with my reason, I filled up my measuring jug to one litre.
I would measure the spoon in millilitres because it is less than one litre.
In fact, I think it's less than half a litre.
I would measure the bucket in litres because I think it is more than one litre.
To fill that bucket, I'm definitely going to be pouring more than one of these jugs.
And then for the swimming pool or the paddling pool, I would measure the pool or paddling pool in litres because to fill that pool it's definitely going to be more than one litre.
And then finally the mug I would measure in millilitres because I estimate that the mug is less than one litre.
Now it's time for your talk task! for your talk task today, I would like you to find which unit would you would use to measure the capacity of these containers.
Would you use millilitres or litres? As you do this, say the sentence out loud.
I would measure the in millilitres or litres because and your reason why.
First of all, the milk carton.
Do you think we'd measure the milk carton in millilitres or litres? Whisper it to the screen now.
Great job.
Now, it depends on the size of the milk carton.
If it's an individual milk carton we'd measure it in millilitres, but if it's a big family size milk carton we'd measure it in litres.
What about the watering can? Millilitres or litres, whisper it to the screen now.
Great, I would measure the watering can in litres because it holds more than one litre.
The egg cup.
Do you think we would measure in millilitres or litres? Whisper it to the screen now.
Great job.
I would measure the egg cup in millilitres because it holds less than one litre.
Finally, the water container.
Would you measure it in millilitres or litres? Whisper it to the screen now.
Great, I agree.
I would measure that big watering container in litres, because it definitely would hold more than one litre.
Now that we've looked at containers that we would use millilitres to measure the contents of the substance, the volume we're going to estimate where we would measure amounts on scales.
Here we are estimating where 550 millilitres is.
Let's look at the scales and the intervals, they are going up.
It is going up in, count with me, 100, 200, 300, 400, 500, 600, 700, 800, 900, 1000.
Did you notice what the scale's going up in? Go on, tell me.
Great job.
The scale is going up in 100's.
We need to find 550.
So here is 500 and 50 isn't marked on the scale.
So we've got to estimate where it is.
And I know that 550 is halfway between 500 and 600.
So the mark would be here for 550.
This is where I estimate the amount.
Can you point to the screen where you would estimate 350 millilitres? I'm going to give you time to think.
Have you got it? Go on, point to the screen now.
Well done, 350 would be here between 300 and 400.
For your independent task today, write the volume of each container, the amount that there is inside in millilitres, then you've got some problems to solve.
Read the questions carefully, look at the items and then write your answers on for each question.
Resume once you're finished.
In this first jug, there is 650 millilitres.
It is halfway between 600 and 700.
In this container, there is 900 millilitres.
It's exact, the line is exactly on 900.
In this container, it is on a mark that's not labelled, but we know the scale's going up in 100's and each mark is halfway between.
Half of 100 is 50.
There is 50 millilitres in this container.
And in the final container, it's halfway between 900 and 1000.
So there is 950 millilitres.
The bucket of water holds more than the water bottle.
The bucket of water holds 1000 millilitres and the water bottle holds 500 millilitres.
I know that two lots of the water bottle would fill the bucket.
This is because half of 1000 millilitres is 500 ml, therefore the bucket of water holds 500 millilitres more than the water bottle.
The capacity of the orange juice carton is, and it tells us here, 150 millilitres.
The orange juice carton when full, has a capacity of 150 millilitres.
The water bottle holds more than the orange juice.
The water bottle holds 500 millilitres and the orange juice holds 150.
The difference between these two amounts is 350 millilitres.
The water bottle holds 350 millilitres more than the orange juice.
Well done on the great lesson.
I hope you've enjoyed yourself and I hope to see you again soon, bye. | 1,631 | 6,985 | {"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} | 4.40625 | 4 | CC-MAIN-2024-22 | latest | en | 0.977366 |
https://csinva.io/notes/ml/evaluation.html | 1,638,658,600,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363125.46/warc/CC-MAIN-20211204215252-20211205005252-00230.warc.gz | 258,290,992 | 6,161 | evaluation
# losses
• define a loss function $\mathcal{L}$
• 0-1 loss: $\vert C-f(X)\vert$ - hard to minimize (combinatorial)
• $L_2$ loss: $[C-f(X)[^2$
• risk = $E_{(x,y)\sim D}[\mathcal L(f(X), y) ]$
• optimal classifiers
• Bayes classifier minimizes 0-1 loss: $\hat{f}(X)=C_i$ if $P(C_i\vert X)=max_f P(f\vert X)$
• KNN minimizes $L_2$ loss: $\hat{f}(X)=E(Y\vert X)$
• classification cost functions
1. misclassification error - not differentiable
2. Gini index: $\sum_{i != j} p_i q_j$
3. cross-entropy: $-\sum_x p(x): \log : \hat p(x)$, where $p(x)$ are usually labels and $\hat p(x)$ are softmax outputs
1. only penalizes target class (others penalized implicitly because of softmax)
2. for binary, $- (p \log \hat p + (1-p) \log (1-\hat p)$
# measures
goodness of fit - how well does the learned distribution represent the real distribution?
• accuracy-based
• accuracy = (TP + TN) / (P + N)
• correct classifications / total number of test cases
• balanced accuracy = 1/2 (TP / P + TN / N)
• denominator is total pos/neg
• recall = sensitivity = true-positive rate = TP / P = TP / (TP + FN)
• what fraction of the real positives do we return?
• specificity = true negative rate = TN / N = TN / (TN + FP)
• what fraction of the real negatives do we return?
• false positive rate = FP / N $= 1 - \text{specificity}$
• what fraction of the predicted negatives are wrong?
• fraction is total predictions
• precision = positive predictive value = TP / (TP + FP)
• what fraction of the prediction positives are true positives?
• negative predictive value = TN / (FN + TN)
• what fraction of predicted negatives are true negatives?
• F-score is harmonic mean of precision and recall: 2 * (prec * rec) / (prec + rec)
• NRI (controversial): compares 2 model’s binary predictions
• curves - easiest is often to just plot TP vs TN or FP vs FN
• roc curve: true-positive rate (recall) vs. false-positive rate
• perfect is recall = 1, false positive rate = 0
• precision-recall curve
• summarizing curves
• AUC: area under (either one) of these curves - usually roc
• concordance = inter-rate reliability
• exact concordance - percentage where cohort is in total agreement (i.e. accuracy)
• Cohen’s kappa coefficient - 0 is uncorrelated, 1 is perfect, negative is inverse correlation
• $\kappa \equiv \frac{p_{o}-p_{e}}{1-p_{e}}=1-\frac{1-p_{o}}{1-p_{e}}$
• $p_o$ is relative observed agreement and $p_e$ is chance expected agreement
• weighted kappa coefficient - used when ordering for predicted labels (being off by more is given bigger weight)
# comparing two things
• odds: $p : \text{not }p$
• odds ratio is a ratio of odds
# cv
• cross validation - don’t have enough data for a test set
• properties
• not good when n < complexity of predictor
• because summands are correlated
• assume data units are exchangeable
• can sometimes use this to pick k for k-means
• data is reused
• types
1. k-fold - split data into N pieces
• N-1 pieces for fit model, 1 for test
• cycle through all N cases
• average the values we get for testing
2. leave one out (LOOCV)
• train on all the data and only test on one
• then cycle through everything
3. random split - shuffle and repeat
4. one-way CV = prequential analysis - keep testing on next data point, updating model
5. ESCV - penalize variance between folds
• regularization path of a regression - plot each coeff v. $\lambda$
• tells you which features get pushed to 0 and when
• for OLS (and maybe other linear models), can compute leave-one-out CV without training separate models
# stability
1. computational stability
• randomness in the algorithm
• perturbations to models
2. generalization stability
• perturbations to data
• sampling methods
1. bootstrap - take a sample
• repeatedly sample from observed sample w/ replacement
• bootstrap samples has same size as observed sample
2. subsampling
• sample without replacement
3. jackknife resampling
• subsample containing all but one of the points
# other considerations
• computational cost
• interpretability
• model-selection criteria
• adjusted $R^2_p$ - penalty
• Mallow’s $C_p$
• $AIC_p$
• $BIC_p$
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# A Problem-Solution Approach - page 29 / 48
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CHAPTER 28 PERFORMANCE TUNING
# Table 28-4. DBCC SHOW_STATISTICS Arguments
Argument
Description
'table name' | 'view name'
_
_
The table or indexed view to evaluate.
target NO_INFOMSGS
The name of the index or named statistics to evaluate.
When designated, NO_INFOMSGS suppresses informational messages.
STAT_HEADER |DENSITY VECTOR | HISTOGRAM [ , n ]
_
Specifying STAT_HEADER, DENSITY_VECTOR, or HISTOGRAM designates which result sets will be returned by the command (you can display one or more). Not designating any of these means that all three result sets will be returned.
Updated
Rows
Rows Sampled
Steps
Density
Average
String
key length
Index
Oct 15 2005
19185
19185
2
0
2
YES
This example demonstrates how to view the statistics information on the Sales.Customer Stats_Customer_CustomerType statistics:
# This returns the following result sets:
, S t a t s _ C u s t o m e r _ C u s t o m e r T y p e )
# 2:32PM
Stats_Customer CustomerType
_
All density
Average Length
Columns
0.5
2
CustomerType
I
0
18484
0
1
S
0
701
0
1
EQ_ROWS
DISTINCT RANGE ROWS
_
_
AVG RANGE ROWS
_
_
RANGE HI KEY
RANGE ROWS
__
_
# How It Works
In the results of this recipe’s example, the All density column indicates the selectivity of a column:
All density
Average Length
Columns
0.5
2
CustomerType
Selectivity refers to the percentage of rows that will be returned given a specific column’s value. A low All density value implies a high selectivity. Columns with high selectivity often make for use- ful indexes (useful to the query optimization process).
In the third result set returned by SHOW_STATISTICS, CustomerType had only two values, I and S (which you can see in the RANGE_HI_KEY in the third result set):
I
0
18484
S
0
701
_ EQ ROWS
_
# ANGE HI KEY
__
_ DISTINCT RANGE ROWS 0 0 _
_ AVG RANGE ROWS 1 1 _
With such a high density of similar values, and low selectivity (one value is likely to return many rows), you can make an educated assumption that an index on this particular column is unlikely to be very useful to SQL Server in generating a query execution plan.
663
Document views 77 Page views 77 Page last viewed Wed Oct 26 04:48:43 UTC 2016 Pages 48 Paragraphs 987 Words 15822 | 620 | 2,369 | {"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} | 2.828125 | 3 | CC-MAIN-2016-44 | longest | en | 0.667886 |
https://slideplayer.com/slide/1386093/ | 1,532,271,965,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593302.74/warc/CC-MAIN-20180722135607-20180722155607-00029.warc.gz | 745,718,816 | 23,209 | # Interference of Waves Section 9.1.
## Presentation on theme: "Interference of Waves Section 9.1."— Presentation transcript:
Interference of Waves Section 9.1
Key Terms Interference Principle of Superposition
Constructive Interference Destructive Interference
Wave Interference Remember that waves are the result of particle vibrations. Particles in a medium are connected by forces. When a single wave passes a particle, it moves in an oval path. Allows the wave to travel in a certain direction. When waves meet, a new wave is generated in a process called interference. The speeds of the combined wave cancel out and the particles move up and down.
Wave Interference Cont’d
The waves pass through each other with no loss of energy. The direction and energy of each wave are preserved. Once they have passed, their characteristics are unchanged.
Principle of Superposition
When two waves meet, the forces of their particles are added together. If the two waves are in phase, the resulting amplitude will be greater than either of the two individual amplitudes. If the two waves are out of phase, the resulting amplitude will be less than either of the two individual amplitudes. Principle of Superposition
Constructive Interference
The process of forming a wave with a larger amplitude when two or more waves combine.
Constructive Interference
Destructive Interference
The process of forming a wave with a smaller amplitude when two or more waves combine
Destructive Interference
Technology Using Interference of Waves
Noise-cancelling headphones generate a wave that is completely out of phase with the sound waves in the environment. When the wave is played in the headphones, the destructive interference of the waves, cancels the ambient noise.
Applying the Principle of Superposition
Practice Problems
Summary The process of generating a new wave when two or more waves meet is called interference. Vibrating particles in a medium react to the sum of all forces on them. Their motion is caused by the sum total of forces on them. The principle of superposition states that, when two waves meet, the resulting amplitude is the sum of the individual amplitudes. Constructive interference occurs when two waves combine and the amplitude of the resulting wave is greater than the amplitudes of all the individual waves. Destructive interference occurs when two waves combine and the amplitude of the resulting wave is less than at least one of the original amplitudes. Humans can design technologies to take advantage of wave properties. An example of such a technology is noise-cancelling headphones.
Homework Principle of Superposition Worksheet Page 419
Questions 1 – 3 Chapter 8 Review Questions Page 408 1-18 Page 409 19-21, 23-27, 29-43 | 570 | 2,761 | {"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} | 3.625 | 4 | CC-MAIN-2018-30 | latest | en | 0.898876 |
https://learn.microsoft.com/pt-br/dotnet/api/system.tuple?view=net-8.0 | 1,726,107,665,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00682.warc.gz | 322,069,550 | 13,314 | # Tuple Class
## Definition
Provides static methods for creating tuple objects.
public ref class Tuple abstract sealed
public static class Tuple
type Tuple = class
Public Class Tuple
Inheritance
Tuple
## Examples
The following example creates an 8-tuple (octuple) that contains prime numbers that are less than 20.
var primes = Tuple.Create(2, 3, 5, 7, 11, 13, 17, 19);
Console.WriteLine("Prime numbers less than 20: " +
"{0}, {1}, {2}, {3}, {4}, {5}, {6}, and {7}",
primes.Item1, primes.Item2, primes.Item3,
primes.Item4, primes.Item5, primes.Item6,
primes.Item7, primes.Rest.Item1);
// The example displays the following output:
// Prime numbers less than 20: 2, 3, 5, 7, 11, 13, 17, and 19
open System
let primes = Tuple.Create(2, 3, 5, 7, 11, 13, 17, 19)
printfn $"Prime numbers less than 20: {primes.Item1}, {primes.Item2}, {primes.Item3}, {primes.Item4}, {primes.Item5}, {primes.Item6}, {primes.Item7}, and {primes.Rest.Item1}" // Prime numbers less than 20: 2, 3, 5, 7, 11, 13, 17, and 19 Dim primes = Tuple.Create(2, 3, 5, 7, 11, 13, 17, 19) Console.WriteLine("Prime numbers less than 20: " + "{0}, {1}, {2}, {3}, {4}, {5}, {6}, and {7}", primes.Item1, primes.Item2, primes.Item3, primes.Item4, primes.Item5, primes.Item6, primes.Item7, primes.Rest.Item1) ' The example displays the following output: ' Prime numbers less than 20: 2, 3, 5, 7, 11, 13, 17, and 19 ## Remarks A tuple is a data structure that has a specific number and sequence of elements. An example of a tuple is a data structure with three elements (known as a 3-tuple or triple) that is used to store an identifier such as a person's name in the first element, a year in the second element, and the person's income for that year in the third element. The .NET Framework directly supports tuples with one to seven elements. In addition, you can create tuples of eight or more elements by nesting tuple objects in the Rest property of a Tuple<T1,T2,T3,T4,T5,T6,T7,TRest> object. Tuples are commonly used in four ways: • To represent a single set of data. For example, a tuple can represent a database record, and its components can represent individual fields of the record. • To provide easy access to, and manipulation of, a data set. • To return multiple values from a method without using out parameters (in C#) or ByRef parameters (in Visual Basic). • To pass multiple values to a method through a single parameter. For example, the Thread.Start(Object) method has a single parameter that lets you supply one value to the method that the thread executes at startup time. If you supply a Tuple<T1,T2,T3> object as the method argument, you can supply the thread's startup routine with three items of data. The Tuple class does not itself represent a tuple. Instead, it is a class that provides static methods for creating instances of the tuple types that are supported by the .NET Framework. It provides helper methods that you can call to instantiate tuple objects without having to explicitly specify the type of each tuple component. Although you can create an instance of a tuple class by calling its class constructor, the code to do so can be cumbersome. The following example uses a class constructor to create a 7-tuple or septuple that contains population data for New York City for each census from 1950 through 2000. // Create a 7-tuple. var population = new Tuple<string, int, int, int, int, int, int>( "New York", 7891957, 7781984, 7894862, 7071639, 7322564, 8008278); // Display the first and last elements. Console.WriteLine("Population of {0} in 2000: {1:N0}", population.Item1, population.Item7); // The example displays the following output: // Population of New York in 2000: 8,008,278 // Create a 7-tuple. let population = Tuple<string, int, int, int, int, int, int>( "New York", 7891957, 7781984, 7894862, 7071639, 7322564, 8008278) // Display the first and last elements. printfn$"Population of {population.Item1} in 2000: {population.Item7:N0}"
// The example displays the following output:
// Population of New York in 2000: 8,008,278
' Create a 7-tuple.
Dim population As New Tuple(Of String, Integer, Integer, Integer, Integer, Integer, Integer) _
("New York", 7891957, 7781984, 7894862, 7071639, 7322564, 8008278)
' Display the first and last elements.
Console.WriteLine("Population of {0} in 2000: {1:N0}",
population.Item1, population.Item7)
' The example displays the following output:
' Population of New York in 2000: 8,008,278
Creating the same tuple object by using a helper method is more straightforward, as the following example shows.
// Create a 7-tuple.
var population = Tuple.Create("New York", 7891957, 7781984, 7894862, 7071639, 7322564, 8008278);
// Display the first and last elements.
Console.WriteLine("Population of {0} in 2000: {1:N0}",
population.Item1, population.Item7);
// The example displays the following output:
// Population of New York in 2000: 8,008,278
// Create a 7-tuple.
let population = Tuple.Create("New York", 7891957, 7781984, 7894862, 7071639, 7322564, 8008278)
// Display the first and last elements.
printfn \$"Population of {population.Item1} in 2000: {population.Item7:N0}"
// The example displays the following output:
// Population of New York in 2000: 8,008,278
' Create a 7-tuple.
Dim population = Tuple.Create("New York", 7891957, 7781984, 7894862, 7071639, 7322564, 8008278)
' Display the first and last elements.
Console.WriteLine("Population of {0} in 2000: {1:N0}",
population.Item1, population.Item7)
' The example displays the following output:
' Population of New York in 2000: 8,008,278
The Create helper methods directly support the creation of tuple objects that have from one to eight components (that is, singletons through octuples). Although there is no practical limit to the number of components a tuple may have, helper methods are not available to create a tuple with nine or more components. To create such a tuple, you must call the Tuple<T1,T2,T3,T4,T5,T6,T7,TRest>.Tuple<T1,T2,T3,T4,T5,T6,T7,TRest> constructor.
Note
For additional information and examples that use tuples, see the documentation for the individual tuple types in the .NET Framework. These are listed in the See Also section at the end of this topic.
## Methods
Creates a new 8-tuple, or octuple. Creates a new 7-tuple, or septuple. Creates a new 6-tuple, or sextuple. Creates a new 5-tuple, or quintuple. Creates a new 4-tuple, or quadruple. Creates a new 3-tuple, or triple. Creates a new 2-tuple, or pair. Creates a new 1-tuple, or singleton. | 1,833 | 6,548 | {"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": 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} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.77919 |
https://rebab.net/change-in-y-over-change-in-x/ | 1,656,134,157,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034170.1/warc/CC-MAIN-20220625034751-20220625064751-00540.warc.gz | 530,348,617 | 6,283 | I understand that this is a reasonable an interpretation and the shows how "fast" \$y\$ values readjust corresponding to \$x\$ values because it"s a ratio, but what I"m asking is, couldn"t the slope have actually been identified as the angle between the line and also the confident \$x\$-axis because that example? and also it would have actually the very same meaning; if the angle was huge (but much less than \$90\$) then that would average the line is steep and \$y\$ values readjust fast corresponding to \$x\$ values...etc. Why is the first meaning better? Is the second one even correct?
an interpretation slope
re-superstructure
point out
monitor
edited Dec 28 "16 in ~ 8:50
Juniven
request Dec 28 "16 at 7:22
Khalid T. SalemKhalid T. Salem
\$endgroup\$
2
1
\$egingroup\$ since there is a deep connection in between the slope and also the derivative \$endgroup\$
–user370967
Dec 28 "16 at 7:25
include a comment |
1
\$egingroup\$
The edge is one thing you might treatment to think about, sure. The rise-over-run is one more thing girlfriend might care to think about. The fact that the word "slope" saw the latter is simply a caprice of history. ~ above the other hand, the fact that the last turned out to it is in an interesting and fruitful thing of investigation is no mere quirk (but also, not that surprising). Periodically we room interested in straight proportionality relations choose \$Y = mX\$, and in those cases, the consistent of proportionality \$m\$ is a organic thing to consider. Ratios are of ubiquitous arithmetic importance, and also that"s all that "slope" comes down to; investigate ratios.
You are watching: Change in y over change in x
But there"s naught wrong through thinking around angles, either. Just because we invest a most time talking about slopes doesn"t average we"re versus thinking about angles. Think about both! Think about everything! rebab.net isn"t one either-or world; you can think around anything, everything, and see what comes of it.
re-publishing
mention
monitor
edited Dec 29 "16 at 2:47
reply Dec 28 "16 in ~ 8:03
Sridhar RameshSridhar Ramesh
\$endgroup\$
include a comment |
1
\$egingroup\$
As rebab.net_QED points out in his comment, the derivative of a function at a allude is same to the steep of the tangent line at the point, so over there is a connection between slopes defined as “rise end run” and rates of change. Over there are other contexts in which the edge of the heat is an ext natural or convenient. Due to the fact that the steep of a heat is same to the tangent the the angle the it makes with the \$x\$-axis, the two definitions are equivalent.
re-superstructure
mention
monitor
answer Dec 28 "16 in ~ 7:40
amdamd
\$endgroup\$
1
1
\$egingroup\$
In the end, it every comes under to the current meaning of steep being more natural than utilizing degrees. Right here are a few thoughts on the subject:
Using angles is dependence on the aspect ratio in between the axes. You can have a line the looks choose it"s \$30^circ\$ in the picture, however it"s really \$89.97^circ\$. This can occur with slope together well, however degrees carry a heavier implication of what something looks like. Usually, through geometric figures, if girlfriend "squash" them, you expropriate that the angles change. You can do that v functions. This is not a great idea, due to the fact that that way that if you attract a graph, it it s okay one angle, and if you"ve created that angle down next to the graph, and also someone provides a poor photocopy, then what"s on there is suddenly invalidated. Therefore the angle has to be inherent in the role alone, i m sorry breaks just how we tink the angles: We accept that if you attract a square, and squash the in one direction, then it"s not a square anymore, and the diagonals room not \$45^circ\$ to the sides.
Also, keep in mind the difference between a line with slope \$89.97^circ\$ and also one v slope \$89.98^circ\$. Since the natural means to "move" is to relocate with constant speed follow me the \$x\$-axis, and not follow me the graph, castle will different fast, while two lines of slope \$30^circ\$ and also \$30.01^circ\$ are more or much less indistinguishable.
Then consider what happens if you change your units. If the systems on the \$x\$-axis is seconds, and the systems on the \$y\$-axis is meters, climate the heat represents what your place is in ~ ny given second, and also the steep your rate in meter / second. To speak you have actually a heat of slope \$1\$ in the coordinate system. If you change the length units come feet, the slope i do not care \$1frac ms cdot 3.28 frac extfeetm = 3.28frac extfeets\$, if the degrees go indigenous \$45^circ\$ to... What, exactly? Insert trigonometry here. Regarded this, what execute the degrees even measure? What"s a organic unit for them? i daresay over there isn"t one.
What about if you add functions? The steep of the sum is the amount of the slopes. What is the edge you obtain if you include a line v angle \$37^circ\$ v a heat of edge \$62^circ\$? It"s not impossible, however it take away a couple of calculations to number out.
And then, of course, the derivatives, and also all the rules we have for those. Shot doing the chain preeminence with degrees. That"s trigonometry galore.
See more: How To Make Brown Dye Minecraft, New Combinations For The Brown Dye
So girlfriend could define slope native degrees, yet you"d only make life difficult for yourself in the long run. | 1,348 | 5,482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-27 | latest | en | 0.935124 |
https://oneclass.com/class-notes/ca/mcmaster/proctech/proctech-2ec3/1903408-proctech-2ec3-lecture-16.en.html | 1,540,177,887,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514497.14/warc/CC-MAIN-20181022025852-20181022051352-00358.warc.gz | 756,254,773 | 90,503 | Class Notes (977,739)
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Lecture 16
# PROCTECH 2EC3 Lecture 16: Week 9
7 Pages
44 Views
2018
Department
Process Automation Technology
Course Code
PROCTECH 2EC3
Professor
Kostas Apostolou
Lecture
16
This preview shows pages 1-2. Sign up to view the full 7 pages of the document.
Week 9: March. 6th, 2018
Problem 1: Steam at
!"#\$
and
%&'()
is expanded through a nozzle to
!##\$
and
*&'()
. Negligible heat is
transferred from the nozzle to its surroundings. The approach velocity of the steam is negligible. Calculate the
exit steam velocity.
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: Go to B.6 at
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At
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and
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it is superheated steam
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Linear Interpolation:
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Open System:
Energy Balance:
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7.25. A fuel oil is burned with air in a boiler furnace. The combustion produces
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W"#\$
. Water enters the boiler tubes as a liquid at
!#\$
and leaves the tubes as
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!#&'()
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Open System: Material coming in and Material going out
Energy Balance:
38
5
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6,+
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T7
5
38
5
,+
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81T80,+
5
j
5
18
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th Week 9: March. 6 , 2018 Problem 1: Steam at 250 and 7 is expanded through a nozzle to 200 and 4 . Negligible heat is transferred from the nozzle to its surroundings. The approach velocity of the steam is negligible. Calculate the exit steam velocity. Open System: = 0 = 0 Energy Balance: + + = = ? 0 + = 1 1 = 0 + = 0 2 2 + = 0 2 Get nd : : Go to B.6 at = 7 , = = At 250 and 7 it is superheated steam Go to B.7 at = 250 ; = 7, Linear Interpolation: = ( ) = 2961 + (7 5 . Go to B.6 at 4 , = = . At 400 and 4 it is superheated steam
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So we can recommend you notes for your school. | 1,563 | 4,067 | {"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-2018-43 | longest | en | 0.718797 |
https://www.hpmuseum.org/forum/archive/index.php?thread-6972.html | 1,571,158,121,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660067.26/warc/CC-MAIN-20191015155056-20191015182556-00302.warc.gz | 905,917,659 | 2,625 | # HP Forums
Full Version: Strange RREF (HP 50g)
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Hello HP 50g friends,
I watched a strange behavier of the RREF command.
If I input the matrix:
Code:
```[[ 7.6440528636 7.6440528636 -300. -53.1753303963 ] [ -10.192070485 -10.192070485 0. 229.289735683 ] [ 61.847 0 0. 8. ]]```
and use the RREF command, I get:
Code:
```[[ 1. 0. 0. .129351464097 ] [ 1. 1.0192070485E14 0. -2.30608094922E15 ] [ 1. -.15384615385 4.61538461549E15 -1.82756879024E15 ]]```
That is something else as aspected, but if I use a second time RREF I get the aspected form of the matrix:
Code:
```[[ 1. 0. 0. 0. ] [ 0. 1. 0. -22.6262264632 ] [ 0. 0. 1. -.395973237876 ]]```
Is there anybody who can explain me the strange behavier? It is not a question of the calculation mode (approxmative or exact)?
Thank you in advance,
sincerely peaceglue
Hello, I don't really know what is happening, but you can do this: with the matrix in level 1, enter the command XQ, this transforms every decimal in the matrix to fractions and sets the calculator in exact mode, then the command RREF and then the command ->NUM.
Hello Juan14,
it seems to a matter of exact or not, in AUR it is described to use RREF only in exact mode. It is a pity, because the exact mode is very slow.
Hello Peaceglue, I was looking at your matrix, and it seems that all the numbers in the matrix must be all reals or all integers, and your first 0 in the the third row is an integer, this somehow force the calculator to go in approximate mode, change it to 0. and run RREF in exact mode and it works.
Hello Juan14,
I replaced the approach using RREF with the Cramer's Rule and get for over 600 solvings of systems with three unknowns (inclusive checking wether DET is zero or not) a calculation time of ca. 6 minutes (with RREF I get 4 times more). That is sufficient fast enough for my purposes.
But anyway, thank you very much for your help.
Greetings
peacecalc
Hey folks,
the hp 50g is a real number crunching machine. Without the DET function but using CROSS and DOT the duration is only less then 4 minutes. wow!
Greetings peacecalc
Reference URL's
• HP Forums: https://www.hpmuseum.org/forum/index.php
• : | 690 | 2,278 | {"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-2019-43 | latest | en | 0.844765 |
https://web.sonoma.edu/users/m/marivani/es210/units/experiment05.shtml | 1,556,244,330,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578747424.89/warc/CC-MAIN-20190426013652-20190426035652-00199.warc.gz | 582,489,535 | 3,536 | ## Decoders and Demultiplexers
### Objective
To design and study decoders and demultiplexers
### Decoder:
1. A 2-4 decoder is a network with two inputs (X0 and X1) and 4 outputs (Y0, Y1, Y2 and Y3). When the inputs are both 0, Y0 = 1 and the other outputs are 0. When the inputs are 0 and 1 respectively, only Y1 is 1. For 10 and 11 inputs only Y2 and Y3 are 1 respectively. Write the truth table. Using NAND gates design a 2-4 decoder. Demonstrate the circuit to your instructor.
2. Study the specification of a 74155 chip, a dual 2-4 decoder. From the truth table in the specs, derive the logical expression of the outputs.
3. Draw the wiring diagram to make the 74155 into a 3-8 decoder. Wire up the circuit and verify the operation.
### Demultiplexer:
A demultiplexer is a combinational circuit with:
n control (or select) lines: C1,..., Cn
2n output lines: f0, f1,..., f(2n)-1
One input line: I
Figure 1 - Demultiplexer
A demultiplexer receives binary information on a single line (I) and transmits this information on one of 2n possible output lines. The selection of a particular output line is controlled by minterms of control lines.
1. Design and implement a 1-4 demultiplexer using NOT and AND gates. Demonstrate the circuit to your instructor.
2. Connect the input I to hi. Write the truth table of this circuit and identify its operations.
3. A 74155 can also be used as a demultiplexer, i.e., it can function like a rotary switch to demultiplex a single input to four different output lines. The select lines control the rotary switch position digitally. Wire up the 74155 as a 1-4 line demux. Connect the outputs to the LED's. Connect the select lines to the switches and change the settings from 00 to 11. Observe what happens.
Figure 2 – 1 to 4 Line Demux
### Design Problems:
1. Using a 3-8 decoder and three 2-input OR gates, design a combinational multiple output system to implement the following functions:
f1(A,B,C) = BC + A'B'C
f2(A,B,C) = BC + ABC'
2. In computer systems the memory address is decoded to select a particular word. Assuming that a computer has 64 words and 6 bits to address the memory, using 74155s as a 3-8 decoder, design a 6 to 64 decoder.
3. Using 74155 as a 3-8 decoder, design a 5 to 32 decoder. | 615 | 2,258 | {"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} | 3.234375 | 3 | CC-MAIN-2019-18 | latest | en | 0.805584 |
https://stromcv.com/factor-calculator-73 | 1,675,947,481,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499966.43/warc/CC-MAIN-20230209112510-20230209142510-00179.warc.gz | 541,604,275 | 6,092 | # Factor calculator\
The Complex Number Factoring Calculator factors a polynomial into imaginary and real parts. Step 2: Click the blue arrow to submit. Choose Factor over the Complex Number from the
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We offer 24/7 support from expert tutors. | 648 | 3,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-06 | latest | en | 0.939116 |
https://www.mail-archive.com/nim-general@thregr.org/msg20638.html | 1,582,757,836,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146562.94/warc/CC-MAIN-20200226211749-20200227001749-00434.warc.gz | 783,178,375 | 5,572 | Re: Suggestions for optimization?
```Incidentally, in the small alphabet case (and realistically anagrams are
usually related to word games over small alphabets) there is a neat trick for
rapid signature computation: map the 26 letters to the first 26 primes (2, 3,
5, 7, .., 97, 101). Then make the signature of a word the product of the
primes. If you never overflow the product this is guaranteed to be a valid
anagram signature by the uniqueness of prime factorization and commutativity of
multiplication. Note that this **removes sorting from the equation entirely**.```
```
In the worst case, that product can start to overflow a 64-bit integer past
about 9 letters (since 101^9 =~ 2^60 and 101^10 > 2^66). However, neither worst
nor average cases really matter since we have a static dictionary of every
possible case. So, it's actually easy to just test a given dictionary to see if
any overflows collide in our exact case. While it is easy to defeat by an
attacker, for "organic/natural" dictionaries this is likely to be very rare
because 2^64 is much more than the square of the number of words in most
dictionaries. So, we are in a regime very far from birthday paradox collisions.
Really, we aren't random, but also really the "load on the address space" is
much less than the square of the number of words - it is the product of the
number of non-product overflowing words and product overflowing words which is
much smaller (assuming most words do not overflow). So, this is really a
somewhat empirical question about words and dictionaries and letters.
With just a naive 'A' .. 'Z' \--> 2 .. 101 mapping and this dictionary:
[https://github.com/jesstess/Scrabble/blob/master/scrabble/sowpods.txt](https://github.com/jesstess/Scrabble/blob/master/scrabble/sowpods.txt)
I got 15_294 words that overflowed (and no collisions!) for 267_751 words. One
can do better, though. At least in English, letters have a very skewed usage
distribution. So, we can make overflows even less likely. You can just look
that frequency up on Wikipedia
([https://en.wikipedia.org/wiki/Letter_frequency](https://en.wikipedia.org/wiki/Letter_frequency))
and sort the first 26 primes by that frequency. Doing that with the above
SOWPODS dictionary, we reduce the number of overflows to 223.
You can do better still with a pre-pass to _measure_ this frequency in the
dictionary, sort the primes list by that measured frequency. I get 175
overflows that way. You can do better still by measuring only the frequency of
letter usage in words _longer than 9 letters_ (where it actually matters to
contain overflow). I get only 136 overflows (over 100x better than the initial
15,000) that way. You might be able to do slightly better by measuring the
frequency of letters in just the 136 overflow words and adjusting, but that is
more work and upsets the apple cart of the prior measurement perhaps requiring
iteration. I think that 136 yields some a priori probability of any collision
of under 2e-12 (under certain not quite right randomness assumptions).
So, while you might worry this "product signature" technique is not applicable
due to overflow, it probably is in a language where anagrams are interesting
Applying all these ideas (except the measurement which I did in a 15 line
Python script for its convenient arbitrary precision arithmetic) in Nim we get:
import strutils,tables,sets #toUpperAscii,Table,HashSet
const prime: array[26, uint] = [ # >.len==9 frequency
7'u, 59, 29, 31, 2, 67, 47, 53, 5, 101, 73, 23, 43,
13, 17, 41, 97, 11, 3, 19, 37, 71, 79, 89, 61, 83 ]
proc product(word: string): uint =
result = 1 #You might worry about overflow, BUT
for ch in word: #..long anagrams are so rare it's ok!
result *= prime[ord(ch) - ord('A')] #223/267751oflow
proc sortByLetter(word: string): string =
var cnt: array[26, int16]
for ch in word: cnt[ord(ch) - ord('A')].inc
for i, c in cnt:
for n in 0 ..< c: result.add chr(ord('A') + i)
proc pBuildQry(dict="words", query: seq[string]) =
var anas = initTable[uint, seq[string]]()
for line in lines(dict):
let word = line.toUpperAscii
for word in query:
let word = word.toUpperAscii
echo word, ":"
try:
for ana in anas[word.product]: echo " ", ana
proc sBuildQry(dict="words", query: seq[string]) =
var anas = initTable[string, seq[string]]()
for line in lines(dict):
let word = line.toUpperAscii
for word in query:
let word = word.toUpperAscii
echo word, ":"
try:
for ana in anas[word.sortByLetter]: echo " ", ana
proc collisions(dict="words") =
var t = initTable[uint, seq[string]]()
for line in lines(dict):
let word = line.toUpperAscii
for product, sigs in t:
if sigs.len > 1:
let set = sigs.toHashSet()
if set.len > 1: echo "collision: ", set
when isMainModule: #You need to nimble install cligen
import cligen #..for this CLI to work.
dispatchMulti([pBuildQry], [sBuildQry], [collisions])
Run
With that same SOWPODS dictionary, I get build times about 1.6x faster with the
product signature than the string-sorted-by-letters-with-counting-sort
signature (129 milliseconds vs 200 milliseconds and 230 ms for algorithm.sort
instead of counting sort). So, at a round figure about 2X as fast as
@lagerratrobe's algorithm. Query times are presumably about a similar speed-up
I don't count the measurement time to decide the order of `primes` in that 129
ms. Honestly, that is kind of an unneeded optimization. I bet you'd have a hard
time finding an organic dictionary defeating this. While no overflows guarantee
no collisions, you almost certainly can get no collisions with many overflows.
It's easy to check as shown above, and you absolutely should. I was mostly just
curious if I could get down to zero overflows.
Of course, these same "measure the data" ideas could also be used on the
Unicode case. If the "working set" of letters over the used dictionary is
smallish like the 26 here and words are short-ish like the at most 15 letters
here. I suspect in any language where anagrams are interesting to people would
have those traits and it would be the fastest approach (at least at query
time). You should, of course, always measure those collisions as above against
your dictionaries. And it would still be faster in some system-wide sense to
pre-compute/save the signature table somewhere, maybe right in the executable
itself with `const` and Nim compile-time execution or otherwise in a data file.
If you're doing this in a more pre-calculation separated way, that also makes
it easier to check for collisions and/or avoid them with an optimal order of
the `primes` array.
``` | 1,697 | 6,568 | {"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-2020-10 | latest | en | 0.929889 |
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CHAPTER 11
11-1. Suppose the firms labor demand curve is given by: w = 20 - 0.01 E, where w is the hourly wage and E is the level of employment. Suppose also that the unions utility function is given by U = w E. It is easy to show that the marginal utility of the wage for the union is E and the marginal utility of employment is w. What wage would a monopoly union demand? How many workers will be employed under the union contract?
Utility maximization requires the absolute value of the slope of the indifference curve equal the absolute value of the slope of the labor demand curve. For the indifference curve, we have that
MU E w = . MUw E
The absolute value of the slope of the labor demand function is 0.01. Thus, utility maximization requires that
w = .01 . E
Substituting for E with the labor demand function, the wage that maximizes utility must solve
w = 0.01 , 2,000 100 w which implies that the union sets a wage of \$10, at which price the firm hires 1,000 workers.
11-2. Suppose the union in problem 1 has a different utility function. In particular, its utility function is given by: U = (w - w*) E where w* is the competitive wage. The marginal utility of a wage increase is still E, but the marginal utility of employment is now w w*. Suppose the competitive wage is \$10 per hour. What wage would a monopoly union demand? How many workers will be employed under the union contract? Contrast your answers to those in problem 1. Can you explain why they are different?
76
Again equate the absolute value of the slope of the indifference curve to the absolute value of the slope of the labor demand curve:
MU E w w* = = 0.01 . MU w E Setting w* = \$10 and using the labor demand equation yields: w 10 = 0.01 . 2,000 100 w Thus, the union demands a wage of \$15, at which price the firm hires 500 workers. In problem 1, the union maximized the total wage bill. In problem 2 the utility function depends on the difference between the union wage and the competitive wage. That is, the union maximizes its rent. Since the alternative employment pays \$10, the union is willing to suffer a cut in employment in order to obtain a greater rent.
11-3. Using the model of monopoly unionism, present examples of economic or political activities that the union can pursue to manipulate the firms elasticity of labor demand. Relate your examples to Marshalls rules of derived demand.
Marshalls rules state that the elasticity of labor demand is lower the 1. lower is the elasticity of substitution; 2. lower is the elasticity of demand for the output; 3. lower is labors share of total costs; and 4. lower is the supply elasticity of other factors of production. Consider two examples: innovations and picket lines. Unions are notoriously bad at allowing firms to introduce (labor saving) innovations in their factories. The long shoremen on the west coast recently struck, because they were unwilling to let cargo crates be identified with bar codes. (The union wanted a union worker to record all movements of crates with pencil and paper.) Thus, the union was pursuing a policy of limiting the supply of other factors of production (rule 4). In a similar vein, when on strike, unions picket the firm in order to decrease the ability of the firm to hire scabs (rule 1).
77
11-4. Suppose the union only cares about the wage and not about the level of employment. Derive the contract curve and discuss the implications of this contract curve.
The utility function U = U(w) implies that the unions indifference curves are horizontal lines, so that the contract curve coincides exactly with the firms labor demand curve (D).
Dollars
U D Employment
11-5. A bank has \$5 million in capital that it can invest at a 5 percent annual interest rate. A group of 50 workers comes to the bank wishing to borrow the \$5 million. Each worker in the group has an outside job available to him or her paying \$50,000 per year. If the group of workers borrows the \$5 million from the bank, however, they can set up a business (in place of working their outside jobs) that returns \$3 million in addition to maintaining the original investment. (a) If the bank has all of the bargaining power (that is, the bank can make a take-it or leave-it offer), what annual interest rate will be associated with the repayment of the loan? What will be each workers income for the year?
If the bank has all of the bargaining power, it will pay each worker exactly their reservation wage, i.e., \$25,000. The total cost of this is \$2.5 million. Thus, the firm will claim the remaining \$500,000 by imposing a 10 percent interest rate as \$500,000 is 10 percent of the original \$5 million.
(b) If the workers have all of the bargaining power (that is, the workers can make a take-it or leaveit offer), what annual interest rate will be associated with the repayment of the loan? What will be each workers income for the year?
If the workers have all of the bargaining power, they will pay the bank its reservation value, i.e., an interest rate of 5 percent. When it does this, the 50 workers receive \$3 million less the 5 percent interest of \$250,000 for a total of \$2.75 million. Split evenly among the 50 workers, this leaves each worker with a yearly income of \$55,000.
78
11-6. Consider a firm that faces a constant per unit price of \$1,200 for its output. The firm hires workers, E, from a union at a daily wage of w, to produce output, q, where q = 2E. Given the production function, the marginal product of labor is 1/E. There are 225 workers in the union. Any union worker who does not work for the firm can find a non-union job paying \$96 per day. (a) What is the firms labor demand function?
The labor demand function, or the marginal revenue product of labor, is MRPE = MR MPE = 1200 / E .
(b) If the firm is allowed to specify w and the union is then allowed to provide as many workers as it wants (up to 225) at the daily wage of w, what wage will the firm set? How many workers will the union provide? How much output will be produced? How much profit will the firm earn? What is the total income of the 225 union workers?
If the firm offers w < \$96, no workers will be provided. This would leave the firm with no output and no profit. The workers would all receive \$96 per day, making their total daily income \$21,600. If the firm offers a wage of w > \$96, all 225 workers will be provided. These 225 workers would produce 30 units of output. The firm would then earn a profit of 30(\$1,200) 225w. Profit, therefore, is maximized when w is minimized subject to the constraint. If the union would supply all 225 workers at a wage of \$96, for example, the firm would offer w = \$96 and earn a daily profit of \$14,400. The total daily income of the 225 workers would remain at \$21,600. If the firm needs to offer strictly more than \$96 per day to attract workers, it would offer a daily wage of \$96.01. All 225 workers would work for the firm, making 30 units of output. The firms daily profit would be \$14,397.75. And the total daily income of the 225 workers would be \$21,602.25.
(c) If the union is allowed to specify w and the firm is then allowed to hire as many workers as it wants (up to 225) at the daily wage of w, what wage will the union set in order to maximize the total income of all 225 workers? How many workers will the firm hire? How much output will be produced? How much profit will the firm earn? What is the total income of the 225 union workers?
79
The spreadsheet looks like the following, where the union specifies the wage, labor demand comes from part (a), and everything else follows naturally:
wage \$96 \$97 \$98 \$99 \$100 \$190 \$191 \$192 \$193 \$194 \$195 Labor Demand 156.25 153.04 149.94 146.92 144.00 39.89 39.47 39.06 38.66 38.26 37.87 Labor Costs \$15,000.00 \$14,845.36 \$14,693.88 \$14,545.45 \$14,400.00 \$7,578.95 \$7,539.27 \$7,500.00 \$7,461.14 \$7,422.68 \$7,384.62 Output 25.00 24.74 24.49 24.24 24.00 12.63 12.57 12.50 12.44 12.37 12.31 Price \$1,200 \$1,200 \$1,200 \$1,200 \$1,200 \$1,200 \$1,200 \$1,200 \$1,200 \$1,200 \$1,200 Revenue \$30,000.00 \$29,690.72 \$29,387.76 \$29,090.91 \$28,800.00 \$15,157.89 \$15,078.53 \$15,000.00 \$14,922.28 \$14,845.36 \$14,769.23 Profit \$15,000.00 \$14,845.36 \$14,693.88 \$14,545.45 \$14,400.00 \$7,578.95 \$7,539.27 \$7,500.00 \$7,461.14 \$7,422.68 \$7,384.62 Union Daily Income \$19,200.00 \$19,353.04 \$19,499.88 \$19,640.77 \$19,776.00 \$22,949.58 \$22,949.90 \$22,950.00 \$22,949.90 \$22,949.60 \$22,949.11
Thus, the union sets a daily wage of \$192. The firm responds by hiring 39.06 workers, who produce 12.5 units of output. The firm earns a daily profit of \$7,500, while the 225 union workers earn a total of \$25,297.92 each day.
11-7. Suppose the unions resistance curve is summarized by the following data. The unions initial wage demand is \$10 per hour. If a strike occurs, the wage demands change as follows: Length of Strike: 1 month 2 months 3 months 4 months 5 or more months Hourly Wage Demanded 9 8 7 6 5
Consider the following changes to the union resistance curve and state whether the proposed change makes a strike more likely to occur, and whether, if a strike occurs, it is a longer strike. (a) The drop in the wage demand from \$10 to \$5 per hour occurs within the span of 2 months, as opposed to 5 months.
If the union is willing to drop its demands very fast, the firm will find it profitable to delay agreement until the wage demand drops to \$5. A strike, therefore, is more likely to occur. If \$5 is the lowest wage the union is willing to accept, the strike would probably last 2 months.
(b) The union is willing to moderate its wage demands further after the strike has lasted for 6 months. In particular, the wage demand keeps dropping to \$4 in the 6th month, \$3 in the 7th month, etc.
If the union is willing to accept even lower wages in the future, some firms will find it optimal to wait the union out. Thus, strikes will be more likely and last longer.
80
(c) The unions initial wage demand is \$20 per hour, which then drops to \$9 after the strike lasts one month, \$8 after 2 months, and so on.
Conditioning on a strike occurring, the length of strike will be unchanged as the resistance curve after the initial demand stays the same. Of course, the probability of a strike occurring increases when the initial demand increases.
11-8. (a) Would you expect unions to be more willing to call a strike during good economic times or bad economics times? Explain.
This is an open question much empirical evidence suggests that strikes are procyclical a conclusion that the model of job search supports. During good economic times, there are many good jobs available, searching for jobs is relatively easy, and the probability of securing a job is quite high. In short, the nonunion option is quite attractive. During such times, therefore, the union may be a tough negotiator, and this tough stance may lead to more strikes being called. (This doesnt explain, though, why the firms are also tough negotiators during such times.) The opposite happens during bad economic times. Namely, jobs are scarce, they are difficult to find, and searching is costly. The non-union option, therefore, is not very attractive. Consequently, the union leadership may be more willing to accept a deal. This softer stance, therefore, may lead to fewer strikes being called.
(b) Does Table 627 of the 2002 U.S. Statistical Abstract provide evidence to support your answer to part (a)? What is the single overriding pattern in this table?
There is some evidence that strikes are more prevalent in good times than bad (compare the 1960s to the early 1970s), but there was much more strike activity in the late 1970s and early 1980s than in the mid to late 1980s. The single most obvious pattern in the table, however, is that as union membership steadily fell over this time period, the level of strike activity also fell. The average percent or working time lost to a strike was about 12 percent, 10 percent, 4 percent, and 2 percent for the 1960s, 1970s, 1980s, and 1990s respectively.
11-9. Suppose the value of the marginal product of labor in the steel industry is VMPE = 100,000 E dollars per year, where E is the number of steel workers. The competitive wage for the workers with the skills needed in steel production is \$30,000 a year, but the industry is unionized so that steel workers earn \$35,000 a year. The steelworkers union is a monopoly union. What is the efficiency cost of the union contract in this industry?
If the steel industry were to pay the competitive wage to its workers, it would employ 70,000 workers, because at this level the VMP of the last employee equals the competitive wage. Under the union wage, however, the industry only hires 65,000 workers. The efficiency cost of the union, therefore, is ()(\$35,000 \$30,000)(\$70,000 \$65,000) = \$125 million per year.
81
11-10. Suppose the economy consists of a union and a non-union sector. The labor demand curve in each sector is given by L = 1,000,000 20w. The total (economy-wide) supply of labor is 1,000,000, and it does not depend upon the wage. All workers are equally skilled and equally suited for work in either sector. A monopoly union sets the wage at \$30,000 in the union sector. What is the union wage gap? What is the effect of the union on the wage in the non-union sector?
In a competitive economy, the wage would be the same in the two sectors, and its value would be such that the total labor demand LD = 2 (1,000,000 20wC) equaled total labor supply. The solution to the equation 2 (1,000,000 20wC) = 1,000,000 is wC = \$25,000. If the union wage is set at \$30,000, the union sector employs 400,000 workers. The remaining 600,000 must be employed in the non-union sector, which will happen if the wage in the non-union sector is (1,000,000 600,000)/20 = \$20,000. Hence, the wage gap between the union and the non-union sectors equals \$10,000, or 50 percent of the non-union wage.
11-11. Consider Table 628 of the 2002 U.S. Statistical Abstract. (a) How many workers were covered by a union contract in 1983? What percent of the workforce was unionized?
In 1983, 20.532 million workers (23.3 percent of all workers) were covered by a union contract in the U.S.
(b) How many workers were covered by a union contract in 2001? What percent of the workforce was unionized?
In 2001, 17.878 million workers (14.8 percent of all workers) were covered by a union contract in the U.S.
(c) Decompose the changes from part (a) to part (b) in terms of public- and private-sector workers and unions.
This dramatic change in union coverage masks an even deeper pattern in union coverage namely that public-sector unions have been growing in absolute numbers and holding their own in percent coverage, while the private-sector unions have experienced sharp decreases in their enrollments. The number of public-sector workers covered by a union contract, for example, increased from just over 7 million in 1983 to almost 8 million in 2001. The percent of public-sector workers who were members of a union held constant over this time span at roughly 37 percent, while the percent of public-sector workers covered by a union contract fell slightly from 45.5 percent in 1983 to 41.7 percent in 2001. On the other hand, the number of private-sector workers covered by a union contract decreased from 13.4 million in 1983 to only 9.9 million in 2001. The percent of private-sector workers covered by a union contract also fell drastically, from 18.5 percent in 1983 to just 9.7 percent in 2001.
82
11-12. Consider table 618 in the 2002 U.S. Statistical Abstract. (a) Calculate the union wage effect. Calculate the union effect on total benefits. Calculate the union effect on total compensation.
The union effects are simply the ratio of the union amount divided by the non-union amount: Wage effect: \$19.33 / \$15.38 = 1.257 percent. Benefit effect: \$10.09 / \$5.41 = 1.865 percent Total compensation effect: \$29.42 / \$20.79 = 1.415 percent.
(b) Note that for most categories, retirement and savings increases total compensation by about 60 to 80 cents per hour, with roughly two-thirds of this expense coming in defined contribution retirement plans. In contrast, retirement and savings adds \$1.64 to the hourly compensation of union workers, and over 70 percent of this comes in the form of defined benefit pension plans, not defined contribution. What is the difference between defined benefit and defined contribution plans? Why might a union prefer (and be able to negotiate) more compensation in defined benefit plans than defined contribution plans?
A defined benefit plan specifies the retirement benefit as a fixed dollar amount. For example, if someone receives 50 percent of their last annual income as their annual pension, this is a defined benefit plan. In contrast, a defined contribution plan specifies the amount of savings into a retirement plan the firm will make. The amount of benefit eventually received by the worker depends on how well the money is invested until retirement. It is generally thought the workers prefer DB plans (though this doesnt need to be the case). DB plans put the risk on the part of the firm, while DC plans put the financial risk on the part of the worker. (The workers at Enron, for example, lost huge amounts of retirement savings not only because they were in a DC plan but also because they were forced to keep most of their contributions as Enron stock.) As unions tend to work in large firms, they may be more able than other workers to negotiate a DB plan.
83 | 4,368 | 17,695 | {"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-2019-43 | latest | en | 0.934285 |
https://forum.knittinghelp.com/t/foll-alt-row/58358 | 1,603,716,289,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891228.40/warc/CC-MAIN-20201026115814-20201026145814-00083.warc.gz | 324,571,464 | 3,714 | Foll alt row
I am doing Debbie Bliss’ teddy bears and I am confused…
“Inc one st at each end of the 3rd row and foll alt row”
Is the “foll alt row” the next row? In other parts of the pattern, it says
"inc at each end of next 2 rows and foll alt row." Huh?
Is this a British thing?
foll alt row = following alternating row. This means that after the increase row you increase on every other row beginning with the second row after the first increase row.
So in this case, since you increased on row 3, you will then increase on row 5 next. Usually, the instructions would say something like increase foll alt row for x rows, not just once.
I’m not sure if it’s a British thing or not. | 173 | 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} | 2.59375 | 3 | CC-MAIN-2020-45 | latest | en | 0.924589 |
https://www.vedantu.com/maths/dividing-by-zero | 1,723,272,989,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00711.warc.gz | 801,571,726 | 36,483 | Courses
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Dividing by Zero
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Last updated date: 09th Aug 2024
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What is Division?
Division is a method of dividing a group of things into equal parts. It is one of the four basic arithmetic operations that give a fair result of sharing. The division's main aim is to see how many equal groups or how many in each group share fairly. It can also be said that division is the inverse operation of multiplication.
Example:
In division, if 12 is divided into 3 equal groups, it will give 4 in each group. This in a mathematical sense can be written as 12/ 3= 4.
• In division, the number which gets divided is called the dividend. The number which it gets divided is called the divisor. The number obtained from dividing is called the quotient while the number left is known as the remainder.
• For example, if we divide 17 with 2, we get 8. Here, 17 is the dividend, 2 is the divisor, 8 is the quotient while 1 is the remainder.
• The product of the quotient and the divisor added to the remainder is always equal to the dividend. This can be written as (Divisor × Quotient) + Remainder = Dividend or (d × Q) + R = D
• For example, if we divide 23 with 2, we get 11. Here, 23 is the dividend, 2 is the divisor, 11 is the quotient while 1 is the remainder. If we follow the above rule then by solving (2 x 11) + 1 we will get 23 which is the dividend. Therefore, the property given above holds true.
• When we divide something by 1, the result will always be the same number. This means that if the divisor is 1, then the quotient will be equal to the dividend.
For example: 10 ÷ 1= 1
• In division, the remainder is always smaller than the divisor.
• Division by zero is considered undefined. (We'll discuss this in detail)
• If the dividend and divisor are the same in the division, then the result will always be 1.
For example: 5 ÷ 5 = 1.
Zero:
Zero is an integer number just before 1. It's an even number that is neither positive nor negative. While zero is considered to be the whole number, it is not a counting number. The value of the zero number is nothing.
Zero Divided by a Number:
Dividing 0 by any number will give us a zero. Zero will never change when you multiply or divide any number by it.
⇒0/x = 0
For example, a person has zero toffees which are to be divided among 7 ( let’s say) children. This means that there is nothing to be shared or distributed among 7 children. If nothing is shared, then no one will get any toffees. Hence, 0 divided by any number gives 0 as the quotient.
Therefore, 0/1 = 0
A Number Divided by Zero:
Never divide any number by zero. We've all been taught this at school, and it's good advice. It's rarely meaningful to divide anything by zero. Dividing by zero does not make sense, because in arithmetic, dividing by zero can also be interpreted as multiplying by zero. Suppose we got an equation, 5/0=X. This also interprets the same equation as 0*X=5. Here, there's no number that could accommodate X to make the equation work.
With reference to the example given above, if we consider 0 by 0 to X, i.e., 0/0=X, it can also be rewritten as 0*X=0, and the problem is that every number works. X could be anything, so this equation is not useful at all. Hence, If we divide by zero, it is considered as "Undefined."
For example, if we have 20 bananas and we want to distribute them evenly to 4 people, then by definition of the division each student would receive 20/4 bananas, i.e. 5 bananas each. If we use the same logic, x/0 means distributing x bananas equally among 0 people. It's completely pointless; there's no rational way of distributing a group of items to 0 people, so we can say it's undefined.
What is Undefined?
Sometimes, when you see "undefined" in maths class, it seems very strange. Mathematicians have never defined what it means to divide by zero. What's the value of that? They didn't do that because they couldn't come up with a good answer. There is no good answer, no good definition. And because of that, any non-zero number, divided by zero, is left "undefined."
Solved Example:
If you have 5 Apples and 5 Friends at your home, how many Apples does each friend get, in a fair share? Everyone will get an Apple each, right?
If you have the same number of apples and no friends at your home, you're partitioning Apples among no people? How can we make sense out of this? This doesn't make sense, and that's what is called undefined.
Did you know?
• The concept of the number zero came in the 7th century much after the invention of other natural numbers. Brahmagupta was the mathematician and astronomer who found the concept of Zero. He also gave rules for addition and subtraction with zero.
• Zero is a real number, integer, Rational, as well as the whole number.
• Zero is always neutral, meaning that there is no such thing as -0 or +0.
• Zero is neither a prime number nor a composite number.
• If zero is added or subtracted to any number, then the number remains the same. But if zero is multiplied with any number then the product is zero.
• The power of any number that is raised by zero is always one.
Conclusion
Highly qualified subject matter experts, curate the study material with utmost care for the needs of a student. The articles, solutions, and sample questions are a great way to conceptual learning. As it is known that mathematics requires practice, students are expected to refer to the material provided is important for better learning. To understand this key requirement and work upon the curriculum keeping that in mind is an important aspect to success. You can easily download the free PDFs and learn constructively.
FAQs on Dividing by Zero
1. Can You Divide a Number by Zero?
Dividing any number by zero does not make sense, because in maths, dividing by zero can be interpreted as multiplying by zero. There's no number that you can multiply by zero to get a non-zero number. There's no solution, so any non-zero number divided by 0 is undefined.
2. Is 0 Divided by 5 Defined or Not?
That is perfectly defined.
0 ÷ 5 = 0
For example, if you have zero apples at your home and there are five people to share, each person will get 0 apples.
3. What is 0/0?
One might argue that 0/0 is 0 because 0 divided by 0 is 0. Another may argue that 0/0 is 1 because anything divided by itself is 1. And that's the problem! Whatever we say 0/0 is equal to, we contradict one or the other key property of numbers. To avoid "breaking maths," we simply say 0/0 is undefined.
4. How is division the inverse of multiplication?
To understand how Division is the inverse of multiplication, consider the following example. Divide a number ‘y’ by 8. According to the inverse rule, multiplication of the result y/8 with the divisor, 8 will give the dividend, y as the result. In mathematical sense,
(y/8) x 8 = y
Therefore, if a number (say y ) is divided by any other number (say 8) , multiplying the quotient by 8 undoes division by 8. This is applicable for any number which is not equal to zero.
Example- Divide 32 by 2.
= 32/2
= 16
Now, multiply the quotient that is 16 with the divisor that is 2.
= 16 x 2
= 32 which is the dividend.
Therefore, we can say that division is the inverse of multiplication. | 1,804 | 7,362 | {"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} | 4.75 | 5 | CC-MAIN-2024-33 | latest | en | 0.935125 |
https://www.lawnsite.com/threads/mulch-sq-ft-to-cubic-ft-conversion.107695/ | 1,513,358,395,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948575124.51/warc/CC-MAIN-20171215153355-20171215175355-00515.warc.gz | 767,475,417 | 31,068 | # mulch sq ft to cubic ft conversion
Discussion in 'Landscape Maintenance' started by contourbs, May 18, 2005.
1. ### contourbsLawnSite Memberfrom fort myers flMessages: 185
i have to lay 2193 sq ft of mulch. what does that equal to in cubic yards. and what would you charge. its only a re mulch
2. ### MowgliLawnSite Memberfrom Lenexa, KansasMessages: 183
If you want to spread it 2" thick, Just devide 2193 by 152. i think it is about 14.6 yards.
3. ### RandysLawnCareLawnSite Memberfrom cary, ILMessages: 6
6.77 Square yards for 1 inch and 13.54 sq Yards for 2 inch. Great mulch coversion site http://www.lumberjax.com/mulch.html#calculator And It depends if they want you to do edging or re trenching so mulch dosent find its way into the yard. and the number or people on crew. Post and il will reply approx price that I would give. and with yardage I would go up and get half a more yard that you need for a remulch. Never know where spots are gonna need it on a remulch. Thanks Randy's Lawn Care
4. ### contourbsLawnSite Memberfrom fort myers flMessages: 185
i used a bed scraper on 3 beds and need to use some weed killer on all of that. it will be 2 inch
5. ### out4nowLawnSite Bronze Memberfrom AZMessages: 1,796
There used to be a sticky with a mulch calculation formula, wonder where that went?
6. ### sheshovelLawnSite Fanaticfrom zone 7 CAMessages: 5,112
One square yard = 27sq feet you can figure this out.
Can't you??
7. ### MarcSmithLawnSite Fanaticfrom Washington DCMessages: 7,157
LxWxH
I make my hieght in yards 1"=.03 yards, 2"=.06 yards 3" .08 yards and I pace out my beds my pace is approx 3'
or you can measure beds in feet and them bed hight in tenths of an inch and then divided by 27 to get from sqft to sqyd
http://www.onlineconversion.com/
I have a small book, actually basicly a pamphlet that cost me a dollar that has several thousand conversion factors. it was made bythe sillcocks-miller Company out of Berkely Height NJ... But I dont think they are in business any more. | 578 | 2,022 | {"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-2017-51 | latest | en | 0.924964 |
https://justaaa.com/physics/69234-a-converging-lens-f1-240-cm-is-located-560-cm-to | 1,675,404,768,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00666.warc.gz | 352,661,384 | 9,119 | Question
# A converging lens (f1 = 24.0 cm) is located 56.0 cm to the left of a...
A converging lens (f1 = 24.0 cm) is located 56.0 cm to the left of a diverging lens (f2 = -28.0 cm). An object is placed to the left of the converging lens, and the final image produced by the two-lens combination lies 19.4 cm to the left of the diverging lens. How far is the object from the converging lens?
here,
for the diverging lens
the focal length , f2 = - 28 cm
the image distance , di2 = - 19.4 cm
let the object distance be do2
using the lens formula
1/f2 = 1/di2 - 1/do2
1/(-28) = 1/(- 19.4) - 1/do2
solving for do2
do2 = - 63.2 cm
for the converging lens
the focal length , f1 = 24 cm
the image distance , di1 = do2 + 56 cm
di1 = - 7.2 cm
let the object distance be do1
using the lens formula
1/f1 = 1/di1 - 1/do1
1/24 = 1 /(-7.2) - 1/do1
do1 = - 5.54 cm
the object is placed 5.54 cm to the left of converging lens
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 345 | 992 | {"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-2023-06 | longest | en | 0.724976 |
http://www.schooltrainer.com/tag/science-answer | 1,656,297,201,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103324665.17/warc/CC-MAIN-20220627012807-20220627042807-00543.warc.gz | 106,636,725 | 7,936 | # Elastic collision of unequal masses
Thursday, July 22nd, 2010
You roll a basketball and a bowling ball directly towards each other at the same speed. Predict what will happen to each after the two collide.
### Solution:
Let’s assume that this is an “elastic collision of unequal masses” (see http://en.wikipedia.org/wiki/Elastic_collision, which will help you visualize this). Let’s also assume that the weight of the basketball is 1 pound and the weight of the bowling ball is 2 pounds, and that the basket ball is traveling from east to west at a speed (v) of 9 miles per hour, and the bowling ball is traveling west to east at 9 miles per hour.
After the collision the bowling ball reverse direction and travel at 3 miles per hour, and the basketball will reverse direction and travel at 15 miles per hour.
# What is inside a flower?
Thursday, July 22nd, 2010
### Soluton:
We found the following picture of what’s inside a flower at
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1 INFO 2950 Intro to Data Science Lecture 17: Power Laws and Big Data Paul Ginsparg Cornell University, Ithaca, NY 29 Oct /25
2 Power Laws in log-log space y = cx k (k=1/2,1,2) log 10 y = k log 10 x +log 10 c sqrt(x) x x**2 100 sqrt(x) x x** /25
3 Zipf s law Now we have characterized the growth of the vocabulary in collections. We also want to know how many frequent vs. infrequent terms we should expect in a collection. In natural language, there are a few very frequent terms and very many very rare terms. Zipf s law (linguist/philologist George Zipf, 1935): The i th most frequent term has frequency proportional to 1/i. cf i 1 i cf i is collection frequency: the number of occurrences of the term t i in the collection. 3/25
4 s law Zipf s law: the frequency of any word is inversely proportional to its rank in the frequency table. Thus the most frequent word will occur approximately twice as often as the second most frequent word, which occurs twice as often as the fourth most frequent word, etc. Brown Corpus: the : 7% of all word occurrences (69,971 of >1M). of : 3.5% of words (36,411) and : 2.9% (28,852) Only 135 vocabulary items account for half the Brown Corpus. The Brown University Standard Corpus of Present-Day American English is a carefully compiled selection of current American English, totaling about a million words drawn from a wide variety of sources... for many years among the most-cited resources in the field. 4/25
5 Zipf s law Zipf s law: The i th most frequent term has frequency proportional to 1/i. cf i 1 i cf is collection frequency: the number of occurrences of the term in the collection. So if the most frequent term (the) occurs cf 1 times, then the second most frequent term (of) has half as many occurrences cf 2 = 1 2 cf and the third most frequent term (and) has a third as many occurrences cf 3 = 1 3 cf 1 etc. Equivalent: cf i = ci k and logcf i = logc+klogi (for k = 1) Example of a power law 5/25
6 Power Laws in log-log space y = cx k (k=1/2,1,2) log 10 y = k log 10 x +log 10 c /sqrt(x) 100/x 100/x** /sqrt(x) 100/x 100/x** /25
7 Zipf s law for Reuters log10 cf log10 rank Fit far from perfect, but nonetheless key insight: Few frequent terms, many rare terms. 7/25
8 more from s law A plot of word frequency in Wikipedia (27 Nov 2006). The plot is in log-log coordinates. x is rank of a word in the frequency table; y is the total number of the words occurrences. Most popular words are the, of and and, as expected. Zipf s law corresponds to the upper linear portion of the curve, roughly following the green (1/x) line. 8/25
9 Power laws more generally E.g., consider power law distributions of the form cr k, describing the number of book sales versus sales-rank r of a book, or the number of Wikipedia edits made by the r th most frequent contributor to Wikipedia. Amazon book sales: cr k, k.87 number of Wikipedia edits: cr k, k 1.7 (More on power laws and the long tail here: Networks, Crowds, and Markets: Reasoning About a Highly Connected World by David Easley and Jon Kleinberg Chpt 18: 9/25
10 Wikipedia edits/month Amazon sales/week / r^{1.7} / r^{.87} Normalization given by the roughly 1 sale/week for the 200,000th ranked Amazon title: 40916r.87 and by the 10 edits/month for the 1000th ranked Wikipedia editor: r User Book rank r 1e+07 1e+06 Wikipedia edits/month Amazon sales/week / r^{1.7} / r^{.87} Long tail: about a quarter of Amazon book sales estimated to come from the long tail, i.e., those outside the top 100,000 bestselling titles e+06 User Book rank r 10/25
11 Another Wikipedia count (15 May 2010) All articles in the English version of Wikipedia, 21GB in XML format (five hours to parse entire file, extract data from markup language, filter numbers, special characters, extract statistics): Total tokens (words, no numbers): T = 1,570,455,731 Unique tokens (words, no numbers): M = 5,800,280 11/25
12 Word frequency distribution follows Zipf s law 12/25
13 rank 1 50 (86M-3M), stop words (the, of, and, in, to, a, is,...) rank 51 3K (2.4M-56K), frequent words (university, January, tea, sharp,...) rank 3K 200K (56K-118), words from large comprehensive dictionaries (officiates, polytonality, neologism,...) above rank 50K mostly Long Tail words rank 200K 5.8M (117-1), terms from obscure niches, misspelled words, transliterated words from other languages, new words and non-words (euprosthenops, eurotrochilus, lokottaravada,...) 13/25
14 Some selected words and associated counts Google Twitter 894 domain domainer 22 Wikipedia Wiki Obama Oprah 3885 Moniker 4974 GoDaddy /25
15 Project Gutenberg (per billion) Over 36,000 items (Jun 2011), average of > 50 new e-books / week the of and to in I that was his he it with is for as had you not be her ,000 th 15/25
16 Bowtie structure of the web A.Broder,R.Kumar,F.Maghoul,P.Raghavan,S.Rajagopalan,S. Stata, A. Tomkins, and J. Wiener. Graph structure in the web. Computer Networks, 33: , Strongly connected component (SCC) in the center Lots of pages that get linked to, but don t link (OUT) Lots of pages that link to other pages, but don t get linked to (IN) Tendrils, tubes, islands # of in-links (in-degree) averages 8 15, not randomly distributed (Poissonian), instead a power law: # pages with in-degree i is 1/i α, α /25
17 Poisson Distribution Bernoulli process with N trials, each probability p of success: ( ) N p(m) = p m (1 p) N m. m Probability p(m) of m successes, in limit N very large and p small, parametrized by just µ = Np (µ = mean number of successes). N! For N m, we have (N m)! = N(N 1) (N m+1) Nm, so ( N) m N! m!, and m!(n m)! Nm p(m) 1 m! Nm( µ ) m ( 1 µ ) N m µ m ( lim 1 µ ) N = e µ µm N N m! N N m! (ignore (1 µ/n) m since by assumption N µm). N dependence drops out for N, with average µ fixed (p 0). The form p(m) = e µµm m! is known as a Poisson distribution (properly normalized: m=0 p(m) = e µ µ m m=0 m! = e µ e µ = 1). 17/25
18 Compare to power law p(m) 1/m /25 Poisson Distribution for µ = 10 p(m) = e 1010m m!
19 Power Law p(m) 1/m 2.1 and Poisson p(m) = e 1010m m! /25
20 Power Law p(m) 1/m 2.1 and Poisson p(m) = e 1010m 1 m! 0.1 (log log scale) e-05 1e-06 1e-07 1e-08 1e /25
21 Power law distributions Slide credit: Dragomir Radev 21/25
22 Examples Moby Dick scientific papers AOL users visiting sites 97 bestsellers AT&T customers on 1 day California /25
23 Moon Solar flares wars ( ) richest individuals 2003 US family names 1990 US cities /25
24 Power law in networks! For many interesting graphs, the distribution over node degree follows a power law film actors telephone call graph networks sexual contacts WWW internet peer-to-peer metabolic network protein interactions exponent! (in/out degree) / / Slide credit: Dragomir Radev 24/25
25 Next Time: More Statistical Methods Peter Norvig, How to Write a Spelling Corrector (See video: The Unreasonable Effectiveness of Data, given 23 Sep 2010.) Additional related references: A. Halevy, P. Norvig, F. Pereira, The Unreasonable Effectiveness of Data, Intelligent Systems Mar/Apr 2009 (copy at resources/unrealdata.pdf) P. Norvig, Natural Language Corpus Data 25/25
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Public Engagement with Research Online Evaluating the impact of research online with Google Analytics The web provides extensive opportunities for raising awareness and discussion of research findings | 8,352 | 33,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2018-17 | latest | en | 0.921515 |
https://www.studypool.com/discuss/362437/what-is-the-normality-requirement-for-ordinary-least-squares-regression?free | 1,481,297,631,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542712.49/warc/CC-MAIN-20161202170902-00199-ip-10-31-129-80.ec2.internal.warc.gz | 1,008,993,128 | 14,096 | ##### What is the normality requirement for ordinary least-squares regression?
Statistics Tutor: None Selected Time limit: 1 Day
A) Both variables must be drawn from samples which are ordinal.
B) The solution must fit a normal distribution.
C)The populations from which the samples come must be normal.
D)
Jan 28th, 2015
the answer is A) Both variables must be drawn from samples which are ordinal.
Jan 28th, 2015
...
Jan 28th, 2015
...
Jan 28th, 2015
Dec 9th, 2016
check_circle | 130 | 488 | {"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-2016-50 | longest | en | 0.874297 |
https://en.commtap.org/language-communication/activities-encourage-use-first-next-and-last-when-describing-sequence-events | 1,642,975,631,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00016.warc.gz | 298,943,071 | 9,831 | ### Categorisation
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# Activities to encourage the use of 'first', 'next' and 'last' when describing a sequence of events.
Description:
These activities will help your child to develop use of the concepts of first, next and last when describing a sequence of events.
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Early years skill: Shape, space and measure Early years typical range: 40-60+m P-scales/Curriculum skill: Maths Shape Space and Measures P-scales/Curriculum level: P8 TAP skill: Expressive Language TAP level: TAP48 Pre/Nat. Curriculum Area: not specified Pre/Nat. Curiculum Standard: not specified
Section: Early Years (0-5yrs) info; Primary (5-11yrs) info; Secondary (11-16yrs) info
Activity/strategy name and materials required How to do the activity Key principles for doing the activity and comments
Sequencing cards
- sets of 3 picture sequencing cards - you will need to source these.
1. Choose a set of three cards and mix them up.
2. Ask your child to sort the pictures in the order that they happened. If your child finds this hard, put them in order together. Match them to the symbols.
3. Describe what is happening in the pictures using the words first, next and last - you can point to the symbols as you say the words.
4. Encourage your child to describe what is happening using the words first, next and last.
If the child finds it difficult to remember whether to put the pictures in order from left to right, or right to left, you can draw a large arrow (→) on the sheet. Numbering the boxes 1-3 can also help.
Photo sequencing
1. Take photographs of your child carrying out different activities. Choose activities that have three steps, e.g. toothpaste on toothbrush, brush teeth, rinse mouth.
2. Print out the photos and carry out the sequencing activity as described in the activity above. Encouarge use of the words first, next and last. | 531 | 2,276 | {"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-2022-05 | latest | en | 0.889144 |
http://www.jiskha.com/display.cgi?id=1247713831 | 1,495,819,239,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608669.50/warc/CC-MAIN-20170526163521-20170526183521-00249.warc.gz | 697,627,210 | 4,190 | Maths
posted by on .
Eleven people, 7 men and 4 women, successfully completed their study at the School of Astronautics. A team of 6 people needs to be selected from them for the next intergalactic travel. In how many ways can 6 astronauts be selected, if at least 2 women and 3 men must be in the team？
thank u!!
• Maths - ,
For at least 2 women and 3 men, the only cases would be
2W,4M = C(4,2) x C(7,4) = 6x35 = 210 , or
3W,3M = C(4,3) x C(7,3) = 4x35 = 140
for a total of 350 ways
(all possible ways would be
0W,6M
1W,5M
2W,4M
3W,3M
4W 2M )
• Maths - ,
Worked out combination of women and men
Either 3 women and 3 men or 4 men and 2 women
Then comination for the 2 Women was n = 4 no. of combos was 6 and r = 2 and for the 3 women n = 4 and r = 3 no. of combos was 4
Follow the formula n!/(n-r)! x (r!)
Same process for men. In each instance 35 combinations for Men.
Then times combination for each set which was (6 x 35) + (4 x 35) = 350
350 combinations for the team | 340 | 985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-22 | latest | en | 0.957796 |
http://petalsofjoy.org/index.php?m=admin&c=index&a=login&dosubmit=1 | 1,611,537,210,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703561996.72/warc/CC-MAIN-20210124235054-20210125025054-00606.warc.gz | 82,982,033 | 15,671 | Write logic content
You may not know you http://www.prismagroup.com.au/professional-case-study-ghostwriter-site-for-phd need it yet, but once you know how to use it, you won't want to live without it Logic in Argumentative Writing: Principles write logic content of Composition Many of the important points of this section are covered in the section on writing Argumentative Essays: Being Logical. It is the business of philosophical logic to extract this knowledge from its concrete integuments, and to render it explicit and pure.". Download the Templates. macbeth essays on imagery You can write it as a function with AND () or as an write logic content operator by linking two statements with &&. I will show you how you can integrate Azure Logic Apps with Azure Storage. Conclusions serve as the thesis of the argument Tutorials Logic provide high quality content writing service with 100% original and unique content, that converts customers and engages your audience. And, if you’re studying the subject, exam tips can come in handy However, freewriting can lead to a complicated process of writing, reverse outlining, revising, rewriting, and editing. Tutorials Logic provide high quality content writing service with 100% original and unique content, that converts customers and engages your audience. Each line contains a string and a number, as below:. When it comes into writing, you need to include logical reasoning as part of it. Create a trigger for when blob is added or modified. Popular Use Cases. decodeBase64(body('Get_blob_content_using_path')) and just the default option like visible in the screenshot. Protection settings can also be documented to. essay on e-waste management
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This entry was posted in Uncategorized. Bookmark the permalink. | 1,151 | 5,825 | {"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-2021-04 | latest | en | 0.892467 |
https://www.enotes.com/homework-help/prove-left-right-sinxcosx-1-cosx-1-cosx-tanx-445619 | 1,498,232,036,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320070.48/warc/CC-MAIN-20170623151757-20170623171757-00623.warc.gz | 877,636,436 | 12,826 | # Prove that the left hand side equals the right hand side: `(sinxcosx)/(1+cosx)` = `(1-cosx)/(tanx)`
durbanville | High School Teacher | (Level 2) Educator Emeritus
Posted on
(There are numerous possibilities for proofs) To prove:
`(sinxcosx)/(1+cos x) = (1-cos x)/tan x`
We will manipulate the LHS (left-hand side). As we have (1+cosx) as the denominator multiply by `(1-cosx)/(1-cos x). ` This will not change the essence of the expression in any way as it is the equivalent of 1. It also suits our purposes (see RHS)
LHS: `therefore (sinxcosx)/(1+cosx) times (1-cos x)/(1-cosx)`
= `((sinxcosx)(1-cos x))/((1+cos x)(1-cosx))`
Simplify this by expanding the brackets regarding the numerator and denominator. The denominator is now a difference of two squares:
= `(sinxcosx- sincos^2x)/(1-cos^2x)`
We are manipulating this to suit our purposes so now factorize using sin (numerator) and the identity `1-cos^2x=sin^2x` in the denominator:
= `((sinx)(cos x-cos ^2 x))/(sin^2 x)`
The sin (numerator) cross cancels with one from the denominator and we factorize the remaining term:
`therefore =(cosx(1-cosx))/(sin x)`
We can rewrite (for our purpses) as:
`cosx/sinx times (1-cosx)/1`
We know that `sinx/cosx = tanx` so it follows that `cosx/sinx=1/tanx`
`therefore = 1/tanx times (1-cosx)/1`
`= (1-cosx)/tanx`
`therefore (sinxcosx)/(1+cosx)= (1-cosx)/tanx`
Therefore LHS=RHS
Sources: | 469 | 1,405 | {"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} | 4.46875 | 4 | CC-MAIN-2017-26 | longest | en | 0.693228 |
https://fr.maplesoft.com/support/help/Maple/view.aspx?path=MathApps/PythagoreanTheorem | 1,721,657,893,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00082.warc.gz | 223,164,855 | 20,395 | Pythagorean Theorem - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
Main Concept The Pythagorean theorem states that any right angle triangle of side lengths a, b, c satisfies where $c$ is the length of the longest side.
Observe the following animation. The area of the light blue region is ${c}^{\mathit{2}}$ at the beginning and ${a}^{\mathit{2}}\mathit{+}{b}^{\mathit{2}}$ at the end, and yet it clearly has not changed. | 134 | 489 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "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} | 2.734375 | 3 | CC-MAIN-2024-30 | latest | en | 0.772778 |
http://yakasee.com/n47A2Pr5/lC2572wD/ | 1,563,868,869,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529007.88/warc/CC-MAIN-20190723064353-20190723090353-00135.warc.gz | 284,878,151 | 11,156 | # One Step Equations Multiplication And Division Math Solving Coloring Worksheet Sign Up As
Published at Friday, 10 May 2019. Color by Number. By .
First off, even though it may seem that children are quite restrained when coloring these coloring pages in terms of shapes, choice of colors and what not, color by number images can really be encouraging for some kids. Namely, there are children who feel frustrated when given too many choices and opportunities to make decisions. For some of them, having to pick colors and plan the way to color a picture might come as torture, making them feel anxious and nervous. That is why these color by number coloring pages are more than welcome: you are already told what to do, therefore you can relax and perform the actions. One more major benefit of color by number coloring pages is that not only young kids learn how to discern colors (in terms of primary and secondary ones), but they also learn how to put numbers in contexts other than counting. Here, numbers are presented as mere symbols that can and will occur anywhere else in real life.
If you don’t want to work with the brushes paint-by-number kits come with, you may want to buy your own at a craft store. In particular, you may want to have a larger brush on hand, not just the very small ones offered in the kits. Some canvases come wrinkled. Spray a light mist of water on the canvas and iron on a low setting on the back side to press out any wrinkles before you begin painting. Paint with just one color at a time, beginning with the largest areas meant for the color. “You’ll notice some shapes have two numbers in them, not just one,” stated ThoughtCo. “This indicates that you need to mix two colors together. Equal proportions should give you a suitable color, but don’t dip your brush from one paint container into the next as you’ll contaminate the colors.”
Since tomorrow is National Coloring Book Day, today seems like a good day to talk about color by number exercises, and some of the principal benefits of color by number activities, and why they are good exercises to give to your kids, either at home or in a classroom setting. So here we go: Color by number exercises encourage creativity, But wait, you might say… color by number exercises give children defined colors and limits… why would this support creativity and imagination? Well, if you have children who don’t naturally want to draw or color, or feel timid doing so, color by number exercises offer a “safe zone” that kids can use to practice working with color and design. This can lead to future drawing, painting, or coloring activities.
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### One Step Equations Multiplication And Division Math Solving Coloring Worksheet Sign Up As
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#### Solving One Step Equations Coloring Worksheet Gallery
99 of 100 by 849 users | 615 | 2,921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-30 | latest | en | 0.964619 |
https://www.stumblingrobot.com/2016/04/24/compute-things-using-binomial-series/ | 1,716,411,378,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00657.warc.gz | 889,752,787 | 15,063 | Home » Blog » Compute some things using the binomial series
# Compute some things using the binomial series
1. Show that the first six terms of the binomial series for are
2. For let be the th coefficient in the binomial series and let denote the remainder after terms, i.e.,
for . Prove that
3. Prove the validity of the identity
Use this identity to compute the first ten decimal places of .
Incomplete.
### One comment
1. S says:
From section 11.15, we know binomial formula for exponent -1/2. From that we get a_n = [1*3*5*…*(2n-1)]/[2^n*n!]*(1/50)^n, and then by using hints one can solve this. | 166 | 609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-22 | latest | en | 0.802106 |
https://everipedia.org/wiki/lang_en/Hertz | 1,723,341,432,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640843545.63/warc/CC-MAIN-20240811013030-20240811043030-00352.warc.gz | 202,735,377 | 26,287 | Everipedia is now IQ.wiki - Join the IQ Brainlist and our Discord for early access to editing on the new platform and to participate in the beta testing.
# Hertz
The hertz (symbol: Hz) is the derived unit of frequency in the International System of Units (SI) and is defined as one cycle per second.[1] It is named after Heinrich Rudolf Hertz, the first person to provide conclusive proof of the existence of electromagnetic waves. Hertz are commonly expressed in multiples: kilohertz (103 Hz, kHz), megahertz (106 Hz, MHz), gigahertz (109 Hz, GHz), terahertz (1012 Hz, THz), petahertz (1015 Hz, PHz), exahertz (1018 Hz, EHz), and zettahertz (1021 Hz, ZHz).
Some of the unit's most common uses are in the description of sine waves and musical tones, particularly those used in radio- and audio-related applications. It is also used to describe the clock speeds at which computers and other electronics are driven. The units are often also used as a representation of Energy, via the photon energy equation, with one Hertz equivalent to h joules.
Hertz
Unit systemSI derived unit
Unit ofFrequency
SymbolHz
Named afterHeinrich Hertz
InSI base unitss
## Definition
The hertz is defined as one cycle per second. The International Committee for Weights and Measures defined the second as "the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom"[2][3] and then adds: "It follows that the hyperfine splitting in the ground state of the caesium 133 atom is exactly 9 192 631 770 hertz, ν(hfs Cs) = 9 192 631 770 Hz." The dimension of the unit hertz is 1/time (1/T). Expressed in base SI units it is 1/second (1/s).
In English, "hertz" is also used as the plural form.[4] As an SI unit, Hz can be prefixed; commonly used multiples are kHz (kilohertz, 103 Hz), MHz (megahertz, 106 Hz), GHz (gigahertz, 109 Hz) and THz (terahertz, 1012 Hz). One hertz simply means "one cycle per second" (typically that which is being counted is a complete cycle); 100 Hz means "one hundred cycles per second", and so on. The unit may be applied to any periodic event—for example, a clock might be said to tick at 1 Hz, or a human heart might be said to beat at 1.2 Hz. The occurrence rate of aperiodic or stochastic events is expressed inreciprocal second or inverse second (1/s or s−1) in general or, in the specific case of radioactive decay, in becquerels.[5] Whereas 1 Hz is 1 cycle per second, 1 Bq is 1 aperiodic radionuclide event per second.
Even though angular velocity, angular frequency and the unit hertz all have the dimension 1/s, angular velocity and angular frequency are not expressed in hertz,[6] but rather in an appropriate angular unit such as radians per second. Thus a disc rotating at 60 revolutions per minute (rpm) is said to be rotating at either 2π rad/s or 1 Hz, where the former measures the angular velocity and the latter reflects the number of complete revolutions per second. The conversion between a frequency f measured in hertz and an angular velocity ω measured in radians per second is
and.
The hertz is an named after Heinrich Hertz. As with every SI unit named for a person, its symbol starts with an upper case letter (Hz), but when written out it follows no special casing, following whatever would contextually befit a common noun; i.e., "hertz" becomes capitalised at the beginning of a sentence and in titles.
## History
The hertz is named after the German physicist Heinrich Hertz (1857–1894), who made important scientific contributions to the study of electromagnetism. The name was established by the International Electrotechnical Commission (IEC) in 1930.[7] It was adopted by the General Conference on Weights and Measures (CGPM) (Conférence générale des poids et mesures) in 1960, replacing the previous name for the unit, cycles per second (cps), along with its related multiples, primarily kilocycles per second (kc/s) and megacycles per second (Mc/s), and occasionally kilomegacycles per second (kMc/s). The term cycles per second was largely replaced by hertz by the 1970s. One hobby magazine, Electronics Illustrated, declared their intention to stick with the traditional kc., Mc., etc. units.[8]
## Applications
### Vibration
Sound is a traveling longitudinal wave which is an oscillation of pressure. Humans perceive frequency of sound waves as pitch. Each musical note corresponds to a particular frequency which can be measured in hertz. An infant's ear is able to perceive frequencies ranging from 20 Hz to 20,000 Hz; the average adult human can hear sounds between 20 Hz and 16,000 Hz.[10] The range of ultrasound, infrasound and other physical vibrations such as molecular and atomic vibrations extends from a few femtohertz[11] into the terahertz range[12] and beyond.
Electromagnetic radiation is often described by its frequency—the number of oscillations of the perpendicular electric and magnetic fields per second—expressed in hertz.
Radio frequency radiation is usually measured in kilohertz (kHz), megahertz (MHz), or gigahertz (GHz).
Light is electromagnetic radiation that is even higher in frequency, and has frequencies in the range of tens (infrared) to thousands (ultraviolet) of terahertz. Electromagnetic radiation with frequencies in the low terahertz range (intermediate between those of the highest normally usable radio frequencies and long-wave infrared light) is often called terahertz radiation. Even higher frequencies exist, such as that of gamma rays, which can be measured in exahertz (EHz). (For historical reasons, the frequencies of light and higher frequency electromagnetic radiation are more commonly specified in terms of their wavelengths or photon energies: for a more detailed treatment of this and the above frequency ranges, see electromagnetic spectrum.)
### Computers
In computers, most central processing units (CPU) are labeled in terms of their clock rate expressed in megahertz (106 Hz) or gigahertz (109 Hz). This specification refers to the frequency of the CPU's master clock signal. This signal is a square wave, which is an electrical voltage that switches between low and high logic values at regular intervals. As the hertz has become the primary unit of measurement accepted by the general populace to determine the performance of a CPU, many experts have criticized this approach, which they claim is an easily manipulable benchmark. Some processors use multiple clock periods to perform a single operation, while others can perform multiple operations in a single cycle.[13] For personal computers, CPU clock speeds have ranged from approximately 1 MHz in the late 1970s (Atari, Commodore, Apple computers) to up to 6 GHz in IBM POWER microprocessors.
Various computer buses, such as the front-side bus connecting the CPU and northbridge, also operate at various frequencies in the megahertz range.
## SI multiples
Submultiples Multiples Value SI symbol Name Value 10−1Hz dHz decihertz 101Hz daHz decahertz 10−2Hz cHz centihertz 102Hz hHz hectohertz 10−3Hz mHz millihertz 103Hz kHz kilohertz 10−6Hz µHz microhertz 106Hz MHz megahertz 10−9Hz nHz nanohertz 109Hz GHz gigahertz 10−12Hz pHz picohertz 1012Hz THz terahertz 10−15Hz fHz femtohertz 1015Hz PHz petahertz 10−18Hz aHz attohertz 1018Hz EHz exahertz 10−21Hz zHz zeptohertz 1021Hz ZHz zettahertz 10−24Hz yHz yoctohertz 1024Hz YHz yottahertz Common prefixed units are in bold face.
Higher frequencies than the International System of Units provides prefixes for are believed to occur naturally in the frequencies of the quantum-mechanical vibrations of high-energy, or, equivalently, massive particles, although these are not directly observable and must be inferred from their interactions with other phenomena. By convention, these are typically not expressed in hertz, but in terms of the equivalent quantum energy, which is proportional to the frequency by the factor of Planck's constant.
Hertz: Unicode characters.[14]
SymbolNameUnicode number
Hertz (Square HZ)U+3390
Kilohertz (Square KHZ)U+3391
Megahertz (Square MHZ)U+3392
Gigahertz (Square GHZ)U+3393
Terahertz (Square THZ)U+3394
• Alternating current
• Bandwidth (signal processing)
• Electronic tuner
• FLOPS
• Frequency changer
• Normalized frequency
• Orders of magnitude (frequency)
• Periodic function
• Radian per second
• Rate
• Unicode CJK Compatibility block which includes common SI units for frequency
## References
[1]
Citation Linkopenlibrary.org"hertz". (1992). American Heritage Dictionary of the English Language (3rd ed.), Boston: Houghton Mifflin.
Oct 1, 2019, 5:00 AM
[2]
Citation Linkwww.bipm.org"SI brochure: Table 3. Coherent derived units in the SI with special names and symbols".
Oct 1, 2019, 5:00 AM
[3]
Citation Linkwww.bipm.org"[Resolutions of the] CIPM, 1964 – Atomic and molecular frequency standards" (PDF). SI brochure, Appendix 1.
Oct 1, 2019, 5:00 AM
[4]
Citation Linkphysics.nist.govNIST Guide to SI Units – 9 Rules and Style Conventions for Spelling Unit Names, National Institute of Standards and Technology
Oct 1, 2019, 5:00 AM
[5]
Citation Linkwww.bipm.org"(d) The hertz is used only for periodic phenomena, and the becquerel (Bq) is used only for stochastic processes in activity referred to a radionuclide." "BIPM – Table 3". BIPM. Retrieved 24 October 2012.
Oct 1, 2019, 5:00 AM
[6]
Citation Linkwww.bipm.org"SI brochure, Section 2.2.2, paragraph 6". Archived from the original on 1 October 2009.
Oct 1, 2019, 5:00 AM
[7]
Citation Linkwww.iec.ch"IEC History". Iec.ch. 15 September 1904. Retrieved 28 April 2012.
Oct 1, 2019, 5:00 AM
[8]
Citation Linkwww.americanradiohistory.comCartwright, Rufus (March 1967). Beason, Robert G. (ed.). "Will Success Spoil Heinrich Hertz?" (PDF). Electronics Illustrated. Fawcett Publications, Inc. pp. 98–99. Retrieved 29 March 2016.
Oct 1, 2019, 5:00 AM
[9]
Citation Linkportal.issn.orgRekdal, Ole Bjørn (1 August 2014). "Academic urban legends". Social Studies of Science. 44 (4): 638–654. doi:10.1177/0306312714535679. ISSN 0306-3127. PMC 4232290.
Oct 1, 2019, 5:00 AM
[10]
Citation Linkwww.mmk.e-technik.tu-muenchen.deErnst Terhardt (20 February 2000). "Dominant spectral region". Mmk.e-technik.tu-muenchen.de. Archived from the original on 26 April 2012. Retrieved 28 April 2012.
Oct 1, 2019, 5:00 AM
[11]
Citation Linkscience.nasa.gov"Black Hole Sound Waves - Science Mission Directorate". science.nasa.go.
Oct 1, 2019, 5:00 AM
[12]
Citation Linkopenlibrary.orgAtomic vibrations are typically on the order of tens of terahertz
Oct 1, 2019, 5:00 AM
[13]
Citation Linkwww.wired.comAsaravala, Amit (30 March 2004). "Good Riddance, Gigahertz". Wired.com. Retrieved 28 April 2012.
Oct 1, 2019, 5:00 AM
[14]
Citation Linkwww.unicode.orgUnicode Consortium (2019). "The Unicode Standard 12.0 – CJK Compatibility ❰ Range: 3300—33FF ❱" (PDF). Unicode.org. Retrieved 24 May 2019.
Oct 1, 2019, 5:00 AM
[16]
Oct 1, 2019, 5:00 AM
[17]
Citation Linkweb.archive.orgNational Research Council of Canada: Optical frequency standard based on a single trapped ion
Oct 1, 2019, 5:00 AM
[18]
Citation Linkweb.archive.orgNational Research Council of Canada: Optical frequency comb
Oct 1, 2019, 5:00 AM
[19]
Citation Linkwww.npl.co.ukNational Physical Laboratory: Time and frequency Optical atomic clocks
Oct 1, 2019, 5:00 AM
[20]
Citation Linkonlinetonegenerator.comOnline Tone Generator
Oct 1, 2019, 5:00 AM
[25]
Citation Linkweb.archive.org"SI brochure, Section 2.2.2, paragraph 6"
Oct 1, 2019, 5:00 AM | 3,080 | 11,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-33 | latest | en | 0.892597 |
https://www.geeksforgeeks.org/rd-sharma-class-8-chapter-1-rational-numbers-exercise-1-1/?ref=rp | 1,675,150,691,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499845.10/warc/CC-MAIN-20230131055533-20230131085533-00561.warc.gz | 789,332,043 | 34,957 | # RD Sharma Class 8 – Chapter 1 Rational Numbers – Exercise 1.1
• Last Updated : 31 Oct, 2020
### Question 1. Add the following rational numbers:
(i) -5 / 7 and 3 / 7
Solution:
(-5 / 7) + 3 / 7
Since denominators are the same hence numerators will be directly added considering their sign.
Therefore, (-5 + 3) / 7 = (-2) / 7
(ii) -15 / 4 and 7 / 4
Solution:
Denominators are same so numerators are directly added.
= (-15) / 4 + 7 / 4
= (-15 + 7) / 4
= (-8) / 4
= (4 * (-2)) / 4
= (-2)
(iii) -8 / 11 and -4 / 11
Solution:
As denominators are the same numerators are added along with their sign.
= (-8) / 11 + (-4) / 11
= (-8 – 4) / 11 (integers with same sign are added)
= (-12) / 11
(iv) 6 / 13 and -9 / 13
Solution:
As, denominators are same numerators are added with their sign.
= 6 / 13 + (-9) / 13
= (6 – 9) / 13 (integers with opposite sign)
= (-3) / 13
### Question 2. Add the following rational numbers:
(i) 3 / 4 and -5 / 8
Solution:
Denominators are different, so we need to take the LCM of denominators to make them into like fractions.
LCM of 4 and 8 = 8
= 3 / 4 + (-5) / 8
= (3 × 2 + (-5)) / 8
= (6 – 5) / 8
= 1 / 8
(ii) 5 / -9 and 7 / 3
Solution:
Denominators are different, so we need to take the LCM of denominators to make them into like fractions.
LCM of 9 and 3 = 9
= (-5 / 9) + 7 / 3
= (-5 + 7 × 3) / 9
= (-5 + 21) / 9
= 16 / 9
(iii) -3 and 3 / 5
Solution:
Denominators are different, so we need to take the LCM of denominators to make them into like fractions.
= (-3) / 1 + 3 / 5
LCM of 1 and 5 = 5
= ((-3) × 5 + 3) / 5
= (-15 + 3) / 5
= (-12) / 5
(iv) -7 / 27 and 11 / 18
Solution:
LCM of 27 and 18
27 = 3 × 3 × 3
18 = 2 × 3 × 3
LCM = 3 × 3 × 3 × 2 = 54
Therefore,
= (-7) / 27 + 11 / 18
= ((-7 × 2 + 11 × 3)) / 54
= (-14 + 33) / 54
= 19 / 54
(v) 31 / -4 and -5 / 8
Solution:
LCM of 4 and 8 = 8
= ((-31 × 2) + (-5)) / 8
= (-62 – 5) / 8
= (-67) / 8
(vi) 5 / 36 and -7 / 12
Solution:
LCM of 36 and 12 is 36
= 5 / 36 + (-7) / 12
= (5 + (-7 × 3)) / 36
= (5 + (-21)) / 36
= (-16) / 36
4 is the common factor theta can be canceled
= (-4) / 9
(vii) -5 / 16 and 7 / 24
Solution:
LCM of 16 and 24
16 = 2 × 2 × 2 × 2
24 = 2 × 2 × 2 × 3
LCM = 2 × 2 × 2 × 2 × 3 = 48
= (-5) / 16 + 7 / 24
= ((-5 × 3) + 7 × 2) / 48
= (-15 + 14) / 48
= (-1) / 48
(viii) 7 / -18 and 8 / 27
Solution:
LCM of 18 and 27
18 = 2 × 3 × 3
27 = 3 × 3 × 3
LCM = 3 × 3 × 3 × 2 = 54
= ((-7 × 3) + 8 × 2) / 54
= (-21 + 16) / 54
= (-5) / 54
### Question 3. Simplify:
(i) 8 / 9 + -11 / 6
Solution:
LCM of 9 and 6
9 = 3 × 3
6 = 2 × 3
LCM = 2 × 3 × 3 = 18
= (8 × 2 + (-11 × 3)) / 18
= (16 – 33) / 18
= (-17) / 18
(ii) 3 + 5 / -7
LCM of 1 and 7 is 7
= (3 × 7 + (-5)) / 7
= (21 – 5) / 7
= 16 / 7
(iii) 1 / -12 + 2 / -15
Solution:
LCM of 12 and 15
12 = 2 × 2 × 3
15 = 3 × 5
LCM = 2 × 2 × 3 × 5 = 60
= ((-1 × 5) + (-2 × 4)) / 60
= (-5 – 8) / 60
= (-13) / 60
(iv) -8 / 19 + -4 / 57
Solution:
LCM of 19 and 57 is 57
= ((-8 × 3) + (-4)) / 57
= (-24 – 4) / 57
= (-28) / 57
(v) 7 / 9 + 3 / -4
Solution:
LCM of 9 and 4 is 36
= (7 × 4 + (-3 × 9)) / 36
= (28 – 27) / 36
= 1 / 36
(vi) 5 / 26 + 11 / -39
Solution:
LCM of 26 and 39
26 = 13×2
39 = 13×3
LCM = 13 × 2 × 3 = 78
= (5 × 3 + (-11 × 2)) / 78
= (15 – 22) / 78
= (-7) / 78
(vii) -16 / 9 + -5 / 12
Solution:
LCM of 16 and 12
9 = 3×3
12 = 2 × 2 × 3
LCM = 3 × 3 × 2 × 2 = 36
= ((-16 × 4) + (-5 × 3)) / 36
= (-64 – 15) / 36
= (-79) / 36
= (-79) / 36
(viii) -13 / 8 + 5 / 36
Solution:
LCM of 8 and 36
8 = 2 × 2 × 2
36 = 2 × 2 × 3 × 3
LCM = 2 × 2 × 2 × 3 × 3 = 72
= ((-13 × 9) + 5 × 2) / 72
= (-117 + 10) / 72
= (-107) / 72
(ix) 0 + -3 / 5
Solution:
0 is the additive identity, if added to any number gives the same number
= (-3) / 5
(x) 1 + -4 / 5
Solution:
LCM of 1 and 5 is 5
= (1 × 5 + (-4)) / 5
= (5 – 4) / 5
= 1 / 5
### Question 4. Add and express the sum as mixed fraction:
(i) -12 / 5 and 43 / 10
Solution:
LCM is 10
= ((-12 × 2 + 43)) / 10
= (-24 + 43) / 10
= 19 / 10
(ii) 24 / 7 and -11 / 4
Solution:
LCM of 7 and 4 is 28
= (24 × 4 + (-11 × 7)) / 28
= (96 – 77) / 28
= 19 / 28
Proper fraction cannot be converted to mixed fraction
(iii) -31 / 6 and -27 / 8
Solution:
LCM of 8 and 6 is 24
= ((-31 × 4) + (-27 × 3)) / 16
= (-124 – 81) / 24
= (-205) / 24
(iv) 101 / 6 and 7 / 8
Solution:
LCM of 8 and 6 is 24
101 / 6 + 7 / 8
= (101 × 4 + 7 × 3) / 24
= (404 + 21) / 24
= 425 / 24
My Personal Notes arrow_drop_up | 2,215 | 4,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2023-06 | latest | en | 0.881236 |
https://web2.0calc.com/questions/any-help_4 | 1,556,283,301,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578770163.94/warc/CC-MAIN-20190426113513-20190426135513-00256.warc.gz | 570,100,941 | 6,035 | +0
# Any help?
+1
240
2
+12
Alyssa changed 2800 Malaysian Ringgits (MYR) TO US dollars (\$) when the exchange was 1 MYR = \$0.325.
At the end of her holiday she had \$210 left.
(a) How many dollars did she spend?
(b) She changed the \$210 for 750 MYR
What was the exchange rate in dollars for 1 MYR?
Any help is very much appreciated!
Mar 3, 2018
#1
+420
+1
well in part (a) first you need to multiply the initial value with the exchange rate and you'll get 910 US dollars
and in part (b) you need to find the answer for 210/750 and you'll get 0.28
Mar 3, 2018
#2
+18138
+1
1 myr = \$ .325
\$.325/ 1 myr = \$x /.myr2800
.325(2800)= \$910
\$910 - 210 = \$700 dollars spent
\$210 / 750myr = \$x / 1 myr
\$0.28 / myr 28 cents per myr or 1 myr = \$.28 Just as DC found!
Mar 3, 2018
edited by Guest Mar 3, 2018 | 314 | 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} | 3.953125 | 4 | CC-MAIN-2019-18 | latest | en | 0.916475 |
https://edurev.in/chapter/21096_Logical-Reasoning | 1,652,929,239,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522741.25/warc/CC-MAIN-20220519010618-20220519040618-00142.warc.gz | 285,645,811 | 49,350 | # Logical Reasoning - Mock Test Series for CLAT 2022 | CLAT | Notes, Videos & Tests
Logical Reasoning is topic-wise collection of Important notes, Topic Wise tests, Video lectures, NCERT Textbook, NCERT Solution, and Previous Year papers is designed in a way where you get a complete chapter-wise package for your preparation of Mock Test Series for CLAT 2022 in one place? Here, the chapter-wise guide is framed by the best teachers having tremendous knowledge in the respective streams, thereby making the Logical Reasoning - Mock Test Series for CLAT 2022 the ultimate study source for the chapter.
## Notes, Videos & Tests you need for Logical Reasoning
Test: Analytical Reasoning- 1 Test | 40 ques | 30 min Test: Analytical Reasoning- 2 Test | 39 ques | 30 min Test: Logical Reasoning - 1 Test | 30 ques | 25 min Test: Logical Reasoning - 2 Test | 30 ques | 25 min Test: Logical Reasoning - 3 Test | 30 ques | 25 min
## Notes for Logical Reasoning - Mock Test Series for CLAT 2022 | CLAT
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Logical Reasoning | Mock Test Series for CLAT 2022 | 462 | 2,044 | {"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-2022-21 | latest | en | 0.88368 |
https://boards.straightdope.com/t/another-quantum-question-diffraction/630105 | 1,600,440,612,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187899.11/warc/CC-MAIN-20200918124116-20200918154116-00526.warc.gz | 318,984,840 | 5,280 | # Another quantum question: diffraction
How does passing particles like photons through a double-slit experiment produce diffraction when the slit itself is in theory a wave function of the atoms it’s built of? How does a wave diffract a wave?
How does it do it? Like that.
It’s not some fundamental immutable law of waves that they can’t ever interact with other waves at all. In this case, some of the waves are charged, and therefore those waves interact with light waves.
Maybe the wave function of the slit has a much smaller wavelength than light, that’s why its wave nature is not apparent?
In quantum mechanics, things have both wave-like and particle-like characteristics, and any either-or description will fail to capture the full phenomenology. So a quantum object can diffract another quantum object just as well as a particle can diffract a wave.
Personally, though, I find it more intuitive to consider the explanation for this behaviour to be given by a modification of probability theory: the ‘probabilities’ (more accurately, the probability amplitudes) behave like waves, i.e. they can cancel out or reinforce one another, contrary to the situation in ordinary probability theory, where the law of total probability holds. Basically, this is due to the fact that probability amplitudes are complex numbers rather than reals, but this is merely a technical detail. What matters is that therefore, the probability that a photon reaches a certain point on the screen (or in space) is not given by the sum of the probabilities that it reaches that point via either one slit. Rather, the probabilities may interfere destructively, and thus, the probability that the photon reaches some point may be lower than the probability that it reaches that point via slit 1 plus the probability that it reaches that point via slit 2; and thus, interference patterns and diffraction occurs, without having to talk about whether the photon is a particle, a wave, or something else entirely.
What I was getting at was classical diffraction assumes that the hole in a barrier that a wave is passing through has infinitely sharp edges- that at any infinitesmal point in space either the wave passes through or is blocked. But if in real life there’s some quantum fuzziness at the edges (or worse, you can’t really think of the barrier and the slit as classically solid), then how does that affect diffraction? Is the interference pattern slightly different than a classical experiment due to these effects?
I believe it’ll be the same interference pattern – the pattern consists of a great many individual photons, which must be averaged over; but in this average, the edges of the slit, even if they’re considered as quantum objects, will be at the same ‘position’ as classically. | 562 | 2,788 | {"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} | 2.6875 | 3 | CC-MAIN-2020-40 | latest | en | 0.945122 |
https://solvedlib.com/n/solve-the-following-assignment-problem-hungarian-method,21303233 | 1,657,077,562,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104660626.98/warc/CC-MAIN-20220706030209-20220706060209-00157.warc.gz | 563,207,634 | 19,614 | # Solve the following Assignment problem (Hungarian Method) Max Z= 60 X1 + 5 X12 + 8 X13 + X14 +
###### Question:
Solve the following Assignment problem (Hungarian Method) Max Z= 60 X1 + 5 X12 + 8 X13 + X14 + 2 X21 + [2 22 + 6 X23 + 5 X24 X31+ 8 X32 + 3 X33 + 9 X34 + 2 X41 + 4 X42 + 6 X43 + 10 X44 s.t. X11 + X12 + X13 + X14 = X21 + X22 + X23 + X24 = X31 ~ X32 + X33 + X34 = X41 + X42 + X43 + X44 = X1l + X21 + X31+ X41 = X12 + X22 ~ X32 + X42 = X13 + X23 + X33 + X43 = X14 + X24 + X34+ X44 = xii = {0, 1}
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3. For the following primal, write the dual and use it to find one Or two lower bounds for the primal value: 2 4-[J x>0, X1 +X2 + X3 = min: Consider this problem: 2 < X1 + 2xz + 9x3 <1, x>0, ~Ix + 9x2 16x3 max: Write the dual, and get an upper bound for the primal maximum: Basic Solution...
##### Rectangular coil with dimensions 0.12 m by 0.29 m has 570 turns of wire. It rotated about long axis in magnetic field of 0.52 T, At what frequency must the coil be rotated fOr generate manimum potential of 130 V?
rectangular coil with dimensions 0.12 m by 0.29 m has 570 turns of wire. It rotated about long axis in magnetic field of 0.52 T, At what frequency must the coil be rotated fOr generate manimum potential of 130 V?... | 623 | 1,753 | {"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": 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} | 4.1875 | 4 | CC-MAIN-2022-27 | latest | en | 0.703964 |
https://www.jiskha.com/questions/1363702/Graysen-has-a-green-number-cube-and-a-white-number-cube-The-faces-of-the-cubes-are | 1,534,666,788,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221214713.45/warc/CC-MAIN-20180819070943-20180819090943-00637.warc.gz | 914,199,180 | 4,483 | # math
Graysen has a green number cube and a white number cube. The faces of the cubes are numbered 1 through 6. Graysen will roll each cube one time. what is the probability that the green cube will land with an even number faceup and the white cube will land with a number greater than 2 faceup
1. pr=prgreen even*prb>2=
= prgreen even*(1-pr(1 or 2))
= 3/6* (1-1/6-1/6)=3/6(4/6)=1/3 check that.
posted by bobpursley
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