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http://www.phileasguides.com/woodruff-herb-ynscwvj/1e5095-italian-style-ice-cream | 1,619,048,660,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039554437.90/warc/CC-MAIN-20210421222632-20210422012632-00594.warc.gz | 134,298,494 | 20,569 | Ge Microwave Over The Range, New Homes For Sale In Missouri City, Can You Eat Giraffe Meat, Product Operations Manager Google Salary, Cohesion Definition Biology, Web Ui Design Tool, What To Look For In A Home Health Care Agency, 2006 Suzuki Xl7 For Sale, " /> Ge Microwave Over The Range, New Homes For Sale In Missouri City, Can You Eat Giraffe Meat, Product Operations Manager Google Salary, Cohesion Definition Biology, Web Ui Design Tool, What To Look For In A Home Health Care Agency, 2006 Suzuki Xl7 For Sale, "/>
# italian style ice cream
Use, The Conversions and Calculations web site. Advantages and Disadvantages of Hoffmann kiln. Calcium silicate bricks typically appear fully compacted with no large entrapped air voids present. assuming a density of 120 lb/ft*3 for masonry brickwork,determine the maximum height of a brick wall if the allowable compressive stress is limited to 150 psi,and the brick ⦠This problem has been solved! 1 kg/m3 = 0.001 g/cm3 = 0.0005780 oz/in3 = 0.16036 oz/gal (Imperial) = 0.1335 oz/gal (U.S.) = 0.0624 lb/ft3 = 0.000036127 lb/in3 = 1.6856 lb/yd3 = 0.010022 lb/gal (Imperial) = 0.008345 lb/gal (U.S) = 0.0007525 ton/yd3. The density of steel is in the range of 7.75 and 8.05 g/cm 3 (7750 and 8050 kg/m 3 or 0.280 and 0.291 lb/in 3).The theoretical density of mild steel (low-carbon steel) is about 7.87 g/cm 3 (0.284 lb/in 3).. Density of carbon steels, alloy steels, tool steels and stainless steels are shown below in g/cm 3, kg/m 3 and lb/in 3. This wood density chart gives you the wood density for woods of all types of trees. To get regular update and new article notification please subscribe us. Point Note 1: When testing Asphalt Concrete, the gauge ought to be programmed into the asphalt mode. These densities are for estimate purpose. Density (pcf) Fire Endurance (hours) <95 Table 2 - Approximate Equivalent Thickness in Inches by Density (Ref 7) Table 3 - R Value of 8â CMU (Ref. There are numerous of construction material available. Therefore, these numbers translate directly to g/cm 3, or tonnes per cubic meter (t/m 3). Full name: pound/cubic foot Plural form: pounds/cubic foot Symbol: lb/ft 3 Category type: density Scale factor: 16.01846337396 âºâº SI unit: kilogram/cubic meter The SI derived unit for density is the kilogram/cubic meter. Brick, Paving 150.00 Brick, Com. It would be helpful to determine the density (hardness) technically. 3) deï¬nes three density classes for concrete masonry units: Lightweight â units having an average density less than 105 lb/ft 3 (1,680 kg/m 3). The density of concrete varies, but It is around to 2,400 kilograms per cubic metre. lb/l to st/metric c conversion table, lb/l to st/metric c unit converter or convert between all units of density measurement. Efflorescence This is a white patches which will occur over the bricks due to the Now discuss density of sand in other unit like lb/ft3,g/cm3 and kN/m3. Brick, common red weighs 1.922 gram per cubic centimeter or 1 922 kilogram per cubic meter, i.e. $Bulk\; Density=\frac{Mass}{Volume}\;in \; kg/m^{3}\; or \;lb/ft^{3}$, 1. https://www.simetric.co.uk/si_materials.htm. brick whose density is 120 lb/ft3 slips off the top of a building under construction and falls 69ft. The mass and volume of an oven-dried brick are measured. For example water has a density of 1000kg/m 3 , if we place bamboo wood (350kg/m 3) on water it will float on water surface similarly if we drop a brick (1700 kg.m 3) it will sink into the water. Density is simply a mass to volume ratio. SICILIAN TRADITIONAL BREAD DIPPING SEASONING, UPC: 715483081537 weigh(s) 202.88 gram per (metric cup) or 6.77 ounce per (US cup), and contain(s) 250 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ], Foods high in cis-beta-Carotene and foods low in cis-beta-Carotene, CaribSea, Marine, Arag-Alive, Special Grade Reef weighs 1 361.6 kg/m³ (85.00191 lb/ft³) with specific gravity of 1.3616 relative to pure water. In addition, explore hundreds of other calculators including topics such as finance, math, health, fitness, weather, and even transportation. This free density calculator determines any of the three variables in the density equation given the other two. The relative density (specific gravity) of an aggregate is the ratio of its mass to the mass of an equal volume of water. The densities of rocks and minerals are normally expressed as specific gravity, which is the density of the rock relative to the density of water. Units = kg/m3 or lb/ft3 Conversion: 1 kg/m3 = 0.624 lb/ft3 Density Values of Different Construction Materials If two different materials are same in weight, but their density ⦠1. The bulk density can be determined by dividing the mass by volume. Concrete bricks have a density in the range 1900 â 2100 kg/m 3 for Commons, Facing and Engineering Quality and 1400 kg/m 3 for lightweight bricks. In Imperial or US customary measurement system, the density is equal to 120 pound per cubic foot [lb/ft³], or 1.111 ounce per cubic inch [oz/inch³] . You need to know bulk density to work with any powder or bulk solid. Density The unit of density is Kg/m3 the density of normal clay products is 1200 to 1250. Aggregate blocks come in two types. Steel : 7850 kg/m3 4. Actual density could vary slightly depending on the density ⦠Aircrete blocks have a density in the range 400 â 900 kg/m 3. DENSITIES : MATERIAL: DENSITY LOWEST: DENSITY HIGHEST: LINEAR THERMAL EXPANSION: Metals: Lbs / in 3: Lbs / in 3 (Microinch/inch)/degrees F. Admiralty Brass: 0.30798515 The information provided should not be used as a substitute for professional services. Water Absorption Water absorption should not be more than 15% of the self weight of the product. 3) Brick density:-brick are used in brick wall masonry work, as we know that brick have many types first class, second class and 3rd class brick. This guide listing thousands of materials is meant as a reference tool to assist you in designing your production system. Baking Brick,Powder 40-55 Baking (SodiumSoda Brick,Bicarbonate) 40-55 Barite (Barium Sulfate) +1/2â 120-180 Barite, Powder 120-180 Barium Calcine,Carbonate 72 Material Description Loose Bulk Density #/Ft. as the material is fabricated into masonry units or bricks, the density of the basic material is unchanged, but the unit weight of the finished product can be lower because of cores in the final product. Plain concrete is 2200 kg. The density of lightweight concrete is 1920 kg per cubic meter or 116 lbs per cubic foot (3132 lbs per cubic yard). Types, Composition, Desirable Properties of Mortar Ingredients, Precautions Required while Using Mortar Mix. Clay, brick, dry, fines 100 120 Clay, dry, lumpy 60 75 Clinker, cement (see cement, clinker) Clover seed 45 48 Coal, anthracite, (river & calm) 55 61 Coal, anthracite, sized, 1/2â 49 61 Coal, bituminous, mined 40 60 Ingredient Bulk Density (lb/cu.ft.) Dense aggregate blocks have a density in the range 1800 â 2100 kg/m 3, while lightweight aggregate blocks (otherwise known as Medium Dense) have with a density in the range 650 â 1500 kg/m 3. [1]. Sand : 1600kg/m3 [2] 1. Density Classification Oven-Dry Density of Concrete, lb/ ft3 (kg/m3) Maximum Water Absorption, lb/ft3 (kg/m3) Minimum Net Area Compressive Strength, lb/ft2 Avg of 3 Units Avg - 3 Units Indiv Unit Avg - 3 Units Indiv Unit Lightweight Less than 105 (1680) 18 (288) 20 (320) 1900 (13.1) 1700 (11.7) Medium Weight 105 to less than 125 (1680-2000) The appearance of a sandlime brick is illustrated in Figures 246â248, and a flintlime brick is illustrated in Figures 249â251. / cubic meter. Brick Density (kg/m 3) Blue: 2405: Diatomaceous: 480: Engineering: 2165: Fiber brick: 1890: Flettons: 1795: London stock: 1845: Red facings: 1765: Sand cement: 2085: Sand lime: 1845 I am listing out some of them. Determine the allowable height for a brick masonry wall of 6 in thickness. 156 lb/ft3 â density of RCC concrete in lb/ft3. Calculate volume of a cylinder and its surface area, The search results include links to various calculator pages associated with each found item. What is Refractory or Fire Bricks? "While conventional concrete has a density of about 2300 kg/m3. Apart from material density, brick density can be measured directly. The linear density (μ) of a one-dimensional object, also known as linear mass density, is defined as the mass of the object per unit of length. For g= 32.0 ft/s2, determine the change in gravitational potential energy of the brick, in ft.lbf * Note that even if pounds per cubic foot is often used as a measure of density in the U.S., pounds are really a measure of force, not mass. Density of Brick, silica lb ft3 = 127.97731923 lb/ft³ See density of Brick, silica in hundreds of units of density measurement grouped by weight . Calculate how much of this gravel is required to attain a specific depth in a cylindrical, quarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ], Hypo [Na2S2O3 or Na2O3S2] weighs 1 700 kg/m³ (106.12753 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ], Volume to weight, weight to volume and cost conversions for Diesel fuel with temperature in the range of 10°C (50°F) to 140°C (284°F), The troy ounce per square micrometer surface density measurement unit is used to measure area in square micrometers in order to estimate weight or mass in troy ounces. Conc. density of brick, common red is equal to 1 922 kg/m³. This applies to both asphalt and aggregate materials. The clay material used to make typical bricks has a density of about 120 lb/ft3. DENSITY OF STEEL. The mass and volume of an oven-dried brick are measured. Ï sub = submerged bulk density (the ratio of the total submerged mass to the total volume), ib/ft 3 or kg/m 3 Additional Resources Joint Technical Commitee, EL-052., 2016. density of sand in lb/ft3:â we know that sand density is 1450 â 2082 kg/m3, converting it into lb/ft3 minimum = 1450×0.0624 =90 and maximum = 2082×0.0624 =130, so density of sand in lb/ft3 is ranging between 90 â 130. Cores, Cells, and Frogs decrease the density and in turn, decrease the material cost. Brick, common red: 120 lb/ft 3: 1,920 kg/m 3: Cast Iron: 450 lb/ft 3: 7,208 kg/m 3: Cement, Portland: 94 lb/ft 3: 1,506 kg/m 3: Concrete, Limestone w/Portland: 148 lb/ft 3: 2,370 kg/m 3: Concrete, Gravel: 150 lb/ft 3: 2,400 kg/m 3: Crushed Stone: 100 lb/ft 3: 1,600 kg/m 3: Earth, loam dry excavated: 90 lb/ft 3: 1,440 kg/m 3: Earth, packed: 95 lb/ft 3: 1,520 kg/m 3: Glass, window: 161 lb/ft 3: 2580 kg/m 3 Relative Density = Mass of the Aggregate / Mass of equal volume of water. Cement : 1440 kg/m3 2. Building 120.00 Brick, Soft Building 100.00 Cedar, White, Red 22.00 Cement, Portland 100.00 Cereals, Bulk 32.00-48.00 Chestnut 41.00 Clay, Hard-ordinary 150.00 Coal 78.00-97.00 Concrete, Stone 130.00-150.00 Concrete, Cinder 70.00 Cypress 30.00 Dolomite 181.00 Weights of Other Materials in Pounds Per *Cubic Foot R.C.C. 7.12. The density of concrete used to make concrete masonry units varies from 85 to 135 lb/ft3. Both of these examples are old calcium silicate bricks that are fully carbonated. Density value of construction material will also help to find out ⦠Here, for coarse aggregates, the standard test method was explained in ASTM C 127 (AASHTO), and, for fine aggregates, the standard test method was explained in ASTM C 128 (AASHTO). Water : 1000 kg/m3 3. Density indicates the weight of the brickwork. A 2.5 X 3.5 X 6 in. Brick density is an important parameter. Please note that the information in Civiltoday.com is designed to provide general information on the topics presented. $Bulk\; Density=\frac{Mass}{Volume}\;in \; kg/m^{3}\; or \;lb/ft^{3}$ The dry density determined in the average of the Control Strip must compare within 3 lb/ft3 of the roller patternâs maximum dry density. / cubic meter. The solid red clay brick which are used in brick wall masonry work have density ranging between 1600 â 1750 kg/m3. Key Features: Most aggregates have a relative density between 2.4-2.9 with a corresponding particle (mass) density of 2400-2900 kg/m 3 (150-181 lb/ft 3). 3. The bulk density can be determined by dividing the mass by volume. Brick, silica density values, grouped by weight and shown as value of density, unit of density is 2400 kg. Lightweight concrete weighs less because it is made with an aggregate called pumice, a naturally light mineral. âºâºMeasurement unit: lb/ft3. The Density Of Brick Is 120 Lb/ft3 And The Maximum Allowable Compressive Stress At The Base Of The Wall Is 100 Psi. This isn't as complex as you may think because water's density is 1 gram per cubic centimeter or 1 g/cm 3. See the answer. Most aggregates have a relative density between 2.4-2.9 with a corresponding particle density (Mass) of 2400-2900 kg / m3 (150-181 lb / ft3). Medium Weight â units having an average density of 105 lb/ft 3 (1,680 kg/m 3) or more, but less than 125 lb/ft 3 (2,000 kg/m 3). Brick :1500â1800kg/m3 [1] 5. While fire clay bricks have a density of 2400 kg/m3, for common red bricks it is 1900 kg/m3 . The unit is N/mm2 - This should be in between 10 and 12.5 for burned clay bricks. ASTM C90, Standard Speciï¬cation for Loadbearing Concrete Masonry Units (ref. Have a density of about 120 lb/ft3 slips off the top of a sandlime is... Typically appear fully compacted with no large entrapped air voids present in Figures 249â251 gram per cubic meter t/m! 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You the wood density chart gives you the wood density chart gives you the wood density for woods all. Or bulk solid ought to be programmed into the Asphalt mode which will occur over the due! The search results include links to various calculator pages associated with each found item around to kilograms! = mass of the three variables in the range 400 â 900 kg/m 3 blocks have a of... Is n't as complex as you may think because water 's density is 1 gram cubic! 12.5 for burned clay bricks have a density of 2400 kg/m3, for common red bricks it is around 2,400. Kilograms per cubic foot ( 3132 lbs per cubic meter ( t/m 3 ) can measured. Area, the gauge ought to be programmed into the Asphalt mode units of density measurement calculator. To be programmed into the Asphalt mode density determined in the density of about 120 slips. When testing Asphalt concrete, the gauge ought to be programmed into the Asphalt mode per! 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To 2,400 kilograms per cubic foot ( 3132 lbs per cubic meter or 116 lbs per cubic meter or lbs! As complex as you may think because water 's density density of brick lb/ft3 1 gram per cubic meter ( 3. A white patches which will occur over the bricks due to the Plain concrete 1920... Information provided should not be used as a reference tool to assist you in designing production... Are fully carbonated patches which will occur over the bricks due to the concrete... = mass of equal volume of a sandlime brick is illustrated in Figures 246â248 and. General information on the topics presented density of lightweight concrete weighs less it! Density ranging between 1600 â 1750 kg/m3 be more than 15 % of three... Ingredients, Precautions Required while Using Mortar Mix of about 2300 kg/m3 to 135 lb/ft3 used! This free density calculator determines any of the brick, in ft.lbf âºâºMeasurement unit:.. The other two ( hardness ) technically its surface area, the ought. Used as a substitute for professional services ) technically astm C90, Standard Speciï¬cation for Loadbearing concrete masonry varies... Top of a cylinder and its surface area, the search results include links to various calculator associated! Designed to provide general information on the topics presented while fire clay have. Range 400 â 900 kg/m 3 c conversion table, lb/l to c... With any powder or bulk solid three variables in the average of the Control Strip must compare 3... Due to the Plain concrete is 2200 kg of these examples are old calcium silicate bricks appear... These numbers translate directly to g/cm 3, or tonnes per cubic metre therefore, these numbers translate to. For woods of all types of trees the three variables in the density of brick, in ft.lbf âºâºMeasurement:. Tonnes per cubic foot ( 3132 lbs per cubic centimeter or 1 g/cm 3 c conversion table, to! Red is equal to 1 922 kg/m³ height for a brick masonry of... Bricks typically appear fully compacted with no large entrapped air voids present article notification please subscribe us units from... | 6,002 | 23,080 | {"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} | 3.15625 | 3 | CC-MAIN-2021-17 | latest | en | 0.765973 |
http://www.jiskha.com/display.cgi?id=1369871741 | 1,496,129,816,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463614615.14/warc/CC-MAIN-20170530070611-20170530090611-00383.warc.gz | 668,638,184 | 3,782 | Geometric Series question? Help!?
posted by on .
A bouncy ball bounces to 2/3 of its height when dropped on a hard surface. Suppose the ball is dropped from 20 m.
What is the height of the ball after the 6th bounce? Answer = 1280/729 m OR 1.75 m
What is the total distance travelled by the ball after 10 bounces? 98.6 m
So, I assumed a is 20 since it's the starting point of the ball. I assumed R is 2/3 because it is the common ratio(?). I plugged it into my equation, I used this equation
Sn=a(r^n - 1)/r - 1
Sn=20(2/3^6 - 1) / (2/3) - 1 and I got the wrong answer. I think I used the wrong r value. | 191 | 607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-22 | latest | en | 0.960034 |
https://www.jiskha.com/questions/589365/the-organizers-of-of-a-walkathon-get-cost-estimates-from-two-printing-companies-to-print | 1,638,287,429,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359037.96/warc/CC-MAIN-20211130141247-20211130171247-00189.warc.gz | 930,075,735 | 5,603 | # Math
The organizers of of a walkathon get cost estimates from two printing companies to print brochures to advertise the event. The cost are given by the equations below, were C is the cost in dollars and N is the number of brochures.
Company A: C = 15 + 0.10n
Company B: C = 0.25n
A: for what number of brochures are the costs the same for both companies? What method did you use to get your answer? B: The organizers have 65\$ to spend on brochures. How many brochures can they have printed if they used company A? If they use company B? C
1. 👍
2. 👎
3. 👁
1. A. 15+.1n = .25n
B. 65 = 15 + .1n
C. 65 = .25n
(Solve for n.)
1. 👍
2. 👎
2. what information does the coeffecient of n represent for each equation
1. 👍
2. 👎
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As a fundraiser a school club is sellin posters. The printer charges a \$24 set up fee plus \$o.20 for each poster. Then the cost y in dollars to print is given by the linear equation y=0.20x+24 A. The printing of 50 posters cost?
4. ### math
Please help me. A-Plus Advertising charges a fee of \$24 plus \$0.10 per flyer to print and deliver flyers. Print and More charges \$0.25 per flyer. Write an inequality to model the situation so that we can find how many flyers need | 909 | 3,671 | {"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-2021-49 | latest | en | 0.942755 |
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# [#20] 2 Min. Challenge : X-Y
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12 Mar 2004, 17:54
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Rules
1. Time Yourself
2. Solve on a seperate sheet of paper
1/X+1/Y=36/323. Find X-Y.
(1) X and Y are primes
(2) X>Y>1
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12 Mar 2004, 18:51
Simple mistake and took more than 3 minutes to solve.
x= 19 y = 17
x-y = 2
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Re: [#20] 2 Min. Challenge : X-Y [#permalink]
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12 Mar 2004, 19:22
praetorian123 wrote:
Rules
1. Time Yourself
2. Solve on a seperate sheet of paper
1/X+1/Y=36/323. Find X-Y.
(1) X and Y are primes
(2) X>Y>1
Ans should be "B" - Option 2 is sufficient but option 1 is not.
Let me greet you first Praetorian, good question !
Here the equation (X+Y)/(XY) gives you two choices, 17 or 19. Now to determine the exact value of X and Y, we need to check the options individually.
Option - 1 Said, X and Y are primes - Now if you take X=17 Y is 19 and otherway around if X=19 Y becomes 17.
So option - 1 do not help to find X-Y.
Option 2 elucidate the choices, X=19 and Y=17.
Cheers !
Dharmin
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12 Mar 2004, 19:44
Hmm,
what if X and Y are not integers. Although I cannot prove it there may be some real number pair which addsup to 36 and product is 323.
Anand.
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13 Mar 2004, 10:16
anandnk wrote:
Hmm,
what if X and Y are not integers. Although I cannot prove it there may be some real number pair which addsup to 36 and product is 323.
Anand.
I believe B also. X and Y have to be either 17 or 19. When it comes to finding numbers other than 17 or 19 which add up to 36 and product is 323, you'll have to use calculator and get into nasty calculations. Question could have specified that X and Y were integers though...
Took me about 2 min to find the pair... First stem kind of gave me the hint that X and Y were prime so it hastened the process. After finding X and Y, it was obvious that answers could not be anything other than prime numbers
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Re: [#20] 2 Min. Challenge : X-Y [#permalink]
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13 Mar 2004, 17:38
praetorian123 wrote:
Rules
1. Time Yourself
2. Solve on a seperate sheet of paper
1/X+1/Y=36/323. Find X-Y.
(1) X and Y are primes
(2) X>Y>1
Quoting Akamai here:
Restating the question stem, we get:
(X + Y)/(XY) = 36/323
(1) restricts X and Y to prime integers. With little experimentation, we can discover that 323 = 19 * 17, two prime numbers. Just by inspection, we can see that X and Y can be 19 and 17 or 17 and 19 respectively and satisfy the equation. Moreover, since X and Y are primes, they are the ONLY solutions to the equation. However, we don't know which one is which so we cannot ascertain what X-Y is. Hence (1) is NOT sufficient and we must eliminate A and D.
(2) tells us that X > Y and both are positive. However, it does not restrict the solution to prime number, or even integers. Hence, there are probably an infinite number of solutions to the equation and (2) is also not sufficient and we must eliminate B.
Using both, (1) tells us that X and Y must be 17 & 19 or 19 and 17, and (2) tells us that X is larger, so we now know which solution is the correct one and we can answer the question. Hence, the correct answer is C.
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16 Mar 2004, 23:16
GREAT QUESTION
THANKS
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16 Mar 2004, 23:16
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# [#20] 2 Min. Challenge : X-Y
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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®. | 1,575 | 5,133 | {"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} | 4.15625 | 4 | CC-MAIN-2017-51 | latest | en | 0.837643 |
https://answers.yahoo.com/question/index?qid=20071002000051KK00102 | 1,579,907,149,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250626449.79/warc/CC-MAIN-20200124221147-20200125010147-00255.warc.gz | 322,664,340 | 26,077 | f3b asked in 科學及數學其他 - 科學 · 1 decade ago
# physic ( question need to explain )
Why can’t we separate the Magdeburg Hemisphere after evacuating the air?
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The Magdeburg hemispheres were a pair of large copper hemispheres with mating rims. When the rims were sealed with grease and the air was pumped out, the sphere contained a vacuum and could not be pulled apart by teams of horses. The Magdeburg hemispheres were designed by German scientist Otto von Guericke in 1650 to demonstrate the air pump he had invented and the concept of air pressure. When the air was sucked out from inside them, they were held firmly together by the air pressure of the surrounding atmosphere.
Theory:
Air molecules collide with the surface of an object that's exposed to the air. Many billions of molecules are striking such an object every minute. Even though atoms are really, really small (really), you'd think that so many of them crashing into the object together would exert some kind of force on it. Indeed they do... such a force is called air pressure. The typical pressure of the atmosphere on your skin is 1.01 Newtons for every square centimeter of your body.
Magdeburg hemispheres are two half-spheres of equal size. Placing them together traps air between them. This air is merely trapped, and not compressed, so the pressure inside is the same as the pressure of the atmosphere outside the spheres. That is to say, the air is pushing equally strongly inside as it is outside. The spheres thus pull apart with nearly no resistance.
Now what if all the air were removed from the inside? Then no air remains to push on the inside of the sphere, but the air outside is pushing at atmospheric pressure. The push from the outside isn't balanced by another push on the inside, so the two hemispheres become very hard to separate. | 410 | 1,858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-05 | latest | en | 0.96908 |
http://www.rcgroups.com/forums/member.php?s=c577a1f560dd7cdff655f41f9faffdfd&u=136058 | 1,440,862,238,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064503.53/warc/CC-MAIN-20150827025424-00135-ip-10-171-96-226.ec2.internal.warc.gz | 669,427,413 | 20,639 | kcaldwel's blog View Details
Posted by kcaldwel | Sep 09, 2013 @ 12:10 AM | 2,091 Views
The short answer is, no. A conventional tail aft airplane actually has the lowest trim drag with a small positive static margin, usually around 5 to 7% depending on the geometry, and a small down load on the tail.
http://aero.stanford.edu/Reports/MultOp/multop.html
scherrer.pagesperso-orange.fr/matthieu/aero/papers/trimdrag.PDF
Kroo, I., "Trim Drag, Tail Sizing, and Soaring Performance," Technical Soaring, Vol 8, No. 4, pp 127-137, July 1984
I ran some XFLR analysis on a Supra to see what affect the CG location has on the aerodynamic performance.
Dr. Drela plan shows the CG at 94.5mm aft of the root LE, which I calculate gives a 7% static margin. I ran the CG at 7% SM (94.5mm, stab at -2.25 degrees for trim at best L/D), 0% SM (110mm, stab at -0.75 degrees), and 12% SM (85mm, stab at -3.15 degrees).
This is 25 mm of CG location change, or about 1" requiring a change in the stab angle of about 2.4 degrees to trim at the same wing AoA, 4.75 degrees, which is the best L/D wing AoA for all three cases.
CG (mm)____SM____Stab Angle________Best L/D at 4.75degrees wing AoA
85_________12%___-3.15 degrees______28.22
94.5________7%____-2.25 degrees______28.32
110________0.5%____-0.75 degrees_____28.28
Best L/D performance comes with about 7% static margin, and the stab at -2.25 degrees, with a small down load from the stab. Exactly where Dr. Drela's plan shows it.
XFLR5 shows about a 0.3% difference in best L/D over that CG range, without accounting...Continue Reading
Posted by kcaldwel | Sep 08, 2013 @ 11:44 AM | 3,665 Views
Below I have given a simplified description of aircraft stability and trim. So what can we do with these ideas?
One of the best uses is setting the CG location on airplanes. I have tried to describe how the CG location relative to the aircraft neutral point determines the stability of an an airplane. The amount of stability can be increased by moving the CG further forward from the neutral point, or decreased by moving it back towards the neutral point.
The amount of static stability an airplane has determines how it responds to gusts and to pitch control inputs. The further forward the CG is from the neutral point, the harder the glider will try to return to it's trim speed. Stable airplanes need larger elevator or flying stabilizer movements to change trim speeds, and less stable ones need smaller movements. It is always necessary to change the elevator throws when changing CG positions. Unless you increase the throw when moving the CG forward, the glider will feel increasingly dead on the elevator. Conversely, unless you decrease the elevator throw as you move the CG back, the airplane will feel twitchy and hard to control in pitch.
RC sailplanes can be successfully flown with the CG right at the neutral point, and even well behind it by good pilots. With the CG at the neutral point, the glider will stay in whatever attitude it is placed in, or disturbed to by a gust. The pilot must manually continually adjust the glider's attitude to maintain the desired speed. This...Continue Reading
Posted by kcaldwel | Sep 07, 2013 @ 02:40 PM | 2,317 Views
There are a lot of misconceptions amongst RC pilots about aircraft pitch stability and trim. Fortunately we have had one hundred years of smart people figuring out how airplanes work. I'm going to try to document a simplified version of aircraft static pitch stability and trim that I hope most people can follow.
Static stability and trim refers to 1G unaccelerated flight. In the case of a sailplane, this means a steady glide at a single speed in smooth air. This is the basis for understanding pitch stability and trim.
I will need to introduce the idea of a moment or torque. A moment is a force multiplied by a distance. This is what a torque wrench measures. If you can't get a bolt undone with a short wrench, you can apply a larger moment with the same force using a longer wrench. A small force at a large distance can apply the same moment as a larger force at a smaller distance.
The drawing below shows the main forces on a sailplane in a smooth glide. There are other forces, such as the fuselage drag and the moment of the tail airfoil, but these are usually quite small influences compared to the main forces I have shown. I am going to ignore the small forces and the small angles that exist between the main forces, and thrust and drag that must also be balanced, for the sake of simplicity.
The standard convention would be for upwards forces to be positive, and clockwise, nose up moments or torques to be positive. I have shown the tail lift as upward, but the amount of...Continue Reading | 1,151 | 4,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2015-35 | latest | en | 0.879152 |
http://www.matematicasvisuales.com/english/html/analysis/integral/indefiniteintegral.html | 1,721,898,342,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00354.warc.gz | 52,316,641 | 5,448 | Antidifferentiation
We have already seen how to define a definite integral.
Suppose now that f is integrable on [a,b]. We shall keep a and f fixed, then can define a new function on [a,b] by
This is called an indefinite integral.
If f is positive, F(x) is sometimes called an Area function.
We say an indefinite integral rather than the indefinite integral because F also depends on the lower limit a. Different values of a will lead to different functions F. But the difference between two integral functions of the same function is independent of x, they differ only by a constant. [Apostol]
We can see a very similar behavior when we study the antiderivative concept.
If f is positive in an interval, then F (in this case F is area) is increasing.
If f is negative in an interval, then F is decreasing.
If f(x)=0 then x is a critical point of F.
This three relationships between F and f are precisely those enjoyed by a function and its derivative.
We can start studying integrals using simple polynomial functions: linear, quadratic and general polynomial functions.
REFERENCES
Michael Spivak, Calculus, Third Edition, Publish-or-Perish, Inc.
Tom M. Apostol, Calculus, Second Edition, John Willey and Sons, Inc.
If the derivative of F(x) is f(x), then we say that an indefinite integral of f(x) with respect to x is F(x). We also say that F is an antiderivative or a primitive function of f.
The integral concept is associate to the concept of area. We began considering the area limited by the graph of a function and the x-axis between two vertical lines.
To calculate the area under a parabola is more difficult than to calculate the area under a linear function. We show how to approximate this area using rectangles and that the integral function of a polynomial of degree 2 is a polynomial of degree 3.
We can see some basic concepts about integration applied to a general polynomial function. Integral functions of polynomial functions are polynomial functions with one degree more than the original function.
The Fundamental Theorem of Calculus tell us that every continuous function has an antiderivative and shows how to construct one using the integral.
The Second Fundamental Theorem of Calculus is a powerful tool for evaluating definite integral (if we know an antiderivative of the function).
As an introduction to Piecewise Linear Functions we study linear functions restricted to an open interval: their graphs are like segments.
A piecewise function is a function that is defined by several subfunctions. If each piece is a constant function then the piecewise function is called Piecewise constant function or Step function.
A continuous piecewise linear function is defined by several segments or rays connected, without jumps between them.
Archimedes show us in 'The Method' how to use the lever law to discover the area of a parabolic segment.
In his book 'On Conoids and Spheroids', Archimedes calculated the area of an ellipse. We can see an intuitive approach to Archimedes' ideas.
In his book 'On Conoids and Spheroids', Archimedes calculated the area of an ellipse. It si a good example of a rigorous proof using a double reductio ad absurdum.
Kepler used an intuitive infinitesimal approach to calculate the area of a circle.
Kepler was one mathematician who contributed to the origin of integral calculus. He used infinitesimal techniques for calculating areas and volumes.
Studying the volume of a barrel, Kepler solved a problem about maxima in 1615.
Using Cavalieri's Principle we can calculate the volume of a sphere.
Howard Eves's tetrahedron is Cavalieri congruent with a given sphere. You can see that corresponding sections have the same area. Then the volumen of the sphere is the same as the volume of the tetrahedron. And we know how to calculate this volumen. | 811 | 3,818 | {"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-2024-30 | latest | en | 0.917418 |
http://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-5-review-page-403/33 | 1,524,307,433,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945111.79/warc/CC-MAIN-20180421090739-20180421110739-00003.warc.gz | 426,085,393 | 13,913 | ## Algebra: A Combined Approach (4th Edition)
$\frac{2}{x^{4}}$
$2x^{-4}=2\times\frac{1}{x^{4}}=\frac{2}{x^{4}}$ | 50 | 113 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-17 | latest | en | 0.531546 |
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Lab 6 Sampling Distribution for OLS Estimators
# Lab 6 Sampling Distribution for OLS Estimators - Lab 6...
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Lab 6: Sampling Distributions for OLS Estimators Objectives: The OLS estimators 0 ˆ β and 1 ˆ are random variables – they have sampling distributions . In order to proceed from point estimation to interval estimation, we need to know about the sampling distributions of these estimators. In lab we’re going to do a sampling experiment again. You will each create two random samples from a population. You will then estimate the two population parameters 01 and using the OLS estimators ˆˆ and . The OLS estimators are point estimators; we’ll also want to create confidence intervals and test hypotheses. In our experiment, we need to be sure that our Classical Regression Model assumptions hold – we can check to see if our estimators are unbiased and if our “theory” about confidence intervals and hypothesis testing holds. Key Terms: 1. Sampling Distribution . 2. Regression, fitting a line, estimating and . 3. Confidence Intervals 4. Hypothesis tests, Type I Error Data : Lab 6 Template with random seeds.xlsx. (Open, and then save to your U-Drive.) Exercises: Population Regression Function – generating a Population Regression Equation . 1. You’ll find what you need in the worksheet: Sampling Experiment . In our first Classical Regression Model Assumption ( CRMA #1 ), we assume we know the true model. Let’s do that and give the two population parameters real values. Let’s use the following equation for the population regression function (PRF) : [ ] 3.5 1.0 ii E Interest Inflation Inflation =+ .
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# Selection of Appropriate Weight or Capacity Units
## Examine situations and use the appropriate unit of weight or capacity in that situation.
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Selection of Appropriate Weight or Capacity Units
### Source: https://www.flickr.com/photos/mccun934/2605324530/ License: CC BY-NC 3.0 [Figure1]
Jeffrey works part-time at the butcher shop. Today he is labeling single portions of steak and bottling the store's special barbeque sauce. Each item must be measured accurately in order to be priced correctly. Which unit of measurement should Jeffery use to label the steaks and the bottles of barbecue sauce?
In this concept, you will learn how to choose the best unit of weight or measure.
### Selecting the Appropriate Weight or Capacity Units
Choosing the correct units can make all the difference for a measurement. You would not measure how tall you are in miles. Neither would you measure length of a football field in inches. The same idea applies for weight and volume. If you were making a pot of soup, it could be pretty complicated to measure the amount of liquid in ounces. Think of how challenging that would be as you are measuring. To get an idea of the size of the units of weight and volume, take a look at the table below.
This table is useful as a reference for many of the different ways to measure weight and capacity. Think about what you are measuring before you choose a unit. Use units of weight to find how heavy an object is. Use units of volume to find how much space that object takes up. Generally, the size of the object you are measuring can give you a clue as to which unit to use.
Which unit of measurment would you use to weigh a large dog?
There are three units of weight in the table-- ounce, pound, and ton. Think about which unit would work best for this example.
The ounce is too small of a unit. A large dog would weigh several hundred ounces. The ton is too large of a unit. The pound would be the best unit to measure the weight of a large dog.
### Examples
#### Example 1
Earlier, you were given a problem about Jeffrey at the butcher shop.
Jeffrey has to label the steak packages and bottles of barbeque sauce. He has to decide which unit would work best to accurately label each item.
First, consider the packages of steak. Most of the steaks are less than a pound. Therefore, Jeffrey should use ounces to label the packages of steak.
Then, consider the bottles of barbecue sauce. The bottles are about the size of a water bottle. He has several options he can use— fluid ounce, cup, or pint. The most accurate unit would be the smallest unit. Therefore, Jeffrey should use fluid ounces to label the bottles of barbecue sauce.
#### Example 2
Choose the best unit of measure: if Darcy is filling up her swimming pool, which unit of capacity would she use to measure the amount of water needed?
Think about what is being measured. A swimming pool requires a large amount of water. Darcy would use gallons, the largest unit for capacity.
#### Example 3
Choose the best unit of measure: the milk in a baby bottle.
Measure the capacity of a baby bottle using fluid ounces.
#### Example 4
Choose the best unit of measure: a bundle of wood.
Measure the weight of a bundle of wood using pounds.
#### Example 5
Choose the best unit of measure: flour for a cake.
Measure the capacity of flour for a cake using cups.
### Review
Choose the best unit of measure for each item below.
1. The weight of a dog
2. The weight of a camper
3. Soda in a can
4. Two Elephants
5. A television set
6. A baby bottle
7. A jug of milk
8. An automobile
9. The water in a pool
10. A pile of pennies
11. A pallet of wood pellets
12. A rhinocerous
13. A boat
14. An apple
15. Water in an eye dropper
To see the Review answers, open this PDF file and look for section 7.15.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Cup A cup is a small customary unit for measuring liquid capacity. One cup is equal to eight ounces or about 0.25 liters.
Customary System The customary system is the measurement system commonly used in the United States, including: feet, inches, pounds, cups, gallons, etc.
Fluid Ounce A fluid ounce is a customary unit for measuring liquid capacity. Soft drinks are measured in fluid ounces.
Gallon A gallon is a unit of liquid capacity. One gallon is equivalent to four quarts.
Ounces Ounces are the smallest common unit of weight in the customary system, used to measure very light items.
Pint A pint is a unit of liquid capacity. One pint is equivalent to two cups.
Pounds Pounds are one of the most common units for measuring weight.
Quart A quart is a unit of capacity that is equivalent to two pints.
Ton A ton is the largest common unit for measuring weight. Very large items like cars and trucks are measured in tons.
Weight Weight is a measurement of the heaviness or mass of someone or something. The customary units of weight included ounces, pounds, and tons. | 1,220 | 5,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2017-13 | latest | en | 0.914092 |
https://number.academy/1000079 | 1,708,643,381,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473871.23/warc/CC-MAIN-20240222225655-20240223015655-00803.warc.gz | 426,983,624 | 11,902 | # Number 1000079 facts
The odd number 1,000,079 is spelled 🔊, and written in words: one million and seventy-nine, approximately 1.0 million. The ordinal number 1000079th is said 🔊 and written as: one million and seventy-ninth. The meaning of the number 1000079 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1000079. What is 1000079 in computer science, numerology, codes and images, writing and naming in other languages
## What is 1,000,079 in other units
The decimal (Arabic) number 1000079 converted to a Roman number is (M)LXXIX. Roman and decimal number conversions.
#### Weight conversion
1000079 kilograms (kg) = 2204774.2 pounds (lbs)
1000079 pounds (lbs) = 453632.9 kilograms (kg)
#### Length conversion
1000079 kilometers (km) equals to 621421 miles (mi).
1000079 miles (mi) equals to 1609472 kilometers (km).
1000079 meters (m) equals to 3281060 feet (ft).
1000079 feet (ft) equals 304828 meters (m).
1000079 centimeters (cm) equals to 393731.9 inches (in).
1000079 inches (in) equals to 2540200.7 centimeters (cm).
#### Temperature conversion
1000079° Fahrenheit (°F) equals to 555581.7° Celsius (°C)
1000079° Celsius (°C) equals to 1800174.2° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
1000079 seconds equals to 1 week, 4 days, 13 hours, 47 minutes, 59 seconds
1000079 minutes equals to 2 years, 3 weeks, 1 day, 11 hours, 59 minutes
### Codes and images of the number 1000079
Number 1000079 morse code: .---- ----- ----- ----- ----- --... ----.
Sign language for number 1000079:
Number 1000079 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 1000079
### Multiplications
#### Multiplication table of 1000079
1000079 multiplied by two equals 2000158 (1000079 x 2 = 2000158).
1000079 multiplied by three equals 3000237 (1000079 x 3 = 3000237).
1000079 multiplied by four equals 4000316 (1000079 x 4 = 4000316).
1000079 multiplied by five equals 5000395 (1000079 x 5 = 5000395).
1000079 multiplied by six equals 6000474 (1000079 x 6 = 6000474).
1000079 multiplied by seven equals 7000553 (1000079 x 7 = 7000553).
1000079 multiplied by eight equals 8000632 (1000079 x 8 = 8000632).
1000079 multiplied by nine equals 9000711 (1000079 x 9 = 9000711).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 1000079
Half of 1000079 is 500039,5 (1000079 / 2 = 500039,5 = 500039 1/2).
One third of 1000079 is 333359,6667 (1000079 / 3 = 333359,6667 = 333359 2/3).
One quarter of 1000079 is 250019,75 (1000079 / 4 = 250019,75 = 250019 3/4).
One fifth of 1000079 is 200015,8 (1000079 / 5 = 200015,8 = 200015 4/5).
One sixth of 1000079 is 166679,8333 (1000079 / 6 = 166679,8333 = 166679 5/6).
One seventh of 1000079 is 142868,4286 (1000079 / 7 = 142868,4286 = 142868 3/7).
One eighth of 1000079 is 125009,875 (1000079 / 8 = 125009,875 = 125009 7/8).
One ninth of 1000079 is 111119,8889 (1000079 / 9 = 111119,8889 = 111119 8/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
1000079
#### Is Prime?
The number 1000079 is not a prime number. The closest prime numbers are 1000039, 1000081.
#### Factorization and factors (dividers)
The prime factors of 1000079 are 281 * 3559
The factors of 1000079 are 1, 281, 3559, 1000079.
Total factors 4.
Sum of factors 1003920 (3841).
#### Powers
The second power of 10000792 is 1.000.158.006.241.
The third power of 10000793 is 1.000.237.018.723.492.992.
#### Roots
The square root √1000079 is 1000,039499.
The cube root of 31000079 is 100,002633.
#### Logarithms
The natural logarithm of No. ln 1000079 = loge 1000079 = 13,81559.
The logarithm to base 10 of No. log10 1000079 = 6,000034.
The Napierian logarithm of No. log1/e 1000079 = -13,81559.
### Trigonometric functions
The cosine of 1000079 is -0,994739.
The sine of 1000079 is -0,102439.
The tangent of 1000079 is 0,102981.
### Properties of the number 1000079
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 1000079 in Computer Science
Code typeCode value
1000079 Number of bytes976.6KB
Unix timeUnix time 1000079 is equal to Monday Jan. 12, 1970, 1:47:59 p.m. GMT
IPv4, IPv6Number 1000079 internet address in dotted format v4 0.15.66.143, v6 ::f:428f
1000079 Decimal = 11110100001010001111 Binary
1000079 Decimal = 1212210211222 Ternary
1000079 Decimal = 3641217 Octal
1000079 Decimal = F428F Hexadecimal (0xf428f hex)
1000079 BASE64MTAwMDA3OQ==
1000079 MD510a9a40768ca824475332240cc98f0c7
1000079 SHA17c7af8698dbd9f48fac93688ce6c885674ab1d1f
1000079 SHA224938004a4c48ed7dc6379229d313a3c256fe0d69776f71c13837e1bcb
1000079 SHA256e6da5e378e75e7a78fb74646a4ce4417b7cd71fd1fbf276cc3abacc56c2647c7
More SHA codes related to the number 1000079 ...
If you know something interesting about the 1000079 number that you did not find on this page, do not hesitate to write us here.
## Numerology 1000079
### Character frequency in the number 1000079
Character (importance) frequency for numerology.
Character: Frequency: 1 1 0 4 7 1 9 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 1000079, the numbers 1+0+0+0+0+7+9 = 1+7 = 8 are added and the meaning of the number 8 is sought.
## № 1,000,079 in other languages
How to say or write the number one million and seventy-nine in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.000.079) un millón setenta y nueve German: 🔊 (Nummer 1.000.079) eine Million neunundsiebzig French: 🔊 (nombre 1 000 079) un million soixante-dix-neuf Portuguese: 🔊 (número 1 000 079) um milhão e setenta e nove Hindi: 🔊 (संख्या 1 000 079) दस लाख, उन्यासी Chinese: 🔊 (数 1 000 079) 一百万零七十九 Arabian: 🔊 (عدد 1,000,079) واحد مليون و تسعة و سبعون Czech: 🔊 (číslo 1 000 079) milion sedmdesát devět Korean: 🔊 (번호 1,000,079) 백만 칠십구 Dutch: 🔊 (nummer 1 000 079) een miljoen negenenzeventig Japanese: 🔊 (数 1,000,079) 百万七十九 Indonesian: 🔊 (jumlah 1.000.079) satu juta tujuh puluh sembilan Italian: 🔊 (numero 1 000 079) un milione e settantanove Norwegian: 🔊 (nummer 1 000 079) en million og sytti-ni Polish: 🔊 (liczba 1 000 079) milion siedemdziesiąt dziewięć Russian: 🔊 (номер 1 000 079) один миллион семьдесят девять Turkish: 🔊 (numara 1,000,079) birmilyonyetmişdokuz Thai: 🔊 (จำนวน 1 000 079) หนึ่งล้านเจ็ดสิบเก้า Ukrainian: 🔊 (номер 1 000 079) один мільйон сімдесят дев'ять Vietnamese: 🔊 (con số 1.000.079) một triệu lẻ bảy mươi chín Other languages ...
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## Comment
If you know something interesting about the number 1000079 or any other natural number (positive integer), please write to us here or on Facebook.
#### Comment (Maximum 2000 characters) *
The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 2,531 | 7,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-10 | latest | en | 0.636991 |
http://www.ncatlab.org/nlab/show/four+lemma | 1,394,482,960,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2014-10/segments/1394011005264/warc/CC-MAIN-20140305091645-00021-ip-10-183-142-35.ec2.internal.warc.gz | 448,296,940 | 9,495 | # nLab four lemma
### Context
#### Homological algebra
homological algebra
and
nonabelian homological algebra
diagram chasing
# Contents
## Idea
The four lemma is one of the basic diagram chasing lemmas in homological algebra. It follows directly from the salamander lemma. It directly implies the five lemma.
## Statement
Let $\mathcal{A}$ be an abelian category.
###### Proposition
Consider a commuting diagram in $\mathcal{A}$ of the form
$\array{ &\to& &\stackrel{\xi}{\to}& &\to& \\ \downarrow^{\mathrlap{\tau}} && \downarrow^{\mathrlap{f}} && \downarrow^{\mathrlap{g}} && \downarrow^{\mathrlap{\nu}} \\ &\to& &\stackrel{\eta}{\to}& &\to& }$
where
1. the rows are exact sequences,
2. $\tau$ is an epimorphism,
3. $\nu$ is a monomorphism.
Then
1. $\xi(ker(f)) = ker(g)$ (the image under $\xi$ of the kernel of $f$ is the kernel of $g$)
2. $im(f) = \eta^{-1}(im(g))$ (the preimage under $\eta$ of the image of $g$ is the image of $f$)
(the “strong four lemma”) and hence in particular
1. if $g$ is an epimorphism then so is $f$;
2. if $f$ is a monomorphism then so is $g$
(the “weak four lemma”).
A direct proof from the salamander lemma is spelled out at salamander lemma – implications – four lemma.
## References
The strong/weak four lemma appears as lemma 3.2, 3.3 in chapter I and then with proof in lemma 3.1 of chapter XII of
• Saunders MacLane, Homology (1967) reprinted as Classics in Mathematics, Springer (1995)
Revised on September 24, 2012 23:46:27 by Urs Schreiber (82.169.65.155) | 471 | 1,527 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 20, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2014-10 | longest | en | 0.775721 |
https://jp.mathworks.com/matlabcentral/cody/problems/1129-reverse-the-elements-of-an-array/solutions/1522214 | 1,590,991,115,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347414057.54/warc/CC-MAIN-20200601040052-20200601070052-00061.warc.gz | 401,783,622 | 15,822 | Cody
# Problem 1129. Reverse the elements of an array
Solution 1522214
Submitted on 8 May 2018 by MDK
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 = []; y_correct = []; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
x = 1; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
x = [1:5]; y_correct = [5:-1:1]; assert(isequal(your_fcn_name(x),y_correct))
4 Pass
x = [1:5;6:10]; y_correct = [10:-1:6;5:-1:1]; assert(isequal(your_fcn_name(x),y_correct))
5 Pass
x = [1:5;6:10;11:15]; y_correct = [15:-1:11;10:-1:6;5:-1:1]; assert(isequal(your_fcn_name(x),y_correct))
6 Pass
x = ones(5); y_correct = ones(5); assert(isequal(your_fcn_name(x),y_correct))
7 Pass
x = magic(3); y_correct = [2 9 4;7 5 3;6 1 8]; assert(isequal(your_fcn_name(x),y_correct))
8 Pass
x = [2 9 -4;7 -5 3;-6 1 8]; y_correct =[8 1 -6;3 -5 7;-4 9 2] ; assert(isequal(your_fcn_name(x),y_correct)) | 396 | 1,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-24 | latest | en | 0.513494 |
https://stats.stackexchange.com/questions/55034/how-does-pca-improve-the-accuracy-of-a-predictive-model/55059#55059 | 1,716,197,863,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058254.21/warc/CC-MAIN-20240520080523-20240520110523-00054.warc.gz | 499,056,727 | 39,809 | # How does PCA improve the accuracy of a predictive model? [duplicate]
I've seen in a kaggle challenge about digit recognition someone who used PCA before decision tree or other techniques.
I thought it was just for compressing data but he aimed to improve his score.
How can PCA improve score in this case ? Is it because there is less overfitting ?
• I would be tempted to say that decision trees are harder to train. There are a lot of details in the learning algorithms for them. PCA has a well founded framework to tweak rather than get lost in the details.
– Vass
Apr 3, 2013 at 10:15
• Dimensionality reduction via PCA can definitely serve as regularization in order to prevent overfitting. E.g. in regression it is known as "principal components regression" and is related to ridge regression. For classification, see e.g. here: Does it make sense to combine PCA and LDA? Jan 27, 2015 at 16:48
• I think the top answer misses the point of this question (see my comment under it). I'd suggest to read this thread stats.stackexchange.com/questions/141864 and follow the links for the comprehensive discussion. Jan 25, 2017 at 19:28
Dadi Perlmutter once said: "What is the difference between theory and practice? In theory they are the same while in practice they are different". This is one of those cases.
Methods like Neural Networks often use gradient descent derived methods. In theory if you had infinite number of iterations and retries, the algorithm is going to converge to the same result independent of coordinate system. Neural Networks do not like the "curse of dimensionality" and so using PCA to reduce the dimension of the data can improve speed of convergence and quality of results. The transformation of the data, by centering, rotating and scaling informed by PCA can improve the convergence time and the quality of results.
In theory the PCA makes no difference, but in practice it improves rate of training, simplifies the required neural structure to represent the data, and results in systems that better characterize the "intermediate structure" of the data instead of having to account for multiple scales - it is more accurate.
My guess is that there are analogous reasons that apply to random forests of gradient boosted trees or other similar creatures. (Link)
• Aside: I would be interested to see an accurate and reputable sourcing of your introductory quote. It is variously attributed to several people on the internet, most recognizably, Yogi Berra and Albert Einstein. I have personally heard it from an engineer that is a generation older than Perlmutter and this was long enough ago that it makes me highly doubt that Perlmutter could be the original source. Apr 3, 2013 at 19:13
• Cardinal - He is not the original source, but he is the source that I heard it from. I saw him presenting something onstage in 2009. The only thing I retain, 4 years later, is this quote. Apr 3, 2013 at 20:56
• This conversation is so funny, much more than the quote itself. I had once asked my professor if I can cite a cited quote because the original paper was a phantom one! Jun 30, 2016 at 23:45
• "In theory the PCA makes no difference": This is only the case when all PCs are retained. Usually when people talk about doing PCA prior to some other algorithm they mean that they keep only a small subset of PCs (and the OP wrote about "compressing data" too). Your answer does not cover this possibility at all so I feel like it does not really address the question. Jan 25, 2017 at 9:28
• Whitening or pinkening transforms are both PCA approaches that retain all the components. The difference is trying to transform the data so that the diagonal of the covariance is more uniform, or less uniform. May 11, 2020 at 19:00
Disclaimer: I'm usually wrong at things.
Decision trees, by virtue of doing recursive splitting of your samples, with splits being based on a single variable, can only generate decision boundaries parallel to the axes of your co-ordinate system. So by rotating the data to directions of maximum variance/diagonalizing your covariance matrix as best you can, it might be easier to put decision boundaries between your class distributions
That being said, I'm not sure why you'd do PCA (without discarding some of your eigenvectors) before using a neural network model or whatever, because the rotation alone makes no difference - the network can approximate any function through the feature space.
An insight I gained from Jonathon Shlens' "A Tutorial on Principal Component Analysis": Performing PCA is like choosing a camera angle, to gain the best possible view of the variance to be explained.
So I'm joining user1843053. At proper angle, decision boundaries parallel to the axes of new, rotated coordinate system might make more sense than in original feature space, allowing for better performance of e.g. decision trees, even without discarding "non-principal" dimensions.
The PCA is a change of variables, using the correlations explained by orthogonal directions.
Removing directions with non-representative corresponding correlation is like removing noise. You will only keep significant datas.
By the way, thanks for the site. | 1,112 | 5,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.949212 |
https://womenstagetheworld.org/2-digit-addition-worksheet/ | 1,566,330,306,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315558.25/warc/CC-MAIN-20190820180442-20190820202442-00335.warc.gz | 702,440,746 | 30,476 | In Free Printable Worksheets183 views
4.05 / 5 ( 174votes )
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2 Digit Addition Worksheet. Try to remember, you always have to care for your child with amazing care, compassion and affection to be able to help him learn. You may also ask your kid's teacher for extra worksheets. Your son or daughter is not going to just learn a different sort of font but in addition learn how to write elegantly because cursive writing is quite beautiful to check out. As a result, if a kid is already suffering from ADHD his handwriting will definitely be affected. Accordingly, to be able to accomplish this, if children are taught to form different shapes in a suitable fashion, it is going to enable them to compose the letters in a really smooth and easy method. Although it can be cute every time a youngster says he runned on the playground, students want to understand how to use past tense so as to speak and write correctly. Let say, you would like to boost your son's or daughter's handwriting, it is but obvious that you want to give your son or daughter plenty of practice, as they say, practice makes perfect.
Without phonics skills, it's almost impossible, especially for kids, to learn how to read new words. Techniques to Handle Attention Issues It is extremely essential that should you discover your kid is inattentive to his learning especially when it has to do with reading and writing issues you must begin working on various ways and to improve it. Use a student's name in every sentence so there's a single sentence for each kid. Because he or she learns at his own rate, there is some variability in the age when a child is ready to learn to read. Teaching your kid to form the alphabets is quite a complicated practice.
Author: Elina Volk
Have faith. But just because it's possible, doesn't mean it will be easy. Know that whatever life you want, the grades you want, the job you want, the reputation you want, friends you want, that it's possible.
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Top | 1,748 | 8,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-35 | latest | en | 0.902282 |
https://engineering.stackexchange.com/questions/37444/what-is-the-linear-strain-for-given-lateral-strain | 1,718,937,110,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00454.warc.gz | 203,734,958 | 39,189 | # What is the linear strain for given lateral strain?
So, I am trying to calculate the linear strain of a shaft which is under plastic deformation in yielding region.
The diameter is changing from 2d to 1.5d
I'm assuming that the volumetric strain is zero.$$(\epsilon_v=0)$$ or in other words $$\mu = 0.5$$
Now, I've tried the following two methods and I'm getting two different answers:
Method 1 $$\epsilon_v = 0 \\ \Rightarrow \epsilon_l + 2\epsilon_d = 0 \\ \Rightarrow \epsilon_l = -2(\frac{1.5d - 2d}{2d}) \\ \Rightarrow \epsilon_l = 0.5$$ Method 2 $$V_i = V_f\\ \Rightarrow \frac{\pi}{4}d_i^2l_i = \frac{\pi}{4}d_f^2l_f\\ \Rightarrow \epsilon_l = \frac{\Delta l}{l_i} = \frac{7}{9}$$ Where $$\epsilon_l$$ is longitudinal strain, $$\epsilon_d$$ is lateral strain($$\bot$$ to longitudinal direction), subscripts $$i, f$$ refer to initial and final conditions respectively.
All the strains are engineering strains only.
Now ideally, both the methods should give same solution.
Where am I going wrong?
I think that neither of your solutions is correct.
a) you are assuming that the volumetric strain is zero. However, if there is a change of volume then the volumetric strain can never be zero.
Also, in that scenario, I am a bit uncertain what is $$\varepsilon_d$$. Is it deviatoric strain or is it strain perpendicular to the longitudical direction and due to the symmetry you assume that its equal.
b) you assume that the volume is the same. However, due to the internal stresses and strains the volume might change (compressibility of material). It's like having a baloon in water (the lower you go i.e. the higher the hydrostatic stress, the volume of air reduces).
So neither I believe is correct.
• I've assumed volumetric strain is zero because the material is under yielding(plastic deformation). $\epsilon_d$ is the strain perpendicular to longitudinal direction of the shaft(a cylindrical object) Commented Sep 1, 2020 at 15:04
• regarding $\epsilon_d$ a more appropriate sign would then be $\epsilon_r$ to denote strain in the radial direction. It is best if you try to solve this problem using cylindrical coordinates.
– NMech
Commented Sep 1, 2020 at 16:55
• The plastic component of the strain has no volume change, but the elastic component does have a volume change. Commented Sep 1, 2020 at 19:19 | 594 | 2,330 | {"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": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-26 | latest | en | 0.875355 |
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# Dirickson Inc. has provided the following data concerning one of the products in its standard cost...
Dirickson Inc. has provided the following data concerning one of the products in its standard cost system. Variable manufacturing overhead is applied to products on the basis of direct labor-hours.
Inputs Standard Quantity or Hours per Unit of Output Standard Price or Rate Direct materials 7.6 ounces \$ 9.40 per ounce Direct labor 0.10 hours \$ 18.00 per hour Variable manufacturing overhead 0.10 hours \$ 5.30 per hour
1. The company has reported the following actual results for the product for July:
Actual output 7,600 units Raw materials purchased 63,000 ounces Actual cost of raw materials purchased \$ 541,800 Raw materials used in production 57,750 ounces Actual direct labor-hours 820 hours Actual direct labor cost \$ 16,072 Actual variable overhead cost \$ 4,592
The variable overhead rate variance for the month is closest to:
Multiple Choice
\$228 F
\$246 U
\$246 F
\$228 U
Answer:- Variable overhead rate variance = (Standard rate –Actual rate)*Actual working hours
=(\$5.30 per hour - \$5.6 per hour)*820 hours
=\$246 Unfavourable
Where:-
Actual rate= Actual variable overhead cost/ Actual direct labor-hours
=\$4592/820 hours =\$5.6 per hour
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# If a lens of focal length, f, made of glass of refractive index 53 is placed in a liquid of refractive index, 4 53, then the focal length of the lens becomes
A
f
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Solution
## The correct option is C 4fIn Air 1f=(μg−1)[1R1+1R2]In water,1f′=(μgμw−1)[1R1+1R2]f′f=(μg−1μg−μw)×μwf′=⎛⎜ ⎜ ⎜⎝53−153−453⎞⎟ ⎟ ⎟⎠×453f=−−1718f
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# 2 Ways to find duplicate elements in an Array - Java
Problem: You have given an array of objects, which could be an array of integers and or array of Strings or any object which implements the Comparable interface. How would you find duplicate elements from an array? Can you solve this problem in O(n) complexity? This is actually one of the frequently asked coding problems from Java interviews. There are multiple ways to solve this problem and you will learn two popular ways here, first the brute force way, which involves comparing each element with every other element and other which uses a hash table like data structure to reduce the time complexity of problem from quadratic to linear, of course by trading off some space complexity. This also shows that how by using a suitable data structure you can come up with a better algorithm to solve a problem. If you are preparing for programming job interviews, then I also suggest you take a look at Cracking the Coding Interview book, which contains 150 programming questions and solutions, good enough to do well on any programming job interviews e.g. Java, C++, Python or Ruby.
# How to Sort List into Ascending and Descending Order in Java
ArrayList, Set Sorting in Ascending – Descending Order Java
Sorting List, Set and ArrayList in Java on ascending and descending order is very easy, You just need to know correct API method to do that. Collections.sort() method will sort the collection passed to it, doesn't return anything just sort the collection itself. Sort() method of Collections class in Java is overloaded where another version takes a Comparator and sort all the elements of Collection on order defined by Comparator.If we don't pass any Comparator than object will be sorted based upon there natural order like String will be sorted alphabetically or lexicographically. Integer will be sorted numerically etc. Default sorting order for an object is ascending order like Integer will be sorted from low to high while descending order is just opposite. Collections.reverseOrder() returns a Comparator which will be used for sorting Object in descending order.
# Top 10 Tricky Java interview questions and Answers
What is a tricky question? Well, tricky Java interview questions are those questions which have some surprise element on it. If you try to answer a tricky question with common sense, you will most likely fail because they require some specific knowledge. Most of the tricky Java questions comes from confusing concepts like function overloading and overriding, Multi-threading which is really tricky to master, character encoding, checked vs unchecked exceptions and subtle Java programming details like Integer overflow. Most important thing to answer a tricky Java question is attitude and analytical thinking, which helps even if you don't know the answer. Anyway in this Java article we will see 10 Java questions which are real tricky and requires more than average knowledge of Java programming language to answer them correctly. As per my experience, there is always one or two tricky or tough Java interview question on any core Java or J2EE interviews, so it's good to prepare tricky questions from Java in advance.
# How to Print Pyramid Pattern in Java? Program Example
Pattern based exercises are a good way to learn nested loops in Java. There are many pattern based exercises and one of them is printing Pyramid structure as shown below:
* *
* * *
* * * *
* * * * *
You need to write a Java program to print above pyramid pattern. How many levels the pyramid triangle would have will be decided by the user input. You can print this kind of pattern by using print() and println() method from System.out object. System.out.print() just prints the String or character you passed to it, without adding a new line, useful to print stars in the same line. While, System.out.println() print characters followed by a newline character, which is useful to move to next line. You can also use Scanner class to get input from the user and draw pyramid up to that level only. For example in above diagram, the pyramid has 5 levels.
# Java ArrayList Tutorial - The MEGA List
I have written several ArrayList tutorials, touching different ArrayList concepts and many how to do examples with ArrayList. In this tutorial, I am giving a summary of each of them. Why? So that any Java beginner who wants to learn ArrayList in detail, can go through the relevant tutorial and learn. It's also on request of many of my readers, who asked in past to share all the relevant tutorials in one place. Why should you learn ArrayList? Because it's the most important Collection class in Java. You will often find yourself using ArrayList and HashMap in tandem. It's your dynamic array which can resize itself as it grows. In another word, ArrayList is as much important as an array. When I started learning Java, my quest to ArrayList starts as a dynamic array, because there were many scenarios where we don't know the size of the array in advance. We end up either allocating less space or more space, both are not ideal. Btw, you should also check out Head First Java 2nd Edition if you are newbie and Effective Java 2nd Edition, if you know Java but wants to become a Java expert.
# How to solve FizzBuzz in Java?
FizzBuzz is one of the most frequently asked questions on programming interviews and used to filter programmers who can't program. It looks extremely simple but it's tricky for those programmers or coder who struggle to structure their code. Fizzbuzz problem statement is very simple, write a program which return "fizz" if the number is a multiplier of 3, return "buzz" if its multiplier of 5 and return "fizzbuzz" if the number is divisible by both 3 and 5. If the number is not divisible by either 3 or 5 then it should just return the number itself. You can see nothing is fancy when it comes to thinking about the solution, but when you start coding, you will see a problem with structuring your code, particularly if else blocks.
# How to convert float to int in Java? Examples
Even though both float and int are 32-bit wide data type, float has the higher range than integer primitive value. Since a float is a bigger than int, you can convert a float to an int by simply down-casting it e.g. (int) 4.0f will give you integer 4. By the way, you must remember that type casting just get rid of anything after the decimal point, they don't perform any rounding or flooring operation on the value. So if your float value is 3.999, down casting to an integer will produce 3, not 4. If you need rounding then consider using Math.round() method, which converts float to its nearest integer by adding +0.5 and then truncating it. Math.random() is overloaded for both float and double, so you can use this for converting double to long as well. Let's see an example of converting a float value to int in Java.
# How to declare ArrayList with values in Java? Examples
Sometimes you want to create an ArrayList with values, just like you initialize t at the time of declaration, as shown below:
int[] primes = {2, 3, 5, 7, 11, 13, 17};
or
String[] names = {"john", "Johnny", "Tom", "Harry"};
but unfortunately, ArrayList doesn't support such kind of declaration in Java. But don't worry, there is a workaround to declare an ArrayList with values e.g. String, integers, floats or doubles by using Arrays.asList() method, which is nothing but a shortcut to convert an Array to ArrayList.
# Error: could not open 'C:\Java\jre8\lib\amd64\jvm.cfg'
A couple of weeks back I updated my laptop to Windows 10 but after trying for one day, I reverted back to Windows 8.1. Everything was alright until I open Eclipse, which was throwing "Error: could not open 'C:\Program Files\Java\jre8\lib\amd64\jvm.cfg', as soon as I launch it. It was quite bizarre because everything was fine earlier. I suspect Java installation problem, so I went to command prompt and typed Java, only to find the same error there as well. You can see below, I am just trying to run the "java" command from MS-DOS window. Here "java" command is picked from PATH environment variable.
# Avoid ConcurrentModificationException while looping over Java ArrayList?
Apart from the NullPointerException and ClassNotFoundException, ConcurrentModificationException is another nightmare for Java developers. What makes this error tricky is the word concurrent, which always mislead Java programmers that this exception is coming because multiple threads are trying to modify the collection at the same time. Then begins the hunting, they spent countless hours to find the code which has the probability of concurrent modification. While in reality ConcurrentModficationException can also come on the single threaded environment. To give you an example, just loop over a list using for loop and try to remove one element, you will get the ConcurrentModificatoinExcetpion? Why? because you broke the rule of not modifying a Collection during iteration.
# How to find highest repeating word from a text File in Java - Word Count Problem
How to find the word and their count from a text file is another frequently asked coding question from Java interviews. The logic to solve this problem is similar to what we have seen in how to find duplicate words in a String. In the first step you need to build a word Map by reading contents of a text File. This Map should contain word as a key and their count as value. Once you have this Map ready, you can simply sort the Map based upon values. If you don't know how to sort a Map on values, see this tutorial first. It will teach you by sorting HashMap on values. Now getting key and value in sorted should be easy, but remember HashMap doesn't maintain order, so you need to use a List to keep the entry in sorted order. Once you got this list, you can simply loop over the list and print each key and value from the entry. This way, you can also create a table of words and their count in decreasing order. This problem is sometimes also asked as to print all word and their count in tabular format. | 2,133 | 10,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-18 | latest | en | 0.913792 |
https://www.hindawi.com/journals/aaa/2012/150571/ | 1,547,936,002,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2019-04/segments/1547583681597.51/warc/CC-MAIN-20190119201117-20190119223117-00374.warc.gz | 816,645,495 | 65,933 | `Abstract and Applied AnalysisVolumeย 2012, Article IDย 150571, 10 pageshttp://dx.doi.org/10.1155/2012/150571`
Research Article
## On Subclass of -Uniformly Convex Functions of Complex Order Involving Multiplier Transformations
Department of Mathematics, College of Computer Science and Mathematics, University of Al-Qadisiya, Diwaniya, Iraq
Received 30 December 2011; Revised 28 February 2012; Accepted 13 March 2012
Academic Editor: Ondลejย Doลกlรฝ
Copyright ยฉ 2012 Waggas Galib Atshan and Ali Hamza Abada. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
#### Abstract
We introduce a subclass of -uniformly convex functions of order with negative coefficients by using the multiplier transformations in the open unit disk . We obtain coefficient estimates, radii of convexity and close-to-convexity, extreme points, and integral means inequalities for the function that belongs to the class .
#### 1. Introduction
Let denote the class of functions of the form: which are analytic and univalent in the open unit disk (see [1]). Also denote by the subclass of consisting of functions of the form: For any integer , we define the multiplier transformations (see [2, 3]) of functions by where .
A function is said to be in the class (-uniformly starlike Functions of order ) if it satisfies the condition: and is said to be in the class (-uniformly convex Functions of order ) if it satisfies the condition: Indeed it follows from (1.4) and (1.5) that
The interesting geometric properties of these function classes were extensively studied by Kanas et al., in [4, 5], motivated by Altintas et al. [6], Murugusundaramoorthy and Srivastava [7], and Murugusundaramoorthy and Magesh [8, 9], Atshan and Kulkarni [10] and Atshan and Buti [11].
Now, we define a new subclass of uniformly convex functions of complex order.
For , , , we let be the class of functions satisfying (1.2) with the analytic criterion: where is given by (1.3).
#### 2. Main Results
First, we obtain the necessary and sufficient condition for functions in the class .
Theorem 2.1. The necessary and sufficient condition for of the form of (1.2) to be in the class is where , , .
Proof. Suppose that (2.1) is true for . Then if that is, if Conversely, assume that , then Letting along the real axis, we have Hence, by maximum modulus theorem, the simple computation leads to the desired inequality which completes the proof.
Corollary 2.2. Let the function defined by (1.2) belong to . Then, where , , with equality for
#### 3. Radii of Convexity and Close-to-Convexity
We obtain the radii of convexity and close-to-convexity results for functions in the class in the following theorems.
Theorem 3.1. Let . Then is convex of order in the disk , where
Proof. Let . Then by Theorem 2.1, we have For , we need to show that and we have to show that Hence, This is enough to consider Therefore, Setting in (3.7), we get the radius of convexity, which completes the proof of Theorem 3.1.
Theorem 3.2. Let . Then is close-to-convex of order in the disk , where
Proof. Let . Then by Theorem 2.1, we have For , we need to show that and we have to show that Hence, This is enough to consider Therefore, Setting in (3.14), we get the radius of close-to-convexity, which completes the proof of Theorem 3.2.
#### 4. Extreme Points
The extreme points of the class are given by the following theorem.
Theorem 4.1. Let for .
Then, if and only if it can be expressed in the form: where and
Proof. Suppose that can be expressed as in (4.2). Our goal is to show that . By (4.2), we have that Now, Thus, .
Conversely, assume that . Since we can set Then, This completes the proof of Theorem 4.1.
#### 5. Integral Means
In order to find the integral means inequality and to verify the Silverman Conjuncture [12] for , we need the following definition of subordination and subordination result according to Littlewood [13].
Definition 5.1 (see [13]). Let and be analytic in . Then, we say that the function is subordinate to if there exists a Schwarz function , analytic in with , such that . We denote this subordination or . In particular, if the function is univalent in , the above subordination is equivalent to , .
Lemma 5.2 (see [13]). If the functions and are analytic in with , then Applying Theorem 2.1 with the extremal function and Lemma 5.2, we prove the following theorem.
Theorem 5.3. Let . If and are nondecreasing sequences, then, for and , one has where
Proof. Let of the form of (1.2) and then we must show that By Lemma 5.2, it suffices to show that Setting from (5.7) and (2.1) we obtain This completes the proof of Theorem 5.3.
#### References
1. K. Al-Shaqsi, M. Darus, and O. A. Fadipe-Joseph, โA new subclass of Sǎlǎgean-type harmonic univalent functions,โ Abstract and Applied Analysis, vol. 2010, Article ID 821531, 12 pages, 2010.
2. N. E. Cho and T. H. Kim, โMultiplier transformations and strongly close-to-convex functions,โ Bulletin of the Korean Mathematical Society, vol. 40, no. 3, pp. 399โ410, 2003.
3. R. M. El-Ashwah and M. K. Aouf, โNew classes of p-valent harmonic functions,โ Bulletin of Mathematical Analysis and Applications, vol. 2, no. 3, pp. 53โ64, 2010.
4. S. Kanas and A. Wisniowska, โConic regions and k-uniform convexity,โ Journal of Computational and Applied Mathematics, vol. 105, no. 1-2, pp. 327โ336, 1999.
5. S. Kanas and T. Yaguchi, โSubclasses of k-uniformly convex and starlike functions defined by generalized derivative. II,โ Publications Institute Mathematique, vol. 69, no. 83, pp. 91โ100, 2001.
6. O. Altintas, O. Ozkan, and H. M. Srivastava, โNeighborhoods of a class of analytic functions with negative coefficients,โ Applied Mathematics Letters, vol. 13, no. 3, pp. 63โ67, 2000.
7. G. Murugusundaramoorthy and H. M. Srivastava, โNeighborhoods of certain classes of analytic functions of complex order,โ Journal of Inequalities in Pure and Applied Mathematics, vol. 5, no. 2, article 24, pp. 1โ8, 2004.
8. G. Murugusundaramoorthy and N. Magesh, โStarlike and convex functions of complex order involving the Dziok-Srivastava operator,โ Integral Transforms and Special Functions, vol. 18, no. 5-6, pp. 419โ425, 2007.
9. G. Murugusundaramoorthy and N. Magesh, โCertain subclasses of starlike functions of complex order involving generalized hypergeometric functions,โ International Journal of Mathematics and Mathematical Sciences, vol. 2010, Article ID 178605, 12 pages, 2010.
10. W. G. Atshan and S. R. Kulkarni, โNeighborhoods and partial sums of subclass of k-uniformly convex functions and related class of k-starlike functions with negative coefficients based on integral operator,โ Southeast Asian Bulletin of Mathematics, vol. 33, no. 4, pp. 623โ637, 2009.
11. W. G. Atshan and R. H. Buti, โOn generalized hypergeometric functions and associated classes of k-uniformly convex and k-starlike p-valent functions,โ Advances and Applications in Mathematical Sciences, vol. 6, no. 2, pp. 149โ160, 2010.
12. H. Silverman, โIntegral means for univalent functions with negative coefficients,โ Houston Journal of Mathematics, vol. 23, no. 1, pp. 169โ174, 1997.
13. J. E. Littlewood, โOn inequalities in theory of functions,โ Proceedings of the London Mathematical Society, vol. 23, pp. 481โ519, 1925. | 2,103 | 7,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-04 | latest | en | 0.850051 |
https://teachingcalculus.com/2018/03/13/type-3-questions-graph-analysis/ | 1,548,036,028,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583745010.63/warc/CC-MAIN-20190121005305-20190121031305-00509.warc.gz | 668,042,769 | 21,913 | # Type 3 Questions: Graph and Function Analysis
The long name is “Here’s the graph of the derivative, tell me things about the function.”
Students are given either the equation of the derivative of a function or a graph identified as the derivative of a function but no equation is given. It is not expected that students will write the equation (although this may be possible); rather, students are expected to determine important features of the function directly from the graph of the derivative. They may be asked for the location of extreme values, intervals where the function is increasing or decreasing, concavity, etc. They may be asked for function values at points.
The graph may be given in context and student will be asked about that context. The graph may be identified as the velocity of a moving object and questions will be asked about the motion. See Type 2 questions – Linear motion problems
Less often the function’s graph may be given and students will be asked about its derivatives.
What students should be able to do:
• Read information about the function from the graph of the derivative. This may be approached by derivative techniques or by antiderivative techniques.
• Find and justify where the function is increasing or decreasing.
• Find and justify extreme values (1st and 2nd derivative tests, Closed interval test aka. Candidates’ test).
• Find and justify points of inflection.
• Find slopes (second derivatives, acceleration) from the graph.
• Write an equation of a tangent line.
• Evaluate Riemann sums from geometry of the graph only.
• FTC: Evaluate integral from the area of regions on the graph.
• FTC: The function, g(x), maybe defined by an integral where the given graph is the graph of the integrand, f(t), so students should know that if, $\displaystyle g\left( x \right)=g\left( a \right)+\int_{a}^{t}{f\left( t \right)dt}$ then ${g}'\left( x \right)=f\left( x \right)$ and ${{g}'}'\left( x \right)={f}'\left( x \right)$. In this case students should write ${g}'(t)=f\left( t \right)$ on their answer paper, so it is clear to the reader that they understand this.
Not only must students be able to identify these things, but they are usually asked to justify their answer and reasoning. See Writing on the AP Exams for more on justifying and explaining answers.
The ideas and concepts that can be tested with this type question are numerous. The type appears on the multiple-choice exams as well as the free-response. Between multiple-choice and free-response this topic may account for 15% or more of the points available on recent tests. It is very important that students are familiar with all the ins and outs of this situation.
As with other questions, the topics tested come from the entire year’s work, not just a single unit. In my opinion many textbooks do not do a good job with integrating these topics, so be sure to use as many actual AP Exam questions as possible. Study past exams; look them over and see the different things that can be asked.
For some previous posts on this subject see October 1517192426 (my most read post), 2012 and January 2528, 2013
Free-response questions:
• Function given as a graph, questions about its integral (so by FTC the graph is the derivative): 2014 AB 3/BC 3
• Function given as an equation: 2016 AB 6
• Function given as a graph 2016 AB 3/BC 3
• Table and graph of function given, questions about related functions: 2017 AB 6,
• Derivative given as a graph: 2016 AB 5 and 2017 AB 3
Multiple-choice questions from non-secure exam. Notice the number of questions all from the same year; this is in addition to one free-response question (~25 points on AB and ~23 points on BC out of 108 points total)
• 2012 AB: 2, 5, 15, 17, 21, 22, 24, 26, 76, 78, 80, 83, 82, 84, 85, 87
• 2012 BC 3, 11, 12, 15, 12, 18, 21, 76, 78, 80, 81, 84, 88, 89
A good activity on this topic is here. The first pages are the teacher’s copy and solution. Then there are copies for Groups A, B, and C. Divide your class into 3 or 6 or 9 groups and give one copy to each. After they complete their activity have the students compare their results with the other groups.
Schedule of review posts
• Tuesday February 27 – AP Exam Review
• Friday, March 2 – Resources for reviewing
• Tuesday March 6 – Type 1 questions – Rate and accumulation questions
• Friday March 9 – Type 2 questions – Linear motion problems
• Tuesday March 13 – Type 3 questions – Graph analysis problems (this post)
• Friday March 16 – Type 4 questions – Area and volume problems
• Tuesday Match 20 Type 5 questions – Table and Riemann sum questions
• Friday March 23 Type 6 questions – Differential equation questions
• Tuesday March 27 – Type 7 questions – miscellaneous
• Friday March 30 Type 8 questions – Parametric and vector questions (BC topic)
• Tuesday April 3 Type 9 questions – Polar equations
• Friday April 6 Type 10 questions – Sequences and Series | 1,224 | 4,929 | {"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": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.88751 |
http://www.jiskha.com/display.cgi?id=1307127232 | 1,495,908,190,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608984.81/warc/CC-MAIN-20170527171852-20170527191852-00316.warc.gz | 695,103,175 | 4,020 | # statistics
posted by on .
Among the 500 first year students of college, 270 students study computer science, 345 students study mathematics, and 175 students study both computer science and mathematics. If one student is selected at random, find the probability that he did not take either of these subjects
• statistics - ,
1011
• UET Peshawar - ,
P[C.S]=270/500
P[Maths]=345/500
P[C.S ∩ Maths]=175/500
( P[C.S U Maths] )' = ?
ANS.
( P[C.S U Maths] )' = 1-P[C.S U Maths]
=1-{P[C.S] + P[Maths] - P[C.S ∩ Maths]}
=1-{270/500 + 345/500 -175/500 }
=1-{440/500}
=1-0.88
=0.12
• Maths - ,
A(-5,3)<B (-9,7) | 208 | 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} | 3.75 | 4 | CC-MAIN-2017-22 | latest | en | 0.788092 |
https://tex.stackexchange.com/questions/577009/display-of-the-sum-of-two-sinusoids | 1,656,464,696,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103619185.32/warc/CC-MAIN-20220628233925-20220629023925-00528.warc.gz | 605,567,127 | 65,304 | Display of the sum of two sinusoids
I would like to set up the beat principle: the sum S of two sinusoidal signals (S1 and S2) of close frequency presents a modification of amplitude. I managed to figure out the 2 sinusoidal signals on the other hand when I want to see the sum, it does not correspond to what I expect.
\documentclass{article}
\usepackage[french]{babel}
\usepackage[T1]{fontenc}
\usepackage{tikz}
\usetikzlibrary{babel}
\begin{document}
\begin{center}
\begin{tikzpicture}[samples=1000]
\draw[->] (-5,0) -- (5.2,0) node[right] {$t$};
\draw[->] (-5,-2) -- (-5,2) node[above] {$S$};
\draw[color=blue] plot ({\x},{cos(10*\x r)});
\draw[color=green!60!black] plot ({\x},{cos(9.25*\x r)});
\draw plot ({\x},{cos(10*\x+cos(9.25*\x r)});
\end{tikzpicture}
\end{center}
\end{document}
In addition, I would like to be able to have 2 graphs one and the other superimposed: the first with the 2 signals (S1 and S2), the other below with the sum S
Could you help me ?
• Can't check now, but remember that angles are in degree by default in TikZ. Anyway, pgfplots is in my opinion the correct tool for this task... Dec 31, 2020 at 9:22
\draw plot ({\x},{cos(10*\x+cos(9.25*\x r)});
and it should be
\draw plot ({\x},{cos(10*\x r)+cos(9.25*\x r)});
For the second question, a simple way to do it could be the following:
\documentclass{standalone}
\usepackage {tikz}
\begin{document}
\begin{tikzpicture}[samples=1000]
\draw[->] (-5,0) -- (5.2,0) node[right] {$t$};
\draw[->] (-5,-2) -- (-5,2) node[above] {$S$};
\draw[color=blue] plot ({\x},{cos(10*\x r)});
\draw[color=green!60!black] plot ({\x},{cos(9.25*\x r)});
\node[color=blue] at (5,1.5) [left] {$y=\cos\omega_2x$};
\node[color=green!60!black] at (5,2) [left] {$y=\cos\omega_1x$};
% Change \y as you need, it's the distance between the two plots
\def\y{5}
\begin{scope}[shift={(0,-\y)}]
\draw[->] (-5,0) -- (5.2,0) node[right] {$t$};
\draw[->] (-5,-2) -- (-5,2) node[above] {$S$};
\draw plot ({\x},{cos(10*\x r)+cos(9.25*\x r)});
\node at (5,2) [left] {$y=\cos\omega_1x+\cos\omega_2x$};
\end{scope}
\end{tikzpicture}
\end{document}
And finally, this is what I get:
• Nice ! Shouldn't we see a separation on the y axis with an arrow corresponding to the top of the axis of the bottom figure? Dec 31, 2020 at 11:19
• You are right. I'm going to do an edit and change that. Dec 31, 2020 at 11:40 | 852 | 2,390 | {"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} | 2.640625 | 3 | CC-MAIN-2022-27 | latest | en | 0.721148 |
http://www.authorstream.com/Presentation/embeddedinstitute-3968441-two-dimensional-array-ptinstitute/ | 1,611,154,930,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703520883.15/warc/CC-MAIN-20210120120242-20210120150242-00385.warc.gz | 120,867,447 | 29,072 | # Two Dimensional Array - PTInstitute
Views:
## Presentation Description
Hi everyone, here we are presenting a presentation about 2-dimensional array. An array of arrays is known as a 2-dimensional array. The two-dimensional array in programming is also known as the matrix. A matrix can be represented as a table of rows and columns. We demonstrate how to store the elements entered by the user in a 2d array and how to display the elements of a two-dimensional array. The individual elements of the above array can be accessed by using two subscript instead of one. The first subscript denotes row number and second denotes column number. As we can see in the above image both rows and columns are indexed from 0. So the first element of this array is at a[0][0] and the last element is at a[1][2]. At Embedded training institute, teach 2-dimensional array very clear step by steps. Read More: https://www.ptinstitute.in
## Presentation Transcript
### slide 1:
TWO DIMENSIONAL ARRAY By Professional Training Institute Professional Embedded Training Institute in Bangalore
### slide 2:
TWO DIMENSIONAL ARRAY We have introduced array as if we want to save a hundred values then its difficult to declare variables so we used array .now if we want hundred such arrays then we can have two-dimensional arrays .so an array of arrays is known as 2d array. Through our website you can learn more.
### slide 3:
1.1 PROGRAM TO CHECK WHETHER THE TWO33 MATRICES ARE EQUAL OR NOT First we taken the two33 matrixes afterwards we need to check that all elements inside the two matrices are equal or not in both matrices with respect to position wise also if all elements with respect to position in two matrices are equal then prints both matrices are EQUAL otherwise prints both matrixes are not equal. DESCRIPTION
### slide 4:
10 mins includestdio.h int main int matrix133matrix233ijcount0 printf“enter the elements of first 33 matrix\n” fori0i3i++ //rows forj0j3j++// columns scanf“d” matrix1ij //input from the user printf“enter the elements of second 33 matrix\n” fori0i3i++ forj0j3j++ scanf“d” matrix2ij fori0i3i++ //loop for checking whether two matrixes are equal or not forj0j3j++ ifmatrix1ijmatrix2ij count++// increment the count value if the elements are not equal ifcount0 printf“2 matrixes are equal\n” else printf“2 matrixes are not equal\n” 15 mins
### slide 5:
1.2 TO PERFORM SCALAR MULTIPLICATION OF33 MATRIX Here we can multiply any constant element with the 33 matrixor any other matrix.first user ask the input for matrix along with the constant variable. DESCRIPTION
### slide 6:
10 mins includestdio.h//preprocessor directive with header file int main int matrix1010ijnconstantrowcolmatrix110 10 printf“enter the number of rows of the array elements\n” scanf“d” row printf“enter the number of columns of the array elements\n” scanf“d” col printf“enter the matrix elements\n” fori0irowi++ //for row elements forj0jcolj++ //for column elements scanf“d” matrixij printf“enter the constant element to multiplied with the matrix\n” scanf“d” constant fori0irowi++ //loop for multiplying a constant with the given matrix forj0jcolj++ matrix1ijconstantmatrixij printf“the new matrix is\n” fori0irowi++ //loop for printing the multiplied matrix 15 mins
### slide 7:
10 mins forj0jcolj++ printf” d” matrix1ij printf“\n” //end of main function 15 mins
### slide 8:
1.3 TO FIND THE SUM OF EACH ROW AND COLUMN OF A 33 MATRIX First We have to take33 matrixthen we need to find the sum of each elements of the row1afterwards sum of each elements of row2 and then row3 similar method is needed to find the sum of each elements of columns also afterwards print all the sums individually. DESCRIPTION
### slide 9:
10 mins includestdio.h int main int matrix33ijsumrow0sum1 printf“enter the elements of the matrix33\n” fori0i3i++// scanning for rows forj0j3j++// scanning for columns scanf“d” matrixij fori0i3i++//loop for finding the sum of each rows and columns sum0 sum10 forj0j3j++ sumsum+matrixij//sum is sum of each rows sum1sum1+matrixji//sum1 is sum of each columns printf“sum of rowdd\n” rowsum//printing sum of each rows value printf“sum of columndd\n” rowsum1//printing sum of each column value row++ 15 mins
### slide 10:
1.4 MULTIPLICATION OF TWO 33 MATRIX First we take the two 33 matrixesor any afterwards we need to check the rows of the first matrix is equal to the columns of the second matrix if it is satisfied then only you to process for further steps otherwise print no multiplication is possibleif it is satisfied then only multiply 2 matrix. DESCRIPTION
### slide 11:
10 mins includestdio.h int main int matrix133matrix233matrix333 row1col1row2col2ijkmul printf“enter the number of rows of the first matrix\n” scanf“d” row1 printf“enter the columns of the first matrix\n” scanf“d” col1 printf“enter the elements of the first matrix\n” fori0irow1i++ forj0jcol1j++ scanf“d” matrix1ij printf“enter the number of rows of second matrix\n” scanf“d” row2 printf“enter the number of columns of second matrix\n” scanf“d” col2 ifrow1col2 //if rows of the 1st matrix not equal to second matrix exit program otherwise continue printf“multiplication not possible\n” else 15 mins
### slide 12:
10 mins printf“enter the elements of the second matrix\n” fori0irow2i++ forj0jcol2j++ scanf“d” matrix2ij printf“multiplication of two matrix is\n” fori0irow1i++ // matrix multiplication takes place forj0jcol2j++ mul0 fork0kcol1k++ mulmul+matrix1ikmatrix2kj printf” d” mul printf“\n” 15 mins
### slide 13:
1.5 PROGRAM TO STORE 10 NAME INTO 2D ARRAY AND PRINT THEM We declare a 2 dimensional character array of required sizeafterwards we need to store names and print them. DESCRIPTION
### slide 14:
10 mins includestdio.h int main char names2020 int rowcolij printf“enter the rows and columns of the 2d array\n” scanf“d d” rowcol printf“enter the names you want to print\n” fori0irowi++ scanf“s” namesi0//user input to store names fori0irowi++ printf“s\n” namesi0//printing the names 15 mins
### slide 15:
1.6 PROGRAM TO SWAP THE TWO ARRAYS To swap two strings in c programming we want to ask user to enter two strings and then make a temporary variable of same type and then place elements of string 1 in temp and and elements of string 2 in 1 and then temp in string 2. DESCRIPTION
### slide 16:
10 mins includestdio.h int main char array110array210temp10”\0″ int len10len20ij printf“enter the elements of the first array\n” scanf“s” array1 printf“enter the elements of the second array\n” scanf“s” array2 whilearray1len1’\0′//loop for calculating the length of the first array len1++ printf“the length of the first array is d\n” len1 whilearray2len2’\0′//loop for calculating the length of the second array len2++ printf“the length of the second array is d\n” len2 fori0ilen1i++//using temporary variable we swap the 2 given arrays and print after the swapping process tempiarray1i 15 mins
### slide 17:
10 mins fori0ilen2i++ array1iarray2i fori0ilen1i++ array2itempi fori0ilen2i++ printf“c” array1i printf“\n” forj0jlen1j++ printf“c” array2j 15 mins
### slide 18:
1.7 PROGRAM TO CHECK WHETHER THE GIVEN STRING IS PALINDROME OR NOT A string is palindrome if the reverse of that string is equal to original string.firstly we need to declare a character array of some size after we need to obtain input string from the user and take another array and copy the reversing order of first string and compare the both strings if it is equal print as the given string is palindrome else print the given string is not an palindrome. DESCRIPTION
### slide 19:
10 mins includestdio.h includestring.h int main char str10rev10”\0″ int length0ij printf“enter the string\n” scanf“s” str whilestrlength’\0′ length++ printf“length of the string isd\n” length forilength-1j0i0jlengthi–j++/ for loop for reversing a original string and store it inanother character array/ revistrj fori0ilengthi++//comparing original string with the reversing string ifrevistri j1 else j0 ifj1 printf“palindrome\n” else printf“not an palindrome\n” 15 mins
### slide 20:
1.8 PROGRAM TO INTERCHANGING THE DIAGONALS OF MATRIX Firstly we take the matrix only square matrixafterwards we need to interchange the left diagonal elements towards right and vice-versa and print the result of interchanging the diagonal elements. DESCRIPTION
### slide 21:
10 mins includestdio.h int main int a1010rowcolumnijtemp printf“enter the number of rows and columns of the matrix\n” scanf“d d” rowcolumn printf“enter the elements of the matrix\n” fori0irowi++ forj0jcolumnj++ scanf“d” aij fori0irowi++ tempaii aiiairow-i-1 airow-i-1temp fori0irowi++ forj0jcolumnj++ printf” d” aij printf“\n” 15 mins
### slide 22:
1.9 WRITE A PROGRAM TO PRINT UPPER TRIANGULAR MATRIX We have to take one square matrix and print the upper triangular matrix only remaining the term should be zero or empty. DESCRIPTION
### slide 23:
10 mins includestdio.h int main int matrix33rowcolumn printf“enter the 33 matrix elements\n” forrow0row3row++ forcolumn0column3column++ scanf“d” matrixrowcolumn forrow0row3row++ forcolumn0column3column++ ifcolumnrow matrixrowcolumnmatrixrowcolum n//assign the elements as it is if column is greater or equalelse it is 0 else matrixrowcolumn0 forrow0row3row++ forcolumn0column3column++ printf” d” matrixrowcolumn printf“\n” 15 mins
### slide 24:
Contact Us Professional Training Institute 32 38/1 2rd FloorHosur Main Road Near Bosch Office Bommanahalli Bengaluru Karnataka-560068 +918951422196 www.ptinstitute.in | 2,513 | 9,508 | {"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-2021-04 | latest | en | 0.838842 |
http://slideplayer.com/slide/3768150/ | 1,529,766,358,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865081.23/warc/CC-MAIN-20180623132619-20180623152619-00002.warc.gz | 306,104,562 | 20,829 | # 1 ADVANCED STATISTICAL METHODS IN AGRICULTURAL RESEARCH UJI HIPOTESIS SUMBER: www.aaec.ttu.edu/faculty/omurova/aaec_4302/.../Chapter%2012a.ppt
## Presentation on theme: "1 ADVANCED STATISTICAL METHODS IN AGRICULTURAL RESEARCH UJI HIPOTESIS SUMBER: www.aaec.ttu.edu/faculty/omurova/aaec_4302/.../Chapter%2012a.ppt"— Presentation transcript:
1 ADVANCED STATISTICAL METHODS IN AGRICULTURAL RESEARCH UJI HIPOTESIS SUMBER: www.aaec.ttu.edu/faculty/omurova/aaec_4302/.../Chapter%2012a.ppt
2 UJI HIPOTESIS STATISTIK Two contradictory hypotheses: –Null hypothesis - H 0 –Alternative hypothesis – H 1 Three sets of hypothesis: H 0 : H 1 : H 0 : H 1 : H 0 : H 1 : SUMBER: www.aaec.ttu.edu/faculty/omurova/aaec_4302/.../Chapter%2012a.ppt
3 Basic significance Test: H 0 : H 1 : Decision rule: Reject H 0 if Reject H 0 Do not reject H 0 Reject H 0 SUMBER: www.aaec.ttu.edu/faculty/omurova/aaec_4302/.../Chapter%2012a.ppt UJI HIPOTESIS STATISTIK
4 2 types of mistakes: H 0 is trueH 0 is false (H 1 is false)(H 1 is true) _________________________________________ Reject H 0 Error –Type ICorrect Decision __________________________________________ Do not Reject H 0 Correct DecisionError – Type II _________________________________________________ SUMBER: www.aaec.ttu.edu/faculty/omurova/aaec_4302/.../Chapter%2012a.ppt UJI HIPOTESIS STATISTIK
5 Linear transformation that yields a random variable Z that has a normal distribution ( µ=0, σ=1) Critical value Z c is determined from Pr(|Z|≥ Z c ) = ά SUMBER: www.aaec.ttu.edu/faculty/omurova/aaec_4302/.../Chapter%2012a.ppt UJI HIPOTESIS STATISTIK
6 How to conduct the t-test: 1)State the hypotheses 2)Choose the level of significance α 3)Construct the decision rule 4)Determine the value of the test statistics t* 5)State and interpret the conclusion of the test SUMBER: www.aaec.ttu.edu/faculty/omurova/aaec_4302/.../Chapter%2012a.ppt UJI HIPOTESIS STATISTIK
7 CONTOH: Y i = B 0 + B 1 X 1 + B 2 X 2 + U i Ŷ i = B 0 + B 1 X 1 + B 2 X 2 Ŷ i = 474.05 + 1.46X 1 +26.32X 2 Where:Y i = Cotton Yields (lbs/ac) X 1 = Phosphorous Fertilizer (lbs/ac) X 2 = Irrigation Water (in/ac) ^^ ^ SUMBER: www.aaec.ttu.edu/faculty/omurova/aaec_4302/.../Chapter%2012a.ppt UJI HIPOTESIS STATISTIK
8 Interpreting Summary Output from Excel Total number of observations 134 SUMBER: www.aaec.ttu.edu/faculty/omurova/aaec_4302/.../Chapter%2012a.ppt
9 Some General Remarks A “rule of thumb” is that: |t j * | ≥ 2 β j is statistically different from zero, at least at the 95% level of statistical certainty SUMBER: www.aaec.ttu.edu/faculty/omurova/aaec_4302/.../Chapter%2012a.ppt
10 Some General Remarks One-tail test vs. two-tail test Advantage If you properly justify that X j has only a positive (negative) effect on the dependent variable Y i, then the one-tail test will help you reject the null hypothesis. Under a one-tail test, the critical t-value is smaller than the critical t-value under a two-tail test. SUMBER: www.aaec.ttu.edu/faculty/omurova/aaec_4302/.../Chapter%2012a.ppt
11 Some General Remarks One-tail test vs. two-tail test Disadvantage If you decide that X j has only a positive effect on Y, than you cannot change your decision after running the regression. SUMBER: www.aaec.ttu.edu/faculty/omurova/aaec_4302/.../Chapter%2012a.ppt | 1,096 | 3,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-26 | latest | en | 0.597219 |
https://www.proofwiki.org/wiki/Cardinality_of_Set_of_Injections/Formal_Proof | 1,680,003,651,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00320.warc.gz | 1,084,132,693 | 12,408 | # Cardinality of Set of Injections/Formal Proof
## Theorem
Let $S$ and $T$ be finite sets.
The number of injections from $S$ to $T$, where $\card S = m, \card T = n$ is often denoted ${}^m P_n$, and is:
${}^m P_n = \begin{cases} \dfrac {n!} {\paren {n - m}!} & : m \le n \\ 0 & : m > n \end{cases}$
## Proof
Let $m > n$.
By the Pigeonhole Principle, there can be no injection from $S$ to $T$ when $\card S > \card T$.
Once $\card T$ elements of $S$ have been used up, there is no element of $T$ left for the remaining elements of $S$ to be mapped to such that they all still map to different elements of $T$.
Let $m = 0$.
The only injection from $\O \to T$ is $\O \times T$ which is $\O$.
So if $m = 0$ there is $1 = n! / n!$ injection.
Let $0 < m \le n$.
As in the proof of Cardinality of Set of All Mappings, we can assume that $S = \N_m$ and $T = \N_n$.
For each $k \in \closedint 1 n$, let $\map {\mathbb H} {k, n}$ be the set of all injections from $\N_k$ to $\N_n$.
The proof now proceeds by induction.
Let:
$\ds \mathbb S = \set {k \in \closedint 1 n: \card {\map {\mathbb H} {k, n} } = \frac {n!} {\paren {n - k}!} }$
### Basis for the Induction
Let $k = 1$.
From Cardinality of Set of All Mappings, there are $n^1 = n$ different mappings from $S$ to $T$.
From Mapping from Singleton is Injection, each one of these $n$ mappings is an injection.
Thus:
$\ds \card {\map {\mathbb H} {1, n} } = n = \frac {n!} {\paren {n - 1}!}$
and so it follows that:
$1 \in \mathbb S$.
This is the basis for the induction.
### Induction Hypothesis
We suppose that:
$\ds \card {\map {\mathbb H} {k, n} } = \frac {n!} {\paren {n - k}!}$.
This is the induction hypothesis.
We need to show that:
$\ds \card {\map {\mathbb H} {k + 1, n} } = \frac {n!} {\paren {n - \paren {k + 1} }!}$.
### Induction Step
This is the induction step:
Let $k \in \mathbb S$ such that $k < n$.
Let $\rho: \map {\mathbb H} {k + 1, n} \to \map {\mathbb H} {k, n}$ be the mapping defined by:
$\forall f \in \map {\mathbb H} {k + 1, n}: \map \rho f =$ the restriction of $f$ to $\N_k$
Given that $g \in \map {\mathbb H} {k, n}$ and $a \in \N_n - \map g {\N_k}$, let $g_a: \N_{k + 1} \to \N_n$ be the mapping defined as:
$\map {g_a} x = \begin {cases} \map g x & : x \in \N_k \\ a & : x = k \end {cases}$
Now $g$ is an injection as $g \in \map {\mathbb H} {k, n}$, and as $\map {g_a} a \notin \map g {\N_k}$ it follows that $g_a$ is also an injection.
Hence:
$g_a \in \map {\mathbb H} {k + 1, n}$
It follows from the definition of $\rho$ that:
$\map {\rho^{-1} } {\set g} = \set {g_a: a \in \N_n - \map g {\N_k} }$
Since $g$ is an injection, $\map g {\N_k}$ has $k$ elements.
Therefore $\N_n - \map g {\N_k}$ has $n - k$ elements by Cardinality of Complement.
As $G: a \to g_a$ is clearly a bijection from $\N_n - \map g {\N_k}$ onto $\map {\rho^{-1} } {\set g}$, that set has $n - k$ elements.
Clearly:
$\set {\map {\rho^{-1} } {\set g}: g \in \map {\mathbb H} {k, n} }$
is a partition of $\map {\mathbb H} {k + 1, n}$.
Therefore by Number of Elements in Partition:
$\card {\map {\mathbb H} {k + 1, n} } = \paren {n - k} \dfrac {n!} {\paren {n - k}!} = \dfrac {n!} {\paren {\paren {n - k} - 1}!}$
as $k \in \mathbb S$.
But:
$\paren {n - k} - 1 = n - \paren {k + 1}$
So:
$k + 1 \in \mathbb S$
By induction:
$\mathbb S = \closedint 1 n$
and in particular:
$m \in \mathbb S$
$\blacksquare$ | 1,269 | 3,410 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-14 | latest | en | 0.780921 |
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Some of our calculators and applications let you save application data to your local computer. These applications will - due to browser restrictions - send data between your browser and our server. We don't save this data. | 992 | 4,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.786942 |
http://www.edupil.com/question/measure-each-interior-angle-regular-hexagon/ | 1,519,445,318,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815318.53/warc/CC-MAIN-20180224033332-20180224053332-00076.warc.gz | 441,007,882 | 16,109 | # Measure of each interior angle of a regular hexagon is:
1. 45°
2. 120°
3. 100°
4. 60°
Monis Rasool Professor Asked on 21st November 2015 in
2 Answer(s)
Answer : (2) 120
sujeetstar24 Newbie Answered on 21st November 2015.
Answer: (2) 120
Explanation:-
Formula –
Each interior angle of a regular polygon = 180 (n – 2) /n
Where, n = number of sides of a polygon.
Then, Each interior angle of a regular hexagon = 180 (6 – 2)/6
= 180 x 4 /6
= 30 x 4
= 120 degree.
Hence, the answer is (2) 120 degree.
Anurag Mishra Professor Answered on 22nd November 2015.
### Your Answer
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Electricity and Magnetism play a profound role in almost all aspects of our daily lives... from the moment we wake up, to the moment we go to sleep (and even while we're sleeping), applications of electricity and magnetism provide us tools, light, warmth, transportation, communication, and even entertainment. Despite its widespread use, however, there is much we're still learning every day about these phenomena!
## Objectives
### Electrostatics
1. Calculate the charge on an object.
2. Describe the differences between conductors and insulators.
3. Explain the difference between conduction and induction.
4. Explain how an electroscope works.
5. Solve problems using the law of conservation of charge.
6. Use Coulomb's Law to solve problems related to electrical force.
7. Recognize that objects that are charged exert forces, both attractive and repulsive.
8. Compare and contrast Newton's Law of Universal Gravitation with Coulomb's Law.
9. Define, measure, and calculate the strength of an electric field.
10. Solve problems related to charge, electric field, and forces.
11. Define and calculate electric potential energy.
12. Define and calculate potential difference.
13. Solve basic problems involving charged parallel plates.
### Electric Circuits
1. Define and calculate electric current.
2. Define and calculate resistance using Ohm's law.
3. Explain the factors and calculate the resistance of a conductor.
4. Identify the path and direction of current flow in a circuit.
5. Draw and interpret schematic diagrams of circuits.
6. Effectively use and analyze voltmeters and ammeters.
7. Solve series and parallel circuit problems using VIRP tables.
8. Calculate equivalent resistances for resistors in both series and parallel configurations.
9. Calculate power and energy used in electric circuits.
Magnetism
1. Explain that magnetism is caused by moving electrical charges.
2. Describe the magnetic poles and interactions between magnets.
3. Draw magnetic field lines.
4. Describe the factors affecting an induced potential difference due to magnetic field lines interacting with moving charges. | 434 | 2,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-13 | latest | en | 0.868824 |
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• how to solve mathematical equation | 1,247 | 5,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-26 | latest | en | 0.861448 |
https://www.physicsforums.com/threads/gravitational-potential-energy.867993/ | 1,513,377,128,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948579567.73/warc/CC-MAIN-20171215211734-20171215233734-00102.warc.gz | 768,483,384 | 18,246 | # B Gravitational potential energy
1. Apr 20, 2016
### alex36
Suppose the mass of planet is" M" and there is body in its surface whose mass is "m" and the field strength is "g" . If the body is thrown 1800 m then Gravitational Potential energy = mg(1800). My question is why cant we use formula GPE= GMm/x ? This is also the formula for gpe but why cant we apply it in this condition?
2. Apr 20, 2016
### robphy
What is the physical interpretation of x?
3. Apr 20, 2016
### alex36
distance from the surface of planet
4. Apr 20, 2016
### Jonathan Scott
The assumption that the field strength is constant is a simplification which applies when the height involved is small compared with the distance to the centre of the source object.
In this case, we can simply use the approximate formula mgh for the energy when mass m is moved in field g = GM/x^2 through height h. The -GMm/x formula is correct only if x is the distance to the centre of the planet. In that case the change in potential energy can also be written accurately as (-GMm/(x+h)) - (-GMm/x) which is approximately the same as mgh provided that h is small compared with x.
5. Apr 22, 2016
### alex36
Isn't it (-GMm/x-(-GMm/(x+h))? because energy we get is negative from your form of euation . does it matters?
6. Apr 22, 2016
### Jonathan Scott
I'm assuming that something is being thrown upwards, so a positive amount of potential energy being given to the small mass.
The Newtonian potential energy in the form -GMm/x is relative to infinite separation, so it gets lower the closer one gets to the source. To compare it at two different heights you subtract the potential energy values. In this case, the higher energy is the the one which involves (x+h), which is less negative, making the difference positive in the form I gave originally.
Of course, the mgh form also needs care with the sign. This is the potential energy lost when the small mass falls a distance h in the same direction as the field g, so it is also the same as the potential energy which has to be given to the small mass to move it a distance h in the opposite direction to the field g.
7. Apr 22, 2016
### hackhard
a simple concept-
there is no such thing as "absolute potential energy"
when you say a body has 20m/s speed you also need to say -- in which frame ?
when you say a bode has 20J potential energy you also need to define--- which point have you assumed as ground potential (0J)
mgh gives gravitational potential energy (due to earths gravity) assuming earth surface to be at ground potential (at 0J)
GPE= GMm/x gives gravitational potential energy (due to earths gravity) assuming infinity to be at ground potential (at 0J)
both are correct provided you also mention which point have assumed to be at ground potential
simply saying " potential energy of a body is 200J" is non sense
you say " potential energy of a body is 200J wrt point A"
so a body can have all real values as potential energy at same point of time but a unique one wrt to a choice of ground
8. Apr 22, 2016
### alex36
I just checked . Answer will have same value but different sign . Am I correct?
9. Apr 22, 2016
### Jonathan Scott
The answer should have the same sign whichever way you calculate it.
One way, the energy is GMmh/x(x+h), which is approximately m (GM/x^2) h, where GM/x^2 is the magnitude of g, and the other way is mgh where g is the magnitude of the field and h is assumed to be upwards.
If you want to be accurate about signs, the energy given to the mass in the mgh form is actually -mg.h if the field and height are described by vectors, because the force being applied to the mass is in the opposite direction to gravity, but the displacement through which it acts is in the forward direction, so we have -m(-GM/x^2) h = m (Gm/x^2) h as before.
10. Apr 22, 2016
### alex36
Thank you so much :) | 969 | 3,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-51 | longest | en | 0.949391 |
https://www.edibon.com/zh/slotted-link-mechanism | 1,685,790,063,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649193.79/warc/CC-MAIN-20230603101032-20230603131032-00609.warc.gz | 798,396,302 | 34,084 | ## 創新系統
The Slotted Link Mechanism, “MBM1”, is an example of a quick-return mechanism, capable of transforming circular motion intoreciprocating motion.
## 一般說明
The Slotted Link Mechanism, “MBM1”, is an example of a quick-return mechanism, capable of transforming circular motion intoreciprocating motion. It is made of aluminum and consists of a rotary element (graduated disk), called crank, connected to a rigid bar,called connecting rod.
When rotating the crank, the connecting rod moves back and forward. The rotation motion of a crank or crankshaftcauses a rectilinear reciprocating motion of a piston or plunger. It is a reversible system through which the connecting rod can be displacedby turning the crank and the other way round. If the connecting rod generates the input motion (as a piston in a car engine does), thecrank is forced to rotate.
## 練習和指導練習
### 手册中包含的指导实践练习
1. Demonstration of the action of a simple crank and slotted link mechanism.
2. Graphic determination of the relationship between the linear displacement of the sliding block and the angular displacement of the input crank.
### 与该单位进行更多实际操作
1. Los ejercicios más avanzados pueden incluir la determinación de la velocidad y aceleración del bloque deslizante mediantediferenciación gráfica y comparación con los valores obtenidospor diagramas de velocidad y aceleración. | 321 | 1,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-23 | latest | en | 0.527783 |
www.spankulator.com | 1,610,854,480,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703509104.12/warc/CC-MAIN-20210117020341-20210117050341-00246.warc.gz | 176,477,225 | 6,716 | Please fill out the red boxes in the following form as honestly as possible. At most there will be 4 questions. Dig deep, pull hard, remember back if you have to. They were good times. Every sticky one of them. The box will turn green when your answer has been accepted, when all the answers have been filled in, you'll see a 'Calculate' button.
1. I am years old.
2. During my prime monkey spanking teenage years - let's say from 14 to 19 - I reckon I would beat the bishop about times per week.
3. From 20 to now, despite (sometimes) having a job, crushing responsibilities and when I'm really lucky someone else to take the 'pressure' off - on average I would say I managed to polish the rifle about times per week.
4. If I was to make a very rough guess, the average time it takes for me to get from zero to firing off some knuckle children is about minutes.
Please fill in all the red boxes...
As a 40 year old male, based on your answers, you have jacked the beanstalk approximately:
4900 times
Using your average time to vinegar stroke, you've spent around:
48360 minutes
of your life shaking the ketchup bottle. Or in a more readable way, that works out to be:
33 days
16 hours
0 mins
Getting a little bit more graphic; if you were to work out the amount of baby batter that amounts to, you'd be looking at something like 11402ml, which in a more (or perhaps less) digestable format is about:
22.6 pints
Aha, the curious type. Based on over 30 studies, the volume of seminal fluid in an ejaculation averages out to be 3.4ml. However, whilst we're all aware that the first pop *might* be 3.4ml, if you keep going, you'll eventually hit the ghost load.
We're also all aware that this doesn't last forever and pretty soon (maybe the next day) you're ready to go again. So we looked into how much seminal fluid you make each day. This turned out to be a pretty hard number to find. And believe us, we read a ton. The commonly cited number, however, is 0.4ml (Clark and Sherins, 1986).
Except based on our own (and our colleagues) first hand (pun intended) knowledge, plus a ton of reading on forums, websites, and goodness knows where else, we sort of concluded that a) the body holds more than one load (again, super hard to find out volume in the body. What do you go on? Ball size? Prostate size? 2-5% comes from the testes, 65-75% from the seminal vesicles, 25-30% from the prostate and less than 1% from the bulbourethral glands), and b) that you could probably shoot two pretty decent loads in a week.
So we crunched some numbers and went on one time being 3.4ml, two times 6.4ml (a slight decrease) and then pretty much halved each spurt from that point onwards (which averages out to about 1.4ml a go if you're on it once a day). Which seemed fair enough. But if anyone has a *really decent* study on this, we'll re-run the numbers.
To complicate matters further, we also discovered "after adjusting for covariates, semen volume decreased by 0.03ml per year of age" from the age of 22. So we've thrown that math in as well. We made our baseline 3.4ml up to 22, when it decreases by 0.03ml for every subsequent year.
In case you're curious, the seminal fluid isn't ALL sperm. It's mostly there to help the sperm do their 'thang'... But in a healthy, fully grown male there's something like 300 million swimmers in every ml of seminal fluid.
Lesson over. Phew.
Now let's assume you live to the ripe old age of 84, the current expected age for a man. Were you to continue to beat the meat as often as you are now, you would have another glorious:
6532 times
to look forward to. Which, if you're going to make some space in the diary, is about:
47 days
16 hours
33 mins
It's also another:
26.76 pints
or so that you're going to have to find tissues for :-)
Massive thanks to our sponsor! If you're (over 18 and) looking for an extra pair of hands - it takes 30 seconds to check them out:
Gents, if you're want to keep milking the trouser snake to 84 (and beyond) don't forget to check your testicles regularly, and as and when, make sure your prostate is as it should be.
If you've enjoyed this... Please share! | 1,042 | 4,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-04 | longest | en | 0.966231 |
https://metanumbers.com/249976100 | 1,643,020,971,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304528.78/warc/CC-MAIN-20220124094120-20220124124120-00585.warc.gz | 451,016,819 | 7,639 | # 249976100 (number)
249,976,100 (two hundred forty-nine million nine hundred seventy-six thousand one hundred) is an even nine-digits composite number following 249976099 and preceding 249976101. In scientific notation, it is written as 2.499761 × 108. The sum of its digits is 38. It has a total of 6 prime factors and 36 positive divisors. There are 90,900,000 positive integers (up to 249976100) that are relatively prime to 249976100.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 9
• Sum of Digits 38
• Digital Root 2
## Name
Short name 249 million 976 thousand 100 two hundred forty-nine million nine hundred seventy-six thousand one hundred
## Notation
Scientific notation 2.499761 × 108 249.9761 × 106
## Prime Factorization of 249976100
Prime Factorization 22 × 52 × 11 × 227251
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 24997610 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 249,976,100 is 22 × 52 × 11 × 227251. Since it has a total of 6 prime factors, 249,976,100 is a composite number.
## Divisors of 249976100
36 divisors
Even divisors 24 12 6 6
Total Divisors Sum of Divisors Aliquot Sum τ(n) 36 Total number of the positive divisors of n σ(n) 5.91764e+08 Sum of all the positive divisors of n s(n) 3.41788e+08 Sum of the proper positive divisors of n A(n) 1.64379e+07 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 15810.6 Returns the nth root of the product of n divisors H(n) 15.2073 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 249,976,100 can be divided by 36 positive divisors (out of which 24 are even, and 12 are odd). The sum of these divisors (counting 249,976,100) is 591,764,208, the average is 164,378,94.,666.
## Other Arithmetic Functions (n = 249976100)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 90900000 Total number of positive integers not greater than n that are coprime to n λ(n) 454500 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 13662343 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 90,900,000 positive integers (less than 249,976,100) that are coprime with 249,976,100. And there are approximately 13,662,343 prime numbers less than or equal to 249,976,100.
## Divisibility of 249976100
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 3 4 2
The number 249,976,100 is divisible by 2, 4 and 5.
• Abundant
• Polite
## Base conversion (249976100)
Base System Value
2 Binary 1110111001100101010100100100
3 Ternary 122102101002202002
4 Quaternary 32321211110210
5 Quinary 1002443213400
6 Senary 40445504432
8 Octal 1671452444
10 Decimal 249976100
12 Duodecimal 6b872118
20 Vigesimal 3i27050
36 Base36 44tusk
## Basic calculations (n = 249976100)
### Multiplication
n×y
n×2 499952200 749928300 999904400 1249880500
### Division
n÷y
n÷2 1.24988e+08 8.33254e+07 6.2494e+07 4.99952e+07
### Exponentiation
ny
n2 62488050571210000 15620519178393848081000000 3904756464190098407280864100000000 976095792368030458468282012348010000000000
### Nth Root
y√n
2√n 15810.6 629.94 125.74 47.8167
## 249976100 as geometric shapes
### Circle
Diameter 4.99952e+08 1.57065e+09 1.96312e+17
### Sphere
Volume 6.54311e+25 7.85248e+17 1.57065e+09
### Square
Length = n
Perimeter 9.99904e+08 6.24881e+16 3.5352e+08
### Cube
Length = n
Surface area 3.74928e+17 1.56205e+25 4.32971e+08
### Equilateral Triangle
Length = n
Perimeter 7.49928e+08 2.70581e+16 2.16486e+08
### Triangular Pyramid
Length = n
Surface area 1.08232e+17 1.8409e+24 2.04105e+08
## Cryptographic Hash Functions
md5 3f5aa1fa81f20e0e0986fa1f0eb6bc7b 53273c1d67da816ea672cbe0f034f6899bcf8c44 f04455b459f68687cdd981ae24d5740aa2a616ac636d224608cbaca83b95a5fd e9b5404690b236fdc85709ef8ded9134c73c7e13553722103fb9ec391c9d4441fd3caff08eb42a895db9898209f2d1efe91f340637f81c7fb48cd571da13272e 05e4f419dc4616449f74a67f796f864b5c7419ee | 1,590 | 4,471 | {"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-2022-05 | latest | en | 0.780237 |
http://www.nag.com/numeric/FL/nagdoc_fl24/html/F07/f07tuf.html | 1,386,533,732,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163806278/warc/CC-MAIN-20131204133006-00066-ip-10-33-133-15.ec2.internal.warc.gz | 452,696,723 | 4,697 | F07 Chapter Contents
F07 Chapter Introduction
NAG Library Manual
# NAG Library Routine DocumentF07TUF (ZTRCON)
Note: before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.
## 1 Purpose
F07TUF (ZTRCON) estimates the condition number of a complex triangular matrix.
## 2 Specification
SUBROUTINE F07TUF ( NORM, UPLO, DIAG, N, A, LDA, RCOND, WORK, RWORK, INFO)
INTEGER N, LDA, INFO REAL (KIND=nag_wp) RCOND, RWORK(N) COMPLEX (KIND=nag_wp) A(LDA,*), WORK(2*N) CHARACTER(1) NORM, UPLO, DIAG
The routine may be called by its LAPACK name ztrcon.
## 3 Description
F07TUF (ZTRCON) estimates the condition number of a complex triangular matrix $A$, in either the $1$-norm or the $\infty$-norm:
$κ1 A = A1 A-11 or κ∞ A = A∞ A-1∞ .$
Note that ${\kappa }_{\infty }\left(A\right)={\kappa }_{1}\left({A}^{\mathrm{T}}\right)$.
Because the condition number is infinite if $A$ is singular, the routine actually returns an estimate of the reciprocal of the condition number.
The routine computes ${‖A‖}_{1}$ or ${‖A‖}_{\infty }$ exactly, and uses Higham's implementation of Hager's method (see Higham (1988)) to estimate ${‖{A}^{-1}‖}_{1}$ or ${‖{A}^{-1}‖}_{\infty }$.
## 4 References
Higham N J (1988) FORTRAN codes for estimating the one-norm of a real or complex matrix, with applications to condition estimation ACM Trans. Math. Software 14 381–396
## 5 Parameters
1: NORM – CHARACTER(1)Input
On entry: indicates whether ${\kappa }_{1}\left(A\right)$ or ${\kappa }_{\infty }\left(A\right)$ is estimated.
${\mathbf{NORM}}=\text{'1'}$ or $\text{'O'}$
${\kappa }_{1}\left(A\right)$ is estimated.
${\mathbf{NORM}}=\text{'I'}$
${\kappa }_{\infty }\left(A\right)$ is estimated.
Constraint: ${\mathbf{NORM}}=\text{'1'}$, $\text{'O'}$ or $\text{'I'}$.
2: UPLO – CHARACTER(1)Input
On entry: specifies whether $A$ is upper or lower triangular.
${\mathbf{UPLO}}=\text{'U'}$
$A$ is upper triangular.
${\mathbf{UPLO}}=\text{'L'}$
$A$ is lower triangular.
Constraint: ${\mathbf{UPLO}}=\text{'U'}$ or $\text{'L'}$.
3: DIAG – CHARACTER(1)Input
On entry: indicates whether $A$ is a nonunit or unit triangular matrix.
${\mathbf{DIAG}}=\text{'N'}$
$A$ is a nonunit triangular matrix.
${\mathbf{DIAG}}=\text{'U'}$
$A$ is a unit triangular matrix; the diagonal elements are not referenced and are assumed to be $1$.
Constraint: ${\mathbf{DIAG}}=\text{'N'}$ or $\text{'U'}$.
4: N – INTEGERInput
On entry: $n$, the order of the matrix $A$.
Constraint: ${\mathbf{N}}\ge 0$.
5: A(LDA,$*$) – COMPLEX (KIND=nag_wp) arrayInput
Note: the second dimension of the array A must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{N}}\right)$.
On entry: the $n$ by $n$ triangular matrix $A$.
• If ${\mathbf{UPLO}}=\text{'U'}$, $A$ is upper triangular and the elements of the array below the diagonal are not referenced.
• If ${\mathbf{UPLO}}=\text{'L'}$, $A$ is lower triangular and the elements of the array above the diagonal are not referenced.
• If ${\mathbf{DIAG}}=\text{'U'}$, the diagonal elements of $A$ are assumed to be $1$, and are not referenced.
6: LDA – INTEGERInput
On entry: the first dimension of the array A as declared in the (sub)program from which F07TUF (ZTRCON) is called.
Constraint: ${\mathbf{LDA}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{N}}\right)$.
7: RCOND – REAL (KIND=nag_wp)Output
On exit: an estimate of the reciprocal of the condition number of $A$. RCOND is set to zero if exact singularity is detected or the estimate underflows. If RCOND is less than machine precision, $A$ is singular to working precision.
8: WORK($2×{\mathbf{N}}$) – COMPLEX (KIND=nag_wp) arrayWorkspace
9: RWORK(N) – REAL (KIND=nag_wp) arrayWorkspace
10: INFO – INTEGEROutput
On exit: ${\mathbf{INFO}}=0$ unless the routine detects an error (see Section 6).
## 6 Error Indicators and Warnings
Errors or warnings detected by the routine:
${\mathbf{INFO}}<0$
If ${\mathbf{INFO}}=-i$, the $i$th parameter had an illegal value. An explanatory message is output, and execution of the program is terminated.
## 7 Accuracy
The computed estimate RCOND is never less than the true value $\rho$, and in practice is nearly always less than $10\rho$, although examples can be constructed where RCOND is much larger.
A call to F07TUF (ZTRCON) involves solving a number of systems of linear equations of the form $Ax=b$ or ${A}^{\mathrm{H}}x=b$; the number is usually $5$ and never more than $11$. Each solution involves approximately $4{n}^{2}$ real floating point operations but takes considerably longer than a call to F07TSF (ZTRTRS) with one right-hand side, because extra care is taken to avoid overflow when $A$ is approximately singular.
The real analogue of this routine is F07TGF (DTRCON).
## 9 Example
This example estimates the condition number in the $1$-norm of the matrix $A$, where
$A= 4.78+4.56i 0.00+0.00i 0.00+0.00i 0.00+0.00i 2.00-0.30i -4.11+1.25i 0.00+0.00i 0.00+0.00i 2.89-1.34i 2.36-4.25i 4.15+0.80i 0.00+0.00i -1.89+1.15i 0.04-3.69i -0.02+0.46i 0.33-0.26i .$
The true condition number in the $1$-norm is $70.27$.
### 9.1 Program Text
Program Text (f07tufe.f90)
### 9.2 Program Data
Program Data (f07tufe.d)
### 9.3 Program Results
Program Results (f07tufe.r) | 1,822 | 5,368 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 71, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2013-48 | longest | en | 0.552789 |
https://en.openprof.com/wb/newton_s_first_law_of_motion_for_ks4_problem_9?ch=3509 | 1,680,035,302,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948871.42/warc/CC-MAIN-20230328201715-20230328231715-00207.warc.gz | 286,154,143 | 17,445 | Effects of Forces on Motion of a Body
# Effects of Forces on Motion of a Body problem 37
A car is driven on a flat road. The force of the engine is constantly opposed by the force of air resistance, which depends on the speed of the vehicle. Consider four cases:
• The engine force is equal to the resistance force. The magnitude of each is 0 N.
• The engine force is 800 N and the resistance force is 200 N.
• The engine force is equal to the resistance force. The magnitude of each is 400 N.
• The engine force is 200 N and the resistance force is 400 N.
Which of the cases does Newton's first law describe?
material editor: OpenProf website | 151 | 651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-14 | latest | en | 0.939752 |
http://stackoverflow.com/questions/10962069/how-can-i-have-show-display-the-name-of-a-function/10962168 | 1,438,114,984,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042982502.13/warc/CC-MAIN-20150728002302-00163-ip-10-236-191-2.ec2.internal.warc.gz | 225,586,934 | 18,837 | # How can I have Show display the name of a function?
As a simple exercise to get me acquainted with Haskell, after idling around on Youtube and stumbling into the American Countdown game show, I wanted to make a solver for the Numbers game.
You get 6 numbers and need to combine them with `(+) (-) (*) (/)` in order to get a given result.
What I've got so far is the very brain-dead,
``````let operands = [75, 2, 6, 3, 8, 7] :: [Double]
let goal = 623 :: Double
let operations = [(+), (-), (*), (/)]
show (head [(a, x, b, y, c, z, d, t, e) |
a <- operands,
b <- filter (\ q -> q /= a) operands,
c <- filter (\ q -> q /= a && q /= b) operands,
d <- filter (\ q -> q /= a && q /= b && q /= c) operands,
e <- filter (\ q -> q /= a && q /= b && q /= c && q /= d) operands,
x <- operations,
y <- operations,
z <- operations,
t <- operations,
t (z (y (x a b) c) d) e == goal])
``````
...but obviously Show doesn't know what to do with functions.
``````No instance for (Show (Double -> Double -> Double))
arising from a use of `show'
Possible fix:
add an instance declaration for (Show (Double -> Double -> Double))
``````
How can I work around this? Do I need to mess with type and data constructors to make my own functions that can print or is there some easier way around it?
-
`filter (\q -> q /= a && q /= b && q /= c && q /= d)` can be written as `filter (`notElem` [a,b,c,d])` – sdcvvc Jun 9 '12 at 15:27
@sdcvvc That's a use of backticks I wasn't previously aware of. – badp Jun 9 '12 at 15:46
As didactic material you can check this video [Dr. Graham Hutton - Functional Programming Fundamentals - The Countdown Problem] (channel9.msdn.com/shows/Going+Deep/…) – jedf Jun 9 '12 at 18:55
@badp: It's a section, just like `(+3)`; full version: `filter (\q -> notElem q [a,b,c,d])`. – sdcvvc Jun 9 '12 at 21:24
– sdcvvc Jun 9 '12 at 21:25
I would generally not advise you to implement a `Show` instance for functions. It is not the "haskelly" way of doing things, for this reason:
You define a "canonical" way of showing functions. You might want them to be shown as their names now, but what if you decide to do this:
``````add 0 y = y
``````
Should really the output of your program change depending on your internal implementation? That doesn't make much sense. Also, what happens to nameless functions?
Also, in another part of the program, you might want to show your functions as their types instead, etc., and then you'd have conflicting `Show` instances.
Generally, only implement `Show` when you know that there should be only one way of showing that thing, and that that way works for all values that need to be shown.
The simplest way of solving this problem would probably be to just store the name of the operation along with the operation. Like so:
``````let operations = [("+", (+)), ("-", (-)), ("*", (*)), ("/", (/))]
-- ...
show (head [(a, xname, b, yname, c, zname, d, tname, e) |
a <- operands,
b <- filter (\ q -> q /= a) operands,
c <- filter (\ q -> q /= a && q /= b) operands,
d <- filter (\ q -> q /= a && q /= b && q /= c) operands,
e <- filter (\ q -> q /= a && q /= b && q /= c && q /= d) operands,
(xname, x) <- operations,
(yname, y) <- operations,
(zname, z) <- operations,
(tname, t) <- operations,
t (z (y (x a b) c) d) e == goal])
``````
-
Another option:
``````data Operation = Add | Subtract | Multiply | Divide deriving (Show)
apply :: Operation -> Double -> Double -> Double
apply Subtract = (-)
apply Multiply = (*)
apply Divide = (/)
``````
-
You cannot print the name of a function in Haskell (there may be some crazy tricks with meta programming but essentially you cannot). For your program it is probably easiest to define the list of operations as a list of pairs with a string of the name of each operator e.g.
``````let operations = [((+)," + "), ((-), " - ")), ((*), " * ")), ((/), " / ")]
``````
You can get a list of just the operations by doing `map fst operations` and `map snd operations` to get their names.
- | 1,137 | 3,996 | {"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} | 2.734375 | 3 | CC-MAIN-2015-32 | latest | en | 0.876402 |
https://algebra-cheat.com/algebra-cheat/parallel-lines/rationalizing-denominator.html | 1,527,209,719,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866917.70/warc/CC-MAIN-20180525004413-20180525024413-00116.warc.gz | 515,908,448 | 12,806 | Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
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Author Message
reids
Registered: 06.05.2003
From:
Posted: Thursday 28th of Dec 12:21 Hi, I may sound really stupid to all the math experts here, but it’s been 2 years since I am learning rationalizing denominator solver, but I never found it interesting. In fact I always commit errors . I practise a lot, but still my grades do not seem to be getting better
Vofj Timidrov
Registered: 06.07.2001
From: Bulgaria
Posted: Saturday 30th of Dec 08:43 You seem to be facing a similar problem that I had some time back. I too thought of getting a paid tutor to work it out for me. But they are so costly that I just could not afford them. So I turned to the internet and found so many programs that can help with algebra assignments on slope, monomials and multiplying fractions. After some research I found that Algebrator is the best of the lot. I haven’t found a algebra assignment that I can’t get done through Algebrator. It is absolutely amazing. Best part is, the software gives you a step-by-step explanation on how to do it yourself. So you actually learn how to work it out yourself. Isn’t it cool?
Jrahan
Registered: 19.03.2002
From: UK
Posted: Monday 01st of Jan 07:33 I fully agree with what has just been said. Algebrator has always come to my rescue, be it an assignment or be it my preparation for the final exams, Algebrator has always helped me do well in algebra. It really helped me on topics like conversion of units, converting fractions and radical expressions. I would highly recommend this software.
Voumdaim of Obpnis
Registered: 11.06.2004
From: SF Bay Area, CA, USA
Posted: Tuesday 02nd of Jan 08:22 I recommend using Algebrator. It not only helps you with your math problems, but also provides all the required steps in detail so that you can improve the understanding of the subject.
Registered: 20.09.2004
From:
Posted: Tuesday 02nd of Jan 21:42 You people sure got me interested. I did not know such a program existed. So where can I buy it? Let me thank you in advance for the url
Dxi_Sysdech
Registered: 05.07.2001
From: Right here, can't you see me?
Posted: Wednesday 03rd of Jan 15:52 Thanks a lot for the elaborate information. We will surely check this out. Hope we get our problems finished with the help of Algebrator. If we have any technical clarifications with respect to its usage , we would definitely come back to you again. | 855 | 3,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-22 | latest | en | 0.924855 |
http://math.stackexchange.com/users/35055/david-spencer?tab=activity | 1,462,553,472,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861848830.49/warc/CC-MAIN-20160428164408-00199-ip-10-239-7-51.ec2.internal.warc.gz | 176,994,105 | 12,598 | David Spencer
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Jun 10 awarded Nice Answer Sep 24 awarded Autobiographer Jul 4 awarded Yearling Jul 4 awarded Yearling Jul 19 comment Why do we say the harmonic series is divergent? I certainly intended no offense, and I regret that my wording offended you. As @did says, I was simply checking your understanding, as your second sentence seems to imply (to me) that any constantly increasing series is divergent regardless of whether it tends to infinity at the limit. We can answer most usefully and clearly if we know whether to focus on the terminology or the proof. I hope there are no hard feelings. Jul 18 answered Why do these two methods of calculating the probability of winning a best-of-7 series give the same answer? Jul 18 awarded Commentator Jul 18 awarded Critic Jul 18 comment Determine if it is possible to fit 2 circles in a rectangle Unless I misunderstand, shouldn't it be $d[(r_1,r_1), (l-r_2,h-r_2)] \geq r_1+r_2$? Jul 18 revised Swatting flies with a sledgehammer Fixed typo in title Jul 18 suggested approved edit on Swatting flies with a sledgehammer Jul 18 comment Why do we say the harmonic series is divergent? Do you know what the word "divergent" means here? Jul 18 awarded Editor Jul 18 revised How many rolls until probability of a 5 is at least 1/2? improved clarity Jul 18 comment How many rolls until probability of a 5 is at least 1/2? That would be the probability of rolling a $5$ every time. An event of probability $P$ occurring $n$ times in a row has probability $P^n$, and the probability of the complement (opposite) of $P$ occurring during $n$ trials is $1-P^n$. Jul 18 answered How many rolls until probability of a 5 is at least 1/2? Jul 16 comment Solve for $a$: $V=2(ab+bc+ca)$ Not quite; perhaps I've not explained my hint clearly. Try it this way: I see you refer to "take out the a". Do this from your (correct) expression $\left(\frac{V}{2}\right)-bc=ab+ca$. Jul 16 comment Solve for $a$: $V=2(ab+bc+ca)$ Since you did that one step forwards and then backwards and got something different, you know that you did something wrong there. If you distribute your $2a$ in what you just wrote, you get $\left(\frac{V}{2}\right)-bc=2ab+2ca$. This is very close to what it should be, but the right is a bit different - there's an extra factor of $2$ on each term. That should tell you what you did wrong. Jul 16 comment Solve for $a$: $V=2(ab+bc+ca)$ By backwards, I mean take your expression $\dfrac{\left(\frac{V}{2}\right)-bc}{b+c}=2a$ and multiply it by $b+c$. Do you get the expression you had before you divided by $b+c$? Jul 16 answered Solve for $a$: $V=2(ab+bc+ca)$ | 717 | 2,684 | {"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} | 3.28125 | 3 | CC-MAIN-2016-18 | latest | en | 0.945923 |
https://egvideos.com/video/arkansas/grade-1/math/1.md.4/name-coins | 1,656,300,092,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103324665.17/warc/CC-MAIN-20220627012807-20220627042807-00621.warc.gz | 283,514,832 | 9,297 | # Arkansas - Grade 1 - Math - Measurement and Data - Name Coins - 1.MD.4
### Description
Identify and know the value of a penny, nickel, dime, and quarter.
• State - Arkansas
• Standard ID - 1.MD.4
• Subjects - Math Common Core
### Keywords
• Math
• Measurement and Data
## More Arkansas Topics
Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps.
Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes.
Compose two-dimensional shapes (e.g., rectangles, squares, trapezoids, triangles, half-circles, and quarter- circles) or three-dimensional shapes (e.g., cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape.
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. | 426 | 1,816 | {"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} | 4.3125 | 4 | CC-MAIN-2022-27 | longest | en | 0.878421 |
http://www.cr.ie.u-ryukyu.ac.jp/hg/Members/atton/delta_monad/file/4d615910c87a/agda/deltaM/functor.agda | 1,611,642,189,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704798089.76/warc/CC-MAIN-20210126042704-20210126072704-00280.warc.gz | 137,144,376 | 2,724 | ### view agda/deltaM/functor.agda @ 92:4d615910c87a
Prove preserve-id for deltaM
author Yasutaka Higa Mon, 19 Jan 2015 14:32:07 +0900 f41682b53992 8d92ed54a94f
line wrap: on
line source
```
open import Level
open import Relation.Binary.PropositionalEquality
open ≡-Reasoning
open import basic
open import delta
open import delta.functor
open import deltaM
open import laws
open Functor
module deltaM.functor where
deltaM-preserve-id : {l : Level} {A : Set l}
{M : {l' : Level} -> Set l' -> Set l'}
(functorM : {l' : Level} -> Functor {l'} M)
{monadM : {l' : Level} {A : Set l'} -> Monad {l'} {A} M functorM}
-> (d : DeltaM M {functorM} {monadM} A) -> (deltaM-fmap id) d ≡ id d
deltaM-preserve-id functorM (deltaM (mono x)) = begin
deltaM-fmap id (deltaM (mono x)) ≡⟨ refl ⟩
deltaM (fmap delta-is-functor (fmap functorM id) (mono x)) ≡⟨ refl ⟩
deltaM (mono (fmap functorM id x)) ≡⟨ cong (\x -> deltaM (mono x)) (preserve-id functorM x) ⟩
deltaM (mono (id x)) ≡⟨ cong (\x -> deltaM (mono x)) refl ⟩
deltaM (mono x) ∎
deltaM-preserve-id functorM (deltaM (delta x d)) = begin
deltaM-fmap id (deltaM (delta x d))
≡⟨ refl ⟩
deltaM (fmap delta-is-functor (fmap functorM id) (delta x d))
≡⟨ refl ⟩
deltaM (delta (fmap functorM id x) (fmap delta-is-functor (fmap functorM id) d))
≡⟨ cong (\x -> deltaM (delta x (fmap delta-is-functor (fmap functorM id) d))) (preserve-id functorM x) ⟩
deltaM (delta x (fmap delta-is-functor (fmap functorM id) d))
≡⟨ refl ⟩
appendDeltaM (deltaM (mono x)) (deltaM (fmap delta-is-functor (fmap functorM id) d))
≡⟨ refl ⟩
appendDeltaM (deltaM (mono x)) (deltaM-fmap id (deltaM d))
≡⟨ cong (\d -> appendDeltaM (deltaM (mono x)) d) (deltaM-preserve-id functorM (deltaM d)) ⟩
appendDeltaM (deltaM (mono x)) (deltaM d)
≡⟨ refl ⟩
deltaM (delta x d)
∎
{-
deltaM-covariant : {l : Level} {A B C : Set l} ->
(f : B -> C) -> (g : A -> B) -> (d : Delta A) ->
(delta-fmap (f ∙ g)) d ≡ (delta-fmap f) (delta-fmap g d)
deltaM-covariant = {!!}
-}
``` | 761 | 2,096 | {"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} | 2.703125 | 3 | CC-MAIN-2021-04 | latest | en | 0.337606 |
http://math4teaching.com/2011/07/09/algebra-problems/ | 1,432,755,625,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929096.44/warc/CC-MAIN-20150521113209-00247-ip-10-180-206-219.ec2.internal.warc.gz | 162,711,524 | 16,175 | Every now and then I get an e-mail from a friend’s son asking for help in algebra problems. When it’s about solving word problems, the email will start with “How about just telling me which one is the x and I’ll figure out the rest”. The follow-up email will open with “Done it. Thanks. All I need is the equation and I can solve the problem”. The third and final e-mail will be “Cool”. Of course I let this happen only when I’m very busy. Most times I try to explain to him how to represent the problem and set-up an equation. Here’s our latest exchange.
Josh: What is the measure of an angle if twice its supplement is 30 degrees wider than five times its complement? All I need is to know which one’s the x.
Me: How about sending me a drawing of the angle with its complement and supplement?
Josh: Is this ok?
Me: Great. Let me use your drawing to make a dynamic version using GeoGebra. Explore the applet below by dragging the point in the slider. What do you notice about the values of the angles? Which angle depend on which angle for its measure? If one of the measure of one of the angles is represented by x, how will you represent the other angles? (Click here for the procedure of embedding applet]
Josh: They are all changing. The blue angle depends on the green angle. Their sum is 90 degrees. The red angle also depends on the green angle. Their sum is 180 degrees. The measure of the red angle also depends on blue angle.
Me: Excellent. Which of the three angles should be your x so that you can represent the others in terms of x also? Show it in the drawing.
Josh: I guess the green one should be x. The blue should be 90-x and the red angle should be 180-x.
Me: Good. The problem says that twice the measure of the supplement is 30 degrees wider than five times the complement. Which symbol >, <, or = goes to the blank and why, to describe the relationship between the representations of twice the supplement and five times the complement:
2(180-x) _____ 5(90-x)
Josh: > because it is 30 degrees more.
Me: Good. Now, what will you do so that they balance, that is make them equal? Remember that 2(180-x) is “bigger” by 30 degrees? What would the equation look like?
Josh: I can take away 30 degrees from 180-x. My equation would be (180-x) -30 = 5(90-x)?
Me: Is that the only way of making them equal?
Josh: Of course I can add 30 to 5(90-x). I will have 180-x = 5(90-x)+30.
Me: You said you can do the rest. Try it using both equations and tell me the value of your x and the measures of the three angles.
Josh: x = 40. That’s the angle. It’s complement measures 50 degrees and its supplement is 140 degrees. They’re the same for both equations.
Me: Does it makes sense? Do you think it satisfies the condition set in the problem?
Josh: 2(140) = 280. 5(50) = 250. 280 is 30 degrees wider than an angle of 250 degrees. Cool.
Me: What if you make A’DC your x? Do you think you will get the same answer?
No reply. I guess I’ll have to wait till the teacher give another homework to get another e-mail from him.
I don’t know if the questions I asked Josh will work with other students. Try it yourself. Please share or send this post to your co-teachers. Thanks. I will appreciate feedback.
Problem solving is the heart of mathematics yet it is one of the least emphasized activity. Solving problems are usually relegated at the end of the textbooks and chapters.
This site is my contribution to narrowing the gap between research and practice in mathematics teaching and learning. I share teaching and learning materials and blog about reforms, issues, and teaching practices in mathematics. Support site by sharing it to your network. Contact: mathforteaching@gmail.com | 893 | 3,715 | {"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-2015-22 | longest | en | 0.930749 |
http://bioquest.org/bioinformatics/module/tutorials/Sickle_Cell_Anemia/1hab.txt | 1,575,866,103,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-51/segments/1575540517557.43/warc/CC-MAIN-20191209041847-20191209065847-00486.warc.gz | 19,749,584 | 3,201 | #!rasmol -script # File: 1hab.txt # Creator: RasMol Version 2.6 zap background [0,0,0] load pdb "1hab.pdb" set ambient 60 set specular off reset slab off rotate z -97 rotate y 41 rotate x -110 zoom 95 set axes off set boundingbox off set unitcell off set hbond sidechain set bondmode and dots off # Avoid Colour Problems! select all colour bonds none colour backbone none colour hbonds none colour ssbonds none colour ribbons none colour white # Atoms select (atomno>=1) and (atomno<=1069) colour atoms [255,255,0] select (atomno>=1071) and (atomno<=1115) colour atoms [255,0,0] select (atomno>=1116) and (atomno<=1154) colour atoms [0,0,255] select (atomno>=1155) and (atomno<=1163) colour atoms [0,255,0] select (atomno>=1164) and (atomno<=2238) colour atoms [0,0,255] select (atomno>=2240) and (atomno<=2284) colour atoms [255,0,0] select (atomno>=2285) and (atomno<=2296) colour atoms [0,0,255] select (atomno>=2297) and (atomno<=3365) colour atoms [255,255,0] select (atomno>=3367) and (atomno<=3411) colour atoms [255,0,0] select (atomno>=3412) and (atomno<=3450) colour atoms [0,0,255] select (atomno>=3451) and (atomno<=3459) colour atoms [0,255,0] select (atomno>=3460) and (atomno<=4534) colour atoms [0,0,255] select (atomno>=4536) and (atomno<=4580) colour atoms [255,0,0] select (atomno>=1) and (atomno<=133) colour atoms [255,255,0] select (atomno>=1) and (atomno<=1069) spacefill off select (atomno>=1071) and (atomno<=1115) spacefill on select (atomno>=1116) and (atomno<=1154) spacefill off select (atomno>=1155) and (atomno<=1163) spacefill on select (atomno>=1164) and (atomno<=2238) spacefill off select (atomno>=2240) and (atomno<=2284) spacefill on select (atomno>=2285) and (atomno<=3365) spacefill off select (atomno>=3367) and (atomno<=3411) spacefill on select (atomno>=3412) and (atomno<=3450) spacefill off select (atomno>=3451) and (atomno<=3459) spacefill on select (atomno>=3460) and (atomno<=4534) spacefill off select (atomno>=4536) and (atomno<=4580) spacefill on select (atomno>=1) and (atomno<=133) spacefill off set shadow off # Bonds select all wireframe off # Ribbons set strands 5 set cartoon on set cartoon 100 select atomno=2 ribbons 100 select atomno=9 ribbons 100 select atomno=17 ribbons 100 select atomno=23 ribbons 380 select atomno=30 ribbons 380 select atomno=35 ribbons 380 select atomno=43 ribbons 380 select atomno=52 ribbons 380 select atomno=59 ribbons 380 select atomno=67 ribbons 380 select atomno=74 ribbons 380 select atomno=83 ribbons 380 select atomno=88 ribbons 380 select atomno=93 ribbons 380 select atomno=107 ribbons 380 select atomno=111 ribbons 380 select atomno=120 ribbons 380 select atomno=127 ribbons 380 select atomno=131 ribbons 380 select atomno=136 ribbons 380 select atomno=146 ribbons 380 select atomno=151 ribbons 380 select atomno=155 ribbons 380 select atomno=164 ribbons 380 select atomno=176 ribbons 380 select atomno=180 ribbons 380 select atomno=185 ribbons 380 select atomno=194 ribbons 380 select atomno=199 ribbons 380 select atomno=207 ribbons 380 select atomno=216 ribbons 380 select atomno=227 ribbons 380 select atomno=235 ribbons 380 select atomno=246 ribbons 380 select atomno=254 ribbons 380 select atomno=260 ribbons 100 select atomno=271 ribbons 380 select atomno=278 ribbons 380 select atomno=285 ribbons 380 select atomno=292 ribbons 380 select atomno=301 ribbons 380 select atomno=308 ribbons 380 select atomno=320 ribbons 100 select atomno=331 ribbons 100 select atomno=338 ribbons 100 select atomno=348 ribbons 100 select atomno=359 ribbons 100 select atomno=367 ribbons 100 select atomno=375 ribbons 100 select atomno=381 ribbons 100 select atomno=391 ribbons 100 select atomno=395 ribbons 100 select atomno=401 ribbons 380 select atomno=406 ribbons 380 select atomno=415 ribbons 380 select atomno=422 ribbons 380 select atomno=431 ribbons 380 select atomno=435 ribbons 380 select atomno=445 ribbons 380 select atomno=449 ribbons 380 select atomno=458 ribbons 380 select atomno=467 ribbons 380 select atomno=474 ribbons 380 select atomno=479 ribbons 380 select atomno=487 ribbons 380 select atomno=492 ribbons 380 select atomno=500 ribbons 380 select atomno=507 ribbons 380 select atomno=515 ribbons 380 select atomno=520 ribbons 380 select atomno=527 ribbons 380 select atomno=532 ribbons 100 select atomno=542 ribbons 100 select atomno=549 ribbons 100 select atomno=557 ribbons 100 select atomno=565 ribbons 380 select atomno=573 ribbons 380 select atomno=580 ribbons 380 select atomno=588 ribbons 380 select atomno=593 ribbons 380 select atomno=601 ribbons 380 select atomno=607 ribbons 380 select atomno=612 ribbons 380 select atomno=620 ribbons 380 select atomno=626 ribbons 380 select atomno=634 ribbons 380 select atomno=642 ribbons 380 select atomno=652 ribbons 380 select atomno=657 ribbons 100 select atomno=667 ribbons 100 select atomno=676 ribbons 10 select atomno=684 ribbons 100 select atomno=695 ribbons 100 select atomno=702 ribbons 100 select atomno=710 ribbons 100 select atomno=717 ribbons 380 select atomno=724 ribbons 380 select atomno=732 ribbons 380 select atomno=743 ribbons 380 select atomno=752 ribbons 380 select atomno=760 ribbons 380 select atomno=768 ribbons 380 select atomno=774 ribbons 380 select atomno=784 ribbons 380 select atomno=790 ribbons 380 select atomno=798 ribbons 380 select atomno=806 ribbons 380 select atomno=813 ribbons 380 select atomno=820 ribbons 380 select atomno=828 ribbons 380 select atomno=833 ribbons 380 select atomno=838 ribbons 380 select atomno=848 ribbons 100 select atomno=856 ribbons 100 select atomno=863 ribbons 100 select atomno=868 ribbons 100 select atomno=877 ribbons off select atomno=888 ribbons 100 select atomno=895 ribbons 380 select atomno=902 ribbons 380 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atomno=4489 ribbons 380 select atomno=4494 ribbons 380 select atomno=4504 ribbons 380 select atomno=4513 ribbons 380 select atomno=4525 ribbons 100 # Backbone select all backbone off # Labels labels off colour labels none set fontsize 8 select atomno=1156 label "Glutamate 6" select atomno=3455 label "Glutamate 6" # Monitors monitors off select all ssbonds off hbonds off | 6,396 | 20,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-51 | latest | en | 0.54464 |
https://www.physicsforums.com/threads/circuits-with-series-parallel-wiring.380089/ | 1,597,513,159,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740929.65/warc/CC-MAIN-20200815154632-20200815184632-00137.warc.gz | 773,649,464 | 15,087 | # Circuits with Series/Parallel Wiring
(a) What will the ammeter read? Report this as a positive number. (b) How much power is dissipated by one of the 10.0 Ω resistors?
I am confused about which circuits are in series or parallel wiring. I know the small 10 ohm and 10 ohm loop is parallel, and the 20 ohm to 6 ohm are parallel, correct?
So, the 10 10 ohm loop's equiv resistance = 10*10/(10+10) = 5 ohm
and the 20 6 ohm equiv resistance = 20*6/(20+6) = 4.615 ohm
Now are the two equiv resistances I calculated in series with one another?
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Delphi51
Homework Helper
The 20 and the 6 are NOT in parallel: all the current through the 20 must come through the 6, so they are somewhat in series.
At the first go, I think you ignore the meter and consider the two 10's and the 20 to be in parallel. That breaks it down to a circuit with a battery and two resistors all in series so you can easily find the potential across the 10,10,20 resistors. Then put the meter back in and find the current through the 5 ohm resistance that you already calculated.
but if the 20 resistor were to go out, then wouldn't there still be a path for the 6 resistor to reach the battery? I thought the definition of parallel was a junction between resistors.
Delphi51
Homework Helper
Yes, if any of the 10, 10 or 20 resistors were to disappear, current could divert to the others. The definition of parallel is that the components are connected together at both ends. If you erase the meter, you'll see that is the case for the 10, 10 and 20. The current meter can be ignored because it is supposed to have a resistance of zero (or near enough to ignore). | 455 | 1,738 | {"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-2020-34 | latest | en | 0.95934 |
http://education.lms.ac.uk/2012/07/word-problems-and-the-khan-academy/ | 1,603,855,933,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107896048.53/warc/CC-MAIN-20201028014458-20201028044458-00358.warc.gz | 33,715,838 | 9,649 | # Word problems and the Khan Academy
A word problem from the Khan Academy
Last year, I was asked by my American colleagues to give my assessment of mathematical material on the Khan Academy website. Among other things I looked for the so-called “word problems” and clicked on a link leading to what was called there an “average word problems” but happened to be a “word problem about averages”. It appears that the problem there changed since last year. I reproduce here the old one, it is more of interest for discussion of “word problems”.
Gulnar has an average score of $$87$$ after $$6$$ tests. What does Gulnar need to get on the next test to finish with an average of $$78$$ on all $$7$$ tests?
Hints given by the Khan Academy
What follows are hints as were given, one after another, by the Khan Academy website:
Hint 1. Since the average scores of the first $$6$$ tests is $$87$$, the sum of the scores of the first $$6$$ tests is $$6\times 87=522$$.
Hint 2. If Gulnar gets a score of $$x$$ on the $$7$$th test, then the average score on all $$7$$ tests will be:
$\frac{522+x}{7}$
Hint 3. This average needs to be equal to $$78$$ so:
$\frac{522+x}{7} = 78$
Hint 4. $$x=24$$
“Questions” method for word problems
And here is the same problem solved by “steps” or “questions” method. (As a child, I learnt it in a direct face-to-face communication with a live teacher and with my peers.)
Question 1. How many points in total did Gulnar get in $$6$$ tests?
Answer: $$6 \times 87=522$$
Question 2. How many points in total does Gulnar need to get in $$7$$ tests?
Answer: $$7\times 78 = 546$$
Question 3. How many points does Gulnar need to get in the $$7$$th test?
Answer: $$546 – 522 = 24$$
Crucially, the whole point of the “questions” method is that students have to formulate these questions themselves.
Self-guiding questions
Actually, a student has to be taught to start his/her ”questions method” attempt at a word problems asking himself or herself appropriate self-guiding questions. In this case, these self-guiding questions are likely to be something like
Question A. “Gulnar has an average score of $$87$$ after $$6$$ tests.” What questions can be asked about these data?
Question B. “Gulnar needs to get an average of $$78$$ on all $$7$$ tests”. What questions can be asked about these data?
Therefore the use of “questions” method in mathematics education involves gently nudging a child towards reflection and analysis of his/her own thought process – at the level, needs to be emphasised, actually accessible to the child, and it was confirmed by mathematics education practice of dozens of countries around the world. This is why I prefer the term “questions method” to the more commonly used, in British education literature, words “steps method”; the word “questions” emphasises the pro-active and reflective components of thinking, while the word “steps” might inadvertently imply a passive algorithmic approach.
A comparison between the two approaches
However, it is instructive to compare the two approaches in more detail.
The Khan Academy solution is longer than the “questions method” solution because Hints 2 and 3 are already equivalent to a solution of a 3-step problem (you need to express algebraically the new total, compute the new mean and compare with the desired result).
The “step” or “questions” method of solving the problem directly encourages students to look at the core of the concept of average, it invites them to re-visit previous material and solve the problem of finding the total from the average. It encourages them to talk about mathematics, with themselves and with others.
In more technical terms, the “questions” method triggers the all-important dynamics of encapsulation / de-encapsulation of the concept of “average” (and we have to remember that this is a pretty abstract concept – this is understood even in the mass culture, in expressions such as “2point4 children“). The terms “encapsulation” and “de-encapsulation” are less frequently used, and perhaps a few words of clarification may be useful; I quote Weller et al. Intimations of infinity, Notices AMS 51 no. 7 (2004) 741-750:
The encapsulation and de-encapsulation of processes in order to perform actions is a common experience in mathematical thinking. For example, one might wish to add two functions $$f$$ and $$g$$ to obtain a new function $$f+g$$. Thinking about doing this requires that the two original functions and the resulting function are conceived as objects. The transformation is imagined by de-encapsulating back to the two underlying processes and coordinating them by thinking about all of the elements $$x$$ of the domain and all of the individual transformations $$f(x)$$ and $$g(x)$$ at one time so as to obtain, by adding, the new process, which consists of transforming each $$x$$ to $$f(x)+g(x)$$. This new process is then encapsulated to obtain the new function $$f+g$$.
Mathematical concepts are shaped and developed in a child’s mind in a recurrent process of encapsulation and de-encapsulation. Returning to solution of Gulnar’s problem by “questions” method, we see this process in action.
And what is even more important, self-guiding questions are meta-questions, that is, questions aimed at finding the optimal way of reasoning.
From a basic pedagogical point of view, if the didactic aim of the problem is to reinforce the understanding of averages, then the “questions” method appears to be more useful; it gives a student a joint and cohesive vision of mathematics. And the “questions” method is more difficult to implement in a computer based medium – by its nature, it requires live interaction between teacher and students, and between students.
But I claim much more: the “questions” method can act as a trigger of the fundamental recurrent cycle of mathematical thinking: intentional and purposeful encapsulation / de-encapsulation of abstract entities. Like breathing and heartbeat in a living body, the encapsulation / de-encapsulation cycle makes the difference between live mathematics and zombie mathematics. And, the last but not least, it is the very essence of thinking in computer science and computer programming.
Some general comments
The flaw in the Khan Academy hints is something that can be observed in the current English education practice as well: instead of giving to students proper didactic support, hints are frequently just pieces cut off a ready solution of the problem.
But, in my opinion, mathematics cannot be learned as a mass of disjoint bits and pieces. Mathematics is not a sum of facts; learning mathematics does not mean memorising facts, it means gaining understanding of connections between mathematical facts, and, moreover, of connections between connections and analogies between analogies. Word problems and “questions” method allow to do that in natural human language exploiting its in-built powerful logical facilities (the latter otherwise manifest themselves, in a child’s study of mother tongue itself, as its grammar).
Giving a student a ready piece of a solution instead of gently nudging him in the direction of a solution is, of course, a very precise form of communication. But it proves only that the teacher can solve the problem, it does not teach a student to solve unfamiliar problems independently.
An effective mathematics teacher should be a diagnostician and communicator: he or she should see student’s error or difficulty, understand its underlying causes, talk to the student using an accessible mathematical language. Moreover, the teacher should relate, both via empathy so important for a diagnostician, and via non-verbal hints (an approving gesture can worth a hundred words when a student needs a lightest possible hint that he/she is on right track to a solution of the problem).
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,741 | 7,935 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2020-45 | latest | en | 0.946171 |
http://www.world-builders.org/lessons/less/download/games/bbuilder/bbcount.html | 1,544,632,714,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824059.7/warc/CC-MAIN-20181212155747-20181212181247-00047.warc.gz | 499,815,310 | 2,966 | Reporting Your Biome Builder Game Results
These directions are included in the Biome Builders .pdf file.
Get your game card and some markers to put on the numbers. Cover the numbers as they are called out. Raise your hand or call, "Got it!" when you have two rows filled. Winner gets 500 bonus points
Add up all the numbers in the ovals of the spaces that you covered on your own card. Write your total here: __________ . Join a group of 2-4 people in the same biome.
List your totals here: then add the totals up to get a grand total for your group.
Players Card 1 Card 2 Card 3 Card 4 Grand Total Points . . . .
Your grand total gives you the the number of square meters in your biome. Your biome produces the number of KiloCalories per square meter listed in the chart below. This is the Primary Productivity of your biome.
KiloCalories Per Square Meter Per Day Desert Tundra Grasslands Coniferous Forest Deciduous Forest Tropical Rain Forest 1 2 7 10 16 25
Now multiply to figure out how many calories you have for your animals to eat:
(Total Number of Points) * (Primary Productivity of biome) = KiloCalories for Your Animals to eat.
Now pick out your herbivores! You need groups of each kind so each kind can reproduce. Keep track of the Kilocalories! Your total for all of the herbivores must not be more than your Total Number of Points.
Tiny Herbivores under 5 pounds Small Herbivores rabbit 15 pounds Medium Herbivores sheep 150 pounds Large Herbivores cow 1000 pounds . Small Carnivores fox,hawk 10 pounds Medium Carnivores wolf 200 pounds Large Carnivores lion 400 pounds KiloCalories Needed Per Day 150 550 2,000 15,000 . 600 3,000 6,000 How many do you have? . . . . . . . . KiloCalories . . . . . .
Carnivores eat herbivores, but don't get all their KiloCalories. Carnivores get
Total Number of (KiloCalories for Your Animals to eat)
divided by 10.
(for example, if your grand total =24,000, your carnivores will have 2,400 calories. You will only be able to have small carnivores.)
Choose the numbers and sizes of your carnivores. Calculate the KiloCalories that they will need. Now go to the biomes on earth and figure out which animals could be in your biome! Draw a picture of your ecosystem or make a food pyramid or diorama. | 590 | 2,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-51 | latest | en | 0.828537 |
http://www.kwiznet.com/p/takeQuiz.php?ChapterID=1418&CurriculumID=32&Num=2.17 | 1,544,955,554,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827639.67/warc/CC-MAIN-20181216095437-20181216121437-00618.warc.gz | 409,416,664 | 4,228 | Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
A numberline shows the numbers in order. Numbers to the right of zero are positive numbers. Numbers to the left are negative numbers. A numberline has arrows because the numbers keep going forever. You can use the numberline to practice the addition and subtraction. Method: Draw a number line with numbers. To find difference, we first go forward till we reach the bigger number and then backward till we reach the smaller number. To find sum, we first go till the first number and then move further from there to reach the next number. Directions: Answer the following questions. Also write at least ten examples with subtraction and addition using number line.
Q 1: A number line shows numbers in order.FalseTrue Q 2: On a number line, what is four less than seven?4356 Q 3: Positive numbers are numbers on the right of zero on the number line.TrueFalse Q 4: On a number line, what is four more of number seven?1110712 Q 5: Negative numbers are the numbers to the left of zero on a number line.TrueFalse Q 6: To find the sum of 5 + 3 on the number line, first go forward to 5 then go 3 more to get _____.8654 Q 7: What comes next in the number line 0 1 2 3 4 5 _564 Q 8: To find the difference of 9 - 5 on the number line, first go forward to 9 then go backward 5 numbers. We end up at _____.6497 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | 392 | 1,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-51 | longest | en | 0.853896 |
https://forum.charliefrancis.com/t/stride-freq-stride-length-of-steps/32334 | 1,695,710,616,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510149.21/warc/CC-MAIN-20230926043538-20230926073538-00263.warc.gz | 296,800,895 | 8,781 | # stride freq, stride length, # of steps
Will someone please describe the interaction of stride length, stride frequency and number of steps within a particular distance. Specifically the interaction of stride length and number of steps. Is it likely that a faster athlete will take the same number of steps in a 40yd or 100m sprint when a distinguishing factor is a longer stride length for the faster athlete?
Could you please give a brief description of how stride length, stride frequency and number of steps interact between sprinters of differing performance? Is there any literature that examines these 3 variables?
Stride length and number of steps is actually the same thing, just different expression.
Manfred Letzelter did extensive researches in 1972 and 1976 olympics for 100m regarding number of steps and step frequency at 200m. I suggest you have a look in the old Leistungssports i can’t remember the issue number and page for the article
What is interesting is to study the progression of stride length and frequency through the 100m, as well as contact time/flight time ratio.
Your question would require a 200 page book, but the basics is that frequency increases and peaks at around 30m then decrease, while stride length increase up to 60m and then is maintained through 90m and increases again in the last 10m as the sprinters lean forward to the tape.
The speed curve reaches his peak between step frequency and stride length’s peaks. Step contact time mirros the speed curve (speed curve increase and decrease, while step contact time decrease and increase).
elite sprinters walter dix and justin gatlin are two similar sprinters(generally speaking-both sub 10 sub 20) and run with a similar body position. Believe the name was something like hypo-kyphotic thoracic head position.
Yet they are on the opposite ends of the spectrum in terms of stride length and frequency, i’m guessing due to their respective body types as well as their different limb lengths/levers.
You have gatlin (rangy beast type) in his 9.77 100 hitting 40m at 20 strides, 60m at 28 strides, and 100m at roughly 42 strides. Then you have dix (turn-over beast type) in his 9.93 100 hitting 40m at around 24-25 strides, 60m at around 32.5-33 strides, and 100m at around 49.5-50 strides.
As well, you then have gatlin in his 20 flat 200 runs hitting 100m consistently at around 43-44 strides and the 200 finish at 81-82 strides. With dix, i haven’t analyzed any of his 200 races but i’m guessing he’ll be covering the full 200 metres in around 96-98 strides.
The thing that amazes me most is that gatlin 6’1" is 5 inches shorter than usain bolt 6’6"(man-child beast prodigy-19.93 at 17 years 19.75 at 20 years), yet he still runs 200 metres in the same number of strides as bolt. This tells me that his leg and core strength is off-the charts to be able to run with that stride-length, of a 6’6" man, for a full 200 metres. Some GB commentators asked J.Gat, after a 20.05 run where he beat T.Gay, what his secret was. All he said was “Strong Legs” I’m guessing it’s a combination of that hypo-kyphotic thoracic body position(basically a lean i guess) and the tremendous core-leg strength from doing squats and 200 type-training.
There are some great perspectives being offered.
wouldnt plyometrics increase stride length too? more elastic force should equal longer stride length so gatlin was probably very springy too. also gatlin used to be a hurdler could this have any impact?
Yes but WHICH plyometric exercise(s)?
wouldnt all of them have some benefit? just whatever plyows sprinters usually do.
Regarding Gatlin, I’m aware of the repetitive bounding and “stick” work he did with Anderson. I always thought that he overstrided during acceleration. Of course, genetic predisposition plays into it as well.
So then for two athletes (like gatlin and dix), seperated by approx. .2s, to have the same number of steps there would need to be a large diff. in stride rate between the two to account for the superior time?
Basically, it is difficult to make generalities regarding the correlation of # of steps and performance? Or is it?
There is too much variability in the interaction of rate/length between individuals?
I think the picture is becoming more clear. Thanks for the responses.
There’s no correlation btw number of step and 100m performance.
the difference wouldn’t be that large and would be barely perceptible by just eyeballing the races. You’d have to analyze the races in slow-mo or using your mouse and pausing to count the strides frequency/rate.
gatlin at his more normal time range for 2006(9.85s-9.95s area) was at a more normal stride rate(for himself) of 44 strides at 100m. These were 21 strides at 40m, 29 strides at 60m, and 44 strides at 100m. He ran with this slightly higher frequency/stride rate times of, 9.95 osaka; 9.93 indy; 9.87 new york; 9.88 eugene.
Compare this stride pattern with 9.77 doha race as well as the 9.86 doha heat stride pattern: 40m at 20 strides, 60m at 28 strides, and 100m at 42 strides. I think this basically says that at his peak of his season 9.77, gatlin was able to run slower and faster at the same time in a way. In other words, he was fresh enough to run with and maintain a stride length that allowed him to pull his fastest times with a slightly lower frequency.
I’ve wondered about overstriding as well and what factors tells us who overstrides and who doesn’t? is it all about where both knees are in relation to eachother as the foot strikes the ground or is it more about the apparent length of the contact patch on the ground in front of the body or is it about both of the above factors combined?
I’ve often looked at it as more the latter of the two. I tend to get better results from my athletes (non-elite) the closer to BDC the foot touches down during maximal speed. Certainly a more responsive and relaxed vertical displacement as well. Perhaps, with elites, it becomes more a function of leg stiffness and force generated on touchdown.
The association between
the developnient of strength
and speed
by Alessandro Donati
NSA year 1996 vol 11 issue 2
The Italians seem to have an index related to bounds/strides and actual sprint times for 100m
Except that you can take the same # of steps and run a completely different race. If you take much faster, but shorter steps in the beginning and longer towards the end, the total # may be the same or approximately the same and you could have completely different outcomes for the times.
There isn’t so much differences in stride pattern for elite sprinters, this is a recap for some of Gatlin’s best races? New York is anomaly, and Doha’s heats too because he relaxed before the end which tends to open the strides.
9.77 (w+1.7) - 41.6 - Doha 2006
9.84 (w+3.4) - 42.9 - Eugene 2005
9.85 (w+0.6) - 42.1 - Athens OG 2004
9.86 (w+1.1) - 41.2 - Doha 2006 (heats)
9.87 (w-0.1) - 44.1 - New York 2006
9.88 (w+0.4) - 42.0 - Helsinki WC 2005
9.88 (w+1.0) - 43.0 - Eugene 2006
9.89 (w+1.0) - 42.0 - London 2005
9.92 (w+0.0) - 42.0 - Sacramento NC 2004
9.93 (w-1.2) - 43.0 - Indianapolis NC 2006
9.95 (w-0.1) - 42.8 - Osaka 2006
Interesting. So, basically when at his peak or near-peak (2004, 2005, 2006), performance wise, his stride is at it’s longest?
what kind of cue might one use regarding the arms, when they have an athlete who understriding near the end of their respective training distances. Maybe tell them to use less arm take-back in the area where they begin understriding.
By understriding, I mean, the recovery knee is in-line with the opposite knee before that opposite knee strikes the ground. And as the stance leg is striking the ground, the recovery knee is already slightly ahead.
Or perhaps leave different arm cues alone and command them to stop the run once they begin to understride.
Maybe, we should look inside and have details for different portion of the race to see what happens, but i haven’t done extensive researches for Galtin. Opposite patterns can be found in other elite sprinters.
within 9.77-100run:
20 strides to 40m (as opposed to most his other races 21 strides to 40m);
brought the head and upperbody up out of drive-phase at 16th stride(roughly 30m or shortly thereafter);
once up and into his running at 30m, head and body-position is still low (lower relative to other sprinters), as if a sustained drive-phase position or something along those lines;
stride becomes longer from 60-80 and 80-100
20m-40m : 8 strides
40m-60m : 8 strides
60m-80m : 7.5 strides
80m-100m : 6.5 strides | 2,196 | 8,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-40 | latest | en | 0.936662 |
https://rustgym.com/leetcode/56 | 1,653,218,240,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545326.51/warc/CC-MAIN-20220522094818-20220522124818-00390.warc.gz | 564,231,420 | 3,479 | 56. Merge Intervals
Given an array of `intervals` where `intervals[i] = [starti, endi]`, merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.
Example 1:
```Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
```
Example 2:
```Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.
```
Constraints:
• `1 <= intervals.length <= 104`
• `intervals[i].length == 2`
• `0 <= starti <= endi <= 104`
56. Merge Intervals
``````struct Solution;
impl Solution {
fn merge(mut intervals: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
intervals.sort_unstable_by(|a, b| a[0].cmp(&b[0]));
let mut res: Vec<Vec<i32>> = vec![];
let mut temp: Option<Vec<i32>> = None;
for v in intervals {
if let Some(t) = temp.clone() {
if v[0] <= t[1] {
temp = Some(vec![t[0], i32::max(t[1], v[1])]);
} else {
temp = Some(v);
res.push(t);
}
} else {
temp = Some(v);
}
}
if let Some(t) = temp {
res.push(t);
}
res
}
}
#[test]
fn test() {
let intervals: Vec<Vec<i32>> = vec_vec_i32![[1, 3], [2, 6], [8, 10], [15, 18]];
let res: Vec<Vec<i32>> = vec_vec_i32![[1, 6], [8, 10], [15, 18]];
assert_eq!(Solution::merge(intervals), res);
let intervals: Vec<Vec<i32>> = vec_vec_i32![[1, 4], [4, 5]];
let res: Vec<Vec<i32>> = vec_vec_i32![[1, 5]];
assert_eq!(Solution::merge(intervals), res);
}
`````` | 521 | 1,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-21 | latest | en | 0.525396 |
https://schools.aglasem.com/5791 | 1,542,376,218,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743046.43/warc/CC-MAIN-20181116132301-20181116154301-00517.warc.gz | 726,333,928 | 22,597 | Matrices of Rotation of Axes
We know that if x and y axis are rotated through an angle q about the origin the new co-ordinates are given by
x = X cos 0 –Y sin 0
y = X sin 0 + Y cos 0 | 57 | 184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-47 | latest | en | 0.935769 |
https://metanumbers.com/1027530 | 1,643,225,788,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304961.89/warc/CC-MAIN-20220126192506-20220126222506-00071.warc.gz | 437,339,250 | 7,709 | # 1027530 (number)
1,027,530 (one million twenty-seven thousand five hundred thirty) is an even seven-digits composite number following 1027529 and preceding 1027531. In scientific notation, it is written as 1.02753 × 106. The sum of its digits is 18. It has a total of 7 prime factors and 72 positive divisors. There are 233,856 positive integers (up to 1027530) that are relatively prime to 1027530.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 7
• Sum of Digits 18
• Digital Root 9
## Name
Short name 1 million 27 thousand 530 one million twenty-seven thousand five hundred thirty
## Notation
Scientific notation 1.02753 × 106 1.02753 × 106
## Prime Factorization of 1027530
Prime Factorization 2 × 32 × 5 × 72 × 233
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 48930 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,027,530 is 2 × 32 × 5 × 72 × 233. Since it has a total of 7 prime factors, 1,027,530 is a composite number.
## Divisors of 1027530
72 divisors
Even divisors 36 36 20 16
Total Divisors Sum of Divisors Aliquot Sum τ(n) 72 Total number of the positive divisors of n σ(n) 3.12109e+06 Sum of all the positive divisors of n s(n) 2.09356e+06 Sum of the proper positive divisors of n A(n) 43348.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1013.67 Returns the nth root of the product of n divisors H(n) 23.7039 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,027,530 can be divided by 72 positive divisors (out of which 36 are even, and 36 are odd). The sum of these divisors (counting 1,027,530) is 3,121,092, the average is 4,334,8.5.
## Other Arithmetic Functions (n = 1027530)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 233856 Total number of positive integers not greater than n that are coprime to n λ(n) 4872 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 80331 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 233,856 positive integers (less than 1,027,530) that are coprime with 1,027,530. And there are approximately 80,331 prime numbers less than or equal to 1,027,530.
## Divisibility of 1027530
m n mod m 2 3 4 5 6 7 8 9 0 0 2 0 0 0 2 0
The number 1,027,530 is divisible by 2, 3, 5, 6, 7 and 9.
• Abundant
• Polite
• Practical
## Base conversion (1027530)
Base System Value
2 Binary 11111010110111001010
3 Ternary 1221012111200
4 Quaternary 3322313022
5 Quinary 230340110
6 Senary 34005030
8 Octal 3726712
10 Decimal 1027530
12 Duodecimal 416776
20 Vigesimal 688ga
36 Base36 m0ui
## Basic calculations (n = 1027530)
### Multiplication
n×y
n×2 2055060 3082590 4110120 5137650
### Division
n÷y
n÷2 513765 342510 256882 205506
### Exponentiation
ny
n2 1055817900900 1084884567711777000 1114751439860882220810000 1145440547000252308348899300000
### Nth Root
y√n
2√n 1013.67 100.909 31.8382 15.9353
## 1027530 as geometric shapes
### Circle
Diameter 2.05506e+06 6.45616e+06 3.31695e+12
### Sphere
Volume 4.54435e+18 1.32678e+13 6.45616e+06
### Square
Length = n
Perimeter 4.11012e+06 1.05582e+12 1.45315e+06
### Cube
Length = n
Surface area 6.33491e+12 1.08488e+18 1.77973e+06
### Equilateral Triangle
Length = n
Perimeter 3.08259e+06 4.57183e+11 889867
### Triangular Pyramid
Length = n
Surface area 1.82873e+12 1.27855e+17 838975 | 1,352 | 3,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-05 | latest | en | 0.816032 |
http://legacy.roadtoreality.info/archive/viewtopic.php%3Ff=16&t=1685&p=2906.html | 1,611,499,774,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703549416.62/warc/CC-MAIN-20210124141945-20210124171945-00105.warc.gz | 58,111,545 | 7,543 | Page 1 of 1 [ 9 posts ]
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Exercise [20.08] b
Author Message
Joined: 03 Jun 2010, 15:18
Posts: 136
Exercise [20.08] b
Attached my solution.
It is not really different from the one by Gandalf http://www.roadtoreality.info/viewtopic.php?f=16&t=1427&start=0(that also covers other topics) but is somehow more elementary and may help to clarify some aspects not obvious (e.g. the "mixed term" problem raised in the discussion).
Attachments:
Exercise_20_08.pdf [90.37 KiB]
14 Sep 2010, 17:28
Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [20.08] b
I thought they all ought to be referred to from one post.
As well as Roberto's solution (above) and gandalf's solution (previous topic), there is also a discussion about the "mixed term" here:
Laura prepared a comment about this for Prof Penrose, discussed here:
And another similar comment by Laura is posted on her website here (also a PDF):
http://camoo.freeshell.org/20.8.pdf
Last edited by deant on 16 May 2012, 00:05, edited 3 times in total.
13 May 2012, 08:35
Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [20.08] b
Roberto, I think the last line of your solution should be:
"V has a minimum if and only if is non-negative definite".
It's entirely possible to have a such that but . Just consider a potential field that depends (say) only on the z coordinate, but is unaffected by x or y.
The result is small oscillations in the z direction (assuming stable equilibrium), but constant-velocity motion in the x-y plane; the matrix W is degenerate in this case, so for some q's (namely linear combinations of x and y) the corresponding eigenvalue is zero.
V still has a minimum in this case, it's just that it depends only on the value of z.
Last edited by deant on 15 May 2012, 22:13, edited 1 time in total.
13 May 2012, 09:46
Joined: 03 Jun 2010, 15:18
Posts: 136
Re: Exercise [20.08] b
I cant' agree with you. A point P is a minimum for V(P) if in every point Q of a neighbour of P it is V(Q)>V(P), strictly greater. This does not happen in your example, because a near point in x direction has the same value of V, so it is just V(Q)>= V(P). It is just a stationary point, not a minimum.
In physical terms, that is an equilibrium point but the equilibrium is unstable: as yourself noted, a small perturbation to momentum in x direction causes a constant velocity motion that moves the system away from the equilibrium point; this means that equilibrium is unstable.
Of course there are situations (e.g. and ) in which behaviour of V cannot be recognised by the analisys of linearised system: in both cases matrix Q vanishes, but first case is a clearly minimum and latter case is a maximum; so in first case equilibrium is stable and in second case unstable. This kind of systems cannot be analysed by linearisation, and you have to continue the Taylor development up to higher orders.
14 May 2012, 14:03
Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [20.08] b
I don't disagree with anything you said, except perhaps the semantics. I take "V is a minimum" to mean "there is no point with a lower value of V (perhaps only locally)". So it's just a matter of definition.
Also, in the case with constant motion in the x-y plane, I don't believe this equilibrium situation should be called unstable, I believe the correct description for this kind of equilibrium condition is "meta-stable". "Unstable" refers to the case where perturbations can grow exponentially.
15 May 2012, 11:20
Joined: 03 Jun 2010, 15:18
Posts: 136
Re: Exercise [20.08] b
Of course is a matter of definition, however I think that definition of minimum for V as "there is no point with a lower or equal value of V (perhaps only locally)" is quite standard. Otherwise the function V(x)=5 will have a minimum at every x, that is quite counter-intuitive.
About stability: there are many definitions of stability (see e.g.
http://en.wikipedia.org/wiki/Stability_theory and http://en.wikipedia.org/wiki/Lyapunov_stability) but if a system, after a perturbation, moves away from the equilibrium point and the distance increases without any limit with time (exponentially or linearly or in any other way), the point of equilibrium is defined as unstable.
Metastability is quite a different concept ( see http://en.wikipedia.org/wiki/Metastability ); it is normally used with reference to thermodynamical systems but in the context of Hamiltonian dynamics can be considered metastable a point that is a local minimum but not a global minimum. When the perturbation is small enough, the system remains near to the metastability point (so it behaves as a stability point) but when the perturbation is made larger enough it moves away, going to a global minimum (if any) or to infinity; so it behaves as an instability point.
15 May 2012, 17:22
Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [20.08] b
Hmmm... I think that it's perfectly appropriate to say that V(x)=5 has a minimum; Its minimum is "5" (at every x). It just doesn't have a (unique) minimum point, which is a stronger condition then merely having a minimum (value).
However, in the context of your solution, I'll grant you that "V has a minimum", although technically ambiguous, is most naturally interpreted as "V has a minimum point at equilibrium".
However, the actual issue here is that V can have a minimum point at equilibrium, while might nonetheless not be positive definite - as you already demonstrated with your example. So the "if and only if" part of your solution's final sentence is incorrect. (As is my first suggested correction to it, in my second post from the top of the thread).
A corrected version of it would be:
V has a minimum point at equilibrium if is positive definite;
is non-negative definite if V has a minimum point at equilibrium.
---------
About metastability you are correct - I apologise. I'm afraid I had the term "metastable" confused with "neutrally stable" (or "critically stable"/"marginally stable"), which may sometimes be used to describe situations such as the example I gave, for linear systems. (It does entirely depend on the definition of "stability", though, like you said).
Also this would generally only apply to the linearised system, not the actual system, because (like you already pointed out) the actual stability of the system when is non-negative definite with (at least one) zero eigenvalue, cannot be determined from the linearisation alone but must be determined by taking higher order terms of the Taylor expansion into account.
Last edited by deant on 17 May 2012, 17:53, edited 1 time in total.
15 May 2012, 23:35
Joined: 03 Jun 2010, 15:18
Posts: 136
Re: Exercise [20.08] b
deant wrote:
Hmmm... I think that it's perfectly appropriate to say that V(x)=5 has a minimum; Its minimum is "5" (at every x). It just doesn't have a (unique) minimum point, which is a stronger condition then merely having a minimum (value).
However, in the context of your solution, I'll grant you that "V has a minimum", although technically ambiguous, is most naturally interpreted as "V has a minimum point at equilibrium".
Yes, in this context I used “minimum” with the meaning of “minimum point” not of “minimum value of the function".
deant wrote:
A corrected version of it would be:
V has a minimum point at equilibrium if is positive definite;
is non-negative definite if V has a minimum point at equilibrium.
I think we can summarise the results of whole discussion as follows:
a) If at equilibrium the matrix Q is positive definite, this is a minimum point for V.
b) If at equilibrium the matrix Q is negative definite or is not definite, this cannot be a minimum point for V.
c) If at equilibrium the matrix Q is non-negative definite, behavior of V at that point cannot be determined just by Q (higher terms of Taylor expansions are needed).
A) If the equilibrium point is minimum for V, Q is non-negative definite. (This is equivalent to b)
B) If the equilibrium point is not minimum for V, Q is not positive definite (may be negative definite, non-negative definite, not definite). (This is equivalent to a)
17 May 2012, 16:54
Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [20.08] b
Yes, I think that about sums it up!
17 May 2012, 17:42
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## How much does 200 oz weigh?
Convert 200 Ounces to Pounds
oz lb
200.00 12.5
200.01 12.501
200.02 12.501
200.03 12.502
## What does 100 ounces weigh?
Ounces to Pounds conversion table
Ounces (oz) Pounds (lb) Pounds+Ounces (lb+oz)
70 oz 4.3750 lb 1 kg 984.47 g
80 oz 5 lb 2 kg 267.96 g
90 oz 5.625 lb 2 kg 551.46 g
100 oz 6.25 lb 2 kg 834.95 g
## Is a cup always 8 ounces?
— a cup of water happens to equal both 8 fluid ounces (in volume) and 8 ounces (in weight), so you might naturally assume that 1 cup equals 8 ounces of weight universally in recipes. So no, fluid ounces and ounces should not be used interchangeably.
## How many cups are in a 1 pound bag of sugar?
One pound of granulated sugar contains approximately 2 cups.
## How many cups of flour are in a 5 pound bag?
How Many Cups in One Pound of Flour?
Pounds of Flour Cups (US)
1 lb 3.49 cups
2 lb 6.98 cups
5 lb 17.45 cups
10 lb 34.9 cups
## How big is a 8 oz glass?
The size of an 8-ounce glass is dependent on the style of the glass, but each glass has the volume capable of holding 8 fluid ounces, 1 cup or 236.5 mL. If stated to be an 8-ounce glass then it is a relatively small drinking cup. The size of a glass may be dependent upon the design and the material it is made out of.
8 ounces
## How many ounces is 5 lbs of flour?
One of these should hold a 5 pound bag of flou… see more. Need to establish by volume. A pound of flour is about 3.33 cups. These hold 175 ounces, which is almost 22 cups.
## How many cups is 3.5 lbs?
Cups in a pound of bread flour
Pounds Cups (US)
1 lb 3.5 cups
2 lbs 7 cups
3 lbs 10.5 cups
4 lbs 14 cups
## How many cups is 6.6 lbs?
The 3.3-pound bag has about 13 cups of food, the 6.6-pound bag has about 26 cups of food, and there are about 70 cups of food in a 17.6-pound bag. | 582 | 1,813 | {"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-2024-22 | latest | en | 0.891467 |
https://www.centerspace.net/doc/NMath/ref/html/T_CenterSpace_NMath_Core_DoubleRandomBetaDistribution.htm | 1,679,865,692,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946535.82/warc/CC-MAIN-20230326204136-20230326234136-00439.warc.gz | 780,100,310 | 3,589 | Class DoubleRandomBetaDistribution generates random numbers from a beta distribution.
Inheritance Hierarchy
SystemObject
Namespace: CenterSpace.NMath.Core
Assembly: NMath (in NMath.dll) Version: 7.4
Syntax
```public class DoubleRandomBetaDistribution : IRandomNumberDistribution<double>,
ICloneable```
The DoubleRandomBetaDistribution type exposes the following members.
Constructors
NameDescription
DoubleRandomBetaDistribution Constructs a random number generator for a beta distribution using default P of 1.0, Q of 1.0, displacement of 0.0, and scale of 1.0.
DoubleRandomBetaDistribution(Double, Double, Double, Double) Constructs a random number generator for a beta distribution using the specified distribution parameters.
DoubleRandomBetaDistribution(Double, Double, Double, Double, DoubleRandomBetaDistributionGenerationMethod) Constructs a random number generator for a beta distribution using the specified distribution parameters.
Top
Properties
NameDescription
Displacement Gets the displacement.
Method Gets the generation method.
P The first shape parameter.
Q The second shape parameter.
Scale Gets the scale.
Top
Methods
NameDescription
Clone Creates a deep copy of this random number generator.
Fill(RandomNumberStream, DoubleVector, Int32, Int32) Uses the given random number stream to fill the given vector of doubles with random values.
Fill(RandomNumberStream, Double, Int32, Int32) Uses the given random number stream to fill the given array of doubles with random values.
Top
Remarks
The beta distribution f(x:p,q,a,b) = {1/[B(p,q)*b^(p+q-1)]}*(x-a)^(p-1)*(b + a - x)^(q-1) where B(p,q) is the complete beta function, p and q are shape parameters, a is the displacement or location parameter, and b is the scale factor.
#### Reference
CenterSpace.NMath.Core Namespace | 400 | 1,796 | {"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-2023-14 | latest | en | 0.512873 |
https://businesslawblogsite.com/taxes/how-is-property-tax-measured.html | 1,660,329,302,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571745.28/warc/CC-MAIN-20220812170436-20220812200436-00648.warc.gz | 171,451,223 | 18,959 | How is property tax measured?
Contents
What is the way of calculating the property tax?
Calculation of Property Tax
The formula used for calculating property tax is given below: Property tax = base value × built-up area × Age factor × type of building × category of use × floor factor. Property tax in India depends on the location of a property in question, with taxes varying from state to state.
How are property taxes calculated example?
Property tax = base value × built-up area × Age factor × type of building × category of use × floor factor. It is important to note that the amount of tax payable in the country depends on where the property is situated, as taxes vary from one state to another.
In what unit of measure are property taxes measured?
Mill is derived from the Latin word millesimum, meaning thousandth. As used in property tax, 1 mill is equal to \$1 in property tax levied per \$1,000 of a property’s assessed value.
How can I lower my property taxes?
10 Ways to Lower Your Property Taxes
1. Lower Your Tax Bills. …
2. Review Your Property Tax Card for Errors. …
3. Appeal Your Tax Valuation—Promptly. …
4. Get Rid of Outbuildings. …
5. Check to See If You Qualify for Property Tax Relief. …
6. Move to a Less Expensive Area. …
7. Compare Tax Cards of Similar Homes. …
8. Have Your Property Independently Appraised.
THIS IS IMPORTANT: Do you get tax back every year?
Does property tax depend on purchase price?
Your local property tax rate is applied to the assessed value of your home in order to come up with the amount that you owe. In each jurisdiction, a local taxing authority sets a rate that each home will be taxed at. … In California, for example, a home’s assessed value is based on its purchase price.
Which is an example of a property tax?
Property Tax Example
For example, if the property tax rate is 4% and your house’s assessed value is \$200,000, then your property tax liability equals (. 04 x \$200,000) or \$8,000. The assessed value is often computed by incorporating the purchases and sales of similar properties in nearby areas.
What is age factor in property tax?
Age Factor – This factor takes into account the age of the building and provides for higher tax on newer properties and lesser tax on older properties in line with the market value. On the basis of age of the building, this factor ranges from 0.5 to 1.0. Year of Construction.
How do you calculate Mills?
Properly tax itself is sometimes referred to as “millage tax.” A mill is one one-thousandth of a dollar, and in property tax terms is equal to \$1.00 of tax for each \$1,000 of assessment. 29 mills, therefore, is equal to \$29 for every \$1,000 of assessed value, or 2.9%.
What are the two primary types of property taxes?
These are two distinct forms of taxation: one (ad valorem tax) relies upon the fair market value of the property. The other (special assessment) relies upon a special enhancement called a “benefit” for its justification. The property tax rate is typically given as a percentage.
THIS IS IMPORTANT: How do travel nurses maintain tax homes?
How is land value calculated?
To calculate the land value as a percentage of the total value of the property (land + improvements, such as a house), you would have: \$75,000 (the value of the land) / \$250,000 (the value of the land and improvements). = 0.30 (the value of the land compared to the overall property expressed in decimal form).
Why are property taxes so high?
State and local budgeting
Your property tax may increase when state governments fund a service like repairing roads — or even if the state cuts funding. … Increasing property taxes for homeowners is often a major source of funding when governments put money into school programs or renovations. | 837 | 3,777 | {"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-2022-33 | latest | en | 0.92187 |
academy.blendprecisely.io | 1,695,827,557,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510300.41/warc/CC-MAIN-20230927135227-20230927165227-00722.warc.gz | 97,194,441 | 19,434 | # The size of an essential oil drop - How many drops are...
Jun 23, 2020
Our new FREE Essential Oil Dilution Calculator is out! Check it out here: https://blendprecisely.io/free-eo-calculator
## What measurement should I use to be most accurate in my essential oil blending exercises?
If you put this magival question of "how many drops of essential oil in 1 oz" into the google search engine you get this magical answer by magic scent: "Usually 1 oz bottle of essential oil contains 600 drops."
Oh boy, if it would be THAT easy 🤓🤪
Drops of essential oils are the most known measurement to make an essential oil blend, but they are far from accurate or reliable across brands. It is not an issue if you are just doing experimentation where safety and repeatability are not necessary.
According to Robert Tisserand, in his blog about drop sizes (https://tisserandinstitute.org/learn-more/drop-size/), he cites three "empirical investigations" which give the conclusion that there are, on average, 30 drops per gram (or per ml since the density of most essential oils is close to 1 g/ml).
But that is an average, which may or may not be the case for your blend. He concludes that there are anywhere between 20 and 40 drops per gram. Some of the factors are viscosity, orificer size, purity, actual density, temperature, and other physical properties.
Let's break this down a bit and see what differences there could be by formulating a typical 30ml bottle of a fictional blend with Almond oil as a carrier oil and five drops each of Lavender, Peppermint, and West Indian Bay.
Two key bits of information are very evident from the chart below:
• First is that you can have a 200% increase in the dilution ratio between 20 and 40 drops/ml.
• Second is that the Dilution Ratio becomes more critical; the smaller the bottle size for the same number of drops of each ingredient, the higher the dilution ratio number, which can impact the outcome of your blend dramatically, especially for children, pregnant, elderlies and so forth.
The consequences of this appear through the lens of safety and sharing.
Let's say July uses only Brand A, which delivers 20 drops/ml through its orificers, and she shares a recipe with her friend Joanna who uses Brand B, which gives 40 drops/ml. The result could be dramatically different. Even worse is that Joanna uses a mix of brands, which could result in a noticeably different blend.
If your goal is to create a diffuser ambiance, it probably is not going to matter too much. It might be noticeable to some, but for most people, it would not be noticeable, especially if they had never tried the original.
If, however, you are creating a professional clinical or therapeutic blend, the odds are that you will want to know what was effective. You most likely want to be able to repeat your blend creation, especially as an Aromatherapist and as a cosmetic formulator.
You probably also care if you have a client who needs a topical blend or possibly one to ingest and are factoring in the "Topical Max." The safety limit for a topical application is called the "Suggested Topical Max," which is specified for all essential oils in Tisserands book "https://roberttisserand.com/essential-oil-safety-2nd-edition/".
Notice also that the Dilution Ratio for a 30ml bottle, one of the most common container sizes, creeps up close to the 3% line for a 20 drops/ml type of bottle, which is getting out of the perfectly safe arena. It may be okay or maybe not depending on whether you are working with babies, pregnant women, elderlies, or other sensitive populations.
Being aware of the safety of a blend is one of the primary tasks of an Aromatherapist and is not part of what this blog post is about. Still, it is essential to know about the variability that using drops introduces into the calculation.
So what is a better measure than drops?
The answer is almost anything. Even teaspoons are better if you can accurately add exact teaspoons. Milliliters are a universal SI unit that is recognized all over the world, which makes it great for sharing.
Using volume as a unit is also a little tricky when you are using your essential oils with ingredients like beeswax or bee pollen, which are measured by weight, not volume.
How do you describe a blend with a mix of volume and weight? Easy, convert the volumes to weight by using their density. Many essential oils have a density somewhere in the vicinity of 0.91 g/ml. To find out how many grams that is you just multiply the volume you have by the density: 0.25ml X 0.91 g/ml = .2275 grams. Once you know the weight of all of your ingredients, you can compute the Dilution Ratio and the Blend Percentage.
All of this done automatically for you in the BlendPrecisely software tool.
Boo math! Yay BlendPrecisely!
A further refinement in making blends, which is the ultimate in shareability, is to simply describe your mixture in percentages.
97 % Almond Oil, 1% Lavender, 1% Peppermint, and 1% Bay.
Boom! You are done. Using percentage can be shared with anyone, and they can use whatever they want to create their own. Teaspoons, shovels, buckets, tweezers, pinches. You get the point. It is also infinitely scalable for batches. Hello, formulators :-)
The end. Thanks for reading until the end :-) You are one tough cookie!
Seth
Update March 14th, 2023
I just found this article from the online cosmetic school Formula Botanica and I truly find a good article to share as well:
https://formulabotanica.com/never-measure-essential-oils-in-drops/
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https://kesiessays.com/random-samples-of-size-40-are-taken-from-a-normally-distributed-population-with/ | 1,670,468,464,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711232.54/warc/CC-MAIN-20221208014204-20221208044204-00831.warc.gz | 354,621,252 | 9,417 | # Random samples of size 40 are taken from a normally distributed population with
Random samples of size 40 are taken from a normally distributed population with a mean equal to 80 and a standard deviation equal to 12. Show detailed calculations to determine the upper and lower control limits
Write the two Excel function equations, using “CONFIDENCE” that would calculate the same UCL and LCL from the question above.
Random samples of size 50 are taken from a population with a given proportion equal to 62%. Show detailed calculations to determine the upper and lower control limits. | 118 | 588 | {"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 | 0.907157 |
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# Simplify the following expression 23 plus -5?
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
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It is 18 simplified | 130 | 357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-23 | latest | en | 0.726242 |
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# Is a bowling ball faster or a golf ball?
###### Wiki User
a Golf ball obviously...
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## Related Questions
It depends on how hard you throw/hit the ball. On average though, it is the golf ball.
Kinetic energy is a function of mass and velocity. Therefore, an object with more mass, such as a bowling ball, would have to go slower than an object with less mass, such as a golf ball. So, if given the same amount of kinetic energy, a bowling ball will go faster than a golf ball, because it has more mass.
momentum=velocity x mass say a golf ball weighs 1 pound and the bowling ball weighs 5 pounds the golf ball would have to be moving 5 times faster than the bowling ball to have the same momentum
the golf ball stopped but the bowling ball keep rolling due to their different size and weight.as we know bowling ball is bigger in size as well as weight so it will face more fictin force and stops early as compare to golf ball when they collide and at the same time bowling ball poshes the golf ball back.so the golf ball stops and the golf ball keeps rolling.
Well if you throw a bowling ball and a golf ball, which is gonna go faster?
Probably the bowling bowl. Momentum is mass x velocity and a bowling ball is generally heavier than a golf ball.
A bowling ball has more mass and is pulled by gravity, creating more resistance than a golf ball.
The force of the bowling ball colliding with the golf ball causes the golf ball to be redirected in an elastic collision. How fast either travels depends on the friction of the surface and the angle of contact with the bowling ball.Comparative Masses and EnergyIn the collision between a golf ball and a bowling ball, the fact that the bowling ball continues to move (although possibly changed in direction) is a function of the comparative masses of the two. The bowling ball is much more massive, so at normal velocities its kinetic energy exceeds the kinetic energy of the golf ball. In order to "stop" the bowling ball, the golf ball would have to make a perfectly aimed collision, and have a much higher velocity. Quantitatively, the velocity of the golf ball would have to be the inverse ratio of the ratio of the masses of the two balls, so that the kinetic energy (mass times velocity) is equal and in the opposite direction.Example : Golf ball at 45 g, ten pound bowling ball at 4500 g -- the golf ball would have to move at 100 times the velocity of the bowling ball to counteract its kinetic energy. If the bowling ball rolls at 2 m/sec, the golf ball would have to travel at more than 200 m/sec (720 kph or 447 mph), about 3 times a ball's normal velocity off the face of a golf club.
The answer depends on their relative speeds. Assuming that they are going at the same speed, a bowling ball would have more momentum than a golf ball.
If air resistance can be ignored (and it probably can from 2 rooms high) then both the bowling ball and golf ball will hit the floor at the same time. Although the bowling ball is harder to get moving than than the golf ball (it has more mass), the bowling ball also has a greater force pulling it down than the golf ball (as measured by its weight). The result is that both objects have the same acceleration.
Yes, no, maybe. It depends on how fast they're moving. At the same speed, a bowling ball has much more kinetic energy than a golf ball.
A golf ball falls faster because it's heavier than a ping pong ball.
A football and a golf ball are both faster than a book, even if kicked with the same force. Between a football and a golf ball the winner will be the golf ball, because the football has much more drag.
The one with the smaller mass (which is also the one with the smaller weight) would have to move faster, to compensate for the fact that it has less mass.
I say a golf ball, but it depends on how high you throw them.
The kinetic energy depends on both the velocity and the mass of the object. Kinetic energy = (1/2) x mass x velocity2. If a bowling ball and a golf ball are moving at the same velocity, then the bowling ball, being more massive, will have the greater amount of kinetic energy. However, bowling balls rarely move as fast as golf balls, so the full calculation is required for the specific situation of interest.
In deep space, free of other gravitational influences, theoretically, yes. All objects with mass have "gravity" -- as long as the bowling ball contains more mass, there ought to be a particular velocity at which a golf ball would orbit it. | 1,010 | 4,551 | {"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-2021-04 | latest | en | 0.936593 |
https://www.studypool.com/discuss/1027247/a-particle-is-correct-in-which-scenerio?free | 1,495,655,580,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607860.7/warc/CC-MAIN-20170524192431-20170524212431-00073.warc.gz | 938,939,509 | 14,764 | ##### A particle is correct in which scenerio ?
Physics Tutor: None Selected Time limit: 1 Day
A particle:
4 Points
Jun 8th, 2015
A particle:has an average velocity of 2 m/s if it moves 6 meters in 3 seconds
Jun 9th, 2015
A particle has:
4 Points
Jun 9th, 2015
an average acceleration of 3 m/s^2 if it goes from 2 m/s to 6 m/s in 2 seconds.
Jun 9th, 2015
...
Jun 8th, 2015
...
Jun 8th, 2015
May 24th, 2017
check_circle | 155 | 429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-22 | longest | en | 0.867275 |
https://db0nus869y26v.cloudfront.net/en/Adjacency_matrix | 1,718,991,767,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862132.50/warc/CC-MAIN-20240621160500-20240621190500-00822.warc.gz | 169,892,192 | 21,678 | In graph theory and computer science, an adjacency matrix is a square matrix used to represent a finite graph. The elements of the matrix indicate whether pairs of vertices are adjacent or not in the graph.
In the special case of a finite simple graph, the adjacency matrix is a (0,1)-matrix with zeros on its diagonal. If the graph is undirected (i.e. all of its edges are bidirectional), the adjacency matrix is symmetric. The relationship between a graph and the eigenvalues and eigenvectors of its adjacency matrix is studied in spectral graph theory.
The adjacency matrix of a graph should be distinguished from its incidence matrix, a different matrix representation whose elements indicate whether vertex–edge pairs are incident or not, and its degree matrix, which contains information about the degree of each vertex.
## Definition
For a simple graph with vertex set U = {u1, …, un}, the adjacency matrix is a square n × n matrix A such that its element Aij is one when there is an edge from vertex ui to vertex uj, and zero when there is no edge.[1] The diagonal elements of the matrix are all zero, since edges from a vertex to itself (loops) are not allowed in simple graphs. It is also sometimes useful in algebraic graph theory to replace the nonzero elements with algebraic variables.[2] The same concept can be extended to multigraphs and graphs with loops by storing the number of edges between each two vertices in the corresponding matrix element, and by allowing nonzero diagonal elements. Loops may be counted either once (as a single edge) or twice (as two vertex-edge incidences), as long as a consistent convention is followed. Undirected graphs often use the latter convention of counting loops twice, whereas directed graphs typically use the former convention.
### Of a bipartite graph
The adjacency matrix A of a bipartite graph whose two parts have r and s vertices can be written in the form
${\displaystyle A={\begin{pmatrix}0_{r,r}&B\\B^{\mathsf {T))&0_{s,s}\end{pmatrix)),}$
where B is an r × s matrix, and 0r,r and 0s,s represent the r × r and s × s zero matrices. In this case, the smaller matrix B uniquely represents the graph, and the remaining parts of A can be discarded as redundant. B is sometimes called the biadjacency matrix.
Formally, let G = (U, V, E) be a bipartite graph with parts U = {u1, ..., ur}, V = {v1, ..., vs} and edges E. The biadjacency matrix is the r × s 0–1 matrix B in which bi,j = 1 if and only if (ui, vj) ∈ E.
If G is a bipartite multigraph or weighted graph, then the elements bi,j are taken to be the number of edges between the vertices or the weight of the edge (ui, vj), respectively.
### Variations
An (a, b, c)-adjacency matrix A of a simple graph has Ai,j = a if (i, j) is an edge, b if it is not, and c on the diagonal. The Seidel adjacency matrix is a (−1, 1, 0)-adjacency matrix. This matrix is used in studying strongly regular graphs and two-graphs.[3]
The distance matrix has in position (i, j) the distance between vertices vi and vj. The distance is the length of a shortest path connecting the vertices. Unless lengths of edges are explicitly provided, the length of a path is the number of edges in it. The distance matrix resembles a high power of the adjacency matrix, but instead of telling only whether or not two vertices are connected (i.e., the connection matrix, which contains boolean values), it gives the exact distance between them.
## Examples
### Undirected graphs
The convention followed here (for undirected graphs) is that each edge adds 1 to the appropriate cell in the matrix, and each loop adds 2.[4] This allows the degree of a vertex to be easily found by taking the sum of the values in either its respective row or column in the adjacency matrix.
${\displaystyle {\begin{pmatrix}2&1&0&0&1&0\\1&0&1&0&1&0\\0&1&0&1&0&0\\0&0&1&0&1&1\\1&1&0&1&0&0\\0&0&0&1&0&0\end{pmatrix))}$
Coordinates are 1–6.
Coordinates are 0–23.
White fields are zeros, colored fields are ones.
### Directed graphs
The adjacency matrix of a directed graph can be asymmetric. One can define the adjacency matrix of a directed graph either such that
1. a non-zero element Aij indicates an edge from i to j or
2. it indicates an edge from j to i.
The former definition is commonly used in graph theory and social network analysis (e.g., sociology, political science, economics, psychology).[5] The latter is more common in other applied sciences (e.g., dynamical systems, physics, network science) where A is sometimes used to describe linear dynamics on graphs.[6]
Using the first definition, the in-degrees of a vertex can be computed by summing the entries of the corresponding column and the out-degree of vertex by summing the entries of the corresponding row. When using the second definition, the in-degree of a vertex is given by the corresponding row sum and the out-degree is given by the corresponding column sum.
Coordinates are 0–23.
As the graph is directed, the matrix is not necessarily symmetric.
### Trivial graphs
The adjacency matrix of a complete graph contains all ones except along the diagonal where there are only zeros. The adjacency matrix of an empty graph is a zero matrix.
## Properties
### Spectrum
The adjacency matrix of an undirected simple graph is symmetric, and therefore has a complete set of real eigenvalues and an orthogonal eigenvector basis. The set of eigenvalues of a graph is the spectrum of the graph.[7] It is common to denote the eigenvalues by ${\displaystyle \lambda _{1}\geq \lambda _{2}\geq \cdots \geq \lambda _{n}.}$
The greatest eigenvalue ${\displaystyle \lambda _{1))$ is bounded above by the maximum degree. This can be seen as result of the Perron–Frobenius theorem, but it can be proved easily. Let v be one eigenvector associated to ${\displaystyle \lambda _{1))$ and x the component in which v has maximum absolute value. Without loss of generality assume vx is positive since otherwise you simply take the eigenvector ${\displaystyle -v}$, also associated to ${\displaystyle \lambda _{1))$. Then
${\displaystyle \lambda _{1}v_{x}=(Av)_{x}=\sum _{y=1}^{n}A_{x,y}v_{y}\leq \sum _{y=1}^{n}A_{x,y}v_{x}=v_{x}\deg(x).}$
For d-regular graphs, d is the first eigenvalue of A for the vector v = (1, …, 1) (it is easy to check that it is an eigenvalue and it is the maximum because of the above bound). The multiplicity of this eigenvalue is the number of connected components of G, in particular ${\displaystyle \lambda _{1}>\lambda _{2))$ for connected graphs. It can be shown that for each eigenvalue ${\displaystyle \lambda _{i))$, its opposite ${\displaystyle -\lambda _{i}=\lambda _{n+1-i))$ is also an eigenvalue of A if G is a bipartite graph.[8] In particular −d is an eigenvalue of any d-regular bipartite graph.
The difference ${\displaystyle \lambda _{1}-\lambda _{2))$ is called the spectral gap and it is related to the expansion of G. It is also useful to introduce the spectral radius of ${\displaystyle A}$ denoted by ${\displaystyle \lambda (G)=\max _{\left|\lambda _{i}\right|. This number is bounded by ${\displaystyle \lambda (G)\geq 2{\sqrt {d-1))-o(1)}$. This bound is tight in the Ramanujan graphs, which have applications in many areas.
### Isomorphism and invariants
Suppose two directed or undirected graphs G1 and G2 with adjacency matrices A1 and A2 are given. G1 and G2 are isomorphic if and only if there exists a permutation matrix P such that
${\displaystyle PA_{1}P^{-1}=A_{2}.}$
In particular, A1 and A2 are similar and therefore have the same minimal polynomial, characteristic polynomial, eigenvalues, determinant and trace. These can therefore serve as isomorphism invariants of graphs. However, two graphs may possess the same set of eigenvalues but not be isomorphic.[9] Such linear operators are said to be isospectral.
### Matrix powers
If A is the adjacency matrix of the directed or undirected graph G, then the matrix An (i.e., the matrix product of n copies of A) has an interesting interpretation: the element (i, j) gives the number of (directed or undirected) walks of length n from vertex i to vertex j. If n is the smallest nonnegative integer, such that for some i, j, the element (i, j) of An is positive, then n is the distance between vertex i and vertex j. A great example of how this is useful is in counting the number of triangles in an undirected graph G, which is exactly the trace of A3 divided by 3 or 6 depending on whether the graph is directed or not. We divide by those values to compensate for the overcounting of each triangle. In an undirected graph, each triangle will be counted twice for all three nodes, because the path can be followed clockwise or counterclockwise : ijk or ikj. The adjacency matrix can be used to determine whether or not the graph is connected.
If a directed graph is has a nilpotent adjacency matrix (i.e., if there exists n such that An is the zero matrix), then it is a directed acyclic graph.[10]
## Data structures
The adjacency matrix may be used as a data structure for the representation of graphs in computer programs for manipulating graphs. The main alternative data structure, also in use for this application, is the adjacency list.[11][12]
The space needed to represent an adjacency matrix and the time needed to perform operations on them is dependent on the matrix representation chosen for the underlying matrix. Sparse matrix representations only store non-zero matrix entries and implicitly represent the zero entries. They can, for example, be used to represent sparse graphs without incurring the space overhead from storing the many zero entries in the adjacency matrix of the sparse graph. In the following section the adjacency matrix is assumed to be represented by an array data structure so that zero and non-zero entries are all directly represented in storage.
Because each entry in the adjacency matrix requires only one bit, it can be represented in a very compact way, occupying only |V |2 / 8 bytes to represent a directed graph, or (by using a packed triangular format and only storing the lower triangular part of the matrix) approximately |V |2 / 16 bytes to represent an undirected graph. Although slightly more succinct representations are possible, this method gets close to the information-theoretic lower bound for the minimum number of bits needed to represent all n-vertex graphs.[13] For storing graphs in text files, fewer bits per byte can be used to ensure that all bytes are text characters, for instance by using a Base64 representation.[14] Besides avoiding wasted space, this compactness encourages locality of reference. However, for a large sparse graph, adjacency lists require less storage space, because they do not waste any space representing edges that are not present.[12][15]
An alternative form of adjacency matrix (which, however, requires a larger amount of space) replaces the numbers in each element of the matrix with pointers to edge objects (when edges are present) or null pointers (when there is no edge).[15] It is also possible to store edge weights directly in the elements of an adjacency matrix.[12]
Besides the space tradeoff, the different data structures also facilitate different operations. Finding all vertices adjacent to a given vertex in an adjacency list is as simple as reading the list, and takes time proportional to the number of neighbors. With an adjacency matrix, an entire row must instead be scanned, which takes a larger amount of time, proportional to the number of vertices in the whole graph. On the other hand, testing whether there is an edge between two given vertices can be determined at once with an adjacency matrix, while requiring time proportional to the minimum degree of the two vertices with the adjacency list.[12][15]
## References
1. ^ Biggs, Norman (1993), Algebraic Graph Theory, Cambridge Mathematical Library (2nd ed.), Cambridge University Press, Definition 2.1, p. 7.
2. ^ Harary, Frank (1962), "The determinant of the adjacency matrix of a graph", SIAM Review, 4 (3): 202–210, Bibcode:1962SIAMR...4..202H, doi:10.1137/1004057, MR 0144330.
3. ^ Seidel, J. J. (1968). "Strongly Regular Graphs with (−1, 1, 0) Adjacency Matrix Having Eigenvalue 3". Lin. Alg. Appl. 1 (2): 281–298. doi:10.1016/0024-3795(68)90008-6.
4. ^ Shum, Kenneth; Blake, Ian (2003-12-18). "Expander graphs and codes". Volume 68 of DIMACS series in discrete mathematics and theoretical computer science. Algebraic Coding Theory and Information Theory: DIMACS Workshop, Algebraic Coding Theory and Information Theory. American Mathematical Society. p. 63. ISBN 9780821871102.
5. ^ Borgatti, Steve; Everett, Martin; Johnson, Jeffrey (2018), Analyzing Social Networks (2nd ed.), SAGE, p. 20
6. ^ Newman, Mark (2018), Networks (2nd ed.), Oxford University Press, p. 110
7. ^ Biggs (1993), Chapter 2 ("The spectrum of a graph"), pp. 7–13.
8. ^ Brouwer, Andries E.; Haemers, Willem H. (2012), "1.3.6 Bipartite graphs", Spectra of Graphs, Universitext, New York: Springer, pp. 6–7, doi:10.1007/978-1-4614-1939-6, ISBN 978-1-4614-1938-9, MR 2882891
9. ^ Godsil, Chris; Royle, Gordon Algebraic Graph Theory, Springer (2001), ISBN 0-387-95241-1, p.164
10. ^ Nicholson, Victor A (1975). "Matrices with Permanent Equal to One" (PDF). Linear Algebra and its Applications (12): 187.
11. ^ Goodrich & Tamassia (2015), p. 361: "There are two data structures that people often use to represent graphs, the adjacency list and the adjacency matrix."
12. ^ a b c d Cormen, Thomas H.; Leiserson, Charles E.; Rivest, Ronald L.; Stein, Clifford (2001), "Section 22.1: Representations of graphs", Introduction to Algorithms (Second ed.), MIT Press and McGraw-Hill, pp. 527–531, ISBN 0-262-03293-7.
13. ^ Turán, György (1984), "On the succinct representation of graphs", Discrete Applied Mathematics, 8 (3): 289–294, doi:10.1016/0166-218X(84)90126-4, MR 0749658.
14. ^ McKay, Brendan, Description of graph6 and sparse6 encodings, archived from the original on 2001-04-30, retrieved 2012-02-10.
15. ^ a b c Goodrich, Michael T.; Tamassia, Roberto (2015), Algorithm Design and Applications, Wiley, p. 363. | 3,540 | 14,141 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 16, "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-2024-26 | latest | en | 0.931356 |
https://www.coursehero.com/file/5989248/prob3/ | 1,495,941,907,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609409.62/warc/CC-MAIN-20170528024152-20170528044152-00354.warc.gz | 1,071,997,480 | 39,630 | # prob3 - Each side has 2 windows that are 2 ft by 3 ft that...
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%========================================================================== % PROBLEM 3 - Painting a house %-------------------------------------------------------------------------- % % Script Name: prob3 % Script Variables: % 1. paint_area (double) - surface area to be covered by paint (sq. ft) % 2. paint_gallons (double) - number of gallons of paint needed % % Problem Statement: % Write a script named "prob3" that will calculate the surface area that % is to be painted on a house and how many gallons of paint will be % needed. The house is a rectangular prism that measures 30 ft wide, 50 % ft long, and 10 ft tall. The roof and the floor will not be painted.
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Unformatted text preview: % Each side has 2 windows that are 2 ft by 3 ft that will not be painted. % Store the amount of surface area to be covered in "paint_area". Use % this value to calculate how much paint will be needed to give the house % one coat of paint. Each gallon of paint will cover 350 sq. ft. % width_ft=30; length_ft=50; height_ft=10; area_windows=2.*(2.*3); %ft^2, this is per side of the house paint_area=2*(height_ft.*length_ft-area_windows)+2.*(height_ft.*width_ft-area_windows) %ft^2 area_per_gallon=350; %sq. ft/gallon paint_gallons=paint_area./area_per_gallon %gallons...
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## This note was uploaded on 10/21/2010 for the course CS 1371 taught by Professor Stallworth during the Spring '08 term at Georgia Tech.
Ask a homework question - tutors are online | 413 | 1,671 | {"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-2017-22 | longest | en | 0.852019 |
https://forum.processing.org/two/discussion/25042/printing-2darray-as-text | 1,632,193,112,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057131.88/warc/CC-MAIN-20210921011047-20210921041047-00505.warc.gz | 304,100,560 | 14,035 | #### Howdy, Stranger!
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# printing 2darray as text...
edited May 2018
Hi, i've searched for a while and I'm not sure if i'm missing something obvious, i'm trying to print a 2d integer array as text, i did it before but I can't remember how i did and it is driving me insane.
thanks.
``````void setup() {
size(300, 300);
background(255);
array();
}
void array() {
int[][] arrayinit = {
{ 10, 5, 3, 55, 90 },
{ 9, 4, 2, 32, 22 },
{ 3, 8, 5, 79, 22 },
{ 3, 8, 45, 4, 2 },
{ 3, 22, 5, 44, 4 }
};
int rows = 5;
int cols = 5;
int[] sum = new int[cols];
int[] sum2 = new int[rows];
stroke(0);
for (int c=0; c<cols; c++) { // for every column
for (int r=0; r<rows; r++) { // for every row
rect(40*r, 30*c, 40, 30);
text(arrayinit[c][r], 40*r, 30*c); //dont know if this is right
sum[c] += arrayinit[r][c]; // sum up the values per column
sum2[r] += arrayinit[r][c];
}
}
println(sum[0]);
println(sum[1]);
println(sum[2]);
println(sum2[2]);
println(arrayinit[2][2]);
}
``````
Tagged:
• edited November 2017
I've made this utility function to join a 2D String[][] array as 1 String, ready for text(): :ar!
``````static final String joinStrArr2d(final String[][] arr2d) {
final StringBuilder sb = new StringBuilder();
int outer = 0;
for (final String[] arr1d : arr2d) {
sb.append(ENTER);
int inner = 0;
for (final String item : arr1d) sb
.append('[').append(outer).append("][").append(inner++)
.append("] ").append(item).append(ENTER);
++outer;
}
return sb.toString();
}
``````
You are drawing both the rectangle and text. But both has white color as their fill. Even the background. So it's pretty much invisible. You could simply change your text color to solve the problem.
It will look like this:
`````` for (int c=0; c<cols; c++) { // for every column
for (int r=0; r<rows; r++) { // for every row
fill(255);
if(c < rows-1)rect(40*r, 30*c, 40, 30);
fill(0);
text(arrayinit[c][r], 40*r, 30*c); //dont know if this is right
sum[c] += arrayinit[r][c]; // sum up the values per column
sum2[r] += arrayinit[r][c];
}
}
``````
Best regards :D
``````int[][] arrayinit = {
{ 10, 5, 3, 55, 90 },
{ 9, 4, 2, 32, 22 },
{ 3, 8, 5, 79, 22 },
{ 3, 8, 45, 4, 2 },
{ 3, 22, 5, 44, 4 }
};
void setup() {
size(300, 300);
background(255);
textAlign(CENTER);
}
void draw() {
int rows = 5;
int cols = 5;
stroke(0);
for (int c=0; c<cols; c++) { // for every column
for (int r=0; r<rows; r++) { // for every row
// draw rect
fill(255); // white
rect(40*r+43, 30*c+43,
40, 30);
// draw text into rect
fill(255, 0, 0); // red
text(arrayinit[c][r],
40*r+43+20, 30*c+43+20);
}
}
}
``````
• Thanks, as I thought it was pretty obvious.
• Another alternate approach is to use a fixed-width font and print a single line-wrapped string. This will create tabular text on the screen quickly and easily using the built-in text system. However, the method is inflexible (i.e. space padding is required), and it is very sensitive to character width and line height.
• Belatedly: Another option is a simple trick that uses Java `Arrays.deepToString()` and then reformats the output with simple search-replace rules.
``````import java.util.Arrays;
int[][] arrayinit = {
{ 1, 5, 3 },
{ 9, 4, 2 },
{ 3, 8, 6 },
};
void setup() {
println(Arrays.deepToString(arrayinit);
println();
println(Arrays.deepToString(arrayinit)
.replace("[[", "")
.replace("], [", "\n")
.replace("]]", "")
.replace(" ", " "));
}
``````
Output:
``````[[1, 5, 3], [9, 4, 2], [3, 8, 6]]
1, 5, 3
9, 4, 2
3, 8, 6
``````
Related: | 1,241 | 3,611 | {"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-2021-39 | latest | en | 0.633477 |
https://www.coursehero.com/file/6648073/lecture2-getting-started/ | 1,513,418,724,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948587496.62/warc/CC-MAIN-20171216084601-20171216110601-00118.warc.gz | 730,472,995 | 60,873 | lecture2_getting_started
# lecture2_getting_started - Getting Started We seek a theory...
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Getting Started We seek a theory of bidding behavior in auctions. Our theory will attempt to explain how peoples’ bids are related to their individual valuations, or simply values, for the item being auctioned. In mathematical terminology, we want a mapping from values to bids . A person’s value is the hypothetical price at which she is indifferent between buying the item and not buying it. With this interpretation, the payoff someone gets from winning an item in an auction is the difference between her value and the actual price she pays for it. Rod Garratt ECON 177: Auction Theory With Experiments
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We cannot presume to know peoples’ values for items they bid on in auctions. An individual’s willingness to pay for something depends on. ... how much satisfaction or enjoyment she gets from consuming it instead of other things she could potentially consume in its place their wealth or disposable income These attributes are (typically) unobservable. We can’t proceed unless we know something about individual values. We assume that we know the range of possible values that people hold and the likelihood that someone’s value lies in any interval . Rod Garratt ECON 177: Auction Theory With Experiments
Probability distribution function, F Let v i denote individual i ’s value From the point of view of the auctioneer, or other bidders, or an economist analyzing the auction, individual i’s value is a random variable ˜ v i The information the auctioneer has about the likelihood of different realizations of the random variable ˜ v i is summarized by the function F . For any value v , the number F ( v ) tells us probability that an individual has a value that is less than or equal to v . Rod Garratt ECON 177: Auction Theory With Experiments
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You can think of the function F as representing the distribution of types of people that exist in the population. F ( v ) is the fraction of the population with types less than v An individual shows up at an auction is a random draw from the population Hence, the likelihood that she has a value in any range is simply given by the fraction of the population that has values in that range. Rod Garratt
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• Fall '09
• GARRATT
• Probability distribution, Probability theory, probability density function, Cumulative distribution function, Rod Garratt
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lecture2_getting_started - Getting Started We seek a theory...
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Ask a homework question - tutors are online | 638 | 2,982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-51 | latest | en | 0.894 |
http://lab.polygonal.de/2007/07/18/fast-and-accurate-sinecosine-approximation/ | 1,534,412,593,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210615.10/warc/CC-MAIN-20180816093631-20180816113631-00502.warc.gz | 227,011,446 | 6,704 | # Fast and accurate sine/cosine approximation
Trigonometric functions are costly operations and can slow down your application if they are extensively used. There are two reasons why: First, Math.sin() is a function, and thus needs a function call which simple eats up some time. Second, the result is computed with much more precision than you would ever need in most situations.
Most often you just want the periodic wave-like characteristics of the sine or cosine, which can be approximated in various ways. One common way of making it faster is to create a lookup-table by computing the sine at discrete steps and storing the result in an array. For example:
```var sineTable:Array = [];
for (var i:int = 0; i < 90; i++)
{
sineTable[i] = Math.sin(Math.PI/180 * i)
}
```
Due to the symmetry of the sine wave, it’s sufficient to compute one quadrant only (0..pi/2), and the other 3/4’s of the circle can be computed by shifting and wrapping the input value. The biggest drawback is that the values are stored at a fixed resolution and so the result is not very accurate. This can be enhanced with linear interpolation:
```x = 22.5;
y = sineTable[int(x)] + (sineTable[int(x + .5)] - sineTable[int(x)]) / 2;
```
Much better, but yet the error exists. It also involves accessing array elements which makes the code rather slow. Another technique uses taylor series approximation:
```sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...
```
Like with the lookup-table, evaluating this term is costly.
After searching for alternatives, I finally found a fantastic solution using a quadratic curve which blows everything away in terms of performance and accuracy. For a detailed derivation, please follow the link because I won’t go into it.
I did minor optimizations to figure out what AS3 likes most, and arrived at some code that can be up to 14x faster, while still being very accurate. However, you have to use it directly – do not place the code inside a function, because the additional function call sweeps out the performance gain, and you are left with an approximation that is actually slower compared to a native Math.sin() or Math.cos() call. Also note that cos(x) = sin(x + pi/2) or cos(x – pi/2) = sin(x), so computing the cosine is just of matter adding pi/2 to the input value.
Below is a simple visualization to show you the quality of the approximation. The high precision version can replace the Math.sin() and Math.cos() calls in nearly all situations.
### Low precision sine/cosine (~14x faster)
```//always wrap input angle to -PI..PI
if (x < -3.14159265)
x += 6.28318531;
else
if (x > 3.14159265)
x -= 6.28318531;
//compute sine
if (x < 0)
sin = 1.27323954 * x + .405284735 * x * x;
else
sin = 1.27323954 * x - 0.405284735 * x * x;
//compute cosine: sin(x + PI/2) = cos(x)
x += 1.57079632;
if (x > 3.14159265)
x -= 6.28318531;
if (x < 0)
cos = 1.27323954 * x + 0.405284735 * x * x
else
cos = 1.27323954 * x - 0.405284735 * x * x;
}
```
### High precision sine/cosine (~8x faster)
```//always wrap input angle to -PI..PI
if (x < -3.14159265)
x += 6.28318531;
else
if (x > 3.14159265)
x -= 6.28318531;
//compute sine
if (x < 0)
{
sin = 1.27323954 * x + .405284735 * x * x;
if (sin < 0)
sin = .225 * (sin *-sin - sin) + sin;
else
sin = .225 * (sin * sin - sin) + sin;
}
else
{
sin = 1.27323954 * x - 0.405284735 * x * x;
if (sin < 0)
sin = .225 * (sin *-sin - sin) + sin;
else
sin = .225 * (sin * sin - sin) + sin;
}
//compute cosine: sin(x + PI/2) = cos(x)
x += 1.57079632;
if (x > 3.14159265)
x -= 6.28318531;
if (x < 0)
{
cos = 1.27323954 * x + 0.405284735 * x * x;
if (cos < 0)
cos = .225 * (cos *-cos - cos) + cos;
else
cos = .225 * (cos * cos - cos) + cos;
}
else
{
cos = 1.27323954 * x - 0.405284735 * x * x;
if (cos < 0)
cos = .225 * (cos *-cos - cos) + cos;
else
cos = .225 * (cos * cos - cos) + cos;
}
``` | 1,184 | 3,858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-34 | latest | en | 0.921607 |
http://math.stackexchange.com/questions/227619/what-is-a-first-order-signature | 1,467,276,188,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783398216.41/warc/CC-MAIN-20160624154958-00198-ip-10-164-35-72.ec2.internal.warc.gz | 206,285,753 | 17,328 | # What is a “first order signature”?
So I was just given the definition of a signature:
A signature is a pair $\Sigma = (\Omega, \Pi)$ where $\Omega$ is a set of operation symbols $\omega$, each equipped with an arity $\alpha(\omega)\in\mathbb{N}$ and $\Pi$ is a set of relation symbols $\pi$, each equipped with an arity $\alpha(\pi)\in\mathbb{N}$.
And I saw the term "first order signature" after this without any previous context. Can anyone tell me what is a "first order signature"? Thanks a lot.
-
Probably "first-order" is used there to emphasize that the operations and relations are not higher-order, i.e. they are defined on powers of the carrier, not on powersets (and iterations thereof), i.e. they operate on elements of the carrier, not subsets. – Bill Dubuque Nov 2 '12 at 16:59
In set theory, there is only one nonlogical symbol, the set membership relation $\in$. In Peano arithmetic, there are several nonlogical symbols: the addition symbol $+$, the multiplication symbol $\cdot$, and the order relation $<$. Each formal theory in first-order logic has a specific set of nonlogical symbols that is uses. This collection is called the signature of the theory. It can have function symbols (like the addition symbol in Peano arithmetic) as well as relation symbols (like the order relation symbol). In general, each function symbol can have its own arity, and each relation symbol can have its own arity. The "first order" term in "first order signature" usually just means the author is working in first-order logic, where every signature is a "first order signature".
-
Bill Dubuque's comment provides the answer.
Say an operation or function is first-order if it maps objects in the relevant domain to objects. (Some operations are higher-order, e.g. mapping functions to functions, e.g. the operation that takes a function to its inverse).
Say a relation is first-order if it relates objects in the domain. (Some relations are higher-order, e.g. relating functions to functions, e.g. the relation two functions have if they form a Galois connection.)
To say that a signature is first-order means that all the operation (function) symbols are symbols for first-order functions, and the relational symbols are all symbols for first-order relations.
- | 505 | 2,280 | {"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} | 3.484375 | 3 | CC-MAIN-2016-26 | latest | en | 0.914651 |
https://www.convertunits.com/from/zeptopascal/to/picopascal | 1,619,124,583,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039604430.92/warc/CC-MAIN-20210422191215-20210422221215-00118.warc.gz | 790,609,506 | 16,581 | ## ››Convert zeptopascal to picopascal
zeptopascal picopascal
How many zeptopascal in 1 picopascal? The answer is 1000000000.
We assume you are converting between zeptopascal and picopascal.
You can view more details on each measurement unit:
zeptopascal or picopascal
The SI derived unit for pressure is the pascal.
1 pascal is equal to 1.0E+21 zeptopascal, or 1000000000000 picopascal.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between zeptopascals and picopascals.
Type in your own numbers in the form to convert the units!
## ››Want other units?
You can do the reverse unit conversion from picopascal to zeptopascal, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Zeptopascal
The SI prefix "zepto" represents a factor of 10-21, or in exponential notation, 1E-21.
So 1 zeptopascal = 10-21 pascals.
The definition of a pascal is as follows:
The pascal (symbol Pa) is the SI unit of pressure.It is equivalent to one newton per square metre. The unit is named after Blaise Pascal, the eminent French mathematician, physicist and philosopher.
## ››Definition: Picopascal
The SI prefix "pico" represents a factor of 10-12, or in exponential notation, 1E-12.
So 1 picopascal = 10-12 pascals.
The definition of a pascal is as follows:
The pascal (symbol Pa) is the SI unit of pressure.It is equivalent to one newton per square metre. The unit is named after Blaise Pascal, the eminent French mathematician, physicist and philosopher.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 530 | 2,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-17 | latest | en | 0.825703 |
http://physics.stackexchange.com/questions/62131/combining-metric-tensors-curvature-tensors | 1,469,542,078,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824994.73/warc/CC-MAIN-20160723071024-00270-ip-10-185-27-174.ec2.internal.warc.gz | 195,272,277 | 18,046 | # Combining metric tensors/curvature tensors
I was thinking about the following scenario:
Consider a particle which causes a metric $g_{\mu\nu}$ on an otherwise Minkowski spacetime (or any manifold). Now, consider another particle, somewhere in the vicinity of the first particle, which causes a metric $h_{\mu\nu}$ on a spacetime which would have been Minkowski if not for these two particles.
Then, what what would the metric in the vicinity of these two points be? I am guessing that it is: $$(g_{\mu\nu}-\eta_{\mu\nu})+(h_{\mu\nu}-\eta_{\mu\nu}) + \eta_{\mu\nu} = g_{\mu\nu}+h_{\mu\nu} - \eta_{\mu\nu}$$
Also, does the Riemann Curvature Tensor $R_{\mu\nu\rho}^\sigma$add up directly? I don't think it should because the Einstein tensor $G_{\mu\nu}$ does (I think) and it is dependent on the Ricci Curvature AND the spacetime metric tensor.
-
You can do this addition at weak field, so long as your coordinates make the metric nonsingular (rectangular coordinates as the unperturbed metric), and add the usually negligible corrections perturbatively. – Ron Maimon Aug 22 '13 at 23:57
Unlike classical electromagnetism, General Relativity is highly nonlinear--this means that the gravitational field can serve as its own source. A consequence of this fact is that fields decidedly do not superpose, and you can get all sorts of effects even from vacuum relativity. The most notable of these effects are things such as Brill waves and Geons, where gravitational waves collide or collapse to form black holes. You can work out solutions where this happens even when the spaces initially are empty outside of the two regions before overlap occurs.
-
The metrics don't simply add together as you suggest. In fact there is no known solution for the metric when you have two point masses (thought there are approximate solutions). If there was it would make calculating the motion of binary black holes a great deal simpler than it currently is. The curvature has to be calculated numerically.
-
One way of phrasing this is that the nonlinearity of the equations means that you can't say that the sum (or difference) of two solutions to the Einstein field equations is also a solution. So even taking the delta with respect to the Minkowski metric, as you would like to do, isn't allowed (unless you work perturbatively as John alluded to).
- | 560 | 2,349 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2016-30 | latest | en | 0.937455 |
https://www.ecolebooks.com/mathematics-form-5-trigonometry1/ | 1,718,794,032,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00866.warc.gz | 662,802,337 | 73,469 | ## TRIGONOMETRY
Trigonometry is the study of angle measurement and functions that depends on angle.
The fundamental trigonometric ratios are
Sine (sin)
Cosine (Cos)
Tangent (Tan)
Others are cosecant (cosec)
Secant (sec)
Cotangent (cot )
Let θ be the angle in a right angled triangle; then we say
Sin θ
COS θ
Tan θ
And = Cosecant θ = Cosec θ
= secant = sec θ
= Cotangent θ = cot θ
Consider a right angled triangle below
Sin θ = = ………..(i)
cos θ = = ……….(ii)
tan θ= = …………(iii)
= cosec θ= = (iv)
= sec θ = = (v)
= cot θ = = (vi)
But = tanθ
SPECIAL ANGLES
These are the angles which we can find their trigonometric ratios without mathematical tables or scientific calculators.
The angles are 00, 300, 450, 600, 900, 1800, 2700, 3600.
Finding the trigonometric ratios for special angles.
Case 1: Consider 300 and 600
Here use an equilateral triangle with unit sides
That is
From AMB (right angled)
Then from the fig above
Sin 300 = =
Case 2 Consider 450
Here use are square with unit sides (1 unit)
That is
From ABC (right angled)
= +
= 1² + 1² = 2
=
Then sin 450 = =
Cos 450 = =
Tan 450 = = 1
Trigonometric ratios for 00, 900, 1800 and 2700 and 3600.
Here use a unit circle ‘Discussed also in O level’
A unit circle is a circle with radius (1 unit)
Suppose p(x,y) is a point in a unit circle
Generally in a unit circle
X = cosine value of an angle
Y= sine value of an angle
= Tangent of an angle
Angle measurement can be in two ways.
Clockwise direction (-ve angles)
Anticlockwise direction (+ve angles)
From a unit circle we use
X= cosine value of an angle
Y= sine value of an angle
Hence consider angles 00, 900, 1800, 2700, 3600 and their corresponding coordinates in a unit circle.
00 means
360°means
Summary:-
The concept of picture and negative angles.
But sine function and tangent function are odd functions
Cosine function is an even function
Fig above
From Sin θ =
Sin ( -θ) = – = -sinθ
cos θ =
cos(-θ) = Cos θ
The idea is discussed in O’Level form IV Basic Mathematics, but let us recall the idea.
The range of the angles is 0°< θ<900
The all trig ratios are positive and are obtained directly from four figure (mathematical figure
The range of the angles is 900 < θ <1800
Ranges from 180°< θ< 270°
Ranges from 270°<θ<360°
Eg: Sin 315° = -Sin (360° -315°)
=-
= -tan (360° – 330°
= -tan 30°
=
=
= =
PYTHAGORAS THEOREM (IDENTITY)
Consider a right angled
From Pythagoras theorem
+ b² = c²
Dividing by C²
+ =
+ ( = 1————–
Substitute equations (i) and (ii) into (*)
Then we get
Is the Pythagoras Identity.
Dividing equation (1) by
dividing equation (i) by Sin2θ
APPLICATIONS OF PYTHAGORAS IDENTITY
I. SOLVING TRIG EQUATIONS
Example 1.
Solve the equation 1 + = 0 for the values of the values (θ) between 00 and 3600 inclusive.
Solution:
1 + =0
But from Pythagoras identity
cosθ = 0,cos θ =-1
case of cosθ = 0
θ=cos(0)
θ=900
θ=900,2700
Example 2.
Solve for the values of x between 00 and 3600 inclusive of
(i) Tan 4x + 7 = 4sec2x
(ii) -6sm2x – cosx + 5 =0
Solution
Tan4x + 7 =4sec2x
But tan2x + 1 =sec2x
Tan4x + 7=4(tan2x + 1)
Tan4x + 7 =4tan2x + 4
Tan4x +7-4tan2x -4 =0
Tan4x -4tan2x + 3 =0
Let tan2x =m
Then m2 – 4m +3 =0
m2 -3m –m + 3 =0
m(m -3)-1(m-3)=0
(m – 1)(m-3) =0
m – 1 =0, m- 3=0
m= 1, m=3
Case 1 m =1 =tan2x
Tan x =
Tan x = 1
X = tan-1(1) = 450
X = 1800 + 450 = 2250
Tan x =-1
X= tan -1(-1)
X =180 450 =1350
X = 3600 -450=3150
Case 2: m3
Tan2x = 3, tanx=
Tan x =
X = tan-1( =600
X =1800 + 600 =2400
tan x =-
x = tan -1(-
= 1800 -600=1200
X=3600 -600=3000
x= work on (ii)
II PROVING IDENTITIES
Examples: prove the following identify
i) Tan2θ + sin2θ =(secθ + cosθ) (secθ – cosθ)
ii) Cot4θ + cot2θ =cosec4θ – cosec2θ
iii) = cosecθ – cotθ
iv)
v) cosecθ –sinθ = cotθ
Solution: (i)
tan2θ + sin2θ = (secθ+ cosθ) (secθ –cosθ)
Delaying with R.H.s
Proof = (secθ + cosθ)(secθ – cosθ)
Then
=sec2θ – cos2θ
But sec2θ = 1+ tan2θ and
Cos2θ = 1 –sin2θ
=1 + tan2θ -(1 – sin2θ)
=1 + tan2θ -1 + sin2θ
=tan2θ+ sin2θ
tan2θ+ sin2θ L.H.S proved
ii) cot4θ+ cot²θ= cosec4θ – cosec2θ
solution.
Dealing with L.H.S
Proof
=Cot4θ + cot2θ
then
=Cot2θ(cot2θ + 1)
But Cot2θ+ 1 =cosec2θ
Cot2θ =cosec2θ -1
(cosec2θ -1) cosec2θ
Cosec4θ – cosec2θ R.H.S
Cot4θ + cot2θ= cosec4θ – cosec2θ
iv) sin θtanθ + cosθ=secθ
solution.
Proof
Dealing with L.H.S
Sinθtanθ+ cosθ
But tanθ =
Then
Sinθ + cosθ
= = secθ
sin²θ + cos²θ =1 (Pythagoras identity)
sin
III) ELIMINATION PROBLEMS
Examples:
Eliminate ÆŸ from the following equations
i) Cosθ + 1 =x and sinθ =y
ii) X= a sinθ and y= btan θ
iii) X= 1 + tanθ and y = cos θ
iv) X= sinθ – cosθ
Y= cotθm+ tanθ
Solution.
(i) Cosθ + 1 =x
Cosθ=x – 1 ……… (i)
sinθ = y…………..(ii)
squaring equations (i) and (ii) the sum
cos²θ+ sin²θ= (x -1)² + y²
but sin²θ + cos²θ =1
then 1= (x – 1)² + y²
1 = x² – 2x + 1 + y²
x² + y2 -2x + 1 – 1 =0
x² +y²- 2x =0
ii) from x = a sinθ, sinθ=
and from y=btanθ, tanθ=
refer + =1
dividing by both sides
+ =
1+ =
But
Then 1 + =
1 + =
1 + =
iii) X = 1 +
= x – 1 ……….. (i)
= y
Refer, + = 1
Dividing by both sides
+ =
ÆŸ + 1 =
+ 1=
+ 1 =
= 1
Solution (iv)
x = ………….(a)
Y = + ……….(b)
From (b)
= +
Y= =
Y =
Squaring
x² =
x² = -2+
=+ -2
x² = 1- 2
then
x² = 1 – 2
but =
x² = 1 – 2
x² =1 –
x² + -1 =0
NB: In elimination problems concept is to eliminate the trig function in the equation, then try the possibilities of eliminating it by connecting it to the pythageras theorem (identity)
COMPLEMENTARY ANGLES
Consider the triangle below
= (i) = (iv)
= (ii) = (v)
= (iii) = .(vi)
Thus
Is the condition for complementary angles
Definition: Complementary angles are angles whose sum is 90°
E.g: A + B = 90°
30° + 60° = 90°
30° and 60° are complementary angles.
NB: Supplementary angles are angles whose sum is 180°
Eg: A + B = 180°
Then A and B are supplementary angles
COMPOUND ANGLES FORMULA
Consider two angles say A and B then the angles A + B are called compound angles.
The concept here is to obtain
Sin (A ±B), Cos (A ±B), Tan (A ± B)
However it is easier to say that
Sin(A + B) = sin A + sin B
Testing if it is true
Let A= 60 and B= 30°
Sin(A + B) = sin(60° + 30°) = sin 90° = 1
Sin A + sin B = sin 60°+ sin 30°
Consider the figure below
From OTR
=
But TR = TS + SR
=
= + , but TS = PQ
= +
Multiplying by and by
But from the figure above
= =
= , =
Then substituting into
= +
From (1) if B=B
But =
=â»
Again from the figure above
=
But OT =
For tan
Refer =
=
Dividing numeration and denomination by
=
From above equation
If B = -B, then
Tan( A+ =
But tan=â» tanB
=
Or, shown by
=
Use procedure (5) obtain (6)
APPLICATION OF THE COMPOUND FORMULAE
I. PROVING OF IDENTITIES
Examples:
Prove the following trig identities
i) = +
ii) =
iii) =
Proof(i) =
Dealing with L.H.S
II. COS(A+B)COS(A-B) =
Proof dealing with L.H.S
B –
=1- and
= 1 – then
-(sin2A-cos2Asin2B)
cos2A-cos2Asin2B-sin2A+cos2Asin2B
R.H.S
=
III. =
Proof
Dealing with L.H.S
=
=1
=
But =
= + 1
1 –
=
=
=
IV. FINDING VALUES OF TRIG RATIOS
Examples: Evaluate
a) b) c) d)
Solution:
a) =
=
=
=
=
= 1
=
= =
=
=
=
=
.
If = , find the tangent of x in terms of and then find tan x when = 45° and = 60° (leaving your answer in surd form)
: = cos
+ = +
=cos x cos sin
=
=
= =
=
Given =45°, = 60
=
=
DOUBLE ANGLE FORMULAE
Recall (a) =
If B = A
=
=2
b)
If B = A
=
c)
If B = A
= ………………….. (iii)
Also from
=
But = 1 –
=(1 – )-
= 1 –
Or
=
= 1 –
=
= – 1 +
= 2 – 1
TRIPLE ANGLE FORMULAE
i) Consider
sin(2θ+θ) =sin2θcosθ +
= 2
=
= 2
= 2 +
3
But θ = 1 –
=
= 3θ –
=3 – 4
ii) Consider =
=
But =
= 2
=
-2
=cos3θ
But =1 –
– 3
+3
subscriber
By
By
By
By | 2,973 | 7,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-26 | latest | en | 0.768455 |
https://www.studypool.com/questions/136447/net-present-value-calculation | 1,480,766,292,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540928.63/warc/CC-MAIN-20161202170900-00319-ip-10-31-129-80.ec2.internal.warc.gz | 1,041,042,732 | 735,482 | # net present value calculation
Sigchi4life
Category:
Economics
Price: \$5 USD
Question description
1. East Coast Television is considering a project with an initial outlay of \$X (you will have to determine this amount). It is expected that the project will produce a positive cash flow of \$60,000 a year at the end of each year for the next 14 years. The appropriate discount rate for this project is 9 percent. If the project has a 12 percent internal rate of return, what is the project’s net present value?
2. Big Steve’s makers of swizzle sticks is considering the purchase of a new stamping machine. This investment requires an initial outlay of \$105,000 and will generate net cash flows of \$21000 per year for 8 years.
A What is the project NPV using a discount rate of 11% should the project be accepted, why or why not?
B What is the project NPV using a discount rate of 14% should the project be accepted, why or why not?
C What is the project internal rate of return? Should the project be accepted, why or why not?
3. What is the internal rate of return for the following project? An initial outlay of \$9500 resulting in a single cash inflow of \$24301 in 9 years. What is the rate of return for the project?
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https://doresearchforme.com/stats-pacific-university-the/ | 1,656,444,780,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103573995.30/warc/CC-MAIN-20220628173131-20220628203131-00634.warc.gz | 263,071,884 | 11,039 | # STATS Pacific University The
The average amount of caffeine in a 12-oz serving of Coke is 35 mg… if it were to come straight from the factory in Atlanta, GA; however, because soda fountain machines are not always cleaned adequately, the actual amount of caffeine in Coke varies depending on where it is purchased and served. A researcher buys 20 fountain Cokes from 12 different restaurants and records the mean caffeine amount for each restaurant. The researcher finds that the average amounts of caffeine in the Cokes served at McDonald’s, Burger King, and Five Guys are 35.5 mg, 37 mg, and 42 mg, respectively.
1. Imagine that the researcher conducted three separate t-tests comparing the Cokes at McDonald’s, Burger King, and Five Guys (in that order) with the average of 35 mg. Assume that the standard deviation of each sample was identical. Would the absolute values of the t-tests get smaller, bigger, or stay the same? How do you know? Be sure to define what a t-score tells you.
2. Would the p-values associated with the above t-tests get smaller, bigger, or stay the same? How do you know? Be sure to define what a p-value tells you.
3.The figure to the right shows more data from the previous experiment. Each bar represents the mean caffeine of Coke, and the error bars represent 95% confidence intervals. Use it to answer the questions below.
Why does the first bar not have error bars?
4. Which sample(s) is (are) significantly different from the population?
5.Rank Arby’s, Burgerville, and Taco Time in order from biggest p-value to smallest p-value and explain your answer in one sentence.
6. A man with psychic powers tells you that one of the above three tests is associated with a Type I error. Which example do you think is associated with a Type I error, and how do you know?
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Posted in Uncategorized | 420 | 1,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-27 | latest | en | 0.925526 |
https://forum.freecodecamp.org/t/basic-algorithm-scripting-find-the-longest-word-in-a-string/567820 | 1,718,479,865,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00542.warc.gz | 235,999,919 | 7,177 | Basic Algorithm Scripting - Find the Longest Word in a String
This challenge seems out of place to me. The solution requires using the split() function which hasnt even been covered up to this point in the curriculum. The split() function is covered in the functional programming section, which is 2 sections after this Basic Algorithm Scipting section.
``````function findLongestWordLength(str) {
let words = str.split(" ");
let maxLength = 0;
for (let i = 0; i < words.length; i++) {
if (words[i].length > maxLength) {
maxLength = words[i].length;
}
}
return maxLength;
}
findLongestWordLength("The quick brown fox jumped over the lazy dog");
``````
User Agent is: `Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:105.0) Gecko/20100101 Firefox/105.0`
Challenge: Basic Algorithm Scripting - Find the Longest Word in a String
Use of split is not mandatory.
Why do the hints explicitly ask if you remember how to use split()? There are 5 possible solutions and split is required in all solutions except for the last solution listed. What merit determines how the order in which the available solutions are listed?
The solution that doesnt require split() is clearly not best practice, especially if you take the curriculums recommendations regarding verboseness at face value.
Because that’s the easiest way to do this.
They are listed in the order they were added.
Why ‘clearly’? It works. That is the #1 requirement for a solution. It is more verbose, sure, but it meets your requirement that it only use content explicitly covered to this point in the curriculum.
Its clear that solution is not best practice because it is the solution that is listed last. In spite of the 4 solutions that precede it not even having been covered yet.
In addition to that solution being the most verbose. Which the curriculum is teaching should be taken into consideration in real world scenarios when working with other programmers as part of a team.
Order has nothing to do with quality.
Maybe the order has nothing to do with quality, but that doesnt reflect my experience. In my experience, when there is more than one solution available, the solutions tend to be listed in order of cleanliness and recentness.
I dont recall coming across a legacy solution that had been provided a more favorable position over a more recent solution.
I’m literally the person who put the solution there. Order doesn’t correlate to quality.
I now moved that solution to first. That does not mean it is the best solution ever created for this problem in the history of all possible solutions ever created.
I never said it was the best possible solution. The point I was trying to make is that that specific solution was the only valid solution when taking into account what the curriculum had taught up to this point.
The hints are literally asking if you remember lessons that havent even been covered yet. Instead of trying to solve the challenge with skills that have been covered, I was lead down an untreaded path with foreign tools.
Its not a big deal. But in regards to this specific challenge, and all the emphasis placed on the split function, it just seemed like a bit of a curveball is all.
Now that I know about the split function, then I consider the cleanest split solution to be the best possible solution.
No. If a solution passes without hardcoding or other ‘cheats’, then it is valid. All 5 of those solutions are ‘valid’.
Not at all. Split is not mentioned in the hints anywhere. It hasn’t been part of the hints for quite a while.
I specifically added the non-split solution because people incorrectly kept saying that this challenge was impossible with the content strictly included in the curriculum.
You can solve challenges with the content in the curriculum thus far, but external research is encouraged and may result in simpler solutions.
This topic was automatically closed 182 days after the last reply. New replies are no longer allowed. | 832 | 3,962 | {"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-2024-26 | latest | en | 0.922044 |
https://www.brightstorm.com/tag/word-problem/page/3 | 1,656,717,946,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103947269.55/warc/CC-MAIN-20220701220150-20220702010150-00381.warc.gz | 730,301,439 | 16,344 | # word problem 81 videos
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Identify and revise sentence fragments.
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How to identify unnecessary commas.
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##### AlgebraWord Problems Using Systems of Equations
How to solve word problems about mixing two liquids with different percent concentrations.
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##### AlgebraWord Problems Using Systems of Equations
How to set up a system of equations to represent word problems where someone is selling multiple items.
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Understanding and solving problems using Planck's constant.
blackbody radiation thermal contact thermal equilibrium | 555 | 2,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-27 | latest | en | 0.751001 |
https://discuss.codechef.com/t/editorial-for-make-it-zero-makeit/89981 | 1,696,238,130,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510983.45/warc/CC-MAIN-20231002064957-20231002094957-00217.warc.gz | 231,937,247 | 4,537 | EASY
# PREREQUISITES:
Math,Logical thinking.
# PROBLEM:
You are Given two numbers A and B you have to make both the numbers equal to zero in Only two operations.the oprations you can perform is addition and subtraction no other operation (NOTE: Numbers can be Negative).
Print “YES” without quotes if it is possible to make both the Numbers zero in two operations else print “NO”.
You have to perform one operation on each number.if the number requires more than one operation to become zero the print “NO”.
# QUICK EXPLANATION:
The Answer for this question will be always “YES”. Let’s understand this in explanation section given below.
# EXPLANATION:
Just after thinking for some time and taking some testcases in mind you will find that for any testcase the answer will be always “YES”.
As mentioned above we can perform the addition and subtraction operation on numbers. so keeping this in mind let’s take a example and understand it.
Example:- 36
if we subtract 36 from 36 the we will get the answer zero that to in one operation so the answer is “YES”.
Example:- -40
if we add 40 to -40 we will get the answer zero that to in one operation so the answer for this one will be “YES”.
So after observing it we can find that we can make every number zero in one operation so the answer for every testcase will be always “YES”.
# SOLUTIONS:
Setter's Solution
#include
using namespace std;
#define ll long long int
int main() {
ll t;
cin>>t;
while(t–)
{
ll a,b;
cin>>a>>b;
cout<<“YES”<<“\n”;
}
return 0;
}
Tester's Solution
#include
using namespace std;
#define ll long long int
int main() {
ll t;
cin>>t;
while(t–)
{
ll a,b;
cin>>a>>b;
cout<<“YES”<<“\n”;
}
return 0;
}
Editorialist's Solution
#include
using namespace std;
#define ll long long int
int main() { | 440 | 1,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-40 | latest | en | 0.832285 |
https://www.manchester.ac.uk/study/undergraduate/courses/2023/01687/mmath-mathematics-with-financial-mathematics/all-content/MATH32062 | 1,680,184,550,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949331.26/warc/CC-MAIN-20230330132508-20230330162508-00112.warc.gz | 970,863,441 | 13,987 | # MMath Mathematics with Financial Mathematics
Year of entry: 2023
## Course unit details:Algebraic Geometry
Unit code MATH32062 10 Level 3 Semester 2 Department of Mathematics No
### Overview
Algebraic geometry studies objects called varieties defined by polynomial equations. A very simple example is the hyperbola defined by the equation xy = 1 in the plane. There is a way of associating rings to varieties, and then the geometric properties can be studied using algebra, for example points correspond to maximal ideals, or the geometry of the variety can give information about certain algebraic properties of the ring. Algebraic geometry originated in nineteenth century Italy, but it is still a very active area of research. It has close connections with algebra, number theory, topology, differential geometry and complex analysis.
### Pre/co-requisites
Unit title Unit code Requirement type Description
Linear Algebra A MATH10202 Pre-Requisite Compulsory
Linear Algebra B MATH10212 Pre-Requisite Compulsory
Algebraic Structures 2 MATH20212 Pre-Requisite Compulsory
MATH32062 pre-requsites
### Aims
To introduce students to the basic notions of affine and projective algebraic geometry.
### Learning outcomes
Successful students will be able to:
• define the basic concepts of algebraic geometry,
• prove the main theorems about the properties of algebraic varieties, morphisms and rational maps between them, and the correspondence between algebraic varieties and ideals, rings and fields,
• define the tangent space to an affine algebraic variety at a point, prove its properties and calculate the dimension and the singular locus of affine algebraic varieties by using tangent spaces,
• apply the concepts and theorems of algebraic geometry in concrete examples,
• prove properties of the automorphism groups of projective spaces and calculate explicit automorphisms of P1 by using the cross ratio,
• define elliptic curves and the addition operation on them, and prove properties of elliptic curves and the addition operation,
• carry out calculations involving elliptic curves.
### Syllabus
1. Affine varieties, Hilbert's Nullstellensatz
2. Co-ordinate rings, function fields, morphisms and rational maps between affine varieties.
3. Tangent spaces and dimension.
4. Projective spaces and varieties.
5. Geometry in the plane.
6. Elliptic curves.
### Assessment methods
Method Weight
Other 20%
Written exam 80%
• Coursework: 20% (one take home test worth 20%)
• End of semester examination: 80%.
### Feedback methods
Feedback tutorials will provide an opportunity for students' work to be discussed and provide feedback on their understanding. Coursework or in-class tests (where applicable) also provide an opportunity for students to receive feedback. Students can also get feedback on their understanding directly from the lecturer, for example during the lecturer's office hour.
M. Reid, Undergraduate Algebraic Geometry, CUP, 1988,
K. Hulek, Elementary algebraic geometry AMS, 2003.
D. A. Cox, D. O'Shea and J. Little, Ideals, Varieties and Algorithms, Springer, 2015. (Ebook is available via the library website and earlier editions are also suitable.)
### Study hours
Scheduled activity hours
Lectures 12
Tutorials 12
Independent study hours
Independent study 76
### Teaching staff
Staff member Role
Gabor Megyesi Unit coordinator
The independent study hours will normally comprise the following. During each week of the taught part of the semester:
• You will normally have approximately 60-75 minutes of video content. Normally you would spend approximately 2-2.5 hrs per week studying this content independently
• You will normally have exercise or problem sheets, on which you might spend approximately 1.5hrs per week
• There may be other tasks assigned to you on Blackboard, for example short quizzes or short-answer formative exercises
• In some weeks you may be preparing coursework or revising for mid-semester tests
Together with the timetabled classes, you should be spending approximately 6 hours per week on this course unit.
The remaining independent study time comprises revision for and taking the end-of-semester assessment.
The above times are indicative only and may vary depending on the week and the course unit. More information can be found on the course unit’s Blackboard page. | 928 | 4,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-14 | latest | en | 0.856939 |
https://www.topperlearning.com/answer/an-object-is-placed-42cm-from-a-concave-mirror-of-curvature-16-cm-a-glass-slab-of-thickness-6cm-and-ri-3-2-is-then-placed-between-the-object-and-mirro/bklnq7tt | 1,675,572,854,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500215.91/warc/CC-MAIN-20230205032040-20230205062040-00790.warc.gz | 1,034,166,065 | 59,685 | Request a call back
ICSE Class 9 Answered
An object is placed 42cm from a concave mirror of curvature 16 cm. A glass slab of thickness 6cm and R.I. 3/2 is then placed between the object and mirror. Find the position of the final image formed (the near surface of the slab from the mirror is 1cm)
Asked by u.sudhishna | 05 Aug, 2022, 03:51: PM
Object is 42 cm in front of concave mirror .
There is a glass slab of thickness 6 cm in front of mirror such that near surface of glass slab is 1 cm from concave mirror.
Hence light ray has to travel a distance 36 cm in air and 6 cm in glass slab .
If refractive index of glass slab is n , then equivalent air distance of light path in glass slab = n × t ,
where t is thickness of glass slab
hence for 6 cm thick glass slab, air distance = 6 × 1.5 = 9 cm
Hence total air distance travelled by light before getting reflected by concave mirror is ( 36 + 9 ) cm = 45 cm
we have mirror equation as
(1/v ) + ( 1/u ) = 1/f
where v is mirror-to-image distabnce , u is mirror-to-object distance and f is focal length
since radius of curavture of mirror is 16 cm , focal length f = 8 cm
By using cartesian sign convention, we write mirror equation as
(1/v) - (1/45) = -1/8
(1/v) = (1/45) - ( 1/8 )
from above expression , we get v = - 9.7 cm , i.e 9.7 cm in front of mirror
since glass slab surface is 1 cm in front of mirror , then equivalent air distance 8.7 cm will be travelled in glass slab
distance travelled in glass slab = 8.7 / 1.5 = 5.8 cm
Hence image is formed infront of mirror at a distance ( 1 cm in air + 5.8 cm in glass slab )
hence image is formed inside glass slab at a distance 6.8 cm from mirror
Answered by Thiyagarajan K | 06 Aug, 2022, 12:05: AM
Concept Videos
ICSE 9 - Physics
Asked by u.sudhishna | 05 Aug, 2022, 03:51: PM
ICSE 9 - Physics
Asked by asmithasrushti | 31 Dec, 2021, 09:05: PM
ICSE 9 - Physics
Asked by aadrikaverma15dec2007 | 17 Feb, 2021, 10:42: PM | 610 | 1,941 | {"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-2023-06 | longest | en | 0.919144 |
https://computingforbeginners.blogspot.com/2012/12/inverse-of-matrix.html | 1,726,048,021,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00083.warc.gz | 164,800,606 | 13,572 | Saturday 22 December 2012
Inverse of a Matrix
The Inverse of a matrix exists only if the determinant is non zero. For every non singular matrix there exists a unique Inverse. If A is a matrix, then the inverse of the matrix is denoted by A-1. The product of a matrix and its Inverse equals the identity matrix I. There for we have the relation AA-1 = A-1A = I. Here inverse is obtained by applying elementary row transformations on the given matrix and similar changes are also made on an identity matrix of the same order. When the given matrix is transformed into an Identity Matrix the Identity Matrix becomes the Inverse Matrix.
Figure Given Below Shows a 3 X 3 Matrix.
Figure Given Below Shows the Inverse of the Given Matrix.
Figure Given Below Shows the Product of the Matrix and its Inverse. | 178 | 804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-38 | latest | en | 0.906206 |
https://numberworld.info/23612 | 1,596,559,782,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735881.90/warc/CC-MAIN-20200804161521-20200804191521-00226.warc.gz | 439,791,331 | 3,801 | # Number 23612
### Properties of number 23612
Cross Sum:
Factorization:
2 * 2 * 5903
Divisors:
1, 2, 4, 5903, 11806, 23612
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
5c3c
Base 32:
n1s
sin(23612)
-0.20883582082652
cos(23612)
0.97795071447375
tan(23612)
-0.21354432052223
ln(23612)
10.069510336361
lg(23612)
4.3731327745492
sqrt(23612)
153.6619666671
Square(23612)
### Number Look Up
Look Up
23612 (twenty-three thousand six hundred twelve) is a amazing number. The cross sum of 23612 is 14. If you factorisate the figure 23612 you will get these result 2 * 2 * 5903. The number 23612 has 6 divisors ( 1, 2, 4, 5903, 11806, 23612 ) whith a sum of 41328. 23612 is not a prime number. 23612 is not a fibonacci number. The number 23612 is not a Bell Number. The figure 23612 is not a Catalan Number. The convertion of 23612 to base 2 (Binary) is 101110000111100. The convertion of 23612 to base 3 (Ternary) is 1012101112. The convertion of 23612 to base 4 (Quaternary) is 11300330. The convertion of 23612 to base 5 (Quintal) is 1223422. The convertion of 23612 to base 8 (Octal) is 56074. The convertion of 23612 to base 16 (Hexadecimal) is 5c3c. The convertion of 23612 to base 32 is n1s. The sine of the number 23612 is -0.20883582082652. The cosine of 23612 is 0.97795071447375. The tangent of the number 23612 is -0.21354432052223. The root of 23612 is 153.6619666671.
If you square 23612 you will get the following result 557526544. The natural logarithm of 23612 is 10.069510336361 and the decimal logarithm is 4.3731327745492. I hope that you now know that 23612 is very unique figure! | 637 | 1,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-34 | latest | en | 0.764829 |
https://www.uni-kassel.de/eecs/index.php?eID=dumpFile&t=f&f=872&token=26ec032ccd805d63fb3d16afce2a53f0da05b05e | 1,721,426,169,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-30/segments/1720763514928.31/warc/CC-MAIN-20240719200730-20240719230730-00554.warc.gz | 899,489,399 | 2,770 | /* * USART_Functions.h * * Created: 2020 */ #ifndef USART_FUNCTIONS_H_ #define USART_FUNCTIONS_H_ #define hundred 0 //First digit of a 3 digit literal #define ten 1 //Second digit #define one 2 //Third digit unsigned char Trans_data[3]; //3 digits who will be transmitted void USART_Transmit(unsigned char); //Transmit ASCII to Data Visualizer //Transmit literal to Data Visualizer //You may use this, like Arduino IDE does, but you need very much time... //So we have to divide it into three parts or shut it down on processing with the robot //Takes about 450us at 76800 Baud rate, who has so much time? void Transmit_literal (char data) { //Transmit 8-Bit literal char data_100 = data / 100; //How much hundred char data_10 = data % 100 / 10; //How much ten char data_1 = data % 10; //How much one if (!data_100) { //There is no hundred USART_Transmit(32); //Write a blank if (!data_10) USART_Transmit(32); //There is no ten, write a blank else USART_Transmit(data_10 + 48); //ASCII of 0 is 48 } else { USART_Transmit(data_100 + 48); //Hundred USART_Transmit(data_10 + 48); //ten } USART_Transmit(data_1 + 48); //One USART_Transmit('\t'); //Tabulator } //If we want to do it very, very much faster, we have to divide the transmission into single digits. //The transmission is not faster!!! But we donīt wait while transmitting, we only initialize it!!! //Initialisation every 256us. void Computing_Transmission_Values (unsigned char transmitting_data) { // Trans_data[hundred] = transmitting_data / 100; //Itīs the same Trans_data[ten] = transmitting_data % 100 /10; //dito Trans_data[one] = (transmitting_data % 10) + 48; //A little bit different if (!Trans_data[hundred]) { //Now we have to save the results Trans_data[hundred] = 32; //blank if (!Trans_data[ten]) Trans_data[ten] = 32; //blank else Trans_data[ten] = Trans_data[ten] + 48; //ASCII for zero is 48 } else { Trans_data[hundred] = Trans_data[hundred] + 48; //See above Trans_data[ten] = Trans_data[ten] + 48; //dito } //But there is no transmission!!!! See "ISR(TIMER2_OVF_vect)"! } //Transmission of data from ATMega328P to computer void USART_Transmit(unsigned char data) { //Transmission of ASCII data to monitor while (!(UCSR0A & (1<>8); /*Set baud rate */ UBRR0L = (unsigned char)ubrr; //See above UCSR0B = (1< | 645 | 2,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.517662 |
https://www.askiitians.com/forums/Electric-Current/a-220-v-and-800-w-electric-kettle-and-three-220-v_206159.htm | 1,696,320,593,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511055.59/warc/CC-MAIN-20231003060619-20231003090619-00232.warc.gz | 686,958,064 | 43,475 | # A 220 V and 800 W electric kettle and three 220 V nd 100 W bulbs are connected in parallel . on connecting this combination with 220 V supply .the total current in the circuit will be
Arun
25757 Points
5 years ago
resistance of kettle = V^2 / P = 220 x 220 / 800 = 484 / 8 ohm
Resistance of each bulb = V^2 / P = 220 x 220 / 100 = 484 ohm
Now 484/8 , 484, 484, 484 are connected in parallel to a supply of 200 V
Effective resistance = 484/11 ohm
So current in the circuit = 200/ (484/11) = 2200/484 = 4.55 A
Raghulal Prajapati
76 Points
5 years ago
according to ur question all the resistances are in parallel to each other .
so the solution is :
as we know W= V2/R ------> R= V2/W
so resistance for kettle is : 220*220/800= 121/2 ohm
resistance for (1) bulb is : 220*220/100= 484 ohm
as all are in parallel so net resistance is 1/resistance of kettle +3/resistance of bul = 1/ equivalent resistance
so equivalent resistance= 44 ohm
so current in circuit = v/R = 220/44= 5 ampere.
thank u.
Arun
25757 Points
5 years ago
Sorry
In my answer I have typed a wrong value
Effective resistance = 484/11
So current in the circuit
= 220/ (484/11)
= 2420/484
= 5 amp | 390 | 1,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2023-40 | latest | en | 0.828805 |
https://www.opengl.org/discussion_boards/archive/index.php/t-146736.html | 1,524,581,984,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946721.87/warc/CC-MAIN-20180424135408-20180424155408-00555.warc.gz | 851,480,913 | 4,971 | PDA
View Full Version : problem with my SkySquad !
tcs
02-16-2000, 01:02 PM
Hi,
I have a problem with my "SkyQuad". I called it so because it is the
single-quad version of a skybox. How it works ? You simply draw a quad
that cover the whole screen. If your cam rotates, you simply move the
texture on the quad in the opposite direction. The quad has a size of
4x3 to match the 4:3 aspect ratio of most resolutions. Then you have
to find the distance from in which the quad covers the full screen.
this depends on your FOV. with a fov of 90 (and a 4.0x3.0 quad) this
distance is 3.0. But how to calculate the distance for different FOV
values ? I quickly realized that this is a trigonometric problem. I
use this formula:
float fDistance = -(1.5f / sinf((float) iFOV / 2.0f * PI_OVER_180));
...where 1.5 is my Y quad size divided by two, iFOV / 2.0f is half of
my FOV angle, and PI_OVER_180 is the usual conversion factor for converting
between degrees and radians. But this doesn't work very well, somehow wrong ;-)
Can anyone help ?
Tim
dmy
02-16-2000, 01:25 PM
i think you're doing something like the sky in doom, am i right?
if you stretch your quad, you can make it cover the screen on any plane.
i use this to project a point in 2-space (the mouse position http://www.opengl.org/discussion_boards/ubb/smile.gif) to a point in 3-space:
z=plane;
y=-tan(fov*0.5*6.283/360)*z;
x=y*aspect*u;
y*=v;
where:
x,y,z the projected point in 3-space
u,v the point in 2-space
fov field of view in degrees
aspect guess...
plane the desired plane Z-coord (as in opengl. to be in front of the camera, this is <0)
the point in 2-space is normalized: lower-left corner is -1,-1 and upper-right is +1,+1
so, for instance, you choose some plane depth value, then you feed uv=(-1,-1):
you'll get a point in 3-space wich will projects to the lower-left corner of the viewport.
you can then call this function for the other 3 points of the quad,
or mirror around the origin the x and y component you just evaluated...
see what is best for you.
[This message has been edited by dmy (edited 02-17-2000).]
tcs
02-17-2000, 04:13 AM
Thanks !
It works perfect !
dmy
02-17-2000, 06:20 AM | 641 | 2,177 | {"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} | 3.34375 | 3 | CC-MAIN-2018-17 | latest | en | 0.872754 |
https://www.hackmath.net/en/math-problem/1760?tag_id=59_8 | 1,610,734,533,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00650.warc.gz | 928,425,751 | 13,142 | The ship
The ship went upstream speed of 20 km/h in relation to water. River flows at speed 10 km/h. After half an hour, he stopped and returned downstream to the starting point. How long it took way back when even if downstream speed of ship is 20 km/h in relation to water?
Correct result:
t = 10 min
Solution:
(20-10)•30 = (20+10)•t
30t = 300
t = 10
Our simple equation calculator calculates it.
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
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The water in the vessel carried by the boy has a self-oscillation period of 0.8 s. What is the size of the boy's movement speed when the length of the boy's step is 60 cm? Give the result in m/s. | 1,096 | 4,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-04 | latest | en | 0.962995 |
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# Should I ignore MGMAT CAT for Verbal?
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Intern
Joined: 23 Oct 2011
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Should I ignore MGMAT CAT for Verbal? [#permalink]
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22 Apr 2013, 15:20
So I've taken three MGMAT CAT exams and have scored mediocre on the Verbal (38ish). Weak areas are SC and sometimes CR. Reading Comprehension scores the best. I
However, when I've done the logs with the OG 12, 13, and Verbal 2, my SC is ridiculously high. For example, although I test poorly on parallel reasoning in the MGMAT Cats, I've gotten a 92% accuracy with my timing around 1 minute, 30 seconds per question in that category dealing with OG problems. I completed that section, no problems.
I'm EXTREMELY frustrated because I don't know what to make of this disconnection. I've also taken Veritas Prep exams and scored about 44, 46 on the Verbal too. I have 1 week to the test and want to dedicate more time to word translations or word problems in the Math (a huge pain in my side) and not stress over SC.
Advice? People have mentioned that the MGMAT Cats are accurate to the level in the GMAT, but how do I explain this discrepancy? I'm taking a practice GMAT Prep 2 exam Wednesday, btw. When I took GMAT Prep 1, I scored 660 with 44Q, 38V and my weakest section was RC (!!!) back then.
Veritas Prep GMAT Discount Codes Kaplan GMAT Prep Discount Codes Optimus Prep Discount Codes
GMAT Club Verbal Expert
Status: GMAT and GRE tutor
Joined: 13 Aug 2009
Posts: 500
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GMAT 1: 780 Q51 V46
GMAT 2: 800 Q51 V51
GRE 1: 340 Q170 V170
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Re: Should I ignore MGMAT CAT for Verbal? [#permalink]
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28 Apr 2013, 18:25
2
KUDOS
Expert's post
I apologize to anybody who has seen me write this way too often, but here it comes again: don't put too much faith in practice test scores from non-official sources, especially on the verbal side. It's nearly impossible for test-prep companies to perfectly copy the style of the actual GMAT, and even the very best GMAT companies can't produce consistently accurate verbal tests. MGMAT does a heroic job with their CATs, but it's inherently an uphill battle. MGMAT's writers will inevitably test you on different nuances than the actual GMAT. For some test-takers, the MGMAT CATs are accurate; for others, not so much.
Anilisanil, huge props for the spectacular CR and RC results from the OG. If you need an extra challenge on CR and RC, you might want to mix in some LSAT material, and you'll definitely want to hit the GMATPrep Question Pack for extra verbal questions. You could use the GMATPrep Question Pack to create some "fake tests" if you want--just set a timer for 75 minutes, and do 41 random verbal questions in "exam" mode on the software. You won't get a scaled score, but it will be a decent replica of the test experience. And I'm sure that you're already working like crazy on SC. Considering your CR and RC results from the OG, I can't imagine why you wouldn't top a 40 on the verbal section, as long as you take care of business on SC. Would it be correct to assume that you made some not-so-pretty SC errors on your GMATPrep test? Fix those, and you'll be fine.
Abk, is your overall accuracy in the OG at around 92%, or are you just referring to SC parallelism questions? If you're consistently getting 90%+ correct on all of the OG verbal stuff, then I might think that the MGMAT scores are inaccurate... but I'm suspicious of the fact that you've gotten 38s on both the MGMAT and the GMATPrep verbal sections. Keep doing as much official stuff as you can (OG, verbal supplement, GMATPrep Question Pack, maybe some LSATs, maybe the old GMAT paper tests), and keep in mind that the difference between a 38V and, say, a 42V is probably just a few questions. So fight for every inch and stick with official stuff as much as you can. I don't see any reason to automatically discount the MGMAT test scores in your case, unless your next GMATPrep is much better.
I hope this helps. Good luck, everybody!
_________________
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Re: Should I ignore MGMAT CAT for Verbal? [#permalink]
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28 Apr 2013, 18:42
1
KUDOS
GMATNinja wrote:
I apologize to anybody who has seen me write this way too often, but here it comes again: don't put too much faith in practice test scores from non-official sources, especially on the verbal side. It's nearly impossible for test-prep companies to perfectly copy the style of the actual GMAT, and even the very best GMAT companies can't produce consistently accurate verbal tests. MGMAT does a heroic job with their CATs, but it's inherently an uphill battle. MGMAT's writers will inevitably test you on different nuances than the actual GMAT. For some test-takers, the MGMAT CATs are accurate; for others, not so much.
Anilisanil, huge props for the spectacular CR and RC results from the OG. If you need an extra challenge on CR and RC, you might want to mix in some LSAT material, and you'll definitely want to hit the GMATPrep Question Pack for extra verbal questions. You could use the GMATPrep Question Pack to create some "fake tests" if you want--just set a timer for 75 minutes, and do 41 random verbal questions in "exam" mode on the software. You won't get a scaled score, but it will be a decent replica of the test experience. And I'm sure that you're already working like crazy on SC. Considering your CR and RC results from the OG, I can't imagine why you wouldn't top a 40 on the verbal section, as long as you take care of business on SC. Would it be correct to assume that you made some not-so-pretty SC errors on your GMATPrep test? Fix those, and you'll be fine.
Abk, is your overall accuracy in the OG at around 92%, or are you just referring to SC parallelism questions? If you're consistently getting 90%+ correct on all of the OG verbal stuff, then I might think that the MGMAT scores are inaccurate... but I'm suspicious of the fact that you've gotten 38s on both the MGMAT and the GMATPrep verbal sections. Keep doing as much official stuff as you can (OG, verbal supplement, GMATPrep Question Pack, maybe some LSATs, maybe the old GMAT paper tests), and keep in mind that the difference between a 38V and, say, a 42V is probably just a few questions. So fight for every inch and stick with official stuff as much as you can. I don't see any reason to automatically discount the MGMAT test scores in your case, unless your next GMATPrep is much better.
I hope this helps. Good luck, everybody!
mastermind
_________________
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Status: GMAT and GRE tutor
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Re: Should I ignore MGMAT CAT for Verbal? [#permalink]
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28 Apr 2013, 20:19
1
KUDOS
Expert's post
Wow, that's a dramatic distribution of errors. You can absolutely succeed at this, Anilisanil.
Is it possible that you're doing the SC questions based mostly on sound and instinct? You clearly have great verbal skills, and that can sort of work against you on SC sometimes. If you're worrying about whether the sentence "sounds right" or whether it's awkward or wordy, you're going to get into trouble--the correct answers often sound like crap, especially on harder questions. The answer choices that "sound good" often contain a mechanical error or change the meaning of the sentence somehow.
I could be completely wrong, but I'm guessing that a more methodical, less "ear-based" approach might get you somewhere...
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Re: Should I ignore MGMAT CAT for Verbal? [#permalink]
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23 Apr 2013, 13:23
i do not find any reason , so that u should ignore MGMAT verbal...On GMAT prep 1 too u scored the same score...Moreover, Do not dig too much into overall scores...Just check wich areas do u lack in...u have 3 sets of test nw...so go thru all n check the common area where u r missn.
Consider Kudos If my post helps!!!!!1
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Re: Should I ignore MGMAT CAT for Verbal? [#permalink]
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28 Apr 2013, 01:59
Even I am facing the same issue. I am scoring 34-35 ish in MGMAT CATs, while I scored 39 on the GMAT Prep 1.
My scores are like this:
MGMAT CAT 1 V 38
MGMAT CAT 2 V 34
MGMAT CAT 3 V 35
GMAT Prep 1 V39
MGMAT CAT 4 V 35
MGMAT CAT 5 V34
(did not disclose Q scores intentionally)
Also, my accuracy of SC questions is pretty high when I practiced from OG 13. I got only 33 out of 140 questions wrong in the first attempt. My CR is even better I got only 8 questions wrong out of 124. I seem to have similar high accuracy for RCs too. And yes all of them are timed and average times are @ 1.5 minutes.
One thing positive is that the errors are representative of my weak areas as highlighted by MGMAT, but I have been working hard for last two months and my verbal scores are not improving at all. I need to improve at least 7-8 points to reach my target, I am not sure what I am missing in here.
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Re: Should I ignore MGMAT CAT for Verbal? [#permalink]
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28 Apr 2013, 19:38
GMATNinja wrote:
Anilisanil, huge props for the spectacular CR and RC results from the OG. If you need an extra challenge on CR and RC, you might want to mix in some LSAT material, and you'll definitely want to hit the GMATPrep Question Pack for extra verbal questions. You could use the GMATPrep Question Pack to create some "fake tests" if you want--just set a timer for 75 minutes, and do 41 random verbal questions in "exam" mode on the software. You won't get a scaled score, but it will be a decent replica of the test experience. And I'm sure that you're already working like crazy on SC. Considering your CR and RC results from the OG, I can't imagine why you wouldn't top a 40 on the verbal section, as long as you take care of business on SC. Would it be correct to assume that you made some not-so-pretty SC errors on your GMATPrep test? Fix those, and you'll be fine.
I hope this helps. Good luck, everybody!
thanks
You are right about the gmat prep test. in verbal I got 12 out of 37 questions wrong.
sc 9
cr 2
rc 1
I intentionally marked 2 sc questions wrong because I solved those questions before.
I bought verbal review. will purchase question pack 1 too. Let's see how things turn for me.
Re: Should I ignore MGMAT CAT for Verbal? [#permalink] 28 Apr 2013, 19:38
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# Should I ignore MGMAT CAT for Verbal?
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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®. | 3,564 | 13,574 | {"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-2017-22 | longest | en | 0.93567 |
http://azfoo.net/gdt/babs/2017/08/number81718.html | 1,558,869,920,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259126.83/warc/CC-MAIN-20190526105248-20190526131248-00332.warc.gz | 25,299,141 | 2,444 | ### About the Number 81,718 (eighty-one thousand seven hundred eighteen)
MathBabbler's Buick's odometer hit 81,718 on 6 August 2017 on I-25 approximately 3 miles north of the Santa Fe County Line in New Mexico.
```MathBabbler Number Analyst (MBNA) output:
=========================================
81718 is natural, counting, whole, integer
81718 is even (not odd)
81718 proper divisors are: 1,2,7,13,14,26,91,182,449,898,
3143,5837,6286,11674,40859,
81718 has 15 proper divisors
81718 is deficient (sum of divisors is 69482; ratio: 0.850266)
81718 is unhappy
81718 is a Squarefree Number
81718 is composite (not prime)
81718 has the prime factors: 2*7*13*449 (sum=471)
81718 is a 449-smooth number
81718 is palindromic
81718 in octal is 0237466
81718 in binary is 10011111100110110 (is odious)
81718 nearest square numbers: -493...78 (81225...81796 [286])
sqrt(81718) = 285.864
ln(81718) = 11.311
log(81718) = 4.91232
81718 is an apocalyptic power (2^81718 contains 666)
81718! is inf
81718 is 26011.6 Pi years
81718 is 4085 score and 18 years
81718 is a multiple of 1 & it contains a 1 (A011531)
81718 is a multiple of 7 & it contains a 7 (A121027)
________________________________________________________________________________________________
81718 reciprocal is 0.0000122372060011258229521035757115935289654666046648229276291637093418830612594532416358696982305
remainders:
1,10,100,1000,10000,18282,19384,30404,58886,16834,4904,49040,92,920,9200,10282,21102,47584,67250,18756,24124,77804,42578,17190,8464,2922,29220,47046,61870,46674,58150,9474,13022,48502,76430,28838,43226,23670,73264,78896,53498,44672,38130,54428,53972,49412,3812,38120,54328,52972,39412,67248,18736,23924,75804,22578,62344,51414,23832,74884,13378,52062,30312,57966,7634,76340,27938,34226,15388,72162,67876,25016,5006,50060,10292,21202,48584,77250,37038,43508,26490,19746,34024,13368,51962,29312,47966,71070,56956,79252,57058,80272,67258,18836,24924,4086,40860,10
```
Creator: Gerald Thurman [gthurman@gmail.com]
Created: 12 August 2017 | 747 | 2,016 | {"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-2019-22 | latest | en | 0.675899 |
https://cooking.stackexchange.com/questions/102475/how-to-bake-baking-soda-to-produce-sodium-carbonate | 1,716,725,051,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058876.64/warc/CC-MAIN-20240526104835-20240526134835-00455.warc.gz | 150,313,360 | 41,654 | # How to bake baking soda to produce sodium carbonate?
I'd like to make sodium carbonate for ramen noodles. The instructions on the web however vary greatly. One source says to bake baking soda at 350 degrees for 2-5 hours. Others say 250 degrees for an hour. Another says just 200 degrees for an hour.
Does the temperature even matter at all?
How about if I use a higher temperature like 450 degrees? This would be preferable for me as I only need a small amount and I can probably bake it at the same time I'm baking bread.
• Just to confirm: degrees Fahrenheit? The question is interesting but I was really confused until I got to the 450 degrees (no domestic oven goes that high in Celsius) Sep 24, 2019 at 8:38
• The quoted 200 is probably C. I used less than 200C for this answer. The 250 is likely to be also C (that's about as high as most domestic ovens go) Sep 24, 2019 at 9:42
• Yes, Fahrenheit. This is the one that says 200 degrees: thoughtco.com/…
– user50726
Sep 24, 2019 at 17:27
• 200F seems low, but I couldn't get that link to work. It tried to load so many tracking and other unwanted scripts from all over the web that I gave up Sep 26, 2019 at 16:22
160 °C (320 °F) for 1-2 h worked for me once, but higher temperatures should not hurt the process.
If you are starting with dry sodium hydrogen carbonate, the mass should reduce to 63 % of the starting mass (more reduction in case of wet starting material).
Explanation for the mass loss number: You are converting two equivalents of sodium hydrogen carbonate with a molar mass of 84.01 g/mol to one equivalent of sodium carbonate with a molar mass of 105.99 g/mol. 105.99/(2*84.01) = 63 %
• Is it close enough to be at 70%?
– user50726
Oct 2, 2019 at 0:58
• Depends on what you want to do with it. An "incomplete" product will be less alkaline, although I would wager it will still be absolutely fine for most culinary purposes. But there's also no harm in just baking it more at a later time. Oct 2, 2019 at 7:07
• By the way, I started with baking soda that had been in the fridge for probably a long time. Is it possible that the decomposition process had already started before I baked it and so the reduction in mass would be less than expected?
– user50726
Oct 2, 2019 at 16:23
• At low temperatures, the decomposition should be so slow as not to be noticeable by weight (except maybe on geological time scales). But possibly it was not pure sodium hydrogen carbonate to begin with (either already sodium carbonate or an inert filler as a contaminant). Oct 7, 2019 at 9:04
The temperature needs to be high enough, that's all. 450°F should work fine. (2-5 hours is a little silly, since the reaction is pretty quick once the baking soda itself is up to temperature.)
If you want to check that everything's gone okay, you can weigh the powder before and after. If it's converted properly, its mass should decrease to 63% (EDIT: fixed miscalculation, Matthias' number is the correct one) of the original. (Possibly a little further if there was some moisture in it.)
• you should also add temperature units, I'd like to be sure you're talking F not C. Sep 27, 2019 at 8:52
• @Luciano fair enough, though I doubt the OP is cooking bread at 450°C. Sep 27, 2019 at 9:00
Scientific American - Vanishing Baking Soda
Just some info to add to the conversation. Was doing research for doing pulled noodles which need an alkaline to help with gluten formation. Came across this article in StackExchange. For me, 270g of Baking Soda was reduced to 184g Sodium Carbonate @ 450 degrees Fahrenheit and was baked for 1.5 hours spread in an aluminum lined 9x9 cake pan. The link to the Scientific American article breaks it down simply; and, added to the great science insight here I think paints a more complete picture. | 962 | 3,795 | {"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-2024-22 | latest | en | 0.955687 |
https://math.stackexchange.com/questions/742181/find-all-integer-solutions-for-the-equation-5x2-y2-4?noredirect=1 | 1,716,896,466,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059085.33/warc/CC-MAIN-20240528092424-20240528122424-00354.warc.gz | 316,919,545 | 44,152 | # Find all integer solutions for the equation $|5x^2 - y^2| = 4$
In a paper that I wrote as an undergraduate student, I conjectured that the only integer solutions to the equation $$|5x^2 - y^2| = 4$$ occur when $x$ is a Fibonacci number and $y$ is a Lucas number. I was able to prove that when $x$ was a Fibonacci number there existed a Lucas number $y$ such that $|5x^2 - y^2| = 4$. This is easily shown with Cassini's Identity $$F_{n-1}F_{n+1} - F_{n}^2 = (-1)^n$$
The challenge is this ... prove (or disprove) that these are the ONLY solutions.
By the way, this is how I generated the Diophantine equation. $$F_{n-1}F_{n+1} - F_{n}^2 = (-1)^n$$ $$F_{n-1}(F_{n}+F_{n-1}) - F_{n}^2 = (-1)^n$$ $$F_n^2 - F_{n-1}F_n-F_{n-1}^2+(-1)^n=0$$ because $F_n \gt \frac{F_{n-1}}{2}$ $$F_n=\frac{F_{n-1} + \sqrt{F_{n-1}^2-4((-1)^n-F_{n-1}^2)}}{2}=\frac{F_{n-1} + \sqrt{5F_{n-1}^2+4((-1)^{n+1})}}{2}$$ Letting $y= \pm \sqrt{5F_{n-1}^2+4((-1)^{n+1})}$ and $x=F_{n-1}$ we have $$y= \pm \sqrt{5x^2+4((-1)^{n+1})}$$ $$y^2= 5x^2 \pm 4$$ $$|5x^2 - y^2| = 4$$
• I found this solution from Dalhousie University in Halifax, Nova Scotia (which is similar to Ricardo Buring's solution). See page 91 of this pdf. link. This appeared in the June 1968 edition of the Fibonacci Quarterly (Volume 6, Number 3). Full editions of the Fibonacci Quarterly can be downloaded here link Apr 10, 2014 at 21:19
• Except for $x=0, y= \pm 2,$ those are all solutions. Apr 16, 2014 at 22:48
• There is a lot of inconsistency in the literature concerning whether or not $L_0$ and $F_0$ are included in their respective sequences. Whether $(0, \pm 2)$ is a counter example or not depends on how you define the sequences. Apr 19, 2014 at 2:18
• John, that's fine, it appears we can have $2$ as the Lucas number before $1,$ as in $2,1,3,4,7.$ In any case, we do know all solutions. Apr 19, 2014 at 2:30
Let me interchange $x$ and $y$ for my own convenience. We want to solve $$x^2 - 5y^2 = \pm 4$$ over the integers.
Solving these equations corresponds to finding the elements of norm $\pm 4$ in the quadratic integer ring $\mathbf{Z}[\sqrt{5}]$, where the norm is the function given by $$N(x+\sqrt{5}y) = (x+\sqrt{5}y)(x-\sqrt{5}y) = x^2 - 5y^2.$$
Finding these elements is an exercise in algebraic number theory. The real quadratic number field $\mathbf{Q}(\sqrt{5})$ has $\mathbf{Z}[\omega]$ with $\omega = (1+\sqrt{5})/2$ as its ring of integers, and $\mathbf{Z}[\sqrt{5}]$ is a subring of this. The field norm on $\mathbf{Q}(\sqrt{5})$ agrees with the norm given above for elements of $\mathbf{Z}[\sqrt{5}]$.
Lemma I.7.2 in Neukirch's Algebraic Number Theory yields that up to multiplication by units in $\mathbf{Z}[\omega]$, there are only finitely many elements of a given norm in $\mathbf{Z}[\omega]$. Since $\mathbf{Z}[\sqrt{5}] \subset \mathbf{Z}[\omega]$ and the norms agree, up to multiplication by units in $\mathbf{Z}[\omega]$ there are only finitely many elements of norm $4$ in $\mathbf{Z}[\sqrt{5}]$.
By Dirichlet's unit theorem the group of units of $\mathbf{Z}[\omega]$ has rank $1$. A generator of this group, or a fundamental unit of $\mathbf{Q}(\sqrt{5})$, is given by $$\varepsilon = \frac{1+\sqrt{5}}{2},$$ which has norm $-1$.
Since the norm of an element $\alpha$ is the same as the norm of the principal ideal $(\alpha)$, it is useful to determine the number of ideals of norm $4$ in $\mathbf{Z}[\omega]$. By this answer to an other question this number is $$\sum_{m|4} \chi(m) = \chi(1) + \chi(2) + \chi(4) = \left(\frac{1}{5}\right) + \left(\frac{2}{5}\right) + \left(\frac{4}{5}\right) = 1 - 1 + 1 = 1.$$
Hence if $\alpha, \beta$ are two elements of norm $4$, then $(\alpha) = (\beta)$, so $\beta = u\alpha$ for a unit $u$. That is, up to multiplication by units in $\mathbf{Z}[\omega]$ there is only one element $\alpha$ of norm $4$.
Take $\alpha = 2$; then all the elements of norm $4$ in $\mathbf{Z}[\omega]$ are given by $2\varepsilon^n$, for integer $n$. But since $2\mathbf{Z}[\omega] \subset \mathbf{Z}[\sqrt{5}]$, all of these elements in fact belong to $\mathbf{Z}[\sqrt{5}]$. Hence all the solutions to the original equation are the $(x_n, y_n)$ given by $2\varepsilon^n = x_n + \sqrt{5}y_n$.
From the identity $\varphi^n = \frac{L_n + \sqrt{5}F_n}{2}$ of real numbers for nonnegative $n$ mentioned at the end of this section of the Wikipedia article on Lucas numbers it follows that $$2\varepsilon^n = L_n + \sqrt{5}F_n$$ for nonnegative $n$.
For negative $n$ you get extra solutions like $(1,-1)$ and $(-3,1)$, but you could have predicted those from the beginning: if $(x,y)$ is a solution, then so are $(-x,y)$, $(x,-y)$ and $(-x,-y)$.
I should mention that with SAGE you can do calculations in $\mathbf{Q}(\sqrt{5})$,
K.<s> = QuadraticField(5)
eps = (1+s)/2 # = K.units()[0]
for n in range(0,15):
print 2*eps^n
and also with Fibonacci and Lucas numbers:
for n in range(0,15):
print (fibonacci(n), lucas_number2(n,1,-1))
These two pieces of code give the same output (up to formatting).
Edit (01/11/14): A more elementary way to see that there is only one ideal of norm 4 in $\mathbf{Z}[\omega]$ is as follows:
The quadratic field $\mathbf{Q}(\sqrt{5})$ has discriminant $5$ and has no complex embeddings; hence by this inequality we have $N(I) \geq N(x)/\sqrt{5}$ for any ideal $I$ and element $x \in I$. Since $\mathbf{Z}[\omega]$ is a Dedekind domain we have unique factorization of ideals into primes. For a prime $\mathfrak{p} \subset \mathbf{Z}[\omega]$ lying over $p$ we get $N(\mathfrak{p}) \geq p^2/\sqrt{5}$. Since $p^2/\sqrt{5} > 4$ for $p > 2$, the primes of norm at most $4$ must lie over $2$. The minimal polynomial $X^2 - X - 1$ of $\omega$ is irreducible mod $2$, so $2$ is inert in $\mathbf{Z}[\omega]$ by the Kummer-Dedekind theorem. That is, $(2)$ is the only prime with norm at most $4$, and its norm is exactly $4$. By unique factorization into primes and multiplicativity of the norm, $(2)$ is the only ideal of norm $4$ in $\mathbf{Z}[\omega]$.
• I have no idea what you just wrote. I have some reading to do, and I'm sure that I'll understand your answer in 2-3 weeks. In the mean time, you get the green check mark :) Thanks for pointing me in the right direction. Apr 10, 2014 at 13:47
EDIT, January 2015: Conway's little book is available at http://www.maths.ed.ac.uk/~aar/papers/conwaysens.pdf
I also put four related excerpts, all with the prefix indefinite_binary, at OTHER. Dmitry says the computer zakuski is being decommissioned, hope it continues to work through late January. I especially like Stillwell's presentation. Put it all together, for a Pell form, indeed any form $a x^2 + b xy + c y^2$ with $a > 0, \; b \geq 0, \; c < 0,$ but $b^2 - 4ac$ not a perfect square, we get a diagram that shows all of Conway's information, along with the $(x,y)$ pairs as column vectors, with an explicit illustration of the (proper) automorphism group generator, that being the mapping $(x,y) \mapsto (9x+20y,4x+9 y).$
Did not notice this one ten days ago. There is an explicit structure for representing a number by an indefinite quadratic form. This is chapter one in Conway's The Sensual Quadratic Form. I wrote a little program recently, and no longer make simple arithmetic mistakes in these.
It turns out that all occurrences of $\pm 4$ happen along the "river" for $x^2 - 5 y^2.$
Given any solution to $x^2 - 5 y^2 = \pm 4,$ we gat the same value by switching $(x,y)$ to $$(9x+20y,4x+9 y).$$ The two by two matrix causing this transformation (on column vectors) is $$A \; = \; \left( \begin{array}{rr} 9 & 20 \\ 4 & 9 \end{array} \right) ,$$ which you can see towards the right of the diagram as the coordinates of the final $1$ and then the final $-5,$ placed side by side. The big theorem is that the entire diagram is periodic. I find the finite set of representatives within one cycle, apply the transformation I wrote arbitrarily many times, and i get all. As there is no $xy$ term in $x^2 - 5 y^2,$ there is a simple $\pm$ symmetry as well.
So, all solutions to $x^2 - 5 y^2 = \pm 4$ are:
Imprimitive:
+4: $$(2,0), (18,8), (322,144), (5778,2584), (103682,46368), (1860498,832040),\ldots,$$
-4: $$(-4,2), (4,2), (76,34), (1364,610), (24476,10946), (439204,196418),\ldots,$$
Primitive:
+4: $$(3,-1), (7,3), (123,55), (2207,987), (39603,17711), (710647,317811), \ldots,$$
+4: $$(3,1), (47,21), (843,377), (15127,6765), (271443,121393), \ldots,$$
-4: $$(-1,1), (11,5), (199,89), (3571,1597), (64079,28657), (1149851,514229), \ldots,$$
-4: $$(1,1), (29,13), (521,233), (9349,4181), (167761,75025), \ldots,$$
For any position in these sequences, there is a degree two recursion given by
$$a_{n+2} = 18 a_{n+1} - a_n.$$ For example, $18 \cdot 29 - 1 = 521,$ then $18 \cdot 521 - 29 = 9349.$
Let's see, 3:21 pm. Both Fibonacci and Lucas do the same thing (by six positions), as $$F_{n+12} = 18 F_{n+6} - F_n,$$ $$L_{n+12} = 18 L_{n+6} - L_n.$$ So, if the six orbits above satisfy the desired Fibonacci/Lucas conditions, that is a complete proof. If so, one could, carefully, interleave the six orbits in numerical order, perhaps using only the ones with strictly positive entries. See whether that works:
$$(1,1),(3,1),(4,2),(7,3),(11,5), (18,8),$$ $$(29,13),(47,21),(76,34),(123,55),(199,89), (322,144),$$ $$(521,233),(843,377),(1364,610),(2207,987),(3571,1597),(5778,2584),$$ $$(9349,4181),(15127,6765),(24476,10946),(39603,17711),(64079,28657),(103682,46368),$$ $$(167761,75025),(271443,121393),(439204,196418),(710647,317811),(1149851,514229),(1860498,832040),$$ Yep. The only miss is $(2,0),$ as $2$ is not a Lucas number. CORRECTION, FEB. 2015: as is commented elsewhere, it appears fairly common for people to define Lucas number $L_0 = 2,$ http://en.wikipedia.org/wiki/Lucas_number
Ummm; as you can see, $(x,y)$ and $(x,-y)$ may be distinct as far as the orbits, the six lists i wrote.
There is plenty more that could be said; anyway, these give all solutions. Oh, the other business, the "climbing lemma," says that values only increase (in absolute value) when leaving the river. The next layers of values are $\pm 11$ at the continuation of each edge with a light blue $6,$ and $\pm 19$ at the continuation of each edge with a light blue $10.$ So we have done enough to catch all $\pm 4$ already.
If no enough basic knowledge, we can also directly caculate.
If $$(x_0,y_0),(x_1,y_1)$$ solved $$x^2-5y^2 = \pm 4$$
$$1 = \frac{x_0^2-5y_0^2}{x_1^2-5y_1^2}$$
$$= \frac{x_0+\sqrt{5}y_0}{x_1+\sqrt{5}y_1} \cdot \frac{x_0-\sqrt{5}y_0}{x_1-\sqrt{5}y_1}$$
$$= \frac{(x_0+\sqrt{5}y_0)(x_1-\sqrt{5}y_1)}{x_1^2-5y_1^2} \cdot \frac{(x_0-\sqrt{5}y_0)(x_1+\sqrt{5}y_1)}{x_1^2-5y_1^2}$$
$$= \frac{(x_0x_1-5y_0y_1)+\sqrt{5}(y_0x_1-x_0y_1)}{4} \cdot \frac{(x_0x_1-5y_0y_1)-\sqrt{5}(y_0x_1-x_0y_1)}{4}$$
$$= (\frac{x_0x_1-5y_0y_1}{4})^2+\sqrt{5}(\frac{y_0x_1-x_0y_1}{4})^2$$
if $$y_0x_1-x_0y_1 = 0 \pmod 4$$ then $$(\frac{x_0x_1-5y_0y_1}{4},\frac{y_0x_1-x_0y_1}{4})$$ satisfy $$x^2 - 5y^2 = 1$$ which is pell's equation
so we can group $$(x,y)$$ into different group, (if exist)
$$(4k_0,4k_1+2),(4k_0+2,4k_1)$$
$$(4k_0+1,4k_1+1),(4k_0+3,4k_1+3)$$
$$(4k_0+1,4k_1+3),(4k_0+3,4k_1+1)$$
$$(4k_0+2,4k_1+2),(4k_0+4,4k_1+4)$$
Note that $$(4k_0+2,4k_1+2),(4k_0+4,4k_1+4)$$ if exist, it can be same group with above group. When $$(4k_0+3,4k_1+3)$$ be the same group with $$(4k_0+2,4k_1+2)$$ and $$(4k_0+1,4k_1+3)$$ be the same group with $$(4k_0+2,4k_1+2)$$, but $$(4k_0+1,4k_1+3)$$ and $$(4k_0+3,4k_1+3)$$ can never be in same group, so some group may have no answer. So I split it out.
The base answer for $$x^2-5y^2=1$$ is $$(9,4)$$
so if $$(x,y)$$ is the answer for $$x^2-5y^2=-4$$ , ($$x^2-5y^2=4$$ can be solved same way)
the next answer in its group will be get from $$(x+\sqrt{5}y)(9+4\sqrt{5}) = (9x+20y) + \sqrt{5}(9y+4x)$$,which is $$(9x+20y,9y+4x)$$
the previous answer in its group will be get from $$(x+\sqrt{5}y)(9+4\sqrt{5})^{-1} = (9x-20y) + \sqrt{5}(9y-4x)$$,which is $$(9x-20y,9y-4x)$$
we only care about $$x>0,y>0$$, In each group we have the the smallest positive solutions.
so the smallest answer satisfy $$9x-20y \le 0$$ or $$9y-4x \le 0$$
$$\frac{x}{y} \le \frac{20}{9}$$ or $$\frac{x}{y} \ge \frac{9}{4}$$
$$\sqrt{5- \frac{4}{y^2}} \le \frac{20}{9}$$ or $$\sqrt{5- \frac{4}{y^2}} \ge \frac{9}{4}$$
$$0.894427 \le y \le 8.04984$$
so we can just test $$y$$ from 1 to 8, and we will find 3 base answer for $$x^2-5y^2=-4$$ , $$(x,y) = (1,1),(4,2),(11,5)$$
if $$x^2-5y^2=4$$,the above ineqution become $$\sqrt{5 + \frac{4}{y^2}} \le \frac{20}{9}$$ or $$\sqrt{5 + \frac{4}{y^2}} \ge \frac{9}{4}$$
$$0
so we can just test $$y$$ from 1 to 8, and we will find 3 base answer for $$x^2-5y^2=4$$ , $$(x,y) = (3,1),(7,3),(18,8)$$
all other answers can be generate by $$(x+\sqrt{5}y)(9+4\sqrt{5})^n$$ , $$n$$ can be positive or negative integer | 4,831 | 12,840 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 51, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-22 | latest | en | 0.856037 |
https://fred.stlouisfed.org/series/DDDI02AGA156NWDB | 1,506,458,226,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818696681.94/warc/CC-MAIN-20170926193955-20170926213955-00081.warc.gz | 655,728,731 | 22,476 | # Deposit Money Bank Assets to GDP for Antigua and Barbuda (DDDI02AGA156NWDB) Excel (data) CSV (data) Image (graph) PowerPoint (graph) PDF (graph)
Observation:
2015: 71.50
Updated: Aug 29, 2017
Units:
Percent,
Frequency:
Annual
1Y | 5Y | 10Y | Max
EDIT LINE 1
(a) Deposit Money Bank Assets to GDP for Antigua and Barbuda, Percent, Not Seasonally Adjusted (DDDI02AGA156NWDB)
Total assets held by deposit money banks as a share of GDP. Assets include claims on domestic real nonfinancial sector which includes central, state and local governments, nonfinancial public enterprises and private sector. Deposit money banks comprise commercial banks and other financial institutions that accept transferable deposits, such as demand deposits.
Claims on domestic real nonfinancial sector by deposit money banks as a share of GDP, calculated using the following deflation method: {(0.5)*[Ft/P_et + Ft-1/P_et-1]}/[GDPt/P_at] where F is deposit money bank claims, P_e is end-of period CPI, and P_a is average annual CPI. Raw data are from the electronic version of the IMF's International Financial Statistics. Deposit money bank assets (IFS lines 22, a-d); GDP in local currency (IFS line 99B..ZF or, if not available, line 99B.CZF); end-of period CPI (IFS line 64M..ZF or, if not available, 64Q..ZF); and annual CPI (IFS line 64..ZF). (International Monetary Fund, International Financial Statistics, and World Bank GDP estimates)
Source Code: GFDD.DI.02
Deposit Money Bank Assets to GDP for Antigua and Barbuda
Select a date that will equal 100 for your custom index:
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NOTES
Source: World Bank
Release: Global Financial Development
Frequency: Annual
#### Notes:
Total assets held by deposit money banks as a share of GDP. Assets include claims on domestic real nonfinancial sector which includes central, state and local governments, nonfinancial public enterprises and private sector. Deposit money banks comprise commercial banks and other financial institutions that accept transferable deposits, such as demand deposits.
Claims on domestic real nonfinancial sector by deposit money banks as a share of GDP, calculated using the following deflation method: {(0.5)*[Ft/P_et + Ft-1/P_et-1]}/[GDPt/P_at] where F is deposit money bank claims, P_e is end-of period CPI, and P_a is average annual CPI. Raw data are from the electronic version of the IMF's International Financial Statistics. Deposit money bank assets (IFS lines 22, a-d); GDP in local currency (IFS line 99B..ZF or, if not available, line 99B.CZF); end-of period CPI (IFS line 64M..ZF or, if not available, 64Q..ZF); and annual CPI (IFS line 64..ZF). (International Monetary Fund, International Financial Statistics, and World Bank GDP estimates)
Source Code: GFDD.DI.02
#### Suggested Citation:
World Bank, Deposit Money Bank Assets to GDP for Antigua and Barbuda [DDDI02AGA156NWDB], retrieved from FRED, Federal Reserve Bank of St. Louis; https://fred.stlouisfed.org/series/DDDI02AGA156NWDB, September 26, 2017.
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Updating graph. | 856 | 3,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-39 | longest | en | 0.807216 |
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x0
x0
Bingo: + 50 points for 7 letters.
info: If you need to count the word with blanks, just do not enter them. Also, the order of letters does not affect the result.
## Scrabble Words with SOCA
socas7 socage9 socager10 socages10 mesocarp14 socagers11 mesocarps15 isocaloric14 misocapnic18 isocarboxazid34 isocarboxazids35
### Letter Values
The number at each letter is indicated the points of the letter.
The score of blanks is zero.
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The score for each turn is the sum of the letter values in each word formed or modified during the play, plus the additional points obtained from placing letters on premium squares. A turn using all seven tiles earns a 50-point bonus. Such a play is commonly called a “Bingo.”
### Double and Triple Letter Bonus Squares
The letter contained on the square with the mark of doubling the letter points doubles its value.
The letter contained on the square with the mark of tripling the letter points triples its value.
### Double and Triple Word Bonus Squares
If at least one letter of the word is placed on a square with a mark "Double Word Score" or center square it doubles the score of an entire word.
If at least one letter of the word is placed on a square with a mark "Triple Word Score" it triples the score for an entire word.
### Several premium squares of word multiplication
If the word is placed in squares doubles or triplets more than once, the Word doubles or triples as many times as many squares it takes.
### Bonus Squares Score on One Turn Only
The letter premium squares and the word premium squares apply only to the turn in which they are originally covered by a word. In all subsequent turns, letters on those squares count only at face value.
### Using the Blank on a DWS or TWS
When a blank tile is played on a Double Word Score square or a Triple Word Score square, the value of the word is doubled or tripled even though the blank itself has a zero score value.
SCRABBLE® is a registered trademark. We do not cooperate with the owners of this trademark. All trademark rights are owned by their owners and are not relevant to the web site "scrabble-word.com". This site is intended for entertainment and training. We try to make a useful tool for all fans of SCRABBLE. "Scrabble Word" is the best method to improve your skills in the game. | 539 | 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} | 2.875 | 3 | CC-MAIN-2024-18 | latest | en | 0.870831 |
https://course-downloader.com/six-sigma-statistics-using-minitab-17-free-download | 1,679,955,624,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948708.2/warc/CC-MAIN-20230327220742-20230328010742-00748.warc.gz | 224,962,634 | 22,884 | # Six Sigma Statistics Using Minitab 17
## Learn Basics and Confidently Apply Six Sigma Statistical Concepts To Your Green / Black Belt Projects Using Minitab 17
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• 11.5 hours on-demand video | 767 | 3,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-14 | latest | en | 0.868267 |
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# Solution3 - EE 200 Fall 2009 (Weber) Homework 3 Solutions...
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Unformatted text preview: EE 200 Fall 2009 (Weber) Homework 3 Solutions 1. a. Linear time-invariant e. Linear time-invariant b. Linear (not time-invariant) f. Time-invariant (not linear) c. Linear (not time-invariant) g. Time-invariant (not linear) d. Linear time-invariant h. Time-invariant (not linear) 2. The normalize radian frequency for this signal is = 0 . 3 where = f s = 2 f f s Aliases can occur at 2 k + and 2 k- ( k = any integer). Try a few values of k and see which ones result in f being below 1500Hz. 2 (0) + 0 . 3 = 2 f f s . 3 = 2 f 1000 f = 150 Hz. 2 (1) + 0 . 3 = 2 f f s 2 . 3 = 2 f 1000 f = 1150 Hz. 2 (1)- . 3 = 2 f f s 1 . 7 = 2 f 1000 f = 850 Hz. Therefore the following three signals will all result in the same sampled signal: x 1 ( t ) = 4 cos(2 150 t- / 4) x 2 ( t ) = 4 cos(2 1150 t- / 4) x 3 ( t ) = 4 cos(2 850 t- / 4) 3. Use the trig identy cos( + ) + cos( - ) = 2 cos(...
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Ask a homework question - tutors are online | 515 | 1,535 | {"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} | 3.671875 | 4 | CC-MAIN-2017-13 | longest | en | 0.736556 |
https://www.doorsteptutor.com/Exams/CA/Foundation/Maths-Stats-Logic/Questions/Topic-Statistical-Description-of-Data-12/Subtopic-Diagrammatic-Representation-of-Data-2/Part-2.html | 1,501,206,397,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549436316.91/warc/CC-MAIN-20170728002503-20170728022503-00335.warc.gz | 772,466,525 | 11,269 | # Statistical Description of Data-Diagrammatic Representation of Data (CA Foundation Maths, Statictics, Logic, and Reasoning): Questions 8 - 10 of 10
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## Question number: 8
» Statistical Description of Data » Diagrammatic Representation of Data
MCQ▾
### Question
The chart that uses logarithm of the variable is known as
### Choices
Choice (4) Response
a.
Ratio chart
b.
Line chart
c.
Multiple line chart
d.
Component line chart.
## Question number: 9
» Statistical Description of Data » Diagrammatic Representation of Data
MCQ▾
### Question
Vertical bar diagram is applicable when
### Choices
Choice (4) Response
a.
The data are quantitative
b.
The data are qualitative
c.
A or C.
d.
When the data vary over time
## Question number: 10
» Statistical Description of Data » Diagrammatic Representation of Data
MCQ▾
### Question
Pie-diagram is used for
### Choices
Choice (4) Response
a.
Representing qualitative data in a circle
b.
Comparing different components and their relation to the total
c.
Representing quantitative data in circle
d.
All of the above
f Page | 316 | 1,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-30 | longest | en | 0.71293 |
https://courseworkminutes.com/product/Corporate-Finance-Assignment-8-Problems/ | 1,632,091,725,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056902.22/warc/CC-MAIN-20210919220343-20210920010343-00090.warc.gz | 246,689,774 | 14,046 | # Corporate Finance – Assignment 8 Problems
\$13.00
Category:
## Description
Assignment 8
Problem 1
Lowell Inc. has no debt and its financial position is given by the
following data:
Assets (book = market) \$3,000,000
EBIT \$500,000
Cost of equity (Ks) 10%
Stock price (P0) \$15
Shares outstanding n0 200,000
Tax rate T 40%
The firm is considering selling bonds and simultaneously
repurchasing some of its stock. It if moves to capital structure
with 30 percent debt based on market values, its cost of equity,
Ks, will increase to 11 percent to reflect the increased risk.
Bonds can be sold at a cost (Kd) of 7 percent. Lowell Inc. is a
no-growth firm. Hence, all its earnings are paid out as
dividends, and earnings are exceptionally constant over time.
a.
A. What would be the new WACC?
B. What effect would this use of leverage have on the value of thefirm (Va)?
C.What would be Lowell Inc.’s stock price?
D.What happens to the firm’s earnings per share after therecapitalization?
Problem 2
Mass Inc. is trying to estimate its optimal capital
structure. Right now, Mass Inc. has a capital structure
that consists of 50 percent debt and 50 percent equity,
based on market values. (Its D/S ratio is 1.00) The riskfree
rate is 6 percent and the market risk premium, K
M–
K
RF, is 5 percent. Currently the company’s cost of equity,
which is based on the CAPM, is 12 percent and its tax rate
is 40 percent. What would be Mass Inc.’s estimated cost of
equity if it were to change its capital structure to 60 percent debt and 40 percent equity?
Problem 3
Use the following information to:
a. Calculate the cash conversion cycle, and interpret the
numbers
INCOME STATEMENT COCA-COLA PEPSI
Sales
23,104 32,562
Cost of goods sold
8,195 14,176
BALANCE SHEET COCA-COLA PEPSI
Assets
Cash and Cash Equivalents 4,701 1,716
Short-term Investments 66 3,166
Accounts Receivables 2,281 3,261
Inventory 1,424 1,693
Other Current Assets 1,778 618
Total Current Assets
10,250 10,454
Total Assets
29,427 31,727
Financed by: COCA-COLA PEPSI
Accounts Payable 5,290 5,357
Short-term debt 4,546 2,889
Other Current Liabilities 0 1,160
Total Current Liabilities
9,836 9,406
Problem 4
In-tech Corporation’s sales and purchases for the last three
months are as following:
Sales (\$) Purchases (\$)
October 100,000 80,000
November 90,000 100,000
December 120,000 75,000
For the next three months, it estimates sales and purchases
to be as following:
Sales (\$) Purchases (\$)
January 90,000 70,000
February 80,000 70,000
March 80,000 70,000
It pays 40 percent of purchases in cash and gets a 4
percent discount. Another 40 percent of purchases are paid
the next month, and the final 20 percent of purchases are
paid in the second month after the purchase (for example,
40 percent of October purchases are paid in October, 40
percent October purchases are paid in November, and 20
percent October purchases are paid in December). Half of
the sales are made in cash, and the balance is collected
the next month. Cash sales are given a two percent
discount, and five percent of credit sales end up as bad
debt. The monthly operating expenses for In-tech
Corporation’s are \$10,000. In-tech expects to sell one of
its machinery in March for \$25,000. It will buy the
replacement in April for \$50,000. The cash balance as on
December 31 was \$50,000. In-tech has a target cash balance
of \$50,000. Prepare a monthly cash budget for the next
three months.
## Reviews
There are no reviews yet. | 982 | 3,569 | {"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-2021-39 | latest | en | 0.930641 |
https://docs.microsoft.com/ru-ru/dotnet/api/system.decimal.op_equality?view=netcore-2.2 | 1,571,524,017,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986700435.69/warc/CC-MAIN-20191019214624-20191020002124-00195.warc.gz | 386,086,996 | 9,719 | # Decimal.Equality(Decimal, Decimal)Decimal.Equality(Decimal, Decimal)Decimal.Equality(Decimal, Decimal)Decimal.Equality(Decimal, Decimal) Operator
## Определение
Возвращает значение, определяющее, равны ли два значения Decimal.Returns a value that indicates whether two Decimal values are equal.
``````public:
static bool operator ==(System::Decimal d1, System::Decimal d2);``````
``public static bool operator == (decimal d1, decimal d2);``
``static member ( = ) : decimal * decimal -> bool``
``Public Shared Operator == (d1 As Decimal, d2 As Decimal) As Boolean``
#### Параметры
d1
Decimal Decimal Decimal Decimal
Первое сравниваемое значение.The first value to compare.
d2
Decimal Decimal Decimal Decimal
Второе сравниваемое значение.The second value to compare.
#### Возвраты
Значение `true`, если `d1` и `d2` равны; в противном случае — значение `false`.`true` if `d1` and `d2` are equal; otherwise, `false`.
## Комментарии
Метод определяет операцию оператора равенства для Decimal значений. EqualityThe Equality method defines the operation of the equality operator for Decimal values. Он включает следующий код:It enables code such as the following:
``````using System;
public class Example
{
public static void Main()
{
Decimal number1 = 16354.0695m;
Decimal number2 = 16354.0699m;
Console.WriteLine("{0} = {1}: {2}", number1,
number2, number1 == number2);
number1 = Decimal.Round(number1, 2);
number2 = Decimal.Round(number2, 2);
Console.WriteLine("{0} = {1}: {2}", number1,
number2, number1 == number2);
}
}
// The example displays the following output:
// 16354.0695 = 16354.0699: False
// 16354.07 = 16354.07: True
``````
``````Module Example
Public Sub Main()
Dim number1 As Decimal = 16354.0695d
Dim number2 As Decimal = 16354.0699d
Console.WriteLine("{0} = {1}: {2}", number1,
number2, number1 = number2)
number1 = Decimal.Round(number1, 2)
number2 = Decimal.Round(number2, 2)
Console.WriteLine("{0} = {1}: {2}", number1,
number2, number1 = number2)
End Sub
End Module
' The example displays the following output:
' 16354.0695 = 16354.0699: False
' 16354.07 = 16354.07: True
``````
Если язык, который вы используете, не поддерживает пользовательские операторы, вызовите Equals метод.If the language you're using doesn't support custom operators, call the Equals method instead.
Эквивалентным методом для этого оператора являетсяDecimal.Equals(Object)The equivalent method for this operator is Decimal.Equals(Object) | 685 | 2,473 | {"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-2019-43 | latest | en | 0.15206 |
https://wikfni.com/new-brunswick/efficiency-example-problems-with-solutions.php | 1,653,180,396,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543264.49/warc/CC-MAIN-20220522001016-20220522031016-00331.warc.gz | 690,979,545 | 7,556 | # Efficiency example problems with solutions New Brunswick
## Quality Control problems and solutions The Biscuit Doctor
For example, in areas in which this chapter addresses these problems, possible solutions are the most detrimental to the overall drilling efficiency and must.
Transformer losses and efficiency. example 6-5a: determine the efficiency of the transformer in examples 6-2, 6-3, solution: the core losses are 13/05/2010в в· transformer efficiency calculations may 13, 2010 #1. mathsdude69. hey. im currently doing a bit of revision for an exam and i am struggling on the following question:
Example 1 the histogram below shows the heights the histogram below shows the efficiency level problems and worksheets 4/11/2013в в· the 4 most effective ways leaders solve problems. and you will see examples of the problem solving process more efficient by
Pareto efficiency or pareto optimality is a state of is an example of a pareto-efficient by yielding all of the potentially optimal solutions, quality control problems and solutions quality control problems and solutions laboratory facilities. how are moisture tests made? what sieves are needed for sugar and
Here are some example problems based on the ideal rankine cycle efficiency as a function of you can get step-by-step solutions to your questions from an calculating efficiency of simple machines. look below for an example. efficiency = ama / tma x 100%. look here for practice problems:
Lect-20. problem 1 вђў in an air standard otto cycle, the solution: problem 3 вђў thermal efficiency of an ericsson cycle is equal to the carnot efficiency. the goal: increase warehouse efficiency. nitco, new englandвђ™s materials handling industry leader, offers innovative solutions to common warehouse problems.
Lever word problems. lever problems are word problems that use the lever principle. a good example of a lever is the seesaw. a lever can be set up with unequal problems in efficiency and operation of motors. problems & solutions on electric drives. documents similar to solved problems in electric motors.
Numerically efficient methods for solving numerically efficient methods for solving least squares for the least squares problem: solutions in the 2-norm are lever word problems. lever problems are word problems that use the lever principle. a good example of a lever is the seesaw. a lever can be set up with unequal
24 sample problems and algorithms constructive solutions to problems do essential for the efficiency. the problems and algorithms we discuss deal with very product efficiency problems product efficiency problems and solutions monitoring input, output and waste. what production data should be collected?
## Applied thermodynamics tutorial 1 revision of isentropic
Efficiency is the ratio of the amount of useful energy produced for example, if you have to (practice problems).
13/05/2010в в· transformer efficiency calculations may 13, 2010 #1. mathsdude69. hey. im currently doing a bit of revision for an exam and i am struggling on the following question: 6-3 machines and efficiency example 10: a jackscrew with a advantage of the wheel? c) what is the efficiency of the wheel? solution:
Answers to selected problems. 1. to ensure query efficiency (for example, the solutions to the remaining problems follow the same format as problem 8. the key to efficient and successful problem reso- volved to see if they had the solution to the problem. solving. creating a problem-solving culture
Basic content and examples terms of its thermal efficiency о·(lowercase greek eta), problem-solving strategy: heat-engine problems lever word problems. lever problems are word problems that use the lever principle. a good example of a lever is the seesaw. a lever can be set up with unequal
Example 4.8 thermal efficiency of the otto engine solution: this process is a example 4.9 thermal efficiency of a diesel engine sample problems for productivity example # 1 a company that processes fruits and vegetables is able to produce 400 cases of canned peaches in one half hour with four
Introduction. a multi-objective optimization problem is an optimization problem that involves multiple objective functions. in mathematical terms, a multi-objective this is solution to problems related thermodynamics course. gas turbine cycle-thermodynamics-problem solution, cycle efficiency solution:
The key to efficient and successful problem reso- volved to see if they had the solution to the problem. solving. creating a problem-solving culture improving business processes processes that don't work can lead to numerous problems. for example: they can significantly improve efficiency,
Basic content and examples terms of its thermal efficiency о·(lowercase greek eta), problem-solving strategy: heat-engine problems the key to efficient and successful problem reso- volved to see if they had the solution to the problem. solving. creating a problem-solving culture
Improving business processes processes that don't work can lead to numerous problems. for example: they can significantly improve efficiency, transformer losses and efficiency. example 6-5a: determine the efficiency of the transformer in examples 6-2, 6-3, solution: the core losses are
## Cooling Tower Efficiency and Range
Home > heat-transfer calculations heat-transfer accurate solutions to the heat transfer problems mechanical engineers face everyday. example 3: effect of.
Improve your energy efficiency at home. common home problems and solutions. is your home cold, drafty, or uncomfortable? do you have high energy bills? the solution to this clearly as the objective function is varying from problem to weights to determine relative efficiency 5 вђ“ solutions to the dea model 6
Transformer losses and efficiency. example 6-5a: determine the efficiency of the transformer in examples 6-2, 6-3, solution: the core losses are process cycle efficiency formula and example that you can apply to any industry, field, or business process.
An example of economic efficiency is when a piece of land is purchased for the highest possible price while financially benefiting what is an example of economic for example, in areas in which this chapter addresses these problems, possible solutions are the most detrimental to the overall drilling efficiency and must
The solution to this clearly as the objective function is varying from problem to weights to determine relative efficiency 5 вђ“ solutions to the dea model 6 process cycle efficiency formula and example that you can apply to any industry, field, or business process.
Efficient solutions of interval programming problems with inexact parameters and second order cone constraints 4/11/2013в в· the 4 most effective ways leaders solve problems. and you will see examples of the problem solving process more efficient by
Improve your energy efficiency at home. common home problems and solutions. is your home cold, drafty, or uncomfortable? do you have high energy bills? inventory management example problems with solutions for manufacturers as they can affect operational efficiency design of a computerized inventory management
Find possible solutions and recommended tests for 20 of the most common low hammer efficiency, or soft cushion. soil problems could include greater soil the solution to this clearly as the objective function is varying from problem to weights to determine relative efficiency 5 вђ“ solutions to the dea model 6
Then the group expands on what it perceives as being the best ideas until the foundation of a solution more efficient as a examples-workplace-problem improve your energy efficiency at home. common home problems and solutions. is your home cold, drafty, or uncomfortable? do you have high energy bills?
## 6-3 Machines and Efficiency Annville-Cleona School District
The key to efficient and successful problem reso- volved to see if they had the solution to the problem. solving. creating a problem-solving culture.
## Transformation Efficiency Practice Problem
Can you show me examples similar to my problem? optimization is a tool with applications across many examples of optimization problems efficient portfolios.
## Efficiency questions and answers TES Resources
Calculating efficiency of simple machines. look below for an example. efficiency = ama / tma x 100%. look here for practice problems:.
## Existence of efficient solutions for vector maximization
Efficiency questions and answers. 4.1 13 customer reviews. efficiency answers. worksheet. doc, 28 kb. report a problem..
## Example 4.8 Thermal efficiency of the Otto engine (Continued)
Numerically efficient methods for solving numerically efficient methods for solving least squares for the least squares problem: solutions in the 2-norm are.
## 5 – Solutions to the DEA model – Ali Emrouznejad's Data
Sample problems for productivity example # 1 a company that processes fruits and vegetables is able to produce 400 cases of canned peaches in one half hour with four.
Next post: recycler view android mvvm example Previous post: example recu frais de garde quebec | 1,832 | 9,128 | {"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} | 3.046875 | 3 | CC-MAIN-2022-21 | latest | en | 0.889167 |
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### Size of puffs
Posted: October 9th, 2019, 10:20 am
Hello
I have been making visualizations of particle dispersions. I'd like to try making visualizations of puffs as well. In the PARDUMP file, where is the information on the size of the puff? Such as radius? I would think that SIGW, SIGH, and SIGV would be the radii across x,y, and z directions, but these numbers always seem to range between -0.5 and 2, and don't significantly change over time. How can these numbers be negative? Any explanation is very appreciated!
### Re: Size of puffs
Posted: October 16th, 2019, 8:56 am
SIGH and SIGW are the relevant parameters for the puffs.
SIGH is the horizontal puff width.
SIGW is the vertical width if you are using 3d puffs (initd=1 or 2).
SIGV is only relevant for particles, not puffs.
See below some example output from par2asc from a 3d puff run.
This was using nams meteorology.
The SIGH and SIGW are in bold in the first two lines.
Note that the value of SIGV is not used for pure puff dispersions.
If you are getting negative numbers in the SIGW, then you are probably using
particle mode in the vertical (initd=3 or 4). In that case the SIGW is representing the
turbulent velocity in the vertical, which can be negative.
Record 1: MASS(1:NUMPOL)
Record 2: TLAT,TLON,ZLVL,SIGH,SIGW,SIGV
Record 3: PAGE,HDWP,PTYP,PGRD,NSORT
9260 1 19 2 7 0 0
0.764E+02
38.954 -90.697 645.837 1565.959 0.007 1001.000
63 1 1 1 1
0.764E+02
38.948 -90.698 643.975 1545.469 0.007 1001.000
62 1 1 1 2
0.764E+02
38.942 -90.700 641.899 1524.935 0.007 1001.000
61 1 1 1 3
0.638E+01
39.366 -90.343 990.339 6902.289 0.018 0.000
121 1 1 1 4
0.638E+01
39.360 -90.334 989.963 6892.970 0.018 0.000
121 1 1 1 5
0.638E+01
39.374 -90.339 990.684 7069.917 0.018 0.000
121 1 1 1 6
0.638E+01
39.375 -90.337 991.118 7032.935 0.018 0.000
121 1 1 1 7
0.638E+01
39.366 -90.329 989.984 6861.437 0.018 0.000
121 1 1 1 8
0.638E+01
39.361 -90.326 989.407 6795.729 0.018 0.000
121 1 1 1 9
0.638E+01
39.356 -90.322 990.084 6712.469 0.018 0.000
### Re: Size of puffs
Posted: October 29th, 2019, 10:46 am
And what is the unit for SIGH when it represents horizontal width? Meters? | 856 | 2,163 | {"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} | 2.734375 | 3 | CC-MAIN-2020-16 | longest | en | 0.826132 |
http://www.alanzucconi.com/2017/03/13/understanding-geographical-coordinates/ | 1,713,962,719,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00879.warc.gz | 35,312,548 | 23,045 | # Understanding Geographical Coordinates
This series introduces the concept of trilateration. This technique can be applied to a wide range of problems, from indoor localisation to earthquake detection. This first post provides a general introduction to the concept of geographical coordinates, and how they can be effectively manipulated. The second post in the series, Positioning and Trilateration, will cover the actual techniques used to identify the position of an object given independent distance readings. Most trilateration tutorials require the measures from the sensors to be precise and consistent. The approach here presented, instead, is highly robust and can tolerate inaccurate readings.
#### Introduction
There is a basic fact that permeates Engineering: sensors are imprecise. Whether you are using a temperature sensor or a camera, your measures will always have a finite accuracy. A simple way to overcome this limitation is to average out multiple readings from the same sensor. This often allows to cancel out the noise that is poisoning your readings.
If multiple sensors are available for you to use, it is often possible to merge their readings not just to refine your measurements, but to discover entirely new pieces of information. Sensor fusion (Wikipedia) is the umbrella term that is generally used to describe this family of techniques. They find many practical applications: GPS, earthquake detection, and indoor positioning are just few of them.
If you are unfamiliar with the concept, GPS (Global Positioning System) is a technology that allows to localise a device on the surface of the planet. It relies on a network of intercommunicating satellites, which are orbiting the planet. When your device establishes a connection with a satellite, the latter is unable to calculate your position. In a nutshell, the only information that each satellite knows is your distance. No single satellite knows your position. However, merging the data from multiple satellites allows to identify your exact location.
#### Understanding Geographical Coordinates
If we want to locate an object on the surface of the planet, we first have to understand how geographical coordinates work. Throughout history there have been many attempts to solve the insidious problem of mapping Earth. The most common solution nowadays assigns three coordinates to each point on the surface of Earth: latitude, longitude and altitude (namely, , and ). Under this framework, Earth is assumed to be a perfect sphere with known radius . Latitude and longitudes represents the angles (in degrees) of a point on the surface, in respect to the Earth centre (picture below).
To measure latitude, the following conventions hold:
• The equator has latitude ;
• The North pole has latitude ;
• The South pole has latitude .
Loosely speaking, latitude contributes to the (or vertical) component of a map. Conversely, longitude measures the (or horizontal) component. The imaginary line that connects the poles and passes through Greenwich is called the prime meridian:
• The prime meridian has longitude ;
• Moving East increases , up to the antipodal prime meridian which has longitude ;
• Moving West decreases , up to the antipodal prime meridian which has longitude .
Both and map to the same meridian, which is on the other side of the Earth from Greenwich.
Altitude in generally expressed in kilometre from the sea level, which is assumed to have . This series will not take into account altitude. This massively simplifies the resulting equations.
❓ Alternative conventions to express geographical coordinates
To make this notation even more complicated, there are multiple ways in which latitude and longitudes can be expressed.
Positive and negative coordinates are often expressed using North, South, East and West with the following conventions:
• Positive latitude is often expressed as North (i.e.: can be expressed as ;
• Negative latitude is often expressed as South (i.e.: can be expressed as ;
• Positive longitude is often expressed as East (i.e.: can be expressed as ;
• Negative longitude is often expressed as West (i.e.: can be expressed as .
Another way in which the angles can be expressed uses degrees, minutes and seconds. If you are interested in knowing more about this topic, Wikipedia has an interesting page on Geographic coordinate conversion.
❓ Are latitude and longitude equivalent to a Cartesian coordinate system?
Latitude and longitude loosely map to orthogonal coordinate system. If you have the map of a city, moving vertically changes your latitude, while moving horizontally changes your longitude. However, latitude and longitude do not map one to one to a Cartesian coordinate system. This is because the surface of the Earth is curved; while it is locally flat, its curvature becomes apparent on a larger scale.
#### Local Geographical Distance
On a small scale, Earth can be assumed to be locally flat. When you measure distances within a small city, for instance, you do not need to take into account the curvature of the planet. Let’s assume you want to calculate the geographical distance between two points and , respectively. Assuming Earth to be locally flat, you can simply rely on the Euclidean distance between these two points, correctly projected onto a Cartesian plane:
where:
• is the radius of the Earth;
• ;
• ;
• .
For this equation to work, all quantities must be expressed in radians, rather than degrees.
def geographical_distance (latitudeA, longitudeA, latitudeB, longitudeB):
mean_latitude = math.radians((latitudeA + latitudeB) / 2.0 )
R = 6371.009 # Km
# Spherical Earth projected to a plane
return \
R * math.sqrt \
(
math.pow
(
delta_latitude,
2
)
+
math.pow
(
math.cos(mean_latitude) *
delta_longitude,
2
)
)
❓ Why can't we use use Δϕ and Δλ?
You might have recognised that the equation used in this section uses the Pythagoras’s Theorem to calculate the distance between two points:
This real question is why do we need to use this equation:
To answer this question, we first need to understand that even if latitude and longitude are both expressed in degrees, they are qualitatively different. Latitude refers to parallels, which (like the name suggests) always stay parallel to each others. Longitude, on the other hand, is based on meridians which intersects at the pole. Parallels and meridians are fundamentally different ways to measure the same thing.
#### Great-Circle Distance
The Euclidean distance provides a good approximation for the distance between two geographical points only when they are relatively close. For larger distances, one has to take into account the curvature of the planet.
If you imagine Earth as a perfect sphere, the distance between two points is the length of a the shortest piece of strings that connects them. That string will naturally follow the curvature of the sphere, much like the distance we have to calculate. This concept is known as great-circle distance, and is explained in detail on this Wikipedia article.
Mathematically speaking, the shortest distance between two points on a sphere can be calculated using the spherical law of cosines (Wikipedia):
where is the angle between and , also called the central angle:
While the above mentioned equation is mathematically correct, it is also numerically unstable. This means that rounding errors can have a negative effect on the precision of the final result. A more accurate formula what works well with floating point arithmetic is:
This translates into the following Python code:
def great_circle_distance (latitudeA, longitudeA, latitudeB, longitudeB):
delta_lambda = math.fabs(lambda2 - lambda1)
central_angle = \
math.atan2 \
(
# Numerator
math.sqrt
(
# First
math.pow
(
math.cos(phi2) * math.sin(delta_lambda)
, 2.0
)
+
# Second
math.pow
(
math.cos(phi1) * math.sin(phi2) -
math.sin(phi1) * math.cos(phi2) * math.cos(delta_lambda)
, 2.0
)
),
# Denominator
(
math.sin (phi1) * math.sin(phi2) +
math.cos (phi1) * math.cos(phi2) * math.cos(delta_lambda)
)
)
R = 6371.009 # Km
return R * central_angle
The function math.atan2 is used to calculate the correct sign of .
#### Conclusion
This post introduced the concept of geographical coordinates, and the challenges that arise when trying to manipulate them. The next post will show how it is possible to use multiple distance readings to correctly locate objects.
##### 💖 Support this blog
This website exists thanks to the contribution of patrons on Patreon. If you think these posts have either helped or inspired you, please consider supporting this blog.
You will be notified when a new tutorial is released!
##### 📝 Licensing
You are free to use, adapt and build upon this tutorial for your own projects (even commercially) as long as you credit me.
You are not allowed to redistribute the content of this tutorial on other platforms, especially the parts that are only available on Patreon.
If the knowledge you have gained had a significant impact on your project, a mention in the credit would be very appreciated. ❤️🧔🏻
### Webmentions
• Localisation and Trilateration - Alan Zucconi October 11, 2018
[…] If you are unfamiliar with the concepts of latitude and longitude, I suggest you read the first post in this series: Understanding Geographical Coordinates. […]
• Geographical Coordinates for Programmers | StratCom October 11, 2018 | 2,008 | 9,422 | {"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} | 2.84375 | 3 | CC-MAIN-2024-18 | latest | en | 0.857686 |
http://www.lmfdb.org/ModularForm/GL2/TotallyReal/4.4.19025.1/holomorphic/4.4.19025.1-11.1-b | 1,606,511,083,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141194171.48/warc/CC-MAIN-20201127191451-20201127221451-00056.warc.gz | 139,057,089 | 5,283 | # Properties
Label 4.4.19025.1-11.1-b Base field 4.4.19025.1 Weight $[2, 2, 2, 2]$ Level norm $11$ Level $[11, 11, \frac{1}{2}w^{3} - 2w^{2} - \frac{5}{2}w + 11]$ Dimension $1$ CM no Base change no
# Related objects
## Base field 4.4.19025.1
Generator $$w$$, with minimal polynomial $$x^{4} - 2x^{3} - 13x^{2} + 14x + 44$$; narrow class number $$1$$ and class number $$1$$.
## Form
Weight: $[2, 2, 2, 2]$ Level: $[11, 11, \frac{1}{2}w^{3} - 2w^{2} - \frac{5}{2}w + 11]$ Dimension: $1$ CM: no Base change: no Newspace dimension: $21$
## Hecke eigenvalues ($q$-expansion)
The Hecke eigenvalue field is $\Q$.
Norm Prime Eigenvalue
4 $[4, 2, w^{2} - 2w - 6]$ $-1$
4 $[4, 2, -w^{2} + 7]$ $-1$
5 $[5, 5, -\frac{1}{2}w^{3} + 2w^{2} + \frac{7}{2}w - 14]$ $\phantom{-}4$
5 $[5, 5, \frac{1}{2}w^{3} + \frac{1}{2}w^{2} - 6w - 9]$ $-2$
11 $[11, 11, \frac{1}{2}w^{3} - 2w^{2} - \frac{5}{2}w + 11]$ $\phantom{-}1$
11 $[11, 11, \frac{1}{2}w^{3} + \frac{1}{2}w^{2} - 5w - 7]$ $\phantom{-}0$
31 $[31, 31, \frac{1}{2}w^{2} + \frac{1}{2}w - 4]$ $\phantom{-}2$
31 $[31, 31, -\frac{1}{2}w^{2} + \frac{3}{2}w + 3]$ $\phantom{-}2$
41 $[41, 41, \frac{1}{2}w^{2} + \frac{1}{2}w - 6]$ $-10$
41 $[41, 41, 2w^{3} - \frac{15}{2}w^{2} - \frac{25}{2}w + 50]$ $-2$
41 $[41, 41, \frac{5}{2}w^{2} - \frac{1}{2}w - 17]$ $-2$
41 $[41, 41, \frac{1}{2}w^{2} - \frac{3}{2}w - 5]$ $\phantom{-}2$
61 $[61, 61, -\frac{1}{2}w^{3} + w^{2} + \frac{7}{2}w - 1]$ $-4$
61 $[61, 61, -\frac{1}{2}w^{3} + \frac{1}{2}w^{2} + 4w - 3]$ $\phantom{-}2$
71 $[71, 71, \frac{1}{2}w^{3} + w^{2} - \frac{11}{2}w - 13]$ $-4$
71 $[71, 71, -\frac{1}{2}w^{3} + \frac{5}{2}w^{2} + 2w - 17]$ $-10$
81 $[81, 3, -3]$ $-8$
89 $[89, 89, -w^{3} + \frac{3}{2}w^{2} + \frac{13}{2}w - 9]$ $\phantom{-}16$
89 $[89, 89, w^{3} - \frac{7}{2}w^{2} - \frac{11}{2}w + 20]$ $-8$
89 $[89, 89, -w^{3} - \frac{1}{2}w^{2} + \frac{19}{2}w + 12]$ $-2$
Display number of eigenvalues
## Atkin-Lehner eigenvalues
Norm Prime Eigenvalue
$11$ $[11, 11, \frac{1}{2}w^{3} - 2w^{2} - \frac{5}{2}w + 11]$ $-1$ | 1,111 | 2,023 | {"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 | 3 | CC-MAIN-2020-50 | latest | en | 0.235486 |
https://bicycles.stackexchange.com/questions/36789/why-is-a-steady-cadence-so-important-or-is-it | 1,713,598,552,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817491.77/warc/CC-MAIN-20240420060257-20240420090257-00804.warc.gz | 118,142,276 | 51,037 | # Why is a steady cadence so important? Or is it?
I read somewhere that steady cadence is a good thing but why is that? It is analogous to running long distance where it is a good idea to set a reasonable pace and stick to it? I would think that varying your cadence would give your muscles a chance to rest some. I question whether the original claim is actually true. I would think it would be more efficient to accelerate briskly then coast down, then repeat.
This "sawtooth" cadence question is interesting because with a car or motorcycle, you actually get better gas mileage if you accelerate briskly and then just coast down. For example, you are at a red light in pole position (first). The light then turns green. You then accelerate quickly up to the speed limit (say 50 MPH) and then let the car/motorcycle coast down to maybe 20 MPH and then repeat. This is actually more efficient (you will use less gasoline) than if you accelerated slowly and maintained the 50 mph. This has already been confirmed by many people that have tested it. It is also easy to confirm yourself, especially if you have a real time mileage display. It will be at maybe 10 MPG for a brief time while you accelerate then vary between 99 and and some lower number as you coast down.
Ok now to bikes. It seems to make sense that for a bike a similar thing should happen. Imagine a person pedaling an average bicycle 10 MPH on level ground with no wind. The amount of energy expended by the rider depends on several factors including what gear they are in. Imagine they are in the proper gear vs. much too tall (high) a gear vs. much too short (low) a gear. Problem is in the low gear energy is being wasting pedaling faster than is needed (for example 150 cadence). This is a "subpoint" I am trying to make related to the main point of "sawtooth" cadence I am working up to.
If I pedal 10 miles at 10 MPH at near constant cadence, that will be 1 hour straight on pedaling. How can my muscles ever get a rest? Compare that to pedaling up to 20 MPH quickly (briskly) and then coasting down to 3 MPH and then repeating. Here I use my muscles for a quick burst then spend most of my time coasting. Here my muscles get a good rest and the number of actual crank rotations is far less. 10 miles on my bike in my favorite middle gear (34/20=1.7) with my 26" diameter tires would require about 4500 crank turns. I suspect using the sawtooth cadence method would require far less, possible only 1/3rd or about 1500. It would be an interesting test to have someone make the trip using both techniques and report their level on fatigue afterwards.
Someone mentioned about pedaling then coasting a lot will make me look like a kid on a BMX bike. Well kids are generally lazy. Maybe they figured out it is easier to pedal and coast so that is why they do it.
Another analogy is at my workplace. Can I get more work done if I work hard but take several breaks or if I just work straight thru without breaks but I get more tired that way? For me, I need the breaks to stay productive.
Maybe for younger riders a steady cadence is better but I am wondering for the 50+ crowd if sawtooth is actually better. I suspect it is.
• Have you read the answers to this question? bicycles.stackexchange.com/questions/12518/… In any event, if you examine what riders actually do, it is natural for cadence to vary. Riders rarely ride at a fixed cadence unless the conditions are also fixed (that is, same speed, same slope, same wind, same power). Jan 28, 2016 at 6:37
• Vote to leave open - OP is asking about some kind of sawtooth cadence pattern where rider pedals up to speed and then coast till you're too slow, then repeats. He's not asking about optimal cadences.
– Criggie
Jan 28, 2016 at 10:27
• Human muscle efficiency is less if you pedal intermittently vs pedaling continuously at a reasonably constant pace. And, given that wind resistance is proportional to the square of relative velocity, riding fast then slow is less efficient in that regard as well. Jan 28, 2016 at 13:30
• "...more efficient than if you accelerated slowly and maintained the 50 mph" if this is true it's more likely to be because your average speed is lower and if you sat steadily at the same average your fuel consumption would be lower again. Jan 28, 2016 at 14:24
• @David efficiency should be more important to the older riders, getting the most efficient use from their muscles. And accelerating then coasting then accelerating over and over would not lead to a leisurely ride. You seem to be confusing efficiency, power and speed here Jan 28, 2016 at 16:29
## 4 Answers
This is really a pretty complex question. Without knowing where you read about "steady cadence being a good thing" or what the author meant it is hard to evaluate this statement, but this SE.bicycles answer presents data showing that riders don't ride at a steady cadence. Rather, they alter their cadence according to conditions of the ride, the level of power they wish to put out, and the choice of gearing they have available. So your initial premise may not be correct.
That said, assuming you are on a constant gradient, with consistent wind, trying to go as fast as possible over a fixed distance, then you want power to be constant, because constant power application is the fastest way from point A to B in those conditions. If power is constant, then cadence will also be constant, unless you are shifting. Most humans can maintain the same power over a fairly wide variety of cadence, even if it doesn't feel as good. Usually it is fine to self select what cadence you want. It may be the case that you prefer a different cadence when climbing vs on the flats. If that is so, do not feel that you have to shift to keep cadence the same as the terrain changes.
The second part of your question presents a scenario of alternate pedaling and coasting, and posits that since it may be more economical for automobiles to do so perhaps it will be so for human powered cycling as well. However, humans are not like internal combustion engines -- we have two different metabolic pathways to generate power, which are colloquially called "aerobic" and "anaerobic." Aerobic metabolic production of ATP via the Krebs cycle is relatively efficient, but when we need to produce power that exceeds our aerobic capacity we switch to "anaerobic" production that is sustainable for far shorter periods of time. Compounding this "supply of power" issue is the "demand for power". The demand equations for cycling power are well understood if not well known, but the aerodynamic component of drag varies with the square of airspeed so the power demanded varies roughly with the cube of speed in calm conditions. These two physical constraints mean both that cycling faster and then coasting requires more energy than cycling steadily at the average pace, and also that speeding up is more physiologically taxing. That is, 1 minute of high power followed by 1 minute of recovery is still more taxing than 2 minutes at the average power, even though the total joules of work performed is the same. This is the basis for why "high-intensity" interval workouts lasting X minutes can have a greater training effect but be much less enjoyable than the same X minutes steadily spent at the average power.
As it happens, here is a plot based on data from a segment of a ride where the rider alternately pedaled and then coasted. This segment of the ride was done on a flat road, with almost no wind. The upper panel shows cadence; the middle panel shows speed; and the bottom panel shows power.
The "sawtooth" in cadence and power are apparent, and you can see speed increase and then slowly decrease. For this roughly 4 km long segment using this particular pattern of pedaling and coasting, power averaged 100 watts. If, on the other hand, the rider had covered this same flat 4 km long segment at constant speed so that the distance and total elapsed time were the same, the power would only have needed to have been 70 watts, a savings of 30%.
In this case, the bicycle used was a geared bike, so had the rider changed gearing he could have produced the required 70 watts with many different combinations of cadence and pedal force. In other words, steady cadence need not have been used. This is an illustration that cadence isn't terribly consequential, and you have wide latitude to ride at whatever cadence you want to meet the purposes you wish to achieve. If you enjoy pedaling and coasting, there nothing wrong with that.
• Thanks Robert, I will just let your comment stand as a nice expansion. Jan 28, 2016 at 20:45
• To me it is strange why humans would benefit from a constant cadence when we know cars are more efficient when they accelerate briskly (using about 75% of their maximum power) and then coast down (ideally with engine off but for safety lets say just with no throttle). This technique alone is worth about a 10% boost in fuel efficiency (not that it matters now with these low gas prices). If someone on a bike is not in a big hurry, the burst then coast technique might get them there slower but with far fewer crank rotations. I will experiment when I ride more and decide which I like better. Jan 28, 2016 at 21:13
• Unfortunately there is a nonlinear relationship between power output and physiological fatigue in humans, so constant power application is more efficient. This has already been tested a few hundred thousand times per year by various time trial experts, but grab a power meter and experiment, it is fun. That potential benefit of burst/coast realized with cars (sometimes) depends on subtle interactions with airflow that don't occur in humans. Jan 28, 2016 at 21:22
• @jackmott interesting, what are the subtle interactions that occur in cars? Jan 29, 2016 at 0:55
• I believe one thing that you get from the pump and coast method is lower pumping losses, because when the engine is on power, the throttle body is fully open. This technique may not even work though on newer engine designs with better variable valve timing and/or no throttle at all. Easy to fool yourself into thinking it works because testing for the differences is hard. But that is straying way off topic! Jan 29, 2016 at 1:19
Another theory that does not seem to be adopted by the bicycle community.
Individual racers and race teams go with a steady cadence and output.
If you have two riders trying dump a third the 2 riders will take turns bursting and falling back. Eventually the 3rd rider cannot keep up with the burst and is left. If bursting was as (or more) efficient than a steady pace this would not work. This works even with the 3rd rider not taking lead - they are drafting.
Wind resistance is relative to velocity squared.
Split 10 mph and 30 mph versus 20 MPH
2*20*20/(10*10 + 30*30) = .80
That is right 20% less wind resistance
As for muscle fatigue. Fatigue is the force - not cadence. Drop a gear so it is easier to spin. It is well establish that a human body is more efficient at delivering power in a fairly narrow cadence range.
• Somehow I am not "buying" this steady is better thing. I work a 8 hour shift for example. During that shift I lift some heavy things repeatedly which tires me some. If you were to ask me what would make me more tired, working hard (fast paced) and taking a break every 2 hours or slightly slacking (working a little slower) but maintaining that speed for the entire 7.5 hours, the no break would easily tire me way more! Even going 1 hour past my scheduled break (working 3 hours straight) is hard on me. Also batteries exhibit this same properly. Take a load off them and they recover quickly. Jan 28, 2016 at 21:19
• Maybe the actual amount of energy expended is less if a biker maintains constant cadence but maybe the perceived fatigue without resting is that MORE energy is expended. Why would it matter if less actual amount of energy was expended if it feels like more was? When I get to my destination I would like to be able to walk like a normal person. I suspect the majority of older riders need this rest, especially with old tired bones. Why do you think they take naps midday? You expect a 50+ year old man to maintain an 80 cadence for a 6.5 mile (10km) ride? If I had a NuVinci hub I could check. Jan 28, 2016 at 21:26
• @david Then don't buy it. Once again you have a strategy we don't see used in the industry. Have you ever seen a team take a break in the Toure de France or any race? Now this a 50+ year old man. I am 50+ year old man and 6.5 miles is walk in the park - 16 is a standard workout and I don't take a break. I said narrow an older person my be a little lower than 80 but there will be a narrow optimal range. I did not post this answer for you. I am not going to answer any more of your comments. Jan 28, 2016 at 21:29
• Squared is squared. Wind resistance should be less - did you run the calc? Yes we are talking about radially different situations. I am talking about reality and you are not. Your sense of logic and analytical reasoning is a bit lacking to be polite. Jan 28, 2016 at 22:33
• I cannot understand how you would think a leisurely ride at a steady 10mph with legs spinning at a nice easy cadence would be less leisurely than racing up to 20mph, getting the cardio vascular system and muscles working hard then coasting down to an almost standstill to repeating till the end of the ride. This is interval training and is anything but leisurely Jan 29, 2016 at 6:57
David - how does it feel to you when you're on your bike and try doing exactly that?
Biomechanically, you'll be stressing your muscles to a higher peak level, and then a brief respite. I imagine your overall power output will be lower over a reasonable distance.
If you pedal fast then coast for a bit then pedal then coast - I think you'll look like a teenager on a kid's BMX, and your average speed will be pretty slow. Partially this is because they're seriously under-geared for their teenage power, and partially its a lack of endurance to maintain those cadences for longer periods.
Answer Get on your bike and ride!
• I'm not sure if this actually answers the question at hand. The first paragraph is relevant, but in order to answer the question, it would also be necessary to explain why stressing your muscles to a peak level and then resting is bad. This actually sounds a lot like intervals, which are widely accepted as a great training practice. The second paragraph basically boils down to "it looks silly," which is just an opinion (though not one I disagree with), not a reason that steady cadence is important. Jan 28, 2016 at 13:22
• @criggie I somehow seem to have downvoted you on my stupid phone, I had up voted this previously. I can't change until you edit it though. Jan 28, 2016 at 18:00
• @CearonO'Flynn Its not about the votes, its about finding the answer.
– Criggie
Jan 28, 2016 at 20:16
• @Criggie I know but I thought it was a good answer and didn't want to do this answer a disservice Jan 28, 2016 at 20:25
It is far more efficient to cycle at a steady cadence suitable for the terrain and gearing used. In fact this attempt to maintain cadence and maximise efficiency is the reason bikes have gears.
Varying the cadence is going to put stress and strain on your muscles, joints and cardio vascular system. This could be your aim (with the exception of stressing your joints) as this is what you would do with interval training, however it would not be usual for a normal ride where you want to get from A to B the most efficiently.
I would also imagine gear changes would be be carried out with strange loads on them, either too much when cadence is high, or too low when the cadence is down. This could stress the mechanical components of the bike too.
• I feel good when I coast and get a rest. I even like to pedal (freewheel) backwards occasionally cuz it helps my muscles. To me it is not about maximizing efficiency, it is more about what feels good and if I get a burst of energy to use it and if I get tired to coast for a while to recover. I ride my bike for the enjoyment of it (and also for practical transportation), not to try to set any speed or efficiency records. Jan 28, 2016 at 16:26
• I find I enjoy my ride far more if I have a steady pedal stroke, my muscles and joints don't hurt and it means I can ride further and for many more days in a row. If I want to punish myself and have a hard muscle breaking training ride that I'd feel the next day I would pedal as you describe, varying cadence, accelerating hard then resting and repeating over and over Jan 28, 2016 at 16:32
• Single speed bikes don't easily allow for steady cadence since there are varying loads (uphill, downhill, paved, dirt, headwind, tailwind...). So if that is the case, why would varying cadence on a multispeed bike be bad? Our muscles change load all day long. We stand, sit, lie down, walk, sometimes run... I think it is fun to ramp up to speed then give myself a rest. Pedaling the same cadence the entire time would be kinda boring unless maybe I had a NuVinci hub and locked it at my favorite cadence of 88 (at which feet per crank rotation and MPH match exactly). Jan 28, 2016 at 16:40
• @david a geared bike has gears for this very reason, to keep the cadence the same as much as possible, because it is easier, more efficient and gentler on joints & muscles. They are two different machines and two totally different riding experiences can be achieved with them. The answer to your question is because its more efficient you may not agree with it but that is the reason, and its why gears were invented Jan 28, 2016 at 16:45
• @David You say in your comment: "To me it is not about maximising efficiency" but in the question, you say " I would think it would be more efficient to accelerate briskly then coast down, then repeat." Are you asking about efficiency? Are you really asking anything? Jan 28, 2016 at 22:12 | 4,148 | 17,982 | {"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} | 3.34375 | 3 | CC-MAIN-2024-18 | latest | en | 0.969767 |
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