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https://books.google.com.mx/books?id=g-BEAAAAIAAJ&lr= | 1,600,776,052,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400205950.35/warc/CC-MAIN-20200922094539-20200922124539-00242.warc.gz | 321,758,555 | 11,097 | An Intermediate Arithmetic: Uniting Mental and Written Exercises in a Natural System of Instruction
Van Antwerp, Bragg & Company, 1876 - 225 páginas
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Página 109 - To divide a whole number by a fraction, — Multiply the dividend by the denominator of the fraction, and divide the product by the numerator.
Página 74 - DIVISION is the process of finding how many times one number is contained in another, or of finding one of the equal parts of a number.
Página 83 - The Greatest Common Divisor of two or more numbers is the greatest number that will exactly divide each of them. Thu4, 18 is the greatest, common divisor of 36 and 54, since it is the greatest number that will divide each of them without a remainder.
Página 58 - If 2 men start from the same place, and travel in opposite directions ; one at the rate of 3| miles in an hour, and the other 4£...
Página 21 - I stands for one ; V for five ; X for ten ; L for fifty ; C for one hundred ; D for five hundred ; M for one thousand.
Página 6 - Multiplication is the process of taking one number as many times as there are units in another.
Página 90 - To reduce a mixed number to an improper fraction, — RULE : Multiply the whole number by the denominator of the fraction, to the product add the numerator, and write the result over the denominator.
Página 85 - The Least Common Multiple of two or more numbers is the least number which is a multiple of each of them; thus, 12 is the least common multiple of 2, 3, and 4.
Página 57 - If 7 men can do a piece of work in 9 days, how many men will it take to do the same work in one day?
Página 108 - RULES. — 1. To divide a fraction by an integer, Divide the numerator or multiply the denominator. 2. To divide a mixed number by an integer, Divide the integral part and then the fraction. | 469 | 1,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-40 | latest | en | 0.810279 |
https://www.jiskha.com/members/profile/posts.cgi?name=Nazish | 1,503,491,105,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886120194.50/warc/CC-MAIN-20170823113414-20170823133414-00099.warc.gz | 924,072,210 | 2,483 | # Posts by Nazish
Total # Posts: 3
physics
First find k.E=1/2mv'2 K.e=1/2 1000x (25)2= E=312500 j Now P=E/t P=312500/10=31250 Now convert in kilo 31250/1000=31.25 thanks | 76 | 171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.621547 |
https://www.askiitians.com/forums/Physical-Chemistry/a-gas-absorbs-a-photon-of-355-nm-and-emits-at-two_81477.htm | 1,686,307,435,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00799.warc.gz | 681,371,124 | 40,358 | # A gas absorbs a photon of 355 nm and emits at two wavelengths. if one of the emissionsis at 680 nm, the other is at
Sunil Kumar FP
8 years ago
energy corresponding to wavelength 355nm=hc/l
where h= planks constant
c=speed of light
l=wavelength
therefore according to the problem,
hc/355*10^-9=hc/680^10^-9 +hc/l*10^-9
1/355=1/680+1/l
l=742nm
Thus the wavelength of the light for other emmission is 742nm
Tanishka Thorat
13 Points
4 years ago
As we know c=v (lambda).. Symbol of not available for lamba
By plank's theory,
E=hv=>E=hc/lambda
Now, using the law of conservation of energy : absorbed energy =emitted energy
hc/lambda=hc/lambda1+hc /lambda 2 =>1/lamdba=1/lambda1+1/lambda 2
=>1/355=1/680+1/lambda2
=>1/lmabda 2=1/133-1/680
=>lambda2=743nm | 262 | 753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2023-23 | latest | en | 0.706202 |
https://us.metamath.org/mpeuni/mmtheorems80.html | 1,713,748,076,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818067.32/warc/CC-MAIN-20240421225303-20240422015303-00165.warc.gz | 518,193,086 | 20,508 | Home Metamath Proof ExplorerTheorem List (p. 80 of 450) < Previous Next > Bad symbols? Try the GIF version. Mirrors > Metamath Home Page > MPE Home Page > Theorem List Contents > Recent Proofs This page: Page List
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Theorem List for Metamath Proof Explorer - 7901-8000 *Has distinct variable group(s)
TypeLabelDescription
Statement
Theoremottpos 7901 The transposition swaps the first two elements in a collection of ordered triples. (Contributed by Mario Carneiro, 1-Dec-2014.)
(𝐶𝑉 → (⟨𝐴, 𝐵, 𝐶⟩ ∈ tpos 𝐹 ↔ ⟨𝐵, 𝐴, 𝐶⟩ ∈ 𝐹))
Theoremrelbrtpos 7902 The transposition swaps arguments of a three-parameter relation. (Contributed by Mario Carneiro, 3-Nov-2015.)
(Rel 𝐹 → (⟨𝐴, 𝐵⟩tpos 𝐹𝐶 ↔ ⟨𝐵, 𝐴𝐹𝐶))
Theoremdmtpos 7903 The domain of tpos 𝐹 when dom 𝐹 is a relation. (Contributed by Mario Carneiro, 10-Sep-2015.)
(Rel dom 𝐹 → dom tpos 𝐹 = dom 𝐹)
Theoremrntpos 7904 The range of tpos 𝐹 when dom 𝐹 is a relation. (Contributed by Mario Carneiro, 10-Sep-2015.)
(Rel dom 𝐹 → ran tpos 𝐹 = ran 𝐹)
Theoremtposexg 7905 The transposition of a set is a set. (Contributed by Mario Carneiro, 10-Sep-2015.)
(𝐹𝑉 → tpos 𝐹 ∈ V)
Theoremovtpos 7906 The transposition swaps the arguments in a two-argument function. When 𝐹 is a matrix, which is to say a function from (1...𝑚) × (1...𝑛) to or some ring, tpos 𝐹 is the transposition of 𝐹, which is where the name comes from. (Contributed by Mario Carneiro, 10-Sep-2015.)
(𝐴tpos 𝐹𝐵) = (𝐵𝐹𝐴)
Theoremtposfun 7907 The transposition of a function is a function. (Contributed by Mario Carneiro, 10-Sep-2015.)
(Fun 𝐹 → Fun tpos 𝐹)
Theoremdftpos2 7908* Alternate definition of tpos when 𝐹 has relational domain. (Contributed by Mario Carneiro, 10-Sep-2015.)
(Rel dom 𝐹 → tpos 𝐹 = (𝐹 ∘ (𝑥dom 𝐹 {𝑥})))
Theoremdftpos3 7909* Alternate definition of tpos when 𝐹 has relational domain. Compare df-cnv 5562. (Contributed by Mario Carneiro, 10-Sep-2015.)
(Rel dom 𝐹 → tpos 𝐹 = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ⟨𝑦, 𝑥𝐹𝑧})
Theoremdftpos4 7910* Alternate definition of tpos. (Contributed by Mario Carneiro, 4-Oct-2015.)
tpos 𝐹 = (𝐹 ∘ (𝑥 ∈ ((V × V) ∪ {∅}) ↦ {𝑥}))
Theoremtpostpos 7911 Value of the double transposition for a general class 𝐹. (Contributed by Mario Carneiro, 16-Sep-2015.)
tpos tpos 𝐹 = (𝐹 ∩ (((V × V) ∪ {∅}) × V))
Theoremtpostpos2 7912 Value of the double transposition for a relation on triples. (Contributed by Mario Carneiro, 16-Sep-2015.)
((Rel 𝐹 ∧ Rel dom 𝐹) → tpos tpos 𝐹 = 𝐹)
Theoremtposfn2 7913 The domain of a transposition. (Contributed by NM, 10-Sep-2015.)
(Rel 𝐴 → (𝐹 Fn 𝐴 → tpos 𝐹 Fn 𝐴))
Theoremtposfo2 7914 Condition for a surjective transposition. (Contributed by NM, 10-Sep-2015.)
(Rel 𝐴 → (𝐹:𝐴onto𝐵 → tpos 𝐹:𝐴onto𝐵))
Theoremtposf2 7915 The domain and range of a transposition. (Contributed by NM, 10-Sep-2015.)
(Rel 𝐴 → (𝐹:𝐴𝐵 → tpos 𝐹:𝐴𝐵))
Theoremtposf12 7916 Condition for an injective transposition. (Contributed by NM, 10-Sep-2015.)
(Rel 𝐴 → (𝐹:𝐴1-1𝐵 → tpos 𝐹:𝐴1-1𝐵))
Theoremtposf1o2 7917 Condition of a bijective transposition. (Contributed by NM, 10-Sep-2015.)
(Rel 𝐴 → (𝐹:𝐴1-1-onto𝐵 → tpos 𝐹:𝐴1-1-onto𝐵))
Theoremtposfo 7918 The domain and range of a transposition. (Contributed by NM, 10-Sep-2015.)
(𝐹:(𝐴 × 𝐵)–onto𝐶 → tpos 𝐹:(𝐵 × 𝐴)–onto𝐶)
Theoremtposf 7919 The domain and range of a transposition. (Contributed by NM, 10-Sep-2015.)
(𝐹:(𝐴 × 𝐵)⟶𝐶 → tpos 𝐹:(𝐵 × 𝐴)⟶𝐶)
Theoremtposfn 7920 Functionality of a transposition. (Contributed by Mario Carneiro, 4-Oct-2015.)
(𝐹 Fn (𝐴 × 𝐵) → tpos 𝐹 Fn (𝐵 × 𝐴))
Theoremtpos0 7921 Transposition of the empty set. (Contributed by NM, 10-Sep-2015.)
tpos ∅ = ∅
Theoremtposco 7922 Transposition of a composition. (Contributed by Mario Carneiro, 4-Oct-2015.)
tpos (𝐹𝐺) = (𝐹 ∘ tpos 𝐺)
Theoremtpossym 7923* Two ways to say a function is symmetric. (Contributed by Mario Carneiro, 4-Oct-2015.)
(𝐹 Fn (𝐴 × 𝐴) → (tpos 𝐹 = 𝐹 ↔ ∀𝑥𝐴𝑦𝐴 (𝑥𝐹𝑦) = (𝑦𝐹𝑥)))
Theoremtposeqi 7924 Equality theorem for transposition. (Contributed by Mario Carneiro, 10-Sep-2015.)
𝐹 = 𝐺 tpos 𝐹 = tpos 𝐺
Theoremtposex 7925 A transposition is a set. (Contributed by Mario Carneiro, 10-Sep-2015.)
𝐹 ∈ V tpos 𝐹 ∈ V
Theoremnftpos 7926 Hypothesis builder for transposition. (Contributed by Mario Carneiro, 10-Sep-2015.)
𝑥𝐹 𝑥tpos 𝐹
Theoremtposoprab 7927* Transposition of a class of ordered triples. (Contributed by Mario Carneiro, 10-Sep-2015.)
𝐹 = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ 𝜑} tpos 𝐹 = {⟨⟨𝑦, 𝑥⟩, 𝑧⟩ ∣ 𝜑}
Theoremtposmpo 7928* Transposition of a two-argument mapping. (Contributed by Mario Carneiro, 10-Sep-2015.)
𝐹 = (𝑥𝐴, 𝑦𝐵𝐶) tpos 𝐹 = (𝑦𝐵, 𝑥𝐴𝐶)
Theoremtposconst 7929 The transposition of a constant operation using the relation representation. (Contributed by SO, 11-Jul-2018.)
tpos ((𝐴 × 𝐵) × {𝐶}) = ((𝐵 × 𝐴) × {𝐶})
2.4.12 Curry and uncurry
Syntaxccur 7930 Extend class notation to include the currying function.
class curry 𝐴
Syntaxcunc 7931 Extend class notation to include the uncurrying function.
class uncurry 𝐴
Definitiondf-cur 7932* Define the currying of 𝐹, which splits a function of two arguments into a function of the first argument, producing a function over the second argument. (Contributed by Mario Carneiro, 7-Jan-2017.)
curry 𝐹 = (𝑥 ∈ dom dom 𝐹 ↦ {⟨𝑦, 𝑧⟩ ∣ ⟨𝑥, 𝑦𝐹𝑧})
Definitiondf-unc 7933* Define the uncurrying of 𝐹, which takes a function producing functions, and transforms it into a two-argument function. (Contributed by Mario Carneiro, 7-Jan-2017.)
uncurry 𝐹 = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ 𝑦(𝐹𝑥)𝑧}
Theoremmpocurryd 7934* The currying of an operation given in maps-to notation, splitting the operation (function of two arguments) into a function of the first argument, producing a function over the second argument. (Contributed by AV, 27-Oct-2019.)
𝐹 = (𝑥𝑋, 𝑦𝑌𝐶) & (𝜑 → ∀𝑥𝑋𝑦𝑌 𝐶𝑉) & (𝜑𝑌 ≠ ∅) (𝜑 → curry 𝐹 = (𝑥𝑋 ↦ (𝑦𝑌𝐶)))
Theoremmpocurryvald 7935* The value of a curried operation given in maps-to notation is a function over the second argument of the original operation. (Contributed by AV, 27-Oct-2019.)
𝐹 = (𝑥𝑋, 𝑦𝑌𝐶) & (𝜑 → ∀𝑥𝑋𝑦𝑌 𝐶𝑉) & (𝜑𝑌 ≠ ∅) & (𝜑𝑌𝑊) & (𝜑𝐴𝑋) (𝜑 → (curry 𝐹𝐴) = (𝑦𝑌𝐴 / 𝑥𝐶))
Theoremfvmpocurryd 7936* The value of the value of a curried operation given in maps-to notation is the operation value of the original operation. (Contributed by AV, 27-Oct-2019.)
𝐹 = (𝑥𝑋, 𝑦𝑌𝐶) & (𝜑 → ∀𝑥𝑋𝑦𝑌 𝐶𝑉) & (𝜑𝑌𝑊) & (𝜑𝐴𝑋) & (𝜑𝐵𝑌) (𝜑 → ((curry 𝐹𝐴)‘𝐵) = (𝐴𝐹𝐵))
2.4.13 Undefined values
Syntaxcund 7937 Extend class notation with undefined value function.
class Undef
Definitiondf-undef 7938 Define the undefined value function, whose value at set 𝑠 is guaranteed not to be a member of 𝑠 (see pwuninel 7940). (Contributed by NM, 15-Sep-2011.)
Undef = (𝑠 ∈ V ↦ 𝒫 𝑠)
Theorempwuninel2 7939 Direct proof of pwuninel 7940 avoiding functions and thus several ZF axioms. (Contributed by Stefan O'Rear, 22-Feb-2015.)
( 𝐴𝑉 → ¬ 𝒫 𝐴𝐴)
Theorempwuninel 7940 The power set of the union of a set does not belong to the set. This theorem provides a way of constructing a new set that doesn't belong to a given set. See also pwuninel2 7939. (Contributed by NM, 27-Jun-2008.) (Proof shortened by Mario Carneiro, 23-Dec-2016.)
¬ 𝒫 𝐴𝐴
Theoremundefval 7941 Value of the undefined value function. Normally we will not reference the explicit value but will use undefnel 7943 instead. (Contributed by NM, 15-Sep-2011.) (Revised by Mario Carneiro, 24-Dec-2016.)
(𝑆𝑉 → (Undef‘𝑆) = 𝒫 𝑆)
Theoremundefnel2 7942 The undefined value generated from a set is not a member of the set. (Contributed by NM, 15-Sep-2011.)
(𝑆𝑉 → ¬ (Undef‘𝑆) ∈ 𝑆)
Theoremundefnel 7943 The undefined value generated from a set is not a member of the set. (Contributed by NM, 15-Sep-2011.)
(𝑆𝑉 → (Undef‘𝑆) ∉ 𝑆)
Theoremundefne0 7944 The undefined value generated from a set is not empty. (Contributed by NM, 3-Sep-2018.)
(𝑆𝑉 → (Undef‘𝑆) ≠ ∅)
2.4.14 Well-founded recursion
Syntaxcwrecs 7945 Declare syntax for the well-founded recursive function generator.
class wrecs(𝑅, 𝐴, 𝐹)
Definitiondf-wrecs 7946* Here we define the well-founded recursive function generator. This function takes the usual expressions from recursion theorems and forms a unified definition. Specifically, given a function 𝐹, a relationship 𝑅, and a base set 𝐴, this definition generates a function 𝐺 = wrecs(𝑅, 𝐴, 𝐹) that has property that, at any point 𝑥𝐴, (𝐺𝑥) = (𝐹‘(𝐺 ↾ Pred(𝑅, 𝐴, 𝑥))). See wfr1 7972, wfr2 7973, and wfr3 7974. (Contributed by Scott Fenton, 7-Jun-2018.) (New usage is discouraged.)
wrecs(𝑅, 𝐴, 𝐹) = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}
Theoremwrecseq123 7947 General equality theorem for the well-founded recursive function generator. (Contributed by Scott Fenton, 7-Jun-2018.)
((𝑅 = 𝑆𝐴 = 𝐵𝐹 = 𝐺) → wrecs(𝑅, 𝐴, 𝐹) = wrecs(𝑆, 𝐵, 𝐺))
Theoremnfwrecs 7948 Bound-variable hypothesis builder for the well-founded recursive function generator. (Contributed by Scott Fenton, 9-Jun-2018.)
𝑥𝑅 & 𝑥𝐴 & 𝑥𝐹 𝑥wrecs(𝑅, 𝐴, 𝐹)
Theoremwrecseq1 7949 Equality theorem for the well-founded recursive function generator. (Contributed by Scott Fenton, 7-Jun-2018.)
(𝑅 = 𝑆 → wrecs(𝑅, 𝐴, 𝐹) = wrecs(𝑆, 𝐴, 𝐹))
Theoremwrecseq2 7950 Equality theorem for the well-founded recursive function generator. (Contributed by Scott Fenton, 7-Jun-2018.)
(𝐴 = 𝐵 → wrecs(𝑅, 𝐴, 𝐹) = wrecs(𝑅, 𝐵, 𝐹))
Theoremwrecseq3 7951 Equality theorem for the well-founded recursive function generator. (Contributed by Scott Fenton, 7-Jun-2018.)
(𝐹 = 𝐺 → wrecs(𝑅, 𝐴, 𝐹) = wrecs(𝑅, 𝐴, 𝐺))
Theoremwfr3g 7952* Functions defined by well-founded recursion are identical up to relation, domain, and characteristic function. (Contributed by Scott Fenton, 11-Feb-2011.)
(((𝑅 We 𝐴𝑅 Se 𝐴) ∧ (𝐹 Fn 𝐴 ∧ ∀𝑦𝐴 (𝐹𝑦) = (𝐻‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑦)))) ∧ (𝐺 Fn 𝐴 ∧ ∀𝑦𝐴 (𝐺𝑦) = (𝐻‘(𝐺 ↾ Pred(𝑅, 𝐴, 𝑦))))) → 𝐹 = 𝐺)
Theoremwfrlem1 7953* Lemma for well-founded recursion. The final item we are interested in is the union of acceptable functions 𝐵. This lemma just changes bound variables for later use. (Contributed by Scott Fenton, 21-Apr-2011.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))} 𝐵 = {𝑔 ∣ ∃𝑧(𝑔 Fn 𝑧 ∧ (𝑧𝐴 ∧ ∀𝑤𝑧 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑧) ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐹‘(𝑔 ↾ Pred(𝑅, 𝐴, 𝑤))))}
Theoremwfrlem2 7954* Lemma for well-founded recursion. An acceptable function is a function. (Contributed by Scott Fenton, 21-Apr-2011.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))} (𝑔𝐵 → Fun 𝑔)
Theoremwfrlem3 7955* Lemma for well-founded recursion. An acceptable function's domain is a subset of 𝐴. (Contributed by Scott Fenton, 21-Apr-2011.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))} (𝑔𝐵 → dom 𝑔𝐴)
Theoremwfrlem3a 7956* Lemma for well-founded recursion. Show membership in the class of acceptable functions. (Contributed by Scott Fenton, 31-Jul-2020.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))} & 𝐺 ∈ V (𝐺𝐵 ↔ ∃𝑧(𝐺 Fn 𝑧 ∧ (𝑧𝐴 ∧ ∀𝑤𝑧 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑧) ∧ ∀𝑤𝑧 (𝐺𝑤) = (𝐹‘(𝐺 ↾ Pred(𝑅, 𝐴, 𝑤)))))
Theoremwfrlem4 7957* Lemma for well-founded recursion. Properties of the restriction of an acceptable function to the domain of another one. (Contributed by Scott Fenton, 21-Apr-2011.) (Revised by AV, 18-Jul-2022.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))} ((𝑔𝐵𝐵) → ((𝑔 ↾ (dom 𝑔 ∩ dom )) Fn (dom 𝑔 ∩ dom ) ∧ ∀𝑎 ∈ (dom 𝑔 ∩ dom )((𝑔 ↾ (dom 𝑔 ∩ dom ))‘𝑎) = (𝐹‘((𝑔 ↾ (dom 𝑔 ∩ dom )) ↾ Pred(𝑅, (dom 𝑔 ∩ dom ), 𝑎)))))
Theoremwfrlem5 7958* Lemma for well-founded recursion. The values of two acceptable functions agree within their domains. (Contributed by Scott Fenton, 21-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴 & 𝑅 Se 𝐴 & 𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))} ((𝑔𝐵𝐵) → ((𝑥𝑔𝑢𝑥𝑣) → 𝑢 = 𝑣))
Theoremwfrrel 7959 The well-founded recursion generator generates a relationship. (Contributed by Scott Fenton, 8-Jun-2018.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺) Rel 𝐹
Theoremwfrdmss 7960 The domain of the well-founded recursion generator is a subclass of 𝐴. (Contributed by Scott Fenton, 21-Apr-2011.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺) dom 𝐹𝐴
Theoremwfrlem8 7961 Lemma for well-founded recursion. Compute the prececessor class for an 𝑅 minimal element of (𝐴 ∖ dom 𝐹). (Contributed by Scott Fenton, 21-Apr-2011.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺) (Pred(𝑅, (𝐴 ∖ dom 𝐹), 𝑋) = ∅ ↔ Pred(𝑅, 𝐴, 𝑋) = Pred(𝑅, dom 𝐹, 𝑋))
Theoremwfrdmcl 7962 Given 𝐹 = wrecs(𝑅, 𝐴, 𝑋) ∧ 𝑋 ∈ dom 𝐹, then its predecessor class is a subset of dom 𝐹. (Contributed by Scott Fenton, 21-Apr-2011.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺) (𝑋 ∈ dom 𝐹 → Pred(𝑅, 𝐴, 𝑋) ⊆ dom 𝐹)
Theoremwfrlem10 7963* Lemma for well-founded recursion. When 𝑧 is an 𝑅 minimal element of (𝐴 ∖ dom 𝐹), then its predecessor class is equal to dom 𝐹. (Contributed by Scott Fenton, 21-Apr-2011.)
𝑅 We 𝐴 & 𝐹 = wrecs(𝑅, 𝐴, 𝐺) ((𝑧 ∈ (𝐴 ∖ dom 𝐹) ∧ Pred(𝑅, (𝐴 ∖ dom 𝐹), 𝑧) = ∅) → Pred(𝑅, 𝐴, 𝑧) = dom 𝐹)
Theoremwfrfun 7964 The well-founded function generator generates a function. (Contributed by Scott Fenton, 21-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴 & 𝑅 Se 𝐴 & 𝐹 = wrecs(𝑅, 𝐴, 𝐺) Fun 𝐹
Theoremwfrlem12 7965* Lemma for well-founded recursion. Here, we compute the value of the recursive definition generator. (Contributed by Scott Fenton, 21-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴 & 𝑅 Se 𝐴 & 𝐹 = wrecs(𝑅, 𝐴, 𝐺) (𝑦 ∈ dom 𝐹 → (𝐹𝑦) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑦))))
Theoremwfrlem13 7966* Lemma for well-founded recursion. From here through wfrlem16 7969, we aim to prove that dom 𝐹 = 𝐴. We do this by supposing that there is an element 𝑧 of 𝐴 that is not in dom 𝐹. We then define 𝐶 by extending dom 𝐹 with the appropriate value at 𝑧. We then show that 𝑧 cannot be an 𝑅 minimal element of (𝐴 ∖ dom 𝐹), meaning that (𝐴 ∖ dom 𝐹) must be empty, so dom 𝐹 = 𝐴. Here, we show that 𝐶 is a function extending the domain of 𝐹 by one. (Contributed by Scott Fenton, 21-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴 & 𝑅 Se 𝐴 & 𝐹 = wrecs(𝑅, 𝐴, 𝐺) & 𝐶 = (𝐹 ∪ {⟨𝑧, (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩}) (𝑧 ∈ (𝐴 ∖ dom 𝐹) → 𝐶 Fn (dom 𝐹 ∪ {𝑧}))
Theoremwfrlem14 7967* Lemma for well-founded recursion. Compute the value of 𝐶. (Contributed by Scott Fenton, 21-Apr-2011.)
𝑅 We 𝐴 & 𝑅 Se 𝐴 & 𝐹 = wrecs(𝑅, 𝐴, 𝐺) & 𝐶 = (𝐹 ∪ {⟨𝑧, (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩}) (𝑧 ∈ (𝐴 ∖ dom 𝐹) → (𝑦 ∈ (dom 𝐹 ∪ {𝑧}) → (𝐶𝑦) = (𝐺‘(𝐶 ↾ Pred(𝑅, 𝐴, 𝑦)))))
Theoremwfrlem15 7968* Lemma for well-founded recursion. When 𝑧 is 𝑅 minimal, 𝐶 is an acceptable function. This step is where the Axiom of Replacement becomes required. (Contributed by Scott Fenton, 21-Apr-2011.)
𝑅 We 𝐴 & 𝑅 Se 𝐴 & 𝐹 = wrecs(𝑅, 𝐴, 𝐺) & 𝐶 = (𝐹 ∪ {⟨𝑧, (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩}) ((𝑧 ∈ (𝐴 ∖ dom 𝐹) ∧ Pred(𝑅, (𝐴 ∖ dom 𝐹), 𝑧) = ∅) → 𝐶 ∈ {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))})
Theoremwfrlem16 7969* Lemma for well-founded recursion. If 𝑧 is 𝑅 minimal in (𝐴 ∖ dom 𝐹), then 𝐶 is acceptable and thus a subset of 𝐹, but dom 𝐶 is bigger than dom 𝐹. Thus, 𝑧 cannot be minimal, so (𝐴 ∖ dom 𝐹) must be empty, and (due to wfrdmss 7960), dom 𝐹 = 𝐴. (Contributed by Scott Fenton, 21-Apr-2011.)
𝑅 We 𝐴 & 𝑅 Se 𝐴 & 𝐹 = wrecs(𝑅, 𝐴, 𝐺) & 𝐶 = (𝐹 ∪ {⟨𝑧, (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩}) dom 𝐹 = 𝐴
Theoremwfrlem17 7970 Without using ax-rep 5189, show that all restrictions of wrecs are sets. (Contributed by Scott Fenton, 31-Jul-2020.)
𝑅 We 𝐴 & 𝑅 Se 𝐴 & 𝐹 = wrecs(𝑅, 𝐴, 𝐺) (𝑋 ∈ dom 𝐹 → (𝐹 ↾ Pred(𝑅, 𝐴, 𝑋)) ∈ V)
Theoremwfr2a 7971 A weak version of wfr2 7973 which is useful for proofs that avoid the Axiom of Replacement. (Contributed by Scott Fenton, 30-Jul-2020.)
𝑅 We 𝐴 & 𝑅 Se 𝐴 & 𝐹 = wrecs(𝑅, 𝐴, 𝐺) (𝑋 ∈ dom 𝐹 → (𝐹𝑋) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
Theoremwfr1 7972 The Principle of Well-Founded Recursion, part 1 of 3. We start with an arbitrary function 𝐺. Then, using a base class 𝐴 and a well-ordering 𝑅 of 𝐴, we define a function 𝐹. This function is said to be defined by "well-founded recursion." The purpose of these three theorems is to demonstrate the properties of 𝐹. We begin by showing that 𝐹 is a function over 𝐴. (Contributed by Scott Fenton, 22-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴 & 𝑅 Se 𝐴 & 𝐹 = wrecs(𝑅, 𝐴, 𝐺) 𝐹 Fn 𝐴
Theoremwfr2 7973 The Principle of Well-Founded Recursion, part 2 of 3. Next, we show that the value of 𝐹 at any 𝑋𝐴 is 𝐺 recursively applied to all "previous" values of 𝐹. (Contributed by Scott Fenton, 18-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴 & 𝑅 Se 𝐴 & 𝐹 = wrecs(𝑅, 𝐴, 𝐺) (𝑋𝐴 → (𝐹𝑋) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
Theoremwfr3 7974* The principle of Well-Founded Recursion, part 3 of 3. Finally, we show that 𝐹 is unique. We do this by showing that any function 𝐻 with the same properties we proved of 𝐹 in wfr1 7972 and wfr2 7973 is identical to 𝐹. (Contributed by Scott Fenton, 18-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴 & 𝑅 Se 𝐴 & 𝐹 = wrecs(𝑅, 𝐴, 𝐺) ((𝐻 Fn 𝐴 ∧ ∀𝑧𝐴 (𝐻𝑧) = (𝐺‘(𝐻 ↾ Pred(𝑅, 𝐴, 𝑧)))) → 𝐹 = 𝐻)
2.4.15 Functions on ordinals; strictly monotone ordinal functions
Theoremiunon 7975* The indexed union of a set of ordinal numbers 𝐵(𝑥) is an ordinal number. (Contributed by NM, 13-Oct-2003.) (Revised by Mario Carneiro, 5-Dec-2016.)
((𝐴𝑉 ∧ ∀𝑥𝐴 𝐵 ∈ On) → 𝑥𝐴 𝐵 ∈ On)
Theoremiinon 7976* The nonempty indexed intersection of a class of ordinal numbers 𝐵(𝑥) is an ordinal number. (Contributed by NM, 13-Oct-2003.) (Proof shortened by Mario Carneiro, 5-Dec-2016.)
((∀𝑥𝐴 𝐵 ∈ On ∧ 𝐴 ≠ ∅) → 𝑥𝐴 𝐵 ∈ On)
Theoremonfununi 7977* A property of functions on ordinal numbers. Generalization of Theorem Schema 8E of [Enderton] p. 218. (Contributed by Eric Schmidt, 26-May-2009.)
(Lim 𝑦 → (𝐹𝑦) = 𝑥𝑦 (𝐹𝑥)) & ((𝑥 ∈ On ∧ 𝑦 ∈ On ∧ 𝑥𝑦) → (𝐹𝑥) ⊆ (𝐹𝑦)) ((𝑆𝑇𝑆 ⊆ On ∧ 𝑆 ≠ ∅) → (𝐹 𝑆) = 𝑥𝑆 (𝐹𝑥))
Theoremonovuni 7978* A variant of onfununi 7977 for operations. (Contributed by Eric Schmidt, 26-May-2009.) (Revised by Mario Carneiro, 11-Sep-2015.)
(Lim 𝑦 → (𝐴𝐹𝑦) = 𝑥𝑦 (𝐴𝐹𝑥)) & ((𝑥 ∈ On ∧ 𝑦 ∈ On ∧ 𝑥𝑦) → (𝐴𝐹𝑥) ⊆ (𝐴𝐹𝑦)) ((𝑆𝑇𝑆 ⊆ On ∧ 𝑆 ≠ ∅) → (𝐴𝐹 𝑆) = 𝑥𝑆 (𝐴𝐹𝑥))
Theoremonoviun 7979* A variant of onovuni 7978 with indexed unions. (Contributed by Eric Schmidt, 26-May-2009.) (Proof shortened by Mario Carneiro, 5-Dec-2016.)
(Lim 𝑦 → (𝐴𝐹𝑦) = 𝑥𝑦 (𝐴𝐹𝑥)) & ((𝑥 ∈ On ∧ 𝑦 ∈ On ∧ 𝑥𝑦) → (𝐴𝐹𝑥) ⊆ (𝐴𝐹𝑦)) ((𝐾𝑇 ∧ ∀𝑧𝐾 𝐿 ∈ On ∧ 𝐾 ≠ ∅) → (𝐴𝐹 𝑧𝐾 𝐿) = 𝑧𝐾 (𝐴𝐹𝐿))
Theoremonnseq 7980* There are no length ω decreasing sequences in the ordinals. See also noinfep 9122 for a stronger version assuming Regularity. (Contributed by Mario Carneiro, 19-May-2015.)
((𝐹‘∅) ∈ On → ∃𝑥 ∈ ω ¬ (𝐹‘suc 𝑥) ∈ (𝐹𝑥))
Syntaxwsmo 7981 Introduce the strictly monotone ordinal function. A strictly monotone function is one that is constantly increasing across the ordinals.
wff Smo 𝐴
Definitiondf-smo 7982* Definition of a strictly monotone ordinal function. Definition 7.46 in [TakeutiZaring] p. 50. (Contributed by Andrew Salmon, 15-Nov-2011.)
(Smo 𝐴 ↔ (𝐴:dom 𝐴⟶On ∧ Ord dom 𝐴 ∧ ∀𝑥 ∈ dom 𝐴𝑦 ∈ dom 𝐴(𝑥𝑦 → (𝐴𝑥) ∈ (𝐴𝑦))))
Theoremdfsmo2 7983* Alternate definition of a strictly monotone ordinal function. (Contributed by Mario Carneiro, 4-Mar-2013.)
(Smo 𝐹 ↔ (𝐹:dom 𝐹⟶On ∧ Ord dom 𝐹 ∧ ∀𝑥 ∈ dom 𝐹𝑦𝑥 (𝐹𝑦) ∈ (𝐹𝑥)))
Theoremissmo 7984* Conditions for which 𝐴 is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 15-Nov-2011.) Avoid ax-13 2386. (Revised by Gino Giotto, 19-May-2023.)
𝐴:𝐵⟶On & Ord 𝐵 & ((𝑥𝐵𝑦𝐵) → (𝑥𝑦 → (𝐴𝑥) ∈ (𝐴𝑦))) & dom 𝐴 = 𝐵 Smo 𝐴
Theoremissmo2 7985* Alternate definition of a strictly monotone ordinal function. (Contributed by Mario Carneiro, 12-Mar-2013.)
(𝐹:𝐴𝐵 → ((𝐵 ⊆ On ∧ Ord 𝐴 ∧ ∀𝑥𝐴𝑦𝑥 (𝐹𝑦) ∈ (𝐹𝑥)) → Smo 𝐹))
Theoremsmoeq 7986 Equality theorem for strictly monotone functions. (Contributed by Andrew Salmon, 16-Nov-2011.)
(𝐴 = 𝐵 → (Smo 𝐴 ↔ Smo 𝐵))
Theoremsmodm 7987 The domain of a strictly monotone function is an ordinal. (Contributed by Andrew Salmon, 16-Nov-2011.)
(Smo 𝐴 → Ord dom 𝐴)
Theoremsmores 7988 A strictly monotone function restricted to an ordinal remains strictly monotone. (Contributed by Andrew Salmon, 16-Nov-2011.) (Proof shortened by Mario Carneiro, 5-Dec-2016.)
((Smo 𝐴𝐵 ∈ dom 𝐴) → Smo (𝐴𝐵))
Theoremsmores3 7989 A strictly monotone function restricted to an ordinal remains strictly monotone. (Contributed by Andrew Salmon, 19-Nov-2011.)
((Smo (𝐴𝐵) ∧ 𝐶 ∈ (dom 𝐴𝐵) ∧ Ord 𝐵) → Smo (𝐴𝐶))
Theoremsmores2 7990 A strictly monotone ordinal function restricted to an ordinal is still monotone. (Contributed by Mario Carneiro, 15-Mar-2013.)
((Smo 𝐹 ∧ Ord 𝐴) → Smo (𝐹𝐴))
Theoremsmodm2 7991 The domain of a strictly monotone ordinal function is an ordinal. (Contributed by Mario Carneiro, 12-Mar-2013.)
((𝐹 Fn 𝐴 ∧ Smo 𝐹) → Ord 𝐴)
Theoremsmofvon2 7992 The function values of a strictly monotone ordinal function are ordinals. (Contributed by Mario Carneiro, 12-Mar-2013.)
(Smo 𝐹 → (𝐹𝐵) ∈ On)
Theoremiordsmo 7993 The identity relation restricted to the ordinals is a strictly monotone function. (Contributed by Andrew Salmon, 16-Nov-2011.)
Ord 𝐴 Smo ( I ↾ 𝐴)
Theoremsmo0 7994 The null set is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 20-Nov-2011.)
Smo ∅
Theoremsmofvon 7995 If 𝐵 is a strictly monotone ordinal function, and 𝐴 is in the domain of 𝐵, then the value of the function at 𝐴 is an ordinal. (Contributed by Andrew Salmon, 20-Nov-2011.)
((Smo 𝐵𝐴 ∈ dom 𝐵) → (𝐵𝐴) ∈ On)
Theoremsmoel 7996 If 𝑥 is less than 𝑦 then a strictly monotone function's value will be strictly less at 𝑥 than at 𝑦. (Contributed by Andrew Salmon, 22-Nov-2011.)
((Smo 𝐵𝐴 ∈ dom 𝐵𝐶𝐴) → (𝐵𝐶) ∈ (𝐵𝐴))
Theoremsmoiun 7997* The value of a strictly monotone ordinal function contains its indexed union. (Contributed by Andrew Salmon, 22-Nov-2011.)
((Smo 𝐵𝐴 ∈ dom 𝐵) → 𝑥𝐴 (𝐵𝑥) ⊆ (𝐵𝐴))
Theoremsmoiso 7998 If 𝐹 is an isomorphism from an ordinal 𝐴 onto 𝐵, which is a subset of the ordinals, then 𝐹 is a strictly monotonic function. Exercise 3 in [TakeutiZaring] p. 50. (Contributed by Andrew Salmon, 24-Nov-2011.)
((𝐹 Isom E , E (𝐴, 𝐵) ∧ Ord 𝐴𝐵 ⊆ On) → Smo 𝐹)
Theoremsmoel2 7999 A strictly monotone ordinal function preserves the membership relation. (Contributed by Mario Carneiro, 12-Mar-2013.)
(((𝐹 Fn 𝐴 ∧ Smo 𝐹) ∧ (𝐵𝐴𝐶𝐵)) → (𝐹𝐶) ∈ (𝐹𝐵))
Theoremsmo11 8000 A strictly monotone ordinal function is one-to-one. (Contributed by Mario Carneiro, 28-Feb-2013.)
((𝐹:𝐴𝐵 ∧ Smo 𝐹) → 𝐹:𝐴1-1𝐵)
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Copyright terms: Public domain < Previous Next > | 14,336 | 29,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-18 | latest | en | 0.641911 |
http://marybourassa.blogspot.com/2016/04/quadratic-visual-patterns.html | 1,490,220,364,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186353.38/warc/CC-MAIN-20170322212946-00238-ip-10-233-31-227.ec2.internal.warc.gz | 232,314,201 | 16,554 | Sunday, 24 April 2016
You all know how much I love Fawn Nguyen's Visual Patterns site. I use them a LOT. They have been part of my warm-ups for years now and have been some of the best moments of my class each week. I have been recreating my warm-ups for my grade 10 applied class this semester (no, I can't leave things alone). I decided to do this so they align more with the curriculum expectations we are working on or provide lagged practice for other expectations. The warm-ups have included quadratic visual patterns for a few weeks now and I decided to step it up a little this past week with with a couple of patterns from Michael Fenton. If you haven't tried these ones before, I encourage you to do so before you scroll down.
We didn't actually work with the colour-coding, instead looked at the squares that overlapped by 1 each time. We worked with the number of circles first, established that this is a quadratic relationship and then found the rule by comparing the "side length" of each square to the step number.
I really also wanted to look at this pattern using the colours as a guide so we started over and found that we ended up with the same simplified rule.
I am totally impressed that some of my students can do these as they are not easy, especially for students who have struggled a lot with math and have trouble making connections. They have shown incredible progress and I love how willing they are to try.
Here is the next one we did:
There is a lot going on with this pattern, but the colours really help show the squares emerging.
I should note that these "warm-ups" took about 45 minutes to work through. It was definitely time well spent.
1 comment:
1. I've been thinking about visual patterns a lot this year and using them extensively in my Alg 1 class. In my experience, a visual pattern is a decent representation of how functions work, but an EXCELLENT framework for building fluency with algebraic expressions.
I ended up teaching my students how to multiply two binomials way before I intended to, mostly because different students would see the pattern differently, get different expressions, and want to know who was right. I needed to teach them how to multiply binomials to prove that multiple answers could all be correct! By the time we got to quadratics, the kids were already fluent.
I am going to use my visual patterns much more extensively in Pre-Algebra next year to help those kids build their fluency with variable expressions. | 526 | 2,494 | {"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} | 4.03125 | 4 | CC-MAIN-2017-13 | longest | en | 0.974621 |
http://www.houseoflearners.com/tutorials/gmat-question-practice | 1,553,192,932,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202530.49/warc/CC-MAIN-20190321172751-20190321194751-00547.warc.gz | 293,388,941 | 4,711 | Free Amazing GMAT Math Practice Questions
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Solving high difficulty level GMAT Sentence Correction questions
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GMAT Data Sufficiency Practice Questions | DS Tutorials | 5 GMAT Sample Questions
Solution to 5 GMAT Sample Questions | GMAT DS Tutorials | Online GMAT Course The question statement in each of these 5 questions is the same. However ... | 1,242 | 5,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-13 | longest | en | 0.807266 |
http://mathhelpforum.com/calculus/112783-true.html | 1,529,495,473,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863518.39/warc/CC-MAIN-20180620104904-20180620124904-00046.warc.gz | 197,462,041 | 10,724 | 1. ## is this true?
$\displaystyle cos(n\Pi)=(-1)^n$
2. Yes. Simply look at the unit circle and you should see this immediately.
3. Of course assuming that $\displaystyle n\in\mathbb{N}$. Otherwise, not so much. If you don't make that restriction you need to say that $\displaystyle \cos(n\pi)=\mathfrak{R}\left((-1)^n\right)$ which is really just a highfalutin way to say $\displaystyle \cos(n\pi)$. This is of course because $\displaystyle (-1)^n=e^{\pi\cdot i\cdot n}=\cos(n\pi)+ i\sin(n\pi)$.
4. I take it $\displaystyle sin(n\pi)=0$ then?
5. Originally Posted by billym
I take it $\displaystyle sin(n\pi)=0$ then?
You don't know that?
6. But when you touch me like this
And you hold me like that
I take it $\displaystyle sin(n\pi)=0$ then?
Whether or not that is true, what I said is still valid. To see this merely note that $\displaystyle \cos(x),\sin(x)$ are both mappings from $\displaystyle \mathbb{R}\mapsto\mathbb{R}$. So $\displaystyle \cos(n\pi),\sin(n\pi)$ is real for any $\displaystyle n\in\mathbb{R}$. Also, it should be apparent that if both $\displaystyle a,b\in\mathbb{R}$ that $\displaystyle \mathfrak{R}\left[a+bi\right]=a$. Consequently, it follows that $\displaystyle \cos(n\pi)=\mathfrak{R}\left[\left(-1\right)^n\right]=\mathfrak{R}\left[e^{n\pi i}\right]=\mathfrak{R}\left[\cos(n\pi)+i\sin(n\pi)\right]=\cos(n\pi)$ | 443 | 1,348 | {"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} | 3.9375 | 4 | CC-MAIN-2018-26 | latest | en | 0.709445 |
http://www.math.grin.edu/~rebelsky/Courses/CS151/2001S/Readings/let.html | 1,512,951,324,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948511435.4/warc/CC-MAIN-20171210235516-20171211015516-00355.warc.gz | 437,406,915 | 5,832 | # Naming Values with Local Bindings
So far we've seen three ways in which a value can be associated with a name in Scheme:
• The names of built-in procedures, such as `cons` and `quotient`, are predefined. When DrScheme starts up, these names are already bound to the procedures they denote.
• The programmer can introduce a new binding by means of a definition. A definition may introduce a new equivalent for an old name, or it may give a name to a newly constructed value.
• When a programmer-defined procedure is called, the parameters of the procedure are bound to the values of the corresponding arguments in the procedure call. Unlike the other two kinds of bindings, parameter bindings are local -- they apply only within the body of the procedure. Scheme discards these bindings when it leaves the procedure and returns to the point at which the procedure was called.
## Redundant Work
There are often times when it seems that you repeat work that should only have to be done once. For example, consider the problem of squaring the average of a list of numbers. We can write
```(* (/ (sum lst) (length lst))
(/ (sum lst) (length lst)))
```
But that's inefficent because we repeat the work of summing the list and computing the length of the list. How can we name the common computation so that we do it only once? If we rely only on the Scheme we know so far, we can write a helper function that takes the average as a parameter.
```;;; Procedure:
;;; square
;;; Parameters:
;;; val, a number
;;; Purpose:
;;; Compute val*val
;;; Produces:
;;; The result of the computation
;;; Preconditions:
;;; val must be a number
;;; Postconditions:
;;; The result is the same "type" of number as val (e.g., if
;;; val is an integer, so is the result; if val is exact,
;;; so is the result).
;;; Citations:
;;; Based on code created by John David Stone dated March 17, 2000
;;; and contained in the Web page
;;; http://www.math.grin.edu/~stone/courses/scheme/procedure-definitions.xhtml
;;; Changes to
;;; Parameter names
;;; Formatting
(define square
(lambda (val)
(* value val)))
```
Now, we can simply write
```(square (/ (sum lst) (length lst)))
```
In that case, `value` names the average of the sum of lst and the length of lst.
But that's a lot of extra work. It's inconvenient to have to write (and document!) a procedure that we're just going to use once.
## Let
Scheme provides `let` expressions as an alternative way to create local bindings. A `let`-expression contains a binding list and a body. The body can be any expression, or sequence of expressions, to be evaluated with the help of the local name bindings. The binding list is a pair of structural parentheses enclosing zero or more binding specifications; a binding specification, in turn, is a pair of structural parentheses enclosing a name and an expression.
Here's the general form of a `let` expression
```(let
((name1 exp1)
(name2 exp2)
...
(namen expn))
body1
body2
...
bodym)
```
When Scheme encounters a `let`-expression, it begins by evaluating all of the expressions inside its binding specifications. Then the names in the binding specifications are bound to those values. Next, the expressions making up the body of the `let`-expression are evaluated, in order. The value of the last expression in the body becomes the value of the entire `let`-expression. Finally, the local bindings of the names are cancelled. (Names that were unbound before the `let`-expression become unbound again; names that had different bindings before the `let`-expression resume those earlier bindings.)
Here's how we'd solve the earlier problem with `let`.
```(let ((average (/ (sum lst) (length lst))))
(* average average))
```
Here's another example of a binding list, taken from a `let`-expression in a real Scheme program:
```(let ((next (car source))
(stuff '()))
...)
```
This binding list contains two binding specifications -- one in which the value of the expression `(car source)` is bound to the name `next`, and the other in which the empty list is bound to the name `stuff`. Notice that binding lists and binding specifications are not procedure calls; their role in a `let`-expression simply to give names to certain values while the body of the expression is being evaluated. The outer parentheses in a binding list are ``structural,'' like the outer parentheses in a `cond`-clause -- they are there to group the pieces of the binding list together.
Using a `let`-expression often simplifies an expression that contains two or more occurrences of the same subexpression. The programmer can compute the value of the subexpression just once, bind a name to it, and then use that name whenever the value is needed again. Sometimes this speeds things up by avoiding such redundancies as the recomputation of values.
In other cases, there is little difference in speed, but the code may be a little clearer. For instance, consider the `remove-all` procedure that removes all copies of a value from a list. In the past, we might have written that procedure as follows.
```;;; Procedure:
;;; remove-all
;;; Parameters:
;;; item, a value
;;; ls, a list of values
;;; Purpose:
;;; Removes all copies of item from ls and its sublists.
;;; Produces:
;;; newls, a list
;;; Preconditions:
;;; ls is a list. It may be empty.
;;; Postconditions:
;;; No values equal to item appear in newls.
;;; Every value not equal to item that appeared in ls also
;;; appears in newls.
;;; Every value that appears in newls also appears in ls.
;;; If a preceded b in ls and neither a nor b equals item,
;;; then a precedes b in newls.
(define remove-all
(lambda (item ls)
(cond
; If the list is empty, removing the element still gives
; us the empty list
((null? ls) null)
; If the first element of the list matches, skip over it.
((equal? item (car ls))
(remove-all item (cdr ls)))
; Otherwise, preseve the first element and remove item
; from the remainder of ls
(else (cons (car ls) (remove-all item (cdr ls)))))))
```
Here is an alternative definition of the `remove-all` procedure which some people find clearer.
```(define remove-all
(lambda (item ls)
; If the list is empty, removing the element still gives
; us the empty list
(if (null? ls) null
(let (
; Name the car of the list first-element
(first-element (car ls))
; Recurse on the rest of the list and name it
; rest-of reulst
(rest-of-result (remove-all item (cdr ls))))
; If the first element of the list matches, skip over it.
(if (equal? first-element item)
rest-of-result
; Otherwise, preserve the first element and attach
; it to the rest.
(cons first-element rest-of-result))))))
```
## Sequencing Bindings with `let*`
Sometimes we may want to name a number of interrelated things. For example, suppose we wanted to square the average of a list of numbers (well, it's something that people do sometimes). Since computing the average involves summing values, we may want to name two different things: the total and the average (mean). We can nest one `let`-expression inside another to name both things.
```(let ((total (+ 8 3 4 2 7)))
(let ((mean (/ total 5)))
(* mean mean)))
```
One might be tempted to try to combine the binding lists for the nested `let`-expressions, thus:
```;; Combining the binding lists doesn't work!
(let ((total (+ 8 3 4 2 7))
(mean (/ total 5)))
(* mean mean))
```
This wouldn't work (try it and see!), and it's important to understand why not. The problem is that, within one binding list, all of the expressions are evaluated before any of the names are bound. Specifically, Scheme will try to evaluate both `(+ 8 3 4 2 7)` and `(/ total 5)` before binding either of the names `total` and `mean`; since `(/ total 5)` can't be computed until `total` has a value, an error occurs. You have to think of the local bindings coming into existence simultaneously rather than one at a time.
Because one often needs sequential rather than simultaneous binding, Scheme provides a variant of the `let`-expression that rearranges the order of events: If one writes `let*` rather than `let`, each binding specification in the binding list is completely processed before the next one is taken up:
```;; Using let* instead of let works!
(let* ((total (+ 8 3 4 2 7))
(mean (/ total 5)))
(* mean mean))
```
The star in the keyword `let*` has nothing to do with multiplication. Just think of it as an oddly shaped letter. It means "do things in sequence, rather than all at once". I have no idea why they've chosen to do that.
## Local Procedures
One can use a `let`- or `let*`-expression to create a local name for a procedure:
```(define hypotenuse-of-right-triangle
(let ((square (lambda (n)
(* n n))))
(lambda (first-leg second-leg)
(sqrt (+ (square first-leg) (square second-leg))))))
```
Regardless of whether `square` is defined outside this definition, the local binding gives it the appropriate meaning within the `lambda`-expression that describes what `hypotenuse-of-right-triangle` does.
## History
February 26, 1997 [John Stone]
• Created
March 17, 2000 [John Stone]
• Last revised.
2 October 2000 [Samuel A. Rebelsky]
Thursday, 22 February 2001 [Samuel A. Rebelsky]
• Some minor formatting changes.
• Added some more introductory text to the section on `let*`.
Friday, 23 February 2001 [Samuel A. Rebelsky]
• Added introductory stuff on naming via procedure calls.
• Extended the `remove-all` example.
This page may be found at `http://www.cs.grinnell.edu/~rebelsky/Courses/CS151/2001S/let.html`. | 2,275 | 9,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-51 | latest | en | 0.878402 |
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December 03, 2020, 08:30:31 am
To live close to great minds is the best kind of education. ..."John Buchan (1875~1940 Scotticsh historian, Governor General of Canada)"
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Author Topic: Pi = 3,14... and LORENZ's reduction (Read 2652 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
tesla2
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« Embed this message on: November 13, 2014, 05:25:40 pm » posted from:Warsaw,Mazowieckie,Poland
author
Engineer and Inventor Poland
http://tesla4.blogspot.com
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tesla2
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« Embed this message Reply #1 on: November 14, 2014, 01:18:15 am » posted from:Warsaw,Mazowieckie,Poland
Object can be wide/narrow
High /short
object-----------------Camera
>>> 30 km/s
^
^ 30 km/s
^
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« Embed this message Reply #2 on: November 14, 2014, 01:23:57 am » posted from:Warsaw,Mazowieckie,Poland
ENERGETIC PROBLEM FOR MICHELSON MORLEY
Light >>> air ----> 30km/s <<< Light
Light is able push objects
http://en.wikipedia.org/wiki/Solar_sail
HOW HEAVY IS AIR ( air = mass m )
How many jouls Light will lost to push air around Earth
30 km/s respect to table please evaluate below picture "now"
and 12 h later after 180 degrees Earth rotation .
How heavy is Mass m left /right ?
IN UNIVERSE EXIST MANY OBJECTS
MORE WIDE Q Equation ...
Q equation must contain "N"objects and "N" speed and "N" forces
Each Vector V can be exchange for many small vectors "Vn" ( 3D directions )
V = v1+v2+v3....................+vn
each v1,v2...vn represent perpendicular direction
to some object N in universe
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« Embed this message Reply #3 on: November 14, 2014, 01:27:20 am » posted from:Warsaw,Mazowieckie,Poland
Discovery : brightness of pictures West and East are not the same
Reason ? Eart's Velocity 30 km/s = 30 000 000 mm/s
( Camera SEE biger/ smaller bulb INVERSE SQUARE LAW respect to p1,2,p3...pn)
(NIKON 5000d remote start, zero outside light ,stative, manual set ,
time 10s , F 8 , Iso 200 - /10 cm to bulb / filtre is important !!!)
first pictures
( brightness - photoshop 10 histogram)
3 pictures west ( -30km/s ) and 3 pictures East (+30 km/s )
http://youtu.be/O9k-zidfJZg
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To live close to great minds is the best kind of education. ..."John Buchan (1875~1940 Scotticsh historian, Governor General of Canada)" | 810 | 2,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-50 | longest | en | 0.672232 |
https://encyclopedia2.thefreedictionary.com/Concave-up | 1,579,342,656,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250592394.9/warc/CC-MAIN-20200118081234-20200118105234-00446.warc.gz | 427,326,718 | 9,991 | # concave function
(redirected from Concave-up)
## concave function
[′kän‚kāv ′fəŋk·shən]
(mathematics)
A function f (x) is said to be concave over the interval a,b if for any three points x1, x2, x3 such that a <>x1<>x2<>x3<>b, f (x2)≥ L (x2), where L (x) is the equation of the straight line passing through the points [x1, f (x1)] and [x3, f (x3)].
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# 1
## Transient Simulations (Restart/Shut-in) Using
CSMFlow
Govind Hegde
September 3
rd
, 2014
2
Restart/Shut-in Simulations Plan
The ultimate goal is to simulate steady state simulations,
followed by a shut-in, followed by a restart and then
We are starting off with the simplest case of a restart
case
Currently facing some problems designing the
simulations using the Hydrodynamic Slug Flow model.
3
Restart Simulations
Previous simulations:
We have an initial U
SG(initial)
, U
SL(initial)
at the inlet of the flowline
U
M
= U
SG(initial)
+ U
SL(initial)
throughout the simulations
U
SG
, U
SL
change during simulation, but U
M
is constant
throughout
Hence, since gas-liquid slip U
S
(H
L
) depends only on the U
M
, it is
a constant function of H
L
4
Restart Simulations
Restart quiescent system with U
M
= 3 m/s, U
SL
= 2 m/s
At time t = 0, fluids in all cells have zero velocity
U
S
(H
L
) = 0
Time = 0 (Initial Conditions)
x = 1 x = 2 x = 3 x = i x = n-2 x = n-1 x = n
x = 0
U
M
= 0
U
SL
= 0
U
M
= 0
U
SL
= 0
U
M
= 0
U
SL
= 0
U
M
= 0
U
SL
= 0
U
M
= 0
U
SL
= 0
U
M
= 0
U
SL
= 0
U
M
= 0
U
SL
= 0
5
Restart Simulations
When we begin the restart simulation, the first cell has
U
M
= source U
M
How do we determine the U
M
of the intermediate cells in
order to calculate slip velocity?
In other words, how does the fluid entering the first cell
travel along the flowline
First timestep (dt=1s), Time = 1s
x = 1 x = 2 x = 3 x = i x = n-2 x = n-1 x = n
x = 0
U
M
= 3
U
SL
= 2
U
M
= ?
U
SL
= ?
U
M
= ?
U
SL
= ?
U
M
= ?
U
SL
= ?
U
M
= 0
U
SL
= 0
U
M
= 0
U
SL
= 0
U
M
= 0
U
SL
= 0
Restart quiescent system with U
M
= 3 m/s, U
SL
= 2 m/s
6
Restart Simulations: Possible Solution
Is it fine to assume that the liquid travels with constant
velocity U
SL
= 2
(or U
M
= 3?), such that the liquid travels
along the line with that velocity like a wave front, and
hence updating the cell to U
M
= 3?
According to this, if the front is assumed to propagate at
liquid superficial velocity, the third and fourth cells will
have U
M
= 3 in the 2
nd
time step.
Timestep = t, dt = 1 s
x = 1 x = 2 x = 3 x = i x = n-2 x = n-1 x = n
x = 0
U
M
= 3 U
M
= ? U
M
= ? U
M
= ? U
M
= 0 U
M
= 0 U
M
= 0
7
Restart Simulations: Questions
Is it fine to assume a constant propagation of wave front
as in the previous slide?
If not, is there a better way of doing this in order to
represent the restart case in a better way?
What other considerations do we have to keep in mind
while simulating restart cases (since we do not consider
pressure and temperature effect on flow) ? | 974 | 2,588 | {"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} | 3.265625 | 3 | CC-MAIN-2019-39 | latest | en | 0.874893 |
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1 HYPOTHESIS TESTING PHILOSOPHY 1 The Philosophy of Hypothesis Testing, Questions and Answers 2006 Samuel L. Baker Question: So I'm hypothesis testing. What's the hypothesis I'm testing? Answer: When you're testing a hypothesis on a regression coefficient, the hypothesis you're testing is that the true coefficient is equal to some value that you specify. For example, you often test the hypothesis that the true value of a coefficient is zero. Question: Wait a minute. I just did a regression and found that the slope coefficient, the coefficient of my X variable, is Isn't that the true value? Answer: Probably not, according to the classical philosophy of statistics. A classical statistician would say that the data you used in your regression were just one of many possible data sets that the true relationship might have produced. Even if the true value of the coefficient were , you'd have to be extraordinarily lucky to find a data set that would give you that number when you ran your regression. Question: I'd like to test the hypothesis that the true value of X's coefficient is zero. I find that the number in my regression output for the t-statistic is bigger than the number in the t table in the 0.05 column at the number of degrees of freedom my model has. Does this mean that I'm 95% sure that the true coefficient is not zero? Answer: It depends on what you mean by "95% sure." Question: Does it mean that the probability is 95% that the true coefficient is not zero? Some books t tables label the column you use to do a twotailed test at the 95% confidence level "0.025." Such a table is for a one-tailed t test. The critical value for a one-tailed test at the 97.5% level is the same as the critical value of a two-tailed test at the 95% level. Answer: Here's that fine point of statistical philosophy again. There's nothing probabilistic or random about the true value. For example, suppose that we want to know the average age of students in the Public Health School. That's a specific non-random number. We don't know what it is, though. If we estimate based on a random sample of students, then it's our estimated value that's random, because we get a different number depending on who happens to be in our sample. Question: That's OK for a random sample from a big population, but what if we have the whole population in our data? Suppose my dependent (Y) variable is the number of infant deaths in each county in South Carolina. Suppose my independent (X) variables are social and economic factors. Your theory has me treat my Y variable as if it were random. But, the number of infant deaths last year in Charleston County, for instance, is not a random number. DHEC counts it carefully, and it's based on the whole population of that county, not a sample. Answer: You have a point there. It's strange to think of reality as random while maintaining that something that exists only in theory, the true value, is absolute. But, that's the classical statistical philosophy. Plato would like it. Question: OK, classical statistical philosopher, what do I mean when I say that I'm 95% confident that some coefficient is not zero?
2 HYPOTHESIS TESTING PHILOSOPHY 2 Answer: You mean this: If the true value of the coefficient were zero, the probability is at least 95% that a regression would have given an estimated value for that coefficient that was closer to zero than the estimated value that you got. Question: That's quite a mouthful. Answer: Yes, and it's stated that way because it's the estimated value that's considered random, not the true value. Question: Still, it takes some work to follow the logic. Answer: Suppose you could run lots of regressions on different data sets, each of which was independently generated by the same true relationship in which the true coefficient of X is 0. (This means that truly there is no relationship.) The more regressions you run, the more likely it is that 95% of the t- statistics you get will be closer to 0 than the critical value in the t table. Question: By the way, what's so special about 95%? Why does everybody use it? Answer: It's just the conventional confidence level. It's reasonably conservative for many purposes. There are two types of errors of inference that you can make testing hypotheses. A Type I error (that's what it's actually called) is rejecting a hypothesis that is really true. A typical Type I error is declaring that a coefficient is not zero, but really it is. The 95% confidence level means that the probability of a Type I error is only 5%. Question: Why not choose a 99% confidence level, or a % confidence level, to make the probability of a Type I error very low? Answer: The lower the probability of a Type I error, the higher the probability of a Type II error. A Type II error is refusing to reject a hypothesis that is really false. This happens if the true coefficient is not zero, but you are unwilling to say it's not zero because your estimated coefficient was not far enough from zero to satisfy you. There is an unavoidable tradeoff here. The 95% confidence level is the usual compromise. Sometimes you'll use the 99% confidence level, if you're a lot more worried about a Type I error than a Type II error. This concern is not just academic. Consider the testing of the safety and efficacy of a new drug. Many lives and many dollars can ride on the selection of an appropriate confidence level.
3 HYPOTHESIS TESTING PHILOSOPHY 3 This table summarizes what Type I and Type II errors are. We do not reject the hypothesis. We say We do reject the hypothesis. Hypothesis is really True We are correct. Type I error False Type II error We are correct. The hypothesis can be true or false. We can refuse to reject the hypothesis, or we can reject the hypothesis. The table above shows the four combinations of those possibilities. If what we say matches the truth, then we are correct. Otherwise, we have made either a Type I or a Type II error. Question: Suppose I want to test the hypothesis that a coefficient equals some number other than zero? Answer: Subtract the hypothesized value from the estimated coefficient you got on your regression output. Divide that difference by the estimated standard error of that coefficient. Compare the quotient (ignoring the minus sign if you have one) with the critical value in the t table. If the quotient is bigger, reject the hypothesis. Question: What's a "confidence interval"? Answer: It's a short cut. It lets you avoid having to repeat the above calculation over and over. Once you compute the confidence interval, you know that you'd reject any hypothesized value that's outside the confidence interval, but not reject any hypothesized value inside the interval. Incidentally, you'll find that the 99% confidence interval is wider than the 95% confidence interval. The bigger confidence interval means a smaller chance of a Type I error, but a bigger chance of a Type II error. Question: Is the probability 95% that the true value lies inside the 95% confidence interval? Answer: Again, the answer is no, not according to classical statistics. The 95% confidence interval you calculate is considered random, because it depends on your data, which are presumed to be one of many possible realities. If, across lots of studies of different things, you make a practice of rejecting hypothesized values because they are outside the 95% confidence intervals, then the tendency is that you'll be right 95% of the time. Question: Classical statistical philosophy is awfully convoluted. Answer: Yes. There is an alternative approach, called Bayesian. This does allow you to apply the concept of probability to true values. However, the classical approach is by far the predominant one, so you should try to understand it so you can converse with other statisticians. Question: I can grind through the mechanics of hypothesis testing, but whole thing seems mysterious. What am I really doing?
4 HYPOTHESIS TESTING PHILOSOPHY 4 Answer: If you accept the idea that your data are the result of a random process, then any number ( statistic ) that you calculate from your data is also the result of the random process. If you make certain assumptions ( hypotheses ) about the random process that generated your data, then you can calculate the distribution of your statistic. You can calculate the probability that your statistic will take on any particular value or be inside any particular range of values. If your statistic is far from its expected value, such that being that far from the expected value is very unlikely, then you reject your hypothesis about how the data were generated. Here's how that works with a coin toss experiment. Flip the coin six times. You can then calculate a statistic, the total number of times it comes up heads in six flips. (This is not a complex calculation. You just count the number of heads and that's your statistic.) Now, hypothesize that the probability that the coin will come up heads is 0.5, and that the flips are independent. Based on your hypothesis -- those assumptions -- you can calculate the distribution of your statistic. It is: Possible values of the statistic, the number of heads in six coin tosses Probability that the statistic is one of those values, assuming that the hypothesis that the coin is fair is true 0 1/ / / / / / / All 1 1 Probability expressed as decimal This distribution has a fat middle, in the sense that the most likely values are towards the middle (2, 3, and 4). The least likely values are towards the tails (0 and 6). If our hypothesis is that the coin is fair, we want to reject that if our statistic is too far out towards either tail. That is, we want to be able to reject the hypothesis if we get either too many heads or too few. To do a two-tailed test, we combine the tails, like this:
5 HYPOTHESIS TESTING PHILOSOPHY 5 Possible values of the statistic, the number of heads in six coin tosses Table for a two-tailed test that the coin is fair Probability that the statistic is one of those values, assuming that the hypothesis that the coin is fair is true 0 or 6 2/64 = Probability that the statistic (the number of heads) is this far or further from its expected value (3). 1 or 5 12/64 = = or 4 30/64 = = the expected value 20/64 = All 1 1 Compare the value of your statistic (number of heads in your actual six tosses) with the table. If your statistic is 0 or 6, you have an outcome which is quite unlikely to happen -- less than a 0.05 probability -- if the hypothesis that the coin is fair is true. You therefore reject the hypothesis and say the coin is not fair. If your statistic is 1 or 5, you cannot reject the hypothesis, because your outcome is not unlikely enough. The probability of being that far from the expected value, 3 heads, is the probability of 1 or 5 plus the probability of 0 or 6, which is , for a total of This is greater than 0.05, your cutoff for rejecting an hypothesis. If you say that the coin is not fair, there is a probability that you are mistaken. That is too high a risk of a mistake for most researchers. The t-test calculation is more complicated, but the idea is the same. You hypothesize that the true coefficient is 0 (or whatever number you want) and you assume that the errors have the normal distribution. The hypothesis, plus the very strong assumption about the errors being normal, imply that the statistic you calculate from the data and call t has a known distribution that you can look up in a table. If, according to the table, the t value you got is very unlikely, then you can conclude that hypothesis is wrong. You reject the hypothesis. Question: We made two statements: (1) that the true coefficient is 0 (or whatever I wanted to test) and (2) that the errors have the normal distribution. When we reject a hypothesis, we always reject (1) and not (2). Why is that? If you aren t sure about (2) then you shouldn t use the t statistic for your hypothesis test. There are other tests that can help you determine whether (2) is appropriate.
6 T TESTS 6 More on T Tests You calculate a special number, called the t-statistic. You calculate it based on the assumption that a certain hypothesis is true. The hypothesis is usually that the true value of a certain one of your regression s coefficients is zero. What does it mean if the coefficient of a variable in a regression is zero? It means that you can leave that variable out of your equation without losing any predictive power. Consider a simple regression, of the form Y = " + \$X + error. If \$ is zero, the \$X term disappears, and the reduces to Y = " + error. Y does not depend on X. Knowing what X is does not help predict what Y is. To calculate the t-statistic for testing whether a particular \$ is zero, we divide our estimate of \$ by its standard error. This measures how steep the regression line is, relative to how far the observed Y values are from the regression line and how spread out the X values are. The t-statistic has no units of measurement. This means that the t-statistic comes out the same regardless of how X and Y are measured. For example, if X and Y are heights of people, as in Francis Galton s original study (to be mentioned in class), we get the same number for the t-statistic if we measure heights in inches or centimeters. Better than that, the t-statistic has a known distribution if the hypothesis that you are testing is true. The printed t-table shows what that distribution is. In any data set with two variables, X and Y, the data for the two variables will almost always correlate at least a little bit, even if there is no true underlying relationship. Suppose you have two rooms. In each room, you dump 1000 coins on the floor. Most likely, one of the rooms will have more heads than the other. A t-test can tell you if the difference is large enough to mean that there s really a difference between the rooms, or if the difference is just by luck. When we compare our calculated t-statistic with the critical value in the t-table, we are asking if the slope estimate we got might have happened just by luck. This could happen if the random errors correlate with X by luck. (Correlate means that there is a linear relationship.) When the errors correlate with X, then the Y values correlate with X, too. We can get fooled into thinking X and Y are related. If there is really no true relationship between X and Y, then random errors will probably work out so that the t-statistic will be close to zero. T-statistics far from zero are unlikely, unless the assumption about there being no true relationship between X and Y is wrong. If the data are such that higher X values are associated with higher Y values, then, when you do a regression with the equation Y = " + \$X + error, the t-statistic for the estimate of \$ will be positive. If lower X values are associated with higher Y values, then the t-statistic will be. The more that the X and Y data are correlated, and the steeper the line is that they seem to lie along, the further the t-statistic will be from zero, either positively or negatively. The table of t-statistic critical values has lots of rows. Theoretically, it could have an infinite number of rows, but my table shows only what will fit on one page. The multitude of rows in the table means that
7 T TESTS 7 there is a family of t distributions, one t distribution for each number of degrees of freedom. (The degrees of freedom is the number of observations minus the number of parameters in the regression equation). Each of these t distributions has a bell shape. When the number of degrees of freedom is small, the bell is wide, so the critical value numbers near the top of the t table are higher. When the number of degrees of freedom is large, the bell is narrower, and the critical value numbers in the t table are smaller. If the number of degrees of freedom is very large, the t distribution becomes indistinguishable from the normal distribution. The t distribution with 50 degrees of freedom. Shaded are the two tails that together have 5% of the total area under the curve. Reject the null hypothesis if the t-statistic is in either tail. Here is a t-distribution bell curve for 50 degrees of freedom. Strictly speaking, this curve is the density function for the t-distribution. T values are on the horizontal axis (here labeled ). The height of the bell curve over any t value is proportional to how likely it is that you would get something close to that t value purely by luck, with no true relationship between X and Y. The curve is highest above where the horizontal axis is 0, indicating that t-statistics near 0 are the most likely to happen by chance. As you move away from zero, in either the + or the - direction, the curve gets lower down. If you move far enough, you get to where a t value that far from zero would happen by luck only 5 times out of 100. Those places are the critical values of the t distribution at the 5% level. They are shown on this graph as the inside edges of the shaded areas. Because the t-statistic s bell curve is symmetrical left and right, The two places have the same number, except that one is positive and one is negative. This is the number that you find in a statistics book s t table, or in our t table in the downloadable file of tables. The portions of the t distribution that are to the right of the critical value in the positive direction, and to the left of the negative of the critical value in the negative direction, are called the tails of the distribution. These are shaded in the diagram above. The t distribution has two tails, one positive and one negative. We use two-tailed t tests in this course. We reject the null hypothesis that \$ is 0 if our calculated t statistic is in either the positive tail or the negative tail. This is appropriate whenever you are not sure in advance whether X is positively or negatively related to Y, so you want to allow for both possibilities. To put it another way, in a two-tailed test we reject the idea that the true \$ = 0 if our t statistic is greater than the critical t value or if it is less than the negative of the critical t value. That is the same as saying that we reject the hypothesis that \$ is 0 if the absolute value of our t-statistic is greater than the critical value. The t-test can be used to test for other possible coefficient values besides 0. The general formula is t-statistic = (\$^ - \$ test )/(standard error of \$^), where \$ test stands for the \$ value we want to test.
8 T TESTS 8 The null hypothesis is that the true \$ is \$ test. If this t-statistic is larger (in absolute value) than the critical value from the t table, reject the hypothesis that the true \$ is \$ test.
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Regression Analysis: A Complete Example This section works out an example that includes all the topics we have discussed so far in this chapter. A complete example of regression analysis. PhotoDisc, Inc./Getty | 9,903 | 45,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-09 | latest | en | 0.932883 |
http://mathhelpforum.com/algebra/18926-help-picture-print.html | 1,501,080,819,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426169.17/warc/CC-MAIN-20170726142211-20170726162211-00388.warc.gz | 203,507,681 | 2,930 | # Help with Picture!
• Sep 13th 2007, 05:10 PM
peachgal
Help with Picture!
http://i4.photobucket.com/albums/y109/mal1cous/math.jpg
The distance along a straight road is measured as shown in the diagram. If the distance between towns A and C is 67 Miles, find the distance between towns A and B.
• Sep 13th 2007, 05:17 PM
Plato
Quote:
Originally Posted by peachgal
The distance along a straight road is measured as shown in the diagram. If the distance between towns A and C is 67 Miles, find the distance between towns A and B.[/SIZE]
$\left( {5x - 2} \right) + \left( {2x + 6} \right) = 67$
• Sep 13th 2007, 05:23 PM
peachgal
Quote:
Originally Posted by Plato
$\left( {5x - 2} \right) + \left( {2x + 6} \right) = 67$
THANKS SO MUCH!!! I did this and I got 9? Is that right? What do I do from here?
• Sep 13th 2007, 05:44 PM
Jhevon
Quote:
Originally Posted by peachgal
THANKS SO MUCH!!! I did this and I got 9? Is that right? What do I do from here?
yes, x = 9. now what is the distance between A and B in terms of x?
• Sep 13th 2007, 05:48 PM
peachgal
Huh? DO you mean, 5(9)-2= 43?????
5x-2?
• Sep 13th 2007, 05:51 PM
Jhevon
Quote:
Originally Posted by peachgal
Huh? DO you mean, 5(9)-2= 43?????
yes. | 428 | 1,212 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-30 | longest | en | 0.928371 |
https://ee.gateoverflow.in/898/gate2018-34 | 1,579,296,785,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250591234.15/warc/CC-MAIN-20200117205732-20200117233732-00191.warc.gz | 421,869,103 | 12,965 | The number of roots of the polynomial, $s^7+s^6+7s^5+14s^4+31s^3+73s^2+25s+200$, in the open left half of the complex plane is
1. $3$
2. $4$
3. $5$
4. $6$ | 73 | 154 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-05 | longest | en | 0.701761 |
https://smallseotools.com/torque-converter/ | 1,726,622,030,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00515.warc.gz | 494,444,737 | 15,522 | # torque converter
To use the torque converter, write the value you want to convert in the “From” field after that select the unit in the “To” field, in which you want to convert. The results will be shown instantly.
From: Dyne-centimeter(dy cm)
To: Dyne-centimeter(dy cm)
## Easily Convert Torque Measurement Unit
If you are here, perhaps you already know what torque is. For those who are not familiar with it, it is the product of force and lever-arm distance that produces rotation. The SI unit for torque is the newton meter, which is dimensionally equal to a joule; however, torque is not expressed in joule.
A Torque converter is a tool that helps you convert units from a particular form of measurement to another form of measurement. It is a tool which is very useful to you. From knowing the rate at which your automatic vehicles work to helping it work and perform better, the usefulness of torque converters cannot be overemphasized.
Online Torque converters help you to generate and calculate the units produced by an engine to different or various units by merely inputting the units online and choosing the standard or form you want to convert it to. There are different torque units of measurement in torque conversion. The conversion could occur to and from Newton-meter, dyne-centimeter, IBF-inch to IBF-foot and krgf-meter.
Torque conversion calculators are the devices that help you transform a unit to another unit. It is highly indispensable in automatic vehicles. While most automatic cars of today come with a torque conversion calculator, you can also do your calculations online to arrive at the unit you desire. That is one of the usefulness of online torque converters.
## Torque Converter Calculator Real-Life Applications
We use torque in different aspects of our everyday lives. It is essential to determine the direction and force required to carry out a particular task. From driving a car or riding a bicycle to using a wrench, torque is involved in making so many things happening around us.
As a matter of fact, torque is bound to be produced, whenever there occurs a turning or rotating movement. Without us realizing, there are many applications of torque happening around us in our everyday life.
If you are in need to convert torque in different units, you can use a performance torque converter tool that will give you different options and accurate value.
### Torque Units Converter Tool– How to Use It
How can I calculate the units? You might ask!
Calculating the unit is relatively simple. There are different Torque measurement unit tools for different purposes. You can calculate the units using any of the online tools or applications. But if you are looking for the best Torque Calculator that will give you accurate conversions don’t look any further.
There are no complicated steps for you to follow; all you need to do is input the unit and select what measurement you want to convert it to. The calculator immediately converts and calculates it for you.
1. Starting by visiting https://smallseotools.com/unit-converter/ and select Torque Converter from the list of tools.
2. Select the input unit and enter the value into the “From” field that you want to convert. Similarly, select the unit in which you want to convert the value.
3. That’s it! You will get the converted value into the “To” field. | 687 | 3,367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-38 | latest | en | 0.923696 |
https://au.mathworks.com/matlabcentral/cody/problems/2869?s_tid=prof_contriblnk | 1,721,522,624,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517544.48/warc/CC-MAIN-20240720235600-20240721025600-00730.warc.gz | 98,465,672 | 21,870 | # Problem 2869. There are 10 types of people in the world
Those who know binary, and those who don't.
The number 2015 is a palindrome in binary (11111011111 to be exact) Given a year (in base 10 notation) calculate how many more years it will be until the next year that is a binary palindrome. For example, if you are given the year 1881 (palindrome in base 10! :-), the function should output 30, as the next year that is a binary palindrome is 1911. You can assume all years are positive integers.
Good luck!!kcul dooG
### Solution Stats
37.6% Correct | 62.4% Incorrect
Last Solution submitted on Jul 20, 2024
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 189 | 745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-30 | latest | en | 0.865121 |
http://mathhelpforum.com/algebra/184376-simplifying-algebraic-fractions.html | 1,526,939,249,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864544.25/warc/CC-MAIN-20180521200606-20180521220606-00403.warc.gz | 192,227,101 | 9,385 | 1. ## Simplifying algebraic fractions
Hello! I have a problem in my textbook that I am working through. It involves simplifying this:
$\displaystyle \frac{8(a^2 - b^2)}{3(a+2)} \div \frac{4(a-b)}{27(a+1)}$
I get as far as this...
$\displaystyle \frac{18(a^2 - b^2)(a+1)}{(a+2)(a-b)}$
Now, according to the book, I can simplify that to:
$\displaystyle \frac{18(a+b)(a+1)}{(a+2)}$
I can see that it works, but I'm not sure why. I don't know what rule they used to do that. Can you please explain why that worked?
2. ## Re: Simplifying algebraic fractions
Originally Posted by joebard
Hello! I have a problem in my textbook that I am working through. It involves simplifying this:
$\displaystyle \frac{8(a^2 - b^2)}{3(a+2)} \div \frac{4(a-b)}{27(a+1)}$
I get as far as this...
$\displaystyle \frac{18(a^2 - b^2)(a+1)}{(a+2)(a-b)}$
Now, according to the book, I can simplify that to:
$\displaystyle \frac{18(a+b)(a+1)}{(a+2)}$
I can see that it works, but I'm not sure why. I don't know what rule they used to do that. Can you please explain why that worked?
You are supposed to know that
$\displaystyle a^2-b^2 = (a+b)(a-b)$
so you can cancel the factor (a - b) | 384 | 1,175 | {"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} | 4.28125 | 4 | CC-MAIN-2018-22 | latest | en | 0.920217 |
https://www.nde-ed.org/Physics/Magnetism/MagneticFieldChar.xhtml | 1,713,913,523,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2024-18/segments/1712296818835.29/warc/CC-MAIN-20240423223805-20240424013805-00554.warc.gz | 815,304,820 | 7,270 | # Magnetic Field Characteristics
### After reading this section you will be able to do the following:
• Describe magnetic fields in different shaped magnets.
• Understand the important properties possessed by magnetic lines of force.
## Magnetic Field In and Around a Bar Magnet
As discussed previously, a magnetic field is a change in energy within a volume of space. The magnetic field surrounding a bar magnet can be seen in the magnetograph below. A magnetograph can be created by placing a piece of paper over a magnet and sprinkling the paper with iron filings. The particles align themselves with the lines of magnetic force produced by the magnet. The magnetic lines of force show where the magnetic field exits the material at one pole and reenters the material at another pole along the length of the magnet. It should be noted that the magnetic lines of force exist in three dimensions but are only seen in two dimensions in the image.
It can be seen in the magnetograph that there are poles all along the length of the magnet but that the poles are concentrated at the ends of the magnet. The area where the exit poles are concentrated is called the magnet's north pole and the area where the entrance poles are concentrated is called the magnet's south pole.
## Magnetic Fields in and around Horseshoe and Ring Magnets
Magnets come in a variety of shapes and one of the more common is the horseshoe (U) magnet. The horseshoe magnet has north and south poles just like a bar magnet but the magnet is curved so the poles lie in the same plane. The magnetic lines of force flow from pole to pole just like in the bar magnet. However, since the poles are located closer together and a more direct path exists for the lines of flux to travel between the poles, the magnetic field is concentrated between the poles.
If a bar magnet was placed across the end of a horseshoe magnet or if a magnet was formed in the shape of a ring, the lines of magnetic force would not even need to enter the air. The value of such a magnet where the magnetic field is completely contained with the material probably has limited use. However, it is important to understand that the magnetic field can flow in loop within a material. (See section on circular magnetism for more information).
## General Properties of Magnetic Lines of Force
Magnetic lines of force have a number of important properties, which include:
• They seek the path of least resistance between opposite magnetic poles. In a single bar magnet as shown to the right, they attempt to form closed loops from pole to pole.
• They never cross one another.
• They all have the same strength.
• Their density decreases (they spread out) when they move from an area of higher permeability to an area of lower permeability.
• Their density decreases with increasing distance from the poles.
• They are considered to have direction as if flowing, though no actual movement occurs.
• They flow from the south pole to the north pole within a material and north pole to south pole in air.
### Review:
1. Magnetic lines of force exist in three dimensions but are only seen in two dimensions in typical depictions.
2. Magnets come in a variety of shapes and one of the more common is the horseshoe (U) magnet.
3. Magnetic field lines never cross one another. | 671 | 3,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.948717 |
https://itprospt.com/num/6361328/the-function-f-k-belaw-cos-bz-cis-plotted-on-the-graph-shownwhat | 1,653,573,724,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662606992.69/warc/CC-MAIN-20220526131456-20220526161456-00532.warc.gz | 383,167,833 | 16,138 | 5
# The function f(k) belaw:cos(bz) + cis plotted on the graph shownWhat are the values of a=and...
## Question
###### The function f(k) belaw:cos(bz) + cis plotted on the graph shownWhat are the values of a=and
The function f(k) belaw: cos(bz) + cis plotted on the graph shown What are the values of a= and
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##### Find the general solution of the differential equationy"2x3 .
Find the general solution of the differential equation y" 2x3 .... | 1,914 | 6,104 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-21 | latest | en | 0.864809 |
https://freewallpaperspot.com/1001653/4th-grade-math-curriculum-california/ | 1,627,526,583,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153814.37/warc/CC-MAIN-20210729011903-20210729041903-00432.warc.gz | 269,785,826 | 13,095 | # 4th Grade Math Curriculum California
Number and Operations – Fractions. The math curriculum is also more advanced.
Fifth Grade I Can Standards Checklists For California Great For Tracking Student Growth Standar Math Standards Common Core Standards Checklist Fifth Grade
### Californias students are held to the same high expectations in mathematics as their national and global peers.
4th grade math curriculum california. Hold your mouse over the name of a skill to view a sample question. This course is aligned with Common Core standards. In online fourth grade classes students develop well-organized compositions explore the history of the United States investigate the forms of energy and apply basic math operations to fractions and decimals.
With IXLs state standards alignments you can easily find unlimited practice questions specifically tailored to each required standard. The California Department of Education CDE Instructional Quality Commission and State Board of Education are commencing the revision process for the Mathematics Framework for California Public Schools. Guidance to ensure success for struggling readers including English learners.
The State Board of Education SBE adopted the California Arts Standards for Public Schools Prekindergarten Through Grade Twelve Arts Standards on January 9 2019To download the publication version of the standards select the link below. Until then you can view a complete list of fourth-grade standards below. Until then you can view a complete list of fourth-grade standards below.
Skills available for California eighth-grade math standards Standards are in black and IXL math skills are in dark green. A complete list of the fourth-grade 1997 California English language arts standards is located on the CDE Content Standards Web page at. 4th Grade Math Curriculum Objectives In 4th grade math instruction should focus on number theory and systems algebraic thinking geometrical figures and objects measurement of length weight capacity time and temperature and data analysis and probability.
Grade 3 – 5 First 10 Days. FocusInstruction is focused on grade-level standards. Grade 4 Overview Operations and Algebraic Thinking.
Fourth Grade Math Curriculum Map. Use the four operations with whole numbers to solve problems. Kindergarten Through Grade Twelve Mathematics FrameworkInformation and updates concerning the revision of the Mathematics Framework will be.
Learn fourth grade matharithmetic measurement geometry fractions and more. Find fourth grade math worksheets and other learning materials for the Common Core State Standards. IXLs skills are aligned to the California Common Core Content Standards and the California Preschool Learning Foundations providing comprehensive coverage of math concepts and applications.
Click on the name of a skill to practice that skill. 2021 Revision of the Mathematics Framework. Although California educators have implemented standards before the CA CCSSM require not only rigorous curriculum and instruction but also conceptual understanding procedural skill and fluency and the ability to apply mathematics.
Browse the fourth grade classes available in our California online schools by clicking on the course descriptions below. Skills available for California fourth-grade science standards IXLs fourth-grade skills will be aligned to the Next Generation Science Standards NGSS soon. Generate and analyze patterns.
Gain familiarity with factors and multiples. This curriculum policy replaces The Ontario Curriculum Grades 18. EnVision MATH California alignment with Common Core State Standards.
Number and Operations in Base Ten. Generalize place value understanding for multi-digit whole numbers. 4th graders will practice tougher multiplication and long division problems learn about angles and lines get introduced to simple algebra as they solve problems with variables and learn to convert different units of measurement.
Curriculum frameworks provide guidance for implementing the content standards adopted by the State Board of Education SBE. California Mathematics Framework Grade Four 191 Standards for Mathematical Content The Standards for Mathematical Content emphasize key content skills and practices at each grade level and support three major principles. A complete list of the fourth-grade CCSS for English language arts with California additions can be found at the end of this section.
Mathematics 2005Beginning in September 2020 all mathematics programs for Grades 1 to 8 will be based on the expectations outlined in this curriculum policy. Operations and Algebraic Thinking. Some of the specific 4th grade math learning targets include.
Number and Operations in Base Ten. Skills available for California fourth-grade math standards IXLs fourth-grade skills will be aligned to the California Common Core Content Standards soon. Our grade 4 math worksheets help build mastery in computations with the 4 basic operations delve deeper into the use of fractions and decimals and introduce the concept of factors.
Frameworks are developed by the Instructional Quality Commission formerly known as the Curriculum Development and Supplemental Materials Commission which also reviews and recommends textbooks and other instructional materials to be adopted by the SBE. Grade 4 math worksheets from K5 Learning.
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#### Author: Bertha Natasha
My name is Bertha Natasha, also known as Bertha, and I am USA, although I currently live in Seattle, WA. Many of my recipes are inspired by healthy food, by my mom's American-Mexican cuisine, who grew up in New Mexico, and also by my granny's kitchen in Texas. | 1,390 | 7,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-31 | latest | en | 0.907604 |
https://www.studypool.com/discuss/466621/help-me-answer-my-math-homework-please?free | 1,481,338,632,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542938.92/warc/CC-MAIN-20161202170902-00429-ip-10-31-129-80.ec2.internal.warc.gz | 1,019,426,656 | 13,928 | Mathematics Tutor: None Selected Time limit: 1 Day
what property of addition is 2+3=3+2?
commutative, associative, or identity.
Apr 8th, 2015
This is called the associative property. It also works for multiplication a * b = b * a where a and b are real numbers and * indicates multiplication.
Apr 8th, 2015
...
Apr 8th, 2015
...
Apr 8th, 2015
Dec 10th, 2016
check_circle | 118 | 377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.925621 |
https://www.physicsforums.com/threads/underpressure-in-a-pot.853971/ | 1,508,199,139,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820487.5/warc/CC-MAIN-20171016233304-20171017013304-00229.warc.gz | 1,188,539,818 | 16,528 | # Underpressure in a pot
1. Jan 25, 2016
### mrlac1212
• Member advised to use the homework template for posts in the homework sections of PF.
A pot is closed tightly with a lid when the temperature of the air in the pot is 80 oC. The temperature of the system then drops to 15 oC. Find the force needed to pull the lid and the pot apart. The diameter of the pot is 20 cm, and the atmospheric pressure is 1,0 bar.
The ANSWER is F = 580 N but I dont know how to get it, I got F = Area x Pressure but so confused about the final pressure.
2. Jan 25, 2016
### Staff: Mentor
Are you familiar with the ideal gas law?
3. Jan 25, 2016
### mrlac1212
yeah right, it in Ideal Gas chapter, with nR = pV/T
4. Jan 25, 2016
### Staff: Mentor
You can find the final pressure in the pot by using this equation. | 232 | 807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-43 | longest | en | 0.910712 |
https://brainmass.com/economics/output-and-costs/marginal-cost-marginal-revenue-383122 | 1,580,314,729,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251799918.97/warc/CC-MAIN-20200129133601-20200129163601-00516.warc.gz | 352,678,355 | 11,984 | Explore BrainMass
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# Marginal Cost & Marginal Revenue
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
The data below are for a competitive business (price-taker).
Output Average Average Average
Fixed Cost Variable Cost Total Costs Marginal Cost
0 \$3,000
1 \$3,000 \$900 \$3,900 \$900
2 \$1,500 \$800 \$2,300 \$700
3 \$1,000 \$700 \$1,700 \$500
4 \$750 \$600 \$1,350 \$300
5 \$600 \$500 \$1,100 \$100
6 \$500 \$550 \$1,050 \$800
7 \$429 \$600 \$1,029 \$900
8 \$375 \$650 \$1,025 \$1,000
9 \$333 \$700 \$1,033 \$1,100
10 \$300 \$750 \$1,050 \$1,200
If the market price is \$1,100 per unit, the company should:
a. six units with profit of \$300.
b. six units with profit of \$3,300.
c. nine units with profit of \$600.
d. nine units with profit of \$3,600.
Thinking it either C or D since that is where MC=MR. But don't know how to calculate the profit! | 290 | 925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-05 | latest | en | 0.798658 |
http://www.ida.liu.se/~732A40/info/courseinfo.en.shtml | 1,432,792,411,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929256.27/warc/CC-MAIN-20150521113209-00154-ip-10-180-206-219.ec2.internal.warc.gz | 521,409,532 | 5,668 | 732A40 Probability Theory
Course information
Course contents:
The course aims to provide the student with a solid understanding of basic results and methods in probability. The topics covered include: major classes of probability distributions, multivariate random variables, conditioning, transforms, order statistics, multivariate normal distributions, and convergence concepts.
Course organisation:
The teaching is comprised of lectures and seminars.
The lectures are devoted to presentations of theories, concepts, and methods.
The seminars are devoted to presentation and discussion of assignments
Examination:
A written final exam. See further under exam.
Course literature:
Gut, A. An intermediate course in probability. 2nd ed. Springer-Verlag, New York, 2009. ISBN 978-1-4419-0161-3
Misprints and corrections may be read at http://www2.math.uu.se/~allan/81misprints.pdf
Course start:
The course starts in late August, but the first lecture will not be held until Wednesday September 3, 2015 at 3.15 in Thomas Bayes
The time before the first lecture should be spent on self-study to fresh up your knowledge in probability theory from previous course(s). My slides (in English) from our third-year bachelor course may be of some help (Slides 1-6 are about probability theory)
Before the first lecture, students should be familiar with at least the following concepts:
• Probabilities of events
• The basic laws of probabilities
• Bayes theorem for events
• Random variables
• Probability distributions
• The most common distributions: normal, chi-square, student t, binomial and poisson.
• Expected value and variance
• Covariance and correlation
• Linear regression
Prerequisities:
Basic skills in probability theory corresponding to an introductory course in statistics.
Calculus (corresponding to a first course)
Useful material:
• The course material is maintained on GitHub. Here is the repository for the course.
• Table with common integrals
• Exams with solutions from a course at KTH. But note that not all exam questions are relevant for this course, in particular the ones for stochastic processes (which we do not cover in this course).
• Wikipedia's page on complex numbers is quite good and compact.
• For whatever it is worth, here are my code snippets for the course.
• CRAN list of distributions in R.
Page responsible: Mattias Villani
Last updated: 2014-09-04 | 510 | 2,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22 | longest | en | 0.874156 |
https://math.answers.com/Q/The_area_of_a_square_is_25_square_yards_find_the_length_of_a_side | 1,708,918,859,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474650.85/warc/CC-MAIN-20240226030734-20240226060734-00332.warc.gz | 387,831,732 | 47,027 | 0
# The area of a square is 25 square yards find the length of a side?
Updated: 10/4/2022
Wiki User
14y ago
Best Answer
All four sides of a square are of equal length. To find the area, you square the length of one side. To find the length of one side from the given area, find the square root of the area. In this case, the square root of 25 is 5. Hence, the length of a side is 5 yards.
Wiki User
14y ago
This answer is:
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Related questions
5 yards
### Find the area of a rectangle that has the length of four yards and a width of 3 yards?
Area = Length x Width so... 4 yards x 3 yards = 12 square yards
### How to find the square yard?
If you mean how do you calculate the square yardage of an area, you multiply the length by the width. If an area is 5 yards long and 5 yards wide, the square yardage is 25 square yards.
### How do you find the area of square whose length is 1 and three quarter yards and width is 2 yards?
You are describing a rectangle NOT a square!
### How do you find square yards if measurements in yards?
Multiply the length times the width.
### The area of a gymnasium floor is 264 square yards the floor is 11yards wide how long is the floor?
The formula for the area of a rectangle is length times width. So given the area, to find the length, divide the area by the width. In this case, 264/11 = 24 yards.
### Find the area of a circle with radius 5 yards?
Area = 78.5 square yards.
9 yards
### How do you find area of a square?
To find the area of a square you multiply its length by its width.
### When the area of a square is 1300 square yards find the dimensions of the square?
the side is 36.055 yards. in order to find the sides of a square with only the area you would need to find the square-root of that data seeing as the sides of the square are identical and are multiplied by each other to get the area.
### How do you find area of a square that has sides of length 96 inches?
how do you find area of a square that has sides of length 96 inches
### How do you find length from area of square?
Take the square root of the area. | 564 | 2,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-10 | latest | en | 0.949681 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition/chapter-7-trigonometric-identities-and-equations-7-7-equations-involving-inverse-trigonometric-functions-7-7-exercises-page-728/10 | 1,545,026,339,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828318.79/warc/CC-MAIN-20181217042727-20181217064727-00538.warc.gz | 867,488,041 | 12,423 | ## Precalculus (6th Edition)
$x=2 \displaystyle \arcsin\frac{y}{3}$
The goal is to isolate x. Divide with 3... $\displaystyle \frac{y}{3}=\sin\frac{x}{2}$ $\qquad$ ...use the definition of arcsin $\displaystyle \frac{x}{2}=\arcsin\frac{y}{3}$ ($\displaystyle \frac{x}{2}$ is the number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ such that $\displaystyle \sin\frac{y}{3}=\frac{x}{2}$) (...If $\displaystyle \frac{x}{2}\in(-\frac{\pi}{2}, \displaystyle \frac{\pi}{2})$, then $x\in(-\pi, \pi)$...) ... isolate the x by multiplying with 2 ... $x=2 \displaystyle \arcsin\frac{y}{3}$ | 219 | 602 | {"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} | 4.03125 | 4 | CC-MAIN-2018-51 | latest | en | 0.395736 |
https://www.wyzant.com/resources/answers/421613/ideal_gas_law_physics | 1,632,168,720,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00682.warc.gz | 1,089,543,391 | 14,485 | Sav J.
# ideal gas law - physics
A monatomic ideal gas expands from point A to point B along the path shown in the figure. What is the change in internal energy of the gas during the process? (The horizontal axis is marked in increments of 1.50 m³.)
(the image of the graph looks like this, but change the increments on the x axis to increments of 1.50 m3 instead of 2 m3: http://www.webassign.net/CJ/p15-24.gif )
By:
Tutor
5.0 (66)
Experienced Physics Teacher for Physics Tutoring
Sav J.
So the pressure for both Pa and Pb are going to be 2*10^5 Pa and Va = 1.5mand Vb = 7.5m3 right?
But would that mean the point made in the graph has no effect?
Im getting an answer of 1.8*10^6 J. Does that seem right?
Report
11/20/17
Arturo O.
For those values of pressure and volume, yes.
Report
11/20/17
Sav J.
Can you explain to me why the middle changes in the graph does take part to effect this problem in anyway? The pressure raising and lowering doesn't matter, only the starting and ending point?
Report
11/20/17
Arturo O.
If the amount of gas is constant, the internal energy of an ideal gas depends only on the temperature of the gas. That is one of the properties of an ideal gas. So the change in internal energy depends only on the change in temperature, and the change in temperature is just the difference between the final and initial temperatures. In addition, recall from thermodynamics that internal energy is an example of a state function. The change in a state function depends only on the initial and final conditions. It does not depend on the path that was followed from one state to the other. An example of a function that is not a state function is work. The work done in going from state A to state B depends on the path from A to B. But the change in the internal energy of an ideal going from from state A to state B depends only on the temperatures of states A and B, as long as the amount of gas is constant.
Report
11/20/17
Sav J.
Oh I see. I'm remembering a lot of these rules from chemistry now. Thank you for all the help!
Report
11/20/17
Arturo O.
You are welcome, Sav.
Report
11/20/17
## Still looking for help? Get the right answer, fast.
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Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 600 | 2,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-39 | latest | en | 0.91805 |
http://physicsteacher.in/2018/09/28/friction-definition/ | 1,568,814,581,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573289.83/warc/CC-MAIN-20190918131429-20190918153429-00286.warc.gz | 151,029,200 | 15,105 | # Friction definition
Friction / Friction definition: If we either slide or attempt to slide a body over a surface, the motion is resisted by a bonding (cold-weld) between the body and the surface.
The resistance is considered to be a force called either the frictional force or simply friction.
This force is directed along the surface, opposite to the direction of the intended motion.
## Friction caused by contact forces – how? – friction definition
The following diagram will help us to understand how contact forces between two surfaces give birth to friction.
See, there is a block on a surface.
If we intend to move the block towards left from right, then the block will exert a downward push (towards SouthWest in this case) on the surface which makes an angle with surface.
Now the surface will also apply an equal and opposite force on the block following Newton’s 3rd law of motion.
We have shown the reaction force as F (contact force in the diagram acting upwards towards North East).
Now if we resolve F along the vertical and horizontal direction, then we find the horizontal component to act opposite to the direction of the intended motion of the block.
So obviously this force is resistive to the intended motion of the block and hence we call it Friction or Frictional force. (labelled in the diagram above)
## Relative motion of the surface below and friction
Now as the block tends to move here from right to left, the surface below again tends to have a relative motion from left to right.
So will that relative motion will also face a resistive friction?
Yes.
How? See here.
Now again we can resolve the downward force exerted on the surface by the block.
(This downward force is acting downwards towards SW, we mentioned about this one in previous section as well)
After resolving we see that, the horizontally resolved component is directing towards left.
This is opposite to the tentative relative motion of the surface.
So this is again a frictional force. But this time it acts on the surface below to resist its relative motion.
What we covered here: This topic answers queries like definition of friction, what is friction, how friction happens between 2 surfaces etc | 446 | 2,218 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-39 | latest | en | 0.941518 |
https://www.calc-online.xyz/distance-between-two-points | 1,721,571,583,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00653.warc.gz | 606,890,732 | 7,520 | # Distance between two points
$\phantom{\rule{0ex}{0ex}}A\left({x}_{a};{y}_{a}\right)\phantom{\rule{0ex}{0ex}}B\left({x}_{b};{y}_{b}\right)\phantom{\rule{0ex}{0ex}}{d}_{AB}=\sqrt{\left({x}_{b}-{x}_{a}{\right)}^{2}+\left({y}_{b}-{y}_{a}{\right)}^{2}}$
A ( ; ) B ( ; )
### Help
To calculate the distance between points A and B just follow the steps below:
• Fill in the form with the coordinates of the points
• Click on the calculate button
Calculator of distance between two points A (xa; ya) and B (xb; yb). | 178 | 514 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-30 | latest | en | 0.687966 |
https://math.answers.com/basic-math/What_is_.015_as_a_fraction | 1,726,148,137,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00337.warc.gz | 349,575,438 | 47,449 | 0
# What is .015 as a fraction?
Updated: 4/28/2022
Wiki User
9y ago
.015 as a fraction is 15/1000. In simplified form it is 3/200.
Wiki User
9y ago
Wiki User
10y ago
-3/20 = -0.15
Wiki User
15y ago
3/20
Anonymous
Lvl 1
4y ago
1.5 | 100 | 245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-38 | latest | en | 0.900869 |
https://healthyalgorithms.com/2010/08/18/applied-approximate-counting-malaria/ | 1,685,878,533,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649741.26/warc/CC-MAIN-20230604093242-20230604123242-00727.warc.gz | 328,400,992 | 24,438 | # Applied Approximate Counting: Malaria
My first first-authored global health paper came out today (I consider it my first “first-authored” paper ever, since the mathematicians I’ve worked with deviantly list authorship in alphabetical order regardless of seniority and contribution). It’s a bit of a mouthful by title: Rapid Scaling Up of Insecticide-Treated Bed Net Coverage in Africa and Its Relationship with Development Assistance for Health: A Systematic Synthesis of Supply, Distribution, and Household Survey Data.
What I find really pleasing about this research paper is the way it continues research I worked on in graduate school, but in a completely different and unexpected direction. Approximate counting is something that my advisor specialized in, and he won a big award for the random polynomial time algorithm for approximating the volume of convex bodies. I followed in his footsteps when I was a student, and I’m still doing approximate counting, it’s just that now, instead of approximating the amount of high-dimensional sand that will fit in an oddly shaped high-dimensional box, I’ve been approximating the number of insecticide-treated bednets that have made it from manufacturers through the distribution supply-chain and into the households of malaria-endemic regions of the world. I’m even using the same technique, Markov-chain Monte Carlo.
I’ve been itching to write about the computational details of this research for a while, and now that the paper’s out, I will have my chance. But for today, I refer you to the PLoS Med paper, and the technical appendix, and the PyMC code on github.
Filed under global health, MCMC
### 4 responses to “Applied Approximate Counting: Malaria”
1. Anand Patil
Abie,
Very nice! This is a great example of the benefits of thoughtful, coherent data fusion in large-scale health assessments. And I’m sure it wasn’t easy to fit… congratulations!
I think I understand everything except the $\eta$ parameter. What does it mean if it’s less than one? Also, why aren’t the household surveys on LLIN stocks affected by it?
Anand
2. Thanks Anand!
$\eta$ and $\alpha$ parameterize my mapping from stock to ownership coverage. It comes from a negative binomial model (and uses the PyMC-preferred parameters). $\eta$ has an interpretation as a sort of “household size factor”, so a value less than one would indicate that something is wrong… I expect posterior values between 3 and 6.
The $\alpha$ parameter is interesting from an inequality perspective. It says something about how closely the nets-per-household matches a Poisson distribution. But I think it’s not safe to draw any conclusions from the data currently available.
Of course, if you really want to know what I’m doing with $\eta$, you can use the source. It’s here, in the formula for a negative binomial r.v. being non-zero.
3. So pretty Abie. I still owe you a drink for a proper toast to your first first authorship! | 631 | 2,951 | {"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": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.933358 |
http://steamcommunity.com/sharedfiles/filedetails/comments/125043795?l=russian | 1,513,292,518,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948551162.54/warc/CC-MAIN-20171214222204-20171215002204-00568.warc.gz | 252,859,178 | 14,399 | Portal 2
Portal 2
Retrocede
Комментариев: 22
bullfrog 18 мая в 0:24
Clever map, and that's a new one on me :). I agree with @Ifairban; I solved it by "one of those, "It's worth a try" moves"
moose9 5 июл. 2016 в 0:55
Nice map, not to difficult, but did make me think a moment.
dire.fcs 17 дек. 2015 в 22:13
I'm sure I didn't solve the right way (never used a panel near the ceiling on the laser emitter area) my solution is a bit ninja but I can enter to the laser emitter area by the angled panel on the ceiling
jandlml 5 ноя. 2015 в 20:51
gotta remember that move
lfairban 19 окт. 2015 в 21:12
Solved it with one of those, "It's worth a try" moves.
CoachHines 22 апр. 2015 в 21:45
Very clever.
wildgoosespeeder 30 дек. 2014 в 20:31
@p0rtalmaster
Me too! It's good! I think what ken.roberts is trying to say is that a fizzling laser cube (before 100% gone) will not redirect but rather block an incoming laser , thus solving a portion of this chamber. I say that trick needs a demonstration or training room. Even moreover, I don't know how I feel about this because any object that can be carried fizzles and can't be used, but this case shows usefullness for the last 1-2 seconds before it is gone.
p0rtalthumper 6 сен. 2013 в 17:29
This test chamber just taught me a valuable lesson...
ken.roberts 29 июн. 2013 в 13:47
This is the first map I have seen that uses the delay of a fizzling cube to solve. Genius !!!
kimist108 18 мая. 2013 в 21:46
Yea me! LOL. I saw a comment made by you very similar to Azorae's comment. It was left as a comment for one of her test chambers.
Pyroguy 15 мая. 2013 в 17:06
How do I get both portals on the laser side while I am on the button side with the cube? I can visualize the lift solution but I can't seem to figure out how to get to it. I get stuck with the fizzler in my way.
baphomet 15 апр. 2013 в 3:02
Very good! It appeared completely impossible at first. In particular, it has a step involving doing something that went against every instinct of mine.
NiclO (- - ) 16 мар. 2013 в 4:54
Great map! Cheers! Wrinkled my brain there for a while.
Gigopler [автор] 18 фев. 2013 в 16:53
k.
Azorae 18 фев. 2013 в 16:52
AgentFire 12 фев. 2013 в 9:40
I loved the idea.
MeatBag 6 фев. 2013 в 21:24
I used pedestal button to respawn cube and move lift down for a while .
Rook 6 фев. 2013 в 16:11
Solution 1: Floor portal cube oscillate in laser to bring lift up and down
Solution 2: Portals aim laser at catcher, cube blocks. Hit pedestal button to respawn and get on lift before the laser hits catcher
Should only have one solution.
Gigopler [автор] 6 фев. 2013 в 8:17
Tell me how you did it.
MeatBag 5 фев. 2013 в 23:18
Good. Very simple solution but I missed it and spent 15 or even more minutes :)
One question: why I need portalable surface under ceil near laser?
Calmly Frenetic 5 фев. 2013 в 19:37
An updated map, preventing the way I originally solved it (and I didn't even tell you how!). Found another way that probably suits the map title, but seems a little dirty.
Sumeron 4 фев. 2013 в 11:36
It feels like a proper and well-functioning test, I just couldn't solve it. I managed to get to the point where I had the laser redirection cube, the laser pointed at the recepticle, I just could not figure out a way to step on the platform and only then engage the laser beam trigger. | 1,015 | 3,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-51 | latest | en | 0.898923 |
http://en.paperblog.com/quick-math-1059390/ | 1,505,977,560,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687702.11/warc/CC-MAIN-20170921063415-20170921083415-00251.warc.gz | 105,687,023 | 14,944 | # Quick Math
By Owlandtwine
My sons have created this thing. It is something of a game and it always begins like this: How old will you be when...?
When I'm six, Theo, you will be eight. When I'm eight you will be ten. When I'm ten you will be twelve. This continues for one or two more minutes, and then it inevitably goes something like this:
When I am twelve, Mom will be forty-six. When I'm eighteen, says Theo, Mom will be fifty...
I listen to their voices and I admire their budding ability to do quick thinking math. Their lovely sugary sounds ping back and forth from mouth to mouth, and I go still. This is funny business to them. They're competing. They're showing off - How old will Mom get?!
There is cotton in my mouth. My heart pauses. I think but do not tell them that hearing their voices roll our ages out before us, as if each passing day isn't going fast enough, is like beating the rug of our life; the dirt and dust shaking out in lines that appear around my eyes, mouth, forehead. Particles mixed with light linger on my no longer taut belly - the one that has stretched and exploded before me in the most beautiful soft roundness, not once but twice, that I could push in with my finger and feel a foot fold, touch heels when they were on my insides. At times there is even an ache and the phantom pain of an arm stretching, a head turning straight towards my soul, a story past. Wasn't that just yesterday's story? Wait, wait, I say, I'm still doing the math.
I am at my computer editing photos. Pandora is set to Joni Mitchell radio. Her voice makes me feel woozy, like I just hit the bottle of my early twenties, when what was time? Theo saddles up next to me and I stop what I'm doing to smell the top of his head; to kiss his thumb sized hole of missing skull bone that pulses in beat with my own heart. He takes my love but dryly says that Joni Mitchell is no Katy Perry. No, I suppose not, I say. But she's just as awesome - in a different way. He raises his eyebrows and gives his head a shake back and forth, his lips fold into a smile. That is his second-grade way of saying, whatever, Mom. Whatever.
Then I find myself alone at the kitchen sink. I'm peeling a mandarin and staring out the window at the blanket of fresh snow covering our world, the bright scent filling me. A winterized goldfinch sits still on a tree branch. Across the way in the park, the geese have returned after a long summer and mild fall; they are huddled together with the last bit of day's light resting on their backs. I am aware of my reshaped space, the softness of it, the feeling of possibility. Still, it is a challenge, I think, to not always be doing the math. At least it is my challenge.
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FASHION, SELF EXPRESSION, SHOPPING | 1,153 | 4,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-39 | latest | en | 0.946942 |
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Sai shared a question:
There are two concentric circles. The radii of the two circles are 100 m and 110 m. A wheel of radius 30 cm rolls on the smaller circle and another wheel rolls on the larger circle. After they have completed one revolution, it is found that the two wheels rolled equal number of times on their respective axes. What is the radius of the other wheel?
(A) 31 cm
(B) 32 cm
(C) 33 cm
(D) 34 cm
Aug 02, 2019 3:19 PM
Created by Sai
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Suggested Mentors | 262 | 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} | 2.515625 | 3 | CC-MAIN-2022-05 | latest | en | 0.872671 |
https://la.mathworks.com/matlabcentral/cody/problems/51-find-the-two-most-distant-points/solutions/2729028 | 1,601,136,549,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400244231.61/warc/CC-MAIN-20200926134026-20200926164026-00192.warc.gz | 460,366,966 | 16,573 | Cody
# Problem 51. Find the two most distant points
Solution 2729028
Submitted on 22 Jul 2020
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 Fail
p = [0 0; 1 0; 2 2; 0 1]; ix_correct = [1 3]; assert(isequal(mostDistant(p),ix_correct))
Error in solution: Line: 11 Column: 4 Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.
2 Fail
p = [0 0; 1 0; 2 2; 0 10]; ix_correct = [2 4]; assert(isequal(mostDistant(p),ix_correct))
Error in solution: Line: 11 Column: 4 Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.
3 Fail
p = [0 0; -1 50]; ix_correct = [1 2]; assert(isequal(mostDistant(p),ix_correct))
Error in solution: Line: 11 Column: 4 Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.
4 Fail
p = [5 5; 1 0; 2 2; 0 10; -100 20; 1000 400]; ix_correct = [5 6]; assert(isequal(mostDistant(p),ix_correct))
Error in solution: Line: 11 Column: 4 Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters. | 398 | 1,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-40 | latest | en | 0.488597 |
https://kupdf.net/download/maths-by-amiya-500-cat-2016_59b23677dc0d60ec02568ee4_pdf | 1,590,676,796,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347399820.9/warc/CC-MAIN-20200528135528-20200528165528-00180.warc.gz | 414,591,284 | 38,959 | # Maths by Amiya 500 CAT 2016
September 8, 2017 | Author: Rahul Sharma | Category: Triangle, Fraction (Mathematics), Euclidean Plane Geometry, Numbers, Geometric Shapes
#### Short Description
Maths by amiya...
#### Description
3E LEARNING
Maths By Amiya 500 CAT 2016 QUANT 500 By :- Amiya
2016
3E LEARNING RANCHI
WWW.3ELEARNING.COM
3E LEARNING - MATHS BY AMIYA 500 CAT 2016
Maths By Amiya 500 (2017 (2017-19) 1. Mohit calculated sum of first N natural number and he found sum is 337 and he knows he counted one number twice or missed one number. then find his minimum possible % error made by him. a. 3% b. 3.99% c. 3.69% d. 3.23% 2. The integers 1,2,….. 30 are written on a board. A person came and erased any two numbers say "a" & "b" and wrote a new number "a+b+2" this process is done by total 29 persons (including first one) then . What is the number left on the board at the end? = (3,5) , T3 = (7,9,11) , T(4) = (13,15,17,19) ..... Then what is the sum of all 3. If T1=(1) , T2= terms of T(10) 4. phi (n) is defined as number of co co-prime prime less than n. If ‘P’ is product of two different prime numbers, whose sum is 1200 then what is the max phi(M) {1,2,3,………..10000}, How many APs can be formed from the elements of S that 5. Consider the set S = {1,2,3,………..1000 start with 1 and end with 100000 and have minimum 3 terms?
6. Total number of integral solutions of 13x - 3y = 1000 for 100< x < 200 7. For how many integral "n" is
is an integer
8. If HCF , LCM and sum of two numbers are 6 , 15 and 23 then find their difference. 9. If 1 7 6 3 , then find the value of f(x - 1). 10. Find the area of the enclosed fig by | x- 5 | FGM and if the area ratio of ∎DEGF to that of ∎BCED is 1:8 then what is ∎DEGF to that of ∎BCED is 1:8 then what is the ratio of AF:FD:DB a. 7: 3:2 b. 2:3:7 c. 2:1:4 d. 1:2:3 e. NoT 197. If in ∆ ABC , points D, F & H are on side AB ; and points E, G & I are on side AC such that ∆ ABC , points D, F & H are on side AB ; and points E, G & I are on side AC such that BC || DE || FG || HI and AH :HF:FD:DB 2:3:5:7 then what is the ratio of BC : DE : FG : HI BC || DE || FG || HI and AH :HF:FD:DB 2:3:5:7 then what is the ratio of BC : DE : FG : HI 198.
qªª«Zªª«rªª«√ªª«⋯ q¬«Z¬«r¬«√¬«⋯
?
199. A number on being divided by ing divided by 6, 8 and 10 successively leaves the remainders successively leaves the remainders 5, 7 and 9 respectively. Find the remainder when the same number is divided by respectively. Find the remainder when the same number is divided by 120. 200. A number on being divided by 7 , 9 and 11 successively leaves the remainders 5, 7 and 9 respectively. Find the remainder when the same number is divided by 693 the remainder when the same number is divided by 693 the remainder when the same number is divided by 693 201.
∗q®¯∗Z®¯∗r®¯∗√®¯ ⋯ °∗q®°Z®°r®°√®⋯ ⋯
?
www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 18
3E LEARNING - MATHS BY AMIYA 500 CAT 2016
202. If
³ q7 3 ∗ Z7 3 ∗ r7 3 ∗ √7 ⋯&µ q7 3 ∗ Z7 3 ∗ r7 3 ∗ √7 … and X-
YR 0 then R ???
203. If P is the set of all possible products of numerical values of f P is the set of all possible products of numerical values of three consecutive dates three consecutive dates e.g.11*12*13 in calendar then , how many elements of P are not divisible by 6. 11*12*13 in calendar then , how many elements of P are not divisible by 6. 11*12*13 in calendar then , how many elements of P are not divisible by 6.
204. If 0 . U 30.U 0 . U ; a 1889 , b1888 then what is c if c is a what is c if c is a positive number. a. 1889 b. 1888 c. a or b d. NoT 205. 1333 1334 1333 3999 ∗ 1333 ∗ 1334 ? ? ?
206. If x-3 , x-4 & fx are factors of 4 & fx are factors of 0 ∗ 68 . the what is the the what is then value of f1 207. What will be the last three digits of the product 5 * 25 * 125 * 625 * 3125 * ………* 5^30
208. AC is the chord of a circle as shown in the fig, BD is perpendicular to AC. Find the length of chord AC if BD 4cm , CD 2cm & radius of circle cm & radius of circle 10 cm.
209. There are many less than 10000 which are perfect square an There are many less than 10000 which are perfect square and tenth place digit is odd. d tenth place digit is odd.
210. What is the probability of choosing a number is Natural number set so that number is a perfect square with same last two digits. perfect square with same last two digits. 211. What is the probability of choosing a number What is the probability of choosing a number whose last two digits are same and it whose last two digits are same and it becomes a prefect square. becomes a prefect square.
212. In a triangle ABC the length of altitude AD is 20 cm and BD:DC 1 : 5 D is point in ABC the length of altitude AD is 20 cm and BD:DC 1 : 5 D is point in between B & C . Find the length of a line segment EF parallel to AD if ratio of area o ∆EFC to that of ∆ABC is 8:15 ∆EFC to that of ∆ABC is 8:15. www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 19
3E LEARNING - MATHS BY AMIYA 500 CAT 2016
213. If ABCD a trapezium , in which CD=15 cm , AB = 3 cm & area of ∆DOC = 25 sq. cm then find the area of trapezium ABCD.
214. If ABCD a trapezium , in which CD=6 cm , AB = 2 cm & area of ∆ ∆ADC = 24 sq. cm then find the area of trapezium ABCD.
215. In a triangle ABC the length of an altitude AD is 8 cm, D is on side BC and this altitude divides the opposite side internally in the ratio 1 : 8. A line segment EF parallel to the altitude which bisects the area of the given triangle ABC (as shown in fig) then what is the ratio of AO:OF.
¸ ? ? ? 216. ∑3¹ 3¹K ·- 1 ∗ - 2¸ a. 2212 b. 2480 c. 2722
d. 2590
www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 20
3E LEARNING - MATHS BY AMIYA 500 CAT 2016
217.
218.
219.
www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 21
3E LEARNING - MATHS BY AMIYA 500 CAT 2016
220.
221.
222. 223. Which one is smallest a. √101 √96 b.√111 111 √106
c.√11 √6
d.√®ª d. ®ª √ªº
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3E LEARNING - MATHS BY AMIYA 500 CAT 2016
224. Which one is smallest b.√123 123 √73 a. √101 √99 225. Which one is largest a. √101 √99 b.√123 123 √73
c.√150 √50
d. 1 √199
c.√150 √50 d.√199 1 1
226. How many statements are How many statements are always correct always correctwhere a, m & n non zero real numbers always correctwhere a, m & n non zero real numbers I. 0» 3 0»∗3 II.0» ∗ 03 0»∗3 III.0» 3 03 » a.
Only one
b. only two b. only two
c. all three
d. None d. None
227. A car and a bus start from opposite end of a national highway. Car starts at 11:00AM A car and a bus start from opposite end of a national highway. Car and reaches opposite end at 2:OOPM same day and bus starts at 10:00AM and reaches opposite at 4:00PM same day same day . then at what time they meet each other. . then at what time they meet each other. 2 such that numerically ³ - & µ - 2 2 228. If X_B Y_B-2 such that numerically then n,B ??? a. 8,10 b. 10,10 b. 10,10 c. 10,8 d. NoT d. NoT
229. If X_B Y_B-2 such that numerically 2 such that numerically ³ - & µ - 2 2 then n,B ??? b. 10,10 b. 10,10 c. 10,8 d. NoT d. NoT a. 8,10 If 50!_10 N_50 then number of trailing 0's in N is ____ 230. If 50!_10 N_50 then number of trailing 0's in N is ____
231. If length of a rectangle is increased by 5.88% then what should be change in area to a rectangle is increased by 5.88% then what should be change in area to keep area constant. b. 5.43% decrease b. 5.43% decrease c. 5.56% decrease d. 6.66%decrease d. 6.66%decrease a. 5.55% decrease
Number of positive integral solutions ordered triplets of 9 bxyzb 20 232. Number of positive integral solutions ordered triplets of 9 bxyzb 20 a. 775 b. 657 c. 885 d. NoT
233. A shopkeeper uses a double pan balance to purchase and to sell his goods. To balance both the pans one has to put 20% more weight on left pan than right pan. If shopkeepers claims that he sells his goods on cost price then find maximum profit claims that he sells his goods on cost price then find maximum profit earned by him if he sold all his items what he purchased. sold all his items what he purchased.
www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 23
3E LEARNING - MATHS BY AMIYA 500 CAT 2016
234.
235.
236.
www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 24
3E LEARNING - MATHS BY AMIYA 500 CAT 2016
237.
238.
239.
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240.
241.
242.
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243.
244. A deck of card has 16 cards which include 4 Aces, 4 Kings, 4 Queens & 245. A deck of card has 16 cards which include 4 Aces, 4 Kings, 4 Queens & 4 Jacks. In How many ways we can arrange these 16 cards such that All Kings should be above All Queens. assume all are of different assume all are of different
246. A shopkeeper claims that he gives 10% discount on his cost price but uses X gm of weight instead of 1000 gm during weight instead of 1000 gm during sell of his goods and earn 12.5% profit, Then X ? sell of his goods and earn 12.5% profit, Then X ?
247. Container A has 10 lit of 30% acid, B has 10 lit of 40% acid & C has 10lit of 60% acid. If 5 lit of A is poured in to B then 5 lit from B to C then finally 5 lit of C is poured in to A. then what would be ratio of be ratio of acid is to water in final solution of A . 248. If 4A ? o 0 then x y ?
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If a car is traveling with the speed of 60 km/hr and after applying break it 0 km/hr and after applying break it 249. If a car is traveling with the speed of stops after 4 m, if the same care is moving with m, if the same care is moving with 120 km/hr then after applying km/hr then after applying break it stops after Xm. Then find X break it stops after Xm. Then find X
250. How many 9 digits numbers are possible by in which all digits are different and number is divisible by 9 a. 18*9! - 1 b. 2*9! b. 2*9! c. 17*8! D. NoT 2? 251. ( – 2 7 , *+,- ? ¿
252. ( 3 3 0 , *+,,- 1= À ?
253. ( 9 24 0 , *+ +,- 4 ? ? =
254. In ∆ ABC, AB4cm, BC6cm & AC5cm. Side AB and AC produced till E & F ∆ ABC, AB4cm, BC6cm & AC5cm. Side AB and AC produced till E & F respectively such that BECF1cm. AD median of respectively such that BECF1cm. AD median of ∆ ABC cuts line EF at G then what is ∆ ABC cuts line EF at G then what is the length of DG.
255. A, B & C start working on a project of BOOK MAKING, in which they have to type pages. They completed this project in 12 days. After completion of the project A , B & C received Rs 1200, Rs 2400 &Rs 3600 as their wages Rs 1200, Rs 2400 &Rs 3600 as their wages respectively, which is proportio , which is proportional to to number of pages they typed Lassuming they worked for equal numbers of hours per dayM. Then in how many days does A alone completes the working same hours a day. dayM. Then in how many days does A alone completes the working same hours a day. 256. If N is a 100 digits largest number and a perfect square of a natural number then which If N is a 100 digits largest number and a perfect square of a natural number then wh digit is 50th digit from left? digit is 50th digit from left? A. 0 B. 8 C. 9 d. 1
257. If A & B are points on circles as shown in fig, and line AB passes through centre of circles O & P. QO & RO are perpendicular on line AB, and AQ & BR meet at C then what is the area of ∆ABC if AO 5 cm & BP 4 cm AO 5 cm & BP 4 cm
258. ( 0. @ 3
· ¸
?
@ 0 **+,- choose best option b. 3 3 b b 2 c. 2 2
d.1 1 b b 0
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259. If two circles cut each other orthogonally then find length of common chord if radii of circles are 3 cm & 4 cm. radii of circles are 3 cm & 4 cm.
solved a quadratic equation and find roots as 3 & 4 & Sood finds roots of same Sood finds roots of same 260. Amiya solved a quadratic equation and find roots as 3 & quadratic equation as 1 & 6 then among options which could be a correct option if it is known that none of them did make mistake in writing coefficient of x^2 did make mistake in writing coefficient of x^2 but made mistake in either coefficient of x or constant term then among options which statement n either coefficient of x or constant term then among options which statement is correct a. Amiya makes mistake in coefficient of x and Sood makes mistake in constant term a. Amiya makes mistake in coefficient of x and Sood makes mistake in constant term b. Amiya makes mistake in constant term and Sood makes mistake in coefficient of x b. Amiya makes mistake in constant term and Sood makes mistake in coefficient of x c. Both of them made mistakes in constant term c. Both of them made mistakes in constant term d. NoT 261. Find the total number of terms in the expansion of Find the total number of terms in the expansion of 1 1 1 ? 1 … 1 262. If P 1 and [ 1 are the factors of are the factors of 1 ? . . . for distinct a and b find maximum value of ab and b find maximum value of ab
263. To complete a piece of work A alone takes 12 days more than A & B together & B alone takes 3 days more than A & B together if C alone takes 3 more days then A & C together then in how many days A, B & C can complete the same work if they work together then in how many days A, B & C can complete the same work if they work together. 264. If Á &  are roots of equation are roots of equation 2 4 6 0 then what would be equation whose then what would be equation whose Ã Ä roots are & Ä
Ã
265. If EN is defined as euler of a number, which is number of co If EN is defined as euler of a number, which is number of co-primes less than equal primes less than equal to N and DN as positive divisor of a natural number N. to N and DN as positive divisor of a natural number N. Then what would be sum of all Then what would be sum of all possible values or E D1200 ??? possible values or E D1200 ??? 266. For how many integral “x” ,
is also an integer
267. If ratio of time taken to complete a piece of work by A, B, C & D alone is 1:2:3:4 and when they all work together take hen they all work together take12 days to complete the same work. If on every ODD to complete the same work. If on every ODD DATE only day only A&C work together and on every EVEN DATE only B & D wor DATE only day only A&C work together and on every EVEN DATE only B & D work together and they started on 01 together and they started on 01-FEB-2016 then on which date total work would be 2016 then on which date total work would be completed. 268. For how many natural “n” less than 100, 5- 6- 3 is divisible by 7
269. In ∆ABC , point D & E lie on BC & F lies on AC. If BD:BE:BC 1:2:3 & AF : FC1:1 then ∆ABC , point D & E lie on BC & F lies on AC. If BD:BE:BC 1:2:3 & AF : FC1:1 then what is the ratio of BO:OP:PF , if O & P are point of intersection of line AD & BF and AE & BF www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 29
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270.
how many different values of x , 1/x has only 4 reoccurring digits after decimal.i.e. 271. For how many different values of x , 1/x has only 4 reoccurring digits after decimal.i.e. 1/x 0.abcdabcdabcd.... Lfew could be same but not allM 1/x 0.abcdabcdabcd.... Lfew could be same but not allM
272. There are three boys Amar, Akbar & Anthony and three girls, Sita, Gita &Sheela. They Amar &Sita started a grou Amar &Sita started a group on 1st of Feb, Akabr joins them on 2ndfeb , Anthony on 4th Feb Gita joins them on 3rd Feb &Sheela on 5th Feb. On 1st , 3rd& 5th Feb total age of boys Feb total age of boys nd th were half of that of girls& on 2 & 4 Feb total age of boys were doub Feb total age of boys were double of that of girls. If age of Sheela is 120 yrs then who is youngest boy and what is his the age??? e of Sheela is 120 yrs then who is youngest boy and what is his the age??? e of Sheela is 120 yrs then who is youngest boy and what is his the age???
273. If red is painted on three sides of a cuboid of dimension 6cm*7cm*8cm and other three sides are painted white and then then by different cuts cuboid is divided in to 336 smaller cubes of volume 1 cubic cm. . If same coloured sides are adjacent to each other, cubes of volume 1 cubic cm. . If same coloured sides are adjacent to each other, then there are how many smaller cubes having minimum one red coloured side. then there are how many smaller cubes having minimum one red coloured side. 274. Ram wrote first 1000 numbers on a word sheet, and then he replaced all 2’s & 7 by 3 and saved the file. Then find the total number of 3’s in saved file. file. Then find the total number of 3’s in saved file.
275. Ram wrote first 1000 numbers on a word sheet, and then he replaced all 2’s & 7 by 3 and saved the file. Then find how many numbers have digit 3. and saved the file. Then find how many numbers have digit 3. 276. Ram wrote first 1000 numbers on a word sheet, and then Ram wrote first 1000 numbers on a word sheet, and then he replaced all 2’s & 7’s by 3 he replaced all 2’s & 7’s by 3 and saved the file. Then find how many numbers have only one 3. and saved the file. Then find how many numbers have only one 3.
277. Ram wrote all first 444 natural numbers side by side , from left hand side and created a big number N 123456..... 444 four hundred forty four big number N 123456..... 444 four hundred forty four. If Mohan del If Mohan deleted all 7's from N then find there are how many digits in N N then find there are how many digits in N 278. If Mark has Rs 5000 in his easy If Mark has Rs 5000 in his easy-recharge wallet, an online mobile recharge portal. Find recharge wallet, an online mobile recharge portal. Find the maximum amount of recharge, which can be possible with his current balance and cash backs, under cash back scheme of 20% cash back on recharge of any amount h back scheme of 20% cash back on recharge of any amount more than or equal to Rs 100 Lassume he is not adding extra amount from any other source except cash backs & he can recharge any integral amountM source except cash backs & he can recharge any integral amountM 279. For how positive integral X For how positive integral X less than 100, X^3 - X^2 is a perfect square. X^2 is a perfect square.
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280. If there are 3 black balls, 6 white balls and 5 green balls. Find the minimum number of picks required to get at number of picks required to get at-least one ball of each colour always. LAssume always. LAssume one can pick only one ball at a timeM one can pick only one ball at a timeM
281. How many obtuse angled integra How many obtuse angled integral sided isosceles triangles are possible if unequal side es are possible if unequal side is 8 cm 282. How many obtuse angled integral sided isosceles triangl How many obtuse angled integral sided isosceles triangles are possible if one side is 5 es are possible if one side is 5 cm Find minimum value of expression 8x 16x 1 283. Find minimum value of expression 284. Find minimum value of expression Find minimum value of expression 4 sec θ sec θ 1
Find the total number of solution pairs of real x,y for below given equation 285. Find the total number of solution pairs of real x,y for below given equation 2 8 15A 10A A 4 6 286. Find the total number of solution pairs of real x,y for below given equation Find the total number of solution pairs of real x,y for below given equation 3 6 72A 4A A 3 4
287. If 3x È 12 then x x ? x = ?
School. 10%,15%, 20%, ,25% & 30% of branch seats Direction: There are only 5 branches in a B Direction: There are only 5 branches in a B-School. 10%,15%, 20%, ,25% & 30% of branch seats are vacant and it is known that each branch has minimum 1/7th of total PGP seats , are vacant and it is known that each branch has minimum 1/7th of total PGP seats , If total vacant seats are x % of total PGP seats then vacant seats are x % of total PGP seats then 288. Find minimum value of x Find minimum value of x Find maximum value of x 289. Find maximum value of x 290. For how many natural numbers less than equal to 100, such that last two digits of its 8th power and 9th power are same. power are same. 291. For how many natural numbers less than equal to 100, such that last three digits of its 5th power and 8th power are same. r are same. 292. 123123123 ...... a 38 digits number mod 13 ???? 123123123 ...... a 38 digits number mod 13 ???? 293. There are how many two digits numbers are possible such that their last two digits of 23rd& 25th powers are same. powers are same. 294. Raman is an intelligent mathematician. He has Rs Raman is an intelligent mathematician. He has Rs830 in his Paytm account. in his Paytm account. Find the maximum amount of transaction transaction which can be possible with his current balance and which can be possible with his current balance and cash backs, under cash back scheme of cash backs, under cash back scheme of 14.28% cash back on recharge of any amount % cash back on recharge of any amount more than or equal to Rs14 14 Lassume he is not adding extra amount from any other Lassume he is not adding extra amount from any other source except cash backs & he can urce except cash backs & he can do transaction in paisa too M www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 31
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Patna, Patiala&Pathankotis7:3:2 & Ratio of number of Pathankotis7:3:2 & Ratio of number of 295. If population ratio of Patna males in Patna, Patiala&Pathankotis 6:2:1 Pathankotis 6:2:1 then which city has maximum female which city has maximum female population and which city has minimum population population and which city has minimum population ? 296. Among the options which one is Among the options which one is largest? a. 13= 18 b. 16 16 19 c. 16 18
d.13 13= 19
297. Arrange A, B & C in ascending order , if A 123^45678 , B 12345^678 & C1234^5678
298. Arrange A, B , C & D in ascending order , if A 1234^5 , B 123^45, C12^345 & D1^2345
Direction :Nakistan is a small country, which has 4 states. Name of states are A,B,C&D and Direction :Nakistan is a small country, which has 4 states. Name of states are A,B,C&D and population of each states is minimum 1/7th of total population of Nakistan. If sex ratio of states of total population of Nakistan. If sex ratio of states A,B,C & D are 600 , 700, 800 & 900 then A,B,C & D are 600 , 700, 800 & 900 then 299. Find minimum possible sex ratio of Nakistan Find minimum possible sex ratio of Nakistan a. 600.17 b. 678 678.61 c. 516.79 d. NoT 300. Find maximum possible sex ratio of Nakistan Find maximum possible sex ratio of Nakistan a. 818.18 b. 828.87 b. 828.87 c. 806.89 d. NoT If John distributed his entire pension into his wife Mona, son William, d 301. If John distributed his entire pension into his wife Mona, son William, daughter Kate, girl-friend Sona and her daughter Sara. If it is known that Mona gets minimum 1/6 friend Sona and her daughter Sara. If it is known that Mona gets minimum 1/6th . William gets minimum 1/8th , Kate gets minimum 1/10th , Sona gets mi , Sona gets minimum 1/6th& Sara gets minimum 1/4th of Ram’s pension. Saving % of their individual sharing of Mona, of Ram’s pension. Saving % of their individual sharing of Mona, William, Kate, Sona& Sara is 30%, 40%, 40% , 50% & 60% .If total saving of all five is X% of total pension of Ram, then among option which could not be val of total pension of Ram, then among option which could not be value of X. ue of X. a. 44.15 % b. 45.83% b. 45.83% c. 48.32% d. 48.89% d. 48.89% 302. There are many natural numbers are possible which is less than 400000 and divisible by 3 but not have any of digits from m 6, 7,8 , 9 < by 3 but not have any of digits from m 6, 7,8 , 9 < 303. How many four digits can be made from the digits m1,2,3,4< How many four digits can be made from the digits m1,2,3,4< that are perfectly divisible that are perfectly divisible by 3? repetition allowed by 3? repetition allowed 304. If Speed of Ram is 20 kmph then he reaches office from home 20 min before his usual time, and if his speed is 10 kmph then he reaches office from home 40 min after his usual time then find distan usual time then find distance between home to office. 305. Raman bought a combined total of 25 bought a combined total of 25 mobiles and tabs. He marked up the . He marked up the mobiles by 20% on their cost price while each while each tabs was marked up by Rs. 2000. He was able to sell was marked up by Rs. 2000. He was able to sell 75% of the mobiles and 2 and 2 tabs and make a profit of Rs. 49,000. The remaining and make a profit of Rs. 49,000. The remaining mobiles and 3 tabs could not be sold by him. Find his overall profit or loss if he gets no return on could not be sold by him. Find his overall profit or loss if he gets no return on unsold items and it is known that a unsold items and it is known that a tabs costs 50% of a monitor. www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 32
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Find the distance in km, for which if we increase our speed by 20 km/hr we if we increase our speed by 20 km/hr we 306. Find the distance in km, for which reach the destination 2 hours earlier but when we reduce our speed by 10 km/hr we take 3 hours more to reach the same destination. we take 3 hours more to reach the same destination.
307. Find the actual time in hour actual time in hour, to cover a distance for which if we increase our spe for which if we increase our speed by "a a" km/hr we reach the destination b" hours earlier but when we reduce our speed by km/hr we reach the destination "b hours earlier but when we reduce our speed by "cc" km/hr we take "d km/hr d" hours hours more to reach the same destination. [4^P[^ [4^P[^ [4^P[^ [4^P[^ a. P^[4 b. P^[4 c. P^[4 d. P^[4 e. NoT 308. When a man covers 2/3rd of distance by car and rest by bus then he takes 12 hours to cover the distance, but if he takes 3/4th of distance by cover the distance, but if he takes 3/4th of distance by car and rest by car then takes 10 and rest by car then takes 10 hours, then find total time to cover same distance, when he hours, then find total time to cover same distance, when he covers half of the distance by covers half of the distance by car and rest by bus.
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If a thief flew with constant speed of "60 km/hr" in a straight road, Direction :If a thief flew with constant speed of "60 km/hr" in a straight road, Direction : and after 2 hours a police and his dog started to chase the thief with speed of "80 km/hr" and "100 km/hr" respectively km/hr" and "100 km/hr" respectively. In this complete journey, Dog touches the Thief . In this complete journey, Dog touches the Thief and comes and touches the Police and move towards Thief and continues the proces and comes and touches the Police and move towards Thief and continues the process, until The Thief is caught. 309. Find total distance travelled by Dog in this process Find total distance travelled by Dog in this process
310. Find total distance travelled by Dog in Find total distance travelled by Dog in this process towards police. this process towards police. 311. Find total distance travelled by Dog in this process towards thief. Find total distance travelled by Dog in this process towards thief.
1 *0-3 … 1 *0-431 *0-44 ;ÌV 7 ? ? ? 312. 1 *0-11 *0-21
313. A boat takes total 4 hours to cover 22 km upstream and 28 km downstream and takes total 6 hours to cover 33 km upstream and 42 km downstream then find speed of hours to cover 33 km upstream and 42 km downstream then find speed of stream.
314. In a 1000 m race A beats B by 250 m and in 1500 m race A beats C by 250 m , then in B Vs C , who win the race and by how much meter if length of track is 2520 m. Vs C , who win the race and by how much meter if length of track is 2520 m. 315. For how many integral values of X integral values of X , |X-1||X2||X-3||X4| b 70 3||X4| b 70
316. Which one is true
¿Ê
? Ë A. W?X ¿Ê
? Ë B. W?X
¿Í
? ¿Ê > WKX ¿Í
? ¿Ê b WKX
317. If Josephus leaves his home for office at a fix time every day , but he reaches his office at different time due to different speeds. different speeds. If his average speed is 60 kmph he reaches If his average speed is 60 kmph he reaches office at 10:45AM , but when his average speed is 40 kmph he reaches at 10:50 AM, what should be his average speed to reach office at 10:47 AM. should be his average speed to reach office at 10:47 AM. 318. If ratio of incomes of A, B & C I'd 5:4:3 & savin If ratio of incomes of A, B & C I'd 5:4:3 & savings ratio of A, B & C is 2:3:4. gs ratio of A, B & C is 2:3:4. If ratio of A's income to C's saving is 10:9 then find the ratio of total expenditure to total savings of all three 319. What would be last 4 digits of What would be last 4 digits of 38 in base 2
320. There are three most active groups on facebook, e three most active groups on facebook, CAT PREPRATION PREPRATION CP, GHANTA CAT GC & MBA. Mentors organised four events named . Mentors organised four events named Chotu Quant Chotu Quant, Motu Quant, Patlu Quant & All Quant. In Chotu Quant, all members of CP & GC participated and average weight of all members of this event is 42 kg In Motu Quant, all members of CP & MBA participated and average weight of all members of CP & MBA participated and average weight of all members of this event is 30 kg is 30 kg In Patlu Quant, all members of MBA & GC participated and average weight of all members of this event is 60 kg is 60 kg www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 34
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& if In ALL Quant, all members of CP, GC & MBA participated and average weight ted and average weight & if In ALL Quant, all members of CP of all members of this event event is X kg
Then what is the range of X. Then what is the range of X. Lassume Facebook has "one member one group" policyM Lassume Facebook has "one member one group" policyM
321. If Raman left his room at X AM and reaches office at Y PM of the same date and Mohan left his room at X AM and reaches office at Y AM of same date. If both stay together and and reaches office at Y AM of same date. If both stay together and work together and time taken by Mohan to reach office is square of that by Raman then what is the ratio of Average Speed of Raman from home to office to that of Mohan. what is the ratio of Average Speed of Raman from home to office to that of Mohan.
322. Raman left his room at 7:00 7:00 AM and reaches office at 12 PM of the same date and PM of the same date and Shobhit left his room at 8:00 8:00 AM and reaches office at 11 AM of same date. If both stay AM of same date. If both stay together and work together together and work together at same place then at what time Shobhit overtakes Raman then at what time Shobhit overtakes Raman.
323. Shobhit & Bitan planned an infinite round race on a circular track of length 100 km. planned an infinite round race on a circular track of length 100 km. Both started race at same point of track, but Bitan cheated Shobhit and started half an hour before the actual race timing and choose Swift , whose speed is 40 km/hr. Shobhit who is a good guy started race at actual time but choose Benz whose speed if 140 km/hr d guy started race at actual time but choose Benz whose speed if 140 km/hr but by mistakes he moves in anticlockwise direction. If at 9:30 AM Bitan reaches starting point for the first time then at what time their 4th meeting would happen if they maintained constant speed throughout the race. ed constant speed throughout the race.
324. Sandeep Patra is a very handsome but a shy guy. Whenever he sees a girl he doubles his speed but start moving in the opposite direction of direction just before seeing girl. If on a day he starts moving with speed If on a day he starts moving with speed of 1 m/s from his home and he sees a girl after of 1 m/s from his home and he sees a girl after every 1sec, then find how far he is from his house just after 11sec when he started. every 1sec, then find how far he is from his house just after 11sec when he started. 325. There is a regular nonagon, whose vertices are alphabets of words DANGEROUS in nonagon, whose vertices are alphabets of words DANGEROUS in same order then what is the value of same order then what is the value of ∠ NGO in degree ???? If ABCDEFGH is a regular octagon then find angle ∠ CDG in degree 326. If ABCDEFGH is a regular octagon then find angle If ABCDEFGHIJ is a regular decagon then find ∠ EAI in degree 327. If ABCDEFGHIJ is a regular decagon then find
328. Find the sum of all internal angles at red dots of all four figures Find the sum of all internal angles at red dots of all four figures
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^...^1 then what is the least value of N natural number what is the least value of N natural number 329. If M 17^16^15^...^1 then for which M mod N is neither o nor 1. Lmod is remainder functionM for which M mod N is neither o nor 1. Lmod is remainder functionM
age of 11 members committee on 31st Dec 2014 was 2014 was 30.27. On 330. If the approx average age of 1st Jan 2017 Ram who is Ram who is one of the member quits the committee and then approx quits the committee and then approx average becomes 29.5. If it is known that . If it is known that , to find average only integral age if 11 years to find average only integral age if 11 years 11 months or 11 years 1 months then in both cases considers as 11 years only is considered then find in which year Ram was bor considered then find in which year Ram was born.
331. Two cities X and Y lie on a straight line. Two men P and Q left simultaneously from same city X towards Y at same time. Both reaches Y and immediately turn around and move towards X. On reaching X, again they turn around and move towards Y. This movement continues indefinitely. nt continues indefinitely. If the distance between X and Y is 60 km and speed of If the distance between X and Y is 60 km and speed of P and Q are 20 Km/hr & 10 Km/hr respectively P and Q are 20 Km/hr & 10 Km/hr respectively . Find the total number of OVERTAKES Find the total number of OVERTAKES in 120 hour after start. 332. Two cities X and Y lie on a straight line. Two men P and Q left s Two cities X and Y lie on a straight line. Two men P and Q left simultaneously from imultaneously from same city X towards Y at same time. Both reaches Y and immediately turn around and move towards X. On reaching X, again they turn around and move towards Y. This movement continues indefinitely. movement continues indefinitely. If the distance between X and Y is 60 km a If the distance between X and Y is 60 km and speed of P and Q are 20 Km/hr & 10 Km/hr respectively P and Q are 20 Km/hr & 10 Km/hr respectively Whenever they meet or overtake they Whenever they meet or overtake they shake hands. Find the total number of handshakes in 60 hour after start shake hands. Find the total number of handshakes in 60 hour after start 333. In a right angled triangle with integral sides, smallest side is four times of difference of In a right angled triangle with integral sides, smallest side is four times of differenc other two sides and sum of all sides is 120 then find area of this triangle other two sides and sum of all sides is 120 then find area of this triangle
334. Two cities Rampur and Sitapur lie on a straight line. Two men A and B left simultaneously from Rampur and Sitapur towards each other. A reaches Sitapur and immediately turns around and move towards Rampur. On reaching Rampur, again he around and move towards Rampur. On reaching Rampur, again he turns around and move towards Sitapur. This movement continues indefinitely. B also travels in a similar manner. travels in a similar manner. The distance between Rampur and Sitapur is 120 km and The distance between Rampur and Sitapur is 120 km and speed of A and B are 40 km/h speed of A and B are 40 km/hr & 10 km/hr respectively. If both started together then in If both started together then in 24 hours find the total number of interactions between them. 24 hours find the total number of interactions between them. 335. Two cities Rampur and Sitapur lie on a straight line. Two men A and B left simultaneously from Rampur and Sitapur towards each other. simultaneously from Rampur and Sitapur towards each other. A reaches Sitapur and A reaches Sitapur and immediately turns around and move towards Rampur. On reaching Rampur, again he turns around and move towards Sitapur. This movement continues indefinitely. B also travels in a similar manner. travels in a similar manner. The distance between Rampur and Sitapur is The distance between Rampur and Sitapur is 60 km and speed of A and B are 20 km/hr & 10 km/hr respectively speed of A and B are 20 km/hr & 10 km/hr respectively. If both started together then in If both started together then in 12 hours find the total number of interactions between them. 12 hours find the total number of interactions between them.
336. Two cities X and Y lie on a straight line. Two men P and Q left simultaneously from X and Y towards each other. P reaches Y and immediately turns around and move towards ards each other. P reaches Y and immediately turns around and move towards X. On reaching X, again he turns around and move towards Y. This movement continues indefinitely. Q also travels in a similar manner. indefinitely. Q also travels in a similar manner. The distance between X and Y is 1000m The distance between X and Y is 1000m and speed of P and Q are 200m/s & 800 m/s respectively are 200m/s & 800 m/s respectively. Whenever they meet or Whenever they meet or overtake they shake hands. overtake they shake hands. www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 36
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337. If in a test there are 5 objective question, each having 4 options with only one option as correct answer. If 1 is awarded for correct answer and no penalty option as correct answer. If 1 is awarded for correct answer and no penalty for wrong or non-attempt. If it is known that 3 is passing marks of this test then , In how If it is known that 3 is passing marks of this test then , In how many ways one can attempt this test so that he passes the test. many ways one can attempt this test so that he passes the test.
338. When I started my today's journey from Hazaribagh to Ranchi. I checked first mile stone it showed Ranchi 100 km , after 40 min I checked another mile stone and found anchi 100 km , after 40 min I checked another mile stone and found Ranchi XY km and after another 20 min when I checked mile stone I realised Ranchi YX km. If I managed a constant speed through my journey, please help me to know speed of my car , since speedometer is not working edometer is not working 339. Mango bite offer - Return 3 empty rappers and get 1 mango bite free Return 3 empty rappers and get 1 mango bite free. If cost of one Return 3 empty rappers and get 1 mango bite free mango bite is Rs 1 and i have Rs 100 then maximum how many mango bite i can eat. mango bite is Rs 1 and i have Rs 100 then maximum how many mango bite i can eat. 340. If P & P2 both are prime numbers P>3 then how many different values of R be If P & P2 both are prime numbers P>3 then how many different val possible if R LP*P2M mod 9 R LP*P2M mod 9 341. ÐÑ Ò Ó Ô , then find PDR , if P,Q & R are distinct prime numbers , then find PDR , if P,Q & R are distinct prime numbers , then find PDR , if P,Q & R are distinct prime numbers
Direction Direction 344 344 - 345 345 :Number of Hair fall is directly proportional to TQ Tension Quotient and TQ is directly proportional to square of pe TQ is directly proportional to square of person integral age ignoring months and days. If at rson integral age ignoring months and days. If at the time of birth a person is considered as tension free 0 TQ and person has 100,000 hairs. the time of birth a person is considered as tension free 0 TQ and person has 100,000 hairs. If at the age of 4 number of hairs a person has is 77,500 then at the age of 4 number of hairs a person has is 77,500 then Total number if hairs at the age of 8 is ____ 342. Total number if hairs at the age of 8 is ____ What is the minimum age of complete baldness 343. What is the minimum age of complete baldness Find ab if a & b are positive integer and a³ - 1 b² 1. 344. Find ab if a & b are positive integer and Find the least value of n for which n output of below given sum is an integer 345. Find the least value of n for which n output of below given sum is an integer n/1*2*3 n/2*3*4 n/3*4*5 ... n/7*8*9 n/1*2*3 n/2*3*4 n/3*4*5 ... n/7*8*9 346. 1³2³3³4³...10³ 1357....n 1357....m, Then find integral 1³2³3³4³...10³ 1357....n 1357....m m & n 347. 1² 4² 7² 10² ..... 40² ??? 1² 4² 7² 10² ..... 40² ??? 348. If average of 11different numbers x1 , x2, x3. .... , x11 is X then what would be average in terms of X if all 11 numbers are multi in terms of X if all 11 numbers are multiplied by 2 then decreased by X. plied by 2 then decreased by X. 349. For different positive integral a, b & c For different positive integral a, b & c ; a² b² c³ then find minimum value of c a² b² c³ then find minimum value of c 350. If N m² , where m is a natural number more than 99 but less than 1000, then for how values of m , N & m are having same last two digits values of m , N & m are having same last two digits 351. Ram and Shyamali each has a fair coin. Ram tossed his coin 5 times and got 4 heads and one tail , Shyamali also tossed her coin 5 times and got 3 heads and 2 tails one tail , Shyamali also tossed her coin 5 times and got 3 heads and 2 tails. One day both meet and both tossed their coins once. What is the probability of getting dif meet and both tossed their coins once. What is the probability of getting different outcomes on each coins ??? outcomes on each coins ??? www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 37
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HCF of sum of all sides of an integral sided right angled triangle is ________ ??? 352. HCF of sum of all sides of an integral sided right angled triangle is ________ ???
HCF of product of all sides of an integral sided right angled triangle is ________ ??? 353. HCF of product of all sides of an integral sided right angled triangle is ________ ???
354. If a ball is dropped from a height of 2000 m and If a ball is dropped from a height of 2000 m and it bounces back but looses 1/9 of it bounces back but looses 1/9 of height after every bounce then find total distance covered by ball till it comes to rest or near to rest 355. Find area of graph in side x² y² 16 but outside |x||y|4 in items of pi Find area of graph in side x² y² 16 but outside |x||y|4 in items of pi
356. Find maximum value of n if n & m are Find maximum value of n if n & m are natural numbers and 90! 20!^n * m 90! 20!^n * m
357. There are how many natural numbers less than equal to 100 , whose product of all factors are square of the number factors are square of the number 358. Number of trailing zeros in C512, 32 Number of trailing zeros in C512, 32 ; Cn,r n!/r!*n-r!
359. There are how many natural There are how many natural numbers less than 1000 , has only odd number of factors numbers less than 1000 , has only odd number of factors total number of factors are odd total number of factors are odd
360. John and Shaw started running from same point , at same time in same direction. John John and Shaw started running from same point , at same time in same direction. maintained a constant speed of 40 km/hr maintained a constant speed of 40 km/hr. But Shaw started with 5km/hr and in ev But Shaw started with 5km/hr and in every next hour he increase his speed by 10 km/hr. next hour he increase his speed by 10 km/hr. Then find after how time from start Shaw Then find after how time from start Shaw will catch John
361. If average of 10 consecutive increasing multiple of 8 is 324 then find 4th smallest term among these 10 terms
362. If N is a two digit largest number whi If N is a two digit largest number which is a factor of 35^10 - 17^10 then N mod 7 ??? 17^10 then N mod 7 ???
Find the sum of all terms of 10th row of below given series 363. Find the sum of all terms of 10th row
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ....................................... .............................................
364.
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365.
internal angles at vertex of given figure 366. Sum of all internal angles at vertex of given figure
367. If in a row Jhon is 23rd from right and Tina is 32nd from left end and there are 5 persons between then , what is the difference between maximum and minimum possible persons in the row 368. If X² YX , XX² AYBX, , XXX² AAYBBX, XXXX² AAAYBBX. Then AB Then AB - Y ??? Where X, XX, XXX are 1, 2, 3 digit numbers so rest are Where X, XX, XXX are 1, 2, 3 digit numbers so rest are
369. If ab bc ca 27 then total number of unordered positive integral solution is ____ then total number of unordered positive integral solution is ____ then total number of unordered positive integral solution is ____ 370. For how many natural N , For how many natural N , N 32^2 is divisible by N4 371. 3 ∗ 9Ê ∗ 27Í ∗ 81 81× … … . ?
372. If ABC & BCA are two three digits perfect squares where A, B & C are different digits then ABC ??? 373. There how many numbers less than 1,00,000 whose sum of digits is 3 There how many numbers less than 1,00,000 whose sum of digits is 3
374. If Sita and Gita daily jog from their home to If Sita and Gita daily jog from their home to park and park to home and then park and park and park to home and then park and so on for two hours, park is 10 km from their home. If they live together and both stats www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 39
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at 7:00 AM, and speed of Sita is 20 km/hr and that of Gita is 30 at 7:00 AM, and speed of Sita is 20 km/hr and that of Gita is 30 km/hr then when km/hr then when Gita overtakes Sita for the first time after ita for the first time after start.
Among the options select correct statements 375. Among the options select correct statements 1. Sum of three different primes can't be divisible by 6 1. Sum of three different primes can't be divisible by 6 2. Product of three different primes can't be divisible by 6 2. Product of three different primes can't be divisible by 6 3. Last two digits of all primes less than 100 is 50 3. Last two digits of all primes less than 100 is 50 4. In an integral sided right angled triangle all sides can't prime. ght angled triangle all sides can't prime.
376. There are how many different integral sided right angled triangles are possible whose There are how many different integral sided right angled triangles ar one side is 25 unit ¿ ¿ ¿ 377. s8 q8 Z8 r8 √8 … ?? ¿
¿
378. If √
√3 then Í ?
379. In how many ways we can rearrange the word In how many ways we can rearrange the word NAGIN such that no vowels be together. NAGIN such that no vowels be together. 380. Find the length of line segment between axes line of a line whose equation is 4x-3y Find the length of line segment between axes line of a line whose equation is 4x 12 381. If A*B*C*D*E*F 60 , where A,B,C,D & E are different integers then what is the maximum value of ABCDEF maximum value of ABCDEF
382. Total number of integral ordered pair solutions for of integral ordered pair solutions for A b 64 64 is a. 175 b. 163 c. 150 d. NoT
383. What if the value of c to get maximum value of What if the value of c to get maximum value of 0 ∗ . ∗ U ? under condition abc under condition abc 108 and all a,b&c are positive numbers. 108 and all a,b&c are positive numbers. 384. If in a ∆ABC, D is midpoint of BC , E is ∆ABC, D is midpoint of BC , E is midpoint of BD , F is midpoint of BE and G is midpoint of BD , F is midpoint of BE and G is midpoint of AF then what is the ratio of area of midpoint of AF then what is the ratio of area of ∆BGF to that of ∆ABC ∆BGF to that of ∆ABC
385. What could be minimum value of N , which COULD satisfy What could be minimum value of N , which COULD satisfy Y 0 . U V where a,b,c & d are distinct distinct natural numbers
386. If few students of section A of PGP transfer to section B then ration of number of nts of section A of PGP transfer to section B then ration of number of students of A to that of B 3:7 but if few students of section B of PGP transfer to section A then ration of number of students of A to that of B 5:2 A then ration of number of students of A to that of B 5:2, then find minimum number , then find minimum number of total students of section A & B. al students of section A & B. 387. If 10 lit of pure alcohol is replaced by 50% of 2 lit alcohol and this process is repeated one more time then find the % of alcohol in final solution. one more time then find the % of alcohol in final solution.
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Total number of terms after expansion of 1 388. Total number of terms after expansion of 389. Total number of terms after expansion of Total number of terms after expansion of A Q
390. How many perfect squares less than 1000 would give remainder 3 when divided by 7 How many perfect squares less than 1000 would give remainder 3 when divided by 7 How many perfect squares less than 1000 would give remainder 7 when divided by 9 391. How many perfect squares less than 1000 would give remainder 7 when divided by 9
392. If C is circum-centre & G is centroid of a G is centroid of a triangle whose sides are 7 , 24 and 25 cm then e 7 , 24 and 25 cm then what is the length of line segment what is the length of line segment CG 393. There are how many days in between next palindromic date DDMMYYYY and last palindromic dateDDMMYYYY if today is 14062016 palindromic dateDDMMYYYY if today is 14062016 394. Find the least value of n more than 4 for which 234567.... n is which 234567.... n is a perfect square a. 15 b. 24 c. 25 d. 26
395. Ram and Mohan started their journey from two opposite extreme end of a tunnel. If Ram started his journey at 8 :00 AM with speed 60 km/hr and Mohan at 7:00 with Ram started his journey at 8 :00 AM with speed 60 km/hr and Mohan at 7:00 with speed 40 km/hr. When both met to each other it is found that one of them covers double distance of other then find the different between maximum or minimum possible length of tunnel. 396. Total number of positive ordered solution of abcd Total number of positive ordered solution of abcd 10 Í ÍÍ Í Í
7
397. Unit digit of
7 is _____ is _____
6666. . . . 398. What is the value of 1.6666 a. 2.66666..... b. 2.565656....... b. 2.565656.......
399. Zx rx √x ⋯ 3xx then x ?
c. 2.77777.....
d. NoT d. NoT
400. What is the minimum value of value of |x1| |x2||x3|
401. What is the minimum value of value of |x| - |x1| |x2|-|x3||x4|
402. If LCM of first N natural numbers is "L" and that of first "N2" natural number is "2*L" then less than 100 how many different N are possible then less than 100 how many different N are possible 403. If 130 20. U V 120. 8.U 40V , 0 ∗ . ∗ U _ 0 then then 4^ ? P[
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24 23A 6A 11Q 2Q 4 42 then 404. If for real x, y & z ; 9 A Q ? 405. What would be remainder What would be remainder when 19
ÊÊ
is divided by 1200
2K Ë ? 406. If 3 9 then
407. Maximum how many vertices could be common in a regular dodecagon 12-sided Maximum how many vertices could be common in a regular dodecagon polygon and regular nonagon 9 nonagon 9-sided polygon
408. Find the maximum value of a2*b3*c4 if 2a3b4c 7 and a , b & c are value of a2*b3*c4 if 2a3b4c 7 and a , b & c are positive numbers 409. Find the Maximum value of a1*b2*c3 if a b c 3 and a , b & c are positive numbers
410. For natural number m, n, x & y if For natural number m, n, x & y if m*x n*y 100 then what would be last digit of m*x n*y 100 then what would be last digit of maximum value of » ∗ A 3 ? ? ?
411. If in a race A beats B by 100 m , B beats C by 100 m and in the same race if A can beats C by "X" m then what is the range of "X" by "X" m then what is the range of "X" a. 100 ³ 200 b. 100 b ³ 200 c. 100 ³ b 200 d. 100 b ³ b 200 e. NoT 412. If abcabc is a 6 digit number and has If abcabc is a 6 digit number and has 20 positive divisors then abc ? positive divisors then abc ?
413. Above image is actual time zone clocks Above image is actual time zone clocks and all are showing their local time all are showing their local time of a same time. If a flight departs from Central at If a flight departs from Central at 02:00 local time at central to city X and arrives at 10:00 and arrives at 10:00 local time at city X and then depart from city X at 11:00 local time at city X to Central and arrives at 09:00 local time at central on same day. Then CITY X is _______ arrives at 09:00 local time at central on same day. Then CITY X is _______ 414. If given figure is a 12 sided regular dodecagon then find angle X in degree is a 12 sided regular dodecagon then find angle X in degree is a 12 sided regular dodecagon then find angle X in degree
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In given fig if AB2unit, BC3unit, CDCA4unit and DA5 unit then find BD 415. In given fig if AB2unit, BC3unit, CDCA4unit and DA5 unit then find BD
If HCF of three natural numbers is same as their LCM then sum of all three numbers are 416. If HCF of three natural numbers is same as their LCM then sum of all three numbers are a. 2*LCM 3*HCF b. 5*HCF - 2*LCM c. HCFLCM c. HCFLCM d. NoT 417. If average of 5 different positive numbers are A and X is average of average of all possible pairs of these 5 numbers then possible pairs of these 5 numbers then a. AX b. g @ ³ c. A b X d. g ³ e. NoT e. NoT
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hour clock as shown in fig, in which outer dial 418. There is a 24 - hour clock as shown in fig, in which outer dial is representing hours and inner dial is minutes and seconds, then find the angle between hands X if time is 20:10. then find the angle between hands X if time is 20:10. 419. If A sold an article to B at 20% profit and B to C at 10% profit t If A sold an article to B at 20% profit and B to C at 10% profit then C sold to A at 20% hen C sold to A at 20% loss then find over all monetary % profit or loss of A. LAssume all transactions happen in \$M 420. For natural numbers x, y if For natural numbers x, y if 6 ∗ 44 A then x y ??? then x y ??? 421. Ù1
Ú Ú Ú⋯
Û ? ? ?
422. Find least palindromic natural number which Find least palindromic natural number which has odd number of digits , divisible by 11 has odd number of digits , divisible by 11 and has 9 as a middle digit. and has 9 as a middle digit. 423. If a palindromic number which has odd number of digits is divisible by 11 then middle digit of this number can not be digit of this number can not be a. 3 b. 2 c. 4 d. 6 e. All possible as a middle digit e. All possible as a middle digit
424. If A increases 10% then 20% then 30% then 40% then it becomes B, But if A increases , A increases 10% then 20% then 30% then 40% then it becomes B, But if A increases , 12% then 22% then 32% then 34% then it becomes C, Then 12% then 22% then 32% then 34% then it becomes C, Then a. BC b. B>C c. BbC d. NoT 425. Find maximum power if 20! Factorial 20 which perfectly divides 202! Factorial 202 Find maximum power if 20! Factorial 20 which perfectly divides 202! Factorial 202 426. Total number of real solutions of x³ sin x Total number of real solutions of x³ sin x 427. s6 2q6 2Z6 2r6 2√6 ⋯ ? a. √2 1
b. √3 3 1
c. 1
d. √3 1
428. Find the sum of common terms/elements of m3,7,11,15,....,103< & m1,7,13,19,....,103< Find the sum of common terms/elements of m3,7,11,15,....,103< & m1,7,13,19,....,103< www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 44
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show nth term of two different APs whose sum of first n terms are 429. If ÜÝ and ÞÝ show nth term of two different APs whose sum of first n terms are denoted by ßÝ and àÝ respectively. If respectively. If ߪª : à Þ¬ . : ªâ 17:11 then find Ü 17:11 º : Þ 430. For natural number x, y and z , For natural number x, y and z , 2 3o 5p 114 then x y - z ? z ?
431. If %profit is same as cost price of article and selling price is 131.25 then find its profit in rupees.
432. Find the concentration % of acid in final solution if 80 lit of 80% acid solution is first replaced by 20 lit water then 40 lit water. replaced by 20 lit water then 40 lit water.
433. If 1000th prime is a four digit number and it is rime is a four digit number and it is equal to 7909 X then X ?? 7909 X then X ?? a. 20 b. 6 c. 8 d. 10 434. Data Sufficiency : What is the cost Data Sufficiency : What is the cost price of the article I. If loss % is same as cost price of the article and selling price is Rs. 24. I. If loss % is same as cost price of the article and selling price is Rs. 24. II. If cost price is 20% of selling price. II. If cost price is 20% of selling price.
2 , 3 , 4 , 5 0-V 6 6
435. Which one is largest among largest among ã
ã
¿ã
Ëã
Àã
436. There are how many four digit number ABCD are possible such that AB, BC, & CD are perfect squares repetition of digits are allowed perfect squares repetition of digits are allowed
437. There are how many three digit numbers are possible say XYZ, if we interchange digits of unit place and hundredth place ZYX then digits of unit place and hundredth place ZYX then resultant number is also a three digit number is also a three digit number and | XYZ - ZYX| is divisible by 7 ZYX| is divisible by 7, such that _ Q 438. What is the last two digits of product of all positive divisors of 1024 digits of product of all positive divisors of 1024 digits of product of all positive divisors of 1024
439. A cone of height X is cut by a plane parallel to the base and at a distance is cut by a plane parallel to the base and at a distance Y from the is cut by a plane parallel to the base and at a distance base, then what is the ratio of what is the ratio of volume of the resulting cone and that of resulting that of resulting the ª « frustum ? where X & Y are pos X & Y are positive integral solution of ä ª â ; where Z is also a positive integer
å
æ
440. If ∆ABC is an equilateral triangle , and point D & E are on sides AB & AC such that , ∆ABC is an equilateral triangle , and point D & E are on sides AB & AC such that , DE||BC, and perimeter of DE||BC, and perimeter of ∆ADE is same as perimeter of quadrilateral BCED then find ∆ADE is same as perimeter of quadrilateral BCED then find ratio of DE:BC.
441. If speed of the group is 1 m/s and that of the joker is 3m/s. Joker starts from back of the last person and moves towards the front one, touches him and turns back, and goes, back to last person. If in this process The Joker covers 16 back to last person. If in this process The Joker covers 16 m when he is going forward m when he is going forward right hand side then find distance traveled by him when he turned back and moves towards left. assume speed of all remains same in entire process and no time loss by The Joker when he was turning back The Joker when he was turning back www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 45
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How many pairs of natural numbers m, n satisfy , where where 20>m 442. How many pairs of natura » 3 K
?
443. A quadratic function fx attains a A quadratic function fx attains a minimum of -5 at x -1. The value of the function at 1. The value of the function at x 0 is -3 . What is the value of f x at x What is the value of f x at x 5?
Direction for 446 to 448 : If the total intake in IIM A , B & C together is 900, and below chart is Direction for 446 to 448 : If the total intake in IIM A , B & C together is 900, and below chart is giving % of Male Mand that of Work giving % of Male Mand that of Work-ExEx in B-Schools. It is also known that 40% of whom Schools. It is also known that 40% of whom are in IIM-C are rest are equally divided between IIM A & IIM C. C are rest are equally divided between IIM A & IIM C. Male (M)
IIM A IIM B IIM C Total
444. What is the % of Male in IIM A What is the % of Male in IIM A
37.037% 33.333% 34.444%
Work Ex (Ex) 55.556% 66.667% 56.667%
445. What is the % of Fresher Fresher Non work Ex in IIM C
446. If in IIM -B , 50% work ex are male then what is the ratio of Male Fresher to Female B , 50% work ex are male then what is the ratio of Male Fresher to Female Work ex in IIM - B
447. One day Bagga with his dog planed to go to Park from their home which is 1200 m from home. Speed of Bagga was 30 m/s and that of his dog was 60 m/s. Both started from same time. His dog reached Park much earlier than Bagga, but dog returned back without wasting any time and came near to Bagga and again run towards Parks and asting any time and came near to Bagga and again run towards Parks and after reaching park came back to Bagga and continue the same process until Bagga reached the Park. If both maintained a constant speed in entire journey and there was no time loss in this process then find is process then find how much meter more did dog move towards park than towards Bagga. 448. If 2 is a root of a quadratic function fx0 and 7*f3f6 then find another roots of fx0
449. If WEIGHT of number a, ab , abc & of number a, ab , abc & abcd a single , double, triple and four digit number abcd a single , double, triple and four digit number respectively is defined by ; respectively is defined by ; WEIGHT a a ; WEIGHT ab ab ; WEIGHTabc bc - a and a and WEIGHTabcd cd-ab ; ab ; then for how many natural number less than 8995, then for how many natural number less than 8995, WIGHT WIGHT WIGHT of that number is 7. of that number is 7.
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450. ABCD is rectangle such that AB12cm and BC5cm, another Rectangle AEFC is such that point D lies on side EF then find the area of rectangle AEFC
451. If log log · 2√ 9¸ 0 then find x
452. If mileage of a car is 17 km per lit If mileage of a car is 17 km per lit with extra 3 lit per hour as wastage as wastage . What should be minimum speed of the car to cover 34 km in 3 lit. minimum speed of the car to cover 34 km in 3 lit. 453. W1 X W1 X W1 X W1 X … … … ? =K=
454.
K
K
K = K
?
K
K
⋯ … … … . ?
455. If 9th term of an AP is 20 20 then sum of first 17 terms of this AP is _____ sum of first 17 terms of this AP is _____
456. If the pendulum of a clock the pendulum of a clock takes 3 seconds to strike 3 o’clock. then h then how much time will it take to strike 9 o’clock? o’clock? 457. W
∗∗K
∗K∗
K∗ ∗
⋯
X ? ? ?
∗∗K
What would be remainder if 1 1 1 ? 1 is divided by is divided by 458. What would be remainder if
3 2
459. If mileage of a car is 15 km per lit with extra km per lit with extra 2 lit per hour as wastage . What should be lit per hour as wastage . What should be minimum speed of the car to cover minimum speed of the car to cover 45 km in 4 lit. 460. If an article was selling at 20% profit , to earn Rs 1320 more profit shopkeeper purchased article at 10% lesser than his cost price and sold it at 10% higher pri purchased article at 10% lesser than his cost price and sold it at 10% higher price than his selling price, then find oldoriginal selling price of his selling price, then find oldoriginal selling price of
461. Find the maximum value of Find the maximum value of a b, if following set of equations has infinite many roots has infinite many roots a-1*x 5*y 140 & 7 & 7*x a1*y b
462. What would be area of triangle formed by axes lines and a line which is perpendicular to a line 3x-2y6 and passing through 2,2 2y6 and passing through 2,2
463. If A is X% of B and B is Y% of A and X is A% of Y and Y is B% of X then what is X% of 20 If A is X% of B and B is Y% of A and X is A% of Y and Y is B% of X then what is X% of 20 464. If aaabcccd is a 8 digit perfect If aaabcccd is a 8 digit perfect square then abcd ???
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465. If ratio of income of X &Y is 3:4 and their expenditure ratio is 4:7 then then whose % saving on income is more whose % saving on income is more a. X b. Y c. can't say d. NoT
466. L√1ML√2ML√3ML√4M.....L√24M ??? ;Where LxM is defined as nearest integer, L √1ML√2ML√3ML√4M.....L√24M ??? ;Where LxM is defined as nearest integer, L3M3, √1ML√2ML√3ML√4M.....L√24M ??? ;Where LxM is defined as nearest integer, L L4.3M4 & L4.51M 5 467. How many integral solutions are possible for L√ M 25 ;Where LxM is defined as ;Where LxM is defined as nearest integer, L3M3, L4.3M4 & L4.51M 5 nearest integer, L3M3, L4.3M4 & L4.51M 5
;Where LxM is defined 468. What is the sum of all integral solutions possible for L√ M 25 ;Where LxM is defined as nearest integer, L3M3, L4.3M4 & L4.51M 5 as nearest integer, L3M3, L4.3M4 & L4.51M 5 469. 4 4 4 . . . . 4 = ;ÌV 17 ? ? ? 470. 4 ∗ 4 ∗ 4 ∗ . . . .∗ 4 = ;ÌV 17 ? ? ?
471. 2= 2 2 . . . . 2 ;ÌV 13 ? ? ?
472. If A,B,C,D,E, & F are points on a circle such that ratio of le If A,B,C,D,E, & F are points on a circle such that ratio of lengths of arc ngths of arc AB , arc BC, arc CD, arc DE, arc EF, arc FA is 1:2:3:4:6:6 then what is the value of angle CDF FA is 1:2:3:4:6:6 then what is the value of angle CDF FA is 1:2:3:4:6:6 then what is the value of angle CDF 473. 1234512345.... 300 digits mod 41 ??? 1234512345.... 300 digits mod 41 ???
474. If A,B,C,D,E, & F are points on a circle such that ratio of lengths of arc AB , arc BC, arc If A,B,C,D,E, & F are points on a circle such that ratio of lengths of arc AB , CD, arc DE, arc EF, arc FA is 1: arc EF, arc FA is 1:1:2:2:3:3 then what is the value of angle then what is the value of angle ACE 475. If fxx-1x5x-9...x61 , then for how many integral "x" fxb0. 9...x61 , then for how many integral "x" fxb0. 9...x61 , then for how many integral "x" fxb0.
476. Ram wrote first "N" natural numbers on a black board then removed one number and found that new average becomes found that new average becomes 21.1 then find which number was removed ? 21.1 then find which number was removed ? 477. If A,B C & D are points on circle and E,F,G & H are mid points of side DA, AB, BC & CD. Find the area of quadrilateral EFGH if AB12cm, BC6 cm, CD8cm & DA 10 cm. AB12cm, BC6 cm, CD8cm & DA 10 cm.
478. What is the area of a right angled triangl What is the area of a right angled triangle whose inradius and circum radius are 10 and e whose inradius and circum radius are 10 and 12 cm respectively 479. What is the area of a right angled triangle whose inradius and circum radius are 4 and 12 cm respectively
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3E LEARNING - MATHS BY AMIYA 500 CAT 2016
480. Inside an equilateral triangle there is a point from which the length of perpendicular on all the three sides are 2 lar on all the three sides are 2cm, 3cm, & 4cm then. what would be area cm then. what would be area of equilateral triangle.
481. If fresh grapes has 90% water content and dry grapes kishmish has 30% water content. In the procedure of making kismish one just need to dry grap content. In the procedure of making kismish one just need to dry grapes, then if a shopkeeper purchased 210kg of grapes at 30 Rs per kg then at what price per kg he should sell kismish to earn 33.33% profit. should sell kismish to earn 33.33% profit.
13x19...x115; Then for how many integral x , fxb 0 Then for how many integral x , fxb 0. 482. If fx x-1x7x-13x19...x115 483. If in the given fig , AD3cm , DF 1cm, FC 2cm & , AD3cm , DF 1cm, FC 2cm & EF || BC then
PçèP 5é ∆deê
find
PçèP 5é ∆ëêì
484. If in the given fig , AD3cm , DF 1cm, FC 2cm & EF || BC then
PçèP 5é ∆deê
find
PçèP 5é ∆def
485. If in the given fig , AD3cm , DF If in the given fig , AD3cm , DF 1cm, FC 2cm & EF || BC
PçèP 5é êìfe êìfe
then find
PçèP 5é ∆def def
486. In the given fig if then find the area of ∆íîï then find the area of íîï
BD:DCCE:EA1:2 and area of ∆AFO is 144 sq. cm, BD:DCCE:EA1:2 and area of ∆AFO is 144 sq. cm,
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3E LEARNING - MATHS BY AMIYA 500 CAT 2016
487. If in the given fig ABCD is parallelogram, O is a point inside the parallelogram. If areas of three triangles are given check fig then find the fourth triangle.
area of
488. In the given triangle D, E & F are mid points of respective sides then find the area ratio of quad find the area ratio of quad BGOF to that of ABC
489. If in the given fig ABCD is quadrilateral and E,F,G H are mid points of respective sides as shown in fig. Areas of three quadrilaterals are given check fig then find the area of fourth quadrilateral.
&
490. If angle ratio of a triangle is 1:2:9 then what is the side ratio of this triangle If angle ratio of a triangle is 1:2:9 then what is the side ratio of this triangle a. 1:2:9 b. 18:9:2 c.√3 ∶ √2 ∶ 2 d. √3 1 ∶ √2 ∶ 2 e. NoT
491. If N the least natural number which is subtracted from 10000*10001*10002*10003 to make it a perfect square then N mod 9 ??? square then N mod 9 ??? 492. Sum of all coefficients of expansion of x1x2x3....x15 is S then S mod 17 of expansion of x1x2x3....x15 is S then S mod 17 ?
493. If an alloy is having Zn, Cd and Fe in the ratio of 3:4:9 and another alloy has the same elements in the ratio of 2:3:4 then find the elements in the ratio of 2:3:4 then find the ratio of these elements in the same order of ratio of these elements in the same order of first alloy and second alloy is mixed in the ratio of 2:3 first alloy and second alloy is mixed in the ratio of 2:3 www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 50
3E LEARNING - MATHS BY AMIYA 500 CAT 2016
Total number of positive integral solution of 10x 7y 7z 100 is ______ 494. Total number of positive integral solution of 10x 7y 7z 100 is ______
495. If añ b ñ c ñ dñ is always divisible by abcd , where a,b,c,&d are natura is always divisible by abcd , where a,b,c,&d are natural is always divisible by abcd , where a,b,c,&d are natura numbers and p is a prime more than 101 then numbers and p is a prime more than 101 then a. a, b , c& d are odd numbers a. a, b , c& d are odd numbers b. c, d , a & b are in AP c. a, b, c & d are in HP d. a, b, c & d are in GP e. NoT
496. If 32A65B is divisible by 72 then what would be remainder if AAABBB is divided by 7, where A & B are two digits both could be same. ere A & B are two digits both could be same. 497. Shopkeeper A : Gives 25% extra and charges 20% less than market price Gives 25% extra and charges 20% less than market price Gives 25% extra and charges 20% less than market price Shopkeeper B: Gives 20% extra and charges 25% less than market price Gives 20% extra and charges 25% less than market price Gives 20% extra and charges 25% less than market price
If both shopkeepers sell same product of same market price then for a customer which shopkeeper is better... ???? shopkeeper is better... ????
498. There is a tilted glass in perfect cylindrical shape with base in perfect cylindrical shape with base radius 7cm and height 10 cm. Glass radius 7cm and height 10 cm. Glass is partially filled with water, in such a way that, that, water level is just touching bottom end and top end of the glass as shown in bottom end and top end of the glass as shown in the figure. Then find the amount of water figure. Then find the amount of water in litter inside the glass.
499. Find the total number of positive integral solution of o p 1 1
500. Amiya and Raman is playing a game Toss laying a game Toss-Toss. Raman wins when tails comes and Toss. Raman wins when tails comes and losses when heads come. Raman gains Rs 10 for tail and loses Rs 10 for head. If Raman wins in first toss then he quits but tries only once more if he losses on the first toss. Then in this game how much could be expected win of Raman. uch could be expected win of Raman.
To get more questions follow
3E LEARNING - MATHS BY AMIYA 500 CAT 2016
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3E LEARNING - MATHS BY AMIYA 500 CAT 2016
SOLUTION :Maths Maths By Amiya 500 (2017(2017 19) 1. Mohit calculated sum of first N natural number and he found sum is 337 and he knows he counted one number twice or missed one number. then find his minimum possible % error made by him. a. 3% b. 3.99% c. 3.69% d. 3.23% Sol: [c] Actual sum should be near to 338 for minimum error, so it should be either 325 or 351 % error on 325 = 12/325*100 = 3.69% % error on 351 = 14/351*100 = 3.99 2. The integers 1,2,….. 30 are written on a board. A person came and erased any two numbers say "a" & "b" and wrote a new number "a+b+2" this pro process cess is done by total 29 persons (including first one) then . What is the number left on the board at the end? Sol: Every time sum would increase by 2 , so total increase = 2*29 = 58 So last number = ∑ « ò¯ = 465+58 = 523 3. If T1=(1) , T2= (3,5) , T3 = (7,9,11) , T(4) = (13,15,17,19) ..... Then what is the sum of all terms of T(10) Answer : 1000 4. phi (n) is defined as number of co co-prime prime less than n. If ‘P’ is product of two different prime numbers, whose sum is 1200 then what is the max phi(M) Sol: phi (n) = n *(1 -1/p)*(1-1/q) 1/q) and so on ... where p and q are primes in prime factors of n. since we need this max so by gó @ ôó ôó ,, value should be close means both prime should be closer to 60 for max phi , so 59 & 61 ; P = 59*61 & phi(P) = 58*60 = =3480. 5. Consider the set S = {1,2,3,………..10000}, How many APs can be formed from the elements of S that start with 1 and end with 10000 and have minimum 3 terms?
Ans: 11 ; a=1 ,Tn=10000 , so 1 + (n (n-1)d = 10000; ; (n-1)d 1)d = 9999 = 3^2*11*101 so 12 factors. n-1 cannot be 1. so 11 6. Total number of integral solutions of 13x - 3y = 1000 for 100< x < 200 Sol: 33 , all numbers which gives remainder 1 when divided by 3 7. For how many integral "n" is
is an integer
Sol: [12] 7 21 7 9 mod (x--3) = 12 , so x-3 3 should be a factor of 12 means 12 values are possible including negative.
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3E LEARNING - MATHS BY AMIYA 500 CAT 2016 8. If HCF , LCM and sum of two numbers are 6 , 15 and 23 then find their difference. Sol: Not a possible case 9. If 1 7 6 3 , then find the value of f(x - 1). Sol: put x = x -2 ,âõ® ®®õ ª¬ 10. Find the area of the enclosed fig by | x- 5 | öõ ö âú öõ – ºú « ®
ª
ª
ª
So 343 216 so 560 integral solution.
13. When a number divided by 6 and 35 remainders are 5 and 7 then what would be remainder when 11 times of the number is divided by 15 a. 8 b. 13 c. 7 d. Data inadequate e. NoT Sol: By CRT 77 is least number for 210 and we want 11*77 mod 15 = 7 14. In a circle AB is diameter of length 34 unit and BC is chord of length 16 unit. If CD is perpendicular on AB such that D lies on AB then what is the length AD. Ans: 450/17 , can do by similarity Ans: ª¯√® ; Area = ÷® ∗ àøÝù
15. What is the area of a rhombus whose one side is 6 unit and one internal angle is 135 degree. 16. If major (longest diagonal) of a rhombus is 6√3 cm and one internal angle of rhombus is 120° then what is the area of rhombus.
Ans :ª¯√
17. 113 100 is divisible by 209 (True / False) ANS: TRUE 18. 113 100 is divisible by 247 (True / False) ANS: TRUE 19. 113 100 is divisible by 143 (True / False) ANS: TRUE 20. 113 100 is divisible by 71 (True / False) ANS: TRUE
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3E LEARNING - MATHS BY AMIYA 500 CAT 2016 21. 113 100 is divisible by 781 (True / False) ANS: TRUE 22. What is the rate per annum if compound interest after 2 years is same as simple interest after 3 years. Ans: 100% Ans: ∶
ò°«
∶ °ª
â
23. At what time between 3&44 both the hands are equi equidistant from figure 6 but not together ª
ª
24. What is the least natural number N for 101^N mod 280 = 1 Ans: N =6 ,cyclicity 280 = 2^3*5*7 ; no need to check for 5 , since 101^n mod 5 = 0 , for all n ; also not for 2^3 since 101^(2m) mod 8 = 1, for all m ; now just check 7 , cyclicityoteuler of 7 is 6 which satisfy all conditions 25. If A takes "a" hours more than time taken to complete a work when A & B work together and B takes "b" hours more than time taken to complete a work when A & B work together. Then what is efficiency ratio of A and B 1. a : b 2. b:a 3. √.: 0 4. √0: . 5. b^2 : a^2 Ans: [3] 26. If X + (1/Y)=1 , Y + (1/Z) = 1 then what is the value of X*Y*Z , if none of X, Y & Z are 0 Ans: -1 ; solve by putting values 27. What is the minimum value of of2123 2451
Ans: ®
ªW
ª X √®
at x = 225
√1! √2! √3! ⋯ … . √1000! ? Ans : 1*7+2*19+3*37+4*61+5*91+6*127+7*169+8*217+9*271+10=6985 28. If [ N ] is greatest integer less than equal to N then
29. What is the coefficient of x^5 in the expansion of (1+x)(2+x)(3+x)(4+x)(5+x)(6+x) Ans: 21 , -(sum (sum of all roots)
30. If P = { x /x : x 15 , \$% -0*890: -8;.,9 < ; then how many subset of P would have at least one prime number Ans : ®ªò ®¬
31. 1 2 3 4 … . . 101 ? Ans :5151
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3E LEARNING - MATHS BY AMIYA 500 CAT 2016 32. ( 2 ; *+,- = … … ?
Ans : 0 33. Find the cubic equation whose roots are one more than roots of 1 0 Ans: x^3-2x^2+2x= 0 ;Put x = x--1 34. What is the minimum value of sinx + cosecx + tanx + cotx Ans :infinity 35. What is the sum of first 1000 terms of series - 1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,.... Ans : 1956 ; 2*1000-44= 44= 1956 ; just check how many one's by checking positions of 1's. 1,3,6,10...990 so 44 36. DS - What is the ratio of Cost Price to Selling Price A. Markup % is 26% more than Profit % and 10% more than Discount % B. Markup is Rs 26 more than Profit (RS) and Rs 10 more than Discount (in RS) Ans: A alone. 37. What would be remainder er if 98! Is divided by 101 ? Ans: 50 38. What is the sum of all internal angles of this star Ans :(11-5×2)×180=180
)...(1+x^99) 39. Coefficient of x⁴ in the expansion of (1+x)(1+x²)(1+x³)(1+x⁴)...(1+x^99) Ans: 2 40. Max number umber of segments we can create on a plane (open or closed ) by the help of a circle , a trianglee and a line is ______ Ans: 12 41. If magnitude of speed in Km/hr of a man is same as rest time in minute in an hour of that man. Find maximum distance covered by man in 5 hours. Ans :x(5-x/12), 5x-x^2/12,5=x/6,x=30, x^2/12,5=x/6,x=30, so 30*(5 30*(5-30/12),150-25*3=75 42. If N is smallest est prime number which is equal to sum of three consecutive prime numbers then what is the sum of digits of N Ans: 4 ; 23 = 5+7+11 www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 56
3E LEARNING - MATHS BY AMIYA 500 CAT 2016
73% are passed in Phy , 89% in Che, 69% 69 in Bio & 43. If in an examination of PCMB , 73 71% in Maths, then minimum how many % people have passed in all four papers. Ans : 2% Work on students who failed in each sub - 27, 11, 31 & 29 , means max 98% fail in at least one so minimum 2% pass in all 44. Total number of positive integral solution of 3x + 2y = 100 such that @ A Ans : 7
45. If in an examination of PCM , 75 75% are passed in Phy , 65% in Che, & 70% 70 in Maths, then minimum how many % people have passed in all three papers. Ans: 10% 46. If in an examination mination of PCM , 60 60% are passed in Phy , 65% in Che, & 70% 70 in Maths, then max how many % people have passed in all three papers. Ans : 47.5% 47. For a natural "n" 2^(12n) - 6^(4n) is div by a. 10 b. 20 c. 50 d. 100 Ans: All
e.All
48. Mohan sells out a toy at 25% profit. Had he purchased at 25% less and sold it for Rs 25 less, then he would have still gained 25%. Find the cost price of toy Ans: 80 49. If sin ∅ cos ∅ then for 0 ∅
Ans :8/3
H
; tan ∅ cot ∅ ? ?
50. Product of first 24 prime number is not divis divisible by a. 391 b. 371 c. 247 d. 279 Ans: [d] 279
e.NoT
51. If product of two sides of an integral sided triangle is 6 then triangle is always I. Acute Angled II. Obtuse Angled III.Right Angled IV. Isosceles V. Equilateral VI. Scalene VII. None of the above is always correct Ans : VII Not ; sides are 2,3,3 & 2,3,4 & 2,3,2 52. How many integral sided isosceles triangle is possible , if sum of two sides is 20 Ans: 31 Sol: If 20 is sum of equal sides, then sides are 10,10,x so total 19 triangles are possible If unequal sides sum id 20 the www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 57
3E LEARNING - MATHS BY AMIYA 500 CAT 2016 Smallest side , 1,2,3,4,5,6, only one iso triangle is possible so 6 possibility 7,8,9 - two iso triangles are possible so again 6 possibilities Total isosceles triangles are = 19 + 6 + 6 = 31 53. What is the maximum area of a triangle - if it is known that sum of any two sides of this triangle is less than equal to 20 unit. Ans: It should be (1/2) *a*bsinC , for max sinC should be 1 , so max area is (1/2)*a*b = 1/2*10*10 = 50 max
54. Fig is a regular octagon then what is the measurement of angle ADH Ans: 22.5 degree
55. Find the max value of n for which 124K 1 is divisible by 53 Ans: 5; 125*50 = 5^5 * k 56. What would be remainder if 344^49 - 1 is divided by 7^5 Ans: 0 57. What is the minimum value of x^2 - 4x +3 Ans: -1 , at x=2 58. What is the minimum value of sin^2 x - 4sin x +3 Ans: 0 , at 90 59. If a , b & c are sides of a right angled triangle and natural numbers then what would be remainder if a*b*c is divisible by 15 Ans: 0 60. (1 - cot1)(1 - cot2)(1 - cot3).....(1 - cot42)(1 - cot43)(1 - cot44) is a. A perfect Square b. A perfect cube c. A perfect square as well as a perfect cube d. Irrational Number e. NoT Ans: [a]; 2^22 , a perfect square 61. If N = (tan46-1)(tan47 -1)(tan48 1)(tan48 -1).....(tan87 -1 )(tan88 -1)(tan89 -1) 1) , then N mod 7 = ? Ans: 2 , 2^22 mod 7 www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 58
3E LEARNING - MATHS BY AMIYA 500 CAT 2016 62. If x + y + z =1 , x^2 + y^2 + z^2 =2 & x^3 + y^3 + z^3 =3 , xyz = ? Ans: 1/6 (x + y + z)^2 = x^2 + y^2 + z^2 +2 (xy + yz + zx) 1 = 2 +2 (xy + yz + zx) =>xy + yz + zx = -1/2 x^3 + y^3 + z^3 - 3xyz = (x + y + z) ( x^2 + y^2 + z^2 - (xy + yz + zx)) 3 - 3xyz = 1(2 + 1/2) = 5/2 3xyz = 1/2 ; xyz = 1/6 63. DS If f(f(x)) = f(f(f(x))) then f(x) = ? A. It is not a constant function B. [f(x)]^2 = f(x^2) Ans: Only A ;If If f(f(x)) = f(f(f(x))) ; It means f(x) = k or f(x) = x A. It is not a constant function so f(x) = x [sufficient] B. [f(x)]^2 = f(x^2) does hold for f(x)=k so f(x) = x , f(x) = 0 and f(x) = 1 , not sufficient only A is sufficient si of rectangle 64. If ratio of perimeter of a rectangle and that of a square is 5:1 and ratio of one side to one side of square is 3:2 then what is the ratio of area of square to that of rectangle. Ans: 4:51 (l+b) : a = 10 : 1 = 20:2 & b / a = 3:2 so a:l:b = 2 :17 :3 ; a^2 : l*b = 4 : 51 65. If ABCDEFGHIJ are a regular polygon then what is meas measurement urement of angle EHA Ans: 72 degree 66. A husband alone can do a piece of work in 60 hours & wife alone in 40 hours, but due to a baby who always destroy their work they together take 16 more hours to complete their work Then in how many days baby alone can destroy all the work. Ans: 60 67. If a + b +c = 5 then what is the maximum value a^2 + 4*b^2 + 9*c^2 Ans: Infinity 68. In a section of PGP18 of IIM IIM-X X the average weight of 30 students is 60 kg. If x new students join the section and average wt of these new students are "70 - X" kgs, then find the maximum possible average weight (approx) of the section after joining new students. a. 60.66 b. 60.73 c. 60.71 d. 60.56 e. NoT Ans: [c] , for x = 5 69. Arrange A, B & C in ascending order If A = pi^(1/pi) , B = e^(1/e) & C = 1 , where pi is 3.14 , e = 2.71 Ans : C,A, Bhttps://youtu.be/4MWD5C5qt84 https://youtu.be/4MWD5C5qt84 70. If log P . log P . log P . 60 then log P . ? ?
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3E LEARNING - MATHS BY AMIYA 500 CAT 2016 Ans: 40
71. If ? 1 is one factor of f(x) = 0 ∗ K . ∗ ? U ∗ V ∗ , ∗ then what is the real roots of f(x)=0 a. 1 b. -1 c. -f/a d. a*b*c*d*e*f e. NoT Ans:[c] ÷ ∗ õò û ∗ õ° ü ∗ õ ý ∗ õ® þ ∗ õ = a(õ° õ õ® õ ª ∗ õ R is real root then by equating constant term R = - f/a 72. There are 6 identical toys and three boys. In how many ways all toys would be distributed among three boys. Ans: 28 ways. (0, 0, 6), (1, 1, 4), (3, 3, 0), (2, 2, 2), (1, 2, 3), (1, 5, 0) and (2, 4, 0) 73. In JhumriTillaiya a Paamwala priced his beedi at 85 paise per beedi, but after budget, he reduced the price of beedi and andand soldbeediof beediof Rs.77.28 in a day. Then what is the total number of beedies he sold in a day day? a.37 b. 47 c. 84 d. 92 Ans: [d] check by options 74. In AB and BC are two chords of a circle, then find length of chord AC C if AB=BC=6cm and radius of circle is 5cm Ans: 9.6 cm 75. If third term of a GP is 4096 andits common ratio is positive.The Product of first 7 terms is less than that of first 6 terms and Product of first 6 terms is greater than that of first 5 terms. then which option is best describing range of common ratio. a. [ 1/32 , 1/16] b. ] 1/32 , 1/16[ c. [ 1/16 , 1/8] d. ] 1/16 , 1/8 [ e. Not Ans: [d] P7 < P6 => P6*T7 < P6 so T7 < 1 P6 < P5 =>P5*T6 > P5 so T6 > 1 T7 = 4096*r^4 < 1 so , r < 1/8 T6 = 4096*r^3 > 1, so , r > 1/6
76. Ans: 23 degree angle by same chord
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3E LEARNING - MATHS BY AMIYA 500 CAT 2016 77. All words having three different vowels and three different consonant are written in dictionary then what would be rank of word BAEI BAEICD Ans: 137 Keeping A fixed in the first place, we have 120 words. Keeping BAC fixed, we have 6 words Keeping BAD fixed, we have 6 words Now comes BAE (CDI, CID, DCI, DIC) after which BAEICD will come. Hence rank = 120+6+6+4+1 = 137 78. In group of 12 people, 4 speak on only Hindi, 5 speak only Tamil and the rest speak both Hindi and Tamil.. In how many ways can the 112 people be arranged in a row such that so are all those who speak only Hindi are together and so are all those who speak only Tamil. Further her all people should be able to converse with both their neighbours. Ans :_HBT_ or _TBH_ 79. If a three hree digit number xyz has only 5 factors then what is the last digit of sum of all factors of six digit number "xyzxyz" . Ans:4 , xyz = 625 then n whose expenditure 80. If Income ratio of A, B & C are 3:7:9 & their saving ratios are 3:1:4 the is maximum ? Ans: CBD 81. If Income ratio of A, B & C are 3:7:9 & their saving ratios are 3:1:4 the then n whose expenditure is minimum? Ans: A Direction for Q 82 & 83 - DOVAFONE has two monthly tariff plans for calling - details are given below Name of Plan P300
Fixed Monthly Charge ` 300
P100
` 100
Benefit 600 min free , after that `1 per min , fractional charge is applied 30 paisa per min, fractional charge is applied
Fractional Charge - If you talk in fraction of min charge would be in fraction of that tariff .eg if extra 1.5 min pay extra ``1.5 in P300 and 45 paisa or 1.66 min then `1.66 1.66 in P300 and 50 paisa in P100 82. If one person talks 500 min per month then which plan is better. Ans: P100 83. If one person talks N min per month and for him both plans cost the same. Then N = ? Ans: Not Possible
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3E LEARNING - MATHS BY AMIYA 500 CAT 2016 84. If ∗ W X W X and f(3) = 82 then what is f(5) = ?
Ans : f(x) = 1 + x^n ; f(3) = 1 + 3^4 = 82 , f(x) = 1 + x^4 f(5) = 1 + 5^4 = 626
85. If angles of a triangle are 30 30° , 60° and 90° then what is the ratio of it's in-radius in to circum radius Ans: r/R = cos 30 + cos 60 + cos 90 - 1=
√ ®
ª ª ®
√ª ®
86. which one is largest among all option options K a. 12 14 b. 13 15? c. 12K 14 d. 15 13? Ans: (a) 87. What is the sum of all external angles of a 8 sided concave polygon. Ans: 1800 ; 180(n+2) 88. If ABCD is a prallelogram. Point E and F are mid points of side BC & CD respectively then what is the area ratioo of Quard AGFD to that of parallelogram ABCD.
Ans :11:20 Solve it by assuming ing a square or rectangle coordinate
on
89. How many numbers are the in the set of first 1000 natural numbers which can be written as sum of two or more consecutive natural numbers. Ans: 990 , all except 2^n terms 90. Total number of integral solution of w^4 + x^4 + y^4 + z^4 = 1797 Ans: 0 , check divisibility by 7 91. If Y √123 ∗ 124 ∗ 125 ∗ 126 1 then what is the digital sum of N (N mod 9) Ans: 1 92. Total number of integral solution ofa³ + b³ + c³ = 43655 Ans : 0
93. If diagonals of a parallelogram are 30cm and 10 cm then among the options which could be sides of parallelogram a. 22 cm & 4 cm b. 20cm & 10 cm c. a & b both possible d. NoT
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3E LEARNING - MATHS BY AMIYA 500 CAT 2016 Ans: [b] - "a" cant be since we can not make 4cm with diagonals (as part of triangle) 94. If A,B& C three cities are in a highway. Distance between two cities A & B is 200 km and city C is 100 km from B & A both. From city "A" a man starts towards "B" with speed of 30 km/hr and on the same time another man start startss from B towards "A" with speed of 20 km/hr. After how much time they exchange their speeds (new speed of man from A and man from B becomes 20 kmph and 30 kmph respectively ) so that they meet at C Sol: After 2 hours 95. If a , b & c are in GP then roots of 0 . U 0 has a. Real b. Equal c. Imaginary d. (b) or (c) Ans: [c]
e. Not
96. If O is centre of the circle, angle DOC = 42° then what is the measurement of angle BFC Ans: 69
97. Sum of two positive integers A &B and its LCM is 455. Then how many unordered pairs of A & B are possible. Sol: Let number be h*x & h*y , h being HCF of numbers. then hx + hy + hxy = 455 Ans: 14; h(x+y+xy)=455 = 5*7*13 So , h = 1, 5, 7, 13 , 5*13 , 7*13 , 5*7 & 5*7*13 if h = 1 , x+y+xy = 455 ; (x+1)(y+1)= 456 sso 5 cases if h = 5 , x+y+xy = 91 ; (x+1)(y+1)= 92 so 2 cases if h = 7 , x+y+xy = 65 ; (x+1)(y+1)= 66 so 2 cases if h = 13 , x+y+xy = 35 ; (x+1)(y+1)= 36 so 3 cases if h = 65 , x+y+xy = 7 ; (x+1)(y+1)= 8 so 2 cases 98. If O is a point inside a parallelogram AB ABCD such that areas of ∆ABO ,∆ ∆BCO & ∆CDO are 12cm^2 , 15 cm^2 and 10 cm^2 then what is the area of ∆ AOD. Ans: 7 cm^2
99. If in ∆ABC ABC , AB = 6 cm , BC = 7 cm and CA = 8 cm , AD is an internal angle bisector of angle A such that point D lies on side BC then le length of AD = ? Ans :AD = 6 cm www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 63
3E LEARNING - MATHS BY AMIYA 500 CAT 2016 where AB = c, BC= a & CA=b í Zû ∗ ü ö ª ûü® úwhere ÷®
100. A34 56B is a 6 digit number in base 12 and when we write this in base 10 then it is divisible by 143 then A * B = ? Ans: 3 , (1,3) 101. If a , b, c, & d are sides of given quadrilateral ABCD then area of ABCD 1. a*b*c*d 2. √0 0∗.∗U∗V 3. 3√0 ∗ . ∗ U ∗ V
Ans: (2)
4. Z
P P4
[^
102. For how many different (0 0 _ . _ U non zero digit ordered combinations (a,b,c) 0[ b 0[ 4 a
Sol:
Case 1 : b =1 , 8*7 = 56
103. In a trapezium ABCD , E & F are mid points of its diagonals and AB || CD . Find length of side CD if AB = 12 cm & EF = 4 cm. Ans: 20 cm. ( a-b/2=EF) 104. In a trapezium PQRS , S & T are mid points of its diagonals and PQ || RS . Find length of side RS if PQ = 12 cm &ST ST = 3 cm. Ans: 6cm or 18 cm 105. If [ x ] denotes greatest integer less than equal to x then 1 1 1 1 1 1 1 1 1 1 \ ]\ ]\ ]\ ] ⋯ ? 3 3 9 3 9 27 3 9 27 81 Ans: 0 106. Three students have each brought his father and mother for admission to a school. The admission head wishes tointerview the nine people one by one, taking care that no child is interviewed before its mother and no husband is interviewed before his wife. wife In howmany different ways can be interviews be arranged? Ans :8*9!/216 [in 2^3 ways father and child can be arranged] 107. In how many ways can we select two squares on a chessboard such that they share either a common side or a common vertex? Ans: 210
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3E LEARNING - MATHS BY AMIYA 500 CAT 2016 108. A person invests Rs.3000 in 3 months fixed deposit scheme of a bank at the rate of 13%. Due to some changes in policies, the rate of interest changed in every three months, after the first period, to 12%, 11% and 10% respectively. Assuming that the person withdraws the interest after every 3 months but continues deposit, how much more interest would that person have earned in oone ne year, if there was no change in the interest rate? a. 50 b. 45 c. 0 d. 180 Ans: [b] 12% , 11% and 10% means reduction of 1% , 2% & 3% =6 % cumulative change (quarterly) Change = 3000*6*1/100*4 = 45 admission ission in health course. The instructor 109. Three wives have each brought his husband for adm wishes to interview all six people one by one, taking care that no wife is interviewed before its husband. In how many different ways can be interviews be arranged? (1) 120 (2) 100 (3) 72 (4) 90 Ans: [4] 90 Total tal ways = Total Ways / ways in which pairs can attached = 6! / (2!*2!*2!) = 90 110. A passenger is planning a trip that involves three connecting buses that leave from Ambikapur, Bokaro and Chandanpur respectively. The first bus leaves Ambikapur every hour, beginning at 8:00 am, and arrives at Bokaro 2.5 hours later. The second bus leaves Bokaro every 20 minutes, beginning at 8:00 am, and arrives at Chandanpur 7/6 hours later. The third bus leaves Chandanpur every hour, beginning at 8:45 am. What is the least total amount of time the passenger must spend between buses if all buses keep to their schedules? a. 25 minutes b. 1 hour 5 minutes c. 1 hour 15 minutes d. 2 hours 20 minutes Ans: 65 min =1 hour 5 minutes Regardless of the time of departure from Ambikapur, arrival at Bokaro will be at 30 minutes past the hour. Buses leave Bokaro on the hour, and at either 20 or 40minutes past the hour. Therefore, the earliest a passenger from Ambikapur could leave Bokaro would be 40 minutes past the hour with a 10 minute wait betweenbuses.. The bus from Bokaro to Chandanpur takes 7/6 hours or 1 hour 10 minutes. A bus taken at 40 minutes past the hour, causing the passenger to have missedthe bus from Chandanpur by 5 minutes.. The passenger therefore has a 55minuteswait, and the least total amount of time the passenger must spend between buses is10 + 55 = 65 minutes or 1 hour 5 minutes. 111. The lengths of the sides CB and CA of a triangle ABC are 4cm and 6cm respectively and the angle C is 120°.. If the line CD bisects the angle C and meets AB at D, then the length of CD is b. 3cm c. 2.4cm d. None of these a. 4 cm
Ans: [c] In Triangle ABC üà ª®« ª®« x = √76 So AD= (2/5)*√76 UÌ% 60
Ë
? o W ∗√ =X
∗?∗o
°® º® õ® ®∗°∗º
; õ íî
, A kl , y = 2.4
Direct Formula = a*b/(a+b) = 4*6/(4+6) if we bisect 120 120° www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 65
3E LEARNING - MATHS BY AMIYA 500 CAT 2016
112. An alloy X consists of 10% Cadmium and 6% zinc and alloy Y consists of 5% Cadmium and 10% zinc. If he needs at least 14 kg of Cadmium and 14 kg of zinc for his h experiment. If alloy X costs 60 paisa per kg and alloy X costs 40 paisa per kg, What is the minimum cost he would occur to fulfil his requirement a. Rs.72 b. Rs.82 c. Rs.92 d. None of these Ans: [c] ; Let a & b are wt of alloy X & Y respectively 0.1a + 0.05b = 14 ....(1) 0.06a + 0.1b = 14 ....(2) a= 100 & b = 80 , so minimum cost = 92 113. What is the unit digit of 1 2 3 ⋯ . 9 a. 0 b. 1 c. 5 d. 9 Ans: [c] 1 9 is divisible by 10 so 2 8 and so on, only 5 left so unit digit = 5 = 5
114. If in a hotel Ram checked in between 1 and 2 o’clock and cheeked out in between 4 and 5 o’clock , if positions of minute hand and hour exchange and maintained same position then at what time did Ram check-out out from hotel? a. 4: 11: 15 b. 4: 11: 45 c 4: 12: 15 d. None of these Ans: [b] no need to solve check time it should be between 4:5 :.. to 4:10:00 115. If numerators of three fractions are in A.P. and their denominators are in G.P and common difference of the numerators is equal to the common ratio of the ddenominators enominators . The product of the first fraction and reciprocal of the second fraction is 6/5 and the product of the second fraction and reciprocal of third fraction is 15/8. If the, then the denominator of the third fraction is a. 9 b. 18 c. 27 d. cannot be determined Ans: [d] , 2/a ; 5/3a; 8/9a Assume numerators be a -d d , a &a+d and denominators be b/d . b& b*d in same order, from question we can get a= 5 , d=3 but cant get b , so cant be determined. 116. Two quadratic equations with positive roots have on onee common root. The sum of the product of all four roots taken two at a time is 143. The equation whose roots are the two different roots is – 14 45 0. The sum of all different roots is a. 20 b. 21 c. 22 d. 24 Ans: [b] , roots are 5,7,9 From – 14 45 0,, we can say two different roots are 9 & 5 , and sum of product of all four roots taken two at a time is 143 so 9*5 + 9*a + 5*a = 143, so a = 7 117. There are 10 pipes that are connected to a tank. Some of them are inlet pipes and the others are outlet pipes. Each of the fill pipes can fill the tank in 6 hours and each of the outlet can www.facebook.com/groups/MBAMathsByAmiya 3E Learning Maths By Amiya Ranchi | 66
3E LEARNING - MATHS BY AMIYA 500 CAT 2016 empty the tank completely in 4 hours. If all the fill pipes and outlet pipes are kept open, an filled tank gets emptied in 10 hours. How many of the 10 pipes are inlets pipes? c. 2 d. None of these a. 8 b. 4 Ans: [d] Not a possible case 118. In how many ways can we select two squares on a chessboard such that they share a common vertex and of same colour? Ans 98 = [4*1 + 24*2 + 36*4]/2 possible value of d e f ⋯ j
119. If A, B , C ,... Z are 26 positive numbers such that A+B+C+....+Z = 13, then find minimum Ans: 26^2/13 = 52
120. If A, B , C & D are sets of few natural numbers then how many ordered set (A,B,C,D) are possible such that g ∪ i ∪ k ∪ l m1,2,3, . . . ,10 | 33,508 | 102,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2020-24 | latest | en | 0.899371 |
https://socratic.org/questions/how-do-you-simplify-10-log-10-19 | 1,586,082,964,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371576284.74/warc/CC-MAIN-20200405084121-20200405114121-00099.warc.gz | 685,720,310 | 6,278 | # How do you simplify 10^(log_10 (19))?
Aug 21, 2015
${10}^{{\log}_{10} \left(19\right)} = 19$
#### Explanation:
It may help to think of it like this:
I don't know the number that is the log base 10 of 19, but I can describe it.
The log base 10 of 19 is the power (exponent) I would need on the 10 (the base), to get 19.
Now I am asked to raise 10 to that power.
What power?
The power I'd need to put on 10 to get 19!
OK, then I'll get 19.
(Like, "What color paint should I buy, if I want to paint my house blue?")
Here's another example:
${10}^{{\log}_{10} 100}$.
This time I do know the log base 10 of the number. it is 2.
Why?
Because 2 is the power I need on 10, to get 100.
OK, now raise 10 to that power, and what do you get?
$100$ | 248 | 751 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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} | 4.1875 | 4 | CC-MAIN-2020-16 | longest | en | 0.953336 |
https://newsbasis.com/what-does-efficiency-mean-in-nba/ | 1,723,352,020,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00530.warc.gz | 318,845,916 | 39,874 | # What does efficiency mean in NBA?
## What does efficiency mean in NBA?
In professional basketball, the most commonly used statistical benchmark for comparing the overall value of players is called efficiency. It is a composite basketball statistic that is derived from basic individual statistics: points, rebounds, assists, steals, blocks, turnovers and shot attempts.
## How is efficiency measured in basketball?
EFF is measured by adding up all of the production stats (points, rebounds, assists, steals, and blocks), subtracting all missed shots (field goals and free throws) and turnovers, then dividing by games played. PER includes fouls drawn as a positive and fouls committed and shots rejected as negatives.
Who has the highest efficiency in the NBA?
Michael Jordan
NBA/ABA
Rank Player PER
1. Michael Jordan* 27.91
2. LeBron James 27.37
3. Anthony Davis 27.09
4. Shaquille O’Neal* 26.43
### What is a differential in basketball?
Point Differential is the numerical gap between points scored and points allowed. Wins in close games tend not to reflect a team’s true skill. Teams that start the season with a better record than their point differential tend to slow down and vice versa.
### What is a good efficiency rating basketball?
A league-average PER is always 15.00, which permits comparisons of player performance across seasons. PER takes into account accomplishments, such as field goals, free throws, 3-pointers, assists, rebounds, blocks and steals, and negative results, such as missed shots, turnovers and personal fouls.
What does efficiency mean in sports?
‘Efficiency’ is a broad-brush term, which refers to how much energy is expended while moving at a given pace – eg through the water when swimming or across the ground when running. Greater efficiency means that less energy is required to sustain a given pace.
## How is NBA vorp calculated?
Therefore, to calculate VORP one must multiply the league’s average runs per out by the player’s total outs; this provides the number of runs an average player would have produced given that certain number of outs to work with. Now multiply that number (of runs) by .
## Was Michael Jordan efficient?
His usage rate was an absurd 45%, but he was efficient, needing just 35 shots to get his 55 points. The rest of the Bulls scored 54 points on 43 shots. That season, Jordan won his ninth scoring title and his fifth NBA championship.
Who is statistically the most clutch NBA player?
LeBron James: Heat Star Statistically Most Clutch Player Since NBA Debut
Player Clutch Playoff FGs since 2003-04
LeBron James 7-of-16
Kobe Bryant 5-of-17
Kevin Durant 5-of-12
Dirk Nowitzki 5-of-12
### What is PS in basketball stats?
Pt: Points per game. Ps: Position.
### What does pa stand for in basketball?
In your league standings, you’ll notice that there are columns for Points For (PF) and Points Against (PA).
Is Player Efficiency Rating accurate?
PER largely measures offensive performance. Hollinger freely admits that two of the defensive statistics it incorporates—blocks and steals (which was not tracked as an official stat until 1973)—can produce a distorted picture of a player’s value and that PER is not a reliable measure of a player’s defensive acumen.
## What is NBA player efficiency?
Highest career player efficiency rating: Michael Jordan (27.91)
• Highest single-season player efficiency rating: Giannis Antetokounmpo (31.90 in 2019-20)
• Highest career playoff player efficiency rating: Michael Jordan (28.60)
• Highest single-season playoff player efficiency rating: Hakeem Olajuwon (38.96 in 1988)
• ## Who averages most points per game NBA?
Wilt Chamberlain holds the all-time records for total points scored (4,029) and points per game (50.4) in a season; both records were achieved in the 1961–62 season. He also holds the rookie records for points per game when he averaged 37.6 points in the 1959–60 season.
What NBA team has the most points?
On December 13, 1983, the Detroit Pistons set the NBA record for most points scored in a game by a team when they scored 186 points in a triple OT game vs the Denver Nuggets. The Pistons won the game, 186-184, in triple OT. That game also set the NBA record for most points scored in a game by both teams (370).
### Who is the best NBA team right now?
San Antonio Spurs. The San Antonio Spurs are an American professional basketball team based in San Antonio, Texas. The Spurs compete in the National Basketball Association as a member of the league’s Western Conference Southwest Division. | 1,011 | 4,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.955868 |
rpom.sporthaus-lang.de | 1,675,676,154,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500334.35/warc/CC-MAIN-20230206082428-20230206112428-00730.warc.gz | 514,908,205 | 21,216 | rm
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# Mathematica derivative online
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• vv is the biggest sale event of the year, when many products are heavily discounted.
• Since its widespread popularity, differing theories have spread about the origin of the name "Black Friday."
• The name was coined back in the late 1860s when a major stock market crashed.
The Derivative Calculator supports computing first, second, , fifth derivatives as well as differentiating functions with many variables (partial derivatives), implicit differentiation and calculating roots/zeros. You learn from calculus that the derivative of a smooth function f ( x ), defined on some interval (a, b), is another function defined by the limit (if it exists) f (x)def = df.
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EDUCATION POINT ONLINE has its own app now. edu email, and click Create Wolfram ID.
## yc
Stay on top of important topics and build connections by joining Wolfram Community groups. .
I then make the substitution ( x, y) = ( x 1 y 1, y 1). This game acquires extra salience in electoral contests that are seen as open.
What is the Derivative in Mathematica? According to the documentation listed about Mathematica, we can apply the derivativefunction to differentiate in calculus. Get a. Step 2 After you have analyzed.
uk on November 25, 2022 by guest [Book] Fractional Differential Equations An Introduction To Fractional. .
## kb
mathematics-of-the-financial-markets-financial-instruments-and-derivatives-modelling-valuation-and-risk-issues 1/1 Downloaded from web. Mathematica contains the function D which will allow you to differentiate a given equation with respect to some variable.
## lm
. The antiderivative calculator allows to integrate online any polynomial.
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> | 382 | 1,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.911934 |
http://www.mathcs.emory.edu/~cheung/Courses/554/Syllabus/3-index/resize-hash.html | 1,542,257,298,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742483.3/warc/CC-MAIN-20181115033911-20181115055911-00423.warc.gz | 477,024,131 | 2,943 | ### Resizing a hash table
• The need to grow a hash table
• Fact:
The performance of a hash index depends on the number of overflows
Example:
• No overflows leads to good performance:
We need only read 1 disk block to find any search key
• Many overflows leads to poor performance:
3 disk blocks
to find some search keys !!!!
• Recall:
• A hash function will map a search key to a value in the range [0 .. (B−1]:
• A typical hash function is computed as follows:
``` Hash( key ) = randomNumberGen( key ) % B ```
Explanation:
• randomNumberGen(key) = computes a random integer number using the key value (as seed)
• The modulo operator "% B" is used to:
Map the integer random number to a value in the range [0 .. (B−1)]
• Dicdatical warning
• Warning:
• The value of B is usually a prime number
(This will spread the keys uniformly over the range [0 .. (B−1)])
• However:
For simplicity, in the following example, I will use: B = 10 (Because computing % 10 is easy)
• Problem caused by changing the size of a Hash Table
• Problem: changing the size of a hash table
• Suppose we use a hash table with size = 10
We use the following simple hash function:
``` Hash(x) = x%10 ```
Sample of the hash table content:
• Suppose we want to increase the hash table size to 20:
We must use a different hash function that covers the wider range !!!
For example:
``` Hash(x) = x % 20 ```
Result:
Some search keys will be hashed into the wrong bucket !!!
Example:
Consequence:
We cannot find (some) the search keys stored in the hash table using the new hash function !!!
• Naive solution to solve the problem of changing the Hash Table size
• Naive (= simple-minded) solution to changing the Hash Table size:
• We re-map all the existing search keys using the new hash function and stored them in the new (size) hash table !!!
Example:
However:
Re-mapping all existing search keys in a hash table is a very expensive operation !!!
• Dynamic Hashing: Techniques to change the size of a hash table
• We will discuss 2 "dynamic" hashing techniques where:
• The size of the hash table can increase (and decrease)
• When the size of hash table is changed:
Only a fraction of the existing search keys need to be re-mapped !!!
• The dynamic hashing technqiues are:
• Extensible hashing:
Extensible hashing can elminate the need/use of overflow block completely HoweverL the size of the hash table increases exponentially....
• Linear hashing:
The size of the hash table increases linearly However: Linear hashing cannot elminate the need/use of overflow block | 603 | 2,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.80767 |
https://examcraze.in/ask/if-the-average-of-m-numbers-is-n2-and-that-of/ | 1,675,734,446,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500368.7/warc/CC-MAIN-20230207004322-20230207034322-00571.warc.gz | 250,718,577 | 25,808 | # If the average of m numbers is n2 and that of…
Are you looking for correct answer of If the average of m numbers is n2 and that of…? Here we have shared detailed answer with explanations.
### If the average of m numbers is n2 and that of n numbers is m2 , then the average of (m + n) numbers is-
1. A. m – n
2. B. mn
3. C. m + n
4. D. $$\frac{m}{n}$$
= $$frac{mn(m + n)}{(m + n)}$$ | 127 | 387 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2023-06 | latest | en | 0.8759 |
https://www.physicsforums.com/threads/net-work-and-kinetic-energy-pushing-a-wagon-to-accelerate-it.1049494/ | 1,719,009,551,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862189.36/warc/CC-MAIN-20240621223321-20240622013321-00852.warc.gz | 820,330,274 | 15,261 | # Net work and kinetic energy (pushing a wagon to accelerate it)
• aqryus
In summary, the conversation discusses the confusion of using different values for work in the kinetic energy equation and questions why friction is not taken into account. The expert confirms that the net work equals the change in kinetic energy and asks for clarification on the answer and the mass of the wagon.
aqryus
Homework Statement
Bill does 500J of work on a wagon, friction does work of -200J. What is the final speed of the wagon if it starts at rest?
Relevant Equations
w=fdcostheta
Ek=1/2mv^2
I'm a little confused because my teacher used Bill's 500J of work for the kinetic energy equation and I don't understand why. I used the net work, so 300J, to find the speed and I'm not sure why that's wrong. Wouldn't friction make the wagon move slower than if there was no friction? So why isn't that accounted for in the kinetic energy equation to find speed? Thank you
You are correct. The net work done on the wagon equals the change in the kinetic energy of the wagon. What did you get for the answer? Was the mass of the wagon specified?
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1K | 431 | 1,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-26 | latest | en | 0.849363 |
https://encyclopediaofmath.org/wiki/Sorgenfrey_topology | 1,659,886,010,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00383.warc.gz | 250,967,141 | 6,972 | Sorgenfrey topology
right half-open interval topology
A topology $\mathfrak{T}$ on the real line $\mathbf R$ defined by declaring that a set $G$ is open in $\mathfrak{T}$ if for any $x\in G$ there is an $\varepsilon_x>0$ such that $[x,x+\varepsilon_x)\subset G$. $\mathbf R$ endowed with the topology $\mathfrak{T}$ is termed the Sorgenfrey line, and is denoted by $\mathbf R^s$.
The Sorgenfrey line serves as a counterexample to several topological properties, see, for example, [a3]. For example, it is not metrizable (cf. also Metrizable space) but it is Hausdorff and perfectly normal (cf. also Hausdorff space; Perfectly-normal space). It is first countable but not second countable (cf. also First axiom of countability; Second axiom of countability). Moreover, the Sorgenfrey line is hereditarily Lindelöf, zero dimensional and paracompact (cf. also Lindelöf space; Zero-dimensional space; Paracompact space). Any compact subset of the Sorgenfrey line is countable and nowhere dense in the usual Euclidean topology (cf. Nowhere-dense set). The Sorgenfrey topology is neither locally compact nor locally connected (cf. also Locally compact space; Locally connected space).
Consider the Cartesian product $X:=\mathbf R^s\times\mathbf R^s$ equipped with the product topology , which is called the Sorgenfrey half-open square topology. Then $X$ is completely regular but not normal (cf. Completely-regular space; Normal space). It is separable (cf. Separable space) but neither Lindelöf nor countably paracompact.
Many further properties of the Sorgenfrey topology are examined in detail in [a1]. Namely, the Sorgenfrey topology is a fine topology on the real line, and $\mathbf R$ equipped with both the Sorgenfrey topology and the Euclidean topology serves as an example of a bitopological space (that is, a space endowed with two topological structures). The Sorgenfrey topology satisfies the condition (tFL) when studying fine limits (if a real-valued function $f$ has a limit at the point $x$ with respect to the Sorgenfrey topology $\mathfrak{T}$ it has the same limit at $x$ with respect to the Euclidean topology when restricted to a $\mathfrak{T}$-neighbourhood of $x$). It has also the $G_\delta$-insertion property (given a subset $A$ of $\mathbf R$, there is a $G_\delta$-subset $G$ of $\mathbf R$ such that $G$ lies in between the $\mathfrak{T}$-interior and the $\mathfrak{T}$-closure of $A$). The Sorgenfrey topology satisfies the so-called essential radius condition: For any point $x$ and any $\mathfrak{T}$-neighbourhood $U_x$ of $x$ there is an "essential radius" $r(x,U_x)>0$ such that whenever the distance of two points $x$ and $y$ is majorized by $\min(r(x,U_x),r(y,U_y))$, then $U_x$ and $U_y$ intersect. The real line $\mathbf R$ equipped with the Sorgenfrey topology and the Euclidean topology is a binormal bitopological space, while $\mathbf R$ with the Sorgenfrey and the density topology is not binormal. See [a1] for answers to interesting questions concerning the class of continuous functions in the Sorgenfrey topology and for functions of the first or second Baire classes.
References
[a1] J. Lukeš, J. Malý, L. Zajíček, "Fine topology methods in real analysis and potential theory" , Lecture Notes in Mathematics , 1189 , Springer (1986) [a2] R.H. Sorgenfrey, "On the topological product of paracompact spaces" Bull. Amer. Math. Soc. , 53 (1947) pp. 631–632 [a3] Steen, Lynn Arthur; Seebach, J.Arthur jun. Counterexamples in topology (2nd ed.) Springer (1978) ISBN 0-387-90312-7 Zbl 0386.54001
How to Cite This Entry:
Sorgenfrey topology. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Sorgenfrey_topology&oldid=37307
This article was adapted from an original article by J. Lukeš (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | 1,038 | 3,860 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-33 | longest | en | 0.859336 |
https://snippets.snyk.io/python/python-program-to-implement-binary-search-without-recursion | 1,695,571,171,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506658.2/warc/CC-MAIN-20230924155422-20230924185422-00779.warc.gz | 568,355,513 | 22,913 | # 10 examples of 'python program to implement binary search without recursion.' in Python
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``22def binary_search_recur(arr, val):23 if arr is None or len(arr) == 0:24 return False25 mid = len(arr)/226 if arr[mid] == val:27 return True28 elif arr[mid] > val:29 return binary_search_recur(arr[:mid], val)30 else:31 return binary_search_recur(arr[mid+1:], val)``
``49def binary_search_1(left, right):50 # 如果选择右中位数，当区间只剩下 2 个元素的时候，51 # 一旦进入 right = mid 这个分支，右边界不会收缩，代码进入死循环52 while left < right:53 # 选择左中位数54 # mid = left + (right - left) // 255 # left 和 right 一般都表示索引，使用 + 且无符号右移，在 left 和 right 都很大的时候，虽然整形溢出，但结果正确56 # 在 Java 中使用 `int mid = (left + right) >>> 1;` ，一定是无符号右移57 mid = (left + right) >> 158 if check(mid):59 # 先写可以排除中位数的逻辑60 left = mid + 161 else:62 # 右边界不能排除，这半边逻辑不用记63 # 逻辑是：上一个分支的另一个边界收缩，但不排除中位数64 right = mid``
``34def binary_search_iterative(array, item):35 # TODO: implement binary search iteratively here36 pass``
``56def test_list_size_2_does_not_exist(self):57 self.assertEqual(linear_search([3, 5], 7), -1)58 self.assertEqual(binary_search_iteratively([3, 5], 7), -1)59 self.assertEqual(binary_search_recursively_in_place([3, 5], 7), -1)60 self.assertFalse(binary_search_recursively_not_in_place([3, 5], 7))``
``45def search(self,key,node = None):4647 if node is None:48 node = self.root4950 if self.root.key == key:51 print "key is at the root"52 return self.root53 54 else:55 if node.key == key :56 print "key exists"57 return node5859 elif key < node.key and node.left is not None:60 print "left"61 return self.search(key,node = node.left)62 63 elif key > node.key and node.right is not None:64 print "right"65 return self.search(key,node = node.right)66 67 else:68 print "key does not exist"69 return None``
``62def binary_search(seq, t, key=None, cmp=None): # bisect module doesn't support key/compare callbacks63 # http://code.activestate.com/recipes/81188-binary-search/64 min = 065 max = len(seq) - 166 if not (cmp or key):67 while True:68 if max < min: return -169 m = (min + max) // 270 k = seq[m]71 if k < t: min = m + 172 elif k > t: max = m - 173 else: return m74 elif key:75 t = key(t)76 while True:77 if max < min: return -178 m = (min + max) // 279 k = key(seq[m])80 if k < t: min = m + 181 elif k > t: max = m - 182 else: return m83 else:84 while True:85 if max < min: return -186 m = (min + max) // 287 s = cmp(seq[m], t)88 if s < 0: min = m + 189 elif s > 0: max = m - 190 else: return m``
``50def test_list_size_2_exists_second_place(self):51 self.assertEqual(linear_search([3, 5], 5), 1)52 self.assertEqual(binary_search_iteratively([3, 5], 5), 1)53 self.assertEqual(binary_search_recursively_in_place([3, 5], 5), 1)54 self.assertTrue(binary_search_recursively_not_in_place([3, 5], 5))``
``40def recursive_search(self, node, data):41 if node is None:42 return None4344 if node.data == data:45 return node4647 if data > node.data:48 self.recursive_search(node.right, data)49 else:50 self.recursive_search(node.left, data)``
``13def binary_search(array, query):14 lo, hi = 0, len(array) - 115 while lo <= hi:16 mid = lo + (hi - lo) // 217 val = array[mid]18 19 # If the element is present at the middle itself 20 21 if val == query:22 return mid23 24 # If element is smaller than mid, then 25 # it can only be present in right subarray 26 27 elif val < query:28 lo = mid + 129 30 # Else the element can only be present 31 # in left subarray 32 33 else:34 hi = mid - 135 36 # We reach here when element is not 37 # present in array 38 39 return None``
``182def search(self, key):183 return self.root.search(key)`` | 1,391 | 4,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-40 | latest | en | 0.327485 |
https://www.tes.com/teaching-resources/shop/closingthegapltd | 1,529,289,288,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859923.59/warc/CC-MAIN-20180618012148-20180618032148-00594.warc.gz | 939,813,460 | 22,051 | #### Year 1- Autumn Term- Block 1 week 1- 2 Place Value Differentiated Sheets White Rose Style
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11 fluency fraction worksheets . Worksheets are differentiated three ways and cover the following objectives for year 1: recognise, find and name a half as one of two equal parts of an object, shape or quantity recognise, find and name a quarter as one of four equal parts of an object, shape or quantity. compare, describe and solve practical problems for capacity and volume [for example, full/empty, more than, less than, half, half full, quarter] For those following whiite rose, these sheets cover summer term, block 2.
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Contains 15 lessons of differentiated fraction work for year 2, and 1 lesson /activity of counting in halves, thirds and quarters which is not differentiated. 16 lessons altogether. Useful for those following white rose- Year 2 spring term block 4 Covers the year 2 objectives: recognise, find, name and write fractions 1/3 , 1/4 , 2/4 and 3/4 of a shape, set of objects or quantity write simple fractions for example, 1/2 of 6 = 3 and recognise the equivalence of 2/4 and 1/2 .
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Contains 5 lessons of decimal work for year 5. Covers the objectives: Lesson 1- Adding decimals within one pictorially Lesson 2- Adding decimals withing one- column addition Lesson 3- Subtracting decimals within one- pictorially Lesson 4- Subtracting decimals within one- column subtraction Lesson 5- adding and subtracting using pictorial images to prove answers. Particularly useful for those following white rose- summer 1, block 1- decimal unit- week 1 of 4 weeks.
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For those following white rose, we have made teaching slides for fluency- Year 1, Spring term, weight and volume (block 4). Visit our facebook page for more resources!
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Three counting maths mats- these can be laminated with targeted questions such as, ‘Which has the least?’ ‘Which has the most?’ ‘If I had one more ___ how many would I have now?’ | 1,764 | 7,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-26 | longest | en | 0.841987 |
https://artscolumbia.org/pie-charts-47905/ | 1,696,330,002,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511075.63/warc/CC-MAIN-20231003092549-20231003122549-00648.warc.gz | 125,673,438 | 23,841 | • Pages 3
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# Pie Charts Analysis Essay (696 words)
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As the name describes, this is done by representing the numerical equivalence of each part as a piece of the whole pie, which in total equates to 100%. The Pennsylvania Department of Health (2001) says that pie charts are a good choice when a relatively small amount of parts, perhaps 3 to 7, need to be represented. With any more it becomes difficult to notice the differences in magnitude; thus, the pie chart loses its simplicity and impact. They can only be used when a total mount is known, one such example would be an election where the total of votes received by all candidates equals 100% of the votes.
Or a budget where the total amount spending is divided in to categories such as labor, facilities costs, advertising, etc which always are a part of the total. However, according to McBride (2003), the pie chart could not he used to show a change in spending through out a period. A pie chart shows data at one instance, like a snapshot and cannot be used to show change in data over time (Para_ 4), With the advent of computers, AD graphs have become somewhat popular, unfortunately a active aspect is that they add complexity to the image and can distort visual proportional value.
It is recommended to stick with flat “20” charts (Para. 6). Pie charts these days are typically generated by computer software, as a result people forget that there is actually a little math involved. To create a pie chart Concordia University (2000) explains that you have to find the total value for the entire category being studied and calculate the percentage for each segment or Para Convert the percentage values for each segment into degrees relative to the 360 degrees in the circle. For example. 12% X 360 degrees = 43 degrees). Then, Draw a circle and divide it into appropriately sized segments.
Sykes and Smith recommend that the sectors of the pie are ordered from largest to the smallest for easier interpretation of the data and they should be drawn in the counter-clockwise direction, (Para. 7) For clarity a complete title of the data being represented is used, along with each slice clearly labeled to identify its characteristics, and the source of the data. Appropriate colors and or shading are also quite valuable in distinguishing the segments. Test the chart in its final armor for readability. Whether it be displayed on a projector, paper or computer screen.
With the assistance tot a coworker, ensure that the audience will be able to clearly interpret the graph, making sure the chart is large and clear enough to read. Based on what you have learned, what can PU point out about the following chart that does not follow best practices’source: Where is earth’s water located? (2003) 2 pie charts were used in the single representation, this could be acceptable if the reader was given free time to study the chart, but the complexity of using en pie chart to further divide a portion Of the Other pie chart would be too confusing for many situations.
Ideally it would be presented separately. Remember to keep it simple. Adding to the confusion is that the same color was used on the 2 different charts, but represent 2 different things. You may have noticed that the chart does not give the actual stats for the “Water usable by humans”. A chart is used to support statistics; and it is best to include the numbers in the chart. The extreme AD effect distorts the visual representation f the pie charts, which is the main reason for using a pie chart in the first place.
Now that you have had the chance to point out some things about a pie chart that can be wrong, what can you find in the next pie chart that was done right’source: Grandfather Economic Report series (2004) Clear title describing the purpose of the chart. Each section is clearly labeled with the statistic included. , Colors were used well to distinguish the pieces. A reasonable # of pieces were used, not overloading the chart. Overall this is an excellent example of what a pie chart should look like with the exception of one el, the pieces were not arranged from largest to smallest in a clockwise manner.
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In case you can't find a relevant example, our professional writers are ready to help you write a unique paper. Just talk to our smart assistant Amy and she'll connect you with the best match. | 1,107 | 5,243 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-40 | latest | en | 0.932528 |
https://www.ademcetinkaya.com/2022/11/nws-stock-price-prediction.html | 1,669,936,651,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710870.69/warc/CC-MAIN-20221201221914-20221202011914-00501.warc.gz | 683,115,870 | 68,735 | News Corp (Class B) Research Report
## Summary
This paper surveys machine learning techniques for stock market prediction. The prediction of stock markets is regarded as a challenging task of financial time series prediction. We evaluate News Corp (Class B) prediction models with Modular Neural Network (Financial Sentiment Analysis) and Ridge Regression1,2,3,4 and conclude that the NWS stock is predictable in the short/long term. According to price forecasts for (n+1 year) period: The dominant strategy among neural network is to Sell NWS stock.
## Key Points
1. Fundemental Analysis with Algorithmic Trading
2. Dominated Move
3. What are main components of Markov decision process?
## NWS Target Price Prediction Modeling Methodology
We consider News Corp (Class B) Stock Decision Process with Modular Neural Network (Financial Sentiment Analysis) where A is the set of discrete actions of NWS stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Ridge Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Financial Sentiment Analysis)) X S(n):→ (n+1 year) $∑ i = 1 n s i$
n:Time series to forecast
p:Price signals of NWS stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## NWS Stock Forecast (Buy or Sell) for (n+1 year)
Sample Set: Neural Network
Stock/Index: NWS News Corp (Class B)
Time series to forecast n: 19 Nov 2022 for (n+1 year)
According to price forecasts for (n+1 year) period: The dominant strategy among neural network is to Sell NWS stock.
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Yellow to Green): *Technical Analysis%
## Adjusted IFRS* Prediction Methods for News Corp (Class B)
1. If the underlyings are not the same but are economically related, there can be situations in which the values of the hedging instrument and the hedged item move in the same direction, for example, because the price differential between the two related underlyings changes while the underlyings themselves do not move significantly. That is still consistent with an economic relationship between the hedging instrument and the hedged item if the values of the hedging instrument and the hedged item are still expected to typically move in the opposite direction when the underlyings move.
2. An entity shall apply this Standard retrospectively, in accordance with IAS 8 Accounting Policies, Changes in Accounting Estimates and Errors, except as specified in paragraphs 7.2.4–7.2.26 and 7.2.28. This Standard shall not be applied to items that have already been derecognised at the date of initial application.
3. There is a rebuttable presumption that unless inflation risk is contractually specified, it is not separately identifiable and reliably measurable and hence cannot be designated as a risk component of a financial instrument. However, in limited cases, it is possible to identify a risk component for inflation risk that is separately identifiable and reliably measurable because of the particular circumstances of the inflation environment and the relevant debt market
4. At the date of initial application, an entity shall assess whether a financial asset meets the condition in paragraphs 4.1.2(a) or 4.1.2A(a) on the basis of the facts and circumstances that exist at that date. The resulting classification shall be applied retrospectively irrespective of the entity's business model in prior reporting periods.
*International Financial Reporting Standards (IFRS) are a set of accounting rules for the financial statements of public companies that are intended to make them consistent, transparent, and easily comparable around the world.
## Conclusions
News Corp (Class B) assigned short-term B2 & long-term Ba1 forecasted stock rating. We evaluate the prediction models Modular Neural Network (Financial Sentiment Analysis) with Ridge Regression1,2,3,4 and conclude that the NWS stock is predictable in the short/long term. According to price forecasts for (n+1 year) period: The dominant strategy among neural network is to Sell NWS stock.
### Financial State Forecast for NWS News Corp (Class B) Stock Options & Futures
Rating Short-Term Long-Term Senior
Outlook*B2Ba1
Operational Risk 4872
Market Risk6179
Technical Analysis8779
Fundamental Analysis5731
Risk Unsystematic3589
### Prediction Confidence Score
Trust metric by Neural Network: 80 out of 100 with 458 signals.
## References
1. Breusch, T. S. A. R. Pagan (1979), "A simple test for heteroskedasticity and random coefficient variation," Econometrica, 47, 1287–1294.
2. J. Filar, L. Kallenberg, and H. Lee. Variance-penalized Markov decision processes. Mathematics of Opera- tions Research, 14(1):147–161, 1989
3. Mnih A, Kavukcuoglu K. 2013. Learning word embeddings efficiently with noise-contrastive estimation. In Advances in Neural Information Processing Systems, Vol. 26, ed. Z Ghahramani, M Welling, C Cortes, ND Lawrence, KQ Weinberger, pp. 2265–73. San Diego, CA: Neural Inf. Process. Syst. Found.
4. E. Altman, K. Avrachenkov, and R. N ́u ̃nez-Queija. Perturbation analysis for denumerable Markov chains with application to queueing models. Advances in Applied Probability, pages 839–853, 2004
5. Athey S, Imbens G. 2016. Recursive partitioning for heterogeneous causal effects. PNAS 113:7353–60
6. Belloni A, Chernozhukov V, Hansen C. 2014. High-dimensional methods and inference on structural and treatment effects. J. Econ. Perspect. 28:29–50
7. D. Bertsekas and J. Tsitsiklis. Neuro-dynamic programming. Athena Scientific, 1996.
Frequently Asked QuestionsQ: What is the prediction methodology for NWS stock?
A: NWS stock prediction methodology: We evaluate the prediction models Modular Neural Network (Financial Sentiment Analysis) and Ridge Regression
Q: Is NWS stock a buy or sell?
A: The dominant strategy among neural network is to Sell NWS Stock.
Q: Is News Corp (Class B) stock a good investment?
A: The consensus rating for News Corp (Class B) is Sell and assigned short-term B2 & long-term Ba1 forecasted stock rating.
Q: What is the consensus rating of NWS stock?
A: The consensus rating for NWS is Sell.
Q: What is the prediction period for NWS stock?
A: The prediction period for NWS is (n+1 year) | 1,699 | 6,887 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-49 | longest | en | 0.832176 |
https://betterlesson.com/lesson/637287/what-is-energy?from=breadcrumb_lesson | 1,571,361,591,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677412.35/warc/CC-MAIN-20191018005539-20191018033039-00553.warc.gz | 375,205,801 | 24,142 | # What is Energy?
6 teachers like this lesson
Print Lesson
## Objective
Students will be able to describe and calculate kinetic energy and potential energy.
#### Big Idea
Students find out what energy is.
20 minutes
The goal of this class is to introduce the concepts of energy and how to solve for kinetic and potential energy. Students start out by reading about energy and obtaining more information about energy (SP8). Then students use computational thinking to solve problems using various energy equations (SP5). All of these skills work together to help lead students to understanding the energy conservation standards HS-PS3-1 and HS-PS3-2.
To start out class, I have students work with the person sitting next to each other during this reading activity. During a paired reading, I have students read one paragraph at a time. Students pull up the Energy Reading on their chromebooks so each student can see the reading. To do the paired reading, I have one partner read the first paragraph out loud. The other partner summarizes verbally the main idea(s) of the paragraph after it has been read. Then I have the partners switch roles to complete the next paragraph, until they have read the entire reading.
When the whole class has finished the reading, I ask students to tell me what the main points of the article are so we can start talking about what energy is. We end up discussing the analogies that are in the article as well as what the bar graph pictures represent in the article. I emphasize here that in each situation there is one or more type of energy that can be calculated.
## Energy Notes Recap
10 minutes
After the discussion about the reading, I have students tell me about they video that they watched for homework about the different types of energy, as seen below.
I make a chart on the board similar to the one that they took notes on (shown below). I make sure to emphasize the difference between each type of energy and I show them another calculation example of each type of energy. Students volunteer their answers and I try to call on as many students as possible to get most of the information as shown in the graphic organizer below. Students should already have this organizer completed from the homework but we discuss this as a recap.
## Energy Problem Set
25 minutes
After a brief review of notes, I have students work with their table groups to complete the Energy Problems WS. I ask students to show all their work and to ask their group members if they need help solving a problem. Since these problems are pretty straight-forward, I try not to answer any questions about how to solve the problem. The expectation is that someone at the table that can help. As students work, I walk around to see which problems students are having trouble with so I can know what types of questions to revisit with the whole class. Overall students were able to use the equations to complete most or all of these problems accurately. The most difficult problems for students are the conceptual questions without numbers. Below is the work from one student on this problem set. If students did not complete the problems, they complete the remaining problems for homework.
## Second Look at Energy Learning Targets
5 minutes
Since we are about halfway through the unit, I ask my students to self-assess on their Unit 5 Energy Learning Target Sheet to see their growth throughout the unit. I also do this to help show students the areas that they still should work on before the test at the end of the unit. My students rate themselves on a scale of 1-4: (1) Novice, (2) Apprentice, (3) Practitioner, (4) Expert. At this stage of the unit I am expecting the 1's to only remain in the targets we have not covered. As I walk around the room I make sure to note who still is marking 1's in targets we have covered so I can give them a little extra help.
## Stoplight Exit Slip
5 minutes
To end class, I use a stoplight exit slip to see where my students are at. Each students takes a post-it note from the resource bin at the center of the table and chooses to stick that note to a red, yellow or green light. They complete the sentence starter on the light they choose.
The red light: Today my learning stopped because...
The yellow light: Today I considered a question, new idea or perspective...
The green light: Today I learned __________ because...
When students complete their sentence they stick the post-it note the appropriate color light which is at the front of the room. I do this closure activity to get an idea of where students are in learning the material. This exit slip is a way that they can anonymously express exactly what they are learning or exactly what they are struggling with. | 968 | 4,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-43 | latest | en | 0.95044 |
www.niupedia.com | 1,642,793,888,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303709.2/warc/CC-MAIN-20220121192415-20220121222415-00507.warc.gz | 948,859,521 | 20,292 | Understanding BEP : Formula, Components, Benefits, and Examples
BEP Definition
What is BEP (Break Even Point)? In accounting and business economics, the definition of BEP is a certain point where expenditure / income and income are in a balanced position (break-even) so that there is no loss or profit.
Another opinion says that the understanding of BEP is a situation where the company's operating activities do not suffer losses and also do not get a profit (break even) because the amount of costs incurred is equal to the amount of income.
Break Even Point analysis technique is used by a company to analyze the projections of how many units are produced or how much money must be received so that the company is at a break-even point or return on capital.
BEP component
Break Even Points (BEP) consists of several components in it. The BEP components are as follows:
1. Fixed Cost
Fixed costs are constant costs if the company carries out production activities or does not carry out production. Examples of fixed costs include; labor salary, machine depreciation costs, equipment costs, and so on.
2. Variable Cost (Variable Cost)
Variable costs are unit costs where they are dynamic depending on the production volume action. If planned production increases, the variable costs will increase. Examples of variable costs; electricity costs, raw material costs, plastic bag costs, and so on.
3. Selling Price
The selling price is the selling price set per unit of goods or services that has been produced by the company.
The purpose of BEP
Every company certainly wants to make a profit from its business activities. To achieve this there are a number of things that can be done related to Break Even Point, as follows :
1. Pressing production and operational costs to the lowest possible level without putting aside quality and quantity so that the company can maintain the product price level.
2. Determine the price of the product in full calculation so that the price of the product matches the desired profit.
3. Increase the volume of activities as much as possible.
The three points above must be done simultaneously because each has an impact on the overall operation of the operation. That is why the profit structure of a company is often described in Break Even Point (BEP) to facilitate understanding the relationship between costs, volume of activities, and profits.
BEP formula
There are two kinds of formulas that can be used for Break Even Point analysis, as follows :
1. BEP in the Unit
BEP = FC / (P - VC)
In this formula we can find out how many units of goods / services must be produced to break even.
Information :
BEP: Break Even Point
FC: Fixed Cost
P: Price per unit
VC: Variable Cost
2. BEP in Dollars
BEP = FC / [1 - (VC / S)]
In this formula we can find out how much Rupiah must be received to get a break-even point. Note: calculations [1- (vc / s)] are also referred to as Margin Contributions Per Unit.
Information :
BEP: Break Even Point
FC: Fixed Cost
VC: Variable Cost
P: Price per unit
S: Sales Volume
That is a brief explanation of the meaning of BEP (Break Even Point), BEP formula, components, benefits, and examples of simple BEP calculations. Hopefully this article is useful for you.
Comment Policy: Silahkan tuliskan komentar Anda yang sesuai dengan topik postingan halaman ini. Komentar yang berisi tautan tidak akan ditampilkan sebelum disetujui.
Buka Komentar | 737 | 3,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.946871 |
http://www.piprime.fr/developpement/venn-diagram_asy/?posts_per_page=-1 | 1,611,563,828,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565376.63/warc/CC-MAIN-20210125061144-20210125091144-00645.warc.gz | 164,188,325 | 8,592 | Asymptote Generalities – fig1860
Category: Asymptote,Examples 2D,GeneralitiesPh. Ivaldi @ 15 h 39 min
(Compiled with Asymptote version 2.14svn-r5318)
```
// Venn diagram // Diagramme de Venn
// Edwards' construction // Construction d'Edwards
import roundedpath;
size(10cm,0);
path [] EdVenn(int n)
{
path [] opath;
if (n>=1)
opath.push(shift(-1.4,-.9)*roundedpath(xscale(2.8)*yscale(.9)*unitsquare,.1));
if (n>=2)
opath.push(shift(0,-.9)*roundedpath(xscale(1.4)*yscale(1.8)*unitsquare,.1));
if (n>=3)
opath.push(scale(.5)*unitcircle);
for (int i=1; i<=n-3; ++i)
{
pair pcle=point(opath[2],1/(2^i)),
ccle=intersectionpoint(pcle--(pcle-dir(opath[2],1/(2^i))), (0,0)--(1,0));
path cle=shift(ccle)*scale(abs(pcle-ccle))*unitcircle;
real[] p1=intersect(cle, opath[2]);
path ocle=subpath(cle,-p1[0],p1[0]);
guide tpath;
real step=360/(2^i), a=0;
for (int j=0; j<2^i; ++j)
{
tpath=tpath..rotate(a)*ocle;
a+=step;
}
opath.push(tpath..cycle);
}
return opath;
}
draw(EdVenn(6));
```
Étiquettes : , , ,
Unofficial package edvenn_pi.asy – fig0100
Category: Asymptote,edvenn_pi.asy,Unofficial packagesPh. Ivaldi @ 3 h 52 min
(Compiled with Asymptote version 2.14svn-r5318)
```
import edvenn_pi;
size(10cm,0);
draw(EdVenn(6));
```
Étiquettes :
Official Asymptote example – venn
Category: Asymptote,Official Gallery One-PagerPh. Ivaldi @ 20 h 57 min
(Compiled with Asymptote version 2.14svn-r5318)
```/* This code comes from The Official Asymptote Gallery */
size(0,150);
pen colour1=red;
pen colour2=green;
pair z0=(0,0);
pair z1=(-1,0);
pair z2=(1,0);
real r=1.5;
path c1=circle(z1,r);
path c2=circle(z2,r);
fill(c1,colour1);
fill(c2,colour2);
picture intersection;
fill(intersection,c1,colour1+colour2);
clip(intersection,c2);
draw(c1);
draw(c2);
label("\$A\$",z1);
label("\$B\$",z2);
pair z=(0,-2);
real m=3;
margin BigMargin=Margin(0,m*dot(unit(z1-z),unit(z0-z)));
draw(Label("\$A\cap B\$",0),conj(z)--z0,Arrow,BigMargin);
draw(Label("\$A\cup B\$",0),z--z0,Arrow,BigMargin);
draw(z--z1,Arrow,Margin(0,m));
draw(z--z2,Arrow,Margin(0,m));
shipout(bbox(0.25cm));
currentpicture.uptodate=true;
```
Étiquettes : , , | 804 | 2,128 | {"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} | 2.953125 | 3 | CC-MAIN-2021-04 | latest | en | 0.3295 |
https://zemnipracelhotak.cz/Portable/3154.html | 1,632,412,486,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057424.99/warc/CC-MAIN-20210923135058-20210923165058-00305.warc.gz | 1,086,547,898 | 11,136 | • ### What is the mix proportions for the M-80 Grade concrete
To prepare a concrete solution, it is necessary to dilute the cement with water in a ratio of 2: 1, that is, the mass of water is about half the weight of cement. Water should not have impurities, dirt, branches and large inclusions this will reduce the strength of the finished concrete.
### List of concrete mix ratio m15 m80 file
proportion of all grade of cement concrete ie. m5 to m80 Civil Jul 14, 2010 proportion of all grade of cement concrete ie. m5 to m80 Civil M15. 1:2:4. M25. (1:1:2). M30. 1:1:2 (Nominal Mix). M35. 3.15 : 2.62 : 2.67. More details » Get Price
### M-80 Grade Pumpable Concrete Civil Engineering Portal
1. M-80 Grade of concrete design for a mean target strength of: 80 + 1.65 x 5 = 88.3 N/mm 2 at 28 day age. 2. With the given set of materials from various trials it was found that with OPC 450 Kg/m 3, Silica Fume 45 Kg/m 3 and W/C+SF ratio of 0.283 gives 28-day average cube compressive strength of
### Water cement ratio formula -Table-Calculation for Mortar
6 行· 12-08-2020· Water cement ratio is the ratio between the weight of water to the weight of cement used
### What is the ratio of m80 grade concrete
What is the ratio of m80 grade concrete Products. As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for any size-reduction requirements including, What is the ratio of m80 grade concrete,
### Water Cement Ratio Water Cement Ratio Calculation W/C
01-11-2020· Recommended Water cement ratio of Works as follows: For ordinary concrete work (sidewalks and driveways): 0.6 to 0.7 If required higher quality concrete is desired: 0.4
### How to Calculate Water Cement Ratio The Concrete Network
Use a maximum .45 water to cement ratio for concrete with severe or very severe sulfate conditions per the 1997 Uniform Building Code (Table 19-A-4) Water permeability increases exponentially when concrete has a water cement ratio greater than .50. Durability increases the
### Water–cement ratio Wikipedia
The water-to-cement ratio is the ratio between the weights of water and cement in a concrete mix. For proper hydration, this ratio (commonly called the w/c ratio) should be about 0.30, assuming no contribution to hydration from external water sources. This article about a civil engineering topic is a stub. You can help Wikipedia by expanding it.
### What is the mix proportions for the M-80 Grade
M80 Grade Concerte :- As we know that, if we are going for mix design of above M20 grade so we have to use the design mix method for finding out the ratios of the Cement, Fine aggregate and Coarse aggregate. So by using of IS456:2000 CODE we can f...
### How to prepare a concrete mix for high strength
Thanks for A2A. Now coming to the question mix design of high strength concrete i.e. above M40 is influenced by properties of cement, sand aggregates & water-cement ratio.To achieve high strength, it is necessary to use lowest possible water-cemen...
### Mix design ratio for m25 to m80 grade of concrete
Mix design ratio for m25 to m80 grade of concrete Products. As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for any size-reduction requirements including, Mix design ratio for m25 to m80 grade of concrete, quarry, aggregate, and different kinds of minerals.
### A NEW MIX DESIGN METHOD FOR HIGH PERFORMANCE
24-08-2016· The mix design of NSC is based primarily on the water-cement (w/c) ratio law first M60, M70, M80 and M90) in the relationships between free water cement ratio and concrete strength for
### Water to cement ratio An Important Strength Factor
30-09-2020· Plain concrete (moderate) Table 5 IS 10252. The Minimum cement content for M15=240 kg/ m3. and Min water cement ratio =water/cement =0.60. =water/240=0.60. Min water requirement for mix=240*.60=120 liters. For Design mix, the W/C Ratio will depend upon the workability, strength requirements. Fig 1 IS 10262.
### What is Water Cement Ratio? Guide & Calculation
Calculation of Water Quantity for Concrete. As you can see from the Chart, the W/C Ratio varies from 0.4 to 0.7 depending on exposure conditions. If we need to calculate Water quantity for concrete, first find the cement content for the volume. Therefore, Required amount of water = 0.5 X 50 kg = 25 litres / 50 kg cement bag.
### meaning of m80 concrete BINQ Mining
08-02-2013· list of concrete mix ratio m15 m80. m15, m20, m25, m30, m40, m45, m50, m60, m70, m75, In Concrete Mix Design Calculations, Water Cement Ratio and Optimum Cement Content and Ideal Aggregate Propotions depends on various phisical conditions.
### Concrete Mix Ratio Concrete Mix Design Concrete
More water equals weaker concrete. The water-cement ratio is important for the concrete to reach its design strength. This is why it is essential to order concrete that has been designed for a particular slump. Adding water onsite to improve workability will weaken the concrete and can cause failure.
### What is Water Cement Ratio? Its Calculation [Complete
Water Quantity Calculation for Concrete. For the calculation of water quantity for concrete, first of all, we have to find the quantity of cement. Just assume that the required volume of cement for the mix is 100 kg and W/C is 0.4. Water Quantity = Water-Cement Ratio x Volume of Cement. Water Quantity = 0.4 x 100 kg = 40 liters / 100 kg cement
### Concrete Mix Ratio: What Is It? What Is 1-2-3?And
16-07-2020· A concrete mix ratio is usually expressed by a set of numbers separated by colons, as is the case with a 1:2:3 ratio. This tells the mixer that they need to add 1 part cement powder, 2 parts sand, and 3 parts aggregate in order to create the desired concrete consistency. Water is not included in these ratios, though, because it can vary more
### List of concrete mix ratio m15 m80 file
List of concrete mix ratio m15 m80 file Products. As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for any size-reduction requirements including, List of concrete mix ratio m15 m80 file, quarry, aggregate, and different kinds of minerals.
### A NEW MIX DESIGN METHOD FOR HIGH PERFORMANCE
24-08-2016· The mix design of NSC is based primarily on the water-cement (w/c) ratio law first M60, M70, M80 and M90) in the relationships between free water cement ratio and concrete strength for
### Water to cement ratio An Important Strength Factor
30-09-2020· Plain concrete (moderate) Table 5 IS 10252. The Minimum cement content for M15=240 kg/ m3. and Min water cement ratio =water/cement =0.60. =water/240=0.60. Min water requirement for mix=240*.60=120 liters. For Design mix, the W/C Ratio will depend upon the workability, strength requirements. Fig 1 IS 10262.
### How to calculate the water-cement ratio for concrete
10-03-2020· According to the ration of water-cement we get strength on any Construction work. Besides the quantity of water in concrete, the quality of water used in concrete have important effects on fresh concrete properties, such as setting time as well as work-ability. Also it have important effects on the strength and durability of concrete.. Water cement ratio table
### How to Calculate Water-Cement Ratio in Design of
29-11-2018· Calculate Water-Cement Ratio for Concrete Mix Design M30 Grade Concrete for R.C.C Work. Cement = OPC 53 Grade, assume 7- Day compressive strength of Cement is 37 N/mm 2. Here, 28 -day compressive strength of concrete = target mean strength= 38.25 N/mm 2.
### Determining the water/cement ratio of hardened
16-06-2021· Water-to-cement (w/c) mass ratio is important because it controls the mechanical properties and durability of hardened concrete. When problems occur and/or non-compliance with the specification is suspected, it is often desirable to be able to determine the w/c ratio.
### Water–cement ratio Wikipedia
The water–cement ratio is the ratio of the weight of water to the weight of cement used in a concrete mix. A lower ratio leads to higher strength and durability, but may make the mix difficult to work with and form. Workability can be resolved with the use of plasticizers or super-plasticizers.. Often, the ratio refers to the ratio of water to cementitious materials, w/cm. Cementitious
### Water-Cement Ratio (W/C) of Concrete online civil
15-11-2014· This means that water makes the concrete workable. It is found theoretically that water required for these two functions is about 0.50 to 0.60 times the weight of cement. This ratio of the amount of water to the amount of cement by weight is termed as the water-cement ratio and the strength and quality of concrete primarily depend upon this ratio.
### Effect of Water Cement Ratio on Strength of Concrete
06-10-2017· Water–cement ratios of 0.45 to 0.60 are more typically used. For higher-strength concrete, lower ratios are used, along with a plasticizer to increase flowability. Too much water will result in segregation of the sand and aggregate components from the cement paste. Also, water that is not consumed by the hydration reaction may leave concrete
### Specification of Concrete for Water Retaining Structures
19-02-2020· Ultimately, the authors opined that not all the cement will hydrate if the water/cement ratio is less than 0.4, even though only half of the water will go into chemical combination. Research carried out by Kim et al (2014) showed that porosity in concrete increased by 150% when the water/cement ratio was increased from 0.45 to 0.6.
### How to prepare a concrete mix for high strength
Thanks for A2A. Now coming to the question mix design of high strength concrete i.e. above M40 is influenced by properties of cement, sand aggregates & water-cement ratio.To achieve high strength, it is necessary to use lowest possible water-cemen...
### Water cement ratio of Concrete !! How to choose W/C
01-09-2020· In Concrete, the water-cement ratio determines the final properties of the concrete. The Water-cement ratio is the most important factor behind the strength of concrete. If a concrete mixture has a high water-cement ratio then it will be workable (easy to work with that concrete) but it will be less stronger as compared to the same concrete with less water-cement ration.
### What is the Right Water-Cement Ratio for Mix Design?
For concrete construction like driveways and sidewalks, a w/c ratio ranging from 0.6 to 0.7 is normally used. The practical range of water-cement ratio ranges from 0.3 to 0.8 that gives stiff and weak concrete respectively. Weak concrete means a fairly wet concrete. A compressive strength about 5600psi can be obtained from a concrete of water
### How to Calculate Water-Cement Ratio in Design of
29-11-2018· Calculate Water-Cement Ratio for Concrete Mix Design M30 Grade Concrete for R.C.C Work. Cement = OPC 53 Grade, assume 7- Day compressive strength of Cement is 37 N/mm 2. Here, 28 -day compressive strength of concrete = target mean strength= 38.25 N/mm 2.
### Water–cement ratio Wikipedia
The water–cement ratio is the ratio of the weight of water to the weight of cement used in a concrete mix. A lower ratio leads to higher strength and durability, but may make the mix difficult to work with and form. Workability can be resolved with the use of plasticizers or super-plasticizers.. Often, the ratio refers to the ratio of water to cementitious materials, w/cm. Cementitious
### Water-Cement Ratio (W/C) of Concrete online civil
15-11-2014· This means that water makes the concrete workable. It is found theoretically that water required for these two functions is about 0.50 to 0.60 times the weight of cement. This ratio of the amount of water to the amount of cement by weight is termed as the water-cement ratio and the strength and quality of concrete primarily depend upon this ratio.
### Determination of water/cement-ratio of concrete
Aalto University, P.O. BOX 11000, 00076 AALTO aalto.fi Abstract of master's thesis 2 Author: Ammad Tauqir Title of thesis: Determination of Water/Cement-Ratio of Concrete Master programme Building Technology Code ENG27 Thesis supervisor: Professor of practice Jouni Punkki Thesis advisor(s): M.Sc Teemu Ojala. Date 09.11.2018 Number of pages 66+14. Language English.
### How To Calculate Water Cement Ratio In Concrete
06-03-2018· After watching this construction video tutorial, you will understand better how to work out proportion of water and cement in concrete. Generally, the water cement ratio remains at under 0.4 to 06 according to IS Code 10262 (2009) toward nominal mix M7.5, M10, M15, M20 and M25. To determine how much water is required in []
### Table 64 water cement ratio for M70 grade of concrete
Table: 6.4 water-cement ratio for M70 grade of concrete Grade of Concrete M70 w/c ratio 0.38 Step4. Calculation of Fly Ash content: Large amounts of powder materials are required to achieve the self compactibility. However, if an excess amount of cement is added, the cost of materials and dry shrinkage will increase. To avoid the above two, a pozzolanic material like flyash (class F) was
### concrete mix design of M60 as per latest IS Code IS
30-03-2020· Water content with using appropriate dosages of super plasticisers =202 X 0.7 =141.4 kg/cum. water cementitious material ratio (w/c ratio) From figure 1 Page no 4 of IS 10262 : 2019 free water cement ratio required for the target strength of 68.25 N/mm2 is 0.28 for OPC 53 grade (curve 3) hence OK. w/c=00.28 | 3,268 | 13,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-39 | latest | en | 0.88436 |
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#### Lab-Aids: Exploring Newton's First Law - Inertia Kit
Using the specially-designed LAB-AIDS‚®/SEPUP track, students investigate inertia by observing a marble„¢s motion. They observe a marble in motion around a circular track then predict what path it will follow when a section of the track is removed, allowing the marble to roll unconstrained across the table top. They discover that once the marble leaves the confines of the track, it moves in a straight line path. Students also investigate how changing the mass of the marble affects the motion of the marble on the track. In Part 2 they complete a reading about inertia and Newton's first law of motion. Developed by SEPUP.
Scientific Concepts:
• An object that is not being subject to a force will continue to move at a constant speed in a straight line
• Friction is a force that causes changes in the speed of an object's motion
• Scientific explanations emphasize evidence, have logically consistent arguments, and use scientific principles, models, and theories
Content List:
• 1 Teacher's Guide
• 28 Student Worksheets and Guides - Parts 1 and 2 | 319 | 1,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-27 | latest | en | 0.855826 |
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Dec9 accepted Exact cover solution Dec8 comment Exact cover solution Rojo, no, they solve that task by other methods, but thanks for your link! Dec8 comment Exact cover solution Pentomino tiling is a task of 2 subtasks: 1) generate all possible pentomino placement in matrix, e.g., generate list of matrices as a result; 2) exact cover algorithm to find all possible pentomino packings. Is it possible to solve in Mathematica? Dec8 comment Exact cover solution Pentomino tiling is one of classic tasks of exact cover algorithm: en.wikipedia.org/wiki/Exact_cover#Pentomino_tiling Dec8 asked Exact cover solution Dec2 accepted Adding another control to Manipulate[MatrixPlot[matrix]] to take specific part of matrix Dec2 comment Adding another control to Manipulate[MatrixPlot[matrix]] to take specific part of matrix wow, thanks for your prompt answer! Dec2 asked Adding another control to Manipulate[MatrixPlot[matrix]] to take specific part of matrix Nov13 awarded Nice Question Nov13 awarded Scholar Nov13 awarded Supporter Nov13 accepted Getting number of binary digits combinations without “forbidden” patterns Nov13 revised Getting number of binary digits combinations without “forbidden” patterns deleted 26 characters in body Nov13 revised Getting number of binary digits combinations without “forbidden” patterns added 95 characters in body Nov13 awarded Editor Nov13 revised Getting number of binary digits combinations without “forbidden” patterns deleted 1 characters in body Nov13 asked Getting number of binary digits combinations without “forbidden” patterns Oct29 awarded Student | 392 | 1,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-18 | longest | en | 0.818735 |
https://gitlab.inria.fr/why3/why3/-/raw/6dc70dbf1ee8d8d2cf4080751794b9c7c99a8314/examples/programs/verifythis_fm2012_lcp/verifythis_fm2012_lcp_LCP_sorted_ge_1.v?inline=false | 1,611,640,954,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-04/segments/1610704798089.76/warc/CC-MAIN-20210126042704-20210126072704-00305.warc.gz | 356,818,148 | 3,033 | (* This file is generated by Why3's Coq driver *) (* Beware! Only edit allowed sections below *) Require Import BuiltIn. Require BuiltIn. Require int.Int. (* Why3 assumption *) Definition unit := unit. Axiom map : forall (a:Type) {a_WT:WhyType a} (b:Type) {b_WT:WhyType b}, Type. Parameter map_WhyType : forall (a:Type) {a_WT:WhyType a} (b:Type) {b_WT:WhyType b}, WhyType (map a b). Existing Instance map_WhyType. Parameter get: forall {a:Type} {a_WT:WhyType a} {b:Type} {b_WT:WhyType b}, (map a b) -> a -> b. Parameter set: forall {a:Type} {a_WT:WhyType a} {b:Type} {b_WT:WhyType b}, (map a b) -> a -> b -> (map a b). Axiom Select_eq : forall {a:Type} {a_WT:WhyType a} {b:Type} {b_WT:WhyType b}, forall (m:(map a b)), forall (a1:a) (a2:a), forall (b1:b), (a1 = a2) -> ((get (set m a1 b1) a2) = b1). Axiom Select_neq : forall {a:Type} {a_WT:WhyType a} {b:Type} {b_WT:WhyType b}, forall (m:(map a b)), forall (a1:a) (a2:a), forall (b1:b), (~ (a1 = a2)) -> ((get (set m a1 b1) a2) = (get m a2)). Parameter const: forall {a:Type} {a_WT:WhyType a} {b:Type} {b_WT:WhyType b}, b -> (map a b). Axiom Const : forall {a:Type} {a_WT:WhyType a} {b:Type} {b_WT:WhyType b}, forall (b1:b) (a1:a), ((get (const b1:(map a b)) a1) = b1). (* Why3 assumption *) Inductive array (a:Type) {a_WT:WhyType a} := | mk_array : Z -> (map Z a) -> array a. Axiom array_WhyType : forall (a:Type) {a_WT:WhyType a}, WhyType (array a). Existing Instance array_WhyType. Implicit Arguments mk_array [[a] [a_WT]]. (* Why3 assumption *) Definition elts {a:Type} {a_WT:WhyType a}(v:(array a)): (map Z a) := match v with | (mk_array x x1) => x1 end. (* Why3 assumption *) Definition length {a:Type} {a_WT:WhyType a}(v:(array a)): Z := match v with | (mk_array x x1) => x end. (* Why3 assumption *) Definition get1 {a:Type} {a_WT:WhyType a}(a1:(array a)) (i:Z): a := (get (elts a1) i). (* Why3 assumption *) Definition set1 {a:Type} {a_WT:WhyType a}(a1:(array a)) (i:Z) (v:a): (array a) := (mk_array (length a1) (set (elts a1) i v)). (* Why3 assumption *) Definition make {a:Type} {a_WT:WhyType a}(n:Z) (v:a): (array a) := (mk_array n (const v:(map Z a))). (* Why3 assumption *) Definition array_bounded(a:(array Z)) (b:Z): Prop := forall (i:Z), ((0%Z <= i)%Z /\ (i < (length a))%Z) -> ((0%Z <= (get1 a i))%Z /\ ((get1 a i) < b)%Z). (* Why3 assumption *) Definition map_eq_sub {a:Type} {a_WT:WhyType a}(a1:(map Z a)) (a2:(map Z a)) (l:Z) (u:Z): Prop := forall (i:Z), ((l <= i)%Z /\ (i < u)%Z) -> ((get a1 i) = (get a2 i)). (* Why3 assumption *) Definition exchange {a:Type} {a_WT:WhyType a}(a1:(map Z a)) (a2:(map Z a)) (i:Z) (j:Z): Prop := ((get a1 i) = (get a2 j)) /\ (((get a2 i) = (get a1 j)) /\ forall (k:Z), ((~ (k = i)) /\ ~ (k = j)) -> ((get a1 k) = (get a2 k))). Axiom exchange_set : forall {a:Type} {a_WT:WhyType a}, forall (a1:(map Z a)), forall (i:Z) (j:Z), (exchange a1 (set (set a1 i (get a1 j)) j (get a1 i)) i j). (* Why3 assumption *) Inductive permut_sub{a:Type} {a_WT:WhyType a} : (map Z a) -> (map Z a) -> Z -> Z -> Prop := | permut_refl : forall (a1:(map Z a)) (a2:(map Z a)), forall (l:Z) (u:Z), (map_eq_sub a1 a2 l u) -> (permut_sub a1 a2 l u) | permut_sym : forall (a1:(map Z a)) (a2:(map Z a)), forall (l:Z) (u:Z), (permut_sub a1 a2 l u) -> (permut_sub a2 a1 l u) | permut_trans : forall (a1:(map Z a)) (a2:(map Z a)) (a3:(map Z a)), forall (l:Z) (u:Z), (permut_sub a1 a2 l u) -> ((permut_sub a2 a3 l u) -> (permut_sub a1 a3 l u)) | permut_exchange : forall (a1:(map Z a)) (a2:(map Z a)), forall (l:Z) (u:Z) (i:Z) (j:Z), ((l <= i)%Z /\ (i < u)%Z) -> (((l <= j)%Z /\ (j < u)%Z) -> ((exchange a1 a2 i j) -> (permut_sub a1 a2 l u))). Axiom permut_weakening : forall {a:Type} {a_WT:WhyType a}, forall (a1:(map Z a)) (a2:(map Z a)), forall (l1:Z) (r1:Z) (l2:Z) (r2:Z), (((l1 <= l2)%Z /\ (l2 <= r2)%Z) /\ (r2 <= r1)%Z) -> ((permut_sub a1 a2 l2 r2) -> (permut_sub a1 a2 l1 r1)). Axiom permut_eq : forall {a:Type} {a_WT:WhyType a}, forall (a1:(map Z a)) (a2:(map Z a)), forall (l:Z) (u:Z), (permut_sub a1 a2 l u) -> forall (i:Z), ((i < l)%Z \/ (u <= i)%Z) -> ((get a2 i) = (get a1 i)). Axiom permut_exists : forall {a:Type} {a_WT:WhyType a}, forall (a1:(map Z a)) (a2:(map Z a)), forall (l:Z) (u:Z), (permut_sub a1 a2 l u) -> forall (i:Z), ((l <= i)%Z /\ (i < u)%Z) -> exists j:Z, ((l <= j)%Z /\ (j < u)%Z) /\ ((get a2 i) = (get a1 j)). (* Why3 assumption *) Definition exchange1 {a:Type} {a_WT:WhyType a}(a1:(array a)) (a2:(array a)) (i:Z) (j:Z): Prop := (exchange (elts a1) (elts a2) i j). (* Why3 assumption *) Definition permut_sub1 {a:Type} {a_WT:WhyType a}(a1:(array a)) (a2:(array a)) (l:Z) (u:Z): Prop := (permut_sub (elts a1) (elts a2) l u). (* Why3 assumption *) Definition permut {a:Type} {a_WT:WhyType a}(a1:(array a)) (a2:(array a)): Prop := ((length a1) = (length a2)) /\ (permut_sub (elts a1) (elts a2) 0%Z (length a1)). Axiom exchange_permut : forall {a:Type} {a_WT:WhyType a}, forall (a1:(array a)) (a2:(array a)) (i:Z) (j:Z), (exchange1 a1 a2 i j) -> (((length a1) = (length a2)) -> (((0%Z <= i)%Z /\ (i < (length a1))%Z) -> (((0%Z <= j)%Z /\ (j < (length a1))%Z) -> (permut a1 a2)))). Axiom permut_sym1 : forall {a:Type} {a_WT:WhyType a}, forall (a1:(array a)) (a2:(array a)), (permut a1 a2) -> (permut a2 a1). Axiom permut_trans1 : forall {a:Type} {a_WT:WhyType a}, forall (a1:(array a)) (a2:(array a)) (a3:(array a)), (permut a1 a2) -> ((permut a2 a3) -> (permut a1 a3)). (* Why3 assumption *) Definition array_eq_sub {a:Type} {a_WT:WhyType a}(a1:(array a)) (a2:(array a)) (l:Z) (u:Z): Prop := (map_eq_sub (elts a1) (elts a2) l u). (* Why3 assumption *) Definition array_eq {a:Type} {a_WT:WhyType a}(a1:(array a)) (a2:(array a)): Prop := ((length a1) = (length a2)) /\ (array_eq_sub a1 a2 0%Z (length a1)). Axiom array_eq_sub_permut : forall {a:Type} {a_WT:WhyType a}, forall (a1:(array a)) (a2:(array a)) (l:Z) (u:Z), (array_eq_sub a1 a2 l u) -> (permut_sub1 a1 a2 l u). Axiom array_eq_permut : forall {a:Type} {a_WT:WhyType a}, forall (a1:(array a)) (a2:(array a)), (array_eq a1 a2) -> (permut a1 a2). Axiom permut_bounded : forall (a1:(array Z)) (a2:(array Z)) (n:Z), ((permut a1 a2) /\ (array_bounded a1 n)) -> (array_bounded a2 n). (* Why3 assumption *) Definition is_common_prefix(a:(array Z)) (x:Z) (y:Z) (l:Z): Prop := (0%Z <= l)%Z /\ (((x + l)%Z <= (length a))%Z /\ (((y + l)%Z <= (length a))%Z /\ forall (i:Z), ((0%Z <= i)%Z /\ (i < l)%Z) -> ((get1 a (x + i)%Z) = (get1 a (y + i)%Z)))). Axiom common_prefix_eq : forall (a:(array Z)) (x:Z), ((0%Z <= x)%Z /\ (x < (length a))%Z) -> (is_common_prefix a x x ((length a) - x)%Z). Axiom common_prefix_eq2 : forall (a:(array Z)) (x:Z), ((0%Z <= x)%Z /\ (x < (length a))%Z) -> ~ (is_common_prefix a x x (((length a) - x)%Z + 1%Z)%Z). Axiom not_common_prefix_if_last_different : forall (a:(array Z)) (x:Z) (y:Z) (l:Z), ((0%Z < l)%Z /\ (((x + l)%Z < (length a))%Z /\ (((y + l)%Z < (length a))%Z /\ ~ ((get1 a (x + (l - 1%Z)%Z)%Z) = (get1 a (y + (l - 1%Z)%Z)%Z))))) -> ~ (is_common_prefix a x y l). Parameter longest_common_prefix: (array Z) -> Z -> Z -> Z. Axiom lcp_spec : forall (a:(array Z)) (x:Z) (y:Z) (l:Z), (((0%Z <= x)%Z /\ (x < (length a))%Z) /\ ((0%Z <= y)%Z /\ (y < (length a))%Z)) -> ((l = (longest_common_prefix a x y)) <-> ((is_common_prefix a x y l) /\ ~ (is_common_prefix a x y (l + 1%Z)%Z))). Axiom lcp_is_cp : forall (a:(array Z)) (x:Z) (y:Z), (((0%Z <= x)%Z /\ (x < (length a))%Z) /\ ((0%Z <= y)%Z /\ (y < (length a))%Z)) -> (is_common_prefix a x y (longest_common_prefix a x y)). Axiom lcp_eq : forall (a:(array Z)) (x:Z) (y:Z), (((0%Z <= x)%Z /\ (x < (length a))%Z) /\ ((0%Z <= y)%Z /\ (y < (length a))%Z)) -> forall (i:Z), ((0%Z <= i)%Z /\ (i < (longest_common_prefix a x y))%Z) -> ((get1 a (x + i)%Z) = (get1 a (y + i)%Z)). Axiom lcp_refl : forall (a:(array Z)) (x:Z), ((0%Z <= x)%Z /\ (x < (length a))%Z) -> ((longest_common_prefix a x x) = ((length a) - x)%Z). (* Why3 assumption *) Inductive ref (a:Type) {a_WT:WhyType a} := | mk_ref : a -> ref a. Axiom ref_WhyType : forall (a:Type) {a_WT:WhyType a}, WhyType (ref a). Existing Instance ref_WhyType. Implicit Arguments mk_ref [[a] [a_WT]]. (* Why3 assumption *) Definition contents {a:Type} {a_WT:WhyType a}(v:(ref a)): a := match v with | (mk_ref x) => x end. (* Why3 assumption *) Definition le(a:(array Z)) (x:Z) (y:Z): Prop := let n := (length a) in (((0%Z <= x)%Z /\ (x < n)%Z) /\ (((0%Z <= y)%Z /\ (y < n)%Z) /\ let l := (longest_common_prefix a x y) in (((x + l)%Z = n) \/ (((x + l)%Z < n)%Z /\ (((y + l)%Z < n)%Z /\ ((get1 a (x + l)%Z) <= (get1 a (y + l)%Z))%Z))))). Axiom le_refl : forall (a:(array Z)) (x:Z), ((0%Z <= x)%Z /\ (x < (length a))%Z) -> (le a x x). Axiom le_trans : forall (a:(array Z)) (x:Z) (y:Z) (z:Z), (((0%Z <= x)%Z /\ (x < (length a))%Z) /\ (((0%Z <= y)%Z /\ (y < (length a))%Z) /\ (((0%Z <= z)%Z /\ (z < (length a))%Z) /\ ((le a x y) /\ (le a y z))))) -> (le a x z). (* Why3 assumption *) Definition sorted_sub(a:(array Z)) (data:(array Z)) (l:Z) (u:Z): Prop := forall (i1:Z) (i2:Z), (((l <= i1)%Z /\ (i1 <= i2)%Z) /\ (i2 < u)%Z) -> (le a (get1 data i1) (get1 data i2)). Axiom sorted_bounded : forall (a:(array Z)) (data:(array Z)) (l:Z) (u:Z) (i:Z), (((l <= i)%Z /\ (i < u)%Z) /\ (sorted_sub a data l u)) -> ((0%Z <= (get1 data i))%Z /\ ((get1 data i) < (length a))%Z). Axiom sorted_le : forall (a:(array Z)) (data:(array Z)) (l:Z) (u:Z) (i:Z) (x:Z), (((l <= i)%Z /\ (i < u)%Z) /\ ((sorted_sub a data l u) /\ (le a x (get1 data l)))) -> (le a x (get1 data i)). Require Import Why3. Ltac ae := why3 "alt-ergo" timelimit 3. (* Why3 goal *) Theorem sorted_ge : forall (a:(array Z)) (data:(array Z)) (l:Z) (u:Z) (i:Z) (x:Z), ((sorted_sub a data l u) /\ ((le a (get1 data u) x) /\ ((l <= i)%Z /\ (i < u)%Z))) -> (le a (get1 data i) x). intros a data l u i x (h1,(h2,(h3,h4))). apply le_trans with (get1 data u). ae. Qed. | 4,133 | 9,769 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-04 | latest | en | 0.293841 |
https://www.physicsforums.com/threads/second-moment-of-area.317018/ | 1,581,902,487,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141460.64/warc/CC-MAIN-20200217000519-20200217030519-00002.warc.gz | 876,765,963 | 15,999 | # Second moment of area ?
second moment of area ???
## Homework Statement
In finding the metacentric height of a floating body ( G-center of gravity ,M-metacentric point ,B-buoyancy point )
so if the object is stable then M should be above G .
GM=BM-BG
i have learned that BM=I/V (where I second moment of are about the horizontal center line ,and V is submerged volume)
so if the object is a cylinder i can find I using Pi*d^(4)/64 ....but can i use the same equation if the cylinder is not a uniform one .
And is it the same thing meant by
1)second moment of area
2)Moment of inertia
Or are they two different values ?
Can they be connected using a another varaible ?
## The Attempt at a Solution
Related Engineering and Comp Sci Homework Help News on Phys.org
Hao
i have learned that BM=I/V (where I second moment of are about the horizontal center line ,and V is submerged volume)
so if the object is a cylinder i can find I using Pi*d^(4)/64 ....but can i use the same equation if the cylinder is not a uniform one .
The second moment of area is similar to the moment of inertia except for one extremely important point:
The moment of inertia includes an integral over density, $$\rho$$.
$$I_{inertia} = \int \rho r^2 dA$$.
The second moment of area does not.
$$I_{area} = \int r^2 dA$$.
Hence, for a uniform object, $$I_{inertia} = \rho I_{area}$$.
Therefore, the second moment of area is independent of the uniformity of the body.
tiny-tim
Homework Helper
moment of inertia ~ second moment of area
Hi Hao!
You're right, and you're wrong!!
For all normal physicists, you're right …
but some structural engineers use the name https://www.physicsforums.com/library.php?do=view_item&itemid=31" for the second moment of area, so they can be the same (and in this case I suspect they are)!
Last edited by a moderator: | 459 | 1,836 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-10 | longest | en | 0.842167 |
https://math.stackexchange.com/questions/989158/a-particular-example-of-a-non-reduced-scheme-with-a-reduced-ring-of-global-sect?noredirect=1 | 1,696,475,257,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00868.warc.gz | 418,491,443 | 35,357 | # A particular example of a non-reduced scheme (with a reduced ring of global sections)?
I am searching for a scheme $X$ which can be obtained by gluing two affine schemes (along open subsets) such that: 1) X is non-reduced; 2) $\Gamma(X,\mathcal O_X)$ is a reduced ring. Any ideas? Thanks in advance.
Hint:
Can you construct a ''fat'' projective line of some sort? Namely, remember that $\text{Spec } k[x]$ is the affine line and we get projective space by gluing together two affine lines. Can you do something similar by fattening up the lines you glue?
My example was the following. Take $\text{Proj } k[x,y,z]/z^2$. I believe that working on affines, this should produce something with only $k$ as global sections. I haven't worked it out in detail however.
• What is your specific example? A fat projective line as a fat ring of global sections, basically because you cannot glue along the fat variable. Oct 24, 2014 at 14:43
• I am also thinking in this direction, but I find it difficult to prevent the nilpotent sections to be glued together and form a nilpotent global section.
– AYK
Oct 24, 2014 at 14:45
• @MartinBrandenburg See my example here. I hope it works, I haven't worked it through in detail as I wrote. Oct 24, 2014 at 14:50
• $\mathrm{Proj}(k[x,y,z]/(z^2)) = D_+(x) \cup D_+(y)$ (since $D_+(z)=\emptyset$) with $D_+(x) = \mathrm{Spec}(k[y/x,z/x]/(z/x)^2)$ and $D_+(y) = \mathrm{Spec}(k[x/y,z/y]/(z/y)^2)$. I don't think that the ring of global sections contains $z$. It has a good chance to be $k$. Oct 24, 2014 at 15:05
• I have computed this and barring mistakes, I believe that this has global sections $k$. Oct 24, 2014 at 15:06 | 489 | 1,660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-40 | latest | en | 0.933554 |
http://www8.cs.umu.se/kurser/TDBA77/VT06/algorithms/BOOK/BOOK4/NODE154.HTM | 1,516,270,281,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887224.19/warc/CC-MAIN-20180118091548-20180118111548-00554.warc.gz | 581,106,280 | 6,290 | Next: Calendrical Calculations Up: Combinatorial Problems Previous: Generating Partitions
## Generating Graphs
Input description: Parameters describing the desired graph, such as the number of vertices n, the number of edges m, or the edge probability p.
Problem description: Generate (1) all or (2) a random or (3) the next graph satisfying the parameters.
Discussion: Graph generation typically arises in constructing test data for programs. Perhaps you have two different programs that solve the same problem, and you want to see which one is faster or make sure that they always give the same answer. Another application is experimental graph theory, verifying whether a particular property is true for all graphs or how often it is true. It is much easier to conjecture the four-color theorem once you have demonstrated 4-colorings for all planar graphs on 15 vertices.
A different application of graph generation arises in network design. Suppose you need to design a network linking ten machines using as few cables as possible, such that the network can survive up to two vertex failures. One approach is to test all the networks with a given number of edges until you find one that will work. For larger graphs, more heuristic approaches, like simulated annealing, will likely be necessary.
Many factors complicate the problem of generating graphs. First, make sure you know what you want to generate:
• Do I want labeled or unlabeled graphs? - The issue here is whether the names of the vertices matter in deciding whether two graphs are the same. In generating labeled graphs, we seek to construct all possible labelings of all possible graph topologies. In generating unlabeled graphs, we seek only one representative for each topology and ignore labelings. For example, there are only two connected unlabeled graphs on three vertices - a triangle and a simple path. However, there are four connected labeled graphs on three vertices - one triangle and three 3-vertex paths, each distinguished by their central vertex. In general, labeled graphs are much easier to generate. However, there are so many more of them that you quickly get swamped with isomorphic copies of the same few graphs.
• What do I mean by random? - There are two primary models of random graphs, both of which generate graphs according to different probability distributions. The first model is parameterized by a given edge probability p. Typically, p=0.5, although smaller values can be used to construct sparser random graphs. In this model a coin is flipped for each pair of vertices x and y to decide whether to add an edge (x,y). All labeled graphs will be generated with equal probability when p=1/2.
The second model is parameterized by the desired number of edges m. It selects m distinct edges uniformly at random. One way to do this is by drawing random (x,y)-pairs and creating an edge if that pair is not already in the graph. An alternative approach to computing the same things constructs the set of possible edges and selects a random m-subset of them, as discussed in Section .
Which of these options best models your application? Probably none of them. Random graphs, by definition, have very little structure. In most applications, graphs are used to model relationships, which are often highly structured. Experiments conducted on random graphs, although interesting and easy to perform, often fail to capture what you are looking for.
An alternative to random graphs is to use ``organic'' graphs, graphs that reflect the relationships among real-world objects. The Stanford GraphBase, discussed below, is an outstanding source of organic graphs. Further, there are many raw sources of relationships electronically available via the Internet that can be turned into interesting organic graphs with a little programming and imagination. Consider the graph defined by a set of WWW pages, with any hyperlink between two pages defining an edge. Or what about the graph implicit in railroad, subway, or airline networks, with vertices being stations and edges between two stations connected by direct service? As a final example, every large computer program defines a call graph, where the vertices represent subroutines, and there is an edge (x,y) if x calls y.
Two special classes of graphs have generation algorithms that have proven particularly useful in practice:
• Trees - Prüfer codes provide a simple way to rank and unrank labeled trees and thus solve all the standard generation problems discussed in Section . There are exactly labeled trees on n vertices, and exactly that many strings of length n-2 on the alphabet .
The key to Prüfer's bijection is the observation that every tree has at least two vertices of degree 1. Thus in any labeled tree, the vertex v incident on the leaf with lowest label is well-defined. We take v to be , the first character in the code. We then delete the associated leaf and repeat the procedure until only two vertices are left. This defines a unique code S for any given labeled tree that can be used to rank the tree. To go from code to tree, observe that the degree of vertex v in the tree is one more than the number of times v occurs in S. The lowest-labeled leaf will be the smallest integer missing from S, which when paired with determines the first edge of the tree. The entire tree follows by induction.
Algorithms for efficiently generating unlabeled rooted trees are presented in the implementation section below.
• Fixed degree sequence graphs - The degree sequence of a graph G is an integer partition where is the degree of the ith highest-degree vertex of G. Since each edge contributes to the degree of two vertices, p is a partition of 2m, where m is the number of edges in G.
Not all partitions correspond to degree sequences of graphs. However, there is a recursive construction that constructs a graph with a given degree sequence if one exists. If a partition is realizable, the highest-degree vertex can be connected to the next highest-degree vertices in G, or the vertices corresponding to parts . Deleting and decrementing yields a smaller partition, which we recur on. If we terminate without ever creating negative numbers, the partition was realizable. Since we always connect the highest-degree vertex to other high-degree vertices, it is important to reorder the parts of the partition by size after each iteration.
Although this construction is deterministic, a semirandom collection of graphs realizing this degree sequence can be generated from G using edge-flipping operations. Suppose edges (x,y) and (w,z) are in G, but (x,w) and (y,z) are not. Exchanging these pairs of edges creates a different (not necessarily connected) graph without changing the degrees of any vertex.
Implementations: The Stanford GraphBase [Knu94] is perhaps most useful as an instance generator for constructing a wide variety of graphs to serve as test data for other programs. It incorporates graphs derived from interactions of characters in famous novels, Roget's Thesaurus, the Mona Lisa, expander graphs, and the economy of the United States. It also contains routines for generating binary trees, graph products, line graphs, and other operations on basic graphs. Finally, because of its machine-independent random number generators, it provides a way to construct random graphs such that they can be reconstructed elsewhere, thus making them perfect for experimental comparisons of algorithms. See Section for additional information.
Combinatorica [Ski90] provides Mathematica generators for basic graphs such as stars, wheels, complete graphs, random graphs and trees, and graphs with a given degree sequence. Further, it includes operations to construct more interesting graphs from these, including join, product, and line graph. Graffiti [Faj87], a collection of almost 200 graphs of graph-theoretic interest, are available in Combinatorica format. See Section .
The graph isomorphism testing program nauty (see Section ), by Brendan D. McKay of the Australian National University, has been used to generate catalogs of all nonisomorphic graphs with up to 11 vertices. This extension to nauty, named makeg, can be obtained by anonymous ftp from bellatrix.anu.edu.au (150.203.23.14) in the directory pub/nauty19.
Nijenhuis and Wilf [NW78] provide efficient Fortran routines to enumerate all labeled trees via Prüfer codes and to construct random unlabeled rooted trees. See Section . A collection of generators for standard families of graphs is included with LEDA (see Section ).
Notes: An extensive literature exists on generating graphs uniformly at random. Surveys include [Gol93, Tin90]. Closely related to the problem of generating classes of graphs is counting them. Harary and Palmer [HP73] survey results in graphical enumeration.
Random graph theory is concerned with the properties of random graphs. Threshold laws in random graph theory define the edge density at which properties such as connectedness become highly likely to occur. Expositions on random graph theory include [ES74, Pal85].
An integer partition is graphic if there exists a simple graph with that degree sequence. Erdős and Gallai [EG60] proved that a degree sequence is graphic if and only if the sequence observes the following condition for each integer r < n:
The bijection between n-2 strings and labeled trees is due to Prüfer [Prü18]. Good expositions on this result include [Eve79a, NW78].
Related Problems: Generating permutations (see page ), graph isomorphism (see page ).
Next: Calendrical Calculations Up: Combinatorial Problems Previous: Generating Partitions
Algorithms
Mon Jun 2 23:33:50 EDT 1997 | 1,982 | 9,763 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.91496 |
https://cognitivecardiomath.com/cognitive-cardio-blog/order-of-operations-sequencing-activity/ | 1,725,974,735,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00160.warc.gz | 153,590,865 | 38,430 | # Free Order of Operations Activity
## Order of Operations Activity for 5th and 6th Grade Math
Are you looking for an easy, effective and FREE order of operations activity to help students practice order of operations in math class?
Early last week, I was trying to think of a different kind of activity to help my middle school math students who needed more reinforcement with order of operations. So I decided to make a sequencing activity.
I hadn’t tried this before, so I wasn’t sure about the best way to design it. But the activity ended up working very well.
This post shares how I put this math activity together and used it with my 6th grade math classes.
## For this order of operations activity, I created 8 different expressions, and then typed out the steps to simplify each expression. You can see what this looks like farther down in the post.
To prepare the activity for the students:
1) I copied the expressions and steps onto different colored papers, so that two expressions would be on the same color.
2) I cut the steps apart into strips, and then put two expressions and their steps (of the same colored paper) into a baggie.
• I figured if I put only one equation in a baggie, the activity would be too simple.
• If I put two expressions of two different colors, it would be too easy.
• So I went with two expressions in the same color. That way they’d have to do some sorting of the expressions.:-)
## Using the Order of Operations Sequencing Activity
I put 3 baggies (6 different expressions and their order of operations steps) into a manila envelope for each group.
• Groups were mostly just partners, with an occasional group of 3.
I typed directions to include in the envelopes, and asked students to do their best to follow those directions before asking for clarification.
• Some of the students worked on this activity, while others completed different activities. So, I needed them to try to work through the directions themselves, before I got to each group to discuss with them.
Some students needed additional instruction, while others did not.
After students put the steps into the correct order, they had to write those steps onto a recording sheet, pictured below.
In each of my math classes, students worked on this activity for about 15-20 minutes.
• Some groups completed all 6 expressions, while others completed only 2-3.
• A few more minutes would have been helpful for those students who didn’t complete as many expressions, but I can revisit the activity with those students this week.
I will definitely use this again next year:)
## More Resources!
If you’re looking for other resources to give your students quality practice with order of operations, I have several in my TPT shop – some print and some digital.
The Math Wheel is great for taking notes when you’re teaching or reviewing this topic.
Order of Operations Footloose is a FREE resource with 30 task cards, student recording sheet, and key.
Truth or Dare Game is a fun way to practice this concept! Students choose Truth or Dare questions that allows them to self-differentiate and think a little more deeply about the concepts. There is a separate digital version of this one as well.
Order of Operations Color by Number is a long-time favorite printable resource that now has digital options!
Order of Operations Color by Answer Pixel Art is digital practice with the pattern/pictures appearing with students’ correct answers.
Google Task Cards and Quiz are the same 30 task cards questions as the Footloose. These cards are self-checking and the Google Forms quiz offers a quick assessment.
## Ellie
### One Comment
Welcome to Cognitive Cardio Math! I’m Ellie, a wife, mom, grandma, and dog ‘mom,’ and I’ve spent just about my whole life in school! With nearly 30 years in education, I’ve taught:
• All subject areas in 4th and 5th grades
• Math, ELA, and science in 6th grade (middle school)
I’ve been creating resources for teachers since 2012 and have worked in the elearning industry for about five years as well! | 861 | 4,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-38 | latest | en | 0.961944 |
https://rip94550.wordpress.com/2010/10/18/regression-1-%E2%80%93-selection-criteria/ | 1,527,051,495,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865450.42/warc/CC-MAIN-20180523043959-20180523063959-00211.warc.gz | 643,917,109 | 29,008 | ## Introduction
I will close my discussion of the (Mathematica®) properties of a regression with “selection criteria” or “measures of improvement”: how do we select the “best” regression?
While I will do a couple of calculations, most numerical work with these criteria will wait until the next regression post.
Mathematica provides 4 such measures: R Squared, Adjusted R Squared, the Akaike Information Criterion (AIC), and the Bayesian Information Criterion (BIC).
In addition, I know almost a dozen more, and I’ll show them to you.
Almost all of them use the error sum of squares ESS; the R Squared and the Adjusted R Squared also use the total sum of squares TSS:
$TSS = (y - \bar{y}).(y-\bar{y})$
$ESS = (y-\hat{y}).(y-\hat{y})$
and the numbers n and k…
n = number of observations
k = number of parameters ($\beta\$s).
## R Squared and Adjusted R Squared
The R Squared is
R2 = 1 – ESS / TSS.
It is 1 if and only if ESS = 0, i.e. there are no errors… y = yhat… the fit is exact at every data point.
This may or may not be a good thing.
In addition, there is a crucial fact pertaining to the use of R Squared as a measure of improvement: if we add a variable to a regession, the R Squared cannot decrease. It doesn’t have to go up, but it cannot get smaller.
In other words, as a measure of improvement, R Squared says we’re never worse off adding a variable. It can never recommend against adding a variable (until we have k > n, at which point the regression will fail, because X’X cannot be inverted).
Adjusted R Squared = 1 – (ESS/n-k) / (TSS/n-1).
As I have said before, it can be interpreted as subtracting the ratio of two estimates of variance. Like the R Squared, it is 1 if there are no errors.
But if we add a variable whose t-statistic is less than 1 (in absolute value), then the Adjusted R Squared will decrease. It does balance complexity (more variables in the model) against smaller total squared error.
In contrast to every other measure we will see here, we want its maximum rather than its minimum.
## The AIC and BIC
The AIC, as Mathematica uses it, appears to be defined as
$n\ Log[\frac{ESS}{n}]+2(k+1)+n Log[2Pi]+n$
It is more than convenient to use rules at this point, so that I can write symbolic equations involving n, k, and ESS without having Mathematica use the numerical values of them.
Let me clear the numerical values… set some rules… then ask for the AIC for the regression (our main one, the Hald data with X1, X2, and X4, which can be found here)… and finally compute it directly using the equation…
Good. The regression AIC agrees with that equation.
(By the way, the //.ps said keep applying the rules until nothing changed. You might have noticed that the rule for ESS was incomplete: it used the numerical value of s^2, but symbolic n and k. That is, while it replaced n and k by numbers, it also introduced n-k. It took another pass to replace n-k by a number.)
The BIC from Mathematica is very similar; it has (k+1) log n where the AIC has 2 (k+1):
$n\ Log[\frac{ESS}{n}]+(k+1) Log[n]+n Log[2Pi]+n$
Let me confirm that that equation agrees with the regression:
As usual, I am happy to see that I know exactly what Mathematica did for me.
I was thrilled to find a table of selection criteria in Ramanathan (see bibliography page). Let me offer you the same thrill, augmented by another set of measures (all those subscripted with c or u, but also including the unsubscripted FPE) found in McQuarrie & Tsai’s “Regression and Time Series Model Selection”, ISBN 981 02 3242 X.
Let
$s^2 = \frac{ESS}{n-k}$
$\Sigma^2 = \frac{ESS}{n}$
You may recognize those as the minimum variance estimate and maximum likelihood estimate of the error variance $\sigma^2\$. And I hope you recognize that $s^2\$ is our “Estimated Variance”.
Then we may define the following:
$SIGMASQ = \Sigma^2\ \frac{1}{(1-\frac{k}{n})} = s^2$
$AIC = \Sigma^2\ e^{2k/n}$
$AICc = \Sigma^2\ e^{\frac{n+k}{n-k-2}}$
$AICu = s^2\ e^{\frac{n+k}{n-k-2}}$
$BIC (= SIC) = \Sigma^2\ n^{k/n}$
$FPE = \Sigma^2\ \frac{n+k}{n-k}$
$FPEu = s^2\ \frac{n+k}{n-k}$
$HQ = \Sigma^2\ log(n)^{2k/n}$
$HQc = s^2\ log(n) ^{\frac{n+k}{n-k-2}}$
$SHIBATA = \Sigma^2\ (1+\frac{2k}{n})$
$GCV = \Sigma^2\ \frac{1}{{(1-\frac{k}{n})}^2}$
$RICE = \Sigma^2\ \frac{1}{1-\frac{2k}{n}}$
We see that every one of them chooses between $s^2\$ or $\Sigma^2\$. Some of them differ in having 2 k / n or (n+k / n-k-2): AIC and AICc, HQ and HQc. Some of them raise something to a power: the various AIC raise e, the two HQ raise log(n), BIC raises n. The FPE, SHIBATA, GCV, and RICE do not raise anything to a power, just multiplying by some function of n and k.
They are all of the form: either $s^2\$ or $\Sigma^2\$ (that is, ESS/n or ESS / n-k) times a function of n and k that penalizes complexity.
SIGMASQ is far and away the simplest. In fact, it is equivalent to the Adjusted R Squared: one increases if and only if the other decreases. It appears that Ramanathan wrote SIGMASQ in terms of $\Sigma^2\$, to be consistent with his other entries, every one of which used $\Sigma^2\$, but SIGMASQ is really just s^2.
This shows us that we could write all of these in terms of $s^2\$ or in terms of $\Sigma^2\$ … but we’d be missing part of of why certain ones were defined. FPEu and AICu are FPE and AIC with $\Sigma^2\$ replaced by $s^2\$. And HQc is to HQ as AICc is to AIC: replace $\Sigma^2\$ by $s^2\$ and replace $\frac{2 k}{n}\$ by $\frac{n+k}{n-k-2}\$.
Let me emphasize that the point of these measures is to assess the combined changes in ESS and k for a set of regressions having the same number of observations (and based on a common set of variables). The point is that n is a constant, while ESS (and $s^2\$ and $\Sigma^2\$) are not.
Why don’t we just stay with the Adjusted R Squared? I tell you, again, without proof, that adding a variable to a regression will raise the Agjusted R Squared if and only if the t-statistic for that variable is greater than 1 (in absolute value). But we might argue that the variable isn’t really significant unless its t-statistic is greater than about 2, which means that the Adjusted R Squared can improve when the regression, in one sense, does not.
We fear, then, that the Adjusted R Squared will “over fit” an equation, and we want something more conservative.
They tell me that the AIC is not sufficiently conservative, so we try other things.
Oh, many of these are named for their inventors. On the other hand, FPE stands for Final Prediction Error – it was invented by Akaike, before he invented the AIC. He also invented the BIC, named the Bayesian Information Criterion. It is also known as the SIC, for Schwarz, who invented it independently. In fact, Ramanathan calls it SCHWARZ, and McQuarrie & Tsai call it SIC, while Mathematica calls it BIC.
HQ stands for Hannan & Quinn, its inventors. SHIBATA and RICE are also named for their inventors.
GCV stands for Generalized Cross-Validation. Cross-validation, in general, refers to the idea of fitting an equation to a subset of the data, and then investigating how well it fits the rest of the data.
## Reconciling the two equations for AIC
There are two other meaures I want to discuss, but first let’s reconcile my (and Ramanathan’s) simple formula for AIC,
$AIC = \Sigma^2\ e^{\frac{2k}{n}}$
with Mathematica’s formula:
$2+2 k+n+n Log[ESS/n]+n Log[2 \pi]$
Drop the three constant terms, getting
$2 k+n Log[ESS/n]$
Subtracting constants will not change which regression has the minimum value. Of course, it will have a different value at the minimum, but if all I want is to know which regression has the minimum AIC, this is as good as the first expression.
Now divide by n. That will have no effect on where the minimum is, either, getting
$\frac{2k}{n}+Log[ESS/n]$
Now take the exponential. That’s a monotonic function, so it won’t move the minimum either.
$e^{\frac{2k}{n}}\ \frac{ESS}{n}$
And that’s my easy-to-use formula for AIC.
Oh, one other piece of business. If you read the Mathematica help system, it will say that the (k+1) term is the number of parameters to be estimated.
But I thought that was k.
Ah. The number of parameters in the fitted model is indeed k, the number of $\beta\$s to be determined. But we are also attempting to estimate the variance $\sigma^2\$ of the errors in the true model, so the number of parameters to be estimated is one more than the number to be fitted: k+1.
Similar algebra will reconcile BIC as Mathematica computes it and as I compute it. And similar algebra will reconcile other expressions in McQuarrie & Tsai with their counterparts in Ramanathan (and my code).
Oh, since TSS depends only on the y’s, and n is a constant, we see that the Adjusted R Squared could have been manipulated to another form which would have a different maximum value but at the same place.
1 – s^2,
which makes it clear that it and s^2 (SIGMASQ) go inversely.
Maybe I should remove s^2 from my list of measures: it really is redundant. I have read that most of the others are not redundant, but I don’t know about all of them.
## PRESS and Mallows’ Cp
Ok. What are the two other measures that I know how to compute?
One, like GCV, is a measure of cross-validation. What if we fit the equation to all but one point? But which point? Why not all of them?
Compute the deletion residuals (which I introduced here)
$ed_i = \frac{e_i}{1-h_i}\$.
Compute the sum of squares, and call it PRESS (Partial Regression ESS):
PRESS = ed.ed
(No, I don’t know why we don’t divide by n-k. But if we divide by n, the resulting statistic is called CV(1), for “cross-validation dropping 1 observation”.)
Easy enough.
That leaves one more: Mallow’s Cp. It differs from all the others in that it requires two regressions; one is a touchstone against which all the others are compared.
We compute the measure in two steps. First, the actual Cp is given by
Cp = ESS / s^2 – (n-2k),
where s^2, however, is the estimated variance of our touchstone regression, but ESS and k come from the specific regression for which we are computing the Cp.
Second, what we want is that regression having the minimum value of
|Cp – k|.
That’s it: we want the regression having Cp closest to k, without regard to sign. (Mallows published the Cp is the early 1970s; in 1995 he suggested that any regression with Cp < k would be worth consideration. But that's not as convenient as minimizing |Cp – k|).
OK, that's how it's computed. But there's one other important consideration.
Look back at the equation for Cp, and suppose we are computing for the touchstone regression. Now we have s^2 = ESS / (n-k) and we get
Cp = n-k – n + 2k = k.
That is, for our touchstone, Cp = k, so |Cp – k| = 0. Cp tells us nothing useful about the touchstone. We want the minimum value of |Cp – k| for all of the regressions except the touchstone.
So we can't learn anything about our touchstone. So it better not be any of the regressions we care about!
In the case of the Hald data, our touchstone will usually be the regression using all 4 variables, and every other measure I compute will tell me that the touchstone is not a candidate for the "best regression".
What bothers me isn't so much that I can't apply Cp to the touchstone, but that the touchstone needs to be a poor enough fit that I don't care to test it.
This won't stop me from computing Cp, but for some of what I do, it won't really be very useful or accurate.
## Generalizing AIC and FPE again
In addition to all these, there are two infinite families of measures which I did not use. If we write
$AIC = \Sigma^2\ E^{\frac{2k}{n}}$
then we could consider it a special case (with $\alpha = 2\$) of
$AIC\alpha = \Sigma^2\ E^{\frac{\alpha k}{n}}\$.
Similarly, we could define
$FPE\alpha = \Sigma^2\ \frac{\alpha k}{n-k}\$
(No, that doesn’t simplify to FPE when ($\alpha = 2\$)… but like so many other different formulae for the same measure, it differs by a constant from the usual FPE. The following expression,
$\Sigma^2\ \frac{2 k}{n-k}\$,
differs from
$FPE = \Sigma^2\ \frac{n+k}{n-k}\$,
by $\Sigma^2\$.
Huh?
As I said, I did not attempt to use these two infinite families of selection criteria. I wonder if there’s some value of $\alpha\$ such that one of these would select the highest Adjusted R Squared but with all t-statistics greater than some cutoff… or maybe an $\alpha\$ so large that it would never recommend adding a variable? Maybe I’ll play around.
In the next post, I’m going to look at regressions using every possible subset of the Hald variables, and we’ll see what these selection criteria tell us.
### One Response to “Regression 1 – Selection Criteria”
1. rip Says:
In a comment under Suggestions, JEP said:
This is a very nice site. I have been using Mathematica’s NonlinearModelFit function and ordering the results according to their AIC & BIC scores. I wanted to calculate my own AIC/BIC values AND compare them to those produced by Mathematica.
I didn’t know what they had done. Your page was a big help in clearing that up for me. I might reference it in a paper unless you know if those formulae are published somewhere within Wolfram’s pages?
Hi JEP,
I’m glad to have been of help. And I wouldn’t mind at all if you cited my presentation of the two equations as confirmation of what Mathematica computes.
My source for the AIC formula was McQuarrie & Tsai, “Regression and Time Series Model Selection”, 981-02-3242-X, p. 21. I modified their SIC (BIC) formula on p. 23 by using (k+1) where they used k.
Then i confirmed that my two equations matched Mathematica’s answers. Mathematica’s brief discussion of AIC and BIC can be found in the tutorial for statistical model analysis, but I do not recall seeing the actual equations anywhere on Wolfram.
BTW, I’m going to copy this comment to the “selection criteria” post, as well as leaving the original under Suggestions.
rip | 3,767 | 13,949 | {"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": 55, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-22 | latest | en | 0.912459 |
https://doczz.net/doc/6831696/worksheet---panther-math--- | 1,660,373,934,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571909.51/warc/CC-MAIN-20220813051311-20220813081311-00652.warc.gz | 224,172,092 | 8,016 | # Worksheet - Panther Math!!!
## Transcription
Worksheet - Panther Math!!!
```13.1 Comparing Two Means Practice
SHOW ALL STEPS FOR FULL CREDIT (#2-4 are “full processes”
1) Improving Sleep It is thought that listening to the sound of the surf will help people get to sleep faster. An
experiment was done using 20 people. For 10 individuals, the sound of surf was piped into their bedroom. For the other
10, no sound was used. The experiment was run for a period of time and the average time it took for them to fall asleep
was monitored and averaged. Following is the data for each group.
With Surf
Without Surf
10.2
11.4
12.5
13.5
13.2
14.5
23.1
22.3
7.5
10
8.9
12.3
12.2
15
13.4
14.4
18.4
19.5
17.5
19.7
a. Check the conditions for performing a two sample t-test for means.
b. What is the standard error of the differences of the two independent samples?
2) Paying for College College financial aid offices expect students to use summer earnings to help pay for college. But
how large are these earnings ? One college studied this question by asking a random sample of students how much they
earned. Omitting students who were not employed, there were 1296 responses. Here are the data in summary form:
Construct a 90% confidence interval for the difference between the mean summer earnings of male and female students.
3) Teaching Reading An educator believes that new reading activities in the classroom will help elementary school
pupils improve their reading ability. She arranges for a third-grade class of 21 students to follow these activities for an 8
week period. A control classroom of 23 third graders follows the same curriculum without the activities. At the end of the
8 weeks, all students are given the Degree of Reading Power (DRP) test, which measures the aspects of reading ability
that the treatment is designed to improve. Here are the data.
Is there good evidence that the new activities improve the mean DRP score?
4) Nicotine and Guinea Pigs Many studies have shown that smoking during pregnancy affects the baby’s health.
Researcher’s investigated the behavior of guinea pig offspring whose mothers had been randomly assigned to receive
either a normal saline or nicotine saline injection throughout pregnancy. Each group consisted of 15 randomly chosen
male and female guinea pigs. At 85 days of age, 10 subjects from each group were randomly chosen to run a maze
and choose a black door rather than a white door at the end of the maze. The number of trials it took each guinea pig
to complete the task successfully with no more than one mistake in two consecutive days was recorded. Here are the
summary statistics on number of trials to successful completion for the nicotine group and the control group:
(Note: Both normal probability plots of the data were linear). what does that tell us!?
Is there a significant difference in the mean number of trials recorded between the treatment group and the control
group?
13.1 Comparing Two Means Practice
SHOW ALL STEPS FOR FULL CREDIT (#2-4 are “full processes”
1) Improving Sleep It is thought that listening to the sound of the surf will help people get to sleep faster. An
experiment was done using 10 people. Some nights the sound of surf was piped into their bedroom and other nights,
no sound was used. The experiment was run for a period of time and the average time it took for them to fall asleep
was monitored and averaged. Following is the data for the 10 people.
With Surf
Without Surf
10.2
11.4
12.5
13.4
13.2
13.2
23.1
22.3
7.5
8.8
8.9
12.3
12.2
14.2
13.4
13.3
18.4
19.9
17.5
19.7
a. Check the conditions for performing a two sample t-test for means.
b. What is the standard error of the differences of the two independent samples?
2) Paying for College College financial aid offices expect students to use summer earnings to help pay for college. But
how large are these earnings ? One college studied this question by asking a random sample of students how much they
earned. Omitting students who were not employed, there were 1296 responses. Here are the data in summary form:
Construct a 90% confidence interval for the difference between the mean summer earnings of male and female students.
3) Teaching Reading An educator believes that new reading activities in the classroom will help elementary school
pupils improve their reading ability. She arranges for a third-grade class of 21 students to follow these activities for an 8
week period. A control classroom of 23 third graders follows the same curriculum without the activities. At the end of the
8 weeks, all students are given the Degree of Reading Power (DRP) test, which measures the aspects of reading ability
that the treatment is designed to improve. Here are the data.
Is there good evidence that the new activities improve the mean DRP score?
4) Nicotine and Guinea Pigs Many studies have shown that smoking during pregnancy affects the baby’s health.
Researcher’s investigated the behavior of guinea pig offspring whose mothers had been randomly assigned to receive
either a normal saline or nicotine saline injection throughout pregnancy. Each group consisted of 15 randomly chosen
male and female guinea pigs. At 85 days of age, 10 subjects from each group were randomly chosen to run a maze
and choose a black door rather than a white door at the end of the maze. The number of trials it took each guinea pig
to complete the task successfully with no more than one mistake in two consecutive days was recorded. Here are the
summary statistics on number of trials to successful completion for the nicotine group and the control group:
(Note: Both normal probability plots of the data were linear). what does that tell us!?
Is there a significant difference in the mean number of trials recorded between the treatment group and the control
group?
```
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More information | 1,359 | 5,982 | {"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} | 3.953125 | 4 | CC-MAIN-2022-33 | latest | en | 0.960498 |
https://math.answers.com/Q/Denominator_and_numerator_of_18_fifths | 1,717,075,234,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00610.warc.gz | 330,778,085 | 48,924 | 0
# Denominator and numerator of 18 fifths?
Updated: 9/16/2023
Wiki User
14y ago
Denominator = 5, numerator = 18
Wiki User
14y ago
Earn +20 pts
Q: Denominator and numerator of 18 fifths?
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Related questions
It is 18.
### How do you turn 3 fifths into a decimal?
Divide the denominator into the numerator.
### What decimal is equal to 3 fifths?
three fifths=0.6 you just divide the numerator by the denominator.
### When you multiply fractions does the denominator change?
Yes, you multiply the numerator by the numerator, and the denominator by the denominator, and then reduce the fraction. For instance:One half of two fifths is1/2 x 2/5 = 2/10 which, when reduced, equals 1/5. Logically, if you have half of two fifths, you have one fifth. If you had half of four fifths, you would have two fifths, and so on.
### How do you simplify 5 and 2 ninths and 2 and 2 fifths?
5 and 2 ninths is 23 ninths. 2 and 2 fifths is 12 fifths. multiply the denominator and the numerator and then add the whole number over the denominator.
### What is one fraction equal to 2 FIfTHS?
2 fifths (not fiths!) = 2/5.Multiply the numerator (top) and the denominator (bottom) of the fraction by any non-zero integer. You will have an equivalent fraction.2 fifths (not fiths!) = 2/5.Multiply the numerator (top) and the denominator (bottom) of the fraction by any non-zero integer. You will have an equivalent fraction.2 fifths (not fiths!) = 2/5.Multiply the numerator (top) and the denominator (bottom) of the fraction by any non-zero integer. You will have an equivalent fraction.2 fifths (not fiths!) = 2/5.Multiply the numerator (top) and the denominator (bottom) of the fraction by any non-zero integer. You will have an equivalent fraction.
### How do you write an equivalent fraction for three fifths?
You can do this by multiplying both the numerator and denominator by a common number. For example, six tenths is an equivalent fraction because the numerator and denominator were both multiplied by 2.
### What number must be added to the denominator of 9 over 16 so that when the numerator is doubled the result is 1?
Doubling the numerator gives 2 × 9 = 18 To have a result of 1, the numerator and denominator must be the same. Thus the denominator needs to be 18. 18 - 16 = 2 → 2 must be added to the denominator.
### What is 54 over 90 simplified?
Expressed as a proper fraction in its simplest form, by dividing the numerator and denominator by 18, 54/90 is equal to 3/5 or three fifths.
### Can four fifths be simplified?
No, if the numerator (4) is 1 number away from the denominator (5) it means it is simplified !
### What is 72 over 90 in simplest form?
Expressed as a proper fraction in its simplest form, by dividing the numerator and denominator by 18, 72/90 is equal to 4/5 or four fifths.
### What is 1 fourth plus 2 fifths?
The key to answering this question is to find your "lowest common denominator." (The denominator is the number below the line of the fraction.) If you multiply the denominators of the two fractions (4 and 5) you will come up with a common denominator (20). That will usually be your lowest common denominator. Then multiply each numerator (the number above the line of the fraction) by the denominator of the other fraction. The numerator of one-fourth is 1; the denominator of two-fifths is 5. Therefore, multiply 1 x 5 and you get 5. The numerator of two-fifths is 2; the denominator of one-fourth is 4. Therefore, multiply 2 x 4 and you get 8. Finally, add the results of your calculations: 5 + 8 = 13 Your new numerator is 13 and your new denominator is 20. | 949 | 3,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.90464 |
https://www.jiskha.com/display.cgi?id=1229465641 | 1,503,394,938,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110573.77/warc/CC-MAIN-20170822085147-20170822105147-00191.warc.gz | 891,504,511 | 3,966 | # Math
posted by .
Can 606/1000 be reduced further than 303/500?
• Math -
if you have a button labeled a b/c
enter 606
hit that button
enter 1000
press =
Voila!!
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https://techcommunity.microsoft.com/t5/excel/how-to-combine-3-if-formulas-with-3-different-value-if-false/td-p/3619938 | 1,708,729,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474470.37/warc/CC-MAIN-20240223221041-20240224011041-00598.warc.gz | 587,678,326 | 58,687 | Copilot for Microsoft 365 Tech Accelerator
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SOLVED
How to combine 3 IF formulas with 3 different value_if_false results into 1 formula
Copper Contributor
How to combine 3 IF formulas with 3 different value_if_false results into 1 formula
I have a matrix, 15 rows (A), 10 columns (B,C,D..). In a result should get a table returning 6 different results.
Need ONLY 1 formula to fit each cell.
Managed to create 3, which now need to be combined - not succeeding on my own.
=IF(AND(\$A2<=5;OR(B\$1<=3;B\$1>=8));13;14)
=IF(AND(\$A2>=6;\$A2<=10;OR(B\$1<=3;B\$1>=8));11;12)
=IF(AND(\$A2>=11;OR(B\$1<=3;B\$1>=8));9;10)
2 Replies
best response confirmed by Ppetu999 (Copper Contributor)
Solution
Re: How to combine 3 IF formulas with 3 different value_if_false results into 1 formula
Hi, this should work like this:
=IF(\$A2<=5,IF(OR(B\$1<=3,B\$1>=8),13,14),IF(\$A2<=10,IF(OR(B\$1<=3,B\$1>=8),11,12),IF(\$A2>=11,IF(OR(B\$1<=3,B\$1>=8),9,10))))
or with ";":
=IF(\$A2<=5;IF(OR(B\$1<=3;B\$1>=8);13;14);IF(\$A2<=10;IF(OR(B\$1<=3;B\$1>=8);11;12);IF(\$A2>=11;IF(OR(B\$1<=3;B\$1>=8);9;10))))
Alternatively, a bit shorter you could do a combination with LOOKUP().
=XLOOKUP(A2,{5;10;11},{13;11;9},9,1)+IF(OR(B\$1<=3,B\$1>=8),0,1)
or with ";":
=XLOOKUP(A2;{5.10.11};{13.11.9};9;1)+IF(OR(B\$1<=3;B\$1>=8);0;1)
Re: How to combine 3 IF formulas with 3 different value_if_false results into 1 formula
Great, the 1st version with ";" works for me well! Thanks a lot!
1 best response
Accepted Solutions
best response confirmed by Ppetu999 (Copper Contributor)
Solution
Re: How to combine 3 IF formulas with 3 different value_if_false results into 1 formula
Hi, this should work like this:
=IF(\$A2<=5,IF(OR(B\$1<=3,B\$1>=8),13,14),IF(\$A2<=10,IF(OR(B\$1<=3,B\$1>=8),11,12),IF(\$A2>=11,IF(OR(B\$1<=3,B\$1>=8),9,10))))
or with ";":
=IF(\$A2<=5;IF(OR(B\$1<=3;B\$1>=8);13;14);IF(\$A2<=10;IF(OR(B\$1<=3;B\$1>=8);11;12);IF(\$A2>=11;IF(OR(B\$1<=3;B\$1>=8);9;10))))
Alternatively, a bit shorter you could do a combination with LOOKUP().
=XLOOKUP(A2,{5;10;11},{13;11;9},9,1)+IF(OR(B\$1<=3,B\$1>=8),0,1)
or with ";":
=XLOOKUP(A2;{5.10.11};{13.11.9};9;1)+IF(OR(B\$1<=3;B\$1>=8);0;1) | 901 | 2,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | longest | en | 0.652642 |
https://bugs.python.org/file46476/siphash13.patch | 1,659,934,164,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-33/segments/1659882570765.6/warc/CC-MAIN-20220808031623-20220808061623-00012.warc.gz | 163,649,831 | 1,377 | diff -r a507882736b2 Python/pyhash.c --- a/Python/pyhash.c Wed Jan 25 22:33:43 2017 +0900 +++ b/Python/pyhash.c Wed Feb 01 18:20:50 2017 +0900 @@ -351,21 +351,19 @@ static PyHash_FuncDef PyHash_Func = {fnv # define ROTATE(x, b) (uint64_t)( ((x) << (b)) | ( (x) >> (64 - (b))) ) #endif -#define HALF_ROUND(a,b,c,d,s,t) \ - a += b; c += d; \ +#define HALF_ROUND(a,b,c,d,s,t) \ + a += b; c += d; \ b = ROTATE(b, s) ^ a; \ d = ROTATE(d, t) ^ c; \ a = ROTATE(a, 32); -#define DOUBLE_ROUND(v0,v1,v2,v3) \ - HALF_ROUND(v0,v1,v2,v3,13,16); \ - HALF_ROUND(v2,v1,v0,v3,17,21); \ - HALF_ROUND(v0,v1,v2,v3,13,16); \ +#define ROUND(v0,v1,v2,v3) \ + HALF_ROUND(v0,v1,v2,v3,13,16); \ HALF_ROUND(v2,v1,v0,v3,17,21); static Py_hash_t -siphash24(const void *src, Py_ssize_t src_sz) { +siphash13(const void *src, Py_ssize_t src_sz) { uint64_t k0 = _le64toh(_Py_HashSecret.siphash.k0); uint64_t k1 = _le64toh(_Py_HashSecret.siphash.k1); uint64_t b = (uint64_t)src_sz << 56; @@ -385,7 +383,7 @@ siphash24(const void *src, Py_ssize_t sr in += 1; src_sz -= 8; v3 ^= mi; - DOUBLE_ROUND(v0,v1,v2,v3); + ROUND(v0,v1,v2,v3); v0 ^= mi; } @@ -404,18 +402,19 @@ siphash24(const void *src, Py_ssize_t sr b |= _le64toh(t); v3 ^= b; - DOUBLE_ROUND(v0,v1,v2,v3); + ROUND(v0,v1,v2,v3); v0 ^= b; v2 ^= 0xff; - DOUBLE_ROUND(v0,v1,v2,v3); - DOUBLE_ROUND(v0,v1,v2,v3); + ROUND(v0,v1,v2,v3); + ROUND(v0,v1,v2,v3); + ROUND(v0,v1,v2,v3); /* modified */ t = (v0 ^ v1) ^ (v2 ^ v3); return (Py_hash_t)t; } -static PyHash_FuncDef PyHash_Func = {siphash24, "siphash24", 64, 128}; +static PyHash_FuncDef PyHash_Func = {siphash13, "siphash13", 64, 128}; #endif /* Py_HASH_ALGORITHM == Py_HASH_SIPHASH24 */ | 702 | 1,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.176635 |
http://www.helpteaching.com/tests/465388/chinese-new-year-probability | 1,513,333,574,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948568283.66/warc/CC-MAIN-20171215095015-20171215115015-00525.warc.gz | 363,994,533 | 7,069 | ##### Print Instructions
NOTE: Only your test content will print.
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# Chinese New Year Probability (Grade 3)
Print Test (Only the test content will print)
## Chinese New Year Probability
1.
You pick one of the items below without looking. How likely is it that you will pick a dragon?
1. certain
2. likely
3. unlikely
4. impossible
2.
You pick one of the items below without looking. How likely is it that you will pick a lantern?
1. certain
2. likely
3. unlikely
4. impossible
3.
You pick one of the items below without looking. How likely is it that you will pick a gold coin?
1. certain
2. likely
3. unlikely
4. impossible
4.
You pick one of the items below without looking. How likely is it that you will pick a dragon?
1. certain
2. likely
3. unlikely
4. impossible
5.
Use the pictures below to answer the questions.
Which item is most likely to be picked at random?
Which item is least likely to be picked at random?
What is the probability of randomly picking coins?
How many more dragons would need to be added to make it equally likely that a dragon or lantern would be picked?
You need to be a HelpTeaching.com member to access free printables. | 312 | 1,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.911261 |
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<sub id="d9npn"><progress id="d9npn"></progress></sub>
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### Linear combination of constrained random variables, and its convergence
I have positive random variables X1, X2, X3, ..., Xn such that their sum=1 (so they are random, subject to constraints that each Xi is positive their sum has to be 1.. so all are fractions). Now, I ...
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# Test: Circle- 2 - JEE MCQ
Test Description
## 30 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Circle- 2
Test: Circle- 2 for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Circle- 2 questions and answers have been prepared according to the JEE exam syllabus.The Test: Circle- 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Circle- 2 below.
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Test: Circle- 2 - Question 1
### The lines 2x – 3y = 5 and 3x – 4y = 7 are diameters of a circle of area 154 sq. units. The equation of the circle is
Detailed Solution for Test: Circle- 2 - Question 1
Test: Circle- 2 - Question 2
### If a be the radius of a circle which touches x-axis at the origin, then its equation is
Detailed Solution for Test: Circle- 2 - Question 2
The equation of the circle with centre at (h,k) and radius equal to a is (x−h)+(y−k)= a2
When the circle passes through the origin and centre lies on x− axis
⇒h = a and k = 0
Then the equation (x−h)2+(y−k)2=abecomes (x−a)2+y2=a2
If a circle passes through the origin and centre lies on x−axis then the abscissa will be equal to the radius of the circle and the y−co-ordinate of the centre will be zero Hence, the equation of the circle will be of the form
(x±a)2+y2=a2⇒x2+a±2ax+y2=a2
=x2 +y±2ax=0 is the required equation of the circle.
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Test: Circle- 2 - Question 3
### The equation of the circle passing through (3, 6) and whose centre is (2, –1) is
Detailed Solution for Test: Circle- 2 - Question 3
(x−2)2+(y+1)2 = r2
(3,6). lies on it
⇒ 1+49=r2
⇒ r2=50
⇒ x2+4−4x+y2+1+2y=50.
⇒ x2+y2−4x+2y−45=0
Test: Circle- 2 - Question 4
The equation of a circle which passes through the three points (3, 0) (1, –6), (4, –1) is
Detailed Solution for Test: Circle- 2 - Question 4
Gen equation of a circle is (x−h)2+(y−k)2=r2 (3,0),(1,−6),(4,−1) circle passes through these points
∴ These three points should satisfy the equation
of circle.
(3−h)2+k2=r2 ___ (I)
(1−h)2 +(−6−k)2 = r2(1−h)2+(6+k)2=r2 ___ (II)
From (I) & (II)
(3−h)2+k2 =(1−h)2+(6+k)2
9+h2−6h+k2 =1+h2 2h+36+k2+12k
9−6h=1−2h+36+12k
9−37=4h+12k
−28=4h+12k
h+3k=−7
h=−7−3k ___ (III)
(4−h)2+(−1−k)2=r2 ___ (IV)
From (I) & (IV)
(3−h)2+k2 = (4−h)2 +(1+k)2
9+h2 −6h+k2
=16+h2 −8h+1+k2+2k
9−6h=17−8h+2k
9−17=−2h+2k
−8=−2h+2k
−h+k=−4
k=−4+h ___ (V)
Put (V) in (III)
h=−7−3(−4+h)
h=−7+12−3h
h=5/4,k=−11/4,r=170/16
Eq of circle: (x−5/4)2 +(y+11/4)2 =170/16
x2+25/16−10x/4+y2 +121/16+22y/4=170/16
Simplifying, we get 2x2+xy2−5x+11y−3=0
Test: Circle- 2 - Question 5
y = √3 + c1 & y = √3 + c2 are two parallel tangents of a circle of radius 2 units, then |c1 – c2| is equal to
Detailed Solution for Test: Circle- 2 - Question 5
For both lines to be parallel tangent the distance between both lines
should be equal to the diameter of the circle
⇒ 4 = |c1−c2|/(1+3)1/2
⇒∣c1−c2∣ = 8
Test: Circle- 2 - Question 6
B and C are fixed point having co-ordinates (3, 0) and (–3, 0) respectively. If the vertical angle BAC is 90º, then the locus of the centroid of the DABC has the equation
Detailed Solution for Test: Circle- 2 - Question 6
Let A (a, b) and G (x. y) Now A, G, O are collinear
⇒ x = (2*0 + a)/3
⇒ a = 3x
and similarly b = 3y
Now (a,b) lies on the circle = 9x2 + 9y2 = 9
=> x2 + y2 = 1
Test: Circle- 2 - Question 7
The area of an equilateral triangle inscribed in the circle x2 + y2 – 2x = 0 is
Detailed Solution for Test: Circle- 2 - Question 7
Test: Circle- 2 - Question 8
The length of intercept on y-axis, by a circle whose diameter is the line joining the points (–4,3) and (12,–1) is
Detailed Solution for Test: Circle- 2 - Question 8
Equation of circle is given by,
Test: Circle- 2 - Question 9
The gradient of the tangent line at the point (a cos a, a sin a) to the circle x2 + y2 = a2, is
Detailed Solution for Test: Circle- 2 - Question 9
Test: Circle- 2 - Question 10
lx + my + n = 0 is a tangent line to the circle x2 + y2 = r2, if
Detailed Solution for Test: Circle- 2 - Question 10
Test: Circle- 2 - Question 11
If y = c is a tangent to the circle x2+y2–2x+2y–2 = 0 at (1, 1), then the value of c is
Detailed Solution for Test: Circle- 2 - Question 11
For line y=c to be tangent to the given circle at point (1,1)
It has to pass through (1,1)
⇒ c = 1
Test: Circle- 2 - Question 12
Line 3x + 4y = 25 touches the circle x2 + y2 = 25 at the point
Detailed Solution for Test: Circle- 2 - Question 12
Equation of tangent at (x1 , y1) to the circle x2 + y2 =25 is given by,
xx1 + yy1 = 25. Now comparing it with 3x+4y = 25
We get required point of contact (3,4)
Test: Circle- 2 - Question 13
The equations of the tangents drawn from the point (0, 1) to the circle x2 + y2 - 2x + 4y = 0 are
Detailed Solution for Test: Circle- 2 - Question 13
Let equation of tangent with slope =m and point (0,1)
(y−1)=m(x−0)⇒y=mx+1
Intersection point
x2+(mx+1)2−2x+4(mx+1)=0
(1+m2)x2+(−2+6m)x+5=0
For y=mx+1 to be tangent, discriminant =0
(6m−2)2−4×5(1+m2)=0
36m2+4−24m−20m2+20=0
16m2−20m+24=0
⇒ 2m2−3m−2=0
(2m+1)(m−2)=0
Test: Circle- 2 - Question 14
The greatest distance of the point P(10, 7) from the circle x2 + y2 – 4x – 2y – 20 = 0 is
Detailed Solution for Test: Circle- 2 - Question 14
Centre and radius of the given circle are C(2,1) and √4+1+20 = 5 respectively.
Now CP=√82+62=10. Hence greatest distance of point P from the given circle is =10+r=15
Test: Circle- 2 - Question 15
The equation of the normal to the circle x2+y2 = 9 at the point is
Detailed Solution for Test: Circle- 2 - Question 15
(x-x1)/(x1+g) = (y-y1)/(y1+g)
[x-1/(2)½]/(1/(2)½) = [y-1/(2)½]/(1/(2)½)
[(2x)½ -1 *(2)½]/(2)½ = [(2x)½ -1 *(2)½]/(2)½
(2x)½ -1 = (2y)½ -1
= x-y = 0
Test: Circle- 2 - Question 16
The parametric coordinates of any point on the circle x2 + y2 – 4x – 4y = 0 are
Detailed Solution for Test: Circle- 2 - Question 16
Test: Circle- 2 - Question 17
The length of the tangent drawn from the point (2, 3) to the circles 2(x2 + y2) – 7x + 9y – 11 = 0.
Detailed Solution for Test: Circle- 2 - Question 17
Test: Circle- 2 - Question 18
Tangents are drawn from (4, 4) to the circle x2 + y2 – 2x – 2y – 7 = 0 to meet the circle at A and B. The length of the chord AB is
Detailed Solution for Test: Circle- 2 - Question 18
Test: Circle- 2 - Question 19
The angle between the two tangents from the origin to the circle (x – 7)2 + (y + 1)2 = 25 equals
Detailed Solution for Test: Circle- 2 - Question 19
Let tangent from origin be y = mx
Using the condition of tangency, we get
The angle between tangents = π/2
Test: Circle- 2 - Question 20
Pair of tangents are drawn from every point on the line 3x + 4y = 12 on the circle x2+ y2 = 4. Their variable chord of contact always passes through a fixed point whose co-ordinates are
Detailed Solution for Test: Circle- 2 - Question 20
Let P(x1,y1) be a point on the line 3x + 4y = 12
Equation of variable chord of contact of P(x1,y1) w.r.t circle x2 + y2 = 4
xx1 + yy1 − 4 = 0 ...(1)
Also 3x1 + 4y1 − 12 = 0
⇒ x1 + 4/3y1 − 4 = 0 ...(2)
Comparing (1) and (2), we get
x = 1; y = 4/3
∴ Variable chord of contact always passes through (1, 4/3)
Test: Circle- 2 - Question 21
The locus of the mid-points of the chords of the circle x2 + y2 – 2x – 4y – 11 = 0 which subtend 60º at the centre is
Detailed Solution for Test: Circle- 2 - Question 21
Let AB be the chord of the circle and P be the midpoint of AB.
It is known that perpendicular from the center bisects a chord.
Thus △ACP is a right-angled triangle.
The equation of the give circle can be written as
(x−1)2+(y−2)2=16
Hence, centre C=(1,2) and radius =r=4 units.
PC=ACsin60degree
= rsin60degree
= 4([2(3)½]/2
= 2(3)1/2 units
Therefore, PC=2(3)1/2
⇒ PC2=12
⇒ (x−1)2+(y−2)2=12
⇒ x2+y2−2x−4y+5=12
⇒ x2+y2−2x−4y−7=0
Test: Circle- 2 - Question 22
The locus of the centres of the circles such that the point (2, 3) is the mid point of the chord 5x + 2y = 16 is
Detailed Solution for Test: Circle- 2 - Question 22
Slope of the given chord = −5/2
Slope of the line joining the midpoint on the chord and the centre of the circle = (3+f)/(2+g)
(5/2)[(3+f)/(2+g)] = −1
⇒ 15 + 5f = 4 + 2g
⇒ Locus is 2x − 5y + 11 = 0
Test: Circle- 2 - Question 23
The equation of the circle having the lines y2 – 2y + 4x – 2xy = 0 as its normals & passing through the point (2, 1) is
Detailed Solution for Test: Circle- 2 - Question 23
The normal line to circle is →y² - 2 y + 4 x -2 xy=0
→ y(y-2) - 2x(y-2)=0
→ (y-2)(y-2x)=0
the two lines are , y=2 and 2 x -y =0
The point of intersection of normals are centre of circle.
→ Put , y=2 in 2 x -y=0, we get
→2 x -2=0
→2 x=2
→ x=1
So, the point of intersection of normals is (1,2) which is the center of circle.
Also, the circle passes through (2,1).
Radius of circle is given by distance formula = [(1-2)² + (2-1)²]½
=(1+1)½ =(2)½
The equation of circle having center (1,2) and radius √2 is
= (x-1)²+(y-2)²=[√2]²
→ (x-1)²+(y-2)²= 2
x²+y² -2x-4y+3 = 0
Test: Circle- 2 - Question 24
A circle is drawn touching the x-axis and centre at the point which is the reflection of (a, b) in the line y – x = 0. The equation of the circle is
Detailed Solution for Test: Circle- 2 - Question 24
Radius of the circle = a
and centre ≡ (b,a)
And circle is touching the x-axis.
The equation of the circle :
x2+y2−2bx−2ay+b= 0
Test: Circle- 2 - Question 25
The number of common tangents of the circles x2 + y2 – 2x – 1 = 0 and x2 + y2 – 2y – 7 = 0
Detailed Solution for Test: Circle- 2 - Question 25
Internally touch ∴ common tangent is one.
Test: Circle- 2 - Question 26
The point from which the tangents to the circles x2 + y2 – 8x + 40 = 0, 5x2 + 5y2 – 25 x + 80 = 0, x2 + y2 – 8x + 16y + 160 = 0 are equal in length is
Detailed Solution for Test: Circle- 2 - Question 26
The Required point is the radical centre of the three given circles. The radical axes of the three circles taken in pairs are 3x - 24 = 0,16y + 120 = 0 and - 3x + 16y + 80 = 0. On solving, the required point is (8, -15/2).
Test: Circle- 2 - Question 27
If the circle x2 + y2 = 9 touches the circle x2 + y2 + 6y + c = 0, then c is equal to
Detailed Solution for Test: Circle- 2 - Question 27
touches the another circle
Now, Central first circle will be
And its radius will be 3 units.
Also,centre of second circle
As both touches each other
So,
Test: Circle- 2 - Question 28
The tangent from the point of intersection of the lines 2x – 3y + 1 = 0 and 3x – 2y –1 = 0 to the circle x2 + y2 + 2x – 4y = 0 is
Detailed Solution for Test: Circle- 2 - Question 28
Test: Circle- 2 - Question 29
The length of the common chord of circles x2 + y2 – 6x – 16 = 0 and x2 + y2 – 8y – 9 = 0 is
Detailed Solution for Test: Circle- 2 - Question 29
Test: Circle- 2 - Question 30
The distance between the chords of contact of tangents to the circle x2 + y2 + 2gx + 2fy + c = 0 from the origin and from the point (g, f) is
Detailed Solution for Test: Circle- 2 - Question 30
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## Mathematics (Maths) for JEE Main & Advanced
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# (Solved): \sum_(m=1)^(\infty ) (a_(m)), a_(1)=1:,a_(m)=(m^(2)-4)a_(m-1) Does series converge ? ...
\sum_(m=1)^(\infty ) (a_(m)), a_(1)=1:,a_(m)=(m^(2)-4)a_(m-1)
Does series converge ?
We have an Answer from Expert | 137 | 321 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-22 | latest | en | 0.609668 |
https://www.askiitians.com/forums/Electrostatics/a-metal-surface-has-a-positive-charge-of-10-9-c-ho_97885.htm | 1,723,146,974,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640740684.46/warc/CC-MAIN-20240808190926-20240808220926-00896.warc.gz | 526,625,890 | 43,419 | # A METAL SURFACE HAS A POSITIVE CHARGE OF 10-9 C HOW MANY ELECTRONS WOULD HAVE BEEN REMOVED FROM THE METAL SURFACE?
Roshan
10 Points
10 years ago
lol. i am also searchng for the answer Lekshmi :P
71 Points
10 years ago
charge=q=10^-9 C , according to quantisation of charge we have, q = ne , n=no. of electrons, e=charge on 1 electron i.e. 1.6 * 10^-19 C, then n=q/e= 6.25 * 10^-29 here * stands for multiply. that`s all. | 145 | 423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-33 | latest | en | 0.706873 |
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Είσοδος
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## ELEMENTARY ARITHMETIC
G. P. QUACKENBOS, A.M. - 1867
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## Arithmetical Examples; Or, Test Exercises for the Use of Advanced Classes
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## Intermediate Arithmetic: For Use in the Common Schools and Academies of the ...
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## The Elements of Written Arithmetic: Combining Analysis and Synthesis ...
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...117. DRY MEASURE is used in measuring grain, fruit, potatoes, salt, charcoal, etc. TABLE. 2 pts. (pt.) make 1 Quart,. qt. 8 Quarts " 1 Peck, pk. 4 Pecks " 1 Bushel, bush. qt pt. pk. 1 = 2 bush. . 1 = 8 = 16 1 = 4 = 32 = 64 SCALE. Descending, 4, 8, 2 ; Ascending, 2,...
## A Practical Arithmetic
George Payn Quackenbos - 1869 - 336 σελίδες
...grain, seeds, vegetables, roots, fruit, salt, coal, and other articles not liquid. TARLE. 2 'pints (pt.) make 1 quart, qt. 8 quarts, 1 peck, pk. 4 pecks, 1 bushel, .... bu. 36 bushels, 1 chaldron, .... chal. IIow many cnbic inches does the beer gallon contain? How many -wine...
## Bryant and Stratton's Commercial Arithmetic, in Two Parts: Designed for the ...
...DRY MEASURE. Used for measuring articles not liquid; as grain, fruit, salt, etc, TABLE. 2 pints (ft.) make 1 quart qt 8 quarts " 1 peck pk. 4 pecks " 1 bushel bu. 36 bushels " 1 chaldron... ch. EQUIVALENTS. ch. to. pk. qt. pt. 1 = 36 = 144 = 1152 = 2304 1 = 4 =...
## The Young Pupils' Arithmetic Objectively Or Synthetically Arranged
David Henry Cruttenden - 1869 - 96 σελίδες
...names are bu-shels, peeks, quaa-t.4, pints, and yilht. 4 gills are equivalent to 1 pint pt. 2 pints " " 1 quart . . . . qt. 8 quarts " " 1 peck .... pk. 4 pecks " " 1 bushel - - - - bu. 5. Bead these ; 931000bu. ; 3pk ; 7ft ; 9pt Igi. в. Eead these; 205102sq.ft; \$95067; £305000 ; 97ct... | 1,482 | 4,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-40 | latest | en | 0.55201 |
http://www.gurufocus.com/term/deb2equity/QSII/Debt%2Bto%2BEquity%2B%2528%2525%2529/Quality%2BSystems%2BInc | 1,490,921,092,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218205046.28/warc/CC-MAIN-20170322213005-00506-ip-10-233-31-227.ec2.internal.warc.gz | 553,188,289 | 27,713 | Switch to:
GuruFocus has detected 7 Warning Signs with Quality Systems Inc \$QSII.
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Quality Systems Inc (NAS:QSII)
Debt-to-Equity
0.08 (As of Dec. 2016)
Quality Systems Inc's current portion of long-term debt for the quarter that ended in Dec. 2016 was \$0.0 Mil. Quality Systems Inc's long-term debt for the quarter that ended in Dec. 2016 was \$25.0 Mil. Quality Systems Inc's total equity for the quarter that ended in Dec. 2016 was \$297.8 Mil. Quality Systems Inc's debt to equity for the quarter that ended in Dec. 2016 was 0.08.
A high debt to equity ratio generally means that a company has been aggressive in financing its growth with debt. This can result in volatile earnings as a result of the additional interest expense.
Definition
Debt to Equity measures the financial leverage a company has.
Quality Systems Inc's Debt to Equity Ratio for the fiscal year that ended in Mar. 2016 is calculated as
Debt to Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt) / Total Equity = (0 + 105) / 269.377 = 0.39
Quality Systems Inc's Debt to Equity Ratio for the quarter that ended in Dec. 2016 is calculated as
Debt to Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt) / Total Equity = (0 + 25) / 297.75 = 0.08
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
In the calculation of Debt to Equity, we use the total of Current Portion of Long-Term Debt and Long-Term Debt divided by Total Equity. In some calculations, Total Liabilities is used to for calculation.
Be Aware
Because a company can increase its Return on Equity by having more financial leverage, it is important to watch the leverage ratio when investing in high Return on Equity companies.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Quality Systems Inc Annual Data
Mar07 Mar08 Mar09 Mar10 Mar11 Mar12 Mar13 Mar14 Mar15 Mar16 deb2equity 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.39
Quality Systems Inc Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 deb2equity 0.00 0.00 0.00 0.00 0.00 0.00 0.39 0.33 0.17 0.08
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 682 | 2,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-13 | latest | en | 0.953469 |
https://electronics.stackexchange.com/questions/254068/reverse-polarity/254070 | 1,719,169,343,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00120.warc.gz | 203,425,052 | 42,728 | # Reverse Polarity
I am working on a project and I want to use a $12\textrm{ V}$ linear actuator which travels in and out by reversing polarity on the 2 lead conductors. I also have a double pole double throw switch to wire in to accomplish this. I would like to have it travel one direction slow and one at full speed.
I believe there should be a way to accomplish this with the use of resistors and possibly another 3 way switch to activate them but can't wrap my head around how the schematic would look. The place that sold this to me suggested a potentiometer and motor controller but I am hopeful there is an easier way. Thank you so much!
• It also depends if your actuator draws a lot of current, more than your resistors are rated. You should be connecting the actuators to the commons of the dpdt switch. Commented Aug 25, 2016 at 3:45
• You can do it crudely with a resistor and a (high current) diode in parallel. One way round, the diode allows high current to flow and drops only about 1V. Test with a resistor first : if the motor won't go slow enough without stalling, the crude approach won't work.
– user16324
Commented Aug 25, 2016 at 9:19
Unless this is a tiny actuator the resistor you would need would be very large. For example let's say your actuator is something like this
The stall current is listed at 3.4A. Say you want to run this at half power so instead of 12V you want 6V to the actuator, using ohms law we can determine the resistor value = Voltage Drop / Current = 6/3.4A = 1.76 Ohms. Ignoring the fact that you can't find a 1.76 Ohm resistor and you would need to settle for something closer the power rating for this would need to be P=I^2*R = 3.4^2*1.76 = 20.34W. (There is an easier way to get to this number but I wanted to show the process) A 20W resistor is big, expensive and requires a heat sink and probably a fan to keep it cool. Also note that you are dumping a lot of power into this resistor to generate heat so from an efficiency standpoint this is very wasteful.
What you want is a DC motor controller which can control the direction of the motor and use an algorithm called Pulse Width Modulation (PWM) to modulate the 12VDC on the output so that the motor can be controlled at nearly any speed with much greater efficiency than a resistor. There are many motor controllers for sale online which you can find at an electronic hobby retailer such as sparkfun or pololu.
• I agree PWM is a good way to go, but there are plenty of 20W resistors that don't need external heatsinks (aside from their cases) or forced air cooling, and there's something to be said for simplicity if you don't mind wasting a little power. Commented Aug 25, 2016 at 5:34
Here's the wiring for the DPDT reversing switch:-
simulate this circuit – Schematic created using CircuitLab
As you can see, a separate wire supplies +12V for forward and reverse. So to make the the actuator go slower in one direction you just have insert something in one of these wires to drop the voltage.
The actuator probably has a DC motor inside. You could use a resistor to slow it down, or several different resistors chosen with a multiway switch. However the motor will draw more current as loading increases, causing the resistor to drop more voltage and slow it down even more. If the load is constant then you can just try different resistor values until you get the speed you want, but if it is variable then actuator speed will vary greatly, possibly even stalling if the voltage drops too much. If the actuator draws a lot of current then the resistor will get hot and needs to be a high power type.
A PWM motor controller solves these problems because it doesn't use resistance to drop the voltage, but instead rapidly switches the motor fully on and off with a variable ratio. The motor sees this as an average voltage that doesn't change with loading, and the controller stays cool because it is either fully on or off and so doesn't waste power.
The power supply can be wired directly to the controller, but its output may not be referenced to either power rail so to get full speed in one direction and controlled speed in the other you should switch both motor wires. The circuit would look like this:-
simulate this circuit
If using a resistor is a tolerable method then you can get different rates by placing a diode across the resistor so that it is bypassed in one direction.
A silicon diode will dissipate about 1 Watt per amp.
The diode will reduce speed somewhat in the fast mode due to the about 1 volt drop across it.
A Schotty diode will drop about 0.5 volt.
Resistors are not great because they reduce the available torque of a DC motor without necessarily reducing the RPM much.
You might be able to use a simple and inexpensive adjustable buck regulator module to reduce the 12V to a suitable voltage for the 'slow' direction, however there is a risk if you switch from fast forward to slow reverse ('plug' the motor) that the motor and gear train inertia would generate enough reverse current to damage the module. A largish Schottky diode on the output of the module would prevent that (parallel reverse-biased), but it still could damage the motor or gear train from the rapid braking.
You can control it all with a center-off DPDT switch. Below is an example. The regulator could be something like an LM2596 if the current rating is suitable. The four optional parts could be added to shut down the regulator when the switch is in the forward or off states- assuming the regulator has an active low enable similar to an LM2596. An LM2596 would be suitable for an actuator with a stall current at 12V of up to several amperes (since you wouldn't be running at the full 12V for the slow direction).
simulate this circuit – Schematic created using CircuitLab
Internal circuit of a typical module (some even have built-in voltmeters) | 1,341 | 5,888 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-26 | latest | en | 0.951328 |
https://www.solutioninn.com/consider-the-system-y1-9451-946x-9491-y2-9452-9492 | 1,702,099,612,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00473.warc.gz | 1,078,765,018 | 25,315 | # Consider the system y1 = 1 + x + 1, y2 = 2 + 2. The disturbances are freely correlated.
## Question:
Consider the system y1 = α1 + βx + ε1, y2 = α2 + ε2. The disturbances are freely correlated. Prove that GLS applied to the system leads to the OLS estimates of α1 and α2 but to a mixture of the least squares slopes in the regressions of y1 and y2 on x as the estimator of β. What is the mixture? To simplify the algebra, assume (with no loss of generality) that x = 0. | 137 | 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 | 3 | CC-MAIN-2023-50 | latest | en | 0.935934 |
https://www.physicsforums.com/threads/dual-representation-and-anti-particles.89846/ | 1,675,677,406,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500334.35/warc/CC-MAIN-20230206082428-20230206112428-00781.warc.gz | 955,653,971 | 17,323 | # Dual Representation and anti-particles
Norman
I am reading ahead for my Group Theory introduction to QFT and I have a question about the dual representation. If the dual representation is the same as the ordinary representation, that is to say the ordinary representation is "real", how do we represent anti-particles in this case? This seems to be the case for SU(2), which is the gauge group for weak interactions.
Thanks,
Ryan
## Answers and Replies
Norman
Anyone have any ideas? Maybe I posted this in the wrong forum... Or is my question completely non-sensical?
Cheers,
Ryan
Norman
Never mind. This is actually a pretty non-sensical question I think. (It is fun having an online conversation with myself) but in case someone else has this crazy question in the future, I will answer with what I think is the correct statement. SU(2) only talks about, say, weak isospin. But to take into account the charge of a particle we would need the group SU(2) X U(1). This group would account for electroweak theory and would allow the leptonic charge to change.
Cheers,
Ryan
Science Advisor
Homework Helper
Norman said:
I am reading ahead for my Group Theory introduction to QFT and I have a question about the dual representation. If the dual representation is the same as the ordinary representation, that is to say the ordinary representation is "real", how do we represent anti-particles in this case? This seems to be the case for SU(2), which is the gauge group for weak interactions.
Thanks,
Ryan
I'm not sure what you mean by requiring the ordinary representation to be "real". With that caveat, here's a stab:
If you take a look at $$\mathcal{SU}(3)$$ and its triplet and dual triplet irreps, you will find that the quantum numbers of the dual are complementary to the quantum numbers of the regular representation. So I'm not really sure what you mean when you say that they are the same. Sure you can rotate one to the other, but the antiparticles carry negated quantum numbers.
Carl
Norman
CarlB said:
I'm not sure what you mean by requiring the ordinary representation to be "real". With that caveat, here's a stab:
If you take a look at $$\mathcal{SU}(3)$$ and its triplet and dual triplet irreps, you will find that the quantum numbers of the dual are complementary to the quantum numbers of the regular representation. So I'm not really sure what you mean when you say that they are the same. Sure you can rotate one to the other, but the antiparticles carry negated quantum numbers.
Carl
SU(3) has a dual, but the dual to SU(2) is equal to the ordinary rep.
Science Advisor
Homework Helper
Norman said:
SU(3) has a dual, but the dual to SU(2) is equal to the ordinary rep.
I see your question now. So your answer to your own question is that if one wants an antiparticle associated with a particle collection that follows an SU(2) symmetry, then one must use a different symmetry to generate the relationship between the particles and antiparticles?
Carl
Science Advisor
Homework Helper
By "dual" u mean "contragradient", i see, now. The fundamental and the contragradient representations of $\mbox{SU(2)}$ are not the same, but equivalent. This means there is a similarity transformation connecting the generators of the two representations. But you can't say "they're the same". Think of the Dirac algebra
$$\left[\gamma_{\mu},\gamma_{\nu}]_{+} =2g_{\mu\nu}\hat{1}_{V}$$
U can't say the Dirac representation is "the same" with the Weyl and the Majorana one.
Daniel.
P.S.Terminology is important.
Norman
CarlB said:
I see your question now. So your answer to your own question is that if one wants an antiparticle associated with a particle collection that follows an SU(2) symmetry, then one must use a different symmetry to generate the relationship between the particles and antiparticles?
Carl
Yes. I believe that is the answer to the question. Someone can correct me if I am wrong.
Norman
dextercioby said:
By "dual" u mean "contragradient", i see, now. The fundamental and the contragradient representations of $\mbox{SU(2)}$ are not the same, but equivalent. This means there is a similarity transformation connecting the generators of the two representations. But you can't say "they're the same". Think of the Dirac algebra
$$\left[\gamma_{\mu},\gamma_{\nu}]_{+} =2g_{\mu\nu}\hat{1}_{V}$$
U can't say the Dirac representation is "the same" with the Weyl and the Majorana one.
Daniel.
P.S.Terminology is important.
Of course. Terminology is always very important. But the lecture notes we were given stated things using the terms I used in my original post. Unfortunately I have to go with what is in there mainly. Oh well, a process is learning.
Cheers,
Ryan
Staff Emeritus
Science Advisor
Gold Member
Even though the representations are equivalent, the weights, i.e., quantum numbers, for the representations have different signs, and hence the connection between particles and antiparticles.
Regards,
George
Norman
George Jones said:
Even though the representations are equivalent, the weights, i.e., quantum numbers, for the representations have different signs, and hence the connection between particles and antiparticles.
Regards,
George
Yes but the weights for SU(2) are just opposite in sign, so changeing their sign maps the positive one onto the negative one and vice versa. This doesn't give you 2 new particles, this brings one particle to another in the same representation. That is just the way I understand it though. I believe this is exactly why a complete theory of weak interactions must incorporate electromagnetic interactions thus giving us a SU(2) X U(1) gauge group. (this may not be WHY EM must be incorporated into Weak interactions, but it is part of the reason I think.)
Science Advisor
Norman I am reading ahead for my Group Theory introduction to QFT and I have a question about the dual representation. If the dual representation is the same as the ordinary representation said:
take a fundamental doublet of fields, u and d, write them as a column su(2)-vector (the {2}). The conjugate doublet of antiparticle fields transforms as a row vector;(u*,d*). Now with aid of the totally antisymmetric tensor (the Levi-Civita SYMBOL) which is invariant under su(2) transformations, we may write the above as a column vector also, with (-d*) as 1st element and (u*) the 2nd.i.e the antiparticle doublet(-d*,u*) transforms exactly as the particle doublet(u,d),and belongs to the same {2}.
regards
sam | 1,520 | 6,484 | {"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} | 2.53125 | 3 | CC-MAIN-2023-06 | latest | en | 0.950822 |
https://docslib.org/doc/4486658/predicate-logic-syntax | 1,701,345,544,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00062.warc.gz | 257,315,135 | 8,542 | <<
Predicate : Syntax
Alice Gao
Lecture 12
CS 245 Logic and Computation Fall 2019 1 / 30 Outline
Learning goals
Symbols
Terms
Formulas
Parse Trees
Revisiting the learning goals
CS 245 Logic and Computation Fall 2019 2 / 30 Learning goals
By the end of this lecture, you should be able to ▶ Define the set of terms inductively. ▶ Define the set of formulas inductively. ▶ Determine whether a in a formula is free or bound. ▶ Prove properties of terms and formulas by structural induction. ▶ Draw the parse tree of a formula.
CS 245 Logic and Computation Fall 2019 3 / 30 The Language of Predicate Logic
▶ Domain: a non-empty set of objects. ▶ Individuals: concrete objects in the domain. ▶ Variables: placeholders for concrete objects in the domain. ▶ Functions: takes objects in the domain as arguments and returns an object of the domain. ▶ Relations: takes objects in the domain as arguments and returns true or false. They describe properties of objects or relationships between objects. ▶ Quantifiers: for how many objects in the domain isthe statement true?
CS 245 Logic and Computation Fall 2019 4 / 30 Outline
Learning goals
Symbols
Terms
Formulas
Parse Trees
Revisiting the learning goals
CS 245 Logic and Computation Fall 2019 5 / 30 Predicate Language 퐿
Eight classes of symbols: ▶ Individual symbols: 푎, 푏, 푐. ▶ Relation symbols: 퐹 , 퐺, 퐻. A special equality symbol ≈ ▶ Function symbols: 푓, 푔, ℎ. ▶ Free variable symbols: 푢, 푣, 푤. ▶ Bound variable symbols: 푥, 푦, 푧. ▶ Connective symbols: ¬, ∧, ∨, →, ↔. ▶ symbol: ∀, ∃. ▶ Punctuation symbols: ‘(’, ‘)’, and ‘,’
CS 245 Logic and Computation Fall 2019 6 / 30 Free and Bound Variables
In a formula ∀푥 퐴(푥) or ∃푥 퐴(푥), the scope of a quantifier is the formula 퐴(푥). A quantifier binds its variable within its scope. An occurrence of a variable in a formula ▶ is bound if it lies in the scope of some quantifier of the same variable. ▶ is free, otherwise.
CS 245 Logic and Computation Fall 2019 7 / 30 Outline
Learning goals
Symbols
Terms
Formulas
Parse Trees
Revisiting the learning goals
CS 245 Logic and Computation Fall 2019 8 / 30 Two Kinds of Expressions
Two kinds of expressions: ▶ A refers to an object in the domain. ▶ A formula evaluates to 1 or 0.
CS 245 Logic and Computation Fall 2019 9 / 30 Terms
The set of terms Term(퐿) is defined below: 1. An individual symbol 푎 standing alone is a term. 2. A free variable symbol 푢 standing alone is a term.
3. If 푡1, … , 푡푛 are terms and 푓 is an 푛-ary function symbol, then 푓(푡1, … , 푡푛) is a term. 4. Nothing else is a term.
CS 245 Logic and Computation Fall 2019 10 / 30 Examples of Terms
Terms: ▶ 푎, 푏, 푐, 푢, 푣, 푤 ▶ 푓(푏), 푔(푎, 푓(푏)), 푔(푢, 푏), 푓(푔(푓(푢), 푏))
A term with no free variable symbols is called a closed term. Which one(s) of the above are closed terms?
CS 245 Logic and Computation Fall 2019 11 / 30 CQ: Which expressions are terms?
Which of the following expressions is a term? If there are multiple correct answers, choose your favourite one. (A) 푤 (B) 푔(푎, 푢) (C) 퐹 (푓(푢, 푣), 푎) (D) 푓(푢, 푔(푣, 푤), 푎) (E) 푔(푢, 푓(푣, 푤), 푎) Individual symbols: 푎 Relation symbols: 퐹 is a binary relation symbol. Function symbols: 푓 is a binary function symbol and 푔 is a 3-ary function symbol. Free variable symbols: 푢, 푣, 푤.
CS 245 Logic and Computation Fall 2019 12 / 30 Defining the set of terms inductively
The set of terms can be inductively defined as follows: ▶ The domain set 푋: The set of finite of symbols of 퐿 ▶ The core set 퐶: The set of all individual symbols and free variable symbols ▶ The set of operations 푃 : The set of all function symbols
CS 245 Logic and Computation Fall 2019 13 / 30 Structural induction on terms
Theorem: Every term has a property 푃 . Proof by structural induction: ▶ Base cases: The term is an individual symbol. The term is a free variable symbol. ▶ Inductive cases:
The term is 푓(푡1, … , 푡푛) where 푓 is an n-ary function and 푡1, ..., 푡푛 are terms. Induction hypotheses: Assume that 푡1, … , 푡푛 all have the property 푃 .
We need to show that 푓(푡1, … , 푡푛) has the property 푃 .
CS 245 Logic and Computation Fall 2019 14 / 30 Outline
Learning goals
Symbols
Terms
Formulas
Parse Trees
Revisiting the learning goals
CS 245 Logic and Computation Fall 2019 15 / 30 Atomic Formulas
The set of atomic formulas Atom(퐿) is defined below: ▶ If 퐹 is an n-ary relation symbol and 푡1, … , 푡푛 (푛 ≥ 1) are terms, then 퐹 (푡1, … , 푡푛) is an . ▶ If 푡1, 푡2 are terms, then ≈ (푡1, 푡2) is an atomic formula. ▶ Nothing else is an atomic formula.
CS 245 Logic and Computation Fall 2019 16 / 30 Examples of Atomic Formulas
Terms: ▶ 푎, 푏, 푐, 푢, 푣, 푤 ▶ 푓(푏), 푔(푎, 푓(푏)), 푔(푢, 푏), 푓(푔(푓(푢), 푏))
Atomic formulas: ▶ 퐹 (푎, 푢, 푓(푏), 푓(푤), 푔(푣, 푓(푎))) ▶ ≈ (푏, 푤)
CS 245 Logic and Computation Fall 2019 17 / 30 Well-Formed Formulas
The set of well-formed formulas Form(퐿) is defined below: 1. An atomic formula is a well-formed formula. 2. If 퐴 is a well-formed formula, then (¬퐴) is a well-formed formula. 3. If 퐴 and 퐵 are well-formed formulas and ⋆ is one of ∧, ∨, →, and ↔, then (퐴 ⋆ 퐵) is a well-formed formula. 4. If 퐴(푢) is a well-formed formula and 푥 does not occur in 퐴(푢), then ∀푥 퐴(푥) and ∃푥 퐴(푥) are well-formed formulas. 5. Nothing else is a well-formed formula.
CS 245 Logic and Computation Fall 2019 18 / 30 Explaining Case 4 of Formulas
If 퐴(푢) is a well-formed formula and 푥 does not occur in 퐴(푢), then ∀푥 퐴(푥) and ∃푥 퐴(푥) are well-formed formulas. ▶ 퐴(푢) is a well-formed formula where 푢 is a free variable in the formula. We want to quantify 푢. ▶ In order to do so, we need to choose a symbol for a bound variable, e.g. 푥. We need to make sure that our choice of the bound variable symbol does not already occur in 퐴(푢).
CS 245 Logic and Computation Fall 2019 19 / 30 Examples for Case 4
▶ We are allowed to generate the formula ∀푦퐹 (푦, 푦). Start with 퐹 (푢, 푢). If we quantify 푢 by replacing it with 푦, we get ∀푦퐹 (푦, 푦).
▶ We are not allowed to generate the formula ∃푦∀푦퐹 (푦, 푦). Start with ∀푦퐹 (푦, 푦). If we want to add the ∃ quantifier, we will need to choose a bound variable symbol that is not 푦 because 푦 already appears in ∀푦 퐹 (푦, 푦). So, there is no way for us to generate ∃푦∀푦퐹 (푦, 푦).
▶ We are allowed to generate the formula ∃푥퐺(푥) ∨ ∀푥퐻(푥). Start with 퐺(푢) and 퐻(푣) separately. We can quantify 푢 by replacing it with 푥 since 푥 does not appear in 퐺(푢)). We get ∃푥퐺(푥). We can quantify 푣 by replacing it with 푥 since 푥 does not appear in 퐻(푣). We get ∀푥퐻(푥). Connecting the two formulas using ∨, we get ∃푥퐺(푥) ∨ ∀푥퐻(푥).
CS 245 Logic and Computation Fall 2019 20 / 30 Examples of Formulas
Well-Formed Formulas: ▶ 퐹 (푎, 푏), ∀푦 퐹 (푎, 푦), ∃푥∀푦 퐹 (푥, 푦) ▶ 퐹 (푢, 푣), ∃푦 퐹 (푢, 푦)
A formula with no free variable symbols is called a closed formula or a sentence. Which formulas above are closed formulas?
CS 245 Logic and Computation Fall 2019 21 / 30 Determine whether a formula is well-formed
Which of the following is a well-formed formula? (A) 푓(푢) → 퐹 (푢, 푣) (B) ∀푥 퐹 (푚, 푓(푥)) (C) 퐹 (푢, 푣) → 퐺(퐺(푢)) (D) 퐺(푚, 푓(푚)) (E) 퐹 (푚, 푓(퐺(푢, 푣))) Individual symbols: 푚. Free Variable Symbols: 푢, 푣. Bound Variable symbols: 푥. Relation symbols: 퐹 and 퐺 are binary relation symbols. Function symbols: 푓 is a unary function.
CS 245 Logic and Computation Fall 2019 22 / 30 Defining the set of formulas inductively
The set of formulas can be inductively defined as follows: ▶ The domain set 푋: The set of finite sequences of symbols of 퐿 ▶ The core set 퐶: The set of all atomic formulas. ▶ The set of operations 푃 :
푓1(푥) = (¬푥) 푓2(푥, 푦) = (푥 ∗ 푦) where ∗ is one of ∧, ∨, →, and ↔. 푓3(퐴(푢)) = ∀푥 퐴(푥), 푓4(퐴(푢)) = ∃푥 퐴(푥) where 푥 does not occur in 퐴(푢).
CS 245 Logic and Computation Fall 2019 23 / 30 Structural induction on formulas
Theorem: Every formula has a property 푃 . Proof by structural induction: ▶ Base cases: The formula is an atomic formula. ▶ Inductive cases: The formula is (¬퐴) where 퐴 is a formula. The formula is (퐴 ∗ 퐵) where 퐴 and 퐵 are formulas and ∗ is a binary connective. The formula is ∀푥 퐴(푥) and ∃푥 퐴(푥) where 퐴(푢) is a formula and 푥 does not occur in 퐴(푢).
CS 245 Logic and Computation Fall 2019 24 / 30 Comparing the Definitions of Well-Formed Formulas
Let’s compare the set of predicate formulas to the set of propositional formulas. Questions to think about: ▶ Which parts of the two definitions are the same? The cases for negation and binary connectives are the same. ▶ Which parts of the two definitions are different? Atomic formulas are different.
▶ Atomic propositional formulas are propositional variables. ▶ Atomic predicate formulas are relations applied to terms. Predicate formulas have one additional case for quantifiers.
CS 245 Logic and Computation Fall 2019 25 / 30 Outline
Learning goals
Symbols
Terms
Formulas
Parse Trees
Revisiting the learning goals
CS 245 Logic and Computation Fall 2019 26 / 30 Parse Trees of Predicate Formulas
▶ The leaves are atomic formulas. ▶ Every quantifier has exactly one child (namely the formula which is its scope). Example: ∀푥(퐹 (푏) → ∃푦(∀푧 퐺(푦, 푧) ∨ 퐻(푓(푢), 푥, 푦)))
CS 245 Logic and Computation Fall 2019 27 / 30 Parse tree
Example: ∀푥(퐹 (푏) → ∃푦(∀푧 퐺(푦, 푧) ∨ 퐻(푓(푢), 푥, 푦))) Parse tree:
∀푥(퐹 (푏) → ∃푦(∀푧 퐺(푦, 푧) ∨ 퐻(푓(푢), 푥, 푦)))
퐹 (푏) → ∃푦(∀푧 퐺(푦, 푧) ∨ 퐻(푓(푢), 푣, 푦))
퐹 (푏) ∃푦(∀푧 퐺(푦, 푧) ∨ 퐻(푓(푢), 푣, 푦))
∀푧 퐺(푤, 푧) ∨ 퐻(푓(푢), 푣, 푤)
∀푧 퐺(푤, 푧) 퐻(푓(푢), 푣, 푤)
퐺(푤, 푢1)
CS 245 Logic and Computation Fall 2019 28 / 30 A few notes on parse trees
1. While constructing the parse tree, when removing a quantifier, we change the bound variable symbol to one of the free variable symbols that hasn’t appeared in the parse tree. For example, When removing ∀푥, we changed 푥 to 푣. When removing ∃푦, we changed 푦 to 푤. 2. Th quantifiers have higher precedence than any other connective. Each quantifier modifies the formula that is immediately after it. For example, ∀푧 modifies 퐺(푤, 푧) instead of 퐺(푤, 푧) ∨ 퐻(푓(푢), 푣, 푤).
CS 245 Logic and Computation Fall 2019 29 / 30 Revisiting the learning goals
By the end of this lecture, you should be able to ▶ Define the set of terms inductively. ▶ Define the set of formulas inductively. ▶ Determine whether a variable in a formula is free or bound. ▶ Prove properties of terms and formulas by structural induction. ▶ Draw the parse tree of a formula.
CS 245 Logic and Computation Fall 2019 30 / 30 | 4,101 | 10,340 | {"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} | 3.6875 | 4 | CC-MAIN-2023-50 | latest | en | 0.768287 |
https://www.chemicalslearning.com/2022/12/types-of-solids-separation-classifier.html | 1,685,557,442,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647409.17/warc/CC-MAIN-20230531182033-20230531212033-00117.warc.gz | 794,250,065 | 49,243 | Types of Solids Separation Classifier in Mechanical Operation
# Types of Solids Separation Classifier in Mechanical Operation
There are a few different types of classifiers that are commonly used. The first type is a rotary classifier, which uses a rotating drum to classify materials. The second type is a vibrating classifier, which uses vibrations to help separate the particles.Â
The third type is a gravity classifier, which relies on gravity to help separate the materials.
Classification is defined as the separation of solid particles into fractions based on their terminal settling velocities. Suppose, for example, that the solid particles to be separated are fed in suspension into a tank of a large cross-sectional area.Â
## Types of classifiers in mineral processingÂ
There are many different types of classifiers in mineral processing, each with its own unique characteristics and applications. Here is a brief overview of some of the most common types:
### Gravity Settling Tank/Gravity classifier
It is the simplest type of classifier. It consists of a large tank with provisions for a suitable inlet and outlet.
A slurry feed enters the tank through an inlet connection. As soon as slurry feed enters the tank, its linear velocity decreases as a result of the enlargement of the cross-sectional area. Solid particles start to settle under the bottom of the tank near the inlet, while the slower-settling particles will be carried farther into the tank before they reach the bottom of the tank. The very fine particles are carried away in the liquid overflow from the tank.Â
Vertical baffles placed at various distances from the inlet within the tank allow for the collection of several fractions according to the terminal falling velocities. Because of the occurrence of considerable overlapping of size, no sharp separation is possible with this classifier.
#### Gravity classifier uses
Gravity classifiers are used in a variety of industries to separate materials based on their specific gravity. Specific gravity is the ratio of the density of a substance to the density of water.Â
Gravity classifiers are typically used to remove light, worthless materials, such as clay and dust, from a product. By doing so, manufacturers can increase the overall purity of their products. Additionally, gravity classifiers can be used to separate products that are of different densities.Â
One common use for a gravity classifier is to remove fine coal from a product stream. By removing the fine coal, power plants can reduce their emissions.Â
Gravity classifiers are also used in the food industry. For example, they can be used to remove shells from nuts. Additionally, they can be used to remove foreign objects from food products.Â
### Cone Classifier/Cone classifier working
A cone Classifier is simply a cone shape is a conical vessel, installed point down, with a discharge launder around the top. The feed is introduced in the form of a suspension through a fed inlet provided at the top. The coarse fraction collects at the point of the cone and is withdrawn periodically or continuously.Â
The fine fraction along with the remaining portion of the liquid is removed from the launder as an overflow. The separation achieved with this unit is only an approximate one. Cone classifiers, though one of the oldest types are still used for relatively crude work because of the low cost of installation. They are used in ore-dressing plants.
### Double cone classifier working principle
This classifier uses hydraulic water for classification (a stream of additional water supplied to a classifier is called hydraulic water)Â Â
The double cone Classifier is shown in the picture. It consists of a conical vessel inside of which is a second hollow cone. The inner cone is slightly larger in angle, arranged apex downward, and is movable in a vertical direction.
The to-be-separated is fed in the form of a suspension to the center of the inner cone. It flows downward through the inner cone and out at a baffle at the bottom of the inner cone and a rising stream of water is fed near the outlet for the coarse material.Â
The solids form the inner cone and then flow through an annulus space between the two cones. Classification action occurs in the annular space, the small particles are carried away in the overflow while the occurs in large Particles settle against the hydraulic water to the bottom and are removed periodically.
### Rake classifier working principle (Mechanical Classifiers)
The rake/mechanical classifier such as the door classifier consists of a rectangular tank with a sloping/inclined bottom. The tank is provided with movable rakes. The feed in the form of a suspension or slurry is introduced continuously near the middle of the tank. The lower end of the overflow from which the fines that are not settled leave with the overflow from which the fines that are not settled leave with the overflow liquid.Â
The heavy material sink to the bottom of the tank. The rakes scrap the settled solids upward along the bottom of the tank toward the top of the tank for discharge from a sand discharge chute. The reciprocating rakes keep the slurry in continuous agitation. The time of the raking stroke is so adjusted that fines do not have time to settle and so remain near the surface of the slurry while the heavy particles have time to settle.
### Spiral classifier working principle
Another mechanical classifier is the spiral classifier. The spiral classifier such as the Akins classifier consists of a semi-cylindrical trough inclined horizontally. The trough is provided with a slow-rotating spiral conveyor and a liquid overflow at the lower end. The spiral conveyor moves the solids which settle to the bottom upward toward the top of the trough.
The slurry is fed continuously near the middle of the trough. The slurry feed rate is so adjusted that fine particles do not have time to settle and are carried out with the overflow liquid. The heavy particles have time to settle, they settle to the bottom of the trough and the spiral conveyor moves the settled solids upward along the floor of the trough toward the top of the trough/the sand product discharge chute.
Note: Rake and spiral classifiers are used for closed-circuit grinding.
### Cyclones Separators Working PrincipaleÂ
A cyclone separator is essentially a settling chamber in which the gravitational separating force is replaced by a much stronger centrifugal separating force.
Cyclones separators are used for the separation of solids from fluids. They offer one of the least expensive means of dust collection from both an operating and an investment viewpoint. They utilize centrifugal force to effect the separation which depends on particle size and on particle density. Thus cyclones are used to effect a separation on the basis of particle size or particle density or both.
In this separator, used for the separation of dust particles or mist from gases, the dust-laden gas is introduced tangentially into a cylindrical vessel at a high velocity (about 30 m/s). Centrifugal force throws the solid particles out against the wall of the vessel and they drop into a conical section of the cyclone and are removed from the bottom opening. The clean gas is taken out through a central outlet at the top.
### Hydrocyclone separator working principle
Hydrocyclone separators may also be used to affect the classification of solids particles suspended in a liquid. In such cases, the commonly used liquid is water.
A liquid cyclone or hydro cyclone has a top cylindrical section and a lower conical section terminating in an apex opening. The top vertical section is covered by a flat plate and is provided with a tangential inlet at the top. The cover has a downward extending pipe to cut the vortex and remove the overflow product since the viscosity of water is much higher than that of a gas, the fluid resistance encountered in this cyclone is greater than that in the cyclone used for dust collection.Â
Therefore the diameter of this cyclone must be smaller in order to get a corresponding greater centrifugal force. Consequently, the diameter of the cylindrical section is less than 375 mm and the cone angle is 16° to 20°. Inlet pressures of the feed to the cyclone lie between 5 to 120 psi.
The slurry feed is pumped into the cylindrical section tangentially. Coarse or heavy solids thrown out against the walls, travel down the sides of the cone section and are discharged in a partially dewatered form at the apex, while the smaller or lighter solids along with the remaining portion of water are removed from the downward extending pipe at the top. Hydrocyclone finds application in degritting operations production, classifying pigments, and ore-dressing practice. Â
#### Hydrocyclone uses
A hydrocyclone is a device that uses centrifugal force to separate solid particles from a liquid or to separate liquid particles from each other. Hydrocyclones are used in a variety of industries, including oil and gas, mining, and manufacturing.Â
Take these Notes is, Orginal Sources:Â Unit Operations-II, KA Gavhane
### What is a cyclone used for?
Cyclones are widely used for the collection of heavy and coarse dust. These units may also be used for separating coarse materials from fine dust.
### What is the difference between a sifter and a classifier?
There are a few key differences between sifters and classifiers. For one, sifters are typically used to remove smaller particles from a mixture, while classifiers are used to separate out larger particles. Additionally, sifters usually have mesh screens that the material is passed through, while classifiers typically have paddles or other mechanisms to help separate the materials. Finally, sifters are often used as part of a larger process (such as flour production), while classifiers are more often used on their own.
Thanks to visit this site. | 1,949 | 9,962 | {"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} | 2.5625 | 3 | CC-MAIN-2023-23 | longest | en | 0.922797 |
http://www.varsitytutors.com/mcat_physical-help/electric-potential-energy | 1,485,043,473,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281263.12/warc/CC-MAIN-20170116095121-00146-ip-10-171-10-70.ec2.internal.warc.gz | 736,356,576 | 34,356 | # MCAT Physical : Electric Potential Energy
## Example Questions
### Example Question #1 : Electric Potential Energy
If the electrical potential energy between two equal charges quadruples, describe the change in the distance between the particles.
The distance was not changed
The distance was quartered
The distance was doubled
The distance was halved
The distance was quartered
Explanation:
Electrical potential energy is given by the equation .
Electrical potential energy is inversely proportional to the distance between the two charges. If the energy is quadrupled, then (the distance between the two equal charges) must have decreased proportionally.
### Example Question #2 : Electric Potential Energy
For the following:
If a charge of magnitude +4e is being held in place 3 nm from a charge of -5e which is also being held in place. What is the potential energy of the system?
Explanation:
Given the equation and plugging in the values of e and k, we get that
It is important to keep in mind that the charge is given in the question and must be incorporated into the formula by multiplying each charge by that value.
### Example Question #1 : Electric Potential Energy
Which of the following is not true regarding electric potential?
Electric potential can be expressed with units of Volts or Joules per Coulomb
A negative charge moving from low potential to high potential will accelerate
When a positive charge moves from a region of low potential to high potential, the electric field does positive work on the charge
All of these are true statements
The positive terminal of a battery has higher electric potential than the negative terminal
When a positive charge moves from a region of low potential to high potential, the electric field does positive work on the charge
Explanation:
A positive test charge will naturally move from high potential to low potential. If it is moved in the opposite direction, then the electric field will do work against its motion (negative work). This be seen from the equation for electric field work:
is the work done by the electric field, is the charge, and is the potential difference. If is positive (the final potential is higher than the initial potential) and is also positive, then work done by the field is negative.
### Example Question #4 : Electric Potential Energy
How much work is required to bring together the three given charges from infinity to the corners of an equilateral triangle of side length 1cm?
Explanation:
Relevant equations:
Step 1: Since the work done to assemble the charges equals their potential energy in this arrangment, find the potential energy between each pair of charges. Work is equal to change in potential energy; since the charges start at infinite distance, initially potential energy is equal to zero.
Charges 1 and 2
Charges 1 and 3
Charges 2 and 3
Step 2: Add together all these potential energies to find the total energy of the arrangement.
### Example Question #4 : Electric Potential Energy
A standard AC outlet is capable of delivering of current. What is the maximum possible voltage given by the outlet. | 636 | 3,156 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.943523 |
http://qbasicnews.com/newforum/thread-10118.html | 1,702,004,560,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100710.22/warc/CC-MAIN-20231208013411-20231208043411-00732.warc.gz | 35,627,223 | 8,507 | Search inside Integers techdude2007 Junior Member Posts: 2 Threads: 2 Joined: May 2008 05-11-2008, 03:44 AM I need to find a function or command that will search inside an Integer. I am trying to see how many times the number 5 would appear in, for example, 448567475. The user enters in a number, and the computer will find how many times the number 5 is in that integer. I tried to use the INSTR command, but it only indicates where the 1st occurance is, not the 2nd or third. Any ideas? roy Junior Member Posts: 33 Threads: 2 Joined: Aug 2007 05-11-2008, 04:19 AM This may not be the shortest way but it works. Code:```DIM a AS LONG CLS REM Your input routine goes here. a = 233454535 b\$ = STR\$(a) FOR i = 1 TO LEN(b\$) IF VAL(MID\$(b\$, i, 1)) = 5 THEN tot = tot + 1 NEXT PRINT "Number of fives = "; tot SYSTEM``` Ralph Senior Member Posts: 544 Threads: 27 Joined: Jan 2005 05-11-2008, 09:34 AM Roy, I think that your solution has the correct approach. I don't think there is another way that would be more practical and logical. Ralph, using QuickBASIC 4.5 and Windows XP Home Edition and Service Pack 2, with HP LaserJet 4L printer. LPG Member Posts: 61 Threads: 11 Joined: Apr 2008 05-11-2008, 09:51 AM (This post was last modified: 05-11-2008, 10:22 AM by LPG.) could you do it like this? Code:```CLS INPUT A\$ S\$ = "5" 1 IF INSTR(A\$,S\$) <> 0 THEN : TOT = TOT + 1 : A\$ = RIGHT\$(A\$,LEN(A\$)-INSTR(A\$,S\$)) : GOTO 1 PRINT "THERE ARE ";TOT;" `"; S\$ ; "`'S" SYSTEM``` This works differently What do you want it for? WHILE RPG\$ <> "complete" : make up silly excuses :WEND Ralph Senior Member Posts: 544 Threads: 27 Joined: Jan 2005 05-11-2008, 11:14 AM Very ingenious LPG, I would not have thought of doing it that way! It is certainly much quicker than the other way! Ralph, using QuickBASIC 4.5 and Windows XP Home Edition and Service Pack 2, with HP LaserJet 4L printer. LPG Member Posts: 61 Threads: 11 Joined: Apr 2008 05-12-2008, 10:54 AM If it is inputting a sring I probably should have done this: Code:```CLS ' 5 finder.v2 NUMBER\$ = "1234567890" IN: INPUT "PLEASE ENTER A NUMBER";A\$ FOR B = 1 TO LEN(A\$) NUM = 0 FOR C = 1 TO 10  IF MID\$(A\$,B,1) = MID\$(NUMBER\$,C,1) THEN NUM = 1 NEXT IF NUM <> 1 THEN PRINT "YOU DIDN'T TYPE A NUMBER." : GOTO IN NEXT S\$ = "5" 1 IF INSTR(A\$,S\$) <> 0 THEN : TOT = TOT + 1 : A\$ = RIGHT\$(A\$,LEN(A\$)-INSTR(A\$,S\$)) : GOTO 1 PRINT "THERE ARE ";TOT;" ";CHR\$(34); S\$ ; CHR\$(34);"'S" SYSTEM``` this checks the user inputs a number. the previous version would count the "5"'s in a phrase. WHILE RPG\$ <> "complete" : make up silly excuses :WEND Ralph Senior Member Posts: 544 Threads: 27 Joined: Jan 2005 05-12-2008, 08:45 PM LPG: I think that your original code is O.K., as it counts any "5" in the inputted string, whether the user enters a number or a string. That was the OP's proposal. Who care if one inputs a string or a number, the question is, "How many characters "5" are there in what I inputted?". Your code answers the question correctly. Ralph, using QuickBASIC 4.5 and Windows XP Home Edition and Service Pack 2, with HP LaserJet 4L printer. « Next Oldest | Next Newest » | 1,027 | 3,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-50 | latest | en | 0.876063 |
https://netlib.org/lapack/explore-html-3.6.1/df/dda/slauu2_8f_source.html | 1,675,558,289,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500158.5/warc/CC-MAIN-20230205000727-20230205030727-00768.warc.gz | 453,205,742 | 6,325 | LAPACK 3.6.1 LAPACK: Linear Algebra PACKage
slauu2.f
Go to the documentation of this file.
1 *> \brief \b SLAUU2 computes the product UUH or LHL, where U and L are upper or lower triangular matrices (unblocked algorithm).
2 *
3 * =========== DOCUMENTATION ===========
4 *
5 * Online html documentation available at
6 * http://www.netlib.org/lapack/explore-html/
7 *
8 *> \htmlonly
10 *> <a href="http://www.netlib.org/cgi-bin/netlibfiles.tgz?format=tgz&filename=/lapack/lapack_routine/slauu2.f">
11 *> [TGZ]</a>
12 *> <a href="http://www.netlib.org/cgi-bin/netlibfiles.zip?format=zip&filename=/lapack/lapack_routine/slauu2.f">
13 *> [ZIP]</a>
14 *> <a href="http://www.netlib.org/cgi-bin/netlibfiles.txt?format=txt&filename=/lapack/lapack_routine/slauu2.f">
15 *> [TXT]</a>
16 *> \endhtmlonly
17 *
18 * Definition:
19 * ===========
20 *
21 * SUBROUTINE SLAUU2( UPLO, N, A, LDA, INFO )
22 *
23 * .. Scalar Arguments ..
24 * CHARACTER UPLO
25 * INTEGER INFO, LDA, N
26 * ..
27 * .. Array Arguments ..
28 * REAL A( LDA, * )
29 * ..
30 *
31 *
32 *> \par Purpose:
33 * =============
34 *>
35 *> \verbatim
36 *>
37 *> SLAUU2 computes the product U * U**T or L**T * L, where the triangular
38 *> factor U or L is stored in the upper or lower triangular part of
39 *> the array A.
40 *>
41 *> If UPLO = 'U' or 'u' then the upper triangle of the result is stored,
42 *> overwriting the factor U in A.
43 *> If UPLO = 'L' or 'l' then the lower triangle of the result is stored,
44 *> overwriting the factor L in A.
45 *>
46 *> This is the unblocked form of the algorithm, calling Level 2 BLAS.
47 *> \endverbatim
48 *
49 * Arguments:
50 * ==========
51 *
52 *> \param[in] UPLO
53 *> \verbatim
54 *> UPLO is CHARACTER*1
55 *> Specifies whether the triangular factor stored in the array A
56 *> is upper or lower triangular:
57 *> = 'U': Upper triangular
58 *> = 'L': Lower triangular
59 *> \endverbatim
60 *>
61 *> \param[in] N
62 *> \verbatim
63 *> N is INTEGER
64 *> The order of the triangular factor U or L. N >= 0.
65 *> \endverbatim
66 *>
67 *> \param[in,out] A
68 *> \verbatim
69 *> A is REAL array, dimension (LDA,N)
70 *> On entry, the triangular factor U or L.
71 *> On exit, if UPLO = 'U', the upper triangle of A is
72 *> overwritten with the upper triangle of the product U * U**T;
73 *> if UPLO = 'L', the lower triangle of A is overwritten with
74 *> the lower triangle of the product L**T * L.
75 *> \endverbatim
76 *>
77 *> \param[in] LDA
78 *> \verbatim
79 *> LDA is INTEGER
80 *> The leading dimension of the array A. LDA >= max(1,N).
81 *> \endverbatim
82 *>
83 *> \param[out] INFO
84 *> \verbatim
85 *> INFO is INTEGER
86 *> = 0: successful exit
87 *> < 0: if INFO = -k, the k-th argument had an illegal value
88 *> \endverbatim
89 *
90 * Authors:
91 * ========
92 *
93 *> \author Univ. of Tennessee
94 *> \author Univ. of California Berkeley
95 *> \author Univ. of Colorado Denver
96 *> \author NAG Ltd.
97 *
98 *> \date September 2012
99 *
100 *> \ingroup realOTHERauxiliary
101 *
102 * =====================================================================
103 SUBROUTINE slauu2( UPLO, N, A, LDA, INFO )
104 *
105 * -- LAPACK auxiliary routine (version 3.4.2) --
106 * -- LAPACK is a software package provided by Univ. of Tennessee, --
107 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
108 * September 2012
109 *
110 * .. Scalar Arguments ..
111 CHARACTER UPLO
112 INTEGER INFO, LDA, N
113 * ..
114 * .. Array Arguments ..
115 REAL A( lda, * )
116 * ..
117 *
118 * =====================================================================
119 *
120 * .. Parameters ..
121 REAL ONE
122 parameter ( one = 1.0e+0 )
123 * ..
124 * .. Local Scalars ..
125 LOGICAL UPPER
126 INTEGER I
127 REAL AII
128 * ..
129 * .. External Functions ..
130 LOGICAL LSAME
131 REAL SDOT
132 EXTERNAL lsame, sdot
133 * ..
134 * .. External Subroutines ..
135 EXTERNAL sgemv, sscal, xerbla
136 * ..
137 * .. Intrinsic Functions ..
138 INTRINSIC max
139 * ..
140 * .. Executable Statements ..
141 *
142 * Test the input parameters.
143 *
144 info = 0
145 upper = lsame( uplo, 'U' )
146 IF( .NOT.upper .AND. .NOT.lsame( uplo, 'L' ) ) THEN
147 info = -1
148 ELSE IF( n.LT.0 ) THEN
149 info = -2
150 ELSE IF( lda.LT.max( 1, n ) ) THEN
151 info = -4
152 END IF
153 IF( info.NE.0 ) THEN
154 CALL xerbla( 'SLAUU2', -info )
155 RETURN
156 END IF
157 *
158 * Quick return if possible
159 *
160 IF( n.EQ.0 )
161 \$ RETURN
162 *
163 IF( upper ) THEN
164 *
165 * Compute the product U * U**T.
166 *
167 DO 10 i = 1, n
168 aii = a( i, i )
169 IF( i.LT.n ) THEN
170 a( i, i ) = sdot( n-i+1, a( i, i ), lda, a( i, i ), lda )
171 CALL sgemv( 'No transpose', i-1, n-i, one, a( 1, i+1 ),
172 \$ lda, a( i, i+1 ), lda, aii, a( 1, i ), 1 )
173 ELSE
174 CALL sscal( i, aii, a( 1, i ), 1 )
175 END IF
176 10 CONTINUE
177 *
178 ELSE
179 *
180 * Compute the product L**T * L.
181 *
182 DO 20 i = 1, n
183 aii = a( i, i )
184 IF( i.LT.n ) THEN
185 a( i, i ) = sdot( n-i+1, a( i, i ), 1, a( i, i ), 1 )
186 CALL sgemv( 'Transpose', n-i, i-1, one, a( i+1, 1 ), lda,
187 \$ a( i+1, i ), 1, aii, a( i, 1 ), lda )
188 ELSE
189 CALL sscal( i, aii, a( i, 1 ), lda )
190 END IF
191 20 CONTINUE
192 END IF
193 *
194 RETURN
195 *
196 * End of SLAUU2
197 *
198 END
subroutine xerbla(SRNAME, INFO)
XERBLA
Definition: xerbla.f:62
subroutine slauu2(UPLO, N, A, LDA, INFO)
SLAUU2 computes the product UUH or LHL, where U and L are upper or lower triangular matrices (unblock...
Definition: slauu2.f:104
subroutine sgemv(TRANS, M, N, ALPHA, A, LDA, X, INCX, BETA, Y, INCY)
SGEMV
Definition: sgemv.f:158
subroutine sscal(N, SA, SX, INCX)
SSCAL
Definition: sscal.f:55 | 2,068 | 5,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.336151 |
http://www.answers.com/topic/food-science?cat=health | 1,527,342,410,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867417.75/warc/CC-MAIN-20180526131802-20180526151802-00030.warc.gz | 332,297,762 | 55,169 | # What are the 5Cs of credit?
5 C's of Credit refer to the factors that lenders of money evaluate to determine credit worthiness of a borrower. They are the following:. 1. Borrower's CHARACTER. 2. Borrow (MORE)
# What does 5c stand for?
The Iphone 5C is Iphone 5Colorful 5c can also stand for thenumber 500 ("c" is the Roman numeral for 100) or for 5 degreesCelsius (centigrade) . +++ . "5c" can not stand fo (MORE)
# What animal is on a 5c coin?
There are multiple animals on 5 cent coins depending on the country and time period such as the Buffalo on the US "buffalo nickel", the Beaver on the Canadian nickel, etc.
# What is -5c plus 9 and how?
You can't tell a thing about -5c+9 until you know what 'c' is. And every time 'c' changes, -5c+9 changes.
In Science
# My son got a 5c in his year 7 science exam is this a good grade?
Yes this is an alright grade but your son should soon start working towards a Level 6B-6A in future exams.
In Volume
# What is 5c in milliliters?
5cc? cc means cubic centimetres which is equal to ml, so 5ml. if you mean cl, then that is equal to 50ml
# What is the answer for 5c equals -75?
The 'answer' is the number that 'c' must be, if 5c is really the same as -75. In order to find out what number that is, you could use 'algebra'. First, write the equatio (MORE)
# How many pixels does the iPhone 5c have?
The iPhone 5c is 640 x 1136 pixels. That is about 326 pixels persquare inch (ppi).
# What is minus 5c in Fahrenheit?
(-5) degrees Celsius = 23 degrees Fahrenheit. Formula: [°F] = [°C] à 9 â 5 + 32
# How many inches is a iPhone 5c?
The screen is 4" big. The height is 4.9", width is 2.33" and thedepth is 0.35" | 510 | 1,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-22 | latest | en | 0.924762 |
https://leanprover-community.github.io/archive/stream/113489-new-members/topic/Practicing.20writing.20proofs.html | 1,718,969,695,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00890.warc.gz | 308,180,041 | 6,254 | ## Stream: new members
### Topic: Practicing writing proofs
#### Adomas Baliuka (Sep 03 2023 at 18:02):
I wrote a proof that subtracting two sums over ranges equals sum over the difference of the ranges. The proof took me a long time and seems extremely ugly. I'm hoping to learn a lot by having someone look at it for 3 seconds and say how to do it in two lines. (Please let me know in case such a question is discouraged here)
lemma bigsum_minus_bigsum (m n : ℕ) (mlen : n ≤ m) (f : ℕ → ℝ) :
∑ k in range m, f k - ∑ h in range n, f h = ∑ h in range (m - n) , f (h + n) := by
induction m generalizing n with
| zero =>
have nz : n = 0 := by linarith
simp [*]
| succ m ih =>
by_cases le : n ≤ m
· rw [minus_succ_eq_succ_minus]
repeat rw [Finset.sum_range_succ]
rw [←ih]
have minsimp : m - n + n = m := by
linarith
rw [minsimp]
have kk : ∑ x in range m, f x = ∑ k in range m, f k := by simp
rw [kk]
have mycomm (a b c : ℝ) : a + b - c = a - c + b := by ring
simp [mycomm]
exact le
exact le
· have eeq : n = m + 1 := by linarith
simp [*]
#### Moritz Firsching (Sep 03 2023 at 18:24):
could be made shorter, but this works:
lemma bigsum_diff (m n : ℕ) (mlen : n ≤ m) (f : ℕ → ℝ) :
∑ k in range m, f k - ∑ h in range n, f h = ∑ h in range (m - n), f (h + n) := by
have := Finset.sum_range_add_sub_sum_range (fun x => f x) n (m - n)
rw [this]
[edited: changed covert to simp only [add_comm]]
#### Adomas Baliuka (Sep 03 2023 at 18:30):
Wow, thanks! It says "expecting term". Is there something missing after convert? Did you know about sum_range_add_sub_sum_range or did you find it by some method I could learn?
#### Yaël Dillies (Sep 03 2023 at 18:33):
Here's a non-cheating proof, using only lemmas simpler than the one you're trying to prove:
import Mathlib.Algebra.BigOperators.Basic
import Mathlib.Data.Real.Basic
open Finset
open scoped BigOperators
lemma bigsum_minus_bigsum (m n : ℕ) (mlen : n ≤ m) (f : ℕ → ℝ) :
∑ k in range m, f k - ∑ h in range n, f h = ∑ h in range (m - n) , f (h + n) := by
rw [←sum_union, union_comm]
· congr
· simp [disjoint_left]
#### Yaël Dillies (Sep 03 2023 at 18:35):
Of course the better way is to use the existing lemma sum_range_add_sub_sum_range , which you could have found by looking for Finsetsumrangerange or Finsetsumrangesubin the docs: you know each part should appear in the name somewhere, and you can try a few permutations out (but typically sum will probably come before range).
#### Moritz Firsching (Sep 03 2023 at 18:39):
ab said:
Wow, thanks! It says "expecting term". Is there something missing after convert? Did you know about sum_range_add_sub_sum_range or did you find it by some method I could learn?
Yes, there was something missing, I changed the original message...
#### Adomas Baliuka (Sep 03 2023 at 19:36):
@Yaël Dillies If you don't mind me asking, how did you think of using these embeddings? Is that a normal technique in some circumstances? It seems very weird and roundabout somehow. You build these embedding objects and then coerce them to functions? (Is there a way to list coercions? I know about #synth but there I need to know input and output type. Is there a way to list instances and filter just by input or output?)
#### Yaël Dillies (Sep 03 2023 at 19:41):
My hand was forced by the way docs#Finset.range_add_eq_union was written. To be quite honest, I would have preferred a lemma of the form range a = range b ∪ (range (b - a)).map (addLeftEmbedding a).
The reason we use those embeddings is because we want to use docs#Finset.map instead of docs#Finset.image. This (somewhat rare pattern) is useful because it lets us carry around the proof of injectivity (or in the case of docs#Finset.sup' nonemptiness, or i ∉ s for docs#Finset.cons, etc...) and spares us a few proof obligations. Typically, I used docs#Finset.sum_map which doesn't require me (re)proving that the function is injective.
#### Moritz Firsching (Sep 03 2023 at 20:19):
lemma bigsum_diff (m n : ℕ) (mlen : n ≤ m) (f : ℕ → ℝ) :
∑ k in range m, f k - ∑ h in range n, f h = ∑ h in range (m - n), f (h + n) := by
have := Finset.sum_range_add_sub_sum_range (fun x => f x) n (m - n)
is better than what I wrote before.
I think @ab is asking about how to go from a name "Finsetsumrangesub" to the name of the theorem and one answer here is to use the search of https://leanprover-community.github.io/mathlib4_docs/. If you already know where to look for approximately, then typing Finset.sum and pressing Ctrl-space will show suggestions.
for the last exact, you can simply type exact? and it should find the (add_sub_of_le mlen).symm part
#### Adomas Baliuka (Sep 04 2023 at 15:11):
@Moritz Firsching Why is this better? It has the same number of lines. I'm trying to understand what kind of proofs and style are considered "good".
#### Ruben Van de Velde (Sep 04 2023 at 15:18):
The have := .. is superfluous in the latest proof
#### Ruben Van de Velde (Sep 04 2023 at 15:22):
I'm not sure I agree it's better, in any case, since the simp only [add_comm] is still somewhat unfortunate. If you want to use convert (which means "I've nearly got the result I need, please show me where it differs from the current goal"), maybe this is nicer:
import Mathlib.Algebra.BigOperators.Basic
import Mathlib.Data.Real.Basic
open BigOperators Finset Nat
lemma bigsum_diff (m n : ℕ) (mlen : n ≤ m) (f : ℕ → ℝ) :
∑ k in range m, f k - ∑ h in range n, f h = ∑ h in range (m - n), f (h + n) := by
convert Finset.sum_range_add_sub_sum_range _ _ _ using 3
#### Ruben Van de Velde (Sep 04 2023 at 15:27):
Or you could do:
import Mathlib.Algebra.BigOperators.Basic
import Mathlib.Data.Real.Basic
open BigOperators Finset Nat
lemma bigsum_diff (m n : ℕ) (mlen : n ≤ m) (f : ℕ → ℝ) :
∑ k in range m, f k - ∑ h in range n, f h = ∑ h in range (m - n), f (h + n) :=
calc
_ = ∑ k in range (n + (m - n)), f k - ∑ k in range n, f k := by rw [add_sub_of_le mlen]
_ = ∑ k in range (m - n), f (n + k) := Finset.sum_range_add_sub_sum_range ..
-- Choose one of the following two lines
-- _ = _ := by simp_rw [add_comm]
-- _ = _ := Finset.sum_congr rfl fun _ _ => by rw [add_comm]
Not sure any of those are necessarily better or worse than the other alternatives
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# Machines X and Y run at different constant rates, and
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Machines X and Y run at different constant rates, and [#permalink] 09 Jul 2008, 05:38
Machines X and Y run at different constant rates, and machine X can complete a certain job in 9 hours. Machine X worked on the job alone for the first 3 hours and the two machines, working together, then completed the job in 4 more hours. How many hours would it have taken machine Y, working alone, to complete the entire job?
(A) 18 (B)13+1/2 (C)7+1/5 (D)4+1/2 (E)3+2/3
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Re: work question [#permalink] 09 Jul 2008, 06:05
X can finish 1/9 of the job in 1 hr.
X works for 3 hrs and finishes 3/9 or 1/3 of the job.
2/3 of the job remains.
Y can finish 1/Y of the job in 1 hr
X and Y finish the remaining 2/3 of the job in 4 hrs.
3/2(1/9 + 1/Y) = 1/4
1/9 + 1/Y = 2/3*1/4
1/9 + 1/Y = 1/6
1/Y = 1/18
This means Y needs 18 hrs to finish the job alone.
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Re: work question [#permalink] 09 Jul 2008, 07:20
The answer is (A) that is 18hrs
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Re: work question [#permalink] 09 Jul 2008, 07:21
2
KUDOS
Here is a non-standard way to look at the work problem:
assign numbers to the "job."
Take this particular question as an example.
Machines X can complete a certain job in 9 hours.
Let the job be: in this case, making 9 pieces of candies
So, Machine X's work rate = 1 candy per hour.
Then the question stated:
Machine X worked on the job alone for the first 3 hours
So, 6 more pieces of candies to go
Next: The two machines, working together, then completed the job in 4 more hours
We know that Machine X can produce 4 pieces of candies in 4 hours
So, Machine Y produces 2 pieces of candies in 4 hours, which is 0.5 piece per hour.
Thus, it will take 18 hours to produce 9 pieces of candies for machine Y alone.
This is what I came up with on the fly during the actual GMAT on a work problem that kinda confused me when I tried with the standard work function way... wondering why I didn't think of this way earlier.....
Nice and easy, with no fraction to confuse you... not to mention that candies are sure yummy
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Re: work question [#permalink] 09 Jul 2008, 13:58
stingraybullray wrote:
X can finish 1/9 of the job in 1 hr.
X works for 3 hrs and finishes 3/9 or 1/3 of the job.
2/3 of the job remains.
Y can finish 1/Y of the job in 1 hr
X and Y finish the remaining 2/3 of the job in 4 hrs.
3/2(1/9 + 1/Y) = 1/4
1/9 + 1/Y = 2/3*1/4
1/9 + 1/Y = 1/6
1/Y = 1/18
This means Y needs 18 hrs to finish the job alone.
could you explain why you flip the 2/3 to get 3/2 ?
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Re: work question [#permalink] 09 Jul 2008, 18:56
I found it easier to set it up like this:
$$4(\frac{1}{9} + \frac{1}{x}) =\frac{6}{9}$$
4 because we're told the two machines work together for 4 hours. 6/9 because that's the same as 2/3 the job remaining as indicated in the stem.
multiply both sides by 1/4 or divide by 4, same operation, different look.
$$\frac{1}{9} + \frac{1}{x} =\frac{6}{9}*\frac{1}{4}$$
$$\frac{1}{9} + \frac{1}{x} =\frac{3}{18}$$
Change 1/9 to 2/18 so i can subtract it from both sides.
$$\frac{2}{18} + \frac{1}{x} =\frac{3}{18}$$
$$\frac{1}{x} =\frac{1}{18}$$
It's easy to see here that x = 18 and with work problems, the denominator is the number of units it takes that worker/machine to complete the portion of the work required.
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Re: work question [#permalink] 10 Jul 2008, 18:13
pmenon wrote:
could you explain why you flip the 2/3 to get 3/2 ?
I used the rate formula RT = W
R = 1/9 + 1/Y; W = 2/3; T = 4
(1/9 + 1/Y)4 = 2/3
3/2(1/9 + 1/Y) = 1/4 -- this is how I got it.
Re: work question [#permalink] 10 Jul 2008, 18:13
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# Machines X and Y run at different constant rates, and
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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®. | 1,949 | 6,110 | {"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 | 4 | CC-MAIN-2015-18 | longest | en | 0.894031 |
https://git.nilu.no/flexpart/flexpart/-/raw/4138764de96fbceafe859f3093cef3641ff9285e/src/centerofmass.f90 | 1,653,608,040,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00595.warc.gz | 337,461,222 | 1,533 | ! SPDX-FileCopyrightText: FLEXPART 1998-2019, see flexpart_license.txt ! SPDX-License-Identifier: GPL-3.0-or-later subroutine centerofmass(xl,yl,n,xcenter,ycenter) ! i i i o o !***************************************************************************** ! * ! This routine calculates the center of mass of n points on the Earth. * ! Input are the longitudes (xl) and latitudes (yl) of the individual * ! points, output is the longitude and latitude of the centre of mass. * ! * ! Author: A. Stohl * ! * ! 24 January 2002 * ! * !***************************************************************************** use par_mod implicit none integer :: n,l real :: xl(n),yl(n),xll,yll,xav,yav,zav,x,y,z,xcenter,ycenter xav=0. yav=0. zav=0. do l=1,n ! Convert longitude and latitude from degrees to radians !******************************************************* xll=xl(l)*pi180 yll=yl(l)*pi180 ! Calculate 3D coordinates from longitude and latitude !***************************************************** x = cos(yll)*sin(xll) y = -1.*cos(yll)*cos(xll) z = sin(yll) ! Find the mean location in Cartesian coordinates !************************************************ xav=xav+x yav=yav+y zav=zav+z end do xav=xav/real(n) yav=yav/real(n) zav=zav/real(n) ! Project the point back onto Earth's surface !******************************************** xcenter=atan2(xav,-1.*yav) ycenter=atan2(zav,sqrt(xav*xav+yav*yav)) ! Convert back to degrees !************************ xcenter=xcenter/pi180 ycenter=ycenter/pi180 end subroutine centerofmass | 377 | 1,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2022-21 | latest | en | 0.35449 |
http://www.engrz.com/2013/12/23/saturated-vapor-and-superheated-vapor/ | 1,513,246,304,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948543611.44/warc/CC-MAIN-20171214093947-20171214113947-00014.warc.gz | 346,773,392 | 12,781 | Home » Mechanical Engineering » Saturated Vapor and Superheated Vapor
# Saturated Vapor and Superheated Vapor
Once boiling starts, the temperature stops rising until the liquid is completely vaporized. That is, the temperature will remain constant during the entire phase-change process if the pressure is held constant. This can easily be verified by placing a thermometer into boiling pure water on top of a stove. At sea level (P 1 atm), the thermometer will always read 100°C if the pan is uncovered or covered with a light lid. During a boiling process, the only change we will observe is a large increase in the volume and a steady decline in the liquid level as a result of more liquid turning to vapor.
Midway about the vaporization line (state 3, Fig. 3–8), the cylinder contains equal amounts of liquid and vapor. As we continue transferring heat, the vaporization process continues until the last drop of liquid is vaporized (state 4, Fig. 3–9). At this point, the entire cylinder is filled with vapor that is on the borderline of the liquid phase. Any heat loss from this vapor will cause some of the vapor to condense (phase change from vapor to liquid). A vapor that is about to condense is called a saturated vapor. Therefore, state 4 is a saturated vapor state. A substance at states between 2 and 4 is referred to as a saturated liquid–vapor mixture since the liquid and vapor phases coexist in equilibrium at these states.
Once the phase-change process is completed, we are back to a single phase region again (this time vapor), and further transfer of heat results in an increase in both the temperature and the specific volume (Fig. 3–10). At state 5, the temperature of the vapor is, let us say, 300°C; and if we transfer some heat from the vapor, the temperature may drop somewhat but no condensation will take place as long as the temperature remains above 100°C (for P = 1 atm). A vapor that is not about to condense (i.e., not a saturated vapor) is called a superheated vapor. Therefore, water at state 5 is a superheated vapor. This constant-pressure phase-change process is illustrated on a T-v diagram in Fig. 3–11.
If the entire process described here is reversed by cooling the water while maintaining the pressure at the same value, the water will go back to state 1, retracing the same path, and in so doing, the amount of heat released willexactly match the amount of heat added during the heating process. In our daily life, water implies liquid water and steam implies water vapor. In thermodynamics, however, both water and steam usually mean only one thing: H2O.
Reference: Thermodynamics – An Engineering Approach
5th Edition
by: Yunus A. Cengel and Michale A. Boles | 608 | 2,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.908379 |
https://crosswordlabs.com/view/shapes-128 | 1,600,827,172,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400209665.4/warc/CC-MAIN-20200923015227-20200923045227-00042.warc.gz | 352,826,889 | 5,007 | ## Shapes
##### Across
1. 3. parallelogram with four congruent sides.
2. 4. a round solid three-dimensional figure with every point on its surface equidistant from its center
3. 7. three-dimensional figure with two congruent circular bases that lie in parallel planes
4. 8. is a three-dimensional figure that has a circular base, a vertex not in the plane of the circle, and a curved lateral surface.
5. 9. the set of all points in a plane that are a given distance, the radius, from a given point
##### Down
1. 1. is a parallelogram with four congruent sides and four right angles.
2. 2. is a three-dimensional al figure whose surfaces, or faces, are polygons.
3. 5. a polyhedron with two congruent and parallel faces, which are called the bases.
4. 6. polyhedron in which one face, the base, is a polygon and the other faces, the lateral faces, are triangles with a common vertex.
5. 10. polyhedron with six faces, each of which is a square | 243 | 943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-40 | latest | en | 0.942698 |
https://mull-overthing.com/how-do-you-describe-the-sampling-distribution-of-the-sample-mean/ | 1,723,170,273,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640751424.48/warc/CC-MAIN-20240809013306-20240809043306-00180.warc.gz | 320,992,796 | 9,081 | MullOverThing
Useful tips for everyday
# How do you describe the sampling distribution of the sample mean?
## How do you describe the sampling distribution of the sample mean?
If the population is normal to begin with then the sample mean also has a normal distribution, regardless of the sample size. For samples of any size drawn from a normally distributed population, the sample mean is normally distributed, with mean μX=μ and standard deviation σX=σ/√n, where n is the sample size.
How is the bias of a sampling distribution measured?
note: The bias of a sampling distribution is measured by computing the distance between the center of the sampling distribution and the population parameter. The precision of an estimator does not depend on the size of the population.
Is the sampling distribution of the sample mean normal?
Now we may invoke the Central Limit Theorem: even though the distribution of household size X is skewed, the distribution of sample mean household size (x-bar) is approximately normal for a large sample size such as 100.
### Which is an example of the sample mean X-bar?
We are now moving on to explore the behavior of the statistic x-bar, the sample mean, relative to the parameter μ (mu), the population mean (when the variable of interest is quantitative). Let’s begin with an example. Birth weights are recorded for all babies in a town. The mean birth weight is 3,500 grams, µ = mu = 3,500 g.
What is the behavior of the statistic P-hat?
So far, we’ve discussed the behavior of the statistic p-hat, the sample proportion, relative to the parameter p, the population proportion (when the variable of interest is categorical).
Is the spread of sample mean related to sample size?
As for the spread of all sample means, theory dictates the behavior much more precisely than saying that there is less spread for larger samples. In fact, the standard deviation of all sample means is directly related to the sample size, n as indicated below. | 410 | 1,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-33 | latest | en | 0.918786 |
https://webot.org/info/en/?search=Block_and_tackle | 1,660,770,773,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573104.24/warc/CC-MAIN-20220817183340-20220817213340-00339.warc.gz | 535,612,238 | 17,422 | # Block and tackle Information
https://en.wikipedia.org/wiki/Block_and_tackle
A block and tackle [1] [2] or only tackle [3] is a system of two or more pulleys with a rope or cable threaded between them, usually used to lift heavy loads.
The pulleys are assembled to form blocks and then blocks are paired so that one is fixed and one moves with the load. The rope is threaded through the pulleys to provide mechanical advantage that amplifies the force applied to the rope. [4]
Hero of Alexandria described cranes formed from assemblies of pulleys in the first century. Illustrated versions of Hero's Mechanica (a book on raising heavy weights) show early block and tackle systems. [5]
## Overview
Various ways of rigging a tackle. All these are "rove to disadvantage" [6] (see below).
A block is a set of pulleys or sheaves mounted on a single frame. An assembly of blocks with a rope threaded through the pulleys is called tackle. The process of threading ropes or cables through blocks is called "reeving", and a threaded block and tackle is said to have been "rove". [7] A block and tackle system amplifies the tension force in the rope to lift heavy loads. They are common on boats and sailing ships, where tasks are often performed manually, as well as on cranes and drilling rigs, where once rove, the tasks are performed by heavy equipment.
In the diagram shown here, the number of rope sections of the tackles shown is as follows:
• Gun tackle: 2
• Luff tackle: 3
• Double tackle: 4
• Gyn tackle: 5
• Threefold purchase: 6
Note that the gun tackle, double tackle and threefold purchase all have the same number of pulleys in both blocks (one, two and three, respectively) whereas the Luff tackle and Gyn tackle have mis-matched blocks with differing numbers of pulleys.
A gun tackle has a single pulley in both the fixed and moving blocks with 2 rope parts (n) supporting the load (FB) of 100 N. The mechanical advantage is 2, requiring a force of only 50 N to lift the load.
A block and tackle is characterized by the use of a single continuous rope to transmit a tension force around one or more pulleys to lift or move a load. Its mechanical advantage is the number of parts of the rope that act on the load. The mechanical advantage of a tackle dictates how much easier it is to haul or lift the load.
If frictional losses are neglected, the mechanical advantage of a block and tackle is equal to the number of parts in the line that either attach to or run through the moving blocks—in other words, the number of supporting rope sections.
An ideal block and tackle with a moving block supported by n rope sections has the mechanical advantage (MA),
${\displaystyle MA={\frac {F_{B}}{F_{A}}}=n,\!}$
where FA is the hauling (or input) force and FB is the load.
Consider the set of pulleys that form the moving block and the parts of the rope that support this block. If there are n of these parts of the rope supporting the load FB, then a force balance on the moving block shows that the tension in each of the parts of the rope must be FB/n. This means the input force on the rope is FA=FB/n. Thus, the block and tackle reduces the input force by the factor n.
A double tackle has two pulleys in both the fixed and moving blocks with four rope parts (n) supporting the load (FB) of 100 N. The mechanical advantage is 4, requiring a force of only 25 N to lift the load.
Ideal mechanical advantage correlates directly with velocity ratio. The velocity ratio of a tackle is the ratio between the velocity of the hauling line to that of the hauled load. A line with a mechanical advantage of 4 has a velocity ratio of 4:1. In other words, to raise a load at 1 metre per second, the hauling part of the rope must be pulled at 4 metres per second. Therefore, the mechanical advantage of a double tackle is 4.
The mechanical advantage of any tackle can be increased by interchanging the fixed and moving blocks so the rope is attached to the moving block and the rope is pulled in the direction of the lifted load. In this case the block and tackle is said to be "rove to advantage." [8]
• "Rove to advantage" – where the pull on the rope is in the same direction as that in which the load is to be moved. The hauling part is pulled from the moving block. [6]
• "Rove to disadvantage" – where the pull on the rope is in the opposite direction to that in which the load is to be moved. The hauling part is pulled from the fixed block. [6]
Diagram 3 shows three rope parts supporting the load W, which means the tension in the rope is W/3. Thus, the mechanical advantage is three-to-one.
By adding a pulley to the fixed block of a gun tackle the direction of the pulling force is reversed though the mechanical advantage remains the same, Diagram 3a. This is an example of the Luff tackle.
The decision of which to use depends on pragmatic considerations for the total ergonomics of working with a particular situation. Reeving to advantage is the most efficient use of equipment and resources. For example, if the load is to be hauled parallel to the ground, reeving to advantage enables the pulling force to be in the direction of the load movement, allowing obstacles to be managed more easily.
Reeving to disadvantage adds an extra sheave to change the direction of the pulling line to a potentially more ergonomic direction, which increases friction losses without improving the velocity ratio. Situations in which reeving to disadvantage may be more desirable include lifting from a fixed point overhead--the additional pulley allows pulling downwards instead of upwards so that the weight of the lifter can offset the weight of the load, or allows pulling sideways, enabling multiple lifters to combine effort.
### Friction
Wooden block on a sailing ship.
The formula used to find the effort required to raise a given weight using a block and fall:
${\displaystyle F_{a}={\frac {L}{N}}{\frac {1}{\textit {eff}}}}$
where ${\displaystyle F_{a}}$ is the force applied to the hauling part of the line (the input force), ${\displaystyle L}$ is the weight of the load (the output force), ${\displaystyle N}$ is the ideal mechanical advantage of the system (which is the same as the number of segments of line extending from the moving block), and ${\displaystyle {\textit {eff}}}$ is the mechanical efficiency of the system (equal to one for an ideal frictionless system; a fraction less than one for real-world systems with energy losses due to friction and other causes). If ${\displaystyle S}$ is the number of sheaves in the purchase, and there is a roughly ${\displaystyle x}$% loss of efficiency at each sheave due to friction, then: [9] [10]
${\displaystyle {\frac {1}{\textit {eff}}}\approx 1+S{\frac {x}{100}}.}$
This approximation is more accurate for smaller values of ${\displaystyle S}$ and ${\displaystyle x}$. [10] A more precise estimate of efficiency is possible by use of the sheave friction factor, ${\displaystyle K}$ (which may be obtainable from the manufacturer or published tables [11]). The relevant equation is: [11]
${\displaystyle {\textit {eff}}={\frac {K^{N}-1}{K^{S}N(K-1)}}.}$
Typical ${\displaystyle K}$ values are 1.04 for roller bearing sheaves and 1.09 for plain bearing sheaves (with wire rope). [11]
The increased force produced by a tackle is offset by both the increased length of rope needed and the friction in the system. In order to raise a block and tackle with a mechanical advantage of 6 a distance of 1 metre, it is necessary to pull 6 metres of rope through the blocks. Frictional losses also mean there is a practical point at which the benefit of adding a further sheave is offset by the incremental increase in friction which would require additional force to be applied in order to lift the load. Too much friction may result in the tackle not allowing the load to be released easily, [notes 1] or by the reduction in force needed to move the load being judged insufficient because undue friction has to be overcome as well.
## Mid-line attachment
When installing a block on an existing line, it is often inconvenient at best to thread the rope through the block to be added.
• Open blocks have a space wide enough between the fixed cheeks to be able to slide the pulley over the rope. These can be extremely small and light while retaining significant strength due to the lack of moving parts.[ citation needed]
• A swing cheek block is a special kind of block which can be opened to engage with a bight, without the necessity to thread the rope through the block or remove the load from the end of the rope. The snatch block is also the load lifting pulley in certain arrangements, such as during use in a recovery operation. [12]
Three snatch blocks showing the ‘snatch’ in the closed, opened, and securely moused
[ citation needed]
Swing-Cheek blocks may be roughly divided into two categories:[ citation needed]
• Swing cheek pulleys: used for light loads or redirection of forces, usually with a single pulley wheel (though multiple sheaves/cheeks are not uncommon) and an attachment point (or several) for a carabiner or sling. The cheeks are not fixed or locked in position aside from the device used to secure them to the load or rigging point.
Examples of use (in an arborocultural setting) include: tail minding/tending, and for setting a rigging point in the tree above the cut to take place - a positive rigging situation.
• Snatch or Impact Blocks: used for heavier loads and more dynamic rigging, the cheeks of these blocks are fixed in place with a pin which locks into the opposite cheek. This pin may form part of the axle for a second pulley, which is secured to the load or rigging point with a soft sling, rather than a solid device such as a shackle. This allows for more even distribution of forces to the faces where the forces will be applied, as opposed to a carabiner or shackle, where the forces are applied more strongly to corners and edges, increasing risk of deformation or damage.
Examples of use (again, in relation to tree care) may include setting a block below the current cut, resulting in a 'negative' rigging situation, in which shock loads can be significant - especially if removing large sections of vertical stem.
## Literature
• M. Oppolzer, T. Wahls: I Like To Move It. Flaschenzüge in der Seiltechnik. Hamburg 2019, ISBN 978-3-9820618-0-1.
• Rescue Technician: Operational Readiness for Rescue Providers. St. Louis, Missouri 1998, ISBN 0-8151-8390-9.
## Notes
1. ^ Friction may mean that the rope in a tackle "bunches" and jams when the force is released if the tackle has too much friction for the load to balance, or that the tackle does not "lower" the load
## References
1. ^ "Tackle" can be in this usage.
2. ^ "Royal Canadian Sea Cadets - Master Lesson Plan - Level T\\\2" (PDF). Archived from the original (PDF) on 2013-03-09. Retrieved 2009-12-27.
3. ^
4. ^ Ned Pelger, ConstructionKnowledge.net
5. ^ A. P. Usher, A History of Mechanical Inventions, Harvard University Press, 1929 (Dover Publications, revised edition, 2011, ISBN 978-0486255934)
6. ^ a b c MacDonald, Joseph A (14 January 2009). Handbook of Rigging: For Construction and Industrial Operations. McGraw-Hill Professional. p. 376. ISBN 978-0-07-149301-7. Tackle may be rigged to advantage - where the pull on the rope is in the same direction as that in which the load is to be moved; or it may be rigged to disadvantage - where the pull on the rope is in the opposite direction of that in which the load is to be moved
7. ^
8. ^ sccheadquarters.com seamanship reference Archived November 11, 2011, at the Wayback Machine
9. ^ Notes on cargo work: Kemp and Young. 3rd Edition. ISBN 0-85309-040-8 Page 4
10. ^ a b Glerum, Jay O. (2007-04-03). Stage rigging handbook (3rd ed.). Southern Illinois University Press. pp. 52–54 (320 pages total). ISBN 978-0-8093-2741-6.
11. ^ a b c
12. ^ Mathews, Lisa (2016-02-02). "What is a Snatch Block?". US Cargo Control Blog. Archived from the original on 2019-12-19. Retrieved 2019-12-19. | 2,861 | 12,057 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.958201 |
https://monashdatafluency.github.io/python-workshop-base/modules/defensive_programming/ | 1,656,872,158,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104248623.69/warc/CC-MAIN-20220703164826-20220703194826-00657.warc.gz | 458,508,974 | 9,014 | ## Defensive Programming
Our previous lessons have introduced the basic tools of programming: variables and lists, file operations, data visualisation, loops, conditionals, and functions. What they haven’t done is show us how to tell whether a program is getting the right answer, and how to tell if it’s still getting the right answer as we make changes to it.
To achieve that, we need to:
• Write programs that check their own operation.
• Write and run tests for widely-used functions.
• Make sure we know what “correct” actually means.
The good news is, doing these things will speed up our programming, not slow it down. As in real carpentry — the kind done with lumber — the time saved by measuring carefully before cutting a piece of wood is much greater than the time that measuring takes.
## Assertions
The first step toward getting the right answers from our programs is to assume that mistakes will happen and to guard against them. This is called defensive programming, and the most common way to do it is to add assertions to our code so that it checks itself as it runs. An assertion is simply a statement that something must be true at a certain point in a program. When Python sees one, it evaluates the assertion’s condition. If it’s true, Python does nothing, but if it’s false, Python halts the program immediately and prints the error message if one is provided. For example, this piece of code halts as soon as the loop encounters a value that isn’t positive:
``````numbers = [1.5, 2.3, 0.7, -0.001, 4.4]
total = 0.0
for n in numbers:
assert n > 0.0, 'Data should only contain positive values'
total += n
print('total is:', total)
``````
``````---------------------------------------------------------------------------
AssertionError Traceback (most recent call last)
<ipython-input-1-091518d2f2e2> in <module>()
3 total = 0.0
4 for n in numbers:
----> 5 assert n > 0.0, 'Data should only contain positive values'
6 total += n
7 print('total is:', total)
AssertionError: Data should only contain positive values
``````
Programs like the Firefox browser are full of assertions: 10-20% of the code they contain are there to check that the other 80–90% are working correctly. Broadly speaking, assertions fall into three categories:
A `precondition` is something that must be true at the start of a function in order for it to work correctly.
A `postcondition` is something that the function guarantees is true when it finishes.
An `invariant` is something that is always true at a particular point inside a piece of code.
For example, suppose we are representing rectangles using a `tuple` of four coordinates `(x0, y0, x1, y1)`, representing the lower left and upper right corners of the rectangle. In order to do some calculations, we need to normalize the rectangle so that the lower left corner is at the origin and the longest side is 1.0 units long. This function does that, but checks that its input is correctly formatted and that its result makes sense:
``````def normalize_rectangle(rect):
'''Normalizes a rectangle so that it is at the origin and 1.0 units long on its longest axis.
Input should be of the format (x0, y0, x1, y1).
(x0, y0) and (x1, y1) define the lower left and upper right corners
of the rectangle, respectively.'''
assert len(rect) == 4, 'Rectangles must contain 4 coordinates'
x0, y0, x1, y1 = rect
assert x0 < x1, 'Invalid X coordinates'
assert y0 < y1, 'Invalid Y coordinates'
dx = x1 - x0
dy = y1 - y0
if dx > dy:
scaled = float(dx) / dy
upper_x, upper_y = 1.0, scaled
else:
scaled = float(dx) / dy
upper_x, upper_y = scaled, 1.0
assert 0 < upper_x <= 1.0, 'Calculated upper X coordinate invalid'
assert 0 < upper_y <= 1.0, 'Calculated upper Y coordinate invalid'
return (0, 0, upper_x, upper_y)
``````
The preconditions on lines 3, 5, and 6 catch invalid inputs:
``````print(normalize_rectangle( (0.0, 1.0, 2.0) )) # missing the fourth coordinate
``````
``````---------------------------------------------------------------------------
AssertionError Traceback (most recent call last)
<ipython-input-3-1b9cd8e18a1f> in <module>()
----> 1 print(normalize_rectangle( (0.0, 1.0, 2.0) )) # missing the fourth coordinate
<ipython-input-2-c94cf5b065b9> in normalize_rectangle(rect)
4 (x0, y0) and (x1, y1) define the lower left and upper right corners
5 of the rectangle, respectively.'''
----> 6 assert len(rect) == 4, 'Rectangles must contain 4 coordinates'
7 x0, y0, x1, y1 = rect
8 assert x0 < x1, 'Invalid X coordinates'
AssertionError: Rectangles must contain 4 coordinates
``````
``````print(normalize_rectangle( (4.0, 2.0, 1.0, 5.0) )) # X axis inverted
``````
``````---------------------------------------------------------------------------
AssertionError Traceback (most recent call last)
<ipython-input-4-325036405532> in <module>()
----> 1 print(normalize_rectangle( (4.0, 2.0, 1.0, 5.0) )) # X axis inverted
<ipython-input-2-c94cf5b065b9> in normalize_rectangle(rect)
6 assert len(rect) == 4, 'Rectangles must contain 4 coordinates'
7 x0, y0, x1, y1 = rect
----> 8 assert x0 < x1, 'Invalid X coordinates'
9 assert y0 < y1, 'Invalid Y coordinates'
10
AssertionError: Invalid X coordinates
``````
The post-conditions on lines 17 and 18 help us catch bugs by telling us when our calculations cannot have been correct. For example, if we normalize a rectangle that is taller than it is wide everything seems OK:
``````print(normalize_rectangle( (0.0, 0.0, 1.0, 5.0) ))
``````
```output
```
(0, 0, 0.2, 1.0)
```
```
but if we normalize one that’s wider than it is tall, the assertion is triggered:
``````print(normalize_rectangle( (0.0, 0.0, 5.0, 1.0) ))
``````
``````---------------------------------------------------------------------------
AssertionError Traceback (most recent call last)
<ipython-input-6-8d4a48f1d068> in <module>()
----> 1 print(normalize_rectangle( (0.0, 0.0, 5.0, 1.0) ))
<ipython-input-2-c94cf5b065b9> in normalize_rectangle(rect)
19
20 assert 0 < upper_x <= 1.0, 'Calculated upper X coordinate invalid'
---> 21 assert 0 < upper_y <= 1.0, 'Calculated upper Y coordinate invalid'
22
23 return (0, 0, upper_x, upper_y)
AssertionError: Calculated upper Y coordinate invalid
``````
Re-reading our function, we realize that line 11 should divide `dy` by `dx` rather than `dx` by `dy`. If we had left out the assertion at the end of the function, we would have created and returned something that had the right shape as a valid answer, but wasn’t. Detecting and debugging that would almost certainly have taken more time in the long run than writing the assertion.
But assertions aren’t just about catching errors: they also help people understand programs. Each assertion gives the person reading the program a chance to check (consciously or otherwise) that their understanding matches what the code is doing.
Most good programmers follow two rules when adding assertions to their code. The first is, fail early, fail often. The greater the distance between when and where an error occurs and when it’s noticed, the harder the error will be to debug, so good code catches mistakes as early as possible.
The second rule is, turn bugs into assertions or tests. Whenever you fix a bug, write an assertion that catches the mistake should you make it again. If you made a mistake in a piece of code, the odds are good that you have made other mistakes nearby, or will make the same mistake (or a related one) the next time you change it. Writing assertions to check that you haven’t regressed (i.e., haven’t re-introduced an old problem) can save a lot of time in the long run, and helps to warn people who are reading the code (including your future self) that this bit is tricky.
### Test-Driven Development
An assertion checks that something is true at a particular point in the program. The next step is to check the overall behavior of a piece of code, i.e., to make sure that it produces the right output when it’s given a particular input. For example, suppose we need to find where two or more time series overlap. The range of each time series is represented as a pair of numbers, which are the time the interval started and ended. The output is the largest range that they all include:
Most novice programmers would solve this problem like this:
1. Write a function `range_overlap`.
2. Call it interactively on two or three different inputs.
3. If it produces the wrong answer, fix the function and re-run that test.
This clearly works — after all, thousands of scientists are doing it right now — but there’s a better way:
1. Write a short function for each test.
2. Write a `range_overlap` function that should pass those tests.
3. If `range_overlap` produces any wrong answers, fix it and re-run the test functions.
Writing the tests before writing the function they exercise is called `test-driven development` (TDD). Its advocates believe it produces better code faster because:
1. If people write tests after writing the thing to be tested, they are subject to confirmation bias, i.e., they subconsciously write tests to show that their code is correct, rather than to find errors.
2. Writing tests helps programmers figure out what the function is actually supposed to do.
Here are three test functions for `range_overlap`:
``````assert range_overlap([ (0.0, 1.0) ]) == (0.0, 1.0)
assert range_overlap([ (2.0, 3.0), (2.0, 4.0) ]) == (2.0, 3.0)
assert range_overlap([ (0.0, 1.0), (0.0, 2.0), (-1.0, 1.0) ]) == (0.0, 1.0)
``````
``````---------------------------------------------------------------------------
NameError Traceback (most recent call last)
<ipython-input-9-dc16b942c085> in <module>()
----> 1 assert range_overlap([ (0.0, 1.0) ]) == (0.0, 1.0)
2 assert range_overlap([ (2.0, 3.0), (2.0, 4.0) ]) == (2.0, 3.0)
3 assert range_overlap([ (0.0, 1.0), (0.0, 2.0), (-1.0, 1.0) ]) == (0.0, 1.0)
NameError: name 'range_overlap' is not defined
``````
The error is actually reassuring: we haven’t written `range_overlap` yet, so if the tests passed, it would be a sign that someone else had and that we were accidentally using their function.
And as a bonus of writing these tests, we’ve implicitly defined what our input and output look like: we expect a list of pairs as input, and produce a single pair as output.
Something important is missing, though. We don’t have any tests for the case where the ranges don’t overlap at all:
``````assert range_overlap([ (0.0, 1.0), (5.0, 6.0) ]) == ???
``````
What should `range_overlap` do in this case: fail with an error message, produce a special value like `(0.0, 0.0)` to signal that there’s no overlap, or something else? Any actual implementation of the function will do one of these things; writing the tests first helps us figure out which is best before we’re emotionally invested in whatever we happened to write before we realized there was an issue.
``````assert range_overlap([ (0.0, 1.0), (1.0, 2.0) ]) == ???
``````
Do two segments that touch at their endpoints overlap or not? Mathematicians usually say “yes”, but engineers usually say “no”. The best answer is “whatever is most useful in the rest of our program”, but again, any actual implementation of `range_overlap` is going to do something, and whatever it is ought to be consistent with what it does when there’s no overlap at all.
Since we’re planning to use the range this function returns as the X axis in a time series chart, we decide that:
1. every overlap has to have non-zero width, and
2. we will return the special value None when there’s no overlap.
`None` is built into Python, and means “nothing here”. (Other languages often call the equivalent value `null` or `nil`). With that decision made, we can finish writing our last two tests:
``````assert range_overlap([ (0.0, 1.0), (5.0, 6.0) ]) == None
assert range_overlap([ (0.0, 1.0), (1.0, 2.0) ]) == None
``````
``````---------------------------------------------------------------------------
NameError Traceback (most recent call last)
<ipython-input-13-42de7ddfb428> in <module>()
----> 1 assert range_overlap([ (0.0, 1.0), (5.0, 6.0) ]) == None
2 assert range_overlap([ (0.0, 1.0), (1.0, 2.0) ]) == None
NameError: name 'range_overlap' is not defined
``````
Again, we get an error because we haven’t written our function, but we’re now ready to do so:
``````def range_overlap(ranges):
'''Return common overlap among a set of [low, high] ranges.'''
lowest = 0.0
highest = 1.0
for (low, high) in ranges:
lowest = max(lowest, low)
highest = min(highest, high)
return (lowest, highest)
``````
(Take a moment to think about why we use `max` to raise `lowest` and `min` to lower `highest`). We’d now like to re-run our tests, but they’re scattered across three different cells. To make running them easier, let’s put them all in a function:
``````def test_range_overlap():
assert range_overlap([ (0.0, 1.0), (5.0, 6.0) ]) == None
assert range_overlap([ (0.0, 1.0), (1.0, 2.0) ]) == None
assert range_overlap([ (0.0, 1.0) ]) == (0.0, 1.0)
assert range_overlap([ (2.0, 3.0), (2.0, 4.0) ]) == (2.0, 3.0)
assert range_overlap([ (0.0, 1.0), (0.0, 2.0), (-1.0, 1.0) ]) == (0.0, 1.0)
``````
We can now test `range_overlap` with a single function call:
``````test_range_overlap()
``````
``````---------------------------------------------------------------------------
AssertionError Traceback (most recent call last)
<ipython-input-16-80290759369d> in <module>()
----> 1 test_range_overlap()
1 def test_range_overlap():
----> 2 assert range_overlap([ (0.0, 1.0), (5.0, 6.0) ]) == None
3 assert range_overlap([ (0.0, 1.0), (1.0, 2.0) ]) == None
4 assert range_overlap([ (0.0, 1.0) ]) == (0.0, 1.0)
5 assert range_overlap([ (2.0, 3.0), (2.0, 4.0) ]) == (2.0, 3.0)
AssertionError:
``````
The first test that was supposed to produce `None` fails, so we know something is wrong with our function. We don’t know whether the other tests passed or failed because Python halted the program as soon as it spotted the first error. Still, some information is better than none, and if we trace the behavior of the function with that input, we realize that we’re initializing `lowest` and `highest` to 0.0 and 1.0 respectively, regardless of the input values. This violates another important rule of programming: always initialize from data.
Fix `range_overlap`. Re-run `test_range_overlap` after each change you make.
## Key points
• Program defensively, i.e., assume that errors are going to arise, and write code to detect them when they do.
• Put assertions in programs to check their state as they run, and to help readers understand how those programs are supposed to work.
• Use preconditions to check that the inputs to a function are safe to use.
• Use postconditions to check that the output from a function is safe to use.
• Write tests before writing code in order to help determine exactly what that code is supposed to do.
``````
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# Differences between a stationary wave and a progressive wave?
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1. State two differences between a stationary wave and a progressive wave in terms of the vibration of the particles.
What exactly is it asking for, I just know that a stationary wave is formed by two progressive wave passing through each other.
2. So it's essentially the difference between a transverse wave and a standing wave.
Transverse waves oscillate perpendicular to the direction of their motion. Hmm..
Standing waves do not transfer energy and they have fixed nodes and anti-nodes so particles only vibrate up and down between the nodes (Half a wavelength).
3. Progressive wave- A wave that propagates transferring energy. Progressive wave can be either a transverse wave or longitudinal disturbance that carries energy Transverse wave are the vibration that travel perpendicular to the direction of propagation. For example if i move a string up and down (direction of propagation) the vibration are going to travel perpendicular to up and down (straight ahead). Longitudinal waves are vibration that travel in the direction of propagation (google it). Finally a standing wave is a wave that does not carry energy and my guide defines it as a progressive wave that is reflected off a boundary (now remember these do no transfer energy so this definition is dodgy.
EDIT: oh so imagine it like this you have a string but one end is fixed, if i wobble the string up and down i am creating a progressive wave but that wave as it move towards the end of the string ... is reflected so the superimpose in some way or another. This is what you are asking to define
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• Level: GCSE
• Subject: Maths
• Word count: 3124
# Number Grid Product Differences Investigation
Extracts from this document...
Introduction
Maths Coursework
## Number Grid Product Differences
This assignment is concerned with product differences in different size matrices for different sized grids.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Here is a 10x10 grid. We are trying to investigate product differences within the 10x10 grid. As you can see I have drawn a 2x2 grid inside the 10x10 matrix. To find out the product difference I do:-
The top left number x the bottom right – the top right x the bottom right.
The product difference I found was 10. This was anywhere on the grid where a 2x2 grid could be done. I can prove this by picking a 2x2 grid and using this clue to work out the product difference. Below is the result of me picking a 2x2 grid from this 10x10 grid and finding out what the product difference was.
48 49 58 59
I did (48x59) – (49x58) = 10. This shows that the product difference is the same anywhere in the grid.
Here are 2 more examples of this that prove that this works anywhere on the grid.
12 13 22 23
I did (12x23) – (13x22) = 10. This shows that the product difference is the same anywhere in the grid.
72 73 82 83
I did (72x83) – (73x82) = 10. This shows that the product difference is the same anywhere in the grid.
Middle
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
The above is a 10x10 matrix with 5x5 matrices in it.
I took the same calculation to work this out which was: -
The top left x the bottom right – the top right x the bottom left.
This is an example from the grid
33 34 35 36 37 43 44 45 46 47 53 54 55 56 57 63 64 65 66 67 73 74 75 76 77
I did the calculation of (33x77) – (37x73) = 160. I had figured out that the product difference was 160 in 5x5 matrices in a 10x10 matrix.
I will do another 2 examples to prove this
23 24 25 26 27 33 34 35 36 37 43 44 45 46 47 53 54 55 56 57 63 64 65 66 67
I did the calculation of (23x67) – (27x63) = 160.
53 54 55 56 57 63 64 65 66 67 73 74 75 76 77 83 84 85 86 87 93 94 95 96 97
I did the calculation of (53x97) – (57x93) = 160.
The examples prove that it works anywhere on the grid.
6x6
I did exactly the same as I have done several times.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
The above is a 10x10 matrix with 6x6 matrices in it.
I took the same calculation to work this out which was: -
The top left x the bottom right – the top right x the bottom left.
This is an example from the grid
35 36 37 38 39 40 45 46 47 48 49 50 55 56 57 58 59 60 65 66 67 68 69 70 75 76 77 78 79 80 85 86 87 88 89 90
I did the calculation of (35x90) – (40x85) = 250. This was the product difference for 6x6 matrices in a 10x10 matrix.
I did 2 more examples to prove this.
25 26 27 28 29 30 35 36 37 38 39 40 45 46 47 48 49 50 55 56 57 58 59 60 65 66 67 68 69 70 75 76 77 78 79 80
I did the calculation of (25x80) – (30x75) = 250. This was the product difference for 6x6 matrices in a 10x10 matrix.
43 44 45 46 47 48 53 54 55 56 57 58 63 64 65 66 67 68 73 74 75 76 77 78 83 84 85 86 87 88 93 94 95 96 97 98
I did the calculation of (43x98) – (48x93) = 250.
This proves that it works anywhere on the grid.
I then decided to see if I could find a formula.
N.b The Nth term is 1 less than the size of the squares
Nth Term 1 2 3 4 5 Sequence 10 40 90 160 250 1st Difference 2nd Difference 10n2 10 40 90 160 250
The second difference is halved and put into the formula. This is because
Now if I wanted to work out the product difference of a 7x7 matrices in a 10x10 matrix I would use the formula of 10n2.
So my calculation would be 10x62which would = 10x 36 which would be 360. And if I know work out the long way this is what the answer would be.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
The above is a 10x10 matrix with 7x7 matrices in it.
I took the same calculation to work this out which was: -
The top left x the bottom right – the top right x the bottom left.
This is an example from the grid
34 35 36 37 38 39 40 44 45 46 47 48 49 50 54 55 56 57 58 59 60 64 65 66 67 68 69 70 74 75 76 77 78 79 80 84 85 86 87 88 89 90 94 95 96 97 98 99 100
Conclusion
The above concludes that the formula 10n2 works in any matrix.
Then I decided to try if could find a rule applied which would work for 3x2 matrices in a 10x10 matrix.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Here is a 10x10 grid. We are trying to investigate product differences within the 10x10 grid. As you can see I have drawn a 3x2 grid inside the 10x10 matrix. To find out the product differences I do: -
The top left number x the bottom right – the top right x the bottom right.
Below is an example to work out
27 28 29 37 38 39
I did the calculation of (27x39) – (29x37) = 20
I tried this 2 more times to see if we had the product.
77 78 79 87 88 89
I did the calculation of (77x89) – (79x87) = 20
44 45 46 54 55 56
I did the calculation of (77x89) – (79x87) = 20
This proved that the formula worked.
Nth Term 1 2 3 4 5 Sequence 20 60 120 200 300 1st Difference 2nd Difference 10(L-1)(W-1) 20 60 120 200 300
N.b The Nth term is 1 less than the size of the squares
I have done this to help me work out the formula.
With the information above put into a table I figured out the general formula using my knowledge gained from the previous work that the general formula is 10(L-1)(W-1). I came to this by firstly using the 2nd product difference and halving it this is because
This would for different sized matrices in different sized matrix. Examples of this would be like a 3x2 in an 8x8 grid etc.
This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.
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# Related GCSE Number Stairs, Grids and Sequences essays
1. ## Number Grid Coursework
- a(a + 10[q - 1] + [p - 1]) (12 + 3)(12 + 40) - 12(12 + 40 + 3) 15 x 52 - 12 x 55 780 - 660 120 (N.B. also = 10 x 3 x 4 = 10[p - 1][q - 1])
2. ## Number Grid Investigation.
( s-1)� Summary With the information I have gathered; tables of results, algebra and sequences have provided me with a broad range of results, enabling me to have a more definitive view of the initial investigation. From the worked examples I have observed that: 10x10 grid: * The smallest square
1. ## Number Grids Investigation Coursework
+ (n - 1) = a + mw - w = a + mw - w + n - 1 If I use these expressions in my original calculation for the difference between the products of the opposite corners, I can prove my formula: (top right x bottom left)
2. ## Algebra Investigation - Grid Square and Cube Relationships
= n+ (Height (h) - 1) x 10 n ~ n+w-1 ~ ~ ~ n+20 ~ n+20+w-1 The above grid simplifies to form: n ~ n+w-1 ~ ~ ~ n+20 ~ n+19+w Stage A: Top left number x Bottom right number = n(n+19+w)
1. ## Number Grid Investigation.
Below are a few examples of 3 X 2 squares taken from a 10 wide grid. I am still using the formula: (TL X BR) - (TR X BL) Example 1. 36 37 38 46 47 48 (36 X 48) - (38 X 46) = 20. Product difference = 20.
2. ## What the 'L' - L shape investigation.
By looking at all of the different grid sizes and their formulae. I can note that they all start with 5L; therefore, my final formula must consist of 5L. Grid Size Final Part Of The Formula Difference 4 by 4 -9 5 by 5 -12 6 by 6 -15 7
1. ## Investigation of diagonal difference.
51 52 53 54 55 56 57 58 59 60 61 62 63 64 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
2. ## Maths - number grid
hope after this I will be able to establish some type of pattern to help me reach my aim of finding an overall formula for any rectangle that could be investigated. 30x55 - 25x60 1650 -1500 Difference =150 66x91 - 61x96 6006 - 5856 Difference = 150 By looking at
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https://www.24houranswers.com/college-homework-library/Business/Economics/41244 | 1,618,431,658,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00284.warc.gz | 711,612,672 | 17,713 | # Model Applications and Economics
## Question
1. A zero-coupon bond is one where the issuer pays no coupons, but pays a single cash-flow at the maturity of the bond. Assume you have 10-year zero-coupon bond that pays \$1,000 at maturity. Assume that the nominal rate of the bond over the next 10 years is constant at 5%, while the inflation rate is 2%.
a) What is the current value of the bond using the nominal interest rate?
b) What is the current value of the bond using the real interest rate as estimated using the Fisher relation approximation?
2. Assume you just deposited \$1,000 into a bank account. The current real interest rate is 2% and inflation is expected to be 6% over the next year.
a) What nominal interest rate would you require from the bank over the next year?
b) How much money will you have at the end of one year?
c) If you are saving to buy a stereo that currently sells for \$1,050, and assuming that the price of the stereo increases at the rate of inflation, will you have enough to buy it?
3. Suppose you are given the utility function U(C, l) = lnC + 2l, where C = consumption and l = leisure, and the production function Y = zK⁰.⁴N⁰.⁶.
Assume that all output is consumed (Y = C), and that hours of leisure/work are normalized as representing fractions of a day (or equivalently, a value of 1 represents 24 hours).
Suppose z = 1 and K = 20.
a) What are the Pareto-optimal hours of work?
b) What are the Pareto-optimal hours of leisure?
c) What is the Pareto-optimal level of output?
4. Given the efficiency wage function:
e(w) = w⁴ – 18w³ + 96w²
What is the optimal wage that the firm should set?
5. Given the production function, Y = zF(K, N) = zKαNβ, what conditions on α and/or β are necessary to achieve the multiple equilibria result as implied by the Keynesian Coordination Model?
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https://www.experts-exchange.com/questions/27870205/How-can-I-add-the-value-in-a-cell-to-a-Sumproduct-formula-in-vba.html | 1,527,027,949,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864968.11/warc/CC-MAIN-20180522205620-20180522225620-00077.warc.gz | 751,098,850 | 19,317 | How can I add the value in a cell to a Sumproduct formula in vba.
How do I add the value of a cell located at RC[-3] on the left of the formula below in the same sheet.
ActiveCell.Offset(0, 1).Range("A1").Select
ActiveCell.FormulaR1C1 = _
"=SUMPRODUCT((SED=RC[-10])*((1& SEC & SEB)+0>=EDATE(R2C2,-11))*((1& SEC & SEB)+0<R[-3]C[-9]),SEI)"
Selection.NumberFormat = "[=0]"""";0.0"
I have tried a couple process without any success could some assist me Please.
Codes-Test.docx
Who is Participating?
VBA ExpertCommented:
Do you just want to add that one value to the result of the formula?
This code will do that.
``````ActiveCell.FormulaR1C1 = _
"=RC[-3] + SUMPRODUCT((SED=RC[-10])*((1& SEC & SEB)+0>=EDATE(R2C2,-11))*((1& SEC & SEB)+0<R[-3]C[-9]),SEI) "
``````
0
Manager, Audit AnalyticsCommented:
Highlighted part below is not a number. hence the error. it retruns true. the problem is in formula not in vba code
ActiveCell.FormulaR1C1 = _
"=SUMPRODUCT((SED=RC[-10])*((1& SEC & SEB)+0>=EDATE(R2C2,-11))*((1& SEC & SEB)+0<R[-3]C[-9]),SEI)"
0
Author Commented:
The above code works if the cell RC[-3] has a value if the cell has not value the formula returns #VALUE.
Can you solve this part of the problem?
THANKS
0
VBA ExpertCommented:
Is there a formula in RC[-3]?
0
Author Commented:
No "RC[-3]" it is just the first stage in the formula see below:
ActiveCell.FormulaR1C1 = _
"=RC[-3] + SUMPRODUCT((SED=RC[-10])*((1& SEC & SEB)+0>=EDATE(R2C2,-11))*((1& SEC & SEB)+0<R[-3]C[-9]),SEI) "
This formula returns #VALUE if "RC[-3]" is empty, the formula works when the cell has a value.
Any solution would be great.
THANKS
0
VBA ExpertCommented:
Does the formula work without RC[-3]?
What is in RC[-3] and what is the active cell when you run the code?
0
Author Commented:
YES, the formula works perfectly for the addition of the 11 months using "EDATE(R2C2-11)" my only issue is that I'm trying to add into the formula the value for the current month. The current month is located at "RC[-3]" on the same row as the formula. The rest of the data is called from another worksheet.
The activecell is Column "K" and the cell that holds the value for "RC[-3] (sample "2.3") is in column "H" on the same row.
I have also tried to use "IFERROR" this works also except when I sort in descending all the blank cells preceed the largest value on the top of the list. I would like the "Blank or Empty" cells to placed at bottom of the list.
THANKS
0
Author Commented:
I've requested that this question be closed as follows:
Accepted answer: 0 points for user2073's comment #a38415822
for the following reason:
The "IFERROR" has worked to solve the problems. Thanks for your assistence it is appreciated.<br /><br />THANKS
0
Author Commented:
This have provided the results I was hopeing for after I included "IFERROR" to the formula. | 866 | 2,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-22 | latest | en | 0.834163 |
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One millisecond is 0.001 of a second
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Joined: 25 Oct 2008
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One millisecond is 0.001 of a second [#permalink]
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24 Jun 2009, 02:41
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One millisecond is 0.001 of a second. The costs for a single run of a computer program are \$1.07 for operating-system overhead, \$0.023 per millisecond of computer time, and \$4.35 for the mounting of a data tape. What is the total of these three costs for 1 run of a program that requires 1.5 seconds of computer time ?
(A) \$7.15
(B) \$8.87
(C) \$28.96
(D) \$35.57
(E) \$39.92
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Re: One millisecond is 0.001 of a second [#permalink]
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24 Jun 2009, 03:39
Op. sys. ovhd. & mount. costs are fixed costs. Comp. time cost is variable cost. Total cost is fixed costs + variable cost.
So total cost is = 1,07 + 4,35 + 1500 x 0,023= 39,92
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Re: One millisecond is 0.001 of a second [#permalink] 24 Jun 2009, 03:39
Display posts from previous: Sort by | 705 | 2,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2018-13 | latest | en | 0.876918 |
https://cs.stackexchange.com/questions/116429/matching-upper-bound-for-online-problems/116433 | 1,624,322,436,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488504969.64/warc/CC-MAIN-20210622002655-20210622032655-00260.warc.gz | 173,170,589 | 35,937 | # Matching Upper Bound for Online Problems
I have an online problem and I want to prove that no deterministic online algorithm is competitive.
Let $$\texttt{ON}$$ be an online algorithm and $$\texttt{OPT}$$ an optimal offline algorithm (an adversary). Given a sequence of $$n$$ inputs $$\sigma=\sigma_1\sigma_2\cdots\sigma_n$$ revealed in an online fashion, let $$v(\texttt{ON}_\sigma)$$ and $$v(\texttt{OPT}_\sigma)$$ be the values achieved respectively by $$\texttt{ON}$$ and $$\texttt{OPT}$$.
What I proved is this: let $$\sigma=\sigma_1\sigma_2$$ be an 2-input sequence. If $$\texttt{ON}$$ acts on $$\sigma_1$$ with action $$a$$, then I could find an input $$\sigma_2$$ such that $$\frac{v(\texttt{OPT}_\sigma)}{v(\texttt{ON}_\sigma)},$$ is unbounded.
The problem is that, if the action $$a$$ changes, then the above ratio could be equal to $$1$$.
Is it sufficient to prove that the ratio is unbounded for some action $$a$$? Or must it be true for any $$a$$? | 286 | 967 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-25 | latest | en | 0.860549 |
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## #1 2006-01-28 09:05:00
ineedhelp
Member
Registered: 2006-01-21
Posts: 7
### check my homework
I have two homework problems that I'm not sure I got right. If someone could tell me if my answer is correct that would be great. If its not right, i'll try to solve it again.
Suppose the sales of a particular brand of appliance satisfy the relationship: S(x) = 140x + 100 where S(x) represents the number of sales in year x, with x = 0 corresponding to 1982. Find the number of sales in 1987.
S(x) = 140(5) + 100
I came up with Sales in 1987 were 800.
Decide whether the pair of lines is parallel, perpendicular, or neither: the line through (3, -5) and (-1, 7) and the line through the points (6, -13) and (-2,11)
The first line slope is 2/2 and and the second line slope is -2/4. I came up with neither, because the the slope would be the same if parallel or reciprocal if perpendicular. In this case they are niether.
Offline
## #2 2006-01-28 09:50:46
mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900
### Re: check my homework
In the second one, the gradients need to be calculated differently.
Between (3, -5) and (-1, 7), there is a gradient of
Similarly, between (6, -13) and (-2, 11), there is a gradient of
They have the same gradient and so are parallel.
Why did the vector cross the road?
It wanted to be normal.
Offline
## #3 2006-01-28 10:23:39
MathsIsFun
Registered: 2005-01-21
Posts: 7,659
### Re: check my homework
And the computer can give you a little help with the straight line graph calculator
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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## #4 2006-01-28 11:05:32
mikau
Member
Registered: 2005-08-22
Posts: 1,504
### Re: check my homework
What the heck is gradients? Oddly my mathbook never mentioned that term.
A logarithm is just a misspelled algorithm.
Offline
## #5 2006-01-28 11:29:12
ineedhelp
Member
Registered: 2006-01-21
Posts: 7
### Re: check my homework
mathsyperson wrote:
In the second one, the gradients need to be calculated differently.
Between (3, -5) and (-1, 7), there is a gradient of
Similarly, between (6, -13) and (-2, 11), there is a gradient of
They have the same gradient and so are parallel.
I see my mistake. I did 7-5 and -1+3 to get 2/2 and 11-13 and -2+6 to get -2/4. I'll fix it. Thanks.
Offline
## #6 2006-01-28 11:55:16
mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900
### Re: check my homework
mikau wrote:
What the heck is gradients? Oddly my mathbook never mentioned that term.
A gradient is just the slope of a line. Well, it's actually the slope of a curve at a given point, but it's most commonly applied to lines. It's basically an less technical way of saying dy/dx.
Why did the vector cross the road?
It wanted to be normal.
Offline | 921 | 2,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-30 | latest | en | 0.902604 |
https://bookdown.org/jarneric/spring_school/1-3-structure-of-the-data.html | 1,722,820,256,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640417235.15/warc/CC-MAIN-20240804230158-20240805020158-00361.warc.gz | 105,765,028 | 4,962 | ## 1.3 Structure of the data
• There are three basic data structures:
1. Cross-sectional data
2. Time-series data
3. Panel data
• Cross-sectional data refers to the multiple observation units during one point in time ($$i=1,2,...,n$$), e.g. income of $$100$$ households in one yer, say $$2004$$
• Time-series refers to a single unit of observation across time ($$t=1,2,...,T$$), e.g. income of a single household for $$5$$ years, say $$2001$$, $$2002$$, $$2003$$, $$2004$$, and $$2005$$
• Panel data include observations for the same set of cross-sectional units $$i$$ at different points in time $$t$$, e.g. income of the same $$100$$ households for $$5$$ years
• Collected data should be always organized in matrix form (observations are presented by rows and variables by columns)
• When collecting the data some issues can emerge:
1. Missing values (abbreviation NA or n/a represents “not applicable”)
2. Measurement errors (collected data may not always present true values)
3. Outliers (extreme values)
Example 1.5 Two tables are given below. Answer the following questions:
1. Which table presents cross-sectionla data, and which one presents time-series data?
2. In which year cross-sectional data are observed?
3. For which audit firm time-series data are observed?
4. Indicate whether quantitative data are continuous or discrete.
TABLE 1.1: Audit firms with respect to the number of employees and net margin (%)
Audit firm Employees Net margin
Deloitte 334800 23.4
PwC 284000 15.6
Erste & Young 398965 20.7
KPMG 227000 13.8
TABLE 1.2: Yearly observations with respect to the number of employees and net margin (%)
Year Employees Net margin
2016 204790 17.3
2017 223600 27.8
2018 287590 16.5
2019 334800 23.4 | 469 | 1,726 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-33 | latest | en | 0.790887 |
https://www.zhygear.com/three-dimensional-solid-modeling-of-involute-straight-bevel-gear/ | 1,721,316,421,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00075.warc.gz | 928,792,655 | 29,574 | # Three dimensional solid modeling of involute straight bevel gear
SolidWorks is a three-dimensional modeling software developed based on Windows system. It has the characteristics of powerful function, easy to learn and practical, and provides a variety of modeling methods. In addition, SolidWorks software provides a variety of 3D software data interface formats, including step, vdafs, IGES, UG, sat, STL, MDT, Parasolid, Pro / E, SolidEdge, inventor and other formats, which are input into parts and assemblies.
At present, some secondary plug-ins, such as Mindy tool set and geartrax, have added automatic generation tools for involute straight bevel gears, which can enable designers to quickly design products and complete assembly. However, this model is simplified, does not consider some details, is not accurate enough, and cannot be used for subsequent CAE analysis and production guidance.
Combined with the dimensional parameters of differential straight bevel gear pair for car, the accurate three-dimensional modeling of involute straight bevel gear is completed by using MATLAB and SolidWorks. The specific modeling process is as follows:
1) Generating spherical involute
According to equation (equation 2.1), it is solved by MATLAB software ϕ The corresponding spherical involute coordinate value in the interval [0,3 π]. In SolidWorks, the function of “curve passing through XYZ point” is used to generate the spherical involute of straight bevel gear tooth profile, as shown in Figure 1.
2) Create a gear tooth solid
Connect the endpoints of the large and small ends of the involute with the origin, and fill the surface to obtain a closed surface. Then, according to the relevant parameters of straight bevel gear, the tooth top cone and tooth root cone are obtained by rotation, and the ruled surface tangent to the two spherical involute surfaces is made to form a multi surface combination, as shown in Fig. 2. Using the surface deletion function to cut the redundant surface, the tooth entity can be obtained, as shown in Figure 3.
3) Complete Gear Solid Modeling
Array tooth entities, rotate the gear contour to obtain the gear matrix, and obtain the three-dimensional solid model of planetary gear and half shaft gear through combination and deletion, as shown in Fig. 4 and Fig. 5.
4) Assemble
In order to facilitate the subsequent dynamic simulation analysis of straight bevel gear system, the assembly of half shaft gear and planetary gear is completed according to the bevel gear assembly conditions:
① Create a new assembly. Insert planetary gear and halfshaft gear.
② Define the gear fit. The matching conditions of straight bevel gear pair mainly include: the axis of half shaft gear is perpendicular to the axis of planetary gear; The base cone apex of the planetary gear coincides with the base cone apex of the half shaft gear; The datum plane of the axis and the end face auxiliary line on the axle shaft gear coincides with the datum plane of the axis and the end face auxiliary line on the planetary gear.
③ Interference check. Use the interference inspection function to check whether the assembly model has interference, so as to ensure that the straight bevel gear assembly has no interference, and obtain the straight bevel gear transmission model that can be used for analysis, as shown in Fig. 6.
It can be seen from this that the three-dimensional modeling of straight bevel gear completed in SolidWorks software can be regarded as a parametric model, so the solid model of straight bevel gear with different relevant parameters can be generated only by modifying the model parameters, such as pressure angle, number of teeth, cone pitch, indexing circle diameter, etc.
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https://thekindergartenconnection.com/space-skip-counting/ | 1,716,793,706,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00168.warc.gz | 481,546,413 | 34,201 | Search
# Space Skip Counting
Rocket ships, stars and planets, oh my! We love our Space themed activities! Is skip counting a skill you are currently working on with your class, or planning it soon? Use these Space Skip Counting printables for your students to work on counting by 2’s, 5’s and 10’s in small groups or math center!
*Pair with our Number Centers & Activities for Pre-K/Kindergarten!
## Space Skip Counting
These FREE printables are suitable for kindergarten or first grade. You can use in small groups, pair children up as partners or place in your math center.
We think it’s fun to provide star-shaped buttons or mini erasers! You might even simply use a marker to help the spaceship to the moon by skip counting in 2’s, 5’s or 10’s!
## Preparation
Print the free sheets in color. Choose which sheets you want to print for working on skip counting in 2’s, 5’s and 10’s. Or print all 3!
Laminate the printable sheets or use a dry-erase pocket so that you can re-use it. It will be a great independent activity for the Math center during your Space unit/theme!
## How to Use
Start counting at the spaceship and move your space-themed manipulative from one star to the next as you count until you reach the moon. This is similar to a Dot-to-Dot activity, but using skip counting instead!
Using small manipulatives helps those who are not that proficient in skip counting yet and might loose their place. It helps with tracking too!
Another hands-on idea is to pack the “path” with the star buttons/erasers! This option is super fun PLUS it also develops those fine motor muscles- picking up small items and placing them júst right!
## Small Group Play
Enlarge the printable and use large star stickers that are numbered the same as the printable.
Give the star stickers to the students to hold on to and count aloud as a group. The stars must then be placed on the right spot as the counting continue.
This develops concentration- having to listen and follow to the counting, and gives opportunity for the teacher to see which students need more practice.
### Grab FREE Printable
Are you ready to start practicing counting by 2’s, 5’s and 10’s with your students? Then click on the large, yellow DOWNLOAD HERE button below!
For more skip counting fun, hop on over to see our FREE Summer Counting by 2’s Activity! | 521 | 2,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.915844 |
https://cstheory.stackexchange.com/questions/53670/what-is-mathrmnc0-uniform-reduction | 1,708,912,238,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474649.44/warc/CC-MAIN-20240225234904-20240226024904-00309.warc.gz | 183,498,108 | 39,294 | # What is $\mathrm{NC}^0$-uniform reduction
I am interesting in strict and right'' formulations of results about $$\mathrm{NC}^1$$-completeness of some languages.
Consider for example Barrington's theorem about $$\mathrm{NC}^1$$-completeness of the problem of multiplying together a sequence of elements of the permutation group $$S_5$$.
I think that this theorem can be formulated by the following way:
This language is complete in the class uniform-$$\mathrm{NC}^1$$ under uniform-$$\mathrm{NC}^0$$-reduction.
I think that we understand what exactly means uniform-$$\mathrm{NC}^1$$: there are several definitions that are equivalent.
But what is the right definition of uniform-$$\mathrm{NC}^0$$-reduction?
By right definition I mean a definition that has standard properties: if $$A \le B$$ and $$B \le C$$ then $$A \le C$$, if $$A \le B$$ and $$B \in \mathrm{NC}^1$$ then $$A \in \mathrm{NC}^1$$.
IMHO my question is very natural and I hope that there is an answer in some books or papers. Could you please give a reference?
• You may consider DLOGTIME projections. Dec 21, 2023 at 13:09
• Why not use the obvious definition? "There exists a uniform family of NC^0 circuits that compute the reduction?" Dec 26, 2023 at 6:19
• In answer to Or Meir's comment: I think that Alexey Milavonov's question is really asking what you mean, when you say "there exists a uniform family of NC^0 circuits". That is, what notion of uniformity should one use? The "obvious" notion (and the notion that I use when I need to refer to "uniform NC^0" is to say that the function that maps 1^n to a description of C_n is computable in Dlogtime-uniform AC^0. But note that some people would find this inelegant, since AC^0 is much more powerful than NC^0, But I don't know of a good alternative. I agree with Emil Jeřábek: are good.. Dec 26, 2023 at 19:08
• By the way, the link to [Barrington, Immerman, Straubing] is more about uniformity for SUBCLASSES of NC^1. The best treatment of the issue of how to define uniform NC^1 that I know is in the textbook by Vollmer. One really wants uniform NC^1 to be the same as AlternatingTime(O(log n)). Simply saying "Dlogtime uniformity" doesn't really work, and you need to make use of the "extended connection language" that was defined first by [Ruzzo]. I suggest looking at Vollmer's discussion of this topic. Dec 26, 2023 at 19:20
• One final comment, following up on Emil Jeřábek's comment: Immerman's textbook also discusses an even more restrictive notion of reduction: quantifier-free projections. These are projections that can be expressed by a first-order formula with no quantifiers. Many "standard" complete problems for various complexity classes are complete even under quantifier-free projections. But building such projections can sometimes be a pain, and it frequently doesn't seem (to me) to be worth the effort of proving hardness under quantifier-free projections. Dec 26, 2023 at 19:26 | 766 | 2,943 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-10 | latest | en | 0.937706 |
http://list.seqfan.eu/pipermail/seqfan/2020-August/072692.html | 1,680,028,790,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948868.90/warc/CC-MAIN-20230328170730-20230328200730-00791.warc.gz | 28,403,275 | 3,030 | # [seqfan] Re: The iterations (x -> 2x-1) and (x -> 2x+1) of primes
M. F. Hasler oeis at hasler.fr
Sun Aug 16 13:45:57 CEST 2020
On Mon, Aug 10, 2020 at 8:47 AM Tomasz Ordowski <tomaszordowski at gmail.com>
wrote:
> The recursion a(n) = 2 a(n-1) - 1 with a(0) = p is given by the formula
> a(n) = (p-1)2^n+1.
> The recursion b(n) = 2 b(n-1) + 1 with b(0) = p is given by the formula
> b(n) = (p+1)2^n-1.
> Problem:
> Are there primes p such that both a(n) and b(n) are all composite for 0 <
> n < p ?
> Conjecture:
> If p is a prime, then there exists 0 < n < p such that a(n) or b(n) is
> prime.
> Try to disprove this strong conjecture!
>
It is certainly difficult to find small counter-examples. The first
iteration k>0 that gives again a prime of the form ...+1 when starting with
the n-th prime is:
apply( a(n)={n=prime(n)-1;for(k=1,oo,ispseudoprime(n<<k+1)&&return(k))},
[1..99])
%1 = [1, 1, 2, 1, 2, 3, 4, 1, 2, 2, 1, 1, 4, 3, 8, 6, 2, 2, 5, 2, 3, 1, 10,
2, 1, 2, 2, 4, 2, 2, 3, 2, 12, 1, 2, 2, 1, 3, 4, 4, 6, 7, 2, 2, 4, 1, 1, 3,
4, 1, 2, 2, 5, 4, 8, 2, 4, 1, 8, 4, 2, 4, 1, 6, 2, 8, 1, 1, 12, 4, 2, 2, 1,
2, 1, 4, 2, 3, 2, 4, 4, 3, 2, 3, 1, 6, 8, 4, 10, 3, 4, 2, 3, 4, 1, 10, 10,
2, 2, 2, 1, 12, 6]
Worth to add to OEIS? (with "0" when no such k exists? Can this occur, if
so for which indices?)
Record values are the following:
m=0;for(n=1,999,m<(m=max(a(n),m))&&printf("a(%d)=%d, ",n,m))
a(1)=1, a(3)=2, a(6)=3, a(7)=4, a(15)=8, a(23)=10, a(33)=12, a(184)=24,
a(262)=56, a(268)=168, a(288)=188, a(298)=678, a(514)=1840, ...
(Are these alway multiples of 4, after 10?)
The -1 version is already there: oeis.org/A257495 = A050412(prime(n))
with first 0 conjectured at prime(42228) = 509203 = A076337(1) : Riesel
number(s).
apply( (a(n)=n=prime(n)+1;for(k=1,oo,ispseudoprime(n<<k-1)&&return(k))),
[1..99])
%3 = [1, 1, 1, 2, 1, 4, 2, 2, 1, 1, 2, 2, 1, 24, 2, 1, 2, 4, 2, 4, 2552, 4,
1, 1, 4, 8, 4, 2, 2, 1, 6, 1, 3, 4, 2, 2, 2, 8, 4, 1, 1, 2, 1, 8, 3, 6, 4,
4, 2, 2, 1, 1, 2, 1, 2, 3, 8, 2, 4, 1, 12, 1, 2, 21, 4, 3, 2, 4, 6, 2, 11,
1, 2, 16, 4, 4, 2, 4, 2, 8, 1, 12, 1, 8, 2, 1, 22, 2, 2, 12, 2, 5, 2, 1, 2,
5, 1, 2, 12]
(13:21) gp > m=0;for(n=1,999,m<(m=max(a(n),m))&&printf("a(%d)=%d, ",n,m))
a(1)=1, a(4)=2, a(6)=4, a(14)=24, a(21)=2552, ... ~> oeis.org/A052340
(considering not only primes)
- Maximilian
PS: I found A078680 = Smallest m > 0 such that n*2^m + 1 is prime, or 0 if
no such m exists,
with first proven zero at n=78557 (which is not a prime, nor +-1 of a
prime).
With the the first a(n) above (...+1 primes) is a(n) = A078680(prime(n)-1).
There's also
A040076 Smallest m >= 0 such that n*2^m+1 is prime, or -1 if no such m
exists.
with other relevant XFREFs:
Cf. A033809 <https://oeis.org/A033809>, A046067 <https://oeis.org/A046067> | 1,447 | 2,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-14 | latest | en | 0.652463 |
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