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# Tutor profile: Artea M.
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Artea M.
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## Questions
### Subject:Basic Math
TutorMe
Question:
Find the c in the following problem: -c +12 = -4c
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Artea M.
Firstly, we have to group all the variable with the same in one side. 12 = -3c Next, in order to solve for c we have to divide 12 by the number in front of c (-3) so c can end up alone. 12/-3=c If asked, we can simplify it to just simply -4=c
### Subject:Calculus
TutorMe
Question:
Take the antiderivative of the following problem ∫x^3(9-x^2)^1/2dx.
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Artea M.
So we cannot just simply use u- substitution in this case. So, we take 9-x^2 and we equal it to 3^2-x^2 because we need the number (a) to be squared. The form would be (a^2 - x^2) and that and (3^2-x^2) fulfills that requirement. x= asin(theta) and then we put in the form - a^2 -a^2sin^2(theta) so that would just simplify to a^2(1-sin^2(theta)) now we figure out the derivative: x=3sin(theta) dx= 3cos(theta) d(theta) Take the integral of ((27sin^3(theta) (9-9sin^2(theta))^1/2*3cos(theta)d(theta) (9-9sin^2theta)^1/2 becomes 3cos(theta) Next, theres 243 ∫sin^3(theta)cos^2(theta) d(theta) 243∫sin^2(theta)cos^2(theta)sin(theta) d(theta) 243∫(1-cos^2(theta))cos^2(theta)sin(theta)d(theta) u=cos(theta) du=-sin(theta) d(theta) -243∫(1-u^2)(u^2)du -243∫(u^2-u^4)du -243(u^3/3 -u^5/5)+C
### Subject:English as a Second Language
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Why does the American government have three branches?
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Artea M.
The Constitution - the law of the country - created by the founding fathers (the people that wrote our government). The three branches - legislative, executive, and judicial - were created so they would keep each other in check and not give too much power to each branch (checks and balances). Most of Europe for example has a parliament that includes most of the government while here it is divided in three.
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https://www.jiskha.com/questions/1071443/the-density-of-liquid-mercury-is-13-6-g-cm-3-how-many-moles-of-mercury-are-there-in-1 | 1,585,832,341,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506959.34/warc/CC-MAIN-20200402111815-20200402141815-00171.warc.gz | 961,105,743 | 4,971 | # chemistry
the density of liquid mercury is 13.6 g/cm^3. how many moles of mercury are there in 1 litre of the metal?
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1. 1000cm^3 * 13.6g/cm^3 * 1mole/200.59g = 67.8 moles
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posted by Steve
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More Similar Questions | 852 | 2,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-16 | latest | en | 0.927571 |
https://codingireland.ie/ViewLesson/microbit-fruit-and-veg-piano-9412/2532 | 1,725,711,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00517.warc.gz | 159,558,115 | 20,552 | Microbit Sensors & Circuits
Intermediate
60 mins
Teacher/Student led
215 points
What you need:
Chromebook/Laptop/PC
Microbit
Crocodile clips
Some fruit & vegetables
# Microbit Fruit and Veg Piano
Discover the electrical conductivity of your body and various fruits and vegetables in this interactive project. You'll create a musical circuit using a Microbit, crocodile clips, and your chosen produce. By following step-by-step instructions, you'll learn to program your Microbit, test your circuits, and finally play your fruit and veg piano.
Learning Goals Learning Outcomes Teacher Notes Lesson Files
### 1 - Introduction
Did you know that your body can conduct electricity? In fact lots of things can conduct electricity, some better than others.
In this project we're going to create an electrical circuit using your body and some fruit and vegetables (which also can conduct electricity) to make your Microbit create some music.
Don't worry this is safe to do, as the amount of electricity we'll be using is very, very low.
Never play with or do experiments with high powered electricity such as plugs and power outlets. This level of electricity is actually dangerous!
### 2 - What you need
For this project you will need:
1. A microbit
2. 4 Crocodile clips
3. 4 pieces of fruit or vegetables such as bananas, apples, carrots etc.
### 3 - Create a new Microbit project
Go to the makecode.microbit.org website and create a new project.
### 4 - Program Pin 0
Microbits have 3 "pins" on them (P0, P1 & P2) that can be used with the GND (ground) pin to create circuits. In a later step we will be attaching the crocodile clips to these pins but for now lets add some come to program what will happen when the circuit is completed.
Add the following code to program P0. You can choose any note or icon you want.
```basic.forever(function () {
if (input.pinIsPressed(TouchPin.P0)) {
music.playTone(262, music.beat(BeatFraction.Quarter))
basic.showIcon(IconNames.Happy)
}
})
```
### 5 - Program Pin 1
Add the following code to program P1. You can choose any note or icon you want, just make them different to P0 and P2.
```basic.forever(function () {
if (input.pinIsPressed(TouchPin.P0)) {
music.playTone(262, music.beat(BeatFraction.Quarter))
basic.showIcon(IconNames.Happy)
}
if (input.pinIsPressed(TouchPin.P1)) {
music.playTone(392, music.beat(BeatFraction.Quarter))
basic.showIcon(IconNames.Angry)
}
})
```
# Unlock the Full Learning Experience
Get ready to embark on an incredible learning journey! Get access to this lesson and hundreds more in our Digital Skills Curriculum. | 622 | 2,602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.870452 |
https://gitlab.math.univ-paris-diderot.fr/letouzey/natded/-/commit/cfcefac48cf57aff38c15f918b9881b03e6ccd24 | 1,660,339,934,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00526.warc.gz | 285,989,523 | 41,785 | Commit cfcefac4 by Samuel Ben Hamou
### Du ménage dans Peano.v et une coquille corrigée dans ZF.v.
parent f0abfe1e
... ... @@ -13,8 +13,8 @@ Local Open Scope eqb_scope. Definition PeanoSign := Finite.to_infinite peano_sign. Definition Zero := Fun "O" []. Definition Succ x := Fun "S" [x]. Notation Zero := (Fun "O" []). Notation Succ x := (Fun "S" [x]). Notation "x = y" := (Pred "=" [x;y]) : formula_scope. Notation "x + y" := (Fun "+" [x;y]) : formula_scope. ... ... @@ -180,31 +180,85 @@ Proof. assumption. Qed. Ltac axiom ax name := Lemma IsAx_adhoc form : check PeanoSign form = true -> FClosed form -> Forall IsAx (induction_schema form ::axioms_list). Proof. intros. apply Forall_forall. intros x Hx. destruct Hx. + simpl. unfold IsAx. right. exists form. split; [ auto | split; [ auto | auto ]]. + simpl. unfold IsAx. left. assumption. Qed. Lemma rec0_rule l G A_x : BClosed (∀ A_x) -> In (induction_schema A_x) G -> Pr l (G ⊢ bsubst 0 Zero A_x) -> Pr l (G ⊢ ∀ (A_x -> bsubst 0 (Succ(#0)) A_x)) -> Pr l (G ⊢ ∀ A_x). Proof. intros. eapply R_Imp_e. + apply R_Ax. unfold induction_schema in H0. unfold BClosed in H. cbn in H. rewrite H in H0. cbn in H0. apply H0. + simpl. apply R_And_i; cbn. * set (v := fresh (fvars (G ⊢ ∀ bsubst 0 Zero A_x)%form)). apply R_All_i with (x:=v). apply fresh_ok. rewrite form_level_bsubst_id. - assumption. - apply level_bsubst. ++ unfold BClosed in H. cbn in H; omega. ++ cbn; omega. * assumption. Qed. (* And tactics to make the proofs look like natural deduction proofs. *) Ltac calc := match goal with | |- Pr ?l (?ctx ⊢ _) => assert (name : Pr l (ctx ⊢ ax)); [ apply R_Ax; compute; intuition | ]; unfold ax in name | |- BClosed _ => reflexivity | |- In _ _ => rewrite <- list_mem_in; reflexivity | |- ~Names.In _ _ => rewrite<- Names.mem_spec; now compute | _ => idtac end. Ltac app_R_All_i t := apply R_All_i with (x := t); [ rewrite<- Names.mem_spec; now compute | ]. Ltac eapp_R_All_i := eapply R_All_i; [ rewrite<- Names.mem_spec; now compute | ]. Ltac inst H l := match l with | [] => idtac | (?u::?l) => apply R_All_e with (t := u) in H; [ inst H l | calc ] end. Ltac sym := apply Symmetry; [ auto | auto | compute; intuition | ]. Ltac axiom ax name := match goal with | |- Pr ?l (?ctx ⊢ _) => assert (name : Pr l (ctx ⊢ ax)); [ apply R_Ax; calc | ]; unfold ax in name end. Ltac ahered := apply Hereditarity; [ auto | auto | compute; intuition | ]. Ltac inst_axiom ax l := let H := fresh in axiom ax H; inst H l; easy. Ltac hered := apply AntiHereditarity; [ auto | auto | compute; intuition | ]. Ltac app_R_All_i t v := apply R_All_i with (x := t); calc; cbn; set (v := FVar t). Ltac eapp_R_All_i := eapply R_All_i; calc. Ltac trans b := apply Transitivity with (B := b); [ auto | auto | auto | compute; intuition | | ]. Ltac sym := apply Symmetry; calc. Ltac thm := unfold IsTheorem; split; [ unfold Wf; split; [ auto | split; auto ] | ]. Ltac ahered := apply Hereditarity; calc. Ltac change_succ := match goal with | |- context[ Fun "S" [?t] ] => change (Fun "S" [t]) with (Succ (t)); change_succ | _ => idtac end. Ltac hered := apply AntiHereditarity; calc. Ltac cbm := cbn; change (Fun "O" []) with Zero; change_succ. Ltac trans b := apply Transitivity with (B := b); calc. Ltac thm := unfold IsTheorem; split; [ unfold Wf; split; [ auto | split; auto ] | ]. Ltac parse term := match term with ... ... @@ -216,15 +270,10 @@ Ltac rec := match goal with | |- exists axs, (Forall (IsAxiom PeanoTheory) axs /\ Pr ?l (axs ⊢ ?A))%type => let form := parse A in exists (induction_schema form :: axioms_list); set (Γ := induction_schema form :: axioms_list); set (rec := induction_schema form); split; [ apply Forall_forall; intros x H; destruct H; [ simpl; unfold IsAx; right; exists form ; split; [ auto | split; [ auto | auto ]] | simpl; unfold IsAx; left; assumption ] | exists (induction_schema form :: axioms_list); set (rec := induction_schema form); set (Γ := rec :: axioms_list); split; [ apply IsAx_adhoc; auto | repeat apply R_Imp_i; eapply R_Imp_e; [ apply R_Ax; unfold induction_schema; cbm; intuition | simpl; apply R_And_i; cbn ] ]; cbm apply rec0_rule; calc ]; cbn (* | _ => idtac *) end. ... ... @@ -235,17 +284,13 @@ Lemma ZeroRight : IsTheorem Intuiti PeanoTheory (∀ (#0 = #0 + Zero)). Proof. thm. rec. + app_R_All_i "y". cbm. sym. axiom ax9 AX9. apply R_All_e with (t := Zero) in AX9; auto. + app_R_All_i "y". cbm. apply R_Imp_i. set (H1 := FVar _ = _). + sym. inst_axiom ax9 [Zero]. + app_R_All_i "y" y. apply R_Imp_i. set (H1 := _ = _). sym. trans (Succ ((FVar "y") + Zero)). - axiom ax10 AX10. apply R_All_e with (t := FVar "y") in AX10; auto. apply R_All_e with (t := Zero) in AX10; auto. trans (Succ (y + Zero)). - inst_axiom ax10 [y; Zero]. - ahered. sym. apply R_Ax. ... ... @@ -256,38 +301,27 @@ Lemma SuccRight : IsTheorem Intuiti PeanoTheory (∀∀ (Succ(#1 + #0) = #1 + Su Proof. thm. rec. + app_R_All_i "x". app_R_All_i "y". cbm. + app_R_All_i "y" y. sym. trans (Succ (FVar "y")). - axiom ax9 AX9. apply R_All_e with (t := Succ (FVar "y")) in AX9; auto. trans (Succ y). - inst_axiom ax9 [Succ y]. - ahered. sym. axiom ax9 AX9. apply R_All_e with (t := FVar "y") in AX9; auto. + app_R_All_i "x". cbm. inst_axiom ax9 [y]. + app_R_All_i "x" x. apply R_Imp_i. app_R_All_i "y". cbm. app_R_All_i "y" y. fold x. set (hyp := ∀ Succ _ = _). trans (Succ (Succ (FVar "x" + FVar "y"))). trans (Succ (Succ (x + y))). - ahered. axiom ax10 AX10. apply R_All_e with (t := FVar "x") in AX10; auto. apply R_All_e with (t := FVar "y") in AX10; auto. - trans (Succ (FVar "x" + Succ (FVar "y"))). inst_axiom ax10 [x; y]. - trans (Succ (x + Succ y)). * ahered. axiom hyp Hyp. apply R_All_e with (t := FVar "y") in Hyp; auto. * axiom ax10 AX10. sym. apply R_All_e with (t := FVar "x") in AX10; auto. apply R_All_e with (t := Succ (FVar "y")) in AX10; auto. inst_axiom hyp [y]. * sym. inst_axiom ax10 [x; Succ y]. Qed. Lemma Comm : IsTheorem Intuiti PeanoTheory ((∀ #0 = #0 + Zero) -> (∀∀ Succ(#1 + #0) = #1 + Succ(#0)) -> ... ... @@ -295,35 +329,27 @@ Lemma Comm : Proof. thm. rec; set (Γ' := _ :: _ :: Γ). + app_R_All_i "x". cbm. app_R_All_i "x". cbm. trans (FVar "x"). + app_R_All_i "x" x. trans x. - sym. assert (Pr Intuiti (Γ' ⊢ ∀ # 0 = # 0 + Zero)). { apply R_Ax. simpl in *; intuition. } apply R_All_e with (t := FVar "x") in H; auto. assert (Pr Intuiti (Γ' ⊢ ∀ # 0 = # 0 + Zero)). { apply R_Ax. calc. } apply R_All_e with (t := x) in H; auto. - sym. axiom ax9 AX9. apply R_All_e with (t := FVar "x") in AX9; auto. + app_R_All_i "y". cbm. inst_axiom ax9 [x]. + app_R_All_i "y" y. apply R_Imp_i. app_R_All_i "x". cbm. trans (Succ (FVar "x" + FVar "y")). app_R_All_i "x" x. trans (Succ (x + y)). - sym. assert (Pr Intuiti ((∀ # 0 + FVar "y" = FVar "y" + # 0) :: Γ' ⊢ ∀ ∀ Succ (#1 + #0) = #1 + Succ (#0))). { apply R_Ax. simpl in *; intuition. } apply R_All_e with (t := FVar "x") in H; auto. apply R_All_e with (t := FVar "y") in H; auto. - trans (Succ (FVar "y" + FVar "x")). assert (Pr Intuiti ((∀ #0 + y = y + #0) :: Γ' ⊢ ∀ ∀ Succ (#1 + #0) = #1 + Succ (#0))). { apply R_Ax. calc. } apply R_All_e with (t := x) in H; auto. apply R_All_e with (t := y) in H; auto. - trans (Succ (y + x)). * ahered. assert (Pr Intuiti ((∀ #0 + FVar "y" = FVar "y" + #0) :: Γ' ⊢ ∀ #0 + FVar "y" = FVar "y" + #0)). { apply R_Ax. apply in_eq. } apply R_All_e with (t := FVar "x") in H; auto. assert (Pr Intuiti ((∀ #0 + y = y + #0) :: Γ' ⊢ ∀ #0 + y = y + #0)). { apply R_Ax. apply in_eq. } apply R_All_e with (t := x) in H; auto. * sym. axiom ax10 AX10. apply R_All_e with (t := FVar "y") in AX10; auto. apply R_All_e with (t := FVar "x") in AX10; auto. inst_axiom ax10 [y; x]. Qed. Lemma Commutativity : IsTheorem Intuiti PeanoTheory (∀∀ #0 + #1 = #1 + #0). ... ... @@ -334,7 +360,7 @@ Proof. * apply ZeroRight. + apply SuccRight. Qed. (** A Coq model of this Peano theory, based on the [nat] type *) Definition PeanoFuns : modfuns nat := ... ... @@ -380,7 +406,7 @@ Proof. unfold PeanoTheory. simpl. unfold PeanoAx.IsAx. intros [IN|(B & -> & CK & CL)]. - compute in IN. intuition; subst; cbn; intros; subst; omega. - compute in IN. calc; subst; cbn; intros; subst; omega. - intros genv. unfold PeanoAx.induction_schema. apply interp_nforall. ... ...
... ... @@ -26,7 +26,7 @@ Definition compat_right := ∀∀∀ (#0 ∈ #1 /\ #1 = #2 -> #0 ∈ #2). Definition ext := ∀∀ (∀ (#2 ∈ #0 <-> #2 ∈ #1) -> #0 = #1). Definition pairing := ∀∀∃∀ (#3 ∈ #2 <-> #3 = #0 \/ #3 = #1). Definition union := ∀∃∀ (#2 ∈ #1 <-> ∃ (#3 ∈ #0 /\ #2 ∈ #3). Definition union := ∀∃∀ (#2 ∈ #1 <-> ∃ (#3 ∈ #0 /\ #2 ∈ #3)). Definition powerset := ∀∃∀ (#2 ∈ #1 <-> ∀ (#3 ∈ #2 -> #3 ∈ #0)). Definition infinity := ∃ (∃ (#1 ∈ #0 /\ ∀ not #2 ∈ #1) /\ ∀ (#3 ∈ #0 -> (∃ (#4 ∈ #0 /\ (∀ (#5 ∈ #4 <-> #5 = #3 \/ #5 ∈ #3)))))). ... ...
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Finish editing this message first! | 3,193 | 8,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-33 | latest | en | 0.630656 |
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`Based on Ray's New Intellectual Arithmetic, this series is recommended for Grade 4. Topics include: review of addition, subtraction, multiplication, and division facts with applications to more difficult problems; finding least common denominators for addition and subtraction of fractions; extensive work with story problems with lesson assignments from Ray’s Intellectual Arithmetic; adding, subtracting, multiplying, and dividing fractions; and ratios.Uses Ray's Intellectual Arithmetic. The Key to the Primary, Intellectual, and Practical is also recommended. The Parent-Teacher Guide is optional. These workbooks have been skillfully written by Dr. Moore expressly for use with the McGuffey Readers and other specific materials published by Mott Media. They are available for grades one through four, with four books for each grade level; They provide clear daily lessons, regular reviews, quizzes, and a pull-out section containing a comprehensive test of mastery for the workbook. Answers for all assignments, quizzes, and tests are included.Dr. Moore wrote these workbooks expressly for use with the Ray’s Arithmetic Series. They are available for grades one through four, with four books for each grade level. They provide clear daily lessons, regular reviews, quizzes, and a pull out section containing a comprehensive test of mastery for that workbook as well as answers for all assignments, quizzes and tests. The general format provides a new lesson on each day, Monday through Thursday, and a test on these concepts on Friday. There are also periodic, multi-concept reviews and tests plus a final test for the entire workbook. As mentioned above, each workbook is designed to be completed in 1/4 of the school year. Although students vary in ability and background, the Series number indicates the suggested grade level while the Book number indicates the 1st, 2nd, 3rd or 4th quarter of the year. Except for a few of the grade 1 workbooks which teach “beginning basics,” each workbook is keyed to one of the Ray’s textbooks. As a child proceeds through the workbooks he/she is referred to the appropriate lesson or exercise in the various texts. While some parents and teachers prefer to personally shepherd their students through the Ray’s Arithmetic textbooks, many prefer the guidance offered by our workbooks.`
# Ray's Arithmetic Workbook Series 4 Book 1
\$7.99Price
bottom of page | 484 | 2,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-26 | latest | en | 0.952406 |
http://mathhelpforum.com/advanced-algebra/74405-matrix-problem-print.html | 1,518,917,853,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811243.29/warc/CC-MAIN-20180218003946-20180218023946-00792.warc.gz | 224,115,624 | 2,651 | # Matrix Problem
$AD\rightarrow 3(\times n \times n) \times 2 \rightarrow 3\times 2 \rightarrow R^{3\times 2},AD=I_3?$, Are you sure ?
$if\ D\in R^{n\times 3},D=[\vec{v}_1,\vec{v}_2,\vec{v}_3], find\ AD=I_3,$
We could do $AD=A[\vec{v}_1,\vec{v}_2,\vec{v}_3]=[A\vec{v}_1,A\vec{v}_2,A\vec{v}_3]=[\vec{e}_1\ \ \vec{e}_2\ \ \vec{e}_3]$
To resolute $A\vec{v}_i=\vec{e}_i$ is easy. | 187 | 376 | {"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": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | longest | en | 0.233319 |
http://www.thenakedscientists.com/forum/index.php?topic=40312.0 | 1,480,956,029,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541773.2/warc/CC-MAIN-20161202170901-00296-ip-10-31-129-80.ec2.internal.warc.gz | 736,288,935 | 16,584 | # The Naked Scientists Forum
### Author Topic: Is a heat index helpful? (Read 6441 times)
#### Geezer
• Neilep Level Member
• Posts: 8328
• "Vive la résistance!"
##### Is a heat index helpful?
« on: 22/07/2011 17:16:45 »
The Eastern US is experiencing some very high temperatures at the moment. The weather people are talking about "heat index" temperatures as well as the actual temperatures.
Are heat index temperatures really helpful, or do they simply confuse people even more? Wouldn't it be better just to quote the air temperature and relative humidity and let people work out the implications for themselves? Is a "heat index" just another attempt at "dumbing down" science?
#### JP
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##### Is a heat index helpful?
« Reply #1 on: 23/07/2011 14:45:00 »
As a warning system to tell people how dangerous the heat/humidity combo is, I think it's better than just telling them the temperature and relative humidity. I certainly can't do the calculation in my head. The formula for it is here:
.
(Source: http://en.wikipedia.org/wiki/Heat_index)
But I agree that the number is pretty useless, since there's no reference for what it means. Perhaps giving the temperature/humidity along with a warning from low->extreme if the heat index is high might be more useful?
#### Geezer
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• "Vive la résistance!"
##### Is a heat index helpful?
« Reply #2 on: 23/07/2011 17:58:56 »
The formula reinforces my point. There are only two variables, T and R. c1 through c9 are a bunch of constants that are supposed (I think) to represent "typical" conditions, which ends up meaning, in total, they are anything but typical!
I think most people are capable of understanding that if the temperature is high, and the relative humidity is high, it's going to feel very unpleasant. I'm wondering if it would not be better to just let people know whether the RH is going to deviate much from the average for a particular temperature.
#### CZARCAR
• Hero Member
• Posts: 686
##### Is a heat index helpful?
« Reply #3 on: 23/07/2011 20:28:23 »
Accuweather? refers to heat index as "feels like" including winter windchill factor. Simple enuff 4me
#### Bored chemist
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##### Is a heat index helpful?
« Reply #4 on: 24/07/2011 14:59:37 »
Great! according to that WIKI page
"Here is a formula[8] for approximating the heat index in degrees Fahrenheit, to within ±1.3 °F. It is useful only when the temperature is at least 80 °F and the relative humidity is at least 40%"
So where I live
http://www.metoffice.gov.uk/climate/uk/averages/19611990/sites/sheffield.html
it is, on average, never useful.
And, according to this
http://www.metoffice.gov.uk/climate/uk/ne/
the temperature has practically never been high enough for the heat index to be meaningful.
On the other hand, the wind-chill factors are often important.
#### damocles
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##### Is a heat index helpful?
« Reply #5 on: 24/07/2011 15:23:44 »
Great! according to that WIKI page
"Here is a formula[8] for approximating the heat index in degrees Fahrenheit, to within ±1.3 °F. It is useful only when the temperature is at least 80 °F and the relative humidity is at least 40%"
So where I live
http://www.metoffice.gov.uk/climate/uk/averages/19611990/sites/sheffield.html
it is, on average, never useful.
And, according to this
http://www.metoffice.gov.uk/climate/uk/ne/
the temperature has practically never been high enough for the heat index to be meaningful.
On the other hand, the wind-chill factors are often important.
That should come as no surprise, Bored Chemist. The residents of Sheffield cannot really complain about uncomfortable heat, though they might sometimes find moderate heat and humidity a little uncomfortable because they are not acclimatized.
But where I come in with a little objection is that this sort of heat index works on the supposition that dry heat is fairly innocuous. When the temperature is >40 deg C (=104 deg F) and the humidity is less than 20%, the conditions raise a whole new and different set of problems, as we found here in Melbourne a couple of years ago when these conditions led to hundreds of bushfire deaths, and at least tens of (unrelated) heat exhaustion deaths in our state. (Americans, consider Death Valley, where I believe the heat is very dry).
People might cook in high humidity high temperature conditions, but they dehydrate readily and dangerously in high temperature low humidity conditions.
« Last Edit: 24/07/2011 15:28:48 by damocles »
#### JP
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##### Is a heat index helpful?
« Reply #6 on: 24/07/2011 16:11:25 »
I think most people are capable of understanding that if the temperature is high, and the relative humidity is high, it's going to feel very unpleasant.
I think you overestimate people. I have a good grasp of temperature and humidity, but the heat index is still more useful to me than temperature and humidity when I'm planning a long run or bike ride outdoors in the heat. What may feel tolerable may actually be just slightly above your body's ability to dissipate heat. While this might not be an issue for a walk outside, it's a huge difference if you're strenuously exercising for an hour.
#### Geezer
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• "Vive la résistance!"
##### Is a heat index helpful?
« Reply #7 on: 24/07/2011 18:34:42 »
I think you overestimate people. I have a good grasp of temperature and humidity, but the heat index is still more useful to me than temperature and humidity when I'm planning a long run or bike ride outdoors in the heat. What may feel tolerable may actually be just slightly above your body's ability to dissipate heat. While this might not be an issue for a walk outside, it's a huge difference if you're strenuously exercising for an hour.
Well, OK, b-b-b-but in that case wouldn't it be better to just assign arbitrary values like the five categories in the table on the Wiki page?
I'm not too keen on the attempt to decide what a temp/RH combination "feels like" in terms of a particular temperature when there are too many variables that will determine what any individual actually does feel. Not only that, but there is an initial supposition about the RH that determines when a temperature "feels like" the actual temperature! That means the heat index temperature should actually be less than the actual temperature when the RH is low (which, as Damocles points out) can be a situation that is just as dangerous as high RH.
(I can see I'm not going to win this argument )
#### CZARCAR
• Hero Member
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##### Is a heat index helpful?
« Reply #8 on: 24/07/2011 18:53:05 »
google Smartvent computerized ventilator which compares in & outdoor humidity so to decide when to ventilate
#### JP
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##### Is a heat index helpful?
« Reply #9 on: 24/07/2011 23:22:16 »
Geezer, I completely agree that the number itself is pretty meaningless. The heat index makes it slightly easier for me to figure out when it's safe to go for a long run or bike ride, but the number is still pretty arbitrary. The number also doesn't have much relation to actual temperatures, as you point out. It would be more useful if they'd just tell you in words how dangerous the heat is if you're doing strenuous exercise.
#### Geezer
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##### Is a heat index helpful?
« Reply #10 on: 25/07/2011 00:05:19 »
Geezer, I completely agree that the number itself is pretty meaningless. The heat index makes it slightly easier for me to figure out when it's safe to go for a long run or bike ride, but the number is still pretty arbitrary. The number also doesn't have much relation to actual temperatures, as you point out. It would be more useful if they'd just tell you in words how dangerous the heat is if you're doing strenuous exercise.
I never pay any attention to the "feels like" temperature here because a) I try to avoid doing anything strenuous; and b) it's seldom very humid here. Do you tend look at the heat index number only, or do you think you look at the actual temp and the heat index and do a sort of tradeoff?
In your case, you are probably not far from the norm that was established to determine the "feels like" values, so it's probably fairly accurate for you. However, a combination that feels like 80°F for you might feel a lot more like 100°F for my pal Fat Bastard
#### JP
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##### Is a heat index helpful?
« Reply #11 on: 26/07/2011 19:16:46 »
Geezer,
I can certainly tell you that I was shocked to learn that the daily high in Singapore was around 30 C, since it felt a lot hotter than that. Of course, the humidity was around 85%. Moving there, I had no idea what 30 C and 85% humidity would feel like. Hearing that the heat index is around 40 C says a bit more to me. In general, I find that for whatever reason I can't tell much about how it will feel just from relative humidity and temperature.
(Actually as an american, 40 C means a trip to a calculator to find out that it's around 104 F.)
As to your question, when I go out to exercise, the most important things to me are temperature, how it feels when I step outside, and if it's sunny. I sometimes check heat index, especially if I'm planning a workout in advance and am relying on the forecast. I almost never check relative humidity except as an afterthought.
#### Geezer
• Neilep Level Member
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• "Vive la résistance!"
##### Is a heat index helpful?
« Reply #12 on: 27/07/2011 07:50:10 »
Well, that pretty much seems to wrap this one up. The heat index is helpful (to JP at least ).
#### JP
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##### Is a heat index helpful?
« Reply #13 on: 27/07/2011 12:18:54 »
But not to Geezer!
So... how about wind chill? :)
#### Geezer
• Neilep Level Member
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• "Vive la résistance!"
##### Is a heat index helpful?
« Reply #14 on: 27/07/2011 17:50:35 »
So... how about wind chill? :)
It's equally dodgy. A bunch of different standards, and quite arbitrary. It would be much more honest just to use a scale of one through five or something similar.
My objection to these things is that they attempt to assign a degree (!) of precision that has little basis.
#### JP
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##### Is a heat index helpful?
« Reply #15 on: 27/07/2011 17:59:07 »
I completely agree. The number has no precision to me, but it's a useful scale in both cases. A warning scale from 1 to 5 would be far more useful.
#### Geezer
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##### Is a heat index helpful?
« Reply #16 on: 27/07/2011 18:15:06 »
If we didn't have so much important work to do here on TNS, we could could propose a new international standard
#### imatfaal
• Neilep Level Member
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• rouge moderator
##### Is a heat index helpful?
« Reply #17 on: 27/07/2011 18:46:40 »
weather in london is so mundane compared to those of you in more exotic climes; basically comes down to two yes/no questions - is it raining? and do I need a coat?
#### Geezer
• Neilep Level Member
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• "Vive la résistance!"
##### Is a heat index helpful?
« Reply #18 on: 27/07/2011 20:22:56 »
weather in london is so mundane compared to those of you in more exotic climes; basically comes down to two yes/no questions - is it raining? and do I need a coat?
Aren't both of those questions pretty much redundant?
#### CliffordK
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##### Is a heat index helpful?
« Reply #19 on: 27/07/2011 20:32:33 »
I spent some time in Missouri.
No AC in my vehicles.
Anyway, I assume the humidity was reasonably constant. So, while 95°F might be different between Missouri and Oregon, it was reasonably representative for comparing day-to-day temperatures in one location.
Wind chill? Of course I would drive with the windows down, and it wasn't bad, except for those moments when a cell phone call would come in and I had to roll up the windows.
Anyway, as long as the temperature kept below 100°F, then it was ok.
I went to Wichita once with 105°F weather, I think. That was plain HOT. The problem is that when you roll down the windows, it just blows warm air at you. But, rolling up the windows doesn't help either.
Anyway, Heat Index might help somewhat in comparing vastly different environments. But, it is no longer fully representative if you can artificially alter the air flow, for example rolling down the window, or using a fan, at which point, the moving air brings you back to needing a true comparison between body temperature and ambient temperature, i.e. just temperature.
Wind Chill in the winter?
Perhaps it is the same concept. Wind speed might be useful. But, as a human we can alter the parameters. So, when you don your winter coat, you are separating the wind from your skin. If you choose to ride your bicycle in 30°F weather... then one is making one's own wind chill.
Anyway, I don't pay any attention to the calculated values.
#### The Naked Scientists Forum
##### Is a heat index helpful?
« Reply #19 on: 27/07/2011 20:32:33 » | 3,449 | 13,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2016-50 | latest | en | 0.902397 |
http://homepages.math.uic.edu/~hanson/pub/SIAMbook/MATLABCodes/mcm09unifnorm.m | 1,369,376,674,000,000,000 | text/plain | crawl-data/CC-MAIN-2013-20/segments/1368704253666/warc/CC-MAIN-20130516113733-00037-ip-10-60-113-184.ec2.internal.warc.gz | 125,382,093 | 1,229 | function mcm09unifnorm % Example Output: FINM 331 Demo for Uniform Monte Carlo; % for uniform rand-deviates on (a,b); clc clear % global a b V % fprintf(' Uniform rand-deviate Monte Carlo:\n'); n = 10000; srtn = sqrt(n); a=-3; b = +2; V = b-a; fprintf('\nn=%i; a=%6.4f; b=%6.4f; V=%6.4f;\n',n,a,b,V); % erfc(x) = 2\sqrt(pi)*int(exp(-t^2),t=x..infty); % erf(x) = 2\sqrt(pi)*int(exp(-t^2),t=0..x); exact = 0.5*(erfc(a/sqrt(2))-erfc(b/sqrt(2))); sig2unif_exact = 2.5/sqrt(pi)*(erf(b)-erf(a))-exact^2; fprintf('\nexact integral = %10.8f;',exact); fprintf('\nsig2unifexact = %9.4e; sigunifexact = %9.4e;'... ,sig2unif_exact,sqrt(sig2unif_exact)); rand('twister',3); U = a+V*rand(1,n); fuv=fu(U); % Monte Carlo estimators: mu_unif = mean(fuv); fprintf('\nmu_unif=%10.8f;',mu_unif); fprintf('\nmu_unif_abserror=%9.4e;',mu_unif-exact); fprintf('\nmu_unif_relerror=%9.4e%%;\n',100*(mu_unif/exact-1)); % Monte Carlo variance estimators: sig2unif = var(fuv); % MATLAB var(x); gives unbiased variance fprintf('\nsig2_unif=%9.4e;',sig2unif); fprintf('\nsig2_unif_abserror=%9.4e;',sig2unif-sig2unif_exact); fprintf('\nsig2unif_relerror=%9.4e%%;\n'... ,100*(sig2unif/sig2unif_exact-1)); % std. errors: se_unif_exact = sqrt(sig2unif_exact)/srtn; se_unif = sqrt(sig2unif)/srtn; fprintf('\nstderr_unif_exact=%9.4e;',se_unif_exact); fprintf('\nstderr_unif=%9.4e;',se_unif); fprintf('\nstderr_unif_diff=%9.4e;\n\n',se_unif-se_unif_exact); % %%%%% function y = fu(x) global V y = V*exp(-x.*x/2)/sqrt(2*pi); %%%%% % % end mcm09unifnorm.m % | 602 | 1,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2013-20 | longest | en | 0.143766 |
http://www.jiskha.com/display.cgi?id=1311376140 | 1,493,538,443,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917124371.40/warc/CC-MAIN-20170423031204-00118-ip-10-145-167-34.ec2.internal.warc.gz | 578,495,814 | 4,137 | # math
posted by on .
d/dx x(1+y)^(1/2) + y(1+x)^(1/2)=0. Can some1 show me the calculation because i didn't get like the answer.
[ans=-1/(1+x)^2 but my calculation=-(sqrt(y+1) (2 sqrt(x+1) sqrt(y+1)+y)/(sqrt(x+1) (2 sqrt(x+1) sqrt(y+1)+x)]
• math - ,
I suppose the question is to solve for y in terms of x, in which case your calculation is not in an explicit form, i.e. it is in the form y=f(x,y).
If the question is to solve for y in terms of x, i.e. the answer represents
y=-1/(1+x)^2,
It should give zero when substituted into the original equation,
d/dx [x(1+y)^(1/2) + y(1+x)^(1/2)]=0
or
d/dx [x(1+y)^(1/2)] + y(1+x)^(1/2)=0
(I do not know which case or if parentheses are missing).
Neither of them equals zero.
So it would be appropriate to check the question for missing parentheses, and the specify the context of the question (solve for y?) etc. before proceeding further. | 292 | 891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.868406 |
https://forums.warframe.com/topic/1231051-status-40-making-status-effectiveness-scale-with-damage/ | 1,708,979,014,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474663.47/warc/CC-MAIN-20240226194006-20240226224006-00257.warc.gz | 265,014,011 | 90,520 | # Status 4.0: Making Status Effectiveness Scale with Damage
## Recommended Posts
Disclaimer: This topic is very dry. Nothing flashy, mostly math and game logic. If you are not up for that then… try? Please? I understand that math and game logic aren’t topics a lot of people enjoy thinking about, but if you ever want something like status to change fundamentally (which I strongly believe it should, and I hope to convince you here to think the same) you gotta wade through this stuff.
If you aren’t up for wading through the dryness, here’s the TL;DR.
TL;DR: Make every status effect scale linearly with the amount of elemental/physical damage inflicted.
The title also sums the premise up nicely. If you are curious how I went about designing a system that works with that premise (and attempt to fill all the holes it creates), a few thousand words await you.
For the most part when I suggest a change or new mechanic it is italicized. Skimming the post for just the section names and italicized sentences should give you a good idea to what is going on.
Please, if there is a part of this proposal you take issue with, do not instantly dismiss the entire proposal. I am only one person, there are definitely issues here. Try and see what my intent is, assume that I am not an idiot, and help me understand where I am mistaken.
In the beginning, there was DoSP
In order to reduce the word count, I made up an acronym for the core concept of this proposal.
DoSP: “Damage of Status Proc” When an attack procs a status effect, the DoSP is the amount of damage of the element that had its status inflicted. This number is NOT affected by health type damage resistances or armor damage reduction.
• Example: If a weapon deals 500 Heat and 800 Viral damage then RNG procs the Heat status effect, the “damage of the status proc” is 500.
“Total DoSP” is the amount of DoSP that an enemy is currently under the effects of. If a status effect has a duration of 6 seconds, after the 6 seconds the Total DoSP will reduce by the DoSP of the given status effect.
• Example: You inflict an initial 500 DoSP attack then a 300 DoSP attack 4 seconds later. For the first four seconds the total DoSP is 500, for the next two seconds the total DoSP is 800, the next four seconds have a total DoSP of 300, and after that the total DoSP is 0.
Status effects are inflicted BEFORE normal damage is inflicted. If a bullet strips armor, the damage of the bullet will consider the armor after being stripped.
Status Proc vs Status Effect: For the purposes of this post a status proc is only defining the action of a status being inflicted. Basically, it is just a yes/no or black/white descriptor. Status effects on the other hand take into account what the status proc did to an enemy, status effects work in shades of grey
All of the effects of status procs can be generalized into two categories, damage and debuff.
Damage Status Procs
These are simple and generally already work with a “have damage matter” system. If a status proc does damage, that damage should be based on the amount damage of the respective damage type the attack had.
• Example: A weapon that deals 500 DoSP will have its damage effect be entirely based on that DoSP. What a novel concept… (no, this isn’t how it currently works)
Damage over time is based on the total DoSP. Multiple status procs of a single type will not create new instances of damage; they will only increase the damage of following instances. This is the exact same result of having multiple instances of status effects but with less UI clutter and a clearer communication of how much DoT you are doing.
Debuff Status Procs
Works by having the effectiveness of a status proc be based on a ratio of the total DoSP to the health (not EHP) of an enemy. This sounds a bit complicated on paper, but in practice I believe it is far more logical than the current status system.
• Example: Let’s start with an enemy with 1000 Health. If you inflict a cold status effect with a DoSP of 100 then the enemy would be slowed by 10% ( 100/1000 = .1 ). But if you inflicted a cold status with a DoSP of 500 then the enemy would be slowed by 50% ( 500/1000 = .5 ). And if you inflict both status effects in sequence (A 100 and 500 DoSP) then the enemy would be slowed by 60%.
Damage status procs already work this way. If the goal is to kill an enemy the effectiveness of a damage dealing status proc is a ratio of the damage of the status effect relative to the health of an enemy. A status proc that deals 10 damage per second on an enemy with 100 health is far more potent at achieving the goal when compared to a 1000 health enemy. Damage status procs are already a function of DoSP and health, it only makes sense for debuff status procs to work the same way.
The effect being relative to the enemy’s health follows a similar logic. If an enemy has ten times the health of another enemy, you expect the healthier enemy to be more resistant to status.
Back to the math, two more acronyms I made up.
SEF: “Status Effect Factor” = (DoSP) / (Health) This is the ratio that determines how powerful a debuff status effect is.
WHS: “Whole Health Status” The effect of a status proc if total DoSP is equal to the health of an enemy.
• Example: Back to the cold status, how that original equation worked is that the WHS was equal to an enemy being slowed by 100%. If, for example, the WHS was only 80%, that would mean that if the 1000 health enemy was inflicted with 500 DoSP it would now only be slowed by 40%.
Multiply the SEF by the WHS to find the effect of a status proc.
The reason SEF exists instead of status effects being based on damage in a vacuum is to make sure that status effects are a consistent experience across the entire game. You don’t want status effects to be completely useless for new players and you don’t want high level weapons to trivialize enemies with “OP” status effects. If an enemy has ten times the health of another enemy, you would expect them to be more resilient to status effects.
To me this just makes a lot more sense. If you use a weapon that hits really hard and inflict a status effect, you would think that the status effect would be more potent than what an SMG would inflict in a single attack. This also naturally makes for a balanced stacking system. A high RoF weapon and low RoF weapon with identical status chances will have the same status “effectiveness” over a period of time. This system automatically balances itself.
Keep in mind that you can deal the full WHS amount of DoSP to an enemy without killing them. This is due to DoSP being unaffected by Armor and/or Shields. The WHS also acts as a cap on the effect of a status proc.
Elemental Status Effects
These are general suggestions, please for the love of god don’t think that this section is the main point of this topic. These are mostly here to give an idea of how the status effects can be converted to the above systems. Again, the purpose of the rework is to propose all status effects to scale with damage, the following are just examples as to how that could be done. If you think some of these suggestions are good, cool. If you think they suck I’d love to hear why it sucks but please don’t use them as proof that the core concept is flawed.
• Heat: Deal 100% of the total DoSP as Heat damage every .5 seconds for 6 seconds. Reduce an enemy’s armor value by 50% of the total DoSP for 6 seconds. Caps at 50% of armor.
• Electricity: A WHS of 100% of damage done to the enemy is chained to enemies in a 8-meter radius as Electricity damage for 6 seconds.
• Cold: Slows the target with a WHS of 100% (fully frozen) for 6 seconds. A WHS of 5 Icicles appear on the target (when the SEF=20% one Icicle will appear, at 40% a second will appear, and etcetera). When an Icicle is hit it deals the total Cold DoSP to the enemy as Finisher damage and destroys the icicle.
• Toxin: Deal 100% of the total DoSP as Toxin damage every .33 seconds for 6 seconds.
• Blast: Enemies in an 8-meter radius have a chance to be inflicted with a Cold or Heat proc relative to the amount of elemental damage of each status type. For example, if the Blast damage is 60% Heat and 90% Cold then there will be a 60% chance that enemies are inflicted with Cold and a 40% chance to be inflicted with Heat. The DoSP of the heat/cold procs is based on the DoSP of each respective damage type.
• Radiation: Creates a radiation field with a WHS range of 15-meters around enemy that deals 50% of the total DoSP every .5 seconds for 6 seconds. Damage from the AoE has a 50% Status Chance. Enemies in range of the field will target the nearest entity.
• Gas: Creates a gas cloud with a WHS range of 15-meters around enemy that deals 100% of the total DoSP every .5 seconds for 6 seconds.
• Magnetic: Pulls enemies in a 8-meter radius towards the target by a WHS of 25m/s for 6 seconds.
• Viral: Amplify damage done to health with a WHS of 500% for 6 seconds.
• Corrosive: Reduce an enemy’s armor value by 100% of the total DoSP for 6 seconds. Caps at stripping 80% of armor.
One more time for the people in the back, these suggestions are not the point of this thread. They are only meant to be examples as to who DoSP can be implemented. A bad suggestion for a specific damage type is not necessarily a sign that the core proposal of this thread is flawed.
Physical Status Mechanics
Physical status is in a weird place due to it being an intrinsic part of weapons that can only be enhanced, not created (outside forced proc mechanics). Of all the things in this proposal this is probably the section I have the most doubt about, but I think that combined with everything else this is the best way to go about it.
Physical status effects are independent of elemental status effects.
• Example: A weapon with a 50% status chance has physical and elemental damage. Every hit now has a 50% chance to proc an elemental status effect and a 50% chance to proc a physical status effect.
Physical status now follows the same DoSP rules as elemental status effects.
This does a few things. First, it makes the base physical status spread of weapons matter again. The status effect of a damage type is just as if not more important than enemy vulnerabilities/resistances to damage types. In addition, it balances the playing field between pure elemental weapons and weapons with physical damage. With emphasis on DoSP pure elemental weapons will gain a great advantage over physical weapons in inflicting status. Elemental weapons being unique is great, but they should not be the general superior status weapon.
Alternatively, don’t do any of this. Have DoSP for physical status effects be based on total physical damage as it is now.
Physical Status Effects
• Slash: Deal 40% of the total DoSP per second for 6 seconds (bypasses armor).
• Impact: Reduce enemy damage by a WHS of 80% and reduce enemy accuracy by a WHS of 70% for 6 seconds. When the SEF > 15% subsequent hits will stagger.
• Puncture: Subsequent shots have a chance to be calculated as “weak point” (2x damage bonus, provides the bonus 2x crit multiplier, bypass shield gating) with a WHS of 150% (A total DoSP above 100% adds a chance to increases the damage bonus to 3x) for 6 seconds.
Forced Status Procs with a Damage Component
If a status proc is forced due to something like Hunter Munitions or a melee stance combo the DoSP will be the base damage of the weapon or the DoSP of the respective damage type, whichever is greater.
Forced Status without a Damage Component
Some forced status effects are done without a damage component. Piercing Roar for example inflicts a Puncture proc without dealing any significant damage. In the case of mechanics that are designed to inflict a status effect without dealing damage their values will be balanced as SEF (Status Effect Factor).
• Example: Piercing Roar inflicts a Puncture proc with a SEF of 50%. This means that any enemy affected by Piercing Roar, regardless of health, will take a puncture effect equal to 50% of Puncture’s WHS (Whole Health Status).
The ability to have forced status effects tied to mechanics that do not deal damage is a valuable tool for DE to have when designing abilities. This type of mechanic could also allow for certain abilities to scale into late game without having to have their damage scale (Frost’s Ice Wave, for example).
UI Representation of Total DoSP
With the current stacking mechanic, we now have a small number that is next to ever status an enemy is inflicted with below their health bar. This transparency in how affected an enemy is by status is a great tool for players, however the above rework would not be able to be represented by a number.
Instead of a number, have the status icon bordered by a circle. A full circle is representative of WHS (the cap for all debuff status effects). Damage status effects already have their potency represented in the UI through the damage numbers that pop up on enemies.
Ease of Modification
If DE were to wish make enemies that are resilient to status effects (like bosses, Liches, etcetera) having all status effects be based on status damage would make it far easier to make blanket changes and effect all status effects proportionally. One way would be to simply give enemies “DoSP Resistance”. This would mean that the initial DoSP dealt to an enemy would be reduced, decreasing the rate by which status effects are inflicted.
• Example: A “DoSP Resistance” of 30% would mean a 100 DoSP Heat proc would only add 70 heat damage to an enemy’s total DoSP.
If DE is more worried about the debuff aspects to status procs, then they could cap WHS to whatever percent they feel adequate.
• Example: Capping WHS at 80% would mean that a viral status would only be able to slow an enemy by up to 400%.
But not only can DE use this to make enemies more resistant, they could use it to make enemies more susceptible to status effects. Imagine if a fire Eximus took double cold DoSP. The ease of modifying the effectiveness of status effects when they are based on DoSP opens the door for new enemy types, new strategies, and new loadout types.
With the latest change to how Liches are affected by status, we got another case of arbitrary numbers having arbitrary consequences. Flat capping the amount of status effects an enemy can be inflicted by only serves to hamstring certain status effects and add another layer of complexity to an already unpredictable system.
Status and Shields
Shield gates did not make shields better, they made things with shields better. Enemies having large shields or even Warframes having large shields was not buffed by adding a shield gate. Grendel makes as much if not more use out of a Shield Gate then a Warframe like Frost.
Shields have an innate 50% DoSP Resistance.
As discussed in the prior section, DoSP gives more tools in DE’s toolbox to tune how everything works together. Enemy shields having innate DoSP resistance would make min/maxing for Corpus fundamentally different than min/maxing for Grineer. And because DoSP is a continuous system, status effects can cleanly carry over to an enemy’s health (and new status effects will add full DoSP) once their shields break.
This change would not benefit Warframes as much as enemies, but it would give shields a bit more of a purpose on Warframes then they currently do.
Disconnecting Status and Critical Hits
*Breaths in*
Have DoSP ignore critical multipliers.
*Breaths out* If this were to be done, many of the proposed numbers for status effects above may need to be adjusted. For me to assemble this concept and present it, I thought it was best to keep it in the scope of the present. It is difficult enough to throw out rough numbers without testing, such a massive change as disconnecting status from crit would make it neigh impossible.
The purpose for this change is to bring down the power of “crit-status hybrid” builds to be more like raw crit and raw status builds. Especially given making all status scale with damage, the power of critical hits would become even more impactful for “status” weapons.
The intention of this rework is to distill the number of variables that effect status to two, status chance and damage. As long as critical hits effect status effectiveness then pure status weapons will remain at a significant disadvantage.
Status Intensity: Status’s version of Critical Damage
Part of the power behind critical mechanics in Warframe is due to it scaling multiplicatively off two stats, critical chance and critical damage. The result of combining these two things is greater than the sum of their parts, leading to the power spike we all know and love with critical weapons.
With status effects being a function of damage, status chance has a lot of similarities with critical chance. And with DoSP being a consistent multiplier for the effectiveness of a status proc, the stage is set for status to get its own version of critical damage that effects all status effects consistently.
Status Intensity: A new weapon stat that exists as a multiplier of DoSP.
• Example: A weapon with 100 and a Status intensity of 1.5x. When the cold status effect is proced the DoSP is 150.
As for balancing this new stat, I would keep the values a bit lower than what Critical Damage can reach. The average base status power for weapons should only be 1. I would not want to see modding Status Intensity be as mandatory for a status weapon as building Critical Damage is for a crit weapon.
Damage Resistances and DoSP
If DoSP and damage are both affected by damage resistances it would lead to enemy resistances/vulnerabilities having a multiplicative effect on each other in the case of damage status procs. There are a few remedies for this, but to be honest I am torn on the right direction.
Option 1: The damage of damage status procs is unaffected by resistances.
This is the quick fix to keep the system working as is. There would only be one instance where enemy resistances are applied to a damage calculation.
Option 2: DoSP ignores enemy resistances.
The damaging effects of status effects (when applicable) are still affected by damage resistances. This would mean damage status procs would have a reduced effectiveness as the damage of the status procs align with the elemental damage, just not with any weird stacking. However, debuff status effects would not be influenced at all by resistances. For example, a Cold status would have the same effectiveness versus a Lancer as a Crewman while a Heat status effect would deal more damage to the Lancer than the Crewman. Big downside here is obviously the inconsistency.
Option 3: Damage resistances effect DoSP but not damage.
The downside to this is that it makes choosing your elements for non-status weapons irrelevant. The benefit of this system could be interesting though, it could open the field to bring different damage types and/or status effects to different factions. You could build Magnetic against Grineer and not get a massive hit to DPS. Your status effects would be incumbered, but they would still be present. It may not even be a bad thing if crit weapons ignored enemy resistances, crit weapons fill the role of an easy, general use weapon. Status weapons would require more min/maxing but if built correctly could exploit enemy weaknesses. Sometimes having a casual option is beneficial.
Option 4: Remove resistances.
Have damage just be damage, the only thing damage types contribute is different status effects. On the surface this is a massive simplification, but it could potentially make for more variation in build crafting if you do not have to worry about being forced into cookie-cutter damage matchups.
This is enough to be its own separate topic and discussion, so I’ll leave it there.
Status 5.0: Death to Status Chance, long live Status Power
As the title of this section insinuates, this is a separate suggestion. Consider this a mental exercise.
Every hit deals a status effect. Rename the Status Chance stat to Status Power. Status Power is a multiplier on DoSP.
• Example: A weapon has 1000 Heat damage and 500 Corrosive damage. The weapon had a Status Chance of 50%, now it has a Status Power of 50%. Every shot from the weapon will deal a heat status proc with a DoSP of 500 and a Corrosive status proc with a DoSP of 250.
In some respects, this is just deleting Status Chance, only leaving the new mechanic of Status Intensity. Status Intensity and Status Power as essentially the same thing, however in the case of this side-proposal they have been merged.
The beauty of making status based on damage is that this rework of the rework would not actually take a lot to do. Status Chance and “Status Power” are basically the exact same thing. On average a weapon using Status Power and one using Status Chance will inflict the exact same DoSP over time. It uses all the same principles but entirely removes RNG from status effects.
Conclusions
I do not think it can be overstated how powerful it would be if status chance were a direct correlation to how good a weapon is at inflicting status effects. A good status weapon will have a high status chance, a bad status weapon will have a low status chance, and this goes for all status types. When DE balances a weapon they do not have to worry about extraneous variables like base fire rate and damage types when balancing the status effectiveness of said weapon. All they must worry about is adjusting a single number.
The reason for removing base damage from the elemental status calculations is to increase the value of modding for elementals. As it stands it is generally the most effective to simply maximize damage output while making sure the element types are correct. By having elemental status effects be only affected by the damage of an element, that means that the elemental mods will add far more value to your build than they currently do. Hopefully to the extent that equipping another elemental mod may have a chance at replacing some “mandatory mods” in niche situations.
If you are concerned with hos this 1:1 stacks up against current status calculations, this is a nerf to any status effect that has less than +100% elemental damage from mods, a buff to everything above +100% elemental damage, and a considerable buff to a weapon with innate elemental damage (for the status effects that use that innate element).
Finally, the changes to critical interactions would work in tandem with everything else to make modding for status a viable alternative, not just addition to, modding for crit. Without the changes to how status effectiveness is calculated making raw status weapons compete with hybrid builds is nearly impossible.
The secret power to this rework comes with the potential of status chance and critical chance being interchangeably powerful. If two weapons, one with a 40% status chance and one with a 40% critical chance, were both roughly the same power level the challenge of balancing weapons would be made astronomically easier. Just imagine what it would do for the modular weapon meta.
Preliminary Q and A
“This is too complicated”
The math that makes something work is not indicative of how complicated a system is to work with. At its core, the way all this works is “if bullet do big damage, bullet do big status” and “If bullet do a lot of little damage, bullet stack little status into big status”. Where the current system is based on an arbitrary stacking mechanic that would require specific knowledge of how everything stacks, the proposal above makes everything simple in concept. Explaining to a player “If it does a lot of damage, it will do a lot of status” is inherently a lot simpler than “Some statuses do damage, some increase effectiveness with more hits. Sometimes the stacking of hits isn’t proportional and there are spikes. Oh, and there are caps on the amount of stacking sometimes, but not always. Those caps also sometimes change depending on the enemy.” I would argue that the proposal above is the most logical and expected way for the system to work for a new player. It doesn’t matter that the equations are complicated, all you need to know to start building for status is that the amount of elemental damage you do makes status better and that status chance makes it happen more often.
“Why doesn’t status duration scale with status effect factor (SEF)”
If status duration scaled with SEF then inflicting a status effect would make subsequent status effects last longer which would make subsequent status effects last longer which would… I think you get the idea. This would mean that the effect of a status proc would be tied to two factors, the DoSP of the new status proc and the total DoSP. When you add multiple scaling factors to anything it is important to make sure that they do not compound with each other.
“You just said compounding factors were bad, why did you suggest status effects that have both range and damage scale?”
Maybe a little bit of hypocrisy, but I think this is a different case. For one these effects still scale with total DoSP, independent of status per second. There also are not any weird companioning effects like there could be if duration scaled with SEF. However, by having two effects that enhance each other it does mean that status effectiveness will not scale linearly with DoSP. Personally, I think that having a few status effects scale better with DoSP than others isn’t all bad if it is kept in check, it adds another layer to modding. In addition to picking status effects for their effects, you may want different effects based on how good of a status weapon you are using. Toxin is a status effect with a consistent value, so it will still have good value on low status weapons. Whereas Gas’s dual scaling means that you don’t get its full value when used on said lower status weapon, with a high status weapon you have the opposite case. Maybe this isn’t the play and the range of a status effect should be unaffected by DoSP, I’d like to hear what yall think.
“Why you nerf crit weapons?”
For status weapons to compete with critical weapons, and for each of those weapons to not be overshadowed by hybrids, multiplicatively stacking multipliers need to go away. This is not intended to be a “nerf” to crit weapons, the larger intention of this change is to allow for buffing status weapons (through buffing status effects) without inadvertently making crit/status hybrids even more powerful.
“You did not explain why the current system is bad”
This is true, my hope was that by reading this proposal the reasons why status stacking is bad would shine through. I could have ranted for a few hundred to thousand words about why status stacking is bad, but I feel like it is easier to explain/understand why status stacking is bad by showing the alternative. Unless status is balanced raw status weapons will NEVER be able to compete with crit consistently. Having to account for fire rate when balancing status is a massive hurdle to DE’s design space.
“That was a lot of words, how long have you been at this?”
In one form or another, since 2017. It has seen a lot of iteration, a couple ground-up rewrites, and many hours being stared at. I need a life. The latest status revision gave me a kick to finish this up… then I sat on it for six months… then [DE]Scott talking about more status changes got me to finally post it.
“This was a waste of time; DE would never do this”
Probably, but it was fun to put together and I can hope.
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Dident read everything but arent there gonna be a problem in the internal engiens calculations due to some weapons being able to produce damage up in the milion?
But i like the idea that it scales of damage
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Wouldn't this make status procs from non scaling damage sources scale backwards? The higher the health of the enemy the harder it is to get the same strength of a status proc. Making it practically impossible to slow down a lvl 9999 enemy with cold status unless you're using scaling damage or forcing a % of the WHS like your suggestion with piercing roar.
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23 minutes ago, Liljeman said:
Wouldn't this make status procs from non scaling damage sources scale backwards? The higher the health of the enemy the harder it is to get the same strength of a status proc. Making it practically impossible to slow down a lvl 9999 enemy with cold status unless you're using scaling damage or forcing a % of the WHS like your suggestion with piercing roar.
I don't think "scale backwards" is a fair way to say it. They just won't scale. If you don't mod a weapon and go against a level 10 then level 100 enemy the damage doesn't scale down, it just did not scale at all. So yes, if you only deal 10 cold DoSP to a level 9999 enemy it will be basically useless. Much like how if you deal 10 damage to a level 9999 enemy it will be basically useless.
You mention Piercing Roar and it seems like you got the gist of how that works. For any ability/mechanic that DE wants to inflict status but not deal (significant) damage then skipping DoSP and going straight to SEF (status effect factor) is how to do that. While there are some abilities that should get this treatment I think there is also an argument that some abilities should be given some sort of scaling damage instead (the damage scaling equation used with Xaku, Grendel, and Vauban has its issues but that is a separate discussion).
At the end of the day this is a nerf to the ability for status effects to infinitely scale with enemies, I won't try to deny that. But I would argue that status effects being independent of enemy scaling is one of the things that makes it nearly impossible to balance the game. The whole point of level scaling is to make things more difficult to deal with, if a core part of our arsenal ignores scaling then issues will always arise.
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A couple of points here. If you're going to have some Status Effects deal damage, then ALL Status Effects need to deal damage (or reduce enemy EHP). For various reasons, Warframe is a game where DPS is king. If players have the choice, they'll always go for the Status Effects that deal damage, thus artificially limiting build variety. Having all Status Effects deal damage (not necessarily to the same extent, just some damage) would get around this problem, as players would then be free to pick the Status Effect they like, rather than having this decided for them by game mechanics. It also solves the issue of Status Guns being quite terrible for damage unless you're cheesing Slash procs or shredding armour - again, a small handful of all possible Status effects.
I'd go one further and simply remove health-type-specific damage resistances altogether, actually, but that's a separate subject. Would solve one of the central issues your proposal is trying to get around, though.
Secondly, can we please face facts and admit that we need "Status Magnitude" as a weapon stat already? If your weapon is a "crit gun," you need to deal with Critical Chance and Critical Damage. Chance determines how likely you are to land a crit, damage determines the magnitude of the crit once it procs. Status, by contrast, has ONLY Status Chance to work with. The magnitude of the actual status effects is fixed, so they almost ALWAYS favour rapid-firing, low-damage weapons. Because you have no means of boosting your status effect, it makes no sense to pick a slow-firing weapon. Yes, your proposal aims to fix this, but it does so via a more "hardcoded" way. Rather than tying status effect magnitude to weapon damage, just move Status Magnitude out to its own stat.
This way you could have weak, slow-firing AoE weapons which nevertheless still deal substantial amounts of Status anyway. You know what's even better? You can tie the amount of damage Status Effects do (as a percentage of that damage component) TO Status Magnitude. That way, even weak guns can still do a lot of Status damage via procs without needing to use one of the, like, three Status Effects that actually scales well. And, if you took my previous suggestion and made all Status effects also deal damage, then that makes Status weapons of ALL damage types into a worthwhile competitor to critguns for actually killing stuff.
I'd personally go one further by removing status/critical chance entirely, have all weakpoint hits be critical hits and all hits in general deal a little bit of status, but that's well outside the scope of this thread. You've likely seen me post it already, as well.
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7 minutes ago, DrBorris said:
I don't think "scale backwards" is a fair way to say it. They just won't scale.
Fair enough, I was arguing with myself if I wanted to go with backwards or not scaling at all anyway.
26 minutes ago, DrBorris said:
But I would argue that status effects being independent of enemy scaling is one of the things that makes it nearly impossible to balance the game. The whole point of level scaling is to make things more difficult to deal with, if a core part of our arsenal ignores scaling then issues will always arise.
I can see your point here, having too easy access to a cc that makes most enemies irrelevant destroys the purpose of scaling entirely.
My fear though is that this system would be a huge nerf to all weapons forcing you to use cheesy tactics much earlier and make abilities the only viable option very quick.
To solve this we would have to discuss what can make an enemy more difficult and how, where and when those difficulties gets added to enemies. Your system seems nice, clean and easy to use for both players and developers but I can't see it work with how the enemy and reward scaling is working right now.
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4 hours ago, Steel_Rook said:
A couple of points here. If you're going to have some Status Effects deal damage, then ALL Status Effects need to deal damage (or reduce enemy EHP). For various reasons, Warframe is a game where DPS is king. If players have the choice, they'll always go for the Status Effects that deal damage, thus artificially limiting build variety. Having all Status Effects deal damage (not necessarily to the same extent, just some damage) would get around this problem, as players would then be free to pick the Status Effect they like, rather than having this decided for them by game mechanics. It also solves the issue of Status Guns being quite terrible for damage unless you're cheesing Slash procs or shredding armour - again, a small handful of all possible Status effects.
I started that section by saying "this isn't the point of this proposal" and I ended the section with "this isn't the point of this proposal". Status effects aren't in a vacuum, in order to deal a status effect you need to deal damage. Especially in this proposal where I emphasize that in order to deal status you need to deal damage. And even in regards to the ones I propose I do believe that they all enable damage.
• Heat: Deals damage and makes enemies susceptible to more damge.
• Electricity: Turns single target damage into AoE damage.
• Cold: Exposes enemies to finisher damage.
• Toxin: Just deals damage
• Blast: See Heat and Cold
• Radiation: Deals damage with a side of CC, also will spread its effect to other enemies (which deals damage)
• Gas: Deals damage in an AoE
• Magnetic: Groups enemies together. Yes, this one does not directly make number go up, but consider that Magnetic/Gas is a thing.
• Viral: Exposes enemies to more damage
• Corrosive: Makes enemies susceptible to more damage.
4 hours ago, Steel_Rook said:
I'd go one further and simply remove health-type-specific damage resistances altogether, actually, but that's a separate subject. Would solve one of the central issues your proposal is trying to get around, though.
Odd that you say this is a separate subject when I have a whole section dedicated to how resistances will work with status. I also say it is a bigger issue that could have its own discussion, but I lay out what the options are including just killing resistances.
4 hours ago, Steel_Rook said:
Secondly, can we please face facts and admit that we need "Status Magnitude" as a weapon stat already? If your weapon is a "crit gun," you need to deal with Critical Chance and Critical Damage. Chance determines how likely you are to land a crit, damage determines the magnitude of the crit once it procs. Status, by contrast, has ONLY Status Chance to work with. The magnitude of the actual status effects is fixed, so they almost ALWAYS favour rapid-firing, low-damage weapons. Because you have no means of boosting your status effect, it makes no sense to pick a slow-firing weapon. Yes, your proposal aims to fix this, but it does so via a more "hardcoded" way. Rather than tying status effect magnitude to weapon damage, just move Status Magnitude out to its own stat.
Did... did you read this thread? For one, I mention this exact mechanic. And secondly THE ENTIRE REASON for this thread to exist is to remove fire rate as a factor in status application entirely. You don't need "status magnitude" (I call it Status Intensity above) in order to level the fire rate playing field, all you need is for all status effects to be an exclusive function of damage. That is the point if this thread, not the suggestions for status effects, the point is to make all status effects based on damage in order to make status chance a consistent and representative factor.
4 hours ago, Steel_Rook said:
I'd personally go one further by removing status/critical chance entirely, have all weakpoint hits be critical hits and all hits in general deal a little bit of status, but that's well outside the scope of this thread. You've likely seen me post it already, as well.
... I really appreciate the time you put into the Forums. There aren't many people that give thoughtful responses and approach so many threads in a logical/civil manner. I haven't been nearly as active as I have in the past but seeing your comments here gives me the hope to occasionally comment myself. That said... did you read the OP? Because I have a section dedicated to this exact thing (well, not the crit part). Furthermore due to the suggested mechanics of DoSP removing status chance in favor of every hit dealing status is extremely easy, the above topic is a foundation that can remolded into many different directions. It feels like you came here with your own opinions on the topic, skimmed to the part you understood at a glance (the status suggestions), then commented without giving the effort to know what I was actually suggesting. It is a lot of words, I know, but could've you just waited till you had time to go through it before you responded? I am sorry if I am misjudging but given your response it doesn't feel like you tried to understand the proposal.
4 hours ago, Liljeman said:
My fear though is that this system would be a huge nerf to all weapons forcing you to use cheesy tactics much earlier and make abilities the only viable option very quick.
To solve this we would have to discuss what can make an enemy more difficult and how, where and when those difficulties gets added to enemies. Your system seems nice, clean and easy to use for both players and developers but I can't see it work with how the enemy and reward scaling is working right now.
It absolutely has that potential, all foundational reworks have that potential. But I hope that the foundation of DoSP is a solid enough one that DE could balance it right. It would be up to finer testing of the values associated with status effects that would determine how good the system actually feels to use.
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This was a lot to digest. Please correct me if I got something wrong.
19 hours ago, DrBorris said:
A status proc that deals 10 damage per second on an enemy with 100 health is far more potent at achieving the goal when compared to a 1000 health enemy. Damage status procs are already a function of DoSP and health, it only makes sense for debuff status procs to work the same way.
I disagree with this approach.
Primary goal of debuff status effects is to debuff, weaken or controll enemies. As such (low damage) utility weapons have a purpose, as they can fill supporte rolles within your loadout. Your concept directly ties CC/utility/support to damage, while damage and support are 2 different domains. Why would I want to CC or weaken a half dead enemy? What if I just want to controll a group of enemies?
Current status stacking is simple and already covers a good number of feats yout try to reinvent. Electric status can deal 0 damage but still CC an enemy. A progressive way to improve existing system would be to create new stacking behaviour (like a new effect after X stacks). You on the other hand had to invent new modiefiers like SEF or WHS to create a workaround around status being tied to enemy health. I am also confused why you are talking about "consistent experience" as the effectiveness of your weapon shifts to the "bad side" over the course of a mission as lvls/HP rise.
I get the impression your intent is to limit status effects against high lvl enemies. If I am correct you achieved this goal.
20 hours ago, DrBorris said:
To me this just makes a lot more sense. If you use a weapon that hits really hard and inflict a status effect, you would think that the status effect would be more potent than what an SMG would inflict in a single attack. This also naturally makes for a balanced stacking system. A high RoF weapon and low RoF weapon with identical status chances will have the same status “effectiveness” over a period of time. This system automatically balances itself.
Weapons already work like that, since harmfull status effects are tied to weapon damage, while weapons damage itself is balanced by firerate. A contestable point is whether you actually need only one hit to inflict a full utility status effect.
20 hours ago, DrBorris said:
Ease of Modification
If DE were to wish make enemies that are resilient to status effects (like bosses, Liches, etcetera) having all status effects be based on status damage would make it far easier to make blanket changes and effect all status effects proportionally. One way would be to simply give enemies “DoSP Resistance”. This would mean that the initial DoSP dealt to an enemy would be reduced, decreasing the rate by which status effects are inflicted.
If DE is more worried about the debuff aspects to status procs, then they could cap WHS to whatever percent they feel adequate.
Current status system offers a more elegant solution, since status effects have a stacking nature, you just need to adjust individual proc limit on an enemy. This allows to create selective weaknesses/resistences to specific status effects, Your idea to adjust WHS would affect all status effects, thus is a less flexible approach.
21 hours ago, DrBorris said:
But not only can DE use this to make enemies more resistant, they could use it to make enemies more susceptible to status effects. Imagine if a fire Eximus took double cold DoSP. The ease of modifying the effectiveness of status effects when they are based on DoSP opens the door for new enemy types, new strategies, and new loadout types.
This idea is already in the game, realized by health type modifiers. What confuses me is that you previously said status should not be influenced by health types, yet you reintroduce the same mechanic under a diffirent name.
21 hours ago, DrBorris said:
With the latest change to how Liches are affected by status, we got another case of arbitrary numbers having arbitrary consequences. Flat capping the amount of status effects an enemy can be inflicted by only serves to hamstring certain status effects and add another layer of complexity to an already unpredictable system.
Bad execution does not mean bad concept. Nothing stopes DE from applying arbitrary numbers to your concept.
21 hours ago, DrBorris said:
Disconnecting Status and Critical Hits
I am really confused by this point, not because you want to disconnect status from crits, but because you previously strenghtened the tie between status to damage in your concept.
I get the idea to separate crit weapons from status weapons from hybrid weapons. However, current misery is primary created by weapons with high crit as well as status stats and forced procs from mods like HM or stance combos.
Second mistake was to cap Corrosive, since potent status weapons cannot reduce enemy defence to 0 and then overload them with otherwise lethal status effects.
Third mistake is that Toxin (not even Toxin status) ignores Shields, so that shielded enemies can be bruteforced with high damage, yes that includes crit and not status, weapons.
Especially points 2 & 3 make high status weapons largely ineffective against enemy defences, while crit weapons have access to high damage as well as powerfull forced procs; not to mention reworked status priority favors low status weapons with forced procs -> crits + Viral meta. And then there is of course status immunnity of VIP targets.
Status revisit advertised to make status weapons more powerfull, but it actually dumpstured them. Damage/status system is awfully managed as a whole.
21 hours ago, DrBorris said:
Status Intensity: Status’s version of Critical Damage
Probably long overdue to add this. Even though bane mods kinda do this already and Empowered Blades alreaedy introduced Status Intensity.
20 hours ago, DrBorris said:
Option 2: DoSP ignores enemy resistances.
The damaging effects of status effects (when applicable) are still affected by damage resistances. This would mean damage status procs would have a reduced effectiveness as the damage of the status procs align with the elemental damage, just not with any weird stacking. However, debuff status effects would not be influenced at all by resistances. For example, a Cold status would have the same effectiveness versus a Lancer as a Crewman while a Heat status effect would deal more damage to the Lancer than the Crewman. Big downside here is obviously the inconsistency.
I do not perceive this as incosistency. Different enemies can be resilient or vulnerable to different effects. I would say this creates the need to diversify your loadout in the first place.
TL;DR: You remove different aspects of the current system and then reintroduce the same ideas under a differnt name. You deem current system faulty based on a bad execution.
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As hesitant as i usually am with nerfing, this time I agree that the crit/status hybrid build is due for a nerf with decoupling status and crit. Otherwise any buff to status effects will just buff hybrid weapons far more than pure status weapons, meaning we are basically just where we started but with more damage.
A bit of a weird case for this however are heavy attack builds on weapons with forced slash procs on heavy attack. They are a weird sort of hybrid and I'm not sure how good they would be if crit didnt affect the slash proc anymore. On the other hand heavy attack builds without good base crit chance arent really a thing currently. Maybe for heavy attacks the proposed status intensity value could be innately higher.
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Abit off topic but I'm curious on how this system would work when enemies deal damage to us. I don't think it would be very enjoyable fighting enemies that caps a very strong status effects on us with 1 bullet. On the other hand I would want a system that's fair: enemies follow the same rules as us for dealing damage with crits, status, headshots etc.
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3 hours ago, ShortCat said:
Primary goal of debuff status effects is to debuff, weaken or controll enemies. As such (low damage) utility weapons have a purpose, as they can fill supporte rolles within your loadout. Your concept directly ties CC/utility/support to damage, while damage and support are 2 different domains. Why would I want to CC or weaken a half dead enemy? What if I just want to controll a group of enemies?
Is it a good thing that we have purely utility weapons in Warframe? Almost every point of progression is about making damage (per second) number go up. If you don't have powerful enough tools to deal with enemies should you be able to deal with enemies?
It would be dishonest to say that a system like this is faultless, but I think the advantages it brings are far more worthwhile than what status stacking brings. Kitguns are the example that keeps coming to my mind. As it stands, balancing kitguns is a mess mostly due to status. The burst/sustained DPS is very well balanced between the different mag size loaders but the effectiveness of status vs. crit is hilariously bad. With status stacking there is no easy fix because not only can we change loaders, we can change fire rate. In addition status is not a linear scaling system so even if fire rate was consistent you wouldn't be able to have it hit punch-for-punch with crit scaling. In theory, if the status effect numbers are balanced right, with the above rework all kitguns could be naturally balanced. It wouldn't take extra tweaking, a 30% status chance is a consistent status effectiveness across the board. It just works.
Now imagine how easy it would be for DE to balance everything else. This is the core to the proposal, the new massive advantage that exists. Status based on hits is not without its advantages but it will never be able to offer the consistency that status based on damage can.
3 hours ago, ShortCat said:
Current status stacking is simple and already covers a good number of feats yout try to reinvent. Electric status can deal 0 damage but still CC an enemy. A progressive way to improve existing system would be to create new stacking behaviour (like a new effect after X stacks). You on the other hand had to invent new modiefiers like SEF or WHS to create a workaround around status being tied to enemy health. I am also confused why you are talking about "consistent experience" as the effectiveness of your weapon shifts to the "bad side" over the course of a mission as lvls/HP rise.
Would you say it is inconsistent for your unmodded Braton to deal less damage to a level 10 enemy than a level 50 one? Of course not. I'm applying the same logic to status. Weapons exist to kill enemies. They kill enemies in a variety of ways but at the end of the day that is what they are there for. A weapons may kill enemies with a side of CC, but it is still to mainly kill enemies.
The core issue I see with status stacking is how it scales with fire rate (a stat that has little correlation to progression) over damage. It also just... stops. You hit an enemy 10 times and then your status is done doing status thing despite you continuing to inflict status. A consistent experience isn't your weapons doing the same things to enemies disconnected to their level rising. A consistent experience is the effects of everything scaling as enemy level scales.
3 hours ago, ShortCat said:
I get the impression your intent is to limit status effects against high lvl enemies. If I am correct you achieved this goal.
To an extent, yes. High level enemies are high level enemies and we should have to approach them as such. I'm not at all suggesting that level 100 enemies should now be able to tank status, that comes down to finer balancing that no one can properly understand without testing, but I do think that a level 500+ should probably be shrugging off your 10 damage cold procs.
3 hours ago, ShortCat said:
Weapons already work like that, since harmfull status effects are tied to weapon damage, while weapons damage itself is balanced by firerate. A contestable point is whether you actually need only one hit to inflict a full utility status effect.
Only two status effects are tied purely to damage. I don't think it is a very good system when you are confined to two status effects if you want to make the best use out of low rate of fire weapons. Wouldn't it be more interesting if I could mod a Daikyu for cold and have it freeze enemies solid that it did not manage to kill?
3 hours ago, ShortCat said:
This idea is already in the game, realized by health type modifiers. What confuses me is that you previously said status should not be influenced by health types, yet you reintroduce the same mechanic under a diffirent name.
I debated leaving that sentence in at that early part. I eventually kept it because I wanted people to forget about as many extraneous variables when going through the core proposal.
Damage resistances are something every enemy has, and they affect, ya know, raw murder damage. DoSP resistance is a bit different in that it does not reduce raw TTK and that every enemy wouldn't necessarily have it. In order to be an enemy in this game you need to have a health type. If you have a health type you have resistances. DoSP resistance isn't something that I suggest with the intention of giving it to every enemy.
3 hours ago, ShortCat said:
Bad execution does not mean bad concept. Nothing stopes DE from applying arbitrary numbers to your concept.
How else is DE supposed to retool the current system? If DE wants to reduce the effectiveness of status on an enemy (which is fair) then reducing the stack cap is the most logical way to go about it in my opinion. This has the disadvantage of not affecting damage procs evenly, but if they wanted to add that they could add a layer of DoT damage reduction. There isn't a way to retool status stacking where it doesn't hit one thing harder than another, at least there isn't without a perfect level of balancing, something I wouldn't expect of anyone.
3 hours ago, ShortCat said:
I am really confused by this point, not because you want to disconnect status from crits, but because you previously strenghtened the tie between status to damage in your concept.
I get the idea to separate crit weapons from status weapons from hybrid weapons. However, current misery is primary created by weapons with high crit as well as status stats and forced procs from mods like HM or stance combos.
I removed the tie between crits and status explicitly because this proposal strengthens the tie between status and damage. Status weapons should be status weapons. Crit weapons are crit weapons. Hybrid weapons will absolutely exist but they will fall in between the two instead of above them.
Having some options with forced status is fine, HM is a band-aid for primary weapons that may need addressing but it doesn't break this proposal. In the case of building for crit and for status you want base damage, a forced slash proc on a crit weapon will be equivalently powerful to a forced proc on a status weapon. To me this makes the idea of balancing even forced procs easier.
3 hours ago, ShortCat said:
Second mistake was to cap Corrosive, since potent status weapons cannot reduce enemy defence to 0 and then overload them with otherwise lethal status effects.
The suggestions weren't supposed to be the point, they are mostly there for example. I kept the corrosive cap because DE went out of their way to cap corrosive in the latest status revision.
3 hours ago, ShortCat said:
Third mistake is that Toxin (not even Toxin status) ignores Shields, so that shielded enemies can be bruteforced with high damage, yes that includes crit and not status, weapons.
Again, not the whole point, and to be honest I was just kinda scared of suggesting anything beyond the status quo with Toxin. Keep in mind the 50% DoSP resistance I suggested for shields, that could possibly also affect Toxin's shield bypass as well. Either way Toxin is a scary discussion I did not want to poke.
3 hours ago, ShortCat said:
Especially points 2 & 3 make high status weapons largely ineffective against enemy defences, while crit weapons have access to high damage as well as powerfull forced procs; not to mention reworked status priority favors low status weapons with forced procs -> crits + Viral meta. And then there is of course status immunnity of VIP targets.
Remember that status effects are calculated on enemy HEALTH, not EHP. Shielded enemies do get some resistance to DoSP, but an armored unit will be susceptible to debuff effects just as much as the Infested.
A benefit I envision with this proposal is that vip targets having status immunity will be far less mandatory if DE wants those targets to stay alive for a few seconds. DE won't have to worry about us instantly getting the max Viral multiplier, instantly slowing them to a crawl, instantly reducing their armor.
3 hours ago, ShortCat said:
Probably long overdue to add this. Even though bane mods kinda do this already and Empowered Blades alreaedy introduced Status Intensity.
I have trouble imagining status intensity being a easy stat to introduce when so many status effects don't have an obvious intensity. And it could easily just lead to things DE would see as problems if we were able to push status effects past the thresholds they put in place for a reason. Status intensity is an easy stat to implement with DoSP, it is just a multiplier on a number that affects every status equally.
3 hours ago, ShortCat said:
I do not perceive this as incosistency. Different enemies can be resilient or vulnerable to different effects. I would say this creates the need to diversify your loadout in the first place.
Maybe, I left all the options in because I did not think it would be fair to pretend I knew which would be best. My bias against the second option is clear but I can see its advantages. I would personally prefer if you took different status effects because of their effects, but a little meta chess does have its value as well.
3 hours ago, ShortCat said:
TL;DR: You remove different aspects of the current system and then reintroduce the same ideas under a differnt name. You deem current system faulty based on a bad execution.
I very much disagree that the faults of status stacking on hits are only in execution. Fire rate being a more important stat in inflicting status effects makes balancing the game a mess at a fundamental level because fire rate isn't a stat that scales with progression. Then having some status effects that scale with damage instead of crits only makes the issue worse. Also, I don't believe you gave fair credit to the things the above system allows for that would be impossible with status stackin on hits, back to the example of kitguns (and the effort to balance all weapons in the game). Imagine how easy it would be for DE to balance weapons if a weapon with a 30% critical chance was nearly equally powerful to a weapon with a 30% status chance given similar base damage. Sure, this could in theory be done in the current system if both weapons had the same fire rate, but how many weapons would this apply to?
2 hours ago, Liljeman said:
Abit off topic but I'm curious on how this system would work when enemies deal damage to us. I don't think it would be very enjoyable fighting enemies that caps a very strong status effects on us with 1 bullet. On the other hand I would want a system that's fair: enemies follow the same rules as us for dealing damage with crits, status, headshots etc.
Damaging status procs are currently the only ones that pose a real threat to us and those already scale with damage. Also due to the astronomical difference between player and enemy EHP I think there is a case that some status effects should be adjusted on a case-by-case basis on how they apply to the player.
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3 hours ago, DrBorris said:
Is it a good thing that we have purely utility weapons in Warframe? Almost every point of progression is about making damage (per second) number go up. If you don't have powerful enough tools to deal with enemies should you be able to deal with enemies?
In my opinion - Yes. We have 3 weapons + 1 Frame in our loadouts. Combining those to achieve a common goal seems like a good design to me.
3 hours ago, DrBorris said:
...kitguns...
Personally, I find Kitguns (or Zaws) bad design, precisely because players have advanced controll over base stats.
3 hours ago, DrBorris said:
Would you say it is inconsistent for your unmodded Braton to deal less damage to a level 10 enemy than a level 50 one? Of course not. I'm applying the same logic to status. Weapons exist to kill enemies. They kill enemies in a variety of ways but at the end of the day that is what they are there for. A weapons may kill enemies with a side of CC, but it is still to mainly kill enemies.
I said it later in my post, but this very point is contestable. Does it make sense to uncover full status effect with only 1 hit? In some cases - yes. It all boils down to the fundamental idea that it is OK for gear to have only utility purpose. You disagree with this stance, but my support for this idea comes from there.
Nevertheless, your concept would put high fire rate, low damage weapons - especially focused on utility status - far behind.
3 hours ago, DrBorris said:
The core issue I see with status stacking is how it scales with fire rate (a stat that has little correlation to progression) over damage. It also just... stops. You hit an enemy 10 times and then your status is done doing status thing despite you continuing to inflict status. A consistent experience isn't your weapons doing the same things to enemies disconnected to their level rising. A consistent experience is the effects of everything scaling as enemy level scales.
Status chance is a more important stat than firerate. Furtheremore, there are ways to scale both.
10 stack limit is the rule from the current system and in some cases it is indeed ill managed. Your proposal however has the same boundaries: at some point weapons will run into an equilibrium of new statuses applied <-> old status expired; it will just vary on weapon to weapon basis, which is, if you ask me, even less consistent or intuitive than a flat status cap.
Furthermore, I also mentioned additional effects after X stacks. One idea of mine for cold is that each proc slows the enemy by % untill an enemy is frozen(=100%) and at this stage the enemy would take increased damage from all sources, while subsequent cold procs would not be wasted, but maintain this frozen state.
3 hours ago, DrBorris said:
Only two status effects are tied purely to damage. I don't think it is a very good system when you are confined to two status effects if you want to make the best use out of low rate of fire weapons. Wouldn't it be more interesting if I could mod a Daikyu for cold and have it freeze enemies solid that it did not manage to kill?
Fire, Toxin, Electricity scale from base damage as well as elemental damage; slash & gas scale from base damage. All harmfull status effects scale from damage. The above statement is not correct.
3 hours ago, DrBorris said:
Damage resistances are something every enemy has, and they affect, ya know, raw murder damage. DoSP resistance is a bit different in that it does not reduce raw TTK and that every enemy wouldn't necessarily have it. In order to be an enemy in this game you need to have a health type. If you have a health type you have resistances. DoSP resistance isn't something that I suggest with the intention of giving it to every enemy.
Then nothing changes in this regard.
3 hours ago, DrBorris said:
How else is DE supposed to retool the current system?
I would start by reducing the total number of elements, banning rainbow builds and let elements fulfill a destinct role. At the same time I would rework enemy health types under one cohesive design plan.
3 hours ago, DrBorris said:
If DE wants to reduce the effectiveness of status on an enemy (which is fair) then reducing the stack cap is the most logical way to go about it in my opinion. This has the disadvantage of not affecting damage procs evenly, but if they wanted to add that they could add a layer of DoT damage reduction. There isn't a way to retool status stacking where it doesn't hit one thing harder than another, at least there isn't without a perfect level of balancing, something I wouldn't expect of anyone.
Within the current system DE has 2 levers, status cap & health type multiplier. Want to adjust a utility proc? - change the cap (here I assume it is possible to assign an individiual cap to each element). Want to adjust proc damage? - adjust health type. Then there is one bonues lever in form of armor/shields, which can "cover" and create 2-3 layered encounters. (Not in the traditional WF way, but kinda like 3 health bars with diffeernt weaknesses. If necessary of course.)
For the record, I am not saying current damage/status system is good, because it is not. I am saying it has a solid basis and its geenral concept (different elements & status stacks & health types) can cover many differnt scenarios.
3 hours ago, DrBorris said:
I removed the tie between crits and status explicitly because this proposal strengthens the tie between status and damage.
That's even worse. Low damage weapons in general would be completely useless, without any option for redemption.
3 hours ago, DrBorris said:
The suggestions weren't supposed to be the point, they are mostly there for example...
I wasn't talking about your status effects in this part, but the actual faults of the damage system.
6 hours ago, ShortCat said:
However, current misery is primary created by weapons with high crit as well as status stats and forced procs from mods like HM or stance combos.
^Theses/first argument
Second mistake was to cap Corrosive, since potent status weapons cannot reduce enemy defence to 0 and then overload them with otherwise lethal status effects.
Third mistake is that Toxin (not even Toxin status) ignores Shields, so that shielded enemies can be bruteforced with high damage, yes that includes crit and not status, weapons.
Especially points 2 & 3 make high status weapons largely ineffective against enemy defences, while crit weapons have access to high damage as well as powerfull forced procs; not to mention reworked status priority favors low status weapons with forced procs -> crits + Viral meta. And then there is of course status immunnity of VIP targets.
^Argumentation
Status revisit advertised to make status weapons more powerfull, but it actually dumpstured them. Damage/status system is awfully managed as a whole.
^Conclusion
3 hours ago, DrBorris said:
I have trouble imagining status intensity being a easy stat to introduce when so many status effects don't have an obvious intensity. And it could easily just lead to things DE would see as problems if we were able to push status effects past the thresholds they put in place for a reason. Status intensity is an easy stat to implement with DoSP, it is just a multiplier on a number that affects every status equally.
Status intensity would be just another conditional +damage mod.
3 hours ago, DrBorris said:
Maybe, I left all the options in because I did not think it would be fair to pretend I knew which would be best. My bias against the second option is clear but I can see its advantages.
I speak in favor of this "color codding" approach because otherwise I see specifically Warframe running into longterm issues. With several hundred weapons available, it becomes difficult to create new meaningfull weapons without feeding power creep. Color codding allows to create weapons with identical stats in all but damage type and those would still have their own niche.
The other solution is to create weapons with perks, like DE is doing. However this increases complexity and in the end development time, since they have to one-up themselves with every new addition. That's called complexity trap.
3 hours ago, DrBorris said:
Imagine how easy it would be for DE to balance weapons if a weapon with a 30% critical chance was nearly equally powerful to a weapon with a 30% status chance given similar base damage.
What is the point? That's like having 2 identical tools, just with diffirent grip colors.
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...Doesn't this just make low damage weapons even worse?
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20 hours ago, DrBorris said:
It feels like you came here with your own opinions on the topic, skimmed to the part you understood at a glance (the status suggestions), then commented without giving the effort to know what I was actually suggesting. It is a lot of words, I know, but could've you just waited till you had time to go through it before you responded? I am sorry if I am misjudging but given your response it doesn't feel like you tried to understand the proposal.
This is partially true, in that I skimmed your OP. I did miss your version of the "status magnitude" idea, this is true. The reason I skimmed, however, is that a majority of your OP is troubleshooting the implications of tying status magnitude to weapon damage and its interactions with shields, armour and health types. It's not that these things aren't relevant, but rather that my counter-proposal almost entirely eliminates them as a consideration. By removing health-type-specific resistances, you remove the major source of balance issues and thus need no further consideration. By tying status magnitude to a separate, independent stat, you don't need to resolve issues between base weapon performance and weapon status viability. If you'd like, I could go through your proposals point-by-point, but there really isn't much there that I specifically disagree with. My post may have come across more confrontational than I'd mean to, but I agree with your general premise here. Scaling Status effect by SOME stat inherent to each individual weapon rather than using the current one-size-fits-all approach is a good thing.
I just happen to have a more radical approach to game changes, in that my proposals often involve retiring entire systems or creating brand new ones. For example, one thing I didn't bring up as I felt my proposal was already radical enough was a reduction in the total damage types in the game. I don't see how DE can make 13 damage types all meaningful, but 4-7 could easily each have its own unique Status effect. Additionally, I don't see a reason to retain the physical/elemental divide between damage types at all. It seems to me that the original paradigm was that all weapons deal physical damage but some weapons (or some mods) have access to these "more special" damage mods. I'd personally rather treat all damage types the same. We don't need 13 damage types, we don't need 3 damage types just for physical damage. We don't need this system grandfathered over from old early 2000s MMOs.
A few parting comments, as well:
On the notion of all status effects dealing damage - I understand what your design is going for. My proposal, however, was a lot more blunt. Rather than trying to come up with different ways in which various Status effects deal damage, I proposed simply putting DOT on all them. The higher your Status Magnitude, the more of your damage goes into the DOT. I'd also go with roughly the same amount of damage for all of them. I'm not opposed to the system you propose, but I feel that the ad hoc complexity of Status effects is currently one of their major downsides.
On the general notion of tying Status Effects to damage - yes, I agree that that's one way of solving core issue. I simply happen to feel that it creates a number of issues of its own. You're already having to compare an abstract "weapon DPS" calculation against enemy health in order to derive a ratio by which to multiply debuff/control status effect magnitude, at least if I'm reading your proposal right. What you're working with here is a fairly static number... Let me give you an example. A "decent" critgun is one with a base critical hit of 25% or up and a critical damage of *2, at least in my assessment. This is irrespective of weapon DPS, enemy EHP or any other considerations. If the weapon has that much, it's decent for crit. The weapon might suck ass anyway, but that's down to other factors - it's still a good critgun within its own context. Similarly, most "status weapons" would end up falling within a relatively narrow band of status effect chance and status effect magnitude. With your calculation, however, you end up having to balance these numbers dynamically against multiple factors and end up with a fluctuating value that doesn't seem to need to fluctuate. To me, it feels like a very roundabout way of determining a ratio which you could simply define as a single stand-alone value and not worry about enemy level, enemy health, weapon damage or anything else of the sort. Unless there's a specific need for a more complex calculation, I tend to find that simpler calculations work better overall.
Again - I get what you're going with. I just feel that the route you've chosen is needlessly complex in terms of implementation and creates an unnecessary amount of extra variables which don't really serve an intended purpose so much as exist as a byproduct. We're both proposing the same thing, effectively, except I propose defining it separately rather than trying to back-calculate it off of existing values. To give you a metaphorical example - think back to how per-pellet Status Chance used to work. Someone at DE wanted to calculate Status chance the same for shotguns as for all other weapons, so attempted to back-calculate per-pellet Status chance off of a generic stat, and the results were spiky and unintended behaviour. It's still a clever calculation that I absolutely love from a mathematical modelling perspective, but I don't think you'll find many who still argue it was the right solution. Keep it simple and you avoid unpredictable behaviour, special-case exceptions and unintended stacking effects. That's really all I'm saying here.
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3 hours ago, ShortCat said:
That's even worse. Low damage weapons in general would be completely useless, without any option for redemption.
3 hours ago, Aldain said:
...Doesn't this just make low damage weapons even worse?
One of my biggest pet peeves with design is using one broken thing to justify another broken thing. If a weapon isn't performing well why not just buff the weapon? Keeping status as is despite what I believe are major issues just to keep a few weapons afloat feels wrong. Especially when the alternative is to just make a number go up a little. A full status rework like this wouldn't mandate another balance pass on weapons but it absolutely should come with one.
Also, when you say "low damage weapons", what do you mean? As it stands very few weapons are actually low damage, what currently makes a weapon bad at dealing damage has more tied to its critical stats. The above proposal is basically giving those status weapons modifiers that can compete with crits, so they no longer need crit to bring them up.
3 hours ago, ShortCat said:
In my opinion - Yes. We have 3 weapons + 1 Frame in our loadouts. Combining those to achieve a common goal seems like a good design to me.
I don't think that is something that fits the pace of Warframe. Even if swap speed is increased to be basically instant it will be pointless to switch weapons when everything dies as quickly as it does. Elite enemies will (hopefully) come around that last more than a second but having status be designed in a way so a couple weapons can have a niche as status applicators against a rare enemy at the expense of every other weapon isn't a good trade in my opinion. Guns are for killing things, Warframes are for manipulating the battlefield. If there is to be overlap then it should come from mechanical differences (like Zakti opening enemies to finishers) not the core damage system.
3 hours ago, ShortCat said:
Personally, I find Kitguns (or Zaws) bad design, precisely because players have advanced controll over base stats.
Okay, but are you really going to just gloss over how useful it would be for status chance to be representative of how good a weapon is at dealing status? From a new player's point of view it makes deciding what to mod for far easier, status number is big so it must be good for status. From DE's perspective it makes things easier because they can experiment with new archetypes for weapons like a cold status bow, or an irradiating sniper. Those things could exist today but it would require them being given new mechanics, with the proposal It Just Works™.
3 hours ago, ShortCat said:
10 stack limit is the rule from the current system and in some cases it is indeed ill managed. Your proposal however has the same boundaries: at some point weapons will run into an equilibrium of new statuses applied <-> old status expired; it will just vary on weapon to weapon basis, which is, if you ask me, even less consistent or intuitive than a flat status cap.
Furthermore, I also mentioned additional effects after X stacks. One idea of mine for cold is that each proc slows the enemy by % untill an enemy is frozen(=100%) and at this stage the enemy would take increased damage from all sources, while subsequent cold procs would not be wasted, but maintain this frozen state.
Doesn't that equilibrium exist with status stacking as well? Me making every status effect suggestion last six seconds wasn't an accident and I probably should have explained some more of the rational behind it. The reason that status effects last six second is because you should have killed that enemy in six seconds. Yes, status will decay, but if the status effects start decaying you should take that as a sign that you aren't geared for the enemy you are shooting at. An enemy living for more than six seconds is a rare occurrence already unless you are pushing the damage system far beyond its limits, occurrences that I don't think should be considered relevant when reworking damage.
Why do you think it is better design that a high fire rate weapon is better at inflicting status? You suggest leaning even further into fire rate being just as important as status chance when calculating how effective a weapon is at inflicting status, how is this better? My quest to "fix" status started three years ago when I was annoyed with how Latron Prime was given a decent status chance but was a bad status weapon due to fire rate, why should my Latron Prime be a bad status weapon?
3 hours ago, ShortCat said:
Status intensity would be just another conditional +damage mod.
So no effect on non-damaging statuses... cool.
3 hours ago, ShortCat said:
What is the point? That's like having 2 identical tools, just with diffirent grip colors.
Can't you say that about every weapon in every game? All they do is eventually enable you to reduce the enemy health to zero, they are basically all the same thing.
• Crit weapons would have high alpha damage and get a considerable damage boost against enemy weakpoints.
• Status weapons are able to dispense their damage in a variety of ways. Debuffing an enemy may make it easier to kill them but it inherently means that the enemy isn't always dead instantly, in the process of killing one enemy you may have hurt their friends as well though.
Just because the number is the same it doesn't mean that they affect enemies the same way. Just above this you talked about "color coding" and how you can give things the same number but different uses.
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6 minutes ago, Steel_Rook said:
-snip-
Wait... with the new quotes collapsing is doing the "-snip-" needed anymore? It feels wrong to not do the -snip- to walls of text...
I get where you are coming from and see how at an extreme high level the proposals are doing the same thing. I don't really see it as working backwards though. This whole concept is based on a very simple idea, if a bullet does big damage the resulting status effect will also big. That is the starting point and while I did use the current status effects as inspiration it wasn't me forcing it all to work together, it just works.
The majority of this thread isn't me trying to make this work with resistances, in the only section where I bring up resistances one of the options I propose is to just get rid of them. Most of the sections are laying out the core rules (like the changes to crit interactions) and exceptions (like forced status procs), things that your proposal needs as well.
One of the early analogies for how the SEF (Status Effect Factor) works conceptually is relevant here. When you apply a damage over time effect the "effectiveness" isn't really the amount of damage it is doing, it is how much closer it is turning the enemy into dead. A 100 DoT against a level 10 and level 100 enemy have vastly different effectiveness because the time they take to die is longer. In effect, the "status effect" of damage over time is the enemy being dead. Every step in between is just a "Status Effect Factor" between 0 and 1. Take that exact logic and apply it to things that aren't damage and you have my proposal, swapping "enemy is dead" with "whole health status" (not incidentally correlated to an enemy having no health).
It is kind of hard to compare to your proposal because it is such a big switch. I remember reading your thread but I don't remember if any status effects had any form of non-damage effect. And if those existed, how did those effects scale? I get that it all looks complicated but I personally don't see it as simpler than the math today. Don't forget that dealing damage in itself is a math equation, I am just applying that math equation to non-damage effects. Making things simple is often a complicated task.
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16 hours ago, DrBorris said:
I get where you are coming from and see how at an extreme high level the proposals are doing the same thing. I don't really see it as working backwards though. This whole concept is based on a very simple idea, if a bullet does big damage the resulting status effect will also big. That is the starting point and while I did use the current status effects as inspiration it wasn't me forcing it all to work together, it just works.
Right, and that's not necessarily BAD. However, I think this is our main point of disagreement - I find that general rule of thumb to be limiting. My primary push here is to ensure that sometimes bullets CAN do big damage but little to no Status and sometimes bullets can do very little damage but still do big Status. Tying Status directly to damage creates a significant design limitation as it unilaterally turns all high-damage weapons into Status guns, including the high-damage ones which were previously critguns. The reason I bring up Criticals so much is that I feel those were done well. You could have really heavy-hitting, high-DPS weapons with good Crit, but not all of them are like that. The Kohm family (Kohm, Kohmak, Twin Kohmak) are pretty good for DPS, but they have low Crit and high Status. Inversely, the Tenora has pretty high DPS and substantial Crit but fairly low Status.
See, I'm not opposed to your principles of game design, necessarily. I like the idea of making slow-firing, heavy-hitting weapons worth using for Status, absolutely. However, I want to leave this as an option for the design team to use, rather than turning it into a hard engine-level limitation. To put it another way, I want slow-firing weapons to be able to have good Status, but I don't want them to HAVE to have good Status across the board by mandate. This is why I propose decoupling Status from Damage - so that you can have one but not the other or both... Or maybe even neither, though the resultant weapon would likely suck. Maybe as a Mk. 1 version? In this, my disagreement here is somewhat ideological, in that I'd rather keep our weapon stats as independent of each other as possible. We can always bind them together at the design level, but binding them together at the engine level is limiting.
16 hours ago, DrBorris said:
It is kind of hard to compare to your proposal because it is such a big switch. I remember reading your thread but I don't remember if any status effects had any form of non-damage effect. And if those existed, how did those effects scale? I get that it all looks complicated but I personally don't see it as simpler than the math today. Don't forget that dealing damage in itself is a math equation, I am just applying that math equation to non-damage effects. Making things simple is often a complicated task.
It's not so much a matter of how complicated the math would be for players to understand, though that is a factor. It's more that complex interactions make it harder for the design team to create a holistic experience without risking odd behaviour under specific semi-rare circumstances. That was the whole problem with per-pellet Status effect chance for weapons with native multi-shot. The system worked reasonably fine up to 95%, even 99% Status Effect. It only broke past 99, which was actually not that common - not a lot of Status weapons could do that, and most needed fairly specific modding setups to do that. Even so, this created the impression in the player base that "100% Status or GTFO." Looking through the Wiki, nearly every shotgun had an entry indicating whether it could achieve 100% Status effect chance before Multishot, and what mods were needed for it. It's pretty safe to say that this wasn't the original designers' intent, but the math worked out to create that edge case which the community then adopted.
But to answer your question on how non-damage status effects would stack - my original proposal was to take a "base" status effect value and then modify this via flat multiplier. For reference, let's look at Crits. Most weapons have a Critical damage multiplier of *2. Accounting for base damage (because that's included in Crit damage), that's a base Critical hit bonus damage multiplier of *1, i.e. the weapon's damage dealt again. A weapon which deals 100 damage with a *2 Crit would deal 100 + 100 damage. The same can be applied to Status Effects. Let's take Cold for example. Let's take Cold, for example, and refactor it a little. Let's say every Cold Status effect reduces enemy "speed" by 5%, up to a maximum of 75% (no more "stacks"). A weapon with a Status Magnitude of *1 would then slow enemies down by 5% per Cold proc, requiring 15 procs to cap. A weapon with a *3 Status Magnitude would deal 15% slow per Cold proc, requiring just 5 procs to cap. That would be a good Status gun. A bad Status gun might have only, say, a 0.4 Status Magnitude, causing it to slow enemies down by just 2% per Cold Proc, which now requires a whole 38 procs to cap.
Now, the above CAN work with Status Chance still in the game, but it can equally as well work with no Status chance - every hit deals Status. At that point, base Status values would probably need to drop, however. The Cold proc I used in the previous example might need to drop to 2-3%, for example. I don't have precise numbers because you're one of the few people to actually take this proposal seriously, so I haven't really had to do precise over-time statistics :) However, I'm convinced that this could work quite well. It's still subject to rate of fire, but so is damage. If need be fast-firing weapons could simply be given lower Status magnitude (or Status chance, if we're keeping that) to compensate. Smart game design always looks at over-time stats rather than base stats anyway.
You can extend this to all of the debuff Status effects, simply scaling the magnitude of each individual proc based on the weapon's Status magnitude off a common base value. Control effects are a little bit more difficult to scale that way, but I'm of the opinion that Control effects ought to be handled differently anyway. Let's take Cold for example again. I'm of the opinion that Cold should eventually freeze enemies solid after a certain threshold. Say once an enemy is slowed down to 75% (i.e. Cold Status cap), they're straight-up frozen and unable to move. You could then keep them frozen as long as the Cold proc is in effect (ostensibly letting players keep them frozen forever) or keep them frozen for a fixed amount of time (modified by Status Magnitude) after which they unfreeze and shed all Cold procs, needing to be re-frozen again. I say this as a lead-up to Radiation and the Confuse effect, in that I believe this should start as something else up until we stack enough Radiation to turn it INTO a confuse. Its current implementation does scale somewhat, but... Well, enemy DPS is far too low to do meaningful work against enemy EHP at even moderate levels (40-ish), so even boosting a single confused enemy's damage up by 500% is not going to do much. Again, I've not put together a holistic list of Status effects and how they scale because it never seemed like anyone cared enough to bother. I can try if you'd like.
And mind you, I'm not suggesting that my system is "simpler" than yours. In fact, it probably requires slightly more care since Status Magnitude would need to be tailored to weapons a bit more than critical damage. The main advantage I see in it is that you CAN tailor Status magnitude to each individual weapon independent of its damage, thus eliminating the danger of special-case exceptions. At least I think so, anyway. Please let me know if you see any obvious holes in my model above, because it feels like I'm forgetting something important.
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14 hours ago, DrBorris said:
One of my biggest pet peeves with design is using one broken thing to justify another broken thing. If a weapon isn't performing well why not just buff the weapon?
Is a weapon like Phantasma broken? It has low damage, but it pulls itself with high status. First few bullets hardly deal any damage, but they inflict status effects to debuff the enemy and afterwards damage climbs really fast. Your concept does not allow such interactions, as by default the best status weapon ist not the one with best status stats, but with highest damage.
15 hours ago, DrBorris said:
Doesn't that equilibrium exist with status stacking as well?
Never denied that. Just pointed out another similarity you previously described as a flaw.
16 hours ago, DrBorris said:
Why do you think it is better design that a high fire rate weapon is better at inflicting status? You suggest leaning even further into fire rate being just as important as status chance when calculating how effective a weapon is at inflicting status, how is this better?
I never implied any of that, high status application atm depends on 3 major variables - firerate, status chance, number of damage instances. Especially since status is allowed to scale past 100% you can inflict more than one status per damage instance. Not to mention, status effects are allowed to be powerfull even with one stack.
16 hours ago, DrBorris said:
So no effect on non-damaging statuses... cool.
You said it yourself, it is difficult to define intensity for some status effects.
Otherwise it is even not much different from +status chance. On the one hand you have higher chance to trigger more weaker effects; on the other hand you trigger less but stronger effects. When I think about it now, status intensity is not really necessary, since results are similar, as long as status is allowed to stack.
16 hours ago, DrBorris said:
All they do is eventually enable you to reduce the enemy health to zero, they are basically all the same thing.
They should not be mechanically identical though.
17 hours ago, DrBorris said:
Status weapons are able to dispense their damage in a variety of ways. Debuffing an enemy may make it easier to kill them but it inherently means that the enemy isn't always dead instantly, in the process of killing one enemy you may have hurt their friends as well though.
Sounds like a utility weapon.
17 hours ago, DrBorris said:
Just because the number is the same it doesn't mean that they affect enemies the same way. Just above this you talked about "color coding" and how you can give things the same number but different uses.
Yes, "color codding" thingy is indeed guilty in this regard. However, weapons operate on a higher layer with hundreds of options. There is no excuse to apply the same logic to a damage system with only 2.5 options.
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5 hours ago, ShortCat said:
Is a weapon like Phantasma broken? It has low damage, but it pulls itself with high status. First few bullets hardly deal any damage, but they inflict status effects to debuff the enemy and afterwards damage climbs really fast. Your concept does not allow such interactions, as by default the best status weapon ist not the one with best status stats, but with highest damage.
Phantasma is not a low damage weapon, with a crit build its non-crit DPS is 17,000, the same as Corinth Prime. The reason it feels low "DPS" is because it lacks crit stats. The above proposal allows all "low damage" (even though they are not actually "low damage", they are just low crit) status weapons to compete regardless of fire rate.
5 hours ago, ShortCat said:
Never denied that. Just pointed out another similarity you previously described as a flaw.
Not sure what flaw you are talking about. Consistency isn't something always being the same, consistency is something being predictable.
5 hours ago, ShortCat said:
I never implied any of that, high status application atm depends on 3 major variables - firerate, status chance, number of damage instances. Especially since status is allowed to scale past 100% you can inflict more than one status per damage instance. Not to mention, status effects are allowed to be powerfull even with one stack.
You may not have implied it but with status stacking on hits it is just a fact. And you bring up >100% status, which is actually an interesting case with status being based on hits. In theory if you want to "fix" debuff hit status on low RoF weapons you could just make a loose inverse correlation between fire rate and status chance. If DE wanted they could give my cold bow a base 100% status chance and then it would be able to inflict a larger cold status in a single hit.
Big problem though, not all status effects are based on hits. On this bow with a massive status chance it would actually be dumb to build it for the debuff status effects because now it will be able to deal absurd damage with the DoTs. This is what I mean by inconsistency, by making one type of weapon good at debuff you over-buff DoTs. On the flip side you have DoTs on high fire rate weapons, this isn't as much of an issue because debuffs are capped but if you want a DoT high fire rate status weapon you will incidentally make the debuff status effects absurd.
Status Chance as a number that represents a weapon dealing status effects is a lie, you have to take into account fire rate as well and even then the verdict varies massively with different elements.
5 hours ago, ShortCat said:
They should not be mechanically identical though.
But... but they aren't. The way you kill an enemy with a status weapon will feel different than killing them with crits. Just because they both scale with damage it doesn't mean they both play the same way.
5 hours ago, ShortCat said:
Sounds like a utility weapon.
I don't even know how to respond to this. Unless you are trying to say that all status weapons are utility weapons. Status weapons and crit weapons are both there to reduce an enemy's health bar to zero but they do it in different ways.
This conversation isn't going anywhere, nor is this thread. Probably should just let it die, a 5k word thesis on status isn't something people seem to care about, I'll take the hint.
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12 hours ago, DrBorris said:
This conversation isn't going anywhere, nor is this thread. Probably should just let it die, a 5k word thesis on status isn't something people seem to care about, I'll take the hint.
Let me summarize my impressions of your concept.
• All status effects are defined by weapon damage,
• Thus low damage weapons are by default terrible at status.
• And restricted on hit effects.
• Additionally, status scales based on damage dealt to total health, thus high lvl enemies will negate all status effects by default.
• This means all status effects move on a scale from good --> bad over the course of a single mission.
• Crits do not interfere with status under the premise to make those 2 more distinct. Yet, you introduce an equivalent of critical damage for status, status intensity.
• Health type midifiers are renamed to to DoSP resistance.
• Status stacks are renamed to WHS.
• UI is changed from obvious stack numbers to vacuous circles.
• There are still 13 elements with overlapping functionality and feature bloat!
Your main idea is to tie status to damage. And you died on that hill.
You remove existing concepts, just to reintroduce them later under a new name with a more convoluted nature, mostly to fix problems you created by tying status to damage. | 20,762 | 97,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.940359 |
http://mathhelpforum.com/statistics/73843-problem-calculating-expected-value.html | 1,527,108,880,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865809.59/warc/CC-MAIN-20180523200115-20180523220115-00060.warc.gz | 173,728,417 | 10,664 | # Thread: problem with calculating expected value
1. ## problem with calculating expected value
Let X be Poisson with Parameter λ
Compute the mean of (1+X)^-1.
thanks
Let X be Poisson with Parameter λ
Compute the mean of (1+X)^-1.
thanks
Start by noting that $\displaystyle E\left(\frac{1}{1+X} \right) = \sum_{x = 0}^{+\infty} \frac{1}{1 + x} \cdot \frac{\lambda^x e^{-\lambda}}{x!} = \sum_{x = 0}^{+\infty}\frac{\lambda^x e^{-\lambda}}{(x+1)!} = \frac{1}{\lambda} \sum_{x = 0}^{+\infty}\frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!}$.
$\displaystyle \sum_{k = 0}^{+\infty}\frac{\lambda^{k} e^{-\lambda}}{k!} = 1$ will be useful in calculating the above sum ....
3. ## thanks but...
first of all thanks for the answer but...
The answer (i have a only the final answer) is λ^-1(1- e^-λ) and
you said that
so isn't the sum above shoudnt be (1- e^-λ) instead of 1???d
thanks again
first of all thanks for the answer but...
The answer (i have a only the final answer) is λ^-1(1- e^-λ) and
you said that
so isn't the sum above shoudnt be (1- e^-λ) instead of 1???d
thanks again
Try to look further than your nose ^^
$\displaystyle \frac{1}{\lambda} \sum_{x = 0}^{+\infty}\frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!}$
Make the change $\displaystyle k=x+1$
and $\displaystyle 1=\sum_{k = 0}^{+\infty}\frac{\lambda^{k} e^{-\lambda}}{k!}=e^{-\lambda}+ \sum_{k=1}^\infty \frac{\lambda^k e^{-\lambda}}{k!}$
$\displaystyle \frac{1}{\lambda} \sum_{x = 0}^{+\infty} \frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!} = \frac{1}{\lambda} \left( \frac{\lambda e^{-\lambda}}{1!} + \frac{\lambda^{2} e^{-\lambda}}{2!} + \, .... \right) = \frac{1}{\lambda} \left( 1 - e^{-\lambda}\right)$ using $\displaystyle \sum_{k = 0}^{+\infty}\frac{\lambda^{k} e^{-\lambda}}{k!} = 1$. | 658 | 1,752 | {"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} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.544794 |
https://socratic.org/questions/how-do-you-solve-frac-x-900-frac-35-15 | 1,601,154,113,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400245109.69/warc/CC-MAIN-20200926200523-20200926230523-00554.warc.gz | 613,418,622 | 6,520 | # How do you solve \frac { x } { 900} = \frac { 35} { 15}?
Dec 1, 2017
$x = 2100$
#### Explanation:
First cross multiply to have a linear equation:
$15 x = 900 \cdot 35$
$15 x = 31500$
To isolate $x$, divide by 15:
$x = \frac{31500}{15}$
$x = 2100$
Dec 1, 2017
The easiest way to do this is to cross multiply and set the equation up like
$15 x = 900 \cdot 35$
#### Explanation:
Think of it like you're multiplying $15$ to both sides, which gets rid of the fraction on the right side (because $\frac{15}{15} = 1$). This also works for the left side when you multiply each side by $900$.
After you get
$15 x = 31500$
simplify it by dividing each side by $15$ to get
$x = 2100$
Dec 2, 2017
$x = 2100$
#### Explanation:
In $\text{ "x/900 = 35/15" }$ you need to isolate $x$
Multiply both sides by $900$ to cancel the fraction with $x$
$\frac{\textcolor{b l u e}{900 \times} x}{900} = \frac{\textcolor{b l u e}{900 \times} 35}{15}$
$\frac{\cancel{900} \times x}{\cancel{900}} = \frac{{\cancel{900}}^{60} \times 35}{\cancel{15}} \text{ } \leftarrow$ simplify
$x = 2100$
In this case I would not use cross-multiplying because it changes a $1 x$ term into a $15 x$ term, which then has to be divided by $15$ | 413 | 1,225 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 24, "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} | 4.65625 | 5 | CC-MAIN-2020-40 | latest | en | 0.817327 |
https://www.iberlibro.com/9781502527400/Practical-Numerical-Methods-Chemical-Engineers-1502527405/plp | 1,513,623,104,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948619804.88/warc/CC-MAIN-20171218180731-20171218202731-00650.warc.gz | 736,961,273 | 17,839 | # Practical Numerical Methods for Chemical Engineers: Using Excel with VBA, 3rd Edition
## Richard A Davis
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This latest edition expands Practical Numerical Methods with more VBA to boost Excel's power for modeling and analysis using the same numerical techniques found in more specialized math software. Visit the companion web site:
www.d.umn.edu/~rdavis/PNM/PNMExcelVBA3
to access all of the book's Excel and VBA files, and learn how to customize your own Excel workbooks with:
1. A refined macro-enabled Excel workbook with a suite of over 170 VBA user-defined functions, macros and user-forms for learning VBA and implementing advanced numerical techniques in Excel.
2. More than 200 example and animation workbook files from the book that demonstrate the power of numerical methods. Customize the example files and macros to tackle your own problems using VBA in Excel.
3. Hundreds of practice problems for self-guided study to sharpen your Excel and VBA skills.
The first chapter sets the stage for problem solving with numerical methods. The next two chapters cover frequently overlooked features of Excel and VBA for implementing numerical methods in Excel, as well as documenting results. The remaining chapters present powerful numerical techniques using Excel and VBA to find roots to algebraic equations, approximate derivatives, optimize, model data by least-squares regression and interpolation, analyze risk and uncertainty, solve integrals & ordinary & partial differential equations:
1. Numerical Methods & Mathematical Modeling: expert problem solving, Primer on Chemical Reaction Engineering
2. Excel: Documentation, Graphing, Worksheet Functions, Input Validation and Formatting, What-if Analysis
3. VBA: Editor and objects, Function and Sub Procedures, Data Types, Structured Programming, Arithmetic and Worksheet Functions, Flow Control, Arrays, Communication, Message and Input Boxes, User Forms, Reading/Writing Files, Debugging, Unit Conversions
4. Linear Equations: Matrix Algebra, Gaussian Elimination and Crout Reduction with Pivoting, Thomas, Cholesky, Power, Jacobi, and Interpolation Methods for Eigenvalues and Eigenvectors, Jacobi and Gauss-Seidel Iteration, Relaxation
5. Taylor Series Analysis: Finite Difference Derivative Approximation, Richardson's Extrapolation, Ridder's algorithm, Sensitivity
6. Nonlinear Equations Root Finding: Methods of Bisection, Regula Falsi, Newton, Secant, Pade, Wegstein, Quasi-Newton, Aitkin/Steffensen, Homotopy, Bairstow (for polynomial roots), Goal Seek and Solver
7. Optimization: Solver, Luus-Jaakola, Quadratic, Golden Section, Powell, Downhill Simplex, Firefly, Constraints, Scaling and Sensitivity
8. Uncertainty and Risk Analysis: Bootstrap, Confidence Intervals, Law of Propagation, Monte Carlo Simulations with Latin Hypercube Sampling
9. Least-squares Regression: Linear, Nonlinear, LINEST, Gauss-Newton, Levenberg-Marquardt, Validation and Assessment, Uncertainty Analysis, Weighted Regression
10. Interpolation: Linear, Newton Divided Difference and Lagrange Polynomials, Rational, Bulirsh-Stoer, Pade, Stineman, Cubic, B, Akima and Constrained Hermite Splines, Bivariate Interpolation
11. Integration: Graphical, Trapezoidal, Midpoint and transformation for Improper Integrals, Romberg, Adaptive Simpson and Gauss-Kronrod, Multiple Integrals by Simpson, Kronrod and Monte Carlo
12. Initial-value Problems: Single Step Euler and Backward Euler, Implicit Trapezoidal for Stiffness, Variable Step Runge-Kutta Cash-Karp, Dormand-Prince, Multi-step Adams-Bashforth-Moulton, Differential-Algebraic Systems
13. Boundary-value Problems and Partial Differential Equations: Shooting, Finite Difference, Orthogonal Collocation, Quasilinearization, Method of Lines, Crank-Nicholson
14. Review: Reference Tables of Excel and VBA Functions, User-defined Functions, Macros, User Forms
"Sinopsis" puede pertenecer a otra edición de este libro.
Richard Davis is a Jean G. Blehart Distinguished Professor of Chemical Engineering at the University of Minnesota Duluth. He earned Ph.D. and B.S. Chemical Engineering degrees from UCSB and BYU, respectively.
Professor Davis has over two decades experience teaching a variety of courses including computational methods, unit operations of momentum, heat and mass transfer, chemical reactor design, engineering economics, bioprocessing and green engineering.
His current teaching and research interests include process modeling and simulation applied to mineral processing, energy conversion, air pollution control, chemical process safety, and environmental management.
Professor Davis serves as the Executive Secretary for the National Chemical Engineering Honor Society Omega Chi Epsilon, and is active in AIChE and ASEE. He is the academic adviser to the local engineering student chapters of Tau Beta Pi, Omega Chi Epsilon, and the Society for Mining, Metallurgy, and Exploration.
He is the recipient of the University’s Outstanding Adviser and Exceptional Teaching awards.
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Destinos, gastos y plazos de envío | 2,474 | 9,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-51 | latest | en | 0.817781 |
https://owenduffy.net/blog/?p=1763 | 1,685,816,391,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649302.35/warc/CC-MAIN-20230603165228-20230603195228-00443.warc.gz | 482,231,645 | 13,252 | # Definition: Current Balun, Voltage Balun
Readers of (How) does this balun work? have asked about the meaning of the terms Current Balun and Voltage Balun. The following is a cut and paste from my old site.
## Definition: Current Balun, Voltage Balun
The terms Current Balun and Voltage Balun are often used. This article explains the key characteristics that determine which best describes a particular balun. The approach taken is to charaterise a balun because of how it behaves externally rather than the internal implementation.
Balun loads are often though of as a simple two terminal device. That is actually quite inadequate, they should be thought of as a three terminal load which can be simplified to a delta or wye equivalent circuit and one terminal of that network is connected to the unbalanced input ‘ground' terminal.
Baluns may appear ideal if they are driving an isolated two terminal load, and there would be no need for a balun if that was the case.
Common mode current is the difference between the output terminal currents, it flows to the unbalanced input ‘ground' terminal.
## Current Balun
An ideal current balun delivers currents that are equal in magnitude and opposite in phase.
A good current balun will approach the ideal condition. It will deliver approximately equal currents with approximately opposite phase, irrespective of the load impedance (including symmetry).
Common mode current will be small.
If the load impedance is not symmetric, then the voltages at each output terminal will not be equal in magnitude and opposite in phase. (Note that for a truly ‘isolated' load, one well represented as a two terminal load, the currents MUST be equal in magnitude and opposite in phase, but the voltages may not be equal in magnitude and opposite in phase.)
A parameter often used to quantify the effect of a current balun is its common mode impedance or choking impedance. An ideal current balun has infinite common mode impedance, a good one has very high common mode impedance (typically thousands of ohms for an effective general purpose balun in an antenna system).
## Voltage Balun
An ideal voltage balun delivers voltages that are equal in magnitude and opposite in phase.
A good voltage balun will approach the ideal condition. It will deliver approximately equal voltages (wrt the input ground) with approximately opposite phase, irrespective of the load impedance (including symmetry).
Common mode voltage ((V1+V2)/2) will be small.
If the load impedance is not symmetric, then the currents flowing in each output terminal will not be equal in magnitude and opposite in phase.
An ideal voltage balun has zero common mode impedance, a good one has very low common mode impedance (ohms).
## Either or neither
A balun cannot be a good Voltage Balun and a good Current Balun at the same time, it is one or the other, or neither.
## Impedance transformation
An ideal balun performs an ideal impedance transformation, nominally 1:1 unless specified otherwise. Practical baluns depart from the ideal, and the departure is often specified as Insertion VSWR.
It is possible to design a balun to not only facilitate the unbalanced to balanced transition, but to perform a nominal impedance transformation (eg 4:1 is common). Voltage Baluns and Current Baluns are both capable of impedance transformation other than nominally 1:1.
## Applications
If the application is one where current balance is important then a current balun is the better choice. For example:
• reducing radiation from an antenna feed line by ensuring that the currents in each feed line conductor are nearly equal but opposite in phase.
If the application is one where voltage balance is important then a voltage balun is the better choice. For example:
• some audio applications where rejection of common mode voltage injected into a source is important.
## An exercise for the reader
Above are two baluns from the ARRL Handbook, neither uses a magnetic core. The second also performs 4:1 impedance transformation. Classify each as a Voltage Balun, Current Balun, or neither. Why? | 839 | 4,116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2023-23 | latest | en | 0.938268 |
https://www.chemicalprocessing.com/automation/instrumentation/article/11307392/deftly-detect-instrument-faults-chemical-processing | 1,721,157,435,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.44/warc/CC-MAIN-20240716183855-20240716213855-00004.warc.gz | 612,512,464 | 45,931 | Deftly Detect Instrument Faults
April 15, 2019
Check for some simple clues to a malfunctioning device
Meter mayhem, I thought to myself. One magnetic flow meter showed 18 gal/min, another read 138 gal/min. However, the magnetic flow meter monitoring the combined flow to a reactor indicated 380 gal/min. Okay, it’s not a mass balance but the temperatures only were 60°F apart. The total flow should be 156 ±3 gal/min. Other flows did go into and out of the feed tank but all were for recirculation except for the one that went to the reactor. The pump for the recirculation had a design flow of 900 gal/min; some people still think you can agitate a 40,000-gal tank with a pump.
Obviously, most of the flow from the large pump went for recirculation, with only about 156 gal/min sent to the reactor. The errant measurement pointed to an instrument problem: magnetic flow meters read high when they become plugged. As I pointed out in a previous column (“Match the Flow Meter to the Service”), flow meters rely on inference: magmeters measure velocity for an assumed density and then infer flow rate. Plugging restricts the pipe diameter and increases the velocity, at least according to the mass flow equation: Mass = Density × Velocity × Area.
[pullquote]
Another indication of a problem with that meter was its trend line. The trends for the 18-gal/min and 138-gal/min meters jumped around. However, the trend line for the large flow meter almost was flat. Often — but not always — that’s a sign of a dead instrument. Sometimes, though, it stems from over-tuning to avoid oscillation of trend lines; that oscillation is a pulse.
Generally, level and temperature measurements don’t have a pulse. If a level measurement is leaping around on a control system faceplate, it usually indicates an electrical problem, not an instrument one. Temperature measurement coupled with flow, such as what you’d see in a furnace or vaporizer, can have a pulse; the trouble is that oscillation from the flow measurement almost certainly will overwhelm the temperature fluctuation.
An excessively high trend in temperature measurement obviously could result from a loose connection (high resistance); this applies to either thermocouples or RTDs. This probably will stand out from the background noise caused by flow measurement oscillation.
Thermocouples fail high, so do RTDs although sometimes the resistance is only somewhat higher than normal. Temperature sensor probe failures often are pretty spectacular (in the sense that high temperature may lead to fire, explosion, reactor breach, pump failure or other catastrophic and unpleasant events).
One good technique for diagnosis is to use the instrument itself. Compare the pump rates into and out of a vessel against the flow calculated from the level trend data. For example, I measured levels of 20.6% and 26.8% only a minute apart for a tank that needed 308.7 gal to alter the level 1%. So, the indicated level change would have required a flow of 1,917 gal/min — stupendously more than the 150-gal/min pump ever could provide!
This wasn’t the only time a level measurement let me down. On one assignment, several hundred pounds of material seemed to appear and disappear in seconds on a faceplate in the control room. The mass measurements were made from level readings. I conducted a level measurement experiment and then looked at the trend data I downloaded: the data showed that nothing could be gleaned from any comparison until the level transmitter gain was dampened.
[callToAction ]
Diagnosing problems with control valves often isn’t too difficult. Control valves are considered unreliable when they don’t do what they’re supposed to do. Look at the response time between the command and the response: a good valve opens quickly while a bad valve sticks, causing the controller to read the failure of the process variable (PV) to respond to change as a need to open the valve further. When the valve finally does spring open, the controller sees the change in PV and tries to crank the valve closed. Because the valve is open now, it slams closed and the process begins again. Often, an engineer will dive into this kind of problem thinking it’s a tuning problem and not a hardware problem, being misled by focusing only on the controller and the measuring instrument.
As I’ve said before, don’t believe information unless you can compare it to other data. Take the time to take the long view.
DIRK WILLARD is a Chemical Processing contributing editor. He recently won recognition for his Field Notes column from the ASBPE. Chemical Processing is proud to have him on board. You can e-mail him at [email protected]
Dirk Willard | Contributing Editor
DIRK WILLARD is a Chemical Processing Contributing Editor.
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# You have a six-sided cube and six cans of paint, each a diff
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You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120
[Reveal] Spoiler: OA
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]
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09 Sep 2013, 08:06
Can someone explain why the answer is B?
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10 Sep 2013, 04:43
Bunuel and other experts,
Can you pls suggest the best approach to solve this problem?
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10 Sep 2013, 06:16
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chetan86 wrote:
You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120
Paint one of the faces red and make it the top face.
5 options for the bottom face.
Now, four side faces can be painted in (4-1)! = 3! = 6 ways (circular arrangements of 4 colors).
Total = 5*6 = 30.
Similar question to practice: a-cube-marked-1-2-3-4-5-and-6-on-its-six-faces-three-89198.html
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]
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10 Sep 2013, 07:56
Bunuel wrote:
chetan86 wrote:
You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120
Paint one of the faces red and make it the top face.
5 options for the bottom face.
Now, four side faces can be painted in (4-1)! = 3! = 6 ways (circular arrangements of 4 colors).
Total = 5*6 = 30.
Similar question to practice: a-cube-marked-1-2-3-4-5-and-6-on-its-six-faces-three-89198.html
Thanks Bunues, great explanation!!
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]
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27 Dec 2013, 20:12
Bunuel wrote:
chetan86 wrote:
You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120
Paint one of the faces red and make it the top face.
5 options for the bottom face.
Now, four side faces can be painted in (4-1)! = 3! = 6 ways (circular arrangements of 4 colors).
Total = 5*6 = 30.
Similar question to practice:
Can you give more detail on how you determined what method to use to solve this problem? I have never heard of the "circular arrangement" technique you used.
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]
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28 Dec 2013, 03:56
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sjmarinov wrote:
Bunuel wrote:
chetan86 wrote:
You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120
Paint one of the faces red and make it the top face.
5 options for the bottom face.
Now, four side faces can be painted in (4-1)! = 3! = 6 ways (circular arrangements of 4 colors).
Total = 5*6 = 30.
Similar question to practice:
Can you give more detail on how you determined what method to use to solve this problem? I have never heard of the "circular arrangement" technique you used.
The number of arrangements of n distinct objects in a row is given by $$n!$$.
The number of arrangements of n distinct objects in a circle is given by $$(n-1)!$$.
From Gmat Club Math Book (combinatorics chapter):
"The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:
$$R = \frac{n!}{n} = (n-1)!$$"
seven-men-and-seven-women-have-to-sit-around-a-circular-92402.html
a-group-of-four-women-and-three-men-have-tickets-for-seven-a-88604.html
the-number-of-ways-in-which-5-men-and-6-women-can-be-seated-94915.html
4-couples-are-seating-at-a-round-tables-how-many-ways-can-131048.html
at-a-party-5-people-are-to-be-seated-around-a-circular-104101.html
seven-family-members-are-seated-around-their-circular-dinner-102184.html
seven-men-and-seven-women-have-to-sit-around-a-circular-11473.html
a-group-of-8-friends-sit-together-in-a-circle-alice-betty-106928.html
seven-men-and-five-women-have-to-sit-around-a-circular-table-98185.html
a-group-of-8-friends-sit-together-in-a-circle-alice-betty-106928.html
find-the-number-of-ways-in-which-four-men-two-women-and-a-106919.html
gmat-club-monday-giveaway-155157.html (700+)
Theory on Combinations: math-combinatorics-87345.html
DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52
Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html
Hope this helps.
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]
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28 Dec 2013, 21:30
This helps a great deal! Thank you Bunuel!
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]
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27 May 2014, 06:19
Bunuel wrote:
chetan86 wrote:
You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120
Paint one of the faces red and make it the top face.
5 options for the bottom face.
Now, four side faces can be painted in (4-1)! = 3! = 6 ways (circular arrangements of 4 colors).
Total = 5*6 = 30.
Similar question to practice: a-cube-marked-1-2-3-4-5-and-6-on-its-six-faces-three-89198.html
I really didn't get how you used circular arrangements here.
We have six different colors and 6 different sides.
How many ways can we paint the cube with each side a different color?
Shouldn't it be 6! ?
Great problem btw
Cheers
J
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]
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28 May 2014, 03:41
2
KUDOS
jlgdr wrote:
Bunuel wrote:
chetan86 wrote:
You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120
Paint one of the faces red and make it the top face.
5 options for the bottom face.
Now, four side faces can be painted in (4-1)! = 3! = 6 ways (circular arrangements of 4 colors).
Total = 5*6 = 30.
Similar question to practice: http://gmatclub.com/forum/a-cube-marked ... 89198.html
I really didn't get how you used circular arrangements here.
We have six different colors and 6 different sides.
How many ways can we paint the cube with each side a different color?
Shouldn't it be 6! ?
Great problem btw
Cheers
J
Painting a cube with six different colors is more complicated than arranging six distinct things in a row.
If you have six distinct colors, say ROYGBP, you could place them in the slots below in 6! ways. Why? Any of the 6 colors could go in the first slot, any of the five remaining could go in the second, and so on. So the total is 6*5*4*3*2*1=6!
_ _ _ _ _ _
Now the cube is a combination of slots and a circular arrangement. Here is my sophisticated diagram for that:
_
_ _ _ _
_
The four slots between the top and bottom 'faces' actually wrap around the whole cube. So, let's say for the four 'slots' you choose ROYG.
ROYG =GROY = YGRO = OYGR = ROYG
All of these arrangements are the same because essentially each one is just a rotation of the cube by 90 degrees, not a different paint job.
So we treat the four 'slots' in the middle the same way we would a circular table. Hence the above solutions: you can choose any color for the top face, you can choose one of the five remaining colors for the bottom face (5 ways), and since the four middle faces are 'in a circle' they can be arranged (4-1)!=3! ways. So the total is 5*3!
The tricky part is that we don't count the ways in which we can choose the color of the first face, since every color is going to be chosen anyway. Essentially, you are finding the ways you can paint the other sides relative to one of painted sides. Otherwise you are including in your total the number of different ways you can look at the cube (which don't constitute a new paint job).
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You have a six-sided cube and six cans of paint, each a diff [#permalink]
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15 Oct 2014, 22:32
Bunuel wrote:
chetan86 wrote:
You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120
Paint one of the faces red and make it the top face.
5 options for the bottom face.
Now, four side faces can be painted in (4-1)! = 3! = 6 ways (circular arrangements of 4 colors).
Total = 5*6 = 30.
@Bunuel
, why haven't the ways to paint the first (top)face have been counted, i.e 6C1 ways to choose from amongst the 6 diffferent paints?? And what does the wording provided in the question in the parentheses entails?
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]
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15 Jan 2015, 22:57
Bunuel wrote:
chetan86 wrote:
You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120
Paint one of the faces red and make it the top face.
5 options for the bottom face.
Now, four side faces can be painted in (4-1)! = 3! = 6 ways (circular arrangements of 4 colors).
Total = 5*6 = 30.
Similar question to practice: a-cube-marked-1-2-3-4-5-and-6-on-its-six-faces-three-89198.html
Hi Bunuel,
Need your help on this. I could understand the circular and bottom face things. that's 5*(4-1)!. But I am not able to understand that top part. If initially no sides are painted, then we could chose 1 among 6 paints right? so shouldn't we multiply 5*(4-1)! with 6, as we have six choices initially?
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]
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15 Jan 2015, 23:29
1
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Expert's post
Vinitkhicha1111 wrote:
Bunuel wrote:
chetan86 wrote:
You have a six-sided cube and six cans of paint, each a different color. You may not mix colors of paint. How many distinct ways can you paint the cube using a different color for each side? (If you can reorient a cube to look like another cube, then the two cubes are not distinct.)
(A) 24
(B) 30
(C) 48
(D) 60
(E) 120
Paint one of the faces red and make it the top face.
5 options for the bottom face.
Now, four side faces can be painted in (4-1)! = 3! = 6 ways (circular arrangements of 4 colors).
Total = 5*6 = 30.
@Bunuel
, why haven't the ways to paint the first (top)face have been counted, i.e 6C1 ways to choose from amongst the 6 diffferent paints?? And what does the wording provided in the question in the parentheses entails?
Why don't we consider the 6 ways in which we can color the top face?
Which face is the top face? All faces are identical. You pick any color and put it on any one side. This can be done in one way only. This is like placing the first person at a round table. All places are identical so place the first one can be put anywhere. Similarly, the first paint can be put on any face of the cube. Now you have a top face (which we have just painted) and a bottom face and 4 identical sides.
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink]
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12 May 2016, 23:40
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Re: You have a six-sided cube and six cans of paint, each a diff [#permalink] 12 May 2016, 23:40
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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®. | 2,000 | 5,554 | {"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} | 2.921875 | 3 | CC-MAIN-2017-04 | latest | en | 0.632895 |
https://nukephysik101.wordpress.com/2016/04/20/energies-in-a-nucleus/ | 1,521,941,650,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651481.98/warc/CC-MAIN-20180325005509-20180325025509-00791.warc.gz | 661,606,401 | 20,726 | There are many kinds of energies, such as single particle energy, potential energy, kinetic energy, separation energy, and Fermi energy. How these energies are related?
I summarized in the following picture that the occupation number as a function of kinetic energy.
Since the nucleus is a highly interactive system, although the temperature of the ground state of a nucleus should be absolute Zero, but the Fermi surface is not sharp but diffusive.
The Fermi energy of a nucleon , which is the maximum kinetic energy of a nucleon, is approximately ~35 MeV.
The potential energy is approximately ~ 50 MeV per nucleon.
There is an additional energy for proton due to Coulomb force, which is a Coulomb barrier.
The separation energy is the different between the potential energy and Fermi energy.
The single particle energy is the energy for each single particle orbit.
The binding energy for a nucleon is the energy requires to set that nucleon to be free, i.e. the energy difference between the single particle energy and the potential energy.
There is a minimum kinetic energy, using the 3D spherical well as an approximation. The n-th root of the spherical Bessel function can give the energy of n-orbit with angular momentum $l$, so that.
$j_{l}(kR) = 0, k^2=\frac{2 m E}{\hbar^2}$
For $l=0$, $j_0(x) = \frac{sin(x)}{x}$, the 1-st root is $x = \pi/2$, then
$k=\frac{\pi}{2R}$
$\frac{2 m E}{\hbar^2} = \frac{\pi^2}{4 R^2}$
use $R = 1.25 A^{1/3}$,
$E = \frac{\pi^2 \hbar^2}{8 m 1.25^2 A^{2/3}}$
use $\hbar c = 200 MeV \cdot fm$, $mc^2 = 940 MeV$
$E = \frac{33.6}{A^{2/3}} MeV$
we can see, for $^{16}O$, the minimum KE, which is the 1s-orbit, is about 5 MeV. | 484 | 1,677 | {"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": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-13 | latest | en | 0.881588 |
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A255728 Decimal expansion of the imaginary part (negated) of the limit (2N -> infinity) of Integral_{1..2N} exp(i*Pi*x)*x^(1/x) dx. 2
6, 8, 4, 0, 0, 0, 3, 8, 9, 4, 3, 7, 9, 3, 2, 1, 2, 9, 1, 8, 2, 7, 4, 4, 4, 5, 9, 9, 9, 2, 6, 6, 1, 1, 2, 6, 7, 1, 0, 9, 9, 1, 4, 8, 2, 6, 5, 4, 9, 9, 9, 4, 3, 4, 3, 2, 2, 6, 3, 0, 3, 7, 7, 1, 3, 8, 1, 5, 3, 0, 5, 8, 1, 2, 4, 9, 7, 6, 6, 3, 8, 1, 5, 0, 9, 5, 9, 8, 3, 4, 2, 1, 2, 7, 2, 1, 4, 7, 8, 6, 7, 2, 2 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,1 LINKS Steven R. Finch, Errata and Addenda to Mathematical Constants, p. 60. EXAMPLE -0.6840003894379321291827444599926611267109914826549994343... MATHEMATICA digits = 103; NIntegrate[x^(1/x)*Sin[Pi*x], {x, 1, Infinity}, WorkingPrecision -> 2*digits] - 1/Pi // RealDigits[#, 10, digits]& // First CROSSREFS Cf. A255727 (real part), A157852. Sequence in context: A054042 A184084 A346402 * A272488 A100608 A335005 Adjacent sequences: A255725 A255726 A255727 * A255729 A255730 A255731 KEYWORD nonn,cons AUTHOR Jean-François Alcover, Mar 04 2015 STATUS approved
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Last modified July 31 06:15 EDT 2021. Contains 346369 sequences. (Running on oeis4.) | 691 | 1,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.592047 |
http://betterlesson.com/common_core/browse/141/ccss-math-content-3-nbt-a-1-use-place-value-understanding-to-round-whole-numbers-to-the-nearest-10-or-100?from=domain_core_lesson_count | 1,487,624,305,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170609.0/warc/CC-MAIN-20170219104610-00440-ip-10-171-10-108.ec2.internal.warc.gz | 27,932,862 | 26,748 | Use place value understanding to round whole numbers to the nearest 10 or 100.
25 Lesson(s)
Colossal Fossils
3rd Grade Math » Unit: Rounding
3rd Grade Math » Unit: Rounding
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3rd Grade Math » Unit: Rounding
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Rounding - Unit Assessment
3rd Grade Math » Unit: Rounding
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3rd Grade Math » Unit: Rounding
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Rounding to the Ten's Place
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Rounding to the Hundred's Place
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Creating a Summer Review Packet (Day 1)
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Happy Hundreds
3rd Grade Math » Unit: Place Value Practice
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Working with concrete representations (blocks) helps students understand the abstraction of regrouping.
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3rd Grade Math » Unit: Place Value
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Common Core Math | 854 | 3,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-09 | latest | en | 0.783973 |
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# Why 'more or less stopped using ISO'?
Started Feb 19, 2014 | Questions thread
Re: Maybe this will help.
bobn2 wrote:
jackdan wrote:
bobn2 wrote:
jackdan wrote:
bobn2 wrote:
crames wrote:
bobn2 wrote:
How does brightness figure into this? That is, CIE brightness, or are you using another definition?
I don't see how the connection between luminance and brightness would cause there to be "no such thing as ISO."
You can calculate a brightness for an sRGB value of 118, since the viewing conditions are defined and you have enough info to plug into a CIECAM02 calculator.
What would be the brightness for an sRGB value of 118, in candela per square metre?
-- hide signature --
Bob
CIECAM02 brightness correlate Q of the 8 bit sRGB value (118,118,118), Q=102.28.
Candelas per square meter are the units for luminance, not the unit-less brightness, so your question doesn't make sense.
You're obviously fishing around for something, without explaining how brightness, that is CIE brightness as discussed in this sub-thread, has anything to do with the existence or non-existence of "ISO."
Apparently "'brightness' is dependent on 'absolute luminance of the stimulus' ".
While you mull things over,
Sorry, you have the mulling the wrong way round. It's you that needs to be mulling. Apparently 'brightness' is dependent on 'absolute luminance of the stimulus' and yet we find that 'brightness' has nothing to do with luminance, so mull that one over.
I found the post where your "Apparently 'brightness' is dependent on 'absolute luminance of the stimulus' " originated, but have not found the origin of "'brightness' has nothing to do with luminance,". Can you point me to that post please?
It came from you, 'candelas per square meter are the units for luminance, not the unit-less brightness'.
-- hide signature --
Bob
I am guessing you thought I was someone else. What I said was, "So although there are no units for brightness there are units for the measurment of the perceived brightness to the human eye." Those units, of course, being for luminance".
I probably have got confused between you and crames.
Crames is the one who know what he is talking about.
So, the units for measurement of the perceived brightness to the human eye are candela per square metre? Really?
Only in so far as "luminance is the measurable quantity which most closely corresponds to brightness."
(Brightness, Luminance, and Confusion. from Information Display March 1993 (vol . 9, iss. 3, pp. 21-24). By Charles P. Halsted )
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Color scheme? Blue / Yellow | 639 | 2,721 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-30 | latest | en | 0.954126 |
http://inform7.com/extensions/Jeremy%20John%20Reeder/Numbers/doc_2.html | 1,516,339,462,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887746.35/warc/CC-MAIN-20180119045937-20180119065937-00088.warc.gz | 169,276,918 | 1,739 | # Numbers
## version 1 by Jeremy John Reeder
Section: Testing
We can test if a number divides another evenly like so:
if (some number) divides (some other number)
if (some number) is divisible by (some other number)
We can also test if a number or other value falls within a certain range:
if N is between (some value) and (some other value)
if N is not between (some value) and (some other value)
Note that it does not matter if the smaller of the two values is put first or last. | 118 | 483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-05 | latest | en | 0.877436 |
https://de.maplesoft.com/support/help/maple/view.aspx?path=lprint&L=G | 1,643,275,827,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00639.warc.gz | 256,412,962 | 31,802 | lprint - Maple Help
# Online Help
###### All Products Maple MapleSim
lprint
linear printing of expressions
Calling Sequence lprint(expr1, expr2, ...) lprint[2](expr1, expr2, ...)
Parameters
expr1, expr2, ... - any expressions
Description
• The procedure lprint returns NULL and prints its arguments in a one-dimensional format.
• The expressions expr1, expr2, ..., each separated from the others by a comma and a space, are printed on a single line.
• Like most procedures in Maple, the arguments to lprint are evaluated before being passed to the procedure.
• In general, the printed form produced by lprint is valid Maple input.
• Maple will print all expressions after normal evaluation using lprint if the interface variable prettyprint is set to $0$.
• Because lprint displays output as a side effect, and returns NULL as the procedure value, the ditto commands, %, %%, and %%%, will not recall the output from lprint.
• The procedure lprint is intended for device independent printing and makes no use of special features of the user interface. For pretty-printing, using a two-dimensional output format, see print. For formatted output, see printf and fprintf.
• Calling lprint[2] instead of lprint produces line-broken and indented output for procedures, but the same output as lprint for all other expressions.
Thread Safety
• The lprint command is thread-safe as of Maple 15.
• For more information on thread safety, see index/threadsafe.
Examples
> $\mathrm{lprint}\left(3{x}^{2},x,\mathrm{Int}\left(\frac{\mathrm{sin}\left(x+y\right)}{x-y},x\right)\right)$
3*x^2, x, Int(sin(x+y)/(x-y),x)
Arguments are evaluated before printing.
> $\mathrm{lprint}\left(\mathrm{int}\left(\frac{\mathrm{sin}\left(x+y\right)}{x-y},x\right)\right)$
Si(x-y)*cos(2*y)+Ci(x-y)*sin(2*y)
Note: In command-line Maple, the output of lprint can be cut and pasted into a Maple session while pretty-printed output cannot be.
> $g≔\frac{{x}^{4}-y}{{y}^{2}-3x}:$
> $\mathrm{lprint}\left(g\right)$
(x^4-y)/(y^2-3*x)
> $\mathrm{print}\left(g\right)$
$\frac{{{x}}^{{4}}{-}{y}}{{{y}}^{{2}}{-}{3}{}{x}}$ (1)
The interface setting of prettyprint determines whether line printing or pretty-printing is used to display the results of normal evaluation.
> $g$
$\frac{{{x}}^{{4}}{-}{y}}{{{y}}^{{2}}{-}{3}{}{x}}$ (2)
> $\mathrm{interface}\left('\mathrm{prettyprint}'=0\right):$
> $g$
(x^4-y)/(y^2-3*x)
lprint[2] line breaks and indents procedures, but prints other expressions in one-dimensional form.
> lprint(proc(x) if x < 0 then x else sqrt(x) end if end proc);
proc (x) if x < 0 then x; else sqrt(x); end if; end proc
> lprint[2](proc(x) if x < 0 then x else sqrt(x) end if end proc);
proc( x ) if x < 0 then x; else sqrt(x); end if; end proc
> $\mathrm{lprint}\left[2\right]\left(g\right)$
(x^4-y)/(y^2-3*x)
>
Compatibility
• The lprint[2] command is new in Maple 2021.
See Also | 872 | 2,948 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 13, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-05 | longest | en | 0.597072 |
profshonle.blogspot.com | 1,369,466,098,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705618968/warc/CC-MAIN-20130516120018-00059-ip-10-60-113-184.ec2.internal.warc.gz | 215,790,413 | 15,693 | ## 23 May 2010
### Approximations and the tools you have
If you are interested in finance then you might be aware of The Rule of 72. The basic idea is if your investment gets interest at some percentage r each year (e.g., a ten percent return implies r=10), then you'll have to wait 72/r years before your investment doubles. That means stocks that have an annual return of about 10 percent will double in 7.2 years. (Note: If you account for inflation, stocks have an annual return of 7%.)
I thought the rule was slick, but when trying to do the math things just didn't add up. You can try constructing the recurrence relation yourself, or you can read moneychimp's explanation. The formula I got was to calculate the log of 2 (two for doubling), base (1 + r/100); it turns out the whole time that the rule of 72 is just an inverse linear approximation to the logarithmic function. But it is mostly right for the ranges people would care about. (For a rate of 5%, the log gives you 14.2 years, while the rule of 72 gives you 14.4; for a rate of 50%, the log gives you 1.7, while the rule of 72 gives you 1.44.)
f(x)=72/x is a really simple function, and when comparing the plots of the two functions it is pretty impressive how close it is:
I like the times when all of the math I've learned, even the basic concepts, become useful in analyzing other things. A computer scientist isn't trained in finance, but knowing that what you've learned can be applied widely can empower you.
I had a physics professor who once quipped "men see parabolic trajectories more often than women do." [So as not to leave it too cryptic, he was referencing stand up urinals.] It's an interesting way of thinking about gravity, and it once made me realize something: A friend in Worcester was showing me his "movie" gun (i.e., it wasn't real) and I considered the scope. I knew that scopes were meant for different target ranges, and it was then that I realized that a scope on a gun is a linear approximation to a parabola. Each scope setting is meant for different ranges, which approximates different parts of the parabola. (Also, they only work with Earth's gravity.) I mention guns in my blog only because, after explaining this to my friend he said something funny: "Man, I should totally take you out shooting; you'd be such a great target... I mean shot."
***
Side note: If investment interests you, check out this piece on index funds by Scott Adams, author of Dilbert. The company referenced, Vanguard, is an excellent one due to its unique corporate structure: Customers are also shareholders in Vanguard, so the company always has the customer's best interests in mind. In our culture it's sometimes considered boorish to talk about money, but somebody's got to tell you about it! Particularly now that pensions are going away and that defined contribution plans are your responsibility, you should empower yourself by knowing as much as you can.
#### 1 comment:
1. This technique is useful to collect information of a huge data.There are many software which provides this tool.I am solving questions related to this topic for my boards examination.
Rate of Change Formula | 710 | 3,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2013-20 | latest | en | 0.97332 |
http://mathhelpforum.com/calculus/31742-reduction-formula.html | 1,481,147,902,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542246.21/warc/CC-MAIN-20161202170902-00403-ip-10-31-129-80.ec2.internal.warc.gz | 169,503,109 | 10,008 | 1. ## Reduction Formula
Given that $I_{n} = \int_{0}^{1}{x^n(1-x)^{\frac{1}{2}}dx}$ where $n \geq 0$, prove that for $n \geq 1$
$I_{n} = \frac{2n}{2n+3} I_{n-1}$
The first thing that comes to mind is $x = \sin^2u$ so $dx = 2 \sin u \cos u du$
giving $I_{n} = \int_{0}^{\frac{\pi}{2}}{sin^{2n} \cos u \ 2 \sin u \cos u \ du }$
$\Rightarrow 2 \int_{0}^{\frac{\pi}{2}}{sin^{2n+1} \cos^2 u \ du }$
my first thought is to use integration by part, with $du = sin^{2n+1} \cos u$ and $v = \cos u$
this doesn't get me very far sadly....
2. Just do integration by parts without the substitution:
$u=x^n$, $dv=\sqrt{1-x}dx$
Thus $du=nx^{n-1}dx$, $v=-\frac{2}{3}(1-x)^{\tfrac{3}{2}}$
and:
$I_{n}=\int_{0}^{1}x^n\sqrt{1-x}\,dx$
$I_{n}=-\frac{2}{3}x^n(1-x)^{\tfrac{3}{2}}|_{x=0}^1-\int_{0}^{1}(nx^{n-1})\cdot\left(-\frac{2}{3}(1-x)^{\tfrac{3}{2}}\right)\,dx$
$I_{n}=\frac{2n}{3}\int_{0}^{1}x^{n-1}(1-x)^{\tfrac{3}{2}}\,dx$
$I_{n}=\frac{2n}{3}\int_{0}^{1}x^{n-1}(1-x)\sqrt{1-x}\,dx$
$I_{n}=\frac{2n}{3}\int_{0}^{1}(x^{n-1}-x^n)\sqrt{1-x}\,dx$
$I_{n}=\frac{2n}{3}\left(\int_{0}^{1}x^{n-1}\sqrt{1-x}\,dx-\int_{0}^{1}x^n\sqrt{1-x}\,dx\right)$
$I_{n}=\frac{2n}{3}(I_{n-1}-I_{n})$
and algebra can do the rest.
--Kevin C. | 595 | 1,208 | {"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": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2016-50 | longest | en | 0.52731 |
http://www.cram.com/flashcards/algebra-1-319756 | 1,524,540,445,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946453.89/warc/CC-MAIN-20180424022317-20180424042317-00417.warc.gz | 382,247,636 | 17,042 | • Shuffle
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### 10 Cards in this Set
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• Back
Two lines are________ if thier slopes are negatice reciprocals Perpendicular Two lines in the same plane that never intersect are______ and have the____slope. Parallel/Same The ratio of the vertical change to the horizontal change is called the___ Slope The y-coordinated of the point at which the graph of a line crosses the verical axis is called the_______ y intercept The form of a line called point-slope form is_______and is used when you are given either___ _____on a line or the___slope and one____on the line. y-y=mx(x-x)/2 points/slope/point The form of a line called slope-intercept form is____ and is used when you are given the____and the____ y=mx+b/slope/y-intercept An equation in the form x=c is a_____ line. vertical An equation in the form y=c horizontal The form of a line called Standard for is________and must not have any_____and A must be_____. Ax+By=C/fractions/positive If I have data given for every year from 1980-2000, interpolation would be estimating for the year___and extrapolating would be estimating for the year____ between 1981-1999/right to left | 405 | 1,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2018-17 | latest | en | 0.816859 |
https://www.neetprep.com/question/69495-Bohrs-model-atomic-radius-first-orbit-r-radiusof-fourth-orbit-isa-rb-rc-r-d-r/126-Physics--Atoms/702-Atoms | 1,590,700,373,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347400101.39/warc/CC-MAIN-20200528201823-20200528231823-00506.warc.gz | 860,304,184 | 63,807 | In Bohr’s model, if the atomic radius of the first orbit is ${\mathrm{r}}_{0}$, then the radius of the fourth orbit is
(a) ${\mathrm{r}}_{0}$ (b) 4${\mathrm{r}}_{0}$
(c) ${\mathrm{r}}_{0}$/16 (d) 16${\mathrm{r}}_{0}$
Concept Questions :-
Bohr's model of atom
To view Explanation, Please buy any of the course from below.
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
If R is the Rydberg’s constant for hydrogen the wave number of the first line in the Lyman series will be
(a) $\frac{\mathrm{R}}{4}$ (b) $\frac{3\mathrm{R}}{4}$
(c) $\frac{\mathrm{R}}{2}$ (d) 2R
Concept Questions :-
Spectral series
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In hydrogen atom, if the difference in the energy of the electron in n =2 and n = 3 orbits is E, the ionization energy of hydrogen atom is
(a) 13.2 E (b) 7.2 E
(c) 5.6 E (d) 3.2 E
Concept Questions :-
Bohr's model of atom
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The first member of the Paschen series in hydrogen spectrum is of wavelength 18,800 Å. The short wavelengths limit of Paschen series is
(a) 1215 Å (b) 6560 Å
(c) 8225 Å (d) 12850 Å
Concept Questions :-
Spectral series
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The ratio of the largest to shortest wavelengths in Lyman series of hydrogen spectra is
(a) $\frac{25}{9}$ (b) $\frac{17}{6}$
(c) $\frac{9}{5}$ (d) $\frac{4}{3}$
Concept Questions :-
Spectral series
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In Bohr model of hydrogen atom, the ratio of periods of revolution of an electron in n =2 and n = 1 orbits is
(a) 2 : 1 (b) 4 : 1
(c) 8 : 1 (d) 16 : 1
Concept Questions :-
Bohr's model of atom
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Difficulty Level:
The ratio of the longest to shortest wavelengths in Brackett series of hydrogen spectra is
(a) $\frac{25}{9}$ (b) $\frac{17}{6}$
(c) $\frac{9}{5}$ (d) $\frac{4}{3}$
Concept Questions :-
Spectral series
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Difficulty Level:
The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true
(a) Its kinetic energy increases and its potential and total energies decrease
(b) Its kinetic energy decreases, potential energy increases and its total energy remains the same
(c) Its kinetic and total energies decrease and its potential energy increases
(d) Its kinetic, potential and total energies decreases
Concept Questions :-
Spectral series
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Difficulty Level:
The ratio of minimum to maximum wavelength in Balmer series is
(a) 5 : 9 (b) 5 : 36
(c) 1 : 4 (d) 3 : 4
Concept Questions :-
Spectral series
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Difficulty Level:
Rutherford’s $\mathrm{\alpha }$-particle experiment showed that the atoms have
(a) Proton (b) Nucleus
(c) Neutron (d) Electrons
Concept Questions :-
Various atomic model
Please attempt this question first.
Difficulty Level: | 1,047 | 3,826 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 17, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-24 | latest | en | 0.753418 |
https://grindskills.com/implementing-ridge-regression-selecting-an-intelligent-grid-for-lambda/ | 1,670,540,922,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711368.1/warc/CC-MAIN-20221208215156-20221209005156-00250.warc.gz | 320,823,242 | 16,427 | # Implementing ridge regression: Selecting an intelligent grid for $\lambda$?
I’m implementing Ridge Regression in a Python/C module, and I’ve come across this “little” problem. The idea is that I want to sample the effective degrees of freedom more or less equally spaced (like the plot on page 65 on the “Elements of Statistical Learning”), i.e., sample:
$$\mathrm{df}(\lambda)=\sum_{i=1}^{p}\frac{d_i^2}{d_i^2+\lambda},$$
where $d_i^2$ are the eigenvalues of the matrix $X^TX$, from $\mathrm{df}(\lambda_{\max})\approx 0$ to $\mathrm{df}(\lambda_{\min})=p$. An easy way to set the first limit is to let $\lambda_{\max}=\sum_i^p d_i^2/c$ (assuming $\lambda_{\max} \gg d_i^2$), where $c$ is a small constant and represents aproximately the minimum degree of freedom that you want to sample (e.g. $c=0.1$). The second limit is of course $\lambda_{\min}=0$.
As the title suggests, then, I need to sample $\lambda$ from $\lambda_{\min}$ to $\lambda_{\max}$ in some scale such that $\mathrm{df}(\lambda)$ is sampled (approximately), say, in $0.1$ intervals from $c$ to $p$…is there an easy way to do this? I thought solving the equation $\mathrm{df}(\lambda)$ for each $\lambda$ using a Newton-Raphson method, but this will add too much iterations, specially when $p$ is large. Any suggestions?
This is a long answer. So, let’s give a short-story version of it here.
• There’s no nice algebraic solution to this root-finding problem, so we need a numerical algorithm.
• The function $\mathrm{df}(\lambda)$ has lots of nice properties. We can harness these to create a specialized version of Newton’s method for this problem with guaranteed monotonic convergence to each root.
• Even brain-dead R code absent any attempts at optimization can compute a grid of size 100 with $p = 100\,000$ in a few of seconds. Carefully written C code would reduce this by at least 2–3 orders of magnitude.
There are two schemes given below to guarantee monotonic convergence. One uses bounds shown below, which seem to help save a Newton step or two on occasion.
Example: $p = 100\,000$ and a uniform grid for the degrees of freedom of size 100. The eigenvalues are Pareto-distributed, hence highly skewed. Below are tables of the number of Newton steps to find each root.
# Table of Newton iterations per root.
# Without using lower-bound check.
1 3 4 5 6
1 28 65 5 1
# Table with lower-bound check.
1 2 3
1 14 85
There won’t be a closed-form solution for this, in general, but there is a lot of structure present which can be used to produce very effective and safe solutions using standard root-finding methods.
Before digging too deeply into things, let’s collect some properties and consequences of the function
$$\newcommand{\df}{\mathrm{df}} \df(\lambda) = \sum_{i=1}^p \frac{d_i^2}{d_i^2 + \lambda} \>.$$
Property 0: $\df$ is a rational function of $\lambda$. (This is apparent from the definition.)
Consequence 0: No general algebraic solution will exist for finding the root $\df(\lambda) – y = 0$. This is because there is an equivalent polynomial root-finding problem of degree $p$ and so if $p$ is not extremely small (i.e., less than five), no general solution will exist. So, we’ll need a numerical method.
Property 1: The function $\df$ is convex and decreasing on $\lambda \geq 0$. (Take derivatives.)
Consequence 1(a): Newton’s root-finding algorithm will behave very nicely in this situation. Let $y$ be the desired degrees of freedom and $\lambda_0$ the corresponding root, i.e., $y = \df(\lambda_0)$. In particular, if we start out with any initial value $\lambda_1 < \lambda_0$ (so, $\df(\lambda_1) > y$), then the sequence of Newton-step iterations $\lambda_1,\lambda_2,\ldots$ will converge monotonically to the unique solution $\lambda_0$.
Consequence 1(b): Furthermore, if we were to start out with $\lambda_1 > \lambda_0$, then the first step would yield $\lambda_2 \leq \lambda_0$, from whence it will monotonically increase to the solution by the previous consequence (see caveat below). Intuitively, this last fact follows because if we start to the right of the root, the derivative is “too” shallow due to the convexity of $\df$ and so the first Newton step will take us somewhere to the left of the root. NB Since $\df$ is not in general convex for negative $\lambda$, this provides a strong reason to prefer starting to the left of the desired root. Otherwise, we need to double check that the Newton step hasn’t resulted in a negative value for the estimated root, which may place us somewhere in a nonconvex portion of $\df$.
Consequence 1(c): Once we’ve found the root for some $y_1$ and are then searching for the root from some $y_2 < y_1$, using $\lambda_1$ such that $\df(\lambda_1) = y_1$ as our initial guess guarantees we start to the left of the second root. So, our convergence is guaranteed to be monotonic from there.
Property 2: Reasonable bounds exist to give “safe” starting points. Using convexity arguments and Jensen’s inequality, we have the following bounds
$$\frac{p}{1+ \frac{\lambda}{p}\sum d_i^{-2}} \leq \df(\lambda) \leq \frac{p \sum_i d_i^2}{\sum_i d_i^2 + p \lambda} \>.$$
Consequence 2: This tells us that the root $\lambda_0$ satisfying $\df(\lambda_0) = y$ obeys
$$\frac{1}{\frac{1}{p}\sum_i d_i^{-2}}\left(\frac{p – y}{y}\right) \leq \lambda_0 \leq \left(\frac{1}{p}\sum_i d_i^2\right) \left(\frac{p – y}{y}\right) \>. \tag{\star}$$
So, up to a common constant, we’ve sandwiched the root in between the harmonic and arithmetic means of the $d_i^2$.
This assumes that $d_i > 0$ for all $i$. If this is not the case, then the same bound holds by considering only the positive $d_i$ and replacing $p$ by the number of positive $d_i$. NB: Since $\df(0) = p$ assuming all $d_i > 0$, then $y \in (0,p]$, whence the bounds are always nontrivial (e.g., the lower bound is always nonnegative).
Here is a plot of a “typical” example of $\df(\lambda)$ with $p = 400$. We’ve superimposed a grid of size 10 for the degrees of freedom. These are the horizontal lines in the plot. The vertical green lines correspond to the lower bound in $(\star)$.
An algorithm and some example R code
A very efficient algorithm given a grid of desired degrees of freedom $y_1, \ldots y_n$ in $(0,p]$ is to sort them in decreasing order and then sequentially find the root of each, using the previous root as the starting point for the following one. We can refine this further by checking if each root is greater than the lower bound for the next root, and, if not, we can start the next iteration at the lower bound instead.
Here is some example code in R, with no attempts made to optimize it. As seen below, it is still quite fast even though R is—to put it politely—horrifingly, awfully, terribly slow at loops.
# Newton's step for finding solutions to regularization dof.
dof <- function(lambda, d) { sum(1/(1+lambda / (d[d>0])^2)) }
dof.prime <- function(lambda, d) { -sum(1/(d[d>0]+lambda / d[d>0])^2) }
newton.step <- function(lambda, y, d)
{ lambda - (dof(lambda,d)-y)/dof.prime(lambda,d) }
# Full Newton step; Finds the root of y = dof(lambda, d).
newton <- function(y, d, lambda = NA, tol=1e-10, smart.start=T)
{
if( is.na(lambda) || smart.start )
lambda <- max(ifelse(is.na(lambda),0,lambda), (sum(d>0)/y-1)/mean(1/(d[d>0])^2))
iter <- 0
yn <- Inf
while( abs(y-yn) > tol )
{
lambda <- max(0, newton.step(lambda, y, d)) # max = pedantically safe
yn <- dof(lambda,d)
iter = iter + 1
}
return(list(lambda=lambda, dof=y, iter=iter, err=abs(y-yn)))
}
Below is the final full algorithm which takes a grid of points, and a vector of the $d_i$ (not $d_i^2$!).
newton.grid <- function(ygrid, d, lambda=NA, tol=1e-10, smart.start=TRUE)
{
p <- sum(d>0)
if( any(d < 0) || all(d==0) || any(ygrid > p)
|| any(ygrid <= 0) || (!is.na(lambda) && lambda < 0) )
stop("Don't try to fool me. That's not nice. Give me valid inputs, please.")
ygrid <- sort(ygrid, decreasing=TRUE)
out <- data.frame()
lambda <- NA
for(y in ygrid)
{
out <- rbind(out, newton(y,d,lambda, smart.start=smart.start))
lambda <- out\$lambda[nrow(out)]
}
out
}
Sample function call
set.seed(17)
p <- 100000
d <- sqrt(sort(exp(rexp(p, 10)),decr=T))
ygrid <- p*(1:100)/100
# Should take ten seconds or so.
out <- newton.grid(ygrid,d) | 2,343 | 8,259 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2022-49 | latest | en | 0.86947 |
https://iim-cat-questions-answers.2iim.com/quant/geometry/geometry-triangles/geometry-triangles_27.shtml | 1,701,353,196,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100227.61/warc/CC-MAIN-20231130130218-20231130160218-00597.warc.gz | 365,214,713 | 13,679 | # CAT Quantitative Aptitude Questions | Geometry Questions for CAT - Triangles
###### CAT Questions | CAT Geometry Questions | Triangle
The question is from CAT Geometry - Triangles. In this problem, it discusses about ratio of areas of equilateral triangle inscribed in circle. CAT Geometry questions are heavily tested in CAT exam. Make sure you master Geometry problems.
Question 27: Consider equilateral triangle T inscribed in circle C, what is ratio of the areas of T and C? Consider Circle C inscribed in equilateral triangle T, what is ratio of the areas of T and C?
1. 3√3:π , 3√3:16π
2. 3√3:4π , 3√3:π
3. √3:π , 3√3:4π
4. √3:π , √3:16π | 182 | 649 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-50 | latest | en | 0.849147 |
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# Meaning
CHAPTER 19
Absorption of overhead is also tenned as levy, recovery, or application of overhead. Cost absorption
refers to the process of absorbing all overhead costs allocated to apportioned over particular cost centre or
production department by the unit produced. Accordingly, the distribution of the overhead cost 10 the cost
centres or cost units is known as Overhead Absorption.
The apportionment of overhead expenses is done by adopting suitable basis such as output, materials,
prime cost, labour hours, machine hours etc. In order to detennine the absorption of overhead in costs of
jobs, products or process, a rate is calculated and it is called as "Overhead Absorption Rate" or "Overhead
Rate." The overhead rate can be calculated as below :
Total Quantity or Value
Different overhead rates are applied based on the features and objectives of the business organization.
The following are the important overhead absorption rates generally employed :
( 1) Actual Overhead Rate
(2) Predetennined Overhead Rate
(3) Blanket Overhead Rate
(4) Multiple Overhead Rate
(5) Nonnal Overhead Rate
(6) Supplementary Overhead Rate
Each of the above overhead absorption rates has been explained in the following pages :
426 A Textbook of Financial Cost and Management Accounting
(1) Actual Overhead Rate: Actual overhead rate as otherwise called the historical rate. This rate is
calculated by dividing the actual overhead absorbed by the actual quantity or value of the base selected for
a particular period. Assuming that overhead rate is calculated on monthly basis, the following formula is
expressed as :
Actual Overhead Rate =
Actual Overhead during the month
x 100
Actual Quantity or Value of the base for the month
(2) Predetermined Overhead Rate: Predetermined Dverhead rate is determined in advance of actual
production and the rate is computed by dividing the budgeted overhead for the accounting period by the
budgeted base for the period. The formula is :
Budgeted Overheads for the Period
Pre-determined Overhead Rate = x 100
Budgeted Base for the Period
(3) Blanket Overhead Rate: Blanket overhead rate is also termed as Single Overhead Rate. A
single overhead rate when computed for the entire factory is known as Blanket Rate. It is calculated as :
Blanket Rate =
Overhead of Entire Factory
Total Quantum of the Base Selected
Single rate may be applied suitably in small concerns and only where a single product is
manufactured.
(4) Multiple Overhead Rate: Multiple overhead rates involve computation of separate rates for each
production department, service department, cost centre, each product or line and for each production
factor. The following formula is used for calculating multiple overhead rate:
Overhead Cost Allocated and Apportioned to Each Cost Centre
Corresponding Base
=
(5) Normal Overhead Rate: Normal Overhead Rate is a predetermined rate calculated with
reference to normal capacity. It is calculated as :
Normal Overhead Rate =
Base at Normal Capacity
(6) Supplementary Overhead Rates: These rates used to carryout adjustment between overhead
absorbed and overhead incurred. These are used in addition to some other rates and is calculated as under:
Supplementary Overhead Rate =
Base Unit or Hours
Methods of Absorption of Overhead
There are number of methods applicable for computing overhead absorption rate. The following are
the various methods of absorbing "Manufacturing Overhead" depending upon the suitable basis selected
for the purpose :
Absorption of Overhead 427
(1) Direct Material Cost Method
(2) Direct Labour Cost Method
(3) Direct Labour Hours Method
(4) Prime Cost Method
(5) Unit of Output Method
(6) Machine Hour Rate Method
(1) Direct Material Cost Method: Under this method, the rate of absorption is calculated on the basis
of direct material cost method. The rate of manufacturing overhead absorption is determined by dividing the
manufacturing overhead by the direct material cost. The result obtained the rate of absorption is expressed as
percentage. Thus, the overhead rate is calculated by the following formula:
Direct Material Percentage Rate
=
Example: 1
-------- x 100
Direct Material Cost
Manufacturing overhead budgeted for 2003 Rs. 20,000
Cost of direct materials Rs. 80,000
Calculation:
20,000
Direct Material Percentage Rate = x 100
80,000
=25%
(2) Direct Labour Cost Method: Direct Labour Cost Method is also termed as Direct Wages Method.
Under this method direct wage rate can be determined by dividing the estimated factory overhead cost
apportioned by the predetermined direct wages, and the result obtained is expressed as a percentage. The
following formula for calculating the percentage rate is :
Percentage of Direct Labour Rate =
Example: 2
------- x 100
Direct Wages
Direct Wages paid in factory during the year 2003, Rs. 10,000
Factory overhead during that the period was Rs. 4,000
4,000
Direct Labour Percentage Rate = x 100 = 40%
10,000
(3) Direct Labour Hours Method: Under this method the rate is determined by dividing the production
overheads by direct labour hours of each department. This method is designed to overcome the objections of
direct labour cost method. This method is most suitable in such industries where the production is carried out
manually or by skilled labours. Thus, the direct labour hour rate will be calculated by applying the following
formula:
428 A Textbook of Financial Cost and Management Accounting
Direct Labour Hour Rate
=
Direct Labour Hours
(4) Prime Cost Method: Under this method, both direct material cost and direct labour cost are taken
into account for determination of recovery rate. The actual or predetermined rate of factory absorption is
computed by dividing actual or budgeted overhead expenses by the aggregate of direct material or direct labour
cost of the department. The following formula is used for calculation of overhead recovery rate:
=
Illustration: 3
Prime Cost
x 100
You are required to find out (1) Direct Material Cost Rate (2) Direct Labour Cost Rate (3) Direct
Labour Hours and (4) Prime Cost Rate from the following particulars :
Total overhead for the period
Total direct labour cost (Direct wages)
Total materials used or Direct material cost
Total direct labour hours
Solution:
Rs. 25,000
Rs. 8,000
Rs. 10,000
Rs. 2,000
(I) Direct Material Cost Rate = ---------------- x 100
=
(2) Direct Labour Cost Rate
=
=
(3) Direct Labour Hours Rate
=
=
(4) Prime Cost Rate
=
=
=
Direct Material Cost
25000
10000
x 100 = 250%
x 100
Direct Wages
25000
x 100 = 312.5%
8000
x 100
Direct Labour Hours
25000
x 100 = Rs.l2.5%
2000
x 100
Prime Cost
25000
x 100
10000 + 8000
25000
x 100 = 138.88%
18000
Illustration: 4
429
The following figures have been extracted from the books of a manufacturing concern. All jobs pass
through the company's two departments:
Direct materials used
Direct labour cost
Direct labour hours
Machine hours
Prod. Dept.
Rs.
6,000
3,000
I,SOO
12,000
10,000
The following information pertains to work order No.555
Direct materials used
Direct labour cost
Direct labour hours
Machine hours
Prod. Dept.
Rs.
240
130
530
510
Finishing Dept.
Rs.
500
1,500
1,200
5,000
2,000
Finishing Dept.
Rs.
20
50
140
50
You are required to prepare a statement showing the different cost results for work order No. 555 under the three
commonly used method.
Solution:
1. Direct Labour Cost Rate =
Production Dept. =
Finishing Dept. =
2. Direct Labour Hour Rate =
Production Dept. =
Finishing Dept. =
3. Machine Hour Rate
=
Production Dept. =
Finishing Dept. =
-------- x 100
Direct Material Cost
I,SOO
--x 100 = 60%
3,000
1,200
x 100 = SO%
1,500
Direct Labour Hours
I,SOO
x 100
--- = 15 paise per hour
12,000
1,200
5,000
= 24 paise per hour
Machine Hours
I,SOO
= IS paise per hour
10,000
1,200
2,000
= 60 paise per hour
430 A Textbook of Financial Cost and Management Accounting
Comparative Statement of Work Order No. 555
Particulars Labour Cost Labour Hour Method Machine Hour Method
Prod. Finish Prod. Finish Prod. Finish
Dept. Dept. Dept. Dept Dept. Dept.
Rs. Rs. Rs. Rs. Rs. Rs.
Materials used 240 20 240 20 240 20
Direct labour 130 50 130 50 130 50
Prime Cost 370 70 370 70 370 70
Factory Overheads 78 40
(i) Direct Labour Cost
[130 x 1 ~ [50 x I:J
(ii) Labour Hours 530 x 15 140 x 24
paise paise
Rs.79.50 Rs.33.60
(iii) Mach.Hours 510 x 18 50 x 60
paise paise
Rs.91.80 Rs.30.00
Total 448 110 449.50 103.60 461.80 100
(5) Unit of Output Method: This method is also termed as Production Unit Method or Cost Unit Rate
Method. Under this method absorption rate is determined on the basis of number of units produced is known
as Cost Unit Rate. The recovery rate is calculated by dividing the actual or budgeted factory overheads by the
number of cost units produced. The formula is :
Cost Unit Rate =
No. of Units Produced
This method is most suitable in such industries where the production of same grade is carried out.
(6) Machine Hour Rate: Machine hour rate means the cost or expenses incurred in running a machine
for one hour. It is one of the scientific methods of absorbing factory expenses where the process of
manufacturing are carried out by machines. Under this method overhead costs are allocated on the basis of
the number of hours a machine or machines are used for a particular job. According to the Institute of Cost
and Management Accountants, England a machine hour rate is "an actual or predetermined rate of cost
apportionment or overhead absorption, which is calculated by dividing the cost to be apportioned or absorbed
by the number of machine hours expended or to be expended."
The machine hour rate is determined by dividing the amount of overhead cost to be apportioned or
absorbed by the number of machine hours. Machine hour rate can be calculated as below :
rate:
Machine Hour Rate = --------
Machine Hours
Calculation Machine Hour Rate: The following steps are required for computing the machine hour
(1) Identify the overhead expenses relating to a specific machine or group of machine in order to
require for computing machine hour rate.
(2) Each machine or group of machine treated as a cost centre.
(3) Manufacturing overhead or machine expenses are grouped into two types:
(a) Fixed or Standing Charges (b) Variable Machine Expenses.
431
(a) Fixed or Standing Charges: Fixed or Standing Charges which remain constant irrespective
of the use of machine. For example, rent, insurance charges, rates, supervision etc.
(b) Variable Machine Expenses: These expenses are variable with use of the machine. For
example, power, depreciation, repairs etc.
(4) An hourly rate of fixed or standing charges will be calculated by totalling of fixed charges and
dividing by the number of normal hours worked by machine.
(5) Normal working hours are calculated by adding the cost relating to non-productive time, i.e.,
normal ideal time for maintenance and setting up etc.
(6) Separate hourly rate for each machine expenses will be calculated.
(7) The total of the standing charges rate and the machine expenses rates per hour will give the
machine hour rate.
Basis for Apportionment of Machine Expenses
The following bases of apportionment of different expenses are required to be considered for the
calculation of machine hour rate :
Expenses
Fixed or Standing Expenses :
(1) Rent and Rates
(2) Heating and Lighting
(3) Supervision
(4) Lubricating Oil and Consumable Stores
(5) Insurance
Machine Expenses:
1. Depreciation
2. Power
3. Repairs
Basis
Floor area occupied by each machine
No. of points used or Floor area or
heating any machine
Time spent on each machine
Machine hours, Past experience or
Capital value.
Insurance value of each machine.
Value of Machine
Horse power of each machine
Cost of repairs spread over its working life
(1) It helps to measure the relative efficiency of different machines.
(2) It facilitates comparison of cost of operating different machines.
(3) It helps to ascertain idle time of machines relating to non-productive time.
(4) It is the most desirable scientific method, where the time factor is taken into account.
(1) It involves more clerical labour in determining the number of machine hours worked.
(2) It does not consider where the expenses not proportional to the working hours of machines.
(3) It is very difficult to measure the machine hours where the works are completed without
operating any machinery.
432
Illustration: 5
A Textbook of Financial Cost and Management Accounting
Calculate machine hour rate of Machine X
Solution:
Consumable stores
Repairs
Heat and light
Rent
Insurance of building
Insurance of machines
Depreciation of machines
Room services
General charges
Normal working hours
Area of sq. fit.
Book value of machines
Rs.
600
800
360
1,200
4,800
800
700
60
90
10,000 hours
100
12,000
Computation of Machine Hour Rate for Machine X
Particulars Total per hour
Rs.
Standing Charges:
Consumable stores 600
Heat and light (360 x 100 1 600) 60
Rent (1200 x 100 1 6(0) 200
Insurance of building (4800 x 100 1 600) 800
Insurance of Machines (800 x 12000 1 32000) 300
Room service (60 x 100 1600) 10
General charge (90 x 100 1 600) 15
Total Standing Charges 1,985
1,985
Standing charges per hour
=
10,000
Machine Expenses:
Repairs (800 1 10,000)
Depreciation of machines (135.481 10,000)
Machine Hour Rate
Working Notes
Rate per hour
Rs.
0.199
0.080
0.014
0.293
(1) Heat and light, rent, insurance of building, room service and general charges have been
distributed on the basis of floor area.
(2) Depreciation of machine has been calculated on the basis of book value of machines and
working hours, i.e., 10,000 x 12,000 (or) 120 : 500 = 6 : 25 .
. . 700 x 6/31 = Rs.135.48
(3) Insurance of machine has been apportioned on the basis of book value of machines.
Illustration: 6
Compute the machine hour rate from the following information :
Cost of Machine
Installation charges }
Estimated scrap value after the
expiry of its life (15 years)
Rent and Rates per month
General lighting per month
Insurance premium for the machine per annum
Repairs and maintenance per month
Power consumption - 10 units per hour
Rate per hour 100 units
Estimated working hours per annum
Supervisor's salary per month
Rs.
1,00,000
10,000
5,000
200
300
960
1,000
20
2,200
600
433
The machine occupies 1,4 th of the total area of the shop. The supervisor is expected to devote 1I5th of his time
for supervising the machine.
Solution:
Computation of Machine Hour Rate
Particulars
Standing Charges :
Rent and Rates (200 x 12 x 1,4)
General lighting (300 x 12 x 1,4)
Repairs and Maintenance
Supervisor's salary (600 x 12 x 115)
Total Standing Charges
4,900
Standing charges per hour = --
2,000
Machine Expenses :
Depreciation (1,00,000+10,000-5,000)
15 x 2,000
Power
Machine Hour Rate
Illustration: 7
Cost of machine Rs. 1,80,000
Freight and installation Rs. 20,000
Working life 10 years
Working hours 4,000 per year
Repair charges 50% of depreciation
Power 10 units per hour @ 10 paise per unit
Lubricating oil @ Rs. 2 per day of 8 hours
Consumable stores @ Rs. 10 per day of 8 hours
Wages of operator @ Rs. 2 per day
Scrap value of machine Rs. 20,000
Per annum Rate Per hour
Rs. Rs.
600
900
960
1,000
1,440
4,900
2.45
3.50
2.00
7.95
434 A Textbook of Financial Cost and Management Accounting
Calculate machine hour rate from the above information :
Solution:
Computation of Machine Hour Rate
Particulars
Standing Charges:
Lubricating oil
Consumable stores (10 x Re.l)
Wages of Operator
Standing charges per day
14
Standing charges per hour --
8
Machine Expenses :
Depreciation
=
=
=
Cost + Freight - Scrap value
Life in hours
1,80,000 + 20,000 - 20,000
4,000 x 10
1,80,000
40,000
50
Repairs 50% of depreciation 4.50 x --- =
100
Power 10 units @ Re. 0.10 each 10 x 0.10 =
Machine Hour Rate
Illustration: 8
=
Per day of 8 hours
Rs.
2
10
2
14
Rate Per hour
Rs.
1.30
4.50
2.25
1.00
9.05
In a factory, a machine is oonsidered to work for 208 hours in a month. It includes maintence time of
8 hours and setup time of 20 hours.
The expense data relating to the machine are as under :
Cost of the machine is Rs.5,00,OOO Life 10 years
Estimate scrap value at the end of life is Rs. 20,000
Repairs and maintenance per Annum Rs. 60,480
Consumable stores per annum Rs. 47,520
Rent of building per annum (The machine under
Reference occupies 1I6th of the area) Rs. 72,000
Supervisor's salary per month
(Common to three machines) Rs. 6,000
Wages of Operator per month per machine Rs. 2,500
General lighting charges per month allocated to the machine Rs. 1,000
Power 25 units per hour at Rs. 2 per unit
Power is required for productive purposes only. Setup time through productive does not require power. The supervisor
and operator are permanent. Repairs and maintenance and consumable stores vary with the running of the machine.
Required:
Calculate Machine Hour Rate for :
(a) Setup Time and
(b) Running Time ..
Solution:
Effective hours
For fixed costs 208 - 8 = 200 hours
For variable costs 208 - 28 = 180 hours
Computation of Machine Hour Rate
Particulars Per month Setup time
Per hour
Rs. Rs.
Standing Charges :
6000 2000
Supervision Rs. --
=
2,000 --
=10
3 200
1,000
1000
General lighting
=
-- =5
200
72,000 6,000 1000
Rent
= = =
1,000 --
=5
12 6 200
Machine Expenses :
Depreciation = 5,00,000 - 20,000
4,80,000 48,000 4000
= = =
4,000 -- = 20
10 12 200
60,480
Repairs =
=
5,040
12
Consumable Stores =
47,520
=
3,960
12
Power = 25 x 2 x 180
9,000
2500
Wages 2,500 --= 12.50
200
Machine Hour Rate 52.50
Illustration: 9
Calculate the machine hour rate from the following informations :
Cost of machine
Scrap value
Repairs and maintenance per month
Standing charges per month
Rs.20,000
Rs.2,000
Rs. 200
Rs. 100
435
Running time
Per hour
Rs.
10
5
5
20
5040
--=28
180
3960
--=22
180
9000
--=50
180
12.50
152.50
436 A Textbook of Financial Cost and Management Accounting
Effective working life
Running time per month
Power used 5 units at 20 paise a unit per hour.
Solution:
10,000 hours
200 hours
Computation of Machine Hour Rate
Particulars
Standing Charges :
Allocated Rs.l00 per month of 200 hours
100
For 200 hours Rs.l00 =
200
Variable Charges :
Cost of machine
Less: scrap
Rs.20,ooo
Rs. 2,000
Depreciation for 10,000 hours =
Hence, for one hour =
Repairs and maintenance
18,000
10,000
Rs.200 per month of 200 hOUrs} =
Power 5 units per hour @ 20 paise
Machine Hour Rate.
IUustration: 10
18,000
200
200
=
=
Rate Per hour
Rs.
0.50
1.80
1.00
1.00
4.30
A department is having 3 machines. The figures indicate the departmental expenses. Calculate the
machine hour rate in respect of these machines from the informations given below:
Depreciation of machinery
Depreciation of building
Repairs to machinery
Insurance of Machinery
Indirect wages
Power
Lighting
Miscellaneous expenditure
Rs.
12,000
2,880
4,000
800
6,000
6,000
800
4,200
36,680
Particulars Machine Machine
A B
Direct Wages Rs.l,200 2,400
Power units 30,000 10,000
No. of workers 4 8
Light points 8 24
Space 400 sq.fit 800 sq.fit
Cost of Machine Rs.3,OO,OOO Rs.l,20,OOO
Hours worked 200 300
Solution:
Computation of Machine Hours Rate
Expenses
Depreciation }
on Machinery
Depreciation }
on Building
MaChine}
Repairs
Insurance
Indirect Wages
Power
Lighting
MiscellaneOUS}
Expenses
Total
Hours worked
Machine hour rate
Working Notes :
Basis:
Basis
Machine
Value
Space
Machine
Value
- do-
No. of workers
Power units
Light points
Direct wages
Direct Wages = 12: 24 : 24 or 1 : 2 : 2
Power units = 3 : 1 : 2
Cost of machine = 30 : 12 : 18
Space = 1 : 2 : 2
Hours worked = 2 : 3 : 3
Light points = 1 : 3 : 6
No. of. workers = 1 : 2 : 2
Illustration: 11
Total
12,000
2,880
4,000
800
6,000
6,000
800
4,200
36,680
200
Rs.70.48
Machine
A
6,000
576
2,000
400
1,200
3,000
80
840
14,096
300
32.77
437
Machine
D
2,400
20,000
8
48
800 sq.fit
Rs. 1,80,000
Machine
B
2,400
1,152
800
160
2,400
1,000
240
1,680
9,832
300
42.51
300
Machine
C
3,600
1,152
1,200
240
2,400
2,000
480
1,680
12,752
From the undernoted data calculate the machine-hour rate of a Mailing Machine.
Cost of Machine
Scrap Value
life 12 years
438 A Textbook of Financial Cost and Management Accounting
Effective Work days
Maintenance & Repairs
Stores consumed
Power Consumption
Supervision Expenses
Idle time estimated
Solution:
Computation of Machine Hour Rate
Effective working days
Total
200 days of 8 hrs
100 days of 6 hrs
7.5% of capital cost
Rs. 1 ,000
Rs.2 per operating hour
1 % of capital cost
Rs.7,500
10%
200 x 8 hours
100 x 6 hours
=
=
1,600 hours
600
2,200 hours
Less : Idle Time estimated 10%
=
220 hours
Net working hours
=
1,980 hours
in a year
Items Basis of
Apportionment
(A) Standing Charges
Depreciation
30,500 - 2,500
12
Maintenance & Repairs 7.5 of capital cost
Stores consumed Actuals
Insurance premium I % of capital cost
Supervision expenses Actuals
Total Standing Charges
(B) Variable cost-power consumption
Machine hour rate (a + b)
Illustration: 12
Amount Rate
per annum per hour
Rs. Rs.
= 2,333.33
= 2,287.40
= 1,000.00
=
305.00
= 7,500.00
13,425.83 6.78
2.00
8.78
Particulars of three machines used in a factory are as under (six week period; 160 hours working) :
Machine X Machine Y Machine Z
Rs. Rs. Rs.
Cost of Machine 10,000 15,000 20,000
No. of workers 2 5 10
Direct wages Rs.3oo Rs. 800 Rs. 1,200
Power Rs.45 Rs.80 Rs. 150
Light points 2 4 6
Area Occupied 100 sq. ft. 250 sq. ft. 400 sq. ft.
The expenses incurred during the period were as follows :
Power
Lighting
Rent and Rates
Depreciation
Repairs
Indirect wages
Canteen expenses
Sundries
Total
Rs.
275
48
450
1,350
1,800
460
51
300
4,734
Compute the machine hour rate for each machine.
Solution:
Compntation of Machine Hour Rate - 160 working hours
Expenses Basis of Total
Apportionment X
Rs. Rs.
Power Actuals 275 45
Lighting Lighting Points 41 8
Rent & Rates Area 450 60
Depreciation Cost of Machine 1,350 300
Repairs Cost of Machine 1,800 400
Indirect wages Direct Wages 460 60
Canteen expo No. of workers 51 6
Sundries Area 300 40
Total (a) 4,734 919
Working hour (b) 160
Machine hours rate
a
--= Rs.5,744
say
b
Rs.5.74
Under Absorption and Over Absorption of Overheads
439
Machine
Y Z
Rs. Rs.
80 150
16 24
150 240
450 600
600 800
160 240
15 30
100 160
1,571 2,244
160 160
9.819 14.025
9.82 14.03
- Absorption of overhead may be based either on the actual rate or predetermined rate. If the actual
rates are used, the costs having been actually incurred and overhead absorbed are equal. But in the case of
predetermined rates, the costs have been determined in advance of incurrence of the overhead expenditure.
This may lead to difference of overhead incurred and overhead absorbed. Such a difference of Overhead
is said to be under absorption of overhead or over absorption of overhead.
According the term over absorption means that the amount of overhead absorption is more than the
actual overhead is said to be over absorption of overhead.
The term under absorption of overhead means that the amount of overhead absorption is less than the
actual overhead incurred is said to be under absorption of overhead.
Causes of Under or Overhead Absorption of Overhead
The following reasons for over and under absorption of overheads :
(1) Actual overhead cost incurred may be more or less than the budgeted overhead.
440 A Textbook of Financial Cost and Management Accounting
(2) Actual machine hours, labour hour and output may be lower or higher than the budgeted or
predetermined base.
(3) Seasonal fluctuations.
(4) Wrong computation of overhead absorption rate, output and machine hours:
(5) Under or Over utilization of production capacity.
Methods of Treatment
The following three important methods may be adopted for overhead adjustment and disposal of over
or under absorption of overheads :
(1) Carrying Over of Overheads
(2) Application or use of supplementary rates
(3) Write off to Costing Profit and Loss Account.
(1) Carrying Over of Overheads: Under this method, the amount of over or under absorption is
carry forward to the next year. This method may be adopted in situation where the normal business cycle
extends for more than one year.
(2) Application of Supplementary Rate: Under this method, the supplementary rate is adopted
when the amount of under or over absorbed overheads is quite large. Supplementary rate is calculated by
dividing the amount of under or over absorbed overheads by the actual base.
Amount of Under or Over Absorbed Overheads
Supplementary Rate =
Actual Base
The supplementary rate may be used as positive supplementary rate or negative supplementary rate.
In the case of positive supplementary rate it is intended to add under absorbed overhead to cost of
production. A negative rate, however, adjusted the cost by deducting the amount of over absorbed
(3) Write otT to Costing Profit and Loss Account: Under this method, if the amount of under or over
absorbed overhead is small it may be written off to Costing Profit and Loss Account. If due to some abnormal
. factors, the amount of under or over absorbed is large it should be transferred to Profit and Loss Account.
Illustration: 13
In a factory, the overheads of a production department are absorbed on the basis of Rs. 18 per
machine hour. The details for the month of October 2002 are as under :
Factory overheads incurred Rs. 16,50,000.
Of the above Rs. 16,50,000
Amount became payable due to an award of labour hour
Prior period expenses booked in the month of October 2002
Actual Machine hours worked
Rs. 2,50,000
Rs. 1,50,000
Rs. 65,000
Actual production was 2,60,000 units, of which 1,95,000 units were sold. On analyzing the reasons it was found
that 40% of the under absorbed overheads was due to defective planning and the rest was attributed to normal cost increase.
How would you treat under absorbed overheads in Cost Accounts?
(C A Inter, Nov. 2002)
Solution:
Under absorbed overhead expenses for the month of Oct. 2002
Total expenses incurred
Less: Amount paid according to labour court award }
(assumed to be non-recurring)
Prior period expenses
Net overhead expenses incurred for the month
Factory overhead absorbed 6,500 hrs x Rs. 18
Treatment of under absorbed overheads in cost account:
2,50,000
1,50,000
(1) 40% due to defective planning. This being abnormal should be debited to P & L :
40
= 80,000 x -- = Rs. 32,000
100
(2) Balance 60% should be distributed over finished goods. Inventory and cost of
60
sales by supplementary rate = 80,000 x -- = Rs. 48,000
100
Under absorbed overheads in Cost Account = Rs. 32,000 + Rs. 48,000
= Rs. 80,000
Finished goods inventory = 48000 x = Rs. 12,000
4
3
Cost of Sales = 48000 x = Rs. 36,000
4
Illustration: 14
Rs.
16,50,000
4,00,000
12,50,COO
11,70,000
Rs.80,OOO
441
The total overhead expenses of a factory are Rs. 4,46,380. Taking into account the normal working of
the factory, overhead was recovered in production at Rs. 1.25 per hour. The actual hours worked were Rs.
2,93,104. How would you proceed to close the books of accounts, assuming that besides 7,800 units
produced of which 7,000 were sold, there were equivalent units in work in progress?
On investigation, it was found that 50% of the unabsorbed overhead was on account of increase in
the cost of indirect materials and indirect labour and the remaining 50% was due to factory inefficiency.
Also give the profit implication of the method suggested.
Solution:
Overhead Recovered from production
(Rs. 293104 x 1.25)
Actual overhead expenses incurred
Amount of under-recovered overhead
Rs.
= 3,66,380
= 4,46,380
= 80,000
(C A Inter, Nov. 2000)
50% of the above amount is due to increase in the cost of indirect material and indirect labour and should be
charged to units produced by means of a supplementary rate.
442 A Textbook of Financial Cost and Management Accounting
No. of total units produced = 7,800 + 200 = 8,000 units
Supplementary rate = 50% of Rs. 80,000 I 8,000 = Rs. 5 per unit
The amount of Rs. 40,000 should be apportioned among cost of sales, finished goods and work in progress at the
rate of Rs.5 per unit.
Cost of sales = 7,000 x Rs. 5
Finished goods = 800 x Rs. 5
Work in progress = 200 x Rs. 5
Rs.
= 35,000
= 4,000
= 1,000
40,000
By using this method, the profit for the period will be reduced by Rs.35,ooo and the value of stock will increase
by Rs.5,OOO.
The balance amount of RsAO,OOO due to factory inefficiency should be charged to Costing Profit and Loss
Account as this is abnormal cost for which the production should not be penalized.
discharge its functions of planning, organizing, controlling, co-ordination and directing. These expenses
are not specifically incurred which cannot be identified with the specific. Thus, the overheads are collected
under a standing order number, allocated and apportioned to various cost centres and units.
The administrative overhead is absorbed under anyone of the following methods:
(1) Transferring to Profit and Loss Account
(2) Apportioning to Works Overheads
(3) Apportioning to Selling Overheads.
Selling and Distribution Overhead : Selling and distribution expenses are incurred for promoting
sales, securing orders, creating demand and distribution of products or output from producers to the
ultimate consumers. The incidence of selling and distribution overheads depends on external factors such
as distance of market, nature of competition etc. which are beyond the control of management. They are
dependent upon customer's behaviour, liking etc. These expenses are the nature of policy costs and hence
not amenable to control. The overhead rate of selling and distribution overheads can be determined by
anyone of the following basis :
(a) A rate per article or unit of production
(b) A percentage on the selling price of each article or production unit
(c) A percentage on the factory cost.
. Treatment of Important Overhead Charges
Expenses on Removal and Reelection of Machine : Such expenses may be incurred due to factors
like change in method of production, an addition or alteration in the factory building, change in flow of
production. All such expenses are treated as production overheads, when amount of such expense is large,
it may be spread over a period of time. If such expenses are incurred due to faulty planning or other
abnormal factor, then they may be charged to Costing Profit and Loss Account.
Training Expenses: Training expenses are part of production, administration and selling &
distribution overheads based on particular employee posted in the department. If such expenses are huge
443
due to high labour turnover, such expenses should be excluded from costs and charged to Costing Profit
and Loss Account.
Packing Expenses: Cost of primary packing necessary for protecting the product or for convenient
handling should become part of prime cost. The cost of packing incurred to facilitate the transportation of
the product from the factory to the customer should become part of distribution cost. In case of special
packing done at the request of the customer the cost of the same should be charged to specific work order
or job. The cost of fancy packing to attract customers is an advertising expenditure. Hence it is to be
treated as selling overhead.
Idle Time Wages: Normal idle time wages is treated as a part of cost of production. Thus in case of
direct workers an allowance for normal idle time is built into labour cost rates. In the case of indirect
works, normal time wages is spread over all the products or jobs through the process of absorption of
factory overhead. Abnormal idle time cost is not included as a part of production cost and is shown as a
separate item in the Costing Profit and Loss Account. So that normal cost are not disturbed.
Overtime Wages: If overtime is resorted to at the desire of the customer, then overtime premium is
charged to concerned job directly. If overtime is required to cope with general production programe for
meeting urgent orders, the overtime premium should be treated as overhead cost of particular department
or cost center which works overtime. If overtime is worked on account of abnormal conditions such as
flood, earthquake etc that should be charged to Costing Profit and Loss Account.
Normal Loss and Abnormal Loss: Treatment of normal and abnormal loss of materials arising
during storage, which inflate the issue price. Normal loss can be charged to stores overheads and also can
be treated as a separate item of overheads to be recovered as a percentage of material consumed. On the
other hand, in the case of abnormal loss, it is charged to Costing Profit and Loss Account. If the loss is due
to error in documentation it should be corrected through adjustment entries.
Idle Capacity Cost: Idle capacity is that part of the capacity of a plant, machine or equipment which
cannot be effectively utilized in production. The idle capacity may arise due to lack of product demand,
non-availability of raw material, shortage of skilled labour, shortage of power etc. Cost associated with idle
capacity are mostly fixed in nature. These costs remain unabsorbed or unrecovered due to under utilization
of plant and service capacity. '
If the idle capacity cost is due to unavoidable reasons a supplementary overhead rate may be used to
recover the idle capacity cost. In this case, the costs are charged to the production capacity utilized.
If the idle capacity is due to avoidable reasons such as faulty planning, etc. the cost should be
charged to Costing Profit and Loss Account.
If the idle capacity cost is due to seasonal factors then the cost should be charged to the cost of
production by inflating overhead rates.
Pre-Production Costs: These are costs incurred during the period when a new factory is in the
process of being established a new project is undertaken or a new product line or product is taken up but
there is no established or formal production to which such costs may be charged. These costs are normally
treated as deferred revenue expenditure and are charged to future production.
Research and Development Cost: These are costs incurred in the discovery of new ideas or
processes by experiment or otherwise and for putting the results of such experiments on a commercial
basis. Research cost defined as the cost of searching for new or improved product, new application of
material or new improved methods, processes, systems or services.
444 A Textbook of Financial Cost and Management Accounting
Development cost is the cost of the process which begins with the implementation of the decision to
use scientific or technical knowledge to produce a new or improved product etc. and ends with the
commencement of formal production of that product by that method.
Cost of Small Tools: Tools purchased may be capitalized and depreciated over life if life is
ascertainable. Revaluation method of depreciation may be used in respect of very small tools of short
effective life. Depreciation may be charged to factory overheads, if tools use can be identified with the
departments. It may be charged to cost of department on the basis of actual issues.
QUESTIONS
1. Explain absorption of overhead
2. What do you understand by overhead rates?
3. Briefly explain the different kinds of overhead absorption rates.
4. Explain the different methods of absorption of overhead.
S. What do you understand by machine hour rate? How it is computed?
6. Briefly explain the methods of treatment of selling and distribution overheads.
7. What do you mean by under absorption and over absorption of overhead? Brief explain the methods of treatment of
under or over absorption of overheads.
8. Indicate the accounting treatment of overhead charges mentioned below :
(a) Idle time wages. (b) Packing expenses. (c) Research and development costs. (d) Cost of small tools.
9. Briefly explain the importance of machine hour rate as a basis for the absorption of factory overheads
10. Compute main hour rate from the following data :
Cost of machine Rs. 1,10,000
Installation charges Rs. 10,000
Estimated scrap value (after IS years) Rs. S,OOO
Rent and rates for the shop Rs. 200 P.M.
General lighting for the shop Rs. 300 P.M.
Insurance premium for the machine Rs. 960 P.a.
Repairs and maintenance Rs. 1000 P.a.
Power consumption 10 units per hour
Rate of power per 100 units Rs. 20
Estimated working hours per annum 2200 which include setting up time of 200 hours.
Shop supervisor's salary per month Rs. 600
The machine occupies 114 of the total area of the shop. The shop supervisor is expected to devote lISth of his time for
supervising the machine.
[Ans : Machine hour rate: Rs.7.9S]
ll. Calculate the machine hour rate from the following information:
Cost of the machine Rs. 19,200
Estimated scrap value Rs. 1,200
Average repairs and maintenance Rs. ISO p.m.
Standing charges allocated Rs. SO p.m.
Effective working life of the machine 10,000 hours
Running time per month 166 hours
Power used by machine
S units per hour at the rate of 19 paise per unit
[Ans : Machine hour rate = Rs. 3.9S]
12. The machine shop of a manufacturing concern has 6 identical machines manned by 6 operators. The total cost of the
machines is Rs. 8,00,000. The following information relates to six monthly period ended 30th September 2003.
Normal available hours per month 208
Absenteeism (without pay) hours per month 18
Leave (with pay) hours per month 20
Normal idle time hours per month 10
Average rate of wages per hour per operator Rs. 2.S0
Production bonus IS% on wages
Power and fuel consumption Rs. 9,000
Supervision and indirect labour Rs. 3,300
Electricity and lighting Rs. 1,200
Repairs and maintenance (per annum) 3% of value of machine
Absorption of Overhead 445
Insurance per annum Rs. 42,000
Depreciation (per annum) 10% of original cost
Allocated factory overheads per annum Rs. 75,670
Calculate machine hour rate
[Ans : Machine hour rate Rs. 25]
13. Universal manufacturing Ltd. have 2 factories. Factory I employs 130 and Factory II employs 150 direct workers. Both
factories work 40 hours per week, and 50 weeks a year.
Overhead Rate are No. I - 25 paise per hour
II - 20 paise per hour
Current overhead expenses No. - I Rs.70,OOO; No. II - Rs.50,OOO. Analyse these figures and state probable causes of
any discrepancy.
[Ans : Factory I Under absorption of overhead expenses Rs. 5,000
Factory II Over absorption of overhead expenses Rs. 10,000]
14. During the year ended 31 st March 2003 the factory overhead costs of three production departments of an organization
are as under:
X Rs.48,950
Y Rs.89,200
Z Rs.64,500
The basis of apportionment of overhead is given below :
Department X - Rs.5 per machine hour for 10,000 hours
Y -75% of Direct Labour Cost of Rs.l,20,000
Z - Rs.4 per piece for 15,000 pieces.
Calculate department-wise under or over absorption of overheads and present the data in a tabular form.
[Ans: Over absorption X - Rs.1050 ; Y - Rs.800; Under absorption Z - Rs.4500]
15. A machine is purchased for cash at Rs.92,OOO. Its working life is estimated to be 18,000 hours after which its scrap
value is estimated at Rs.2,ooo. It is assumed from past experience that:
(1) The machine will work for 1,800 hours annually.
(2) The repair charges will be Rs. 10,800 during the whole period of life of the machine.
(3) The power consumption will be 5 units per hour at Rs. 2 per unit.
(4) Other annual standing charges are estimate to be :
(a) Rent of department (machine occupies 1I5th of the place) Rs. 7,800
(b) Light (12 points in the department; 2 points engaged in machine) Rs. 2,880
(c) Foreman's salary (1/4th of his time is occupied in the machine) Rs. 60,000
(d) Insurance premium (fire) for machine Rs. 360
(e) Cotton waste Rs. 600
Find out machine hour rate on the basis of the above data for allocation of the works expenses to all jobs for which the
machine is used. .
[Ans : Machine hour rate: Rs. 25.60]
16. Calculate the machine hour rate for machine Q from the following data:
Cost of the machine Rs. 51,000
Estimated life 20 years of 2400 hour each
Established repairs for life Rs. 12,000
Power consumption per hour 10 units
Rate for power 5 paise per unit
Insurance ~ % per annum
Machine charges Rs. 30 per month
The machine is kept in a rented shed and there is one supervisor. The machine occupies 1I41h of his time for this
machine. Rent for the shed is Rs. 400 per month. Supervisor's salary is Rs. 500 per month. Electricity charges for the
Rent is Rs. 50 per month. Half the electricity charges are to be borne by this machine.
[Ans : Machiner hour rate Rs. 3.45; Standing charges per hour Rs. 1.51, Variable cost per hour Rs. 1.94]
17. From the following particulars, calculate the machine hour rate for a drilling machine:
Rs.
Cost of the drilling Machine 42,000
Estimated scrap value 2,000
Estimated working life 10 years of 2000 hours each
Running time for 4 weekly period 150 hours
Estimated repairs for life 10,000
Standing charges allocated to this machine for 4 weekly period 300
Power consumption per hour 5 units 10 paise per unit
[Ans : Machine hour rate per hour Rs. 5]
446 A Textbook of Financial Cost and Management Accounting
18. The following is the budget of superb engineering works for the year 2003 :
Rs.
Direct labour cost 98,000
Direct labour hours 1,55,000
Machine hours 50,000
(a) From the above figures prepare the overhead application rate using the following methods :
(a) Direct Labour Hour (b) Direct Labour Cost (c) Machine hour
(b) Prepare a comparative statement of cost, showing the result of application of each of the above rates to Job
No. 555 from the undermentioned data.
Direct material cost Rs. 45
Direct labour: wages Rs. 40
Direct labour: hours 40
Machine hours 30
[Ans : Overhead application rate: (a) Rs. 40 per labour hour (b) 63.27% (c) Rs. 1.24 per machine hour]
Comparative Statement Cost:
Direct labour hour method Rs. III
Direct labour cost method Rs. 126.63
Machine hour method Rs. 132.20
19. Calculate the machine hour rate for machine X from the following information:
Cost of the machine Rs.16,OOO
Estimated scrap value Rs. 1,000
Effective working life 10000 hours
Running time per hour-weekly period 160 hours
Average cost of repairs and maintence charges per four-weekly period Rs.120
Standing charges allocated to machine X per four-weekly period Rs.4O
Power used by the machine 4 units per hour at a cost of 5 paise per hour
[Ans : Machine-hour rate Rs. 2.55]
20. From the following information, compute the machine-hour rate in respect of a machine
Cost of the machine Rs. 55,000
Estimated scrap value Rs. 3,400
Effective working life 10000 hours
Repairs estimated over usual life of machine Rs.7,500
Standing charges of shop for four week period Rs. 8,550
Hours worked in four weekly period Rs. 1,200
Number of machine in shop 30
Powers used each machine, per hour 5 units
Cost of power per unit 5 paise.
21. Compute machine hour rate from the information given below:
Cost of machine Q Rs. 1,35,000
Life of the machine 10 years
Estimated scrap value (after 10 years) Rs. 19,800
Working hours 1,800
Insurance per annum Rs. 450
Cotton wastes per annum Rs. 750
Rent per dept._per annum Rs. 9,750
Foreman's salary per annum Rs. 75,000
Lighting for dept. (per annum) Rs. 3,600
Repairs for entire life Rs. 1,440
Machine Q occupies 1/5
0h
of the area and foreman devotes 1/4
0h
of his time to the machine. The machine has two light
points out of the total 12 for lighting in the department.
[Ans : Machine hour rate Rs. 27.20]
22. A machine costing Rs. 20,000 is expected to run for 10 years at the end of which its scrap value is estimated to be Rs.
2000. Installation charges Rs. 200. Repairs for 10 years life is estimated to be Rs. 1800 and the machine is expected to
run for 2190 hours in a year. Its power consumption would be 15 units per hour at Rs. 5 for per 100 units. The machine
occupies 1I4
oh
of the area of the department and has two points out of total ten for lighting. The foreman has to devote
about 1/3"' of his time to this machine. The rent for this department is Rs. 300 p.m. and charges for lighting 80 p.m.
The foreman is paid a salary of Rs. 960 p.m. Find out the hourly rate, assuming insurance is @ 1 % p.a. and expenses
on oil etc. are Rs. 9 per month.
[Ans : Machine hour rate Rs. 4.059]
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###### CAT Questions | Algebra | Inequalities - Cubic Inequalities
CAT Questions
The question is about finding the solution of a cubic inequality. Our task is to find the solution of the cubic inequality. Inequalities are crucial to understand many topics that are tested in the CAT exam. Having a good foundation in this subject can help us tackle questions in Coordinate Geometry, Functions, and most importantly in Algebra. A range of CAT questions can be asked based on this simple concept.
Question 2: Solve the inequality x3 – 5x2 + 8x – 4 > 0.
1. (2, ∞)
2. (1, 2) ∪ (2, ∞)
3. (-∞, 1) ∪ (2, ∞)
4. (-∞, 1)
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##### Method of solving this CAT Question from Algebra - Inequalities : We all know (or should soon) how to solve a quadratic inequality. How about a cubic? Try to see if you can factorise the equation, and see how the equation behaves around the roots.
Let a, b, c be the roots of this cubic equation
a + b + c = 5
ab + bc + ca = 8
abc = 4
This happens when a = 1, b = 2 and c = 2 {This is another approach to solving cubic equations}.
The other approach is to use polynomial remainder theorem
If you notice, sum of the coefficients = 0
=> P(1) = 0
=> (x - 1) is a factor of the equation. Once we find one factor, we can find the other two by dividing the polynomial by (x - 1) and then factorizing the resulting quadratic equation.
(x - 1) (x - 2) (x - 2) > 0
Let us call the product (x - 1) (x - 2) (x - 2) as a black box.
If x is less than 1, the black box is a –ve number
If x is between 1 and 2, the black box is a +ve number
If x is greater than 2, the black box is a +ve number
Since we are searching for the regions where the black box is a +ve number, the solution is as follows: 1 < x < 2 OR x > 2
The question is "Solve the inequality x3 – 5x2 + 8x – 4 > 0."
##### Hence the answer is "(1, 2) ∪ (2, ∞)"
Choice B is the correct answer.
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Email: prep@2iim.com | 762 | 2,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-18 | latest | en | 0.916766 |
https://www.physicsforums.com/threads/for-z-x-iy-find-the-relationship-between-x-and-y.855332/ | 1,508,815,327,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187827853.86/warc/CC-MAIN-20171024014937-20171024034937-00486.warc.gz | 961,705,439 | 16,434 | # For z = x+iy find the relationship between x and y
1. Feb 2, 2016
### zigzag7
1. The problem statement, all variables and given/known data
For z = x+iy find the relationship between x and y so that (Imz2) / z2 = -i
2. The attempt at a solution
I attempted this in a few different ways (i.e. looking at the exponential and trig forms of complex numbers)... I settled on simple FOIL which gave me the following:
(x+iy)^2 = x^2 + i2xy - y^2
The imaginary part is 2xy; so:
2xy / (x^2 + i2xy - y^2) = -i <-- from original problem
from here, multiplying -i by the denominator gives:
2xy = -ix^2 + 2xy + iy^2
Cancel out 2xy to get zero on the left side, and factor out i, leaving:
x^2 = y^2
However.... this does not seem to produce any solutions resulting in -i.
For example, if x=2 and y=2, z^2 = 8i ... for x=-2 and y=2, z^2 = -8i ...
The problem is, there never seems to be a sign change for Im(z^2) over z^2.
Is this problem flawed, or am I missing something obvious...?
2. Feb 2, 2016
### Staff: Mentor
What is wrong with those examples?
Im(8i)=8. What is 8/(8i)?
Im(-8i)=-8. What is -8/(-8i)?
3. Feb 2, 2016
### zigzag7
Ah, I didn't realize that 1/i equals -i.
Thanks for the quick response! | 399 | 1,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-43 | longest | en | 0.909405 |
http://www.wyzant.com/answers/3175/can_0_be_divided_into_anything | 1,369,117,175,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699730479/warc/CC-MAIN-20130516102210-00027-ip-10-60-113-184.ec2.internal.warc.gz | 811,598,686 | 7,797 | Search 69,756 tutors
# Can 0 be divided into anything?
Another way to see why this is true: try dividing by a very small number. For example: 2/0.001 = 2000. Now make that denominator even smaller: 2/0.0001 = 20,000. And smaller: 2/0.0000001 = 20,000,000.
As you can probably tell by now, as the denominator gets smaller and shrinks towards zero, the quotient (the answer) grows larger, towards infinity. So technically speaking, you can say that dividing anything by zero = infinity, and that's "undefined" (can't put a number on it) in math.
Listen to your teacher, and try inputing values into your calculator!!!!!!!!!!!
Here is an even simpler way to see that you can't divide by zero.
Let x be a real number. and define y=x/0. By definition of division, we get 0y = x. The left hand side of this last equation equals 0 because of multiplication by zero. Thus this equation will be only true if x = 0. This means that for x ≠ 0, the expresson x/0 is automatically undefined.
In the special case of x = 0, the last equation, 0y = x holds true, for all real y. Therefore if we consider defining 0/0, we find that all real numbers work equally well. As a result, the expression 0/0 is called indeterminate. By themselves with no context, indeterminate numbers are undefined.
There are seven expressions total that are recognized as indeterminate:
0 / 0 ∞ / ∞ 0 · ∞ ∞ - ∞ ∞0 1∞ 00
You will learn them when you take calculus, as they arrise when taking limits.
No, zero cannot be divided in to anything. Whenever you see a zero in the denominator, your answer is "No Real Solution" | 442 | 1,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2013-20 | latest | en | 0.906694 |
https://physics.stackexchange.com/questions/692520/intuition-for-the-delta-m-0-pm-1-dipole-selection-rule | 1,723,232,757,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640768597.52/warc/CC-MAIN-20240809173246-20240809203246-00758.warc.gz | 360,220,754 | 40,615 | # Intuition for the $\Delta m = 0,\pm 1$ dipole selection rule
The selection rules for hydrogen are: $$\Delta l=\pm1$$ and $$\Delta m=0,\pm1$$. The first makes intuitive sense because of the conservation of angular momentum and the fact that a photon has spin 1. But does there exist an intuitive explanation for the $$m$$ selection rule as well?
• I think that your intuitive argument might be misleading as a photon can carry higher values of orbital angular momentum: en.wikipedia.org/wiki/Orbital_angular_momentum_of_light Commented Feb 3, 2022 at 18:06
• @Mauricio but as far as I understand photons with different orbital angular momentum can only created artificially which should be here no problem. Commented Feb 3, 2022 at 18:34
• Write the interaction Hamiltonian between atom and photon. The conservation (selection rule) is the result of interaction Hamiltonian.
– ytlu
Commented Feb 3, 2022 at 20:22
• The reasoning for the selection rules for $\Delta m$ is similar as that for $\Delta\ell$. The angular momentum of the photon also has projections. For example $q_\mathrm{ph}=0$ and $\pm 1$ are associated with linearly polarized light and left/right circular polarized light. If you have non polarized light, you have to sum over all contributions of the projection quantum number. To get better intuitive understanding you need to know some angular momentum algebra (the selection rules follow directly from the 3j Symbols).
– Paul
Commented Feb 4, 2022 at 19:21
The selection rule for $$m$$ corresponds to the conservation of the $$J_z$$ component of the angular momentum, where $$z$$ is the chosen quantization axis.
• The $$\Delta m=0$$ possibility occurs when the system is driven by light which is linearly polarized along the quantization axis, in which case each photon has total angular momentum $$1$$ but nevertheless has $$J_z=0$$.
• The $$\Delta m=\pm1$$ case happens when the system is driven by light which is circularly polarized in the plane orthogonal to the quantization axis, in which case each photon has angular momentum component $$J_z=\pm 1$$ that must be absorbed by the system. | 505 | 2,120 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-33 | latest | en | 0.900643 |
http://math.nakouzia.com/111 | 1,566,117,635,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313747.38/warc/CC-MAIN-20190818083417-20190818105417-00286.warc.gz | 125,015,546 | 92,558 | # algebraic, question 3, grade 9
21/06/2016
3)
Given A(x) = ${(3x+2)^2}-{(x-2)^2}$ and B(x) = $2{x^2}-8-(x+2)(-2x+6)$
1. Write A(x) as a product of 3 factors then deduce its roots.
2. Factorize B(x)
3. Expand B(x) and solve B(x)+20 = 0.
4. let F(x) = $\frac{{A(x)-2B(x)}}{{A(x)}}$
i) Find the domain of definition of F(x).
ii) Verify that F(x) = $\frac{5}{{4x}}$
iii) Solve F(x) = $\frac{{- 5}}{8}$
iv) Evaluate $F(\sqrt3)$
5) Consider E(x) = $(a-b){x^3}+2a{x^2}-{b^2}x+x$
Find a&b if E(x) is identical to 0. | 238 | 526 | {"found_math": true, "script_math_tex": 7, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2019-35 | latest | en | 0.540623 |
https://stats.stackexchange.com/questions/107932/r-analyzing-relationship-between-two-or-more-binary-variables | 1,685,968,125,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652116.60/warc/CC-MAIN-20230605121635-20230605151635-00376.warc.gz | 568,962,819 | 39,278 | # R - Analyzing Relationship Between Two (or more) Binary Variables
Say I have two vectors:
Action.Taken = c(0,1,0,0,1,1,0,1,0)
Success = c(0,0,0,1,0,1,0,1,0)
The first tells me whether or not a specific action was taken in a trial and the second tells me whether or not that trial succeeded. How would I go analyzing these two vectors to answer the following question: Does taking action (Action.Taken = 1) affect whether or not success is had (Success = 1)? I'd like some measure of significance as an regression/hypothesis testing.
I'm looking for an answer that I can implement using R. I am also quite new to stats, so it would be nice if someone could give me a relatively simple, straightforward answer/example.
Thanks!
• Consider the chi-square test for independence. Jul 14, 2014 at 21:37
• And the correlation coefficient (phi) that can be computed based on Chi-Square. Jul 14, 2014 at 21:45
• Yeah, I've actually tried the prop.test function, which returns, among other things, the p-value given the null hypothesis that two (or more) proportions aren't different. I think phi is just as simple as cor(dat), but I don't know what/how much that tells me. Jul 14, 2014 at 21:51
contingency = table(Action.Taken, Success)
chisq.test(contingency) | 334 | 1,262 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-23 | latest | en | 0.952834 |
https://gmatclub.com/forum/if-n-is-a-positive-integer-and-r-is-the-remainder-when-68277.html | 1,531,869,002,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589932.22/warc/CC-MAIN-20180717222930-20180718002930-00094.warc.gz | 655,570,502 | 49,337 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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if n is a positive integer and r is the remainder when
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if n is a positive integer and r is the remainder when [#permalink]
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02 Aug 2008, 07:53
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if n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24. what is the value of r?
(1) n is not divisible by 2
(2) n is not divisible by 3
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Re: DS: remainder (GMATPREP 1) [#permalink]
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02 Aug 2008, 08:17
selvae wrote:
You are right, selvae.
Could you provide explanation?
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Re: DS: remainder (GMATPREP 1) [#permalink]
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02 Aug 2008, 08:37
1
n-1 and n+1 are consecutive even numbers (since n is not a factor of 2 and 3 so n is odd)
Difference between n-1 and n+1 is 2, so either of the one will be divisible by 2 and other will be divisible by 4,
since they are consecutive even numbers.
Also since 3 is not a factor of n, either n-1 or n+1 should be divisible by 3,
so we got 2*4*3 = 24.
so we will get reminder as 0 always.
try with number 5 which is the minimum possible number....
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Re: DS: remainder (GMATPREP 1) [#permalink]
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02 Aug 2008, 08:39
selvae wrote:
n-1 and n+1 are consecutive even numbers (since n is not a factor of 2 and 3 so n is odd)
Difference between n-1 and n+1 is 2, so either of the one will be divisible by 2 and other will be divisible by 4,
since they are consecutive even numbers.
Also since 3 is not a factor of n, either n-1 or n+1 should be divisible by 3,
so we got 2*4*3 = 24.
so we will get reminder as 0 always.
try with number 5 which is the minimum possible number....
Wow, nice one. Thanks!!
+1 for you
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Re: DS: remainder (GMATPREP 1) [#permalink]
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02 Aug 2008, 12:12
judokan wrote:
if n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24. what is the value of r?
(1) n is not divisible by 2
(2) n is not divisible by 3
consider (1) => say n=3 => n^2-1=8 => remainder 8
also say n=5 => remainder =0 => insufficient
consider (2) => say n=2 => remainder =3
say n=4 => remainder =15 => insufficient
consider (1) and (2) => n is not div by 2 and 3 ,also not div by 6,4 etc
hence say n is div 5 => n=5 => remainder=0
n=10 => remainder =6 => insufficient
IMO E
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Re: DS: remainder (GMATPREP 1) [#permalink]
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02 Aug 2008, 18:55
spriya wrote:
judokan wrote:
selvae wrote:
n-1 and n+1 are consecutive even numbers (since n is not a factor of 2 and 3 so n is odd)
Difference between n-1 and n+1 is 2, so either of the one will be divisible by 2 and other will be divisible by 4,
since they are consecutive even numbers.
Also since 3 is not a factor of n, either n-1 or n+1 should be divisible by 3,
so we got 2*4*3 = 24.
so we will get reminder as 0 always.
try with number 5 which is the minimum possible number....
OOPs good explanation again i messed up taking n=10 as example
selvae,where do u practice Quant from
Wow, nice one. Thanks!!
+1 for you
I think the key of this one is first to list out the numbers for n,
1,2,3,4,5,6,7,8,9,10 ...
and then eliminate, as n is not a multiple of 2,
1,3,5,7,9,
then you will see the remainder changing among different value of n.
so not siufficient
if we add (2)
then the value of n could be
1,5,7,11,13,17
you will see all these value n will give (n+1)(n-1) divisbile by 24
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Re: DS: remainder (GMATPREP 1) [#permalink] 02 Aug 2008, 18:55
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http://forums.wolfram.com/mathgroup/archive/2007/May/msg01580.html | 1,596,606,527,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735909.19/warc/CC-MAIN-20200805035535-20200805065535-00258.warc.gz | 40,299,807 | 11,223 | Re: Re: Ellipse equation simplification on Mathematica:
```On 30 May 2007, at 18:29, Narasimham wrote:
> On May 29, 2:19 pm, Andrzej Kozlowski <a... at mimuw.edu.pl> wrote:
>> On 29 May 2007, at 12:15, Andrzej Kozlowski wrote:
>>
>>
>>
>>
>>
>>> On 29 May 2007, at 10:53, Andrzej Kozlowski wrote:
>>
>>
>>>> On 28 May 2007, at 14:00, Narasimham wrote:
>>
>>>>> On May 19, 1:54 pm, Andrzej Kozlowski <a... at mimuw.edu.pl> wrote:
>>>>>> On 18 May 2007, at 19:06, Narasimham wrote:
>>
>>
>>>>>>> threa...
>>
>>>>>>> Constant[c,d th,ph] ;
>>
>>>>>>> (* th, ph are spherical cords of tip of tube *) ;
>>
>>>>>>> cp = Cos[ph] ; sp = Sin[ph] ; cth = Cos[th] ; sth = Sin[th] ;
>>
>>>>>>> (* earlier typo corrected *)
>>
>>>>>>> d1 = Sqrt[(x + d cp cth + c )^2 + ( y + d cp sth )^2 + (d sp)
>>>>>>> ^2 ]
>>
>>>>>>> d2 =Sqrt[(x - d cp cth - c )^2 + ( y - d cp sth )^2 + (d sp)
>>>>>>> ^2 ]
>>
>>>>>>> FullSimplify[ d1 + d2 + 2 d - 2 a == 0] ;
>>
>>>>>>> When d = 0, algebraic/trigonometric simplification brings about
>>>>>>> common ellipse form:
>>
>>>>>>> (x/a)^2 + y^2/(a^2-c^2) = 1
>>
>>>>>>> Request help for bringing to standard form involving constants
>>>>>>> a,c
>>>>>>> and the new tube length constant d.
>>
>>>>>>> Regards,
>>>>>>> Narasimham
>>
>>>>>> I don't think such a form exists. Consider the following.
>>
>>>>>> id1 = {d1^2 - ((x + d*cp*cth + c)^2 + (y + d*cp*sth)^2 + (d*sp)
>>>>>> ^2),
>>>>>> d2^2 - ((x - d*cp*cth - c)^2 + (y - d*cp*sth)^2 + (d*sp)^2),
>>>>>> sp^2 + cp^2 - 1, sth^2 + cth^2 - 1};
>>
>>>>>> id = Prepend[id1, d1 + d2 + 2 d - 2 a];
>>
>>>>>> Now consdier first the case of the ellipse:
>>
>>>>>> d = 0;
>>
>>>>>> gr = GroebnerBasis[id, {x, y, a, c}, {cp, sp, cth, sth, d1,
>>>>>> d2},
>>>>>> MonomialOrder -> EliminationOrder]
>>>>>> {-a^4 + c^2 a^2 + x^2 a^2 + y^2 a^2 - c^2 x^2}
>>
>>>>>> This tells us that
>>
>>>>>> First[%] == 0
>>>>>> -a^4 + c^2*a^2 + x^2*a^2 + y^2*a^2 - c^2*x^2 == 0
>>
>>>>>> is the equation of the ellipse, and this can be easily brought to
>>>>>> standard form by hand. But now consider your "general" case:
>>
>>>>>> Clear[d]
>>>>>> gr = GroebnerBasis[id, {x, y, a, c, d}, {cp, sp, cth, sth, d1,
>>>>>> d2},
>>>>>> MonomialOrder -> EliminationOrder]
>>>>>> {}
>>
>>>>>> This means that elimination cannot be performed and no "standard
>>>>>> form"
>>>>>> of the kind you had in mind exists. Unless of course there is
>>>>>> a bug
>>>>>> in GroebnerBasis (v. unlikely) or I have misunderstood what
>>>>>> in mind.
>>
>>>>>> Andrzej Kozlowski
>>
>>>>> I checked for case of tube parallel to x- or y-axis produces
>>>>> ellipses
>>>>> and suspected validity even in 3-D general case.
>>
>>>>> Narasimham
>>
>>>> OK., now I see that I misundertood you and you wrote that cd,th,
>>>> ph (and presumably a) are supposed to be constants, so you do not
>>>> wish to eliminate them. But now one can easily prove that what you
>>>> get is not, in general, an ellipse. In this situation Groebner
>>>> basis works and you can obtain a rather horrible quartic equation
>>
>>>> id1 = {d1^2 - ((x + d*cp*cth + c)^2 + (y + d*cp*sth)^2 + (d*sp)^2),
>>>> d2^2 - ((x - d*cp*cth - c)^2 + (y - d*cp*sth)^2 + (d*sp)^2),
>>>> sp^2 + cp^2 - 1, sth^2 + cth^2 - 1};
>>
>>>> id = Prepend[id1, d1 + d2 + 2*d - 2*a];
>>
>>>> v = GroebnerBasis[id, {x, y, a, c, cp, cth}, {sth, sp, d1, d2},
>>>> MonomialOrder -> EliminationOrder][[1]];
>>
>>>> First[v] == 0
>>
>>>> is the equation (I prefer not to include the output here).
>>
>>>> Looking at v see that the non zero coefficients are only the free
>>>> coefficient, the coefficients of x^2, y^2, x^2 y^2, x^4 and y^4.
>>>> So only in some cases you will get a quadratic (for example when
>>>> the quartic happens to be a perfect square as in the case d=0, or
>>>> when the free coefficient vanishes, as in the trivial case a=d,
>>>> or when the coefficients of 4-degree terms vanish). One can work
>>>> out all the cases when gets a quadratic but it is also easy to
>>>> find those when one does not. For example, taking both th and ph
>>>> to be 60 degrees (so that cth and cph are both 1/2) we get:
>>
>>>> v /. {cp -> 1/2, cth -> 1/2, d -> 4, a -> 2, c -> 1}
>>
>>>> y^4 - 48 x^2 y^2 + 120 y^2 + 3600
>>
>>>> This is certianly is not the equation of an ellipse.
>>
>>>> Andrzej Kozlowski
>>
>>> Sorry; that last example was a bad one, because with these
>>> parameters (d>a) the original equations do not have solutions. The
>>> point is that the equations that we get after elimination will have
>>> more solutions than the ones we start with, but all the solutions
>>> of the original ones will satisfy the new ones. So to see that we
>>> do not normally get an ellipse we need a different choice of
>>> parameters, with a>d. So take
>>
>>> w = First[v ]/. {cp -> 1/2, cth -> 1/2, sp -> Sqrt[3]/2, sth -> Sqrt
>>> [3]/2, d -> 1,
>>> a -> 3, c -> 1}
>>
>>> (1521*x^4)/256 + (2229*y^2*x^2)/128 - (117*x^2)/4 + (3721*y^4)/256
>>> - (183*y^2)/4 + 36
>>
>>> Now, this is clearly not the equation of an ellipse. We can see now
>>> the relationship between this and your original equation. With the
>>> above values of the parameters your equation takes the form:
>>
>>> eq = d1 + d2 + 2*d - 2*a == 0 /. {cp -> 1/2, cth -> 1/2, sp -> Sqrt
>>> [3]/2,
>>> sth -> Sqrt[3]/2, d -> 1, a -> 3,
>>> c -> 1}
>>> Sqrt[(x - 5/4)^2 + (y - Sqrt[3]/4)^2 + 3/4] +
>>> Sqrt[(x + 5/4)^2 + (y + Sqrt[3]/4)^2 + 3/4] - 4 == 0
>>
>>> So now look at the graph:
>>
>>> gr1=ContourPlot[
>>> Sqrt[(x - 5/4)^2 + (y - Sqrt[3]/4)^2 + 3/4] +
>>> Sqrt[(x + 5/4)^2 + (y + Sqrt[3]/4)^2 + 3/4] - 4 == 0, {x, -5,
>>> 5}, {y, -5,
>>> 5}]
>>
>>> Looks like a nice ellipse, right? Unfortunately it is only a part
>>> of the graph of
>>
>>> w =First[v]
>>
>>> gr2 = ContourPlot[Evaluate[w == 0], {x, -5, 5}, {y, -5, 5}]
>>
>>> The picture is not quite convincing, because the latter contour
>>> plot is not very accurate but when you see them together:
>>
>>> Show[gr1, gr2]
>>
>>> the relationship becomes obvious. In any case, what you get a
>>> quartic curve that looks of course like an ellipse but isn't.
>>
>>> Andrzej Kozlowski
>>
>> Sorry, a small correction is needed. I wrote:
>>
>>> w=First[v]
>>
>>> gr2 = ContourPlot[Evaluate[w == 0], {x, -5, 5}, {y, -5, 5}]
>>
>> Ignore the first line (w=First[v]) in the above since it will prevent
>> the plot from working. Just evaluate the second line with the value
>> of w defined earlier.
>>
>> Andrzej Kozlowski
>
> Thanks Andrzej so much for all the cumbersome & involved
> verifications.
>
> Ok,are we now in a situation to find out what exhaustively all
> situations/conditions give ellipses?
>
> Should coefficients of third and fourth order term vanish,or should it
> have repeated roots etc.?
>
> At least don't we get ellipses for tube center when th = 0,?
> (inclusive when the tube is parallel or perpendicular to x- axis, th =
> 0,ph = 0,Pi/2? For this case new a,b need to be defined in terms of
> c,a and d).
>
> Regards,
> Narasimham
>
>
Yes, indeed you do get ellipses in various special cases. For
example, as you say, when th=0 you can easily find ellipses that are
solutions. In fact it could be that all solutions in this case are
ellipses. (see below).
Here is how one can investigate this.
cth = 1; sth = 0;
id1 = {d1^2 - ((x + d*cp*cth + c)^2 + (y + d*cp*sth)^2 + (d*sp)^2),
d2^2 - ((x - d*cp*cth - c)^2 + (y - d*cp*sth)^2 + (d*sp)^2),
sp^2 + cp^2 - 1};
id = Prepend[id1, d1 + d2 + 2*d - 2*a];
v = GroebnerBasis[id, {x, y, a, c, cp}, { sp, d1, d2},
MonomialOrder -> EliminationOrder][[1]]
w = Collect[v, {x, y}, Simplify]
(-y^2)*(a - d)^2 + (a^2 - 2*d*a - c*(c + 2*cp*d))*(a - d)^2 +
(-a^2 + 2*d*a + c^2 + (cp^2 - 1)*d^2 + 2*c*cp*d)*x^2
We got a quadratic which we hope is the equation of an ellipse. For
this to be an ellipse one of two things has to happen. Either the
coefficients of x^2 and y^2 are both positive and the free
coefficient non-positive or the coefficients of x^2 and y^2 are both
negative and the free coefficient non-negative. Lets write down two
conditions:
cond1 = Reduce[
0 <= cp <= 1 &&
Coefficient[w, x^2] <
0 && Coefficient[w, y^2] < 0 && (w /. {x -> 0, y -> 0})
>= 0 &&
a >= d];
cond2 = Reduce[
0 <= cp <= 1 &&
Coefficient[w, x^2] >
0 && Coefficient[w, y^2] > 0 && (w /. {x -> 0, y -> 0})
<= 0 &&
a >= d];
cond = Reduce[cond1 || cond2];
Now, let's ask Mathematica to find an example satisfying cond:
ex1 = Flatten[FindInstance[cond, {d, a, cp, c}]]
{d -> 9800/149, a -> 149, cp -> 0, c -> 51}
Let's see the equation:
Simplify[(v /. ex1) == 0]
96040000*x^2 + 153784801*y^2 == 0
well, yes, this is an ellipse but a rather trivial one. Let's try to
make it find something with a non zero value of cp
ex2 = Flatten[FindInstance[cond && a != 0 && 0 < cp < 1, {d, a, cp, c}]]
{d -> -1, a -> 1, cp -> 1/2, c -> -1}
Simplify[(v /. ex2) == 0]
(7*x^2)/4 + 4*y^2 == 4
This time we got a nice ellipse. So certainly, in the case th=0 you
get lots of elliptical solutions.
The question is: are all possible solutions in this case elliptical?
I tried telling Mathematica to find a solution for which the
following holds:
ex3 = Flatten[FindInstance[Not[cond] && a!=0 && 0 <= cp <=1, {d, a, c}]]
but since after quite a long time Mathematica was still unable to
find an example and as I need to use it for another purpose I had to
abort. So I suspect the case th==0 all solutions are elliptical (and
the same for ph==0) but Reduce will probably not be able to prove it
(or FindInstance find a counterexample). I can't spend any more time
on this but I think you can now see what can be (and what probably
can't) be done with Mathematica in connection with this problem.
```
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In this HackerRank Maximizing Mission Points problem solution we have given d_lat, d_long, and the definitions for n cities. we need to find and print the maximum possible total points that Xander can earn on a mission.
## Problem solution in Python.
```from collections import namedtuple
from bisect import bisect_left
import sys
Place = namedtuple('Place', 'lat, long, height, points')
chunkplaces={} # places get inserted into lists contained here, grouped by keys of their locations
chunkvals={} # holds values
giant = False
def getkey(place, off_lat=0, off_long=0):
return ((place.lat // d_lat + off_lat) * 200011) + place.long // d_long + off_long # unique for n<=200000
def recordvalue(place, val):
if val < 0:
return # not worth going here; no need to track
key = getkey(place)
if key not in chunkplaces:
chunkplaces[key] = []
chunkvals[key] = []
if giant:
if len(chunkvals[key]) == 0:
chunkvals[key].append(-val)
chunkplaces[key].append(place)
else:
if val < -chunkvals[key][0]:
return
else:
chunkvals[key][0] = -val
chunkplaces[key][0] = place
else:
i = bisect_left(chunkvals[key], -val)
chunkplaces[key].insert(i, place)
chunkvals[key].insert(i, -val)
# print(i, val, [val for val in chunkvals[key]])
def getbestinchunk(place, key, best):
# find best suitable match in chunk
if key not in chunkvals:
return 0
for i, (cand, val) in enumerate(zip(chunkplaces[key], chunkvals[key])):
# print("evaluating %s"%cand)
if -val < best:
# this is the best we have, but it's not as good as we've seen other places; abort
return 0
if abs(place.lat - cand.lat) <= d_lat \
and abs(place.long - cand.long) <= d_long :
# and cand.height > place.height: # height is given, assuming they're unique
# suitable, return it
return -val
# no suitable match
return 0
def getbest(place):
# find best match in this and neighboring chunks, then pick the best
best = 0 # always have the option to stop here
for i in [0,1,-1]:
for j in [0,1,-1]:
key = getkey(place, i, j)
ret = getbestinchunk(place, key, best)
if ret > best:
best = ret
return best
def calculatevalue(place):
val = place.points + getbest(place)
recordvalue(place, val)
return val
if __name__ == "__main__":
n, d_lat, d_long = input().strip().split(' ')
n, d_lat, d_long = [int(n), int(d_lat), int(d_long)]
places = []
if d_lat == 200000:
giant = True
for a0 in range(n):
latitude, longitude, height, points = input().strip().split(' ')
latitude, longitude, height, points = [int(latitude), int(longitude), int(height), int(points)]
places.append(Place(latitude, longitude, height, points))
places.sort(key=lambda p: -p.height) # compute highest first
best = 0
for p in places:
ret = calculatevalue(p)
if ret > best:
best = ret
print(best)
```
{"mode":"full","isActive":false}
## Problem solution in Java.
```import java.util.*;
public class Solution1 {
public static class Point {
public int l;
public int r;
public int xt;
public int yt;
public long tot;
public Point(int l, int r, int xt, int yt) {
this.l = l;
this.r = r;
this.xt = xt;
this.yt = yt;
this.tot = 0;
}
}
static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
int n = scanner.nextInt(), x = scanner.nextInt(), y = scanner.nextInt();
HashMap<Integer, ArrayList<Point>> blocks = new HashMap<>();
long maxPoints = Long.MIN_VALUE;
int MAXIMUM = 200000;
Point[] points = new Point[MAXIMUM];
Arrays.fill(points, null);
for (int i = 0; i < n; i++) {
int l = scanner.nextInt(), r = scanner.nextInt(),xt = scanner.nextInt(), yt = scanner.nextInt();
Point point = new Point(l, r, xt, yt);
points[xt - 1] = point;
}
for (int i = 0; i < MAXIMUM; i++) {
Point curPoint = points[i];
if (points[i] != null) {
int blockNumber = points[i].l / x;
Point curMax = null;
Point max = null;
ArrayList<Point> prevBlock = getBlock(blockNumber - 1, blocks);
ArrayList<Point> curBlock = getBlock(blockNumber, blocks);
ArrayList<Point> nextBlock = getBlock(blockNumber + 1, blocks);
if (prevBlock != null) {
curMax = findMax(prevBlock, curPoint, x, y);
max = curMax;
}
curMax = findMax(curBlock, curPoint, x, y);
if (max == null) {
max = curMax;
} else {
if (curMax != null && curMax.tot >= max.tot) {
max = curMax;
}
}
curMax = findMax(nextBlock, curPoint, x, y);
if (max == null) {
max = curMax;
} else {
if (curMax != null && curMax.tot >= max.tot) {
max = curMax;
}
}
curPoint.tot = (max != null ? max.tot + curPoint.yt : curPoint.yt);
addPoint(curBlock, curPoint, 0, curBlock.size() - 1);
if (maxPoints < curPoint.tot) {
maxPoints = curPoint.tot;
}
}
}
if (maxPoints == Long.MIN_VALUE) {
System.out.println(0);
} else {
System.out.println(maxPoints);
}
}
private static ArrayList<Point> getBlock(int blockNumber, HashMap<Integer, ArrayList<Point>> blocks) {
if (blockNumber < 0) {
return null;
}
ArrayList<Point> block = blocks.get(blockNumber);
if (block == null) {
block = new ArrayList<>();
blocks.put(blockNumber, block);
}
return block;
}
private static Point findMax(ArrayList<Point> block, Point point, int ld, int rd) {
for (int i = block.size(); i > 0; i--) {
Point prevPoint = block.get(i - 1);
if (Math.abs(prevPoint.r - point.r) <= rd
&& Math.abs(prevPoint.l - point.l) <= ld) {
return prevPoint;
}
}
return null;
}
private static void addPoint(ArrayList<Point> block, Point point, int left, int right) {
final long value = point.tot;
if (block.isEmpty()) {
} else if (right - left <= 1) {
long leftValue = block.get(left).tot;
long rightValue = block.get(right).tot;
if (value < leftValue) {
} else if (value >= leftValue && value <= rightValue) {
} else {
if (block.size() - 1 == right) {
} else {
int index = right + 1;
}
}
} else {
int middle = (right + left) / 2;
long middleValue = block.get(middle).tot;
if (middleValue <= value) {
} else {
}
}
}
}
```
{"mode":"full","isActive":false}
## Problem solution in C++.
```#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200000;
const long long INF = 1e15;
long long tree[MAXN * 4];
void makeTree() {
for (int i = 1; i < MAXN * 4; ++i) {
tree[i] = -INF;
}
}
void update(int x, long long val, int u = 1, int l = 1, int r = MAXN) {
if (l == r) {
tree[u] = val;
return;
}
int m = (l + r) / 2;
if (x <= m) {
update(x, val, 2 * u, l, m);
}
else {
update(x, val, 2 * u + 1, m + 1, r);
}
tree[u] = max(tree[2 * u], tree[2 * u + 1]);
}
long long query(int x, int y, int u = 1, int l = 1, int r = MAXN) {
if (x > r or y < l) {
return -INF;
}
if (x <= l and r <= y) {
return tree[u];
}
int m = (l + r) / 2;
return max(query(x, y, 2 * u, l, m), query(x, y, 2 * u + 1, m + 1, r));
}
struct Point {
int x, y, z, w;
long long dp;
bool operator < (const Point &o) const {
return z < o.z;
}
};
Point p[MAXN + 1];
long long DP[MAXN + 1];
int X_LIM, Y_LIM;
void merge(int l, int r) {
int m = (l + r) / 2;
vector<pair<int, int> > left;
vector<pair<int, int> > right;
for (int i = l; i <= m; ++i) {
left.push_back({p[i].x, p[i].z});
}
for (int i = m + 1; i <= r; ++i) {
right.push_back({p[i].x, p[i].z});
}
sort(left.begin(), left.end());
sort(right.begin(), right.end());
int left_l = 0;
int left_r = -1;
for (auto u : right) {
int i = u.second;
int x = u.first;
while (left_r + 1 < left.size() and left[left_r + 1].first - x <= X_LIM) {
++left_r;
int who = left[left_r].second;
update(p[who].y, p[who].dp);
}
while (left_l < left.size() and x - left[left_l].first > X_LIM) {
int who = left[left_l].second;
update(p[who].y, -INF);
++left_l;
}
int yLow = max(1, p[i].y - Y_LIM);
int yHigh = min(MAXN, p[i].y + Y_LIM);
p[i].dp = max(p[i].dp, p[i].w + query(yLow, yHigh));
}
while (left_l <= left_r) {
int who = left[left_l].second;
update(p[who].y, -INF);
++left_l;
}
}
void solve(int l, int r) {
if (l == r) {
p[l].dp = max(p[l].dp, (long long)p[l].w);
return;
}
int m = (l + r) / 2;
solve(l, m);
merge(l, r);
solve(m + 1, r);
}
int main() {
makeTree();
int n;
scanf("%d %d %d", &n, &X_LIM, &Y_LIM);
for (int i = 0; i < n; ++i) {
scanf("%d %d %d %d", &p[i].x, &p[i].y, &p[i].z, &p[i].w);
p[i].dp = -INF;
}
sort(p, p + n);
for (int i = 0; i < n; ++i) {
p[i].z = i;
}
solve(0, n - 1);
long long ans = 0;
for (int i = 0; i < n; ++i) {
ans = max(ans, p[i].dp);
}
printf("%lld\n", ans);
}
```
{"mode":"full","isActive":false}
## Problem solution in C.
```#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct _tree
{
int *y;
long long *a;
int size;
int N;
int *left_idx;
int *right_idx;
} tree;
int diff(int x,int y);
long long max(long long x,long long y);
void init(int n,int *N);
long long range_sum( int i, int j);
void updatea(int i);
void build(int v);
void merge(tree *t,tree *x,tree *y);
int get_i(int*a,int num,int size);
int med(int*a,int size);
long long query(int v);
void update(int v,int idx);
int x1,x2,y1,y2,N,tl[800000],tr[800000];
long long val,*tt;
int lat[200000]={0},lon[200000]={0},poi[200000],tla[200000]={0};
tree t[800000];
int main()
{
int n,x,y,c,i,j,max_idx=-1,a,b,C,d;
long long max=0,dp;
scanf("%d%d%d",&n,&x,&y);
for(i=c=0;i<n;i++)
{
scanf("%d%d%d%d",&a,&b,&C,&d);
if(d>0)
c++;
lat[C-1]=a;
lon[C-1]=b;
poi[C-1]=d;
tla[a-1]=b;
}
if(!c)
{
printf("0");
return 0;
}
tl[1]=0;
tr[1]=199999;
build(1);
for(i=199999;i>=0;i--)
if(lat[i])
{
if(max_idx!=-1 && diff(lat[max_idx],lat[i])<=x && diff(lon[max_idx],lon[i])<=y)
{
dp=max;
}
else
{
x1=lat[i]-x-1;
x2=lat[i]+x-1;
y1=lon[i]-y;
y2=lon[i]+y;
dp=query(1);
}
if(dp>0)
dp+=poi[i];
else
dp=poi[i];
if(dp>max){
max=dp;
max_idx=i;
}
if(dp>0){
x1=lat[i]-1;
y1=lon[i];
val=dp;
update(1,-1);
}
}
printf("%lld",max);
return 0;
}
int diff(int x,int y)
{
return (x<y)?(y-x):(x-y);
}
long long max(long long x,long long y)
{
return (x>y)?x:y;
}
void init(int n,int *N)
{
(*N) = 1;
while( (*N) < n ) (*N) *= 2;
}
long long range_sum( int i, int j)
{
long long ans = 0;
for( i += N, j += N; i <= j; i = ( i + 1 ) / 2, j = ( j - 1 ) / 2 )
{
if( i % 2 == 1 ) ans = max(ans,tt[i]);
if( j % 2 == 0 ) ans = max(ans,tt[j]);
}
return ans;
}
void updatea(int i)
{
int j;
for( j = i + N; j; j /= 2 ) tt[j] = max(tt[j],val);
}
void build(int v)
{
if(tl[v]==tr[v])
{
t[v].size=1;
t[v].y=(int*)malloc(t[v].size*sizeof(int));
t[v].a=(long long*)malloc(4*t[v].size*sizeof(long long));
memset(t[v].a,0,4*t[v].size*sizeof(long long));
t[v].y[0]=tla[tl[v]];
init(t[v].size,&t[v].N);
}
else
{
int tm=(tl[v]+tr[v])/2;
tl[2*v]=tl[v];
tl[2*v+1]=tm+1;
tr[2*v]=tm;
tr[2*v+1]=tr[v];
build(v*2);
build(v*2+1);
merge(&t[v],&t[v*2],&t[v*2+1]);
}
return;
}
void merge(tree *t,tree *x,tree *y)
{
int i=0,j=0;
t->size=x->size+y->size;
t->y=(int*)malloc(t->size*sizeof(int));
t->left_idx=(int*)malloc(t->size*sizeof(int));
t->right_idx=(int*)malloc(t->size*sizeof(int));
t->a=(long long*)malloc(t->size*sizeof(long long)*4);
memset(t->a,0,t->size*sizeof(long long)*4);
init(t->size,&t->N);
while(i<x->size || j<y->size)
{
if(i==x->size){
t->y[i+j]=y->y[j];
t->left_idx[i+j]=i-1;
t->right_idx[i+j]=j;
j++;
}
else if(j==y->size)
{
t->y[i+j]=x->y[i];
t->left_idx[i+j]=i;
t->right_idx[i+j]=j-1;
i++;
}
else if(x->y[i]<=y->y[j])
{
t->y[i+j]=x->y[i];
t->left_idx[i+j]=i;
t->right_idx[i+j]=j-1;
i++;
}
else{
t->y[i+j]=y->y[j];
t->left_idx[i+j]=i-1;
t->right_idx[i+j]=j;
j++;
}
}
return;
}
int get_i(int*a,int num,int size)
{
if(size==0)
{
return 0;
}
if(num>med(a,size))
{
return get_i(&a[(size+1)>>1],num,size>>1)+((size+1)>>1);
}
else
{
return get_i(a,num,(size-1)>>1);
}
}
int med(int*a,int size)
{
return a[(size-1)>>1];
}
long long query(int v)
{
int i,j;
if(tl[v]>x2 || tr[v]<x1 || !t[v].a[1])
return 0;
if(tl[v]>=x1 && tr[v]<=x2){
i=get_i(t[v].y,y1,t[v].size);
j=get_i(t[v].y,y2+1,t[v].size)-1;
if(j<i)
return 0;
N=t[v].N;
tt=t[v].a;
return range_sum(i,j);
}
else if(tl[v]!=tr[v])
return max(query(2*v),query(2*v+1));
return 0;
}
void update(int v,int idx)
{
if(tl[v]<=x1 && tr[v]>=x1)
{
if(idx==-1)
idx=get_i(t[v].y,y1,t[v].size);
N=t[v].N;
tt=t[v].a;
updatea(idx);
if(tl[v]!=tr[v])
{
update(v*2,t[v].left_idx[idx]);
update(v*2+1,t[v].right_idx[idx]);
}
}
return;
}```
{"mode":"full","isActive":false} | 4,138 | 11,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2023-06 | latest | en | 0.660228 |
https://discourse.julialang.org/t/using-split-ode-solvers-within-method-of-steps/94569 | 1,686,265,156,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00401.warc.gz | 248,000,341 | 6,199 | # Using Split ODE Solvers within Method of Steps
I am interesting in simulating delayed reaction-diffusion equations of the form
u_t = \Delta u + f(u,u(t-\tau))
in a 2-D spatial domain of modest size (10,000 equations) and a non-stiff nonlinearity (for now). My current progress involves optimizing and benchmarking the ODE function and algorithm/settings choices for the undelayed problem, which splitting methods seem well-suited for at some tolerances and integration lengths. As I understand it, the extension to a DDEProblem more or less just involves rewriting your ODE function to incorporate the history function, then MethodOfSteps just wraps some adaptive time-stepping integrator and makes sure that the integration resolves the discontinuities well. Therefore, I expect at least some of the optimization for the non-delayed case to transfer to the delayed case.
Of course I can probably get away with just using a fully implicit method in MethodOfSteps, but I’d at least like to know if it’s possible to use a splitting method within MethodOfSteps. I will eventually be using a variant of this problem in some Bayesian inference, so trying eke out as much performance as possible is preferable. As I currently understand, to solve a DDE, you create a DDEProblem from a function of the state u and history function h, then the solver alg is just MethodOfSteps(ode_alg_of_choice()). For a SplitODEProblem, I have to give it two functions/a linear operator + function representing the stiff and non-stiff RHS terms then call a split algorithm in solve. It’s not clear to me from my own understanding and from browsing the many examples in the DE.jl documentation on how these two methods can interface (if at all). I’m not at the stage yet to provide a MWE of the delayed problem, but I wanted to know if there was some general framework to stitch together these two features for future use.
Understanding this potential code interface is important for future extensions of this problem such as
• Introduction of a stiff, but easily separable, component to the nonlinearity
• Galerkin projection onto a lower dimensional subspace (problem size decreases drastically but matrices/tensors become dense)
• Increasing tolerances to allow for faster short-time integration
It’s in theory possible. I’m not sure if it currently dues the wrapping correctly to keep the special form, but in theory it should be possible. It probably needs an issue. This is something I personally haven’t tried myself.
Thanks for the quick response. I haven’t gotten to the stage of actually implementing it yet either, so within the next couple of days I will test some naive methods of wrapping the split solver and post here with the results. For the full-order system, this actually isn’t too big of a deal since the Jacobian of the complete system is so sparse that fully implicit methods still perform alright. The actual PDE I am solving contains multiple fields interacting so the bandwidth is too large for sparse direct solvers, but GMRES handles it pretty well.
For the eventual projected system though, I envision splitting algorithms will provide a marked increase over the fully-implict methods since the Jacobian will be dense and stiff, but one could prefactor the LU decomposition of the stiff part for use in a splitting method. I could still use this LU as a preconditioner to the Jacobian solve in a fully implicit method, but splitting would be ideal I think. | 714 | 3,473 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-23 | latest | en | 0.919243 |
https://www.gradesaver.com/textbooks/science/physics/physics-10th-edition/chapter-6-work-and-energy-focus-on-concepts-page-165/22 | 1,685,986,387,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652149.61/warc/CC-MAIN-20230605153700-20230605183700-00001.warc.gz | 856,993,755 | 11,787 | ## Physics (10th Edition)
According to the work-energy theorem, $$W_{nc}=E_f-E_0$$ $$W_{air}=\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)$$ The skydiver's velocity remains constant, so $\frac{1}{2}m(v_f^2-v_0^2)=0$ We have $h_f-h_0=-325m$ (Since the skydiver falls downward, her final height is less than her initial height): $$W_{air}=-mg\Delta h \\ W_{air}=-(92kg)(9.81 m/s^2)(325m)\\ W_{air}=-2.93\times10^5J$$ | 164 | 407 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-23 | latest | en | 0.690217 |
http://slideplayer.com/slide/3810771/ | 1,506,402,008,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818694719.89/warc/CC-MAIN-20170926032806-20170926052806-00699.warc.gz | 305,537,935 | 21,350 | # CPS 290.4/590.4 Crowdsourcing Societal Tradeoffs Vincent Conitzer Markus Brill Rupert Freeman
## Presentation on theme: "CPS 290.4/590.4 Crowdsourcing Societal Tradeoffs Vincent Conitzer Markus Brill Rupert Freeman"— Presentation transcript:
CPS 290.4/590.4 Crowdsourcing Societal Tradeoffs Vincent Conitzer conitzer@cs.duke.edu Markus Brill brill@cs.duke.edu Rupert Freeman rupert@cs.duke.edu
Motivating question 1 bag of landfill trash How to determine x? is as bad as using x gallons of gasoline Other examples: clearing an acre of forest, fishing a ton of bluefin tuna, causing the average person to sit in front of a screen for another 5 minutes a day, …
Example decision scenario Benevolent government would like to get old inefficient cars off the road But disposing of a car and building a new car has its own energy and other costs Perhaps also benefits to the economy Which cars should the government aim to get off the road? Even energy costs are not directly comparable, e.g., perhaps gasoline contributes to energy dependence, coal does not
Inconsistent tradeoffs can result in inefficiency Agent 1: 1 gallon = 3 bags = -1 util I.e., agent 1 feels she should be willing to sacrifice up to1 util to reduce trash by 3, but no more Agent 2: 1.5 gallons = 1.5 bags = -1 util Agent 3: 3 gallons = 1 bag = -1 util Cost of reducing gasoline by x is x 2 utils for each Cost of reducing trash by y is y 2 for each What will they do individually? (A little calculus…) What would be a better solution for all involved?
Pigovian taxes Taxes intended to discourage undesirable behavior More precisely: pay the cost your activity imposes on society (the externality of your activity) If we decided using 1 gallon of gasoline came at a cost of \$x to society, so we charged a tax of \$x on each gallon… … then people would only use a gallon if they gained at least \$x from doing so (net of purchase cost), which is societally ideal Isn’t this better than (say) sales tax? But where would we get x? Arthur Cecil Pigou
One approach: let’s vote! So what should the outcome be…? Average? Median? Note average(2,4,10)=5.33 but average(1/2,1/4,1/10)=.2833 (not 1/5.33) Is this the right way to vote? Who even gets to vote? Where/how? x should be 10x should be 4x should be 2
Perhaps more reasonable… E.g., due to missing information (what kind of trash?) or plain uncertainty How should we aggregate these? x should be between 9 and 11 x should be between 2 and 6 x should be between 0 and 4
Voting more than once… How to prevent this manipulation? x should be 10x should be 4x should be 2 median x should be 10 median
Consistency (for an individual)
A paradox: individual consistency may lead to societal inconsistency But we want our societal tradeoffs to be consistent… Otherwise we’ll get inefficiency or incomprehensibility
Decomposition Maybe we should reason about these components directly… How? How do we determine the components? gasoline use energy dependence carbon dioxide in atmosphere objectivesubjective
Another paradox gasoline use energy dependence global warming landfill trash Everyone agrees GU contributes 1 to GW, 1 to ED Voter 1: 1 GW = 2 LT, 1 ED = 1 LT (so 1 GU = 3 LT) Voter 2: 1 GW = 1 LT, 1 ED = 2 LT (so 1 GU = 3 LT) Voter 3: 1 GW = 1 LT, 1 ED = 1 L (so 1 GU = 2 LT) Median on attributes would conclude 1 GW = 1 ED = 1 LT (so 1 GU = 2 LT), but median directly gives 1 GU = 3 LT
Objective vs. subjective tradeoffs Some tradeoffs seem objective… Measuring carbon dioxide emissions … others subjective… Global warming vs. energy dependence Separate process? Who determines which is which? How to bring expert knowledge to bear?
Prediction markets
Locality Should we always obtain a single global tradeoff? Right tradeoff may depend on where we are … though then arguably they’re not the same things being traded off Different entities (say, countries) may wish to reach their tradeoffs independently I may care only about the opinions of my neighbors in my social network
Some topics to cover Social choice theory Voting Judgment aggregation Voting in pursuit of the “truth” Strategic agents Game theory Mechanism design Prediction markets Preference elicitation | 1,025 | 4,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-39 | longest | en | 0.887774 |
https://www.futurestarr.com/blog/other/160-lbs-to-kg | 1,660,010,892,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00566.warc.gz | 702,645,952 | 23,437 | FutureStarr
160 Lbs to Kg
# 160 Lbs to Kg
via GIPHY
### Lbs to Kg
via GIPHY
160 lb to kg, 160 lb in kg, 160 lb to Kilogram, 160 lb in Kilogram, 160 lbs to kg, 160 lbs in kg, 160 Pound to kg, 160 Pound in kg, 160 Pound to Kilogram, 160 Pound in Kilogram, 160 Pounds to Kilogram, 160 Pounds in Kilogram, 160 lbs to Kilogram, 160 lbs in Kilogram, 160 lbs to Kilograms, 160 lbs in Kilograms, 160 lb to Kilograms, 160 lb in Kilograms. Lbs to Kg ConversionsWelcome to lbs to kg, our category containing the posts about a specific mass conversion of pounds to kilograms. For each conversion of the unit international avoirdupois pound, an imperial and customary United States unit of measurement, you can find the x lbs to kg formula and the value in kg, along with instructions on how to conduct the math. Because our results are rounded to 3 decimal digits, each article also contains a pound to kg converter with higher precision. We then provide you with the equivalent in the metric unit kilogram for Troy, London, metric, Merchant’s as well as Tower pounds. In the context of changing x pounds to kilograms, we also discuss the frequently asked question, and tell you where to find the inverse calculation and further information on the units mentioned in our articles. Finally, every posts also comes with a feedback form to ask a question. Test it, for example, by entering 160 pounds into kilos, from 160 lbs to kg, or 160 lbs convert to kg, among many others terms which you can look up using the custom search engine in combination with our conversions. Here we show you the weight conversion 160 lbs to kg for some pound measurements (units of mass) which are no longer in official use, except for precious metals like silver and gold which are measured in Troy ounces. (Source:What is 160 lbs converted to kg? Learn the answer to this conversion or try our easy to use calculator below to convert any value of lbs to kg with the most accurate results.
A POWERFUL TOOL for converting pounds to kilograms (pounds to kg). How to convert 160 Pounds (lbs) to kilograms (kg). How Heavy Is 160 Pounds in Kilograms? How do I convert lbs to kg? How Many Pounds in a Kilogram? If you looked for 160 lbs to kg online and found this page, you’re in luck as we are here to explain how the conversion works. To answer the question, 160 lbs. is equal to 72.57 kilograms (kg). But if you want to know more about facts about pounds and kilograms, then we suggest. Convert 160lbs to kg, 160 lbs to kg conversion, 160 pounds to kilos calculator, 160 lbs to kg converter, easy way to convert 160 pounds to kilograms, 160 pounds equals how many kg, you can use our calculator to convert 160lbs to kg fast (Source: 7kgtolbs.com)
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August 08, 2022 | Sami Ullah | 1,049 | 3,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-33 | latest | en | 0.889215 |
https://astarmathsandphysics.com/a-level-maths-notes/s2/3770-the-continuity-correction.html | 1,534,924,388,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219692.98/warc/CC-MAIN-20180822065454-20180822085454-00579.warc.gz | 614,083,111 | 12,343 | ## The Continuity Correction
The normal distribution is a continuous distribution,but it can in fact also be used to model the distribution of adiscrete random variable that fits the normal distribution. We dothis by introducing the continuity correction.
If we want to find the probability that a randomvariable takes the value k then we find
If we want to find the probability that a randomvariable takes any value less than or equal to k then we find
If we want to find the probability that a randomvariable takes any value less than k then we find
Notice the difference in the above two expressions. Because X maybe equal to k in the first of these two expressions,we haveinsteadof
If we want to find the probability that a randomvariable takes any value greater than or equal to k then we find
If we want to find the probability that a random variable takesany value greater than k then we find
Notice the difference in the above two expressions. Becausemaybe equal tointhe first of these two expressions,we haveinsteadof
Example: The number of worms perofsoil is normally distributed with mean 51.4 and variance 20. Find theprobability that the number of worms inis60.
To findso
To findso | 253 | 1,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | longest | en | 0.854825 |
https://nz.education.com/worksheets/third-grade/multiplication-facts/CCSS-Math-Content-4/ | 1,611,211,006,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522242.73/warc/CC-MAIN-20210121035242-20210121065242-00116.warc.gz | 478,167,771 | 23,347 | # Search Our Content Library
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Multiplication Facts
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Division Facts: Fives
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Division Facts: Fives
Learners work on their division facts for 5 in this practice worksheet.
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Learners work on their division facts for 9 in this practice worksheet.
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Properties of Multiplication: Commutative
This worksheet introduces children to the commutative property, then uses a series of prompts to help them practice it.
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In this worksheet, students will be asked to identify multiplication and division facts.
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Multiplication Strategies: Multiplying by Multiples of 10
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Multiplication Strategies: Multiplying by Multiples of 10
Dive into multiplication with this worksheet that introduces the multiplication strategy of multiplying one-digit numbers by multiples of ten. | 234 | 1,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.822192 |
https://testbook.com/question-answer/gold-is-12-times-as-heavy-as-aluminum-and-copper-i--5fb678011b5c16e438f74ab2 | 1,713,799,765,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818312.80/warc/CC-MAIN-20240422144517-20240422174517-00201.warc.gz | 491,432,247 | 46,628 | # Gold is 12 times as heavy as aluminum and copper is 5 times as heavy as aluminum.In what ratio gold and copper should be mixed to get an alloy 8 times of aluminum.
This question was previously asked in
SSC GD Previous Paper 25 (Held On: 3 March 2019 Shift 3)_English
View all SSC GD Constable Papers >
1. 2 : 1
2. 3 : 4
3. 1 : 2
4. 4 : 3
Option 2 : 3 : 4
Free
PYST 1: SSC GD कांस्टेबल - हिंदी (16 नवंबर 2021 Shift 1 को आयोजित)
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25 Questions 50 Marks 15 Mins
## Detailed Solution
Given:
Gold is 12 times as heavy as aluminum and copper is 5 times as heavy as aluminum.
Calculation:
Let the ratio of gold and copper that taken be x : y
According to the question,
⇒ $$\frac{{12x + 5y}}{{x + y}} = \frac{8}{1}$$
⇒ 12x + 5y = 8x + 8y
⇒ 4x = 3y
⇒ x/y = 3/4
∴ Gold and copper are taken in the ratio 3 : 4. Alternate Method
Allegation method use in this question,
∴ Gold and copper are taken in the ratio 3 : 4. | 335 | 936 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 4.3125 | 4 | CC-MAIN-2024-18 | latest | en | 0.851014 |
https://kr.mathworks.com/matlabcentral/profile/authors/11748319?page=2 | 1,660,418,475,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571982.99/warc/CC-MAIN-20220813172349-20220813202349-00502.warc.gz | 345,970,481 | 3,584 | 해결됨
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Use a timetable to analyze a train timetable (Part 4)
You are analyzing a train timetable (you have some time to kill since you just missed your train!) What is the average daily tri...
3년 이상 전
해결됨
Use a timetable to analyze a train timetable (Part 3)
Oh no, you missed your train to Boston (again?)! How many times can you miss that train today? (In other words, how many Boston ...
3년 이상 전
해결됨
Use a timetable to analyze a train timetable (Part 2)
Oh no, you missed your train to Boston! Find the departure time of the next available train by analyzing the train timetable. Yo...
3년 이상 전
해결됨
Use a timetable to analyze a train timetable (Part 1)
Oh no, you missed your train to Boston! Find the departure time of the next available train by analyzing the train timetable. Th...
3년 이상 전
해결됨
Words Count: A Cell Array Approach
Given an input character vector consisting of words, punctuation marks, white spaces, and possibly newline characters (\n), arra...
3년 이상 전
해결됨
Combine the first and last names
MATLAB R2016 provides a rich set of functions to work with string arrays. In this problem, you will be given two string arrays o...
3년 이상 전
해결됨
Words Count: A String Array Approach
Given an input character vector consisting of words, punctuation marks, white spaces, and possibly newline characters (\n), arra...
3년 이상 전
해결됨
String Array Basics, Part 4: Convert String Array with Missing Values to Cell Array
<http://www.mathworks.com/help/matlab/characters-and-strings.html String array> and cell array are two types of containers for s...
3년 이상 전
해결됨
String Array Basics, Part 3: Convert Cell Array with Missing Values to String Array
<http://www.mathworks.com/help/matlab/characters-and-strings.html String array> and cell array are two types of containers for s...
3년 이상 전
해결됨
Use R2016b Text Manipulations to Fix These Addresses (Part 3)
We have a series of addresses like the following which we'd like to reformat. Each of the addresses lacks a space and a comma ...
3년 이상 전
해결됨
Use R2016b Text Manipulations to Fix These Addresses (Part 2)
We have a series of addresses like the following which we'd like to reformat. All the addresses are in the Boston area of Mass...
3년 이상 전
해결됨
Use R2016b Text Manipulations to Fix These Addresses (Part 1)
We have a series of addresses like the following which we'd like to reformat. Can you remove the latitude and longitude from the...
3년 이상 전
해결됨
Word Counting and Indexing
You are given a list of strings, each being a list of words divided by spaces. Break the strings into words, then return a maste...
3년 이상 전
해결됨
Matrix indexing with two vectors of indices
Given a matrix M and two index vectors a and b, return a row vector x where x(i) = M(a(i),b(i)).
3년 이상 전
해결됨
Return elements unique to either input
Given two numeric inputs a and b, return a row vector that contains the numbers found in only a or only b, but not both. For ex...
3년 이상 전
해결됨
Return unique values without sorting
If the input vector A is [42 1 1], the output value B must be the unique values [42 1] The *values of B are in the s...
3년 이상 전 | 1,348 | 5,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-33 | latest | en | 0.593718 |
http://fitness.stackexchange.com/questions/9728/how-do-i-measure-vo2max-at-home-without-specialized-equipment/9730 | 1,469,378,580,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824113.35/warc/CC-MAIN-20160723071024-00310-ip-10-185-27-174.ec2.internal.warc.gz | 90,161,105 | 18,771 | # How do I measure Vo2Max at home without specialized equipment?
I'm about to start the Tabata routine to increase my cardio endurance.
The best I've found is this calculator.
Are those results accurate? If not, please suggest an alternative.
-
They are accurate as far as they go, but they are rough estimates at best.
VO2 max is a measurement of how much oxygen you are actually using at maximal exertion. The only way to accurately measure this is by using specialized equipment with a mask that captures all exhalations, and then measures the amount of oxygen and carbon dioxide that is being exhaled to calculate how much oxygen you are actually using. Check with your local university/college kinesiology programs, some of them offer VO2 testing, or if you have a local triathlon/endurance coaching center, many of them offer VO2 testing for anywhere from \$50 - \$200 US dollars.
Also, VO2 max is really more an indicator of potential than it is anything else. You can raise your VO2 max by doing nothing more than losing weight, since part of the calculation is based on body weight. Say you have a 220 pound (100 kg) man who has a VO2 max of 50ml/kg/min, or a VO2 max of 5 liters (5000 ml). This man does absolutely nothing except diet to lose 44 lbs, so he is now at 80 kg. His VO2 max is now 62.5. You're not any more fit, you've just lost weight. (This is a somewhat simplistic example, there are other factors that go into this in most cases.)
-
I want to measure vo2max b/c I'm trying to improve my cardio ability (using HIIT approaches) just so I can keep up with the other folks in my Mt. Biking club. – Clay Nichols Nov 16 '12 at 15:53
I understand that. Honestly though, if you want to keep up with the other cyclists, the best way to do that is more time on the bike, in the hurt zones. I am a bicycle racer and triathlete and part time coach, and that is the best way to improve in a sport is to do more of that sport. – JohnP Nov 17 '12 at 1:31
@JohnP In your example, you suppose that an athlete who merely loses weight would increase his V02max just by losing weight... that isn't accurate. VO2 max is really just the volume of O2 consumed per minute. It is often reported in terms of ml/kg/min to compare different sized athletes, but the 220 pound athlete doesn't magically gain fitness just by losing weight. His V02 would likely stay the same, but his V02/kg would change. – lecrank Jan 25 at 20:47
@lecrank - I'm not suggesting that fitness increases (I state that in my last sentence). But, if you lose weight, your VO2Max will automatically increase, simply because it's a divisor in the equation. It's an indicator of potential, nothing more. If you want a better metric, vVO2Max (Velocity or speed at VO2 max) is much better. – JohnP Jan 25 at 21:00
You could try the Cooper test, which is basically "run at a steady pace for 12 minutes and look at the table". I've always found it sufficient for my purposes (to see how much I improved in a given time period).
-
On estimating your VO2Max number IN your home, and without specialist equipment, you could simply use the Sorensen formula:
VO2max = 15.3 * ( MaximumHeartRateBMP / BasalHeartRateBMP)
where:
MaximumHeartRateBMP = 210 - 0.5*AgeYears
[ACMS Guidelines for Exercise Testing and Prescription]
Alternatively use one of the Step tests: Queens, YMCA, Harvard, or the Rockport 1 mile walk, Cooper 12min run, Kindle's 12 min run, 1.5 mile run.... test. Unfortunately the VO2max number you get out from any of these tests is impacted by effort put in, so the numbers can fluctuate significantly e.g.
- | 886 | 3,602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2016-30 | latest | en | 0.967695 |
https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-5-section-5-4-polynomials-in-several-variables-exercise-set-page-375/34 | 1,537,543,195,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157216.50/warc/CC-MAIN-20180921151328-20180921171728-00180.warc.gz | 748,209,458 | 13,310 | ## Introductory Algebra for College Students (7th Edition)
$30x^{4}y-54x^2y^{2}$
Using the Distributive Property, the product of the given expression, $6x^2y(5x^2-9y) ,$ is \begin{array}{l}\require{cancel} 6x^2y(5x^2)+6x^2y(-9y) \\\\= 30x^{2+2}y-54x^2y^{1+1} \\\\= 30x^{4}y-54x^2y^{2} .\end{array} | 140 | 298 | {"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} | 4.21875 | 4 | CC-MAIN-2018-39 | longest | en | 0.6203 |
https://community.appsheet.com/t/i-could-use-some-assistance-with-calculating/4516 | 1,558,941,335,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262029.97/warc/CC-MAIN-20190527065651-20190527091651-00163.warc.gz | 418,811,310 | 4,659 | # I could use some assistance with calculating ...
(Thomas Cunningham) #1
I could use some assistance with calculating a client’s age (at arrest) based on their date of birth and the date they are arrested, these dates being in two different tables. Within the Client Table (ClientTbl) there is static information such as: ID Number, Last Name, First Name, Sex, Date of Birth, etc… Within the Court Table (CourtTbl) there are fields like: ID Number, Court Date, Arrest Date, Charge, Age, etc…. Within “Age” field in the Court Table, I would like to calculate the clients age at time of arrest based on these two dates. I have included an example below:
ClientTbl: Sample Data ID Number Last Name First Name Sex
DateOfBirth
Ter124
Doe
John
Male 12/05/1974
CourtTbl: Sample Data ID Number CourtDate ArrestDate Charge
Age
Ter124
04/13/2018 04/12/2018 DUI
???
Obviously, this client would be 43 years old, but that is the formula that I am having trouble with because the dates are in to different tables and I need to match ID Numbers. Thank you in advance for your assistance.
@Thomas_Cunningham I assume your ID Number is the primary key of your Client table. In that case, make the ID Number of your Court table a ref column. Then you can de-reference DateOfBirth column from the Court table. help.appsheet.com - References Between Tables
If you need help calculating the age in years based on the date of birth, see this thread: https://plus.google.com/+MultitechVisions/posts/QKfzMzkDWxH References Between Tables help.appsheet.com
(Thomas Cunningham) #3
Thank you for the information and I checked out De-reference and the video. I think I understand how it could work, but being new to this, I’m still a bit lost. Any other suggestions or hints? Thank you!
Here’s an example app: https://www.appsheet.com/samples/A-basic-demo-of-table-references?appGuidString=6ffad040-b04a-4325-aa73-ace5df8ba1b5 appsheet.com - Table References - A basic demo of table references Table References - A basic demo of table references appsheet.com
(Thomas Cunningham) #5
I’m afraid that i am not getting it. I have looked at the app that you suggested and there doesn’t seem to be similar formulas there that i could use to teach myself the steps needed.
The ID field in the ClientTbl is the primary field and the ID field in the CourtTbl is a ref.
I would think it would be something like: Any(Select(CourtTbl[ID], “ArrestDate” - ClientTbl[ID], “DateOfBirth”) | 593 | 2,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-22 | latest | en | 0.874017 |
https://proxies-free.com/javascript-build-a-stamp-duty-calculator-for-various-types-of-buyers/ | 1,606,800,430,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141652107.52/warc/CC-MAIN-20201201043603-20201201073603-00157.warc.gz | 426,054,553 | 8,296 | # javascript – Build a stamp duty calculator for various types of buyers
I had a task where I had to build a stamp duty calculator, which calculates the tax or stamp duty first-time buyers(`ftb`) returning buyers (`rtb`), and second-home buyers (`shb`) had to pay when purchasing the home.
These were the tax bands, although I think some of the percentages may be slightly wrong.
``````First Time Buyers
0 to £300,000 - 0%
£300,001 to £925,000 - 5%
£925001 to £1,500,000 - 10%
£1,500,000 and above - 12%
0 to £125,000 - 0%
£125,001 to £250,000 - 2%
£250,001 to £925,000 - 5%
£925,001 to £1,500,000 - 10%
£1,500,000 and above - 12%
0 to £125,000 - 3%
£125,001 to £250,000 - 5%
£250,001 to £925,000 - 8%
£925,001 to £1,500,000 - 13%
£1,500,000 and above - 15%
``````
Here is my solution, I look forward to the feedback.
``````const stampDutyCalc = (price) => {
let stampdutyObject = {
ftb: 0,
rtb: 0,
shb: 0,
};
return stampdutyObject;
};
if (price <= 300000) {
stampdutyObject.ftb += 0;
} else if (price > 300000 && price <= 925000) {
stampdutyObject.ftb += price * 0.05;
} else if (price > 925000 && price <= 1500000) {
stampdutyObject.ftb += (price - 925000) * 0.1 + (925000 - 300000) * 0.05;
} else {
stampdutyObject.ftb +=
(price - 1500000) * 0.12 +
(1500000 - 925000) * 0.1 +
(925000 - 300000) * 0.05;
}
}
if (price <= 125000) {
stampdutyObject.rtb += 0;
} else if (price > 125000 && price <= 250000) {
stampdutyObject.rtb += price * 0.02;
} else if (price > 250000 && price <= 925000) {
stampdutyObject.rtb += (price - 250000) * 0.05 + (250000 - 125000) * 0.02;
} else if (price > 925000 && price <= 1500000) {
stampdutyObject.rtb +=
(price - 925000) * 0.1 +
(925000 - 250000) * 0.05 +
(250000 - 125000) * 0.02;
} else {
stampdutyObject.rtb +=
(price - 1500000) * 0.12 +
(1500000 - 925000) * 0.1 +
(925000 - 250000) * 0.05 +
(250000 - 125000) * 0.02;
}
}
if (price <= 125000) {
stampdutyObject.shb = price * 0.03;
} else if (price > 125000 && price <= 250000) {
stampdutyObject.shb += (price - 125000) * 0.05 + 125000 * 0.03;
} else if (price > 250000 && price <= 925000) {
stampdutyObject.shb +=
(price - 250000) * 0.08 + (250000 - 125000) * 0.05 + 125000 * 0.03;
} else if (price > 925000 && price <= 1500000) {
stampdutyObject.shb +=
(price - 925000) * 0.13 +
(925000 - 250000) * 0.08 +
(250000 - 125000) * 0.05 +
125000 * 0.03;
} else {
stampdutyObject.shb +=
(price - 1500000) * 0.15 +
(1500000 - 925000) * 0.13 +
(925000 - 250000) * 0.08 +
(250000 - 125000) * 0.0 +
125000 * 0.03;
}
}
if (price <= 125000) {
stampdutyObject.rtb += 0;
} else if (price > 125000 && price <= 250000) {
stampdutyObject.rtb += price * 0.02;
} else if (price > 250000 && price <= 925000) {
stampdutyObject.rtb += (price - 250000) * 0.05 + (250000 - 125000) * 0.02;
} else if (price > 925000 && price <= 1500000) {
stampdutyObject.rtb +=
(price - 925000) * 0.1 +
(925000 - 250000) * 0.05 +
(250000 - 125000) * 0.02;
} else {
stampdutyObject.rtb +=
(price - 1500000) * 0.12 +
(1500000 - 925000) * 0.1 +
(925000 - 250000) * 0.05 +
(250000 - 125000) * 0.02;
}
}
`````` | 1,233 | 3,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.727405 |
http://stackoverflow.com/questions/439814/prime-factor-of-300-000-000-000/440215 | 1,438,394,026,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988399.65/warc/CC-MAIN-20150728002308-00251-ip-10-236-191-2.ec2.internal.warc.gz | 216,518,126 | 33,010 | # Prime factor of 300 000 000 000?
I need to find out the prime factors of over 300 billion. I have a function that is adding to the list of them...very slowly! It has been running for about an hour now and i think its got a fair distance to go still. Am i doing it completly wrong or is this expected?
Edit: Im trying to find the largest prime factor of the number 600851475143.
Edit: Result:
{
List<Int64> ListOfPrimeFactors = new List<Int64>();
Int64 Number = 600851475143;
Int64 DividingNumber = 2;
while (DividingNumber < Number / DividingNumber)
{
if (Number % DividingNumber == 0)
{
Number = Number/DividingNumber;
}
else
DividingNumber++;
}
listBox1.DataSource = ListOfPrimeFactors;
}
}
-
Please clarify with proper language what is the precise problem you are trying to solve. What you explain could have at least five or six possible meanings. Also, do you have any sort of upper bound? 10^40000000000 is "over 300 billion" and might indeed take hours... :) – Daniel Daranas Jan 13 '09 at 17:12
Do you really mean that you're trying to find the prime numbers over 300 000 000 000? (as a prime factor is something else entirely) – Rowland Shaw Jan 13 '09 at 17:14
Are you trying to solve project Euler's problem 3 ? – Perpetualcoder Jan 13 '09 at 17:14
is this a homework assignment? – Anthony Potts Jan 13 '09 at 17:17
(1) Read and fully understand en.wikipedia.org/wiki/Prime_number (2) Read and fully understand en.wikipedia.org/wiki/Prime_factor (3) Read and fully understand en.wikipedia.org/wiki/Integer_factorization (4) Start thinking about algorithms (5) Choose one and code it – Daniel Daranas Jan 14 '09 at 9:37
Are you remembering to divide the number that you're factorizing by each factor as you find them?
Say, for example, you find that 2 is a factor. You can add that to your list of factors, but then you divide the number that you're trying to factorise by that value.
Now you're only searching for the factors of 150 billion. Each time around you should start from the factor you just found. So if 2 was a factor, test 2 again. If the next factor you find is 3, there's no point testing from 2 again.
And so on...
-
No... erm explain please =] – xoxo Jan 13 '09 at 17:05
Ohh ok i get it! Thanks alot! – xoxo Jan 13 '09 at 17:14
Look into researching recursion. – EBGreen Jan 13 '09 at 17:15
Recursion for prime factoring? – Eclipse Jan 13 '09 at 17:16
The process is recursive, but I wouldn't implement it as a recursive function or you'll get... a stackoverflow ;) Use iteration. – Dave Swersky Jan 13 '09 at 17:40
Finding prime factors is difficult using brute force, which sounds like the technique you are using.
Here are a few tips to speed it up somewhat:
• Start low, not high
• Don't bother testing each potential factor to see whether it is prime--just test LIKELY prime numbers (odd numbers that end in 1,3,7 or 9)
• Don't bother testing even numbers (all divisible by 2), or odds that end in 5 (all divisible by 5). Of course, don't actually skip 2 and 5!!
• When you find a prime factor, make sure to divide it out--don't continue to use your massive original number. See my example below.
• If you find a factor, make sure to test it AGAIN to see if it is in there multiple times. Your number could be 2x2x3x7x7x7x31 or something like that.
• Stop when you reach >= sqrt(remaining large number)
Edit: A simple example: You are finding the factors of 275.
1. Test 275 for divisibility by 2. Does 275/2 = int(275/2)? No. Failed.
2. Test 275 for divisibility by 3. Failed.
3. Skip 4!
4. Test 275 for divisibility by 5. YES! 275/5 = 55. So your NEW test number is now 55.
5. Test 55 for divisibility by 5. YES! 55/5 = 11. So your NEW test number is now 11.
6. BUT 5 > sqrt (11), so 11 is prime, and you can stop!
So 275 = 5 * 5 * 11
Make more sense?
-
What do you mean 'divide it out'? And im not sure how the square root affects it? – xoxo Jan 13 '09 at 17:13
If you haven't found a factor by the sqrt of a number, you're not going to find any no matter how far you go. This makes a HUGE difference when checking larger numbers. – Eclipse Jan 13 '09 at 17:15
Why do you say "at least all odd numbers ending in 1,3,7 or 9"? – PEZ Jan 13 '09 at 18:54
You do have to balance that waste with the effort needed to skip every 5th number. You'd probably get more benefit by skipping every 3rd odd number after 3. Note that from the computers point of view, there's nothing special about numbers that end in 5 in decimal. – Eclipse Jan 13 '09 at 20:04
@Grace, to your latter question, yes, I think so. There are two issues here: "programming" and "solving a real world problem". In your case the real world problem is "prime number" and the very first thing you should do is study the theory about them. Then think about algorithms. – Daniel Daranas Jan 14 '09 at 9:29
Factoring big numbers is a hard problem. So hard, in fact, that we rely on it to keep RSA secure. But take a look at the wikipedia page for some pointers to algorithms that can help. But for a number that small, it really shouldn't be taking that long, unless you are re-doing work over and over again that you don't have to somewhere.
For the brute-force solution, remember that you can do some mini-optimizations:
• Check 2 specially, then only check odd numbers.
• You only ever need to check up to the square-root of the number (if you find no factors by then, then the number is prime).
• Once you find a factor, don't use the original number to find the next factor, divide it by the found factor, and search the new smaller number.
• When you find a factor, divide it through as many times as you can. After that, you never need to check that number, or any smaller numbers again.
• If you do all the above, each new factor you find will be prime, since any smaller factors have already been removed.
-
The number in the question is neither large nor hard to factor. – Sam Meldrum Jan 13 '09 at 17:12
When we say factoring big numbers is hard, we mean numbers above approximately 2^512. RSA and similar probably use primes up around 2^2048. The number in the OP is only about 2^39... much, much smaller – abelenky Jan 14 '09 at 0:47
Here is an XSLT solution!
This XSLT transformation takes 0.109 sec.
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xmlns:saxon="http://saxon.sf.net/"
xmlns:f="http://fxsl.sf.net/"
exclude-result-prefixes="xs saxon f"
>
<xsl:import href="../f/func-Primes.xsl"/>
<xsl:output method="text"/>
<xsl:template name="initial" match="/*">
<xsl:sequence select="f:maxPrimeFactor(600851475143)"/>
</xsl:template>
<xsl:function name="f:maxPrimeFactor" as="xs:integer">
<xsl:param name="pNum" as="xs:integer"/>
<xsl:sequence select=
"if(f:isPrime(\$pNum))
then \$pNum
else
for \$vEnd in xs:integer(floor(f:sqrt(\$pNum, 0.1E0))),
\$vDiv1 in (2 to \$vEnd)[\$pNum mod . = 0][1],
\$vDiv2 in \$pNum idiv \$vDiv1
return
max((f:maxPrimeFactor(\$vDiv1),f:maxPrimeFactor(\$vDiv2)))
"/>
</xsl:function>
</xsl:stylesheet>
This transformation produces the correct result (the maximum prime factor of 600851475143) in just 0.109 sec.:
6857
The transformation uses the f:sqrt() and f:isPrime() defined in FXSL 2.0 -- a library for functional programming in XSLT. FXSL is itself written entirely in XSLT.
f:isPrime() uses Fermat's little theorem so that it is efficient to determine primeality.
-
Way neat! Too bad the problem 3 forum is closed so you can't post this there. I don't think I've seen an XSLT solution there. – PEZ Jan 15 '09 at 13:56
@PEZ: What is the problem 3 forum? – Dimitre Novatchev Jan 15 '09 at 13:59
Problem 3 at Project Euler (projecteuler.net/index.php?section=problems&id=3). When you register at that site you get an answer box there and if you submit the correct one (which you obviously will) you'll get access to a forum where people have discussed the problem and their solutions. – PEZ Jan 15 '09 at 15:00
Though the problem was presented long ago so the forum is "full" and they have closed it for new entries. – PEZ Jan 15 '09 at 15:01
@PEZ: Thank you for mentioninh ProjectEuler. I already solved 28 problems and the only programming language I used has been XSLT. – Dimitre Novatchev Jan 23 '09 at 3:47
One last thing nobody has mentioned, perhaps because it seems obvious. Every time you find a factor and divide it out, keep trying the factor until it fails.
64 only has one prime factor, 2. You will find that out pretty trivially if you keep dividing out the 2 until you can't anymore.
-
\$ time factor 300000000000 > /dev/null
real 0m0.027s
user 0m0.000s
sys 0m0.001s
You're doing something wrong if it's taking an hour. You might even have an infinite loop somewhere - make sure you're not using 32-bit ints.
-
The key to understanding why the square root is important, consider that each factor of n below the square root of n has a corresponding factor above it. To see this, consider that if x is factor of n, then x/n = m which means that x/m = n, hence m is also a factor.
-
I wouldn't expect it to take very long at all - that's not a particularly large number.
Could you give us an example number which is causing your code difficulties?
-
600851475143. i cut it in half (thats what i thought i remembered from school) to make it 300425737571. I get rid of the numbers divisible by 2, 3, 5 and 7. Then i get rid of any numbers that cannot be divided into the number and find any primes left... – xoxo Jan 13 '09 at 17:12
"600851475143. i cut it in half (thats what i thought i remembered from school) to make it 300425737571." Sorry???? If you don't know the basic semantics of the problem, you shouldn't even try. – Daniel Daranas Jan 13 '09 at 17:17
No Daniel, thanks but i think ill give this one a go! – xoxo Jan 13 '09 at 17:18
600851475143 and 300425737571 each have four distinct prime factors. They share none in common. @Daniel Daranas is suggesting that you understand why "cutting [an odd number] in half" doesn't work. For example, you may try the same with 27, which "cuts" to 13 but shares no factors with 13. – A. Rex Jan 14 '09 at 9:00
@A. Rex thank you for your clarification. Yes, what I was saying is basically that you can "try" any approaches, but it helps if they are mathematically correct. Dividing the number by 7.65 and then taking its logarithm certainly won't help. – Daniel Daranas Jan 14 '09 at 9:22
Here's one site where you can get answers: Factoris - Online factorization service. It can do really big numbers, but it also can factorize algebraic expressions.
-
The fastest algorithms are sieve algorithms, and are based on arcane areas of discrete mathematics (over my head at least), complicated to implement and test.
The simplest algorithm for factoring is probably (as others have said) the Sieve of Eratosthenes. Things to remember about using this to factor a number N:
• general idea: you're checking an increasing sequence of possible integer factors x to see if they evenly divide your candidate number N (in C/Java/Javascript check whether N % x == 0) in which case N is not prime.
• you just need to go up to sqrt(N), but don't actually calculate sqrt(N): loop as long as your test factor x passes the test x*x<N
• if you have the memory to save a bunch of previous primes, use only those as the test factors (and don't save them if prime P fails the test P*P > N_max since you'll never use them again
• Even if you don't save the previous primes, for possible factors x just check 2 and all the odd numbers. Yes, it will take longer, but not that much longer for reasonable sized numbers. The prime-counting function and its approximations can tell you what fraction of numbers are prime; this fraction decreases very slowly. Even for 264 = approx 1.8x1019, roughly one out of every 43 numbers is prime (= one out of every 21.5 odd numbers is prime). For factors of numbers less than 264, those factors x are less than 232 where about one out of every 20 numbers is prime = one out of every 10 odd numbers is prime. So you'll have to test 10 times as many numbers, but the loop should be a bit faster and you don't have to mess around with storing all those primes.
There are also some older and simpler sieve algorithms that a little bit more complex but still fairly understandable. See Dixon's, Shanks' and Fermat's factoring algorithms. I read an article about one of these once, can't remember which one, but they're all fairly straightforward and use algebraic properties of the differences of squares.
If you're just testing whether a number N is prime, and you don't actually care about the factors themselves, use a probabilistic primality test. Miller-Rabin is the most standard one, I think.
-
When I do x*x<N instead of x<sqrt(N) it takes longer for the number in question (in Python). Maybe sqrt(N) brings the comparison down to integer types that compare faster? – PEZ Jan 13 '09 at 21:21
if N is fixed maybe the interpreter is "smart" enough to evaluate it only once... – Jason S Jan 13 '09 at 22:03
In another thread in factorization I ran a test on the Sieve--it absolutely STUNK, even when put up against a crude algorithm that didn't consider all the tricks listed here. (Not that I'm sure all the tricks are good--the cost may be higher than what they save!) – Loren Pechtel Jan 14 '09 at 1:12
Jason, the thing is that I don't compute sqrt(N) very often, only when I find a factor. It's the comparison that takes place often. Maybe it's much (my tests indicate that it's much) more efficient to compare p<sqrt_N than p_times_p<N. – PEZ Jan 14 '09 at 8:55
I spent some time on this since it just sucked me in. I won't paste the code here just yet. Instead see this factors.py gist if you're curious.
Mind you, I didn't know anything about factoring (still don't) before reading this question. It's just a Python implementation of BradC's answer above.
On my MacBook it takes 0.002 secs to factor the number mentioned in the question (600851475143).
There must obviously be much, much faster ways of doing this. My program takes 19 secs to compute the factors of 6008514751431331. But the Factoris service just spits out the answer in no-time.
-
@PEZ: What is the problem 3 forum? – Dimitre Novatchev Jan 15 '09 at 14:11
It's a forum on the Project Euler site (projecteuler.net) that you get access to once you've submitted the correct answer to the problem we're discussing here. Check it out. But I warn you; It's highly addictive! – PEZ Jan 15 '09 at 15:21
@PEZ: Many thanks for mentioning ProjectEuler. I already solved 28 problems and the only programming language I used has been XSLT. – Dimitre Novatchev Jan 23 '09 at 3:57
That's amazing. Can't you put your solutions up on gist.github.com or some such. I'd like to see some more XSLT solutions to problems I've solved. And the last problem was launched last sat so if you solve that one you can probably post on the project forum. They're launching a new problem tomorrow. – PEZ Jan 23 '09 at 9:05
The specific number is 300425737571? It trivially factors into 131 * 151 * 673 * 22567. I don't see what all the fuss is...
-
Here's some Haskell goodness for you guys :)
primeFactors n = factor n primes
where factor n (p:ps) | p*p > n = [n]
| n `mod` p /= 0 = factor n ps
| otherwise = p : factor (n `div` p) (p:ps)
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
Took it about .5 seconds to find them, so I'd call that a success.
-
He specifically asked for hints, not code. – EvilTeach Jan 13 '09 at 19:04
Evilteach is too picky: the please no code was edited later – blabla999 Jan 13 '09 at 20:25
He? I am a she! and no i did ask for no code in the original question but its not the language i am using so its no bother! – xoxo Jan 14 '09 at 8:37
If you don't sieve, it's less code and faster: primeFactors n = factor n (2:[3,5..]) – A. Rex Jan 14 '09 at 9:28
the backticks around mod and div didn't show up. Anybody else who's trying to use this code, put `backticks` around mod and div. @A. Rex: faster always? or just in the given case? Would the sieve help more as the input number gets bigger? – LarsH Aug 18 '10 at 9:08
It shouldn't take that long, even with a relatively naive brute force. For that specific number, I can factor it in my head in about one second.
You say you don't want solutions(?), but here's your "subtle" hint. The only prime factors of the number are the lowest three primes.
-
No there are more then that! The specific number is 300425737571 so go for it... – xoxo Jan 13 '09 at 17:09
I almost don't want to know the answer to this... but why do you need the factor of JUST this specific number? – EdgarVerona Jan 13 '09 at 20:04
It's for projecteuler. – recursive Jan 13 '09 at 21:29
You could use the sieve of Eratosthenes to find the primes and see if your number is divisible by those you find.
-
Yes thats what im working from lol – xoxo Jan 13 '09 at 17:19
a sieve is extremely slow compared to just testing every odd number. – zaratustra Jan 13 '09 at 18:14
You only need to check it's remainder mod(n) where n is a prime <= sqrt(N) where N is the number you are trying to factor. It really shouldn't take over an hour, even on a really slow computer or a TI-85.
-
Sighs no i think its just my lack of prime number knowledge! – xoxo Jan 13 '09 at 17:20
Before I became a programmer I was a PhD student in Number Theory, and it looks like there are a number of other people who are in to that too. As mentioned in other posts prime numbers are what keeps everything secure. If you get the time and it interests you, check it out. – Anthony Potts Jan 14 '09 at 17:04
Your algorithm must be FUBAR. This only takes about 0.1s on my 1.6 GHz netbook in Python. Python isn't known for its blazing speed. It does, however, have arbitrary precision integers...
import math
import operator
def factor(n):
"""Given the number n, to factor yield a it's prime factors.
factor(1) yields one result: 1. Negative n is not supported."""
M = math.sqrt(n) # no factors larger than M
p = 2 # candidate factor to test
while p <= M: # keep looking until pointless
d, m = divmod(n, p)
if m == 0:
yield p # p is a prime factor
n = d # divide n accordingly
M = math.sqrt(n) # and adjust M
else:
p += 1 # p didn't pan out, try the next candidate
yield n # whatever's left in n is a prime factor
def test_factor(n):
f = factor(n)
n2 = reduce(operator.mul, f)
assert n2 == n
def example():
n = 600851475143
f = list(factor(n))
assert reduce(operator.mul, f) == n
print n, "=", "*".join(str(p) for p in f)
example()
# output:
# 600851475143 = 71*839*1471*6857
(This code seems to work in defiance of the fact that I don't know enough about number theory to fill a thimble.)
-
You lucked out with your choice: 71 shows up fairly quickly and that greatly reduces the time spent. Try factoring this one: 12073531234081300153 – Jason S Jan 13 '09 at 18:46
Yes, naturally, even this fairly stupid algorithm would be better if it didn't waste its time trying values of p that are not prime. However, I should point out that 600851475143 is specifically mentioned in the question, so it wasn't luck, as such. ;-) – bendin Jan 13 '09 at 20:44
Just to expand/improve slightly on the "only test odd numbers that don't end in 5" suggestions...
All primes greater than 3 are either one more or one less than a multiple of 6 (6x + 1 or 6x - 1 for integer values of x).
-
Semi-prime numbers of that size are used for encryption, so I am curious as to what you exactly want to use them for.
That aside, there currently are not good ways to find the prime factorization of large numbers in a relatively small amount of time.
-
300 billion is less than 40 bits. Encryption primes are closer to 2048 bits... – flussence Jan 13 '09 at 17:13
True on the less than 40 bits part, my mistake. However, encryption primes iirc are usually 512 bits each (1024 bits total for the resulting key). You use two, multiply them, and get a semi-prime (a number with only two prime factors). – Matthew Brubaker Jan 13 '09 at 17:25 | 5,479 | 20,165 | {"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} | 3.75 | 4 | CC-MAIN-2015-32 | latest | en | 0.900217 |
http://us.metamath.org/mpegif/mapdh8ac.html | 1,526,887,274,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863967.46/warc/CC-MAIN-20180521063331-20180521083331-00095.warc.gz | 302,410,116 | 10,903 | Mathbox for Norm Megill < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > mapdh8ac Unicode version
Theorem mapdh8ac 32307
Description: Part of Part (8) in [Baer] p. 48. (Contributed by NM, 13-May-2015.)
Hypotheses
Ref Expression
mapdh8a.h
mapdh8a.u
mapdh8a.v
mapdh8a.s
mapdh8a.o
mapdh8a.n
mapdh8a.c LCDual
mapdh8a.d
mapdh8a.r
mapdh8a.q
mapdh8a.j
mapdh8a.m mapd
mapdh8a.i
mapdh8a.k
mapdh8ac.f
mapdh8ac.mn
mapdh8ac.eg
mapdh8ac.ee
mapdh8ac.x
mapdh8ac.y
mapdh8ac.z
mapdh8ac.t
mapdh8ac.yn
mapdh8ac.ew
mapdh8ac.w
mapdh8ac.yw
mapdh8ac.xy
mapdh8ac.wz
mapdh8ac.xz
Assertion
Ref Expression
mapdh8ac
Distinct variable groups: ,, ,, , ,, ,, , ,, ,, ,, ,, , ,, , ,, , ,, ,, ,, ,, ,, ,,
Allowed substitution hints: (,) () (,) () (,) () () (,) () () () (,,) (,) () (,,) () () () (,,) (,,) () () () ()
Proof of Theorem mapdh8ac
StepHypRef Expression
1 mapdh8a.h . . 3
2 mapdh8a.u . . 3
3 mapdh8a.v . . 3
4 mapdh8a.s . . 3
5 mapdh8a.o . . 3
6 mapdh8a.n . . 3
7 mapdh8a.c . . 3 LCDual
8 mapdh8a.d . . 3
9 mapdh8a.r . . 3
10 mapdh8a.q . . 3
11 mapdh8a.j . . 3
12 mapdh8a.m . . 3 mapd
13 mapdh8a.i . . 3
14 mapdh8a.k . . 3
15 mapdh8ac.f . . 3
16 mapdh8ac.mn . . 3
17 mapdh8ac.eg . . 3
18 mapdh8ac.ew . . 3
19 mapdh8ac.x . . 3
20 mapdh8ac.y . . 3
21 mapdh8ac.w . . 3
22 mapdh8ac.t . . 3
23 mapdh8ac.yw . . 3
24 mapdh8ac.xy . . 3
25 mapdh8ac.yn . . 3
261, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25mapdh8ab 32306 . 2
27 mapdh8ac.ee . . 3
28 mapdh8ac.z . . 3
29 mapdh8ac.wz . . 3
30 mapdh8ac.xz . . 3
311, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 27, 19, 21, 28, 22, 29, 30, 25mapdh8ab 32306 . 2
3226, 31eqtrd 2462 1
Colors of variables: wff set class Syntax hints: wn 3 wi 4 wa 359 wceq 1652 wcel 1725 wne 2593 cvv 2943 cdif 3304 cif 3726 csn 3801 cpr 3802 cotp 3805 cmpt 4253 cfv 5440 (class class class)co 6067 c1st 6333 c2nd 6334 crio 6528 cbs 13452 c0g 13706 csg 14671 clspn 16030 chlt 29879 clh 30512 cdvh 31607 LCDualclcd 32115 mapdcmpd 32153 This theorem is referenced by: mapdh8ad 32308 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1555 ax-5 1566 ax-17 1626 ax-9 1666 ax-8 1687 ax-13 1727 ax-14 1729 ax-6 1744 ax-7 1749 ax-11 1761 ax-12 1950 ax-ext 2411 ax-rep 4307 ax-sep 4317 ax-nul 4325 ax-pow 4364 ax-pr 4390 ax-un 4687 ax-cnex 9030 ax-resscn 9031 ax-1cn 9032 ax-icn 9033 ax-addcl 9034 ax-addrcl 9035 ax-mulcl 9036 ax-mulrcl 9037 ax-mulcom 9038 ax-addass 9039 ax-mulass 9040 ax-distr 9041 ax-i2m1 9042 ax-1ne0 9043 ax-1rid 9044 ax-rnegex 9045 ax-rrecex 9046 ax-cnre 9047 ax-pre-lttri 9048 ax-pre-lttrn 9049 ax-pre-ltadd 9050 ax-pre-mulgt0 9051 This theorem depends on definitions: df-bi 178 df-or 360 df-an 361 df-3or 937 df-3an 938 df-tru 1328 df-fal 1329 df-ex 1551 df-nf 1554 df-sb 1659 df-eu 2284 df-mo 2285 df-clab 2417 df-cleq 2423 df-clel 2426 df-nfc 2555 df-ne 2595 df-nel 2596 df-ral 2697 df-rex 2698 df-reu 2699 df-rmo 2700 df-rab 2701 df-v 2945 df-sbc 3149 df-csb 3239 df-dif 3310 df-un 3312 df-in 3314 df-ss 3321 df-pss 3323 df-nul 3616 df-if 3727 df-pw 3788 df-sn 3807 df-pr 3808 df-tp 3809 df-op 3810 df-ot 3811 df-uni 4003 df-int 4038 df-iun 4082 df-iin 4083 df-br 4200 df-opab 4254 df-mpt 4255 df-tr 4290 df-eprel 4481 df-id 4485 df-po 4490 df-so 4491 df-fr 4528 df-we 4530 df-ord 4571 df-on 4572 df-lim 4573 df-suc 4574 df-om 4832 df-xp 4870 df-rel 4871 df-cnv 4872 df-co 4873 df-dm 4874 df-rn 4875 df-res 4876 df-ima 4877 df-iota 5404 df-fun 5442 df-fn 5443 df-f 5444 df-f1 5445 df-fo 5446 df-f1o 5447 df-fv 5448 df-ov 6070 df-oprab 6071 df-mpt2 6072 df-of 6291 df-1st 6335 df-2nd 6336 df-tpos 6465 df-undef 6529 df-riota 6535 df-recs 6619 df-rdg 6654 df-1o 6710 df-oadd 6714 df-er 6891 df-map 7006 df-en 7096 df-dom 7097 df-sdom 7098 df-fin 7099 df-pnf 9106 df-mnf 9107 df-xr 9108 df-ltxr 9109 df-le 9110 df-sub 9277 df-neg 9278 df-nn 9985 df-2 10042 df-3 10043 df-4 10044 df-5 10045 df-6 10046 df-n0 10206 df-z 10267 df-uz 10473 df-fz 11028 df-struct 13454 df-ndx 13455 df-slot 13456 df-base 13457 df-sets 13458 df-ress 13459 df-plusg 13525 df-mulr 13526 df-sca 13528 df-vsca 13529 df-0g 13710 df-mre 13794 df-mrc 13795 df-acs 13797 df-poset 14386 df-plt 14398 df-lub 14414 df-glb 14415 df-join 14416 df-meet 14417 df-p0 14451 df-p1 14452 df-lat 14458 df-clat 14520 df-mnd 14673 df-submnd 14722 df-grp 14795 df-minusg 14796 df-sbg 14797 df-subg 14924 df-cntz 15099 df-oppg 15125 df-lsm 15253 df-cmn 15397 df-abl 15398 df-mgp 15632 df-rng 15646 df-ur 15648 df-oppr 15711 df-dvdsr 15729 df-unit 15730 df-invr 15760 df-dvr 15771 df-drng 15820 df-lmod 15935 df-lss 15992 df-lsp 16031 df-lvec 16158 df-lsatoms 29505 df-lshyp 29506 df-lcv 29548 df-lfl 29587 df-lkr 29615 df-ldual 29653 df-oposet 29705 df-ol 29707 df-oml 29708 df-covers 29795 df-ats 29796 df-atl 29827 df-cvlat 29851 df-hlat 29880 df-llines 30026 df-lplanes 30027 df-lvols 30028 df-lines 30029 df-psubsp 30031 df-pmap 30032 df-padd 30324 df-lhyp 30516 df-laut 30517 df-ldil 30632 df-ltrn 30633 df-trl 30687 df-tgrp 31271 df-tendo 31283 df-edring 31285 df-dveca 31531 df-disoa 31558 df-dvech 31608 df-dib 31668 df-dic 31702 df-dih 31758 df-doch 31877 df-djh 31924 df-lcdual 32116 df-mapd 32154
Copyright terms: Public domain W3C validator | 3,130 | 5,616 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-22 | latest | en | 0.089624 |
https://www.velocityreviews.com/threads/this-is-kinda-buggin-me-now-anyone-able-to-figure-itout.222466/ | 1,591,235,038,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436828.65/warc/CC-MAIN-20200604001115-20200604031115-00165.warc.gz | 933,559,966 | 12,214 | # This is kinda buggin me now... anyone able to figure itout?
Discussion in 'Computer Support' started by AJ, Apr 22, 2005.
1. ### AJGuest
3 people go to a hotel and get a room. The night clerk charged them £30
rent. So they each paid £10 and went to their room.
The manager comes in and says to the clerk that there is a special on and
the room was only £25. So he sent the porter up to the room with 5 pound
coins. Seeing as the 3 couldn't split £5 coins between the 3 of them they
gave the porter £2 and each took £1 back, meaning they paid £9 each for the
room.
Sooo where is the other £1?
they paid £9 each for the room... = £27
gave the porter £2 =£29
missing £1 is where?
AJ, Apr 22, 2005
2. ### GuestGuest
So why are you adding 2 to 27? The hotel still only got 25. The porter
got 5 which he split into 3 and 2. All those values are negative with
respect to what the hotel got paid (30 - 5 = 30 - 3 - 2). You are
subtracting 3 and then adding 2. Wrong. You subtract 3 (to get what
the patrons paid in total) AND you also *subtract* 2 to get down to the
25 that the hotel got paid. The porter stole the 2 dollars. You are
trying to pretend that 27 + 2 is what the hotel got paid. Wrong. The
hotel got paid 27 - 2.
Guest, Apr 22, 2005
3. ### Bill PGuest
Dollars? Who said anything about dollars?
Bill P, Apr 22, 2005
4. ### GuestGuest
Ooooh, excuse me for accidentally putting "dollars" into one of those
sentences in my reply. So when you wash your hair, you actually follow
those instructions of "wash, rinse, repeat" indefinitely? I wasn't
about to waste my time figuring out how to insert the pound character
and had intended to leave all values as numeric only. The point was to
prove the failed logic of adding 2 when it was supposed to get
substracted.
Guest, Apr 22, 2005
5. ### Bill PGuest
Lighten up ;-)
Bill P, Apr 22, 2005
6. ### GuestGuest
It you can't take shit, don't toss it.
Guest, Apr 22, 2005
7. ### ScraggyGuest
Try that yourself.
Scraggy, Apr 22, 2005
8. ### GuestGuest
He tossed first. Mommy, mommy, they're picking on me, mommy.
Guest, Apr 22, 2005
9. ### pcbutts1Guest
The Hotel got 25
The Porter got 2
The extra 3 went to the guests 1 each.
The guests paid a total of 27, 25 to the hotel and a 2 to the porter as a
tip.
pcbutts1, Apr 22, 2005
10. ### iamgilbertGuest
they didnt pay 9 pounds each...u only persume they did....
they actually paid £8.33 each... (£25/3 = £8.3333333)
see?
iamgilbert, Apr 22, 2005
11. ### ScraggyGuest
<vbg>
Scraggy, Apr 22, 2005
12. ### Arnold P. FurshlugginerGuest
Your inability to comprehend simple arithmetic is astounding. Perhaps
you should consult a mushroom, which obviously has a higher "interlect"
than yourself.
Arnold P. Furshlugginer, Apr 23, 2005 | 841 | 2,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.972549 |
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https://bedtimemath.org/fun-math-teeth-in-world/ | 1,725,777,997,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00200.warc.gz | 119,016,963 | 20,972 | # The Toothy Truth
How many teeth are out there in the world? Well, of the 8 billion people out there, about 7 billion have their final 28 teeth, or 32 with wisdom teeth. Let’s call it 30 – that gives us 210 billion teeth. Then if about 1 billion kids still growing teeth have 10 each, that’s another 10 billion. But what about all our toothy animal friends – dogs, cats, beavers, raccoons…and sharks, who have 50-100 teeth at any time. There are trillions of teeth out there – and probably not enough toothbrushes!
Wee ones: Look in a mirror. How many teeth can you see? Count as many as you can!
Little kids: If you have 4 teeth on top and 4 teeth on the bottom facing them, how many teeth do you have? Bonus: If you then lose 2 teeth, grow 3 more teeth, lose another tooth, and then grow 2 more, how many teeth do you have now?
Big kids: If there are 210 billion grown-up teeth out there and 10 billion kid teeth, how many teeth is that? Bonus: If you start with 2 teeth and end up with 32 teeth, how many times does your number of teeth double?
The sky’s the limit: If at a party every kid has 10 teeth and every grown-up has 28, how many of each must you have if there are 114 teeth in the room?
Wee ones: Different for everyone…see how many you can count!
Little kids: 8 teeth. Bonus: 10 teeth.
Big kids: 220 billion teeth. Bonus: 4 times: to 4 teeth, then 8, then 16, then 32.
The sky’s the limit: 3 kids and 3 grown-ups. Since the numbers ends in 4, and since kid teeth don’t affect the last digit (we keep adding 10), we need 3 grown-ups to give us 84 grown-up teeth. Then we need 3 more kids to add the 30 to get us to 114.
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how much cement in 1m3 of concrete - Grinding Mill China
Typically, 1m3 of concrete is made up of 350Kg of cement, 700Kg of sand, 1,200Kg of chippings and 150 Litres of water … » More detailed how much cement for 1m3 m20 – Gulin Mining
how to calculate 1m3 of concrete for m20 m25 m30
How to Calculate Quantities of Cement, Sand and ... - Happho. Watch video· A simple yet more accurate method (DLBD method) of calculating cement, sand and aggregate for Nominal Concrete mix M15, M20, M25 and M30 grade Concrete A simple yet more accurate method (DLBD method) of calculating cement, sand and aggregate for Nominal Concrete mix M15, M20, M25 and M30 grade Concrete
how we can calculate weight of cement in 1m3 m75 concrete
weight of 1m3 of gravel aggregate and water for 1m3 concrete? how we can calculate weight of cement in 1m3 water for 1m3 ... 2012· m30 grade concrete 1m3 weight ... pre: mining iron ore plant by zenith
Concrete Mix Design Calculation - M20, M25, M30 ...
Concrete Mix Design Calculation for M20, M25, M30 Concrete with Procedure & Example Concrete mix design is the process of finding right proportions of cement, sand and aggregates for concrete to achieve target strength in structures.
What is the procedure to calculate cement bags in 1 cubic ...
The water cement ratio = 0.45Now we willcalculate the volume of concretethat can be produced with one bag of cement (i.e. 50 kg cement) for the mass proportions of concrete materials.Thus, theabsolute volume of concretefor 50 kg ofcement =[image]Thus, for the proportion of mix considered, with on3 bag of cement of 50 kg, 0.1345 m3 of concrete ...
m20 mix ratio how to use cement in 1m3 - refloresta-bahia.org
Aggregate Replacement in Concrete in Indian Context ... study concerns mainly on the possible use of stone waste in construction .... aggregates in concrete mix. .... proportion is done in Table 6. ... aggregate for 1m3 M20 grades of concrete.
how do we calculate quantity of cement, sand and ...
how do we calculate quantity of cement, sand and aggregates in 1 m3 of M30 grade concrete ?.. Answer / subilash .s(civil engineer) for designing 1m3 concrete of M25 mix the ratio is 1:1:2
CONCRETE GRADE: M5 = 1:4:8 M10= 1:3:6 M15= 1:2:4 M20= 1:1 ...
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Quantities of Materials Per Cubic Meter of Concrete Mix ...
Quantity of materials such as cement, sand, coarse aggregates and water required per cubic meter of concrete and mortar for different mix proportions varies with the mix design of the concrete …
What is the cement content for 1m3 of M20,M25 concrete?
What is the cement content for 1m3 of M20,M25 concrete?.. Answer / mohammedazharuddin. QUANTITY OF CEMENT IN 1CUM FOR M20,M25 IS 320KG AND 340KG ... M10-3.4nos of bags of cement M15-6.2 M20-6.8 M25-7.5 M30-8.5 M35-9 M40-10. Is This Answer Correct ? 1 Yes : 6 No : Post New Answer. More Civil Engineering Interview Questions.
cement requrement for 1m3 m30 concreat - educationcare.in
Concrete Mix Design As Per Indian Standard Code Concrete Mix Design ... mixes into a number of grades as M10, M15, M20, M25, M30, M35 and M40. ... smaller is the cement requirement for a particular water-cement ratio, ... volume of concrete = gross volume (1m3 ) minus the volume of entrapped air Sc...
material quantity for 1m3 of concrete for m20 grade ...
Posts Related to material quantity for 1m3 of concrete for m20 grade » quantity of cement,sand,coarse aggregate use in design mix m15,m20,m25,m30 grade concrete » what is the quantity of aggregate and cement in 40 grade concrete » how much quantity cement should be used in m10 grade concrete
how we calculate of Sand, cement and aggregate of M20 ...
M20 (1 cement :1.5 sand :3 stone/brick aggregate). To determine the proportions you have to perform mix design, for this you have to find out the sp.gr. of cement, CA, FA, and water cement ratio ...
water for 1m3 concrete for m30 grade - geetaschool.in
aggregate required for 1m3 concrete m30 - gyptech.in. water for 1m3 concrete for m30 grade - Hence 200 Litres of Water is required for 1m3 of M20 sand and aggregates in 1 m3 of M30 grade concrete ratio of cement [More info] calculate the rates for 1m3 concrete grade 20.
Quantity and Rate Analysis for Reinforced Concrete ...
The Quantity of materials like sand, cement and coarse aggregates vary with mix design such as M15 (1:2:4), M20 (1:1.5:3), M25, M30 etc.. Here we will see the rate analysis for 1m 3 of reinforced concrete.
m30 grade concrete 1m3 weight - University Courses
m30 grade how much sand and aggregate : 4.7/5 · m30 grade concrete 1m3 weight BINQ : 4.8/5 · To Calculate Quantity Of Sand And to calculate quantity of sand and aggregate in m30 quantity of aggregate required for m20 concrete cement sand aggregate required for 1m3 concrete …
how to calculate 1m3 of concrete for m20 m25 m30
m30 grade concrete 1m3 weight - BINQ Mining. Nov 13, 2012· how do we calculate quantity of cement, sand and aggregates in … for designing 1m3 concrete of M25 mix the ratio is … 1 m3 of M30 grade concrete ...
-cement consumption m15 grade concrete 1m3-
Posts Related to material quantity for 1m3 of concrete for m20 grade ... » consumption of cement in m30 grade concrete » how to calculate cement in concrete m15.
SUSTAINABLE USE OF RECYCLED AGGREGATE IN CONCRETE - …
For this investigation, the concrete Grade M25, M30 &M35 for the samples was used. The detailed mix designs of different grads of concrete are given below. Table 1.Mix proportion for 1m3 M25 Concrete W/C ratio Water Cement Sand Aggregate 0.41 231.56 kg 480.87 kg 645.87 kg 1161.44 kg ...
M-30 Mix Designs as per IS-10262-2009 - Civil Engineering
Following table shows the M-30 Mix Designs as per IS-10262-2009, hope this helps all civil engineers here. M-30 CONCRETE MIX DESIGN . As per IS 10262-2009 & MORT&H . A-1. ... My design mix m30 cement-390kg 20mm – 566 kg 10mm – 488 kg corse sand – 757 kg water – 217 kg
Civil Work - CONCRETE GRADE: M5 = 1:4:8 M10= 1:3:6 M15 ...
See more of Civil Work on Facebook. Log In. or. Create New Account. See more of Civil Work on Facebook. Log In. ... M30 ( 1 : 2 : 2.87) Cement : 380 Kg/ M 3 20 mm Jelly : 654 Kg/ M 3 12.5 mm Jelly : 436 Kg/ M 3 ... 1m3 Conrete = 0.9 m3 Jelly + 0.55 m3 Sand + 0.225 m3 BRICK: Weight = 3.17 - …
How to Calculate Quantities of Cement, Sand and ... - Happho
With constant research and development in the field of cement technology and its manufacturing process,a M20 mix of "1:1.5:3″(by volume) would be too rich,over engineered and uneconomical (~7.5 bags of cement per cum) and will ultimately result into a M30 concrete and above, the reason being latest generation of 53 grade OPC cement is ...
Concrete Mix Design As Per Indian Standard Code
Concrete Mix Design. Introduction. The process of selecting suitable ingredients of concrete and determining their relative amounts with the objective of producing a concrete of the required, strength, durability, and workability as economically as possible, is termed the concrete mix design.
weight of one wet cubic meter m30 concrete « BINQ Mining
BINQ Mining > Ore Process > weight of one wet cubic meter m30 concrete; Print. weight of one wet cubic meter m30 concrete. Posted at:June 6, 2013[ 4.9 - 2469 Ratings] What is the unit weight of concrete – Wiki Answers. ... Typically, 1m3 of concrete is made up of 350Kg of cement, ...
weight of 1 m3 of m25 concrete - alphaom.co.in
in concrete grade m30 how much cement required - GV… How much cubic meter agg required for m25 grade concrete for 40 cement bags Products. As a leading ... m30 grade how much sand and aggregate for 1m3 – … | 2,110 | 7,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-22 | longest | en | 0.842435 |
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Physics 505 Fall 2005 Homework Assignment #7 — Solutions Textbook problems: Ch. 4: 4.10 Ch. 5: 5.3, 5.6, 5.7 4.10 Two concentric conducting spheres of inner and outer radii a and b , respectively, carry charges ± Q . The empty space between the spheres is half-filled by a hemispherical shell of dielectric (of dielectric constant ±/± 0 0, as shown in the figure. a Q - Q + b a ) Find the electric field everywhere between the spheres. This is a somewhat curious problem. It should be obvious that without any dielectric the electric field between the spheres would be radial ~ E = Q 4 π± 0 ˆ r r 2 We cannot expect this to be unmodified by the dielectric. However, we note that the radial electric field is tangential to the interface between the dielectric and empty region. Thus the tangential matching condition E k 1 = E k 2 is automatically satisfied. At the same time there is no perpendicular component to the interface, so there is nothing to worry about for the D 1 = D 2 matching condition. This suggests that we guess a solution of the radial form ~ E = A ˆ r r 2 where A is a constant to be determined. This guess is perhaps not completely obvious because one may have expected the field lines to bend into or out of the dielectric region. However, we could also recall that parallel fields do not get bent across the dielectric interface. We may use the integral form of Gauss’ law in a medium to determine the above constant A I ~ D · ˆ n da = Q ± 0 A r 2 (2 πr 2 ) + ±A r 2 (2 πr 2 ) = Q
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or A = Q/ 2 π ( ± + ± 0 ). Hence ~ E = Q 2 π ( ± + ± 0 ) ˆ r r 2 Note that 1 2 ( ± + ± 0 ) may be viewed as the average permittivity in the volume between the spheres. b ) Calculate the surface-charge distribution on the inner sphere. The surface-charge density is given by σ = D ± ± r = a where either D = ± 0 E or D = ±E depending on region. This gives σ = ± ± + ± 0 Q 2 πa 2 ; dielectric side ± 0 ± + ± 0 Q 2 πa 2 ; empty side (1) Note that the total charge obtained by integrating σ over the surface of the inner sphere gives Q as expected. c ) Calculate the polarization-charge density induced on the surface of the dielectric at r = a . The polarization charge density is given by ρ pol = -∇· ~ P where ~ P = ± 0 χ e ~ E = ( ± - ± 0 ) ~ E . Since the surface of the dielectric at r = a is against the inner sphere, we can take the polarization to be zero inside the metal (‘outside’ the dielectric). Gauss’ law in this case gives σ pol = - P ± ± r = a = - ( ± - ± 0 ) E ± ± r = a = - ± - ± 0 ± + ± 0 Q 2 πa 2 Note that when this is combined with (1), the total (free and polarization) charge
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This note was uploaded on 11/21/2011 for the course PHY 505 taught by Professor Stephens,p during the Fall '08 term at SUNY Stony Brook.
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hw07a - Physics 505 Homework Assignment #7 Solutions...
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Feb21 awarded Yearling Dec21 revised Show that $\left| \mathbb{P}(X=m)-\mathbb{P}(Y=m) \right| \le \mathbb{P}(Y\ne X)$ edited the title and the probability notation Dec21 suggested approved edit on Show that $\left| \mathbb{P}(X=m)-\mathbb{P}(Y=m) \right| \le \mathbb{P}(Y\ne X)$ Dec20 revised Does $\theta(n)$ = $1/x$ make any sense? edited the notation Dec20 suggested approved edit on Does $\theta(n)$ = $1/x$ make any sense? Dec20 awarded Constituent Dec15 awarded Caucus Oct2 revised How to prove the eigenvalues of tridiagonal matrix? corrected spelling Oct2 suggested approved edit on How to prove the eigenvalues of tridiagonal matrix? Sep30 awarded Explainer Jul2 awarded Curious Feb21 awarded Yearling Dec30 comment Evaluate $\sum_{n=1}^{\infty} \frac{\ln n}{(n+1)!}$ You may use Stirling's approximation. Dec28 asked Sources on Inverse Spectral Theory Dec6 comment Sufficient statistic for uniform distribution Check the Fisher–Neyman factorization theorem. Nov19 revised Show $f(x) = (x^4+x^2+1)/(x^3+1)$ is $O(x)$ added 1 characters in body Nov19 revised Show $f(x) = (x^4+x^2+1)/(x^3+1)$ is $O(x)$ deleted 1 characters in body Nov18 revised Show $f(x) = (x^4+x^2+1)/(x^3+1)$ is $O(x)$ added 50 characters in body Nov18 answered Show $f(x) = (x^4+x^2+1)/(x^3+1)$ is $O(x)$ Nov16 awarded Informed | 446 | 1,421 | {"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} | 3.421875 | 3 | CC-MAIN-2015-18 | longest | en | 0.762551 |
https://stackoverflow.com/questions/6921348/can-anyone-perhaps-teach-me-how-to-further-optimize-this-print-up-to-the-nth-pr/6927027 | 1,603,384,122,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879673.14/warc/CC-MAIN-20201022141106-20201022171106-00643.warc.gz | 540,463,179 | 30,994 | Can anyone perhaps teach me how to further optimize this 'print up to the nth prime number' script? [closed]
I'm a 17 year old getting started with programming with the help of the Python programming language.
I've been seeking to optimize this algorithm, perhaps by eliminating one of the loops, or with a better test to check for prime numbers.
Trying to calculate and display 100000 prime numbers has the script pausing for about 6 seconds as it populates the list with primes before the primes list is returned to the console as output.
I've been experimenting with using
``````print odd,
``````
to simply print every found prime number, which is faster for smaller inputs like n = 1000, but for n = 1000000 the list itself prints much faster (both in the python shell and in the console).
Perhaps the entire code/algorithm should be revamped, but the script should remain essentially the same: The user types in the number of prime numbers to be printed (n) and the script returns all prime numbers up to the nth prime number.
``````from time import time
odd = 1
primes = [2]
n = input("Number of prime numbers to print: ")
clock = time()
def isPrime(number):
global primes
for i in primes:
if i*i > number:
return True
if number%i is 0:
return False
while len(primes) < n:
odd += 2
if isPrime(odd):
primes += [odd]
print primes
clock -= time()
print "\n", -clock
raw_input()
``````
I might wanna rewrite the whole script to use a sieve like the Sieve of Atkin: http://en.wikipedia.org/wiki/Sieve_of_Atkin
However, I am simply a beginner at Python (or even at programming: I started writing code only 2 weeks ago) and it would be quite a challenge for me to figure out how to code a Sieve of Atkin algorithm in Python.
I wish a google hacker out there would hand hold me through stuff like this :(
• This is a great question, but I think it's better suited at codereview.stackexchange.com. Stack Overflow is mostly for specific programming questions that have definitive answers. – templatetypedef Aug 3 '11 at 3:38
You could use prime sieve, and with a simple twist:
1. Define the first prime 2 as you do, set the largest number reached (`max`) to 2;
2. Generate a list of `n` consecutive numbers from `max+1` to `max+n`;
3. Use sieve with the primes on this list. When sieving, set the beginning number for each prime to the smallest number in the list that could be divided by the prime;
4. If the amount is not reacher, goto 2.
This way, you could control the length of the list, and as the length grows larger, the speed will be faster. However, this is a total rework of the algorithm, and is harder to program.
Here's a sample code, which is quite crude, but this only takes less than 70% time of the original:
``````from math import sqrt
from time import time
primes = [2]
max = 3
n = input("Number of prime numbers to print: ")
r=2
clock = time()
def sieve(r):
global primes
global max
s = set(range(max,max+r))
for i in primes:
b=max//i
if (b*i<max):
b=b+1
b=b*i
while b<=max+r-1:
if b in s:
s.remove(b)
b=b+i
for i in s:
primes.append(i)
while len(primes) < n:
r=primes[-1]
sieve(r)
max=max+r
primes=primes[0:n]
print primes
clock -= time()
print "\n", -clock
raw_input()
``````
There are many ways to improve this, this just shows the notion of the approach.
Also, this can blow up the memory when the number is large. I used the dynamic limit try to somewhat relieve this.
And if you are really curious (and fearless), you could look at the more complicated implementations in various open source projects. One example is Pari/GP, which is written in C++, and is blazing fast (I tested 1 to 50000000 in less than 1 min, if I remember correctly). Translating them to Python may be hard, but will be helpful, perhaps not just for yourself;-)
One simple optimizations which could be applied without hacking the code completely.
• the i*i on every prime gets very wasteful as the list gets longer. Instead calculate the square root of i outside the loop and test against this value inside the loop.
However square root is itself and expensive calculation and the majority of candidate numbers will be rejected as divisible by one of the lower primes (3,5,7) so this turns out to be not such a good optimization (pessimization?). But we don't actually need to be that precise and a simple check that the prime is less than one third of the value has a similar effect without the computational cost of the square root calculation, but, at the expense of a relatively few unnecessary test.
• I just tried calculating sqrt(number) outside the loop and then testing the elements in primes[] against sqrt(number), but my script is still as slow. If only there was a way to get rid of that nasty pause after entering a large value for n. – Sweetgirl17 Aug 3 '11 at 4:06
As was already said by Ziyao Wei I'd also try a Sieve implementation. The only thing I'd improve is to use the Prime number theorem as a starting point for the used size.
Computing the inverse function isn't straightforward in pure python, but an iterative approach should be good enough and that way you could get a pretty good idea how large the sieve would have to be. Since I don't really remember the proofs for the theorem in detail and it's 6am in the morning here, someone else will have to chip in to say if the theorem guarantees any certain upper boundary that could be used to allow using the simple sieve without having to worry about growing it. iirc that's sadly not the case.
As already mentioned, the presented algorithm cannot be improved significantly. If a fast solution is requested then the Eratosthenes sieve is appropriate. The size `x` of the sieve can be estimated using `n >= x/(ln x + 2)` if `x >= 55`. This equation can be solved using the Newton's iteration. The presented algorithm is about 10 times faster the original:
``````def sieveSize(n):
# computes x such that pi(x) >= n (assumes x >= 55)
x = 1.5 * n # start
y = x - n * math.log(x) - 2 * n
while abs(y) > 0.1:
derivative = 1 - n/x
x = x - y / derivative
y = x - n * math.log(x) - 2 * n
return int(x) + 1
def eratosthenes(n):
# create a string flags: flags[i]=='1' iff i prime
size = sieveSize(n)
flags = ['1'] * size # start with: all numbers are prime
flags[0] = flags[1] = '0' # 0 and 1 are not primes
i = 0
while i * i < size:
if flags[i] == '1':
for j in range(i * i, size, i):
flags[j] = '0'
i += 1
return flags
def primes(n):
flags = eratosthenes(n)
prims = []
for i in range(0, len(flags)):
if flags[i] == '1':
prims.append(i)
return prims
prims = primes(100000)
``````
Any number that ends in 5, other than 5, is not a prime. So you can put a statement that skips any number ending in 5 that is greater than 5.
• That requires converting the number to a decimal representation, which will take far more time than it saves. The naive primality testing algorithm will stop as soon as it divides the number by 5, but converting it to decimal will keep dividing the number mod 10 until all decimal digits have been determined. – user57368 Aug 3 '11 at 6:08 | 1,801 | 7,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.912035 |
https://newpathworksheets.com/math/grade-1/using-number-line/missouri-common-core-standards | 1,569,264,318,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514577478.95/warc/CC-MAIN-20190923172009-20190923194009-00479.warc.gz | 598,906,728 | 8,460 | ## ◂Math Worksheets and Study Guides First Grade. Using Number Line
### The resources above correspond to the standards listed below:
#### Missouri Common Core Standards
MO.CC.NBT.1. Number and Operations in Base Ten
Extend the counting sequence.
NBT.1.1. Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral.
Understand place value.
NBT.1.3. Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <.
Use place value understanding and properties of operations to add and subtract.
NBT.1.4. Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten.
MO.CC.OA.1. Operations and Algebraic Thinking | 265 | 1,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-39 | latest | en | 0.870439 |
https://www.hindawi.com/journals/amp/2014/694580/ | 1,653,308,447,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00604.warc.gz | 934,774,403 | 142,453 | / / Article
Research Article | Open Access
Volume 2014 |Article ID 694580 | https://doi.org/10.1155/2014/694580
Ayşe Betül Koç, Musa Çakmak, Aydın Kurnaz, "A Matrix Method Based on the Fibonacci Polynomials to the Generalized Pantograph Equations with Functional Arguments", Advances in Mathematical Physics, vol. 2014, Article ID 694580, 5 pages, 2014. https://doi.org/10.1155/2014/694580
# A Matrix Method Based on the Fibonacci Polynomials to the Generalized Pantograph Equations with Functional Arguments
Accepted31 Jul 2014
Published13 Aug 2014
#### Abstract
A pseudospectral method based on the Fibonacci operational matrix is proposed to solve generalized pantograph equations with linear functional arguments. By using this method, approximate solutions of the problems are easily obtained in form of the truncated Fibonacci series. Some illustrative examples are given to verify the efficiency and effectiveness of the proposed method. Then, the numerical results are compared with other methods.
#### 1. Introduction
Many phenomena in applied branches that fail to be modeled by the ordinary differential equations can be described by the delay differential equations. Many researchers have studied different applications of those equations in variety of applied sciences such as biology, physics, economy, and electrodynamics (see [14]). Pantograph equations with proportional delays play an important role in this context. The existence and uniqueness of the analytic solutions of the multipantograph equation are investigated in [5]. A numerical approach to multipantograph equations with variable coefficients is also studied in [6]. An extension of the multipantograph equation is known to be the generalized pantograph equation with functional arguments defined as under the mixed conditions where proportional delay-, constant delay-, , , and are real and/or complex coefficients, and the coefficients of th order unknown function- and known are the analytical functions defined in the interval .
In recent years, many researchers have developed different numerical approaches to the generalized pantograph equations as variational iteration method [7], differential transform approach [8], Taylor method [9], collocation method based on Bernoulli matrix [10], and Bessel collocation method [11]. In this study, we investigate a collocation method based on the Fibonacci polynomial operational matrix for the numerical solution of the generalized pantograph equation (1). Even the Fibonacci numbers have been known for a long time; the Fibonacci polynomials are very recently defined to be an important agent in the world of polynomials [12, 13]. Compared to the methods of the orthogonal polynomials, the Fibonacci approach has proved to give more precise and reliable results in the solution of differential equations [14].
This study is organized as follows. In the second part, a short review of the Fibonacci polynomials is presented. A Fibonacci operational matrix for the solution of the pantograph equation is developed in Section 3. Some numerical examples are given in Section 4 to illustrate efficiency and effectiveness of the method.
#### 2. Operational Matrices of the Fibonacci Polynomials
The Fibonacci polynomials are determined by following general formula [12, 13]: with and . Now, we will mention some matrix relations in terms of Fibonacci polynomials.
##### 2.1. Fibonacci Series Expansions
To obtain an expansion form of the analytic solution of the pantograph equation, we use the Fibonacci collocation method as follows.
Suppose that (1) has a continuous function solution that can be expressed in the Fibonacci polynomials Then, a truncated expansion of -Fibonacci polynomials can be written in the vector form where the Fibonacci row vector and the unknown Fibonacci coefficients column vector are given, respectively, by
##### 2.2. Matrix Relations of the Derivatives
The th order derivative of (5) can be written as where , , and is the coefficient vector of the polynomial approximation of th order derivative. Then, there exists a relation between the Fibonacci coefficients as where is operational matrix for the derivative defined by [14] Making use of (7) and (9) yields
#### 3. Solution Procedure for the Pantograph Differential Equations
Let us recall the th order linear pantograph differential equation, The first step in the solution procedure is to define the collocation points in the domain, so that Then, collocating problem (12) at the points in (13) yields The system (14) can, alternatively, be rewritten in the matrix form where Therefore, the th order derivative of the unknown function at the collocation points can be written in the matrix form as or, equivalently, To express the functional terms of (1) as in the form (5), let we put instead of in the relation (18) and then obtain or where are Fibonacci operational matrices corresponding to the coefficients . Therefore, replacing (18) and (20) in (15) gives the fundamental matrix equation for the problem (12) as which corresponds to a system of algebraic equations for the unknown Fibonacci coefficients , . In other words, when we denote the expression in the sum by , for and , we get Thus, the augmented matrix of (22) becomes
On the other hand, in view of (11), the conditions (2) can be taken into account by forming the following matrix equation: where Therefore, the augmented matrix of the specified conditions is Consequently, (23) together with (26) can be written in the new augmented matrix form This form can also be achieved by replacing some rows of the matrix (23) by the rows of (26) or adding those rows to the matrix (23) provided that . Finally, the vector (thereby vector of the coefficients ) is determined by applying some numerical methods designed especially to solve the system of linear equations. On the other hand, when the singular case appears, the least square methods are inevitably available to reach the best possible approximation. Therefore, the approximated solution can be obtained. This would be the Fibonacci series expansion of the solution to the problem (12) with the specified conditions.
##### 3.1. Accuracy of the Results
We can, now, proceed with a short accuracy analysis of the problem in a similar way to [18]. As the truncated Fibonacci series expansion is an approximate solution of (1) with (2), it must satisfy the following equality for , : or When ( is any integer) is prescribed, the truncation limit is increased until the difference at each of the collocation points becomes smaller than the desired value .
#### 4. Numerical Results
In this part, three illustrative examples are given in order to clarify the findings of the previous section. The errors of the proposed method are compared with those of the errors that occurred in the solutions by some other methods in Tables 13 for two sample examples. It is noted here that the number of collocation points in the examples is indicated by the capital letter .
Present method Exponential approach [15] Taylor polynomial approach [16] 0.2 0.4 0.6 0.8 1.0
Present method Taylor polynomial approach [16] Taylor method [9] 0.2 0 0.4 0 0.6 0 0.8 1.0
Present method Taylor matrix method [6] Boubaker matrix method [17] 0.2 0.4 0.6 0.8 1.0
Example 1 (see [10, 11]). Consider the following linear pantograph type problem equation: with the initial conditions
The exact solution of this problem is known to be . When the solution procedure in Section 3 is applied to the problem, the solution of the linear algebraic system gives the numerical approximation of the solution to the problem. It is noteworthy that the method reaches the exact solution even for .
Example 2 (see [9, 15, 16]). Now, consider the following equation with variable coefficient given by and the condition
The exact solution is also known to be . A comparison of the absolute errors of the proposed approach Taylor method [16] and the exponential approach [15] is given in Table 1 for . Another comparison of the present method with the methods of Taylor polynomials [9, 16] is also given for and in Table 2. These results show that the Fibonacci approach has better accuracy, at least one decimal place, than the other methods.
Example 3 (see [6, 17]). Finally, let us consider the pantograph equation with variable coefficients and the condition which has the exact solution . Computed results are compared with the results of the Taylor [6] and Boubaker [17] matrix methods in Table 3.
#### Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
#### Acknowledgments
This study was supported by Research Projects Center (BAP) of Selcuk University. The authors would like to thank Selcuk University and TUBITAK for their support. Also, the authors would like to thank the editor and referees for their valuable comments and remarks which led to a great improvement of the article. They have denoted here that a minor part of this study was presented orally at the “2nd International Eurasian Conference on Mathematical Sciences and Applications (IECMSA-2013),” Sarajevo, August, 2013.
#### References
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3. M. Dehghan and F. Shakeri, “The use of the decomposition procedure of Adomian for solving a delay differential equation arising in electrodynamics,” Physica Scripta, vol. 78, no. 6, Article ID 065004, 2008.
4. J. R. Ockendon and A. B. Tayler, “The dynamics of a current collection system for an electric locomotive,” Proceedings of the Royal Society of London A, vol. 322, pp. 447–468, 1971. View at: Publisher Site | Google Scholar
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6. M. Sezer, S. Yalcinbas, and N. Sahin, “Approximate solution of multi pantograph equation with variable coefficients,” Journal of Computational and Applied Mathematics, vol. 214, no. 2, pp. 406–416, 2008. View at: Publisher Site | Google Scholar
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8. Y. Keskin, A. Kurnaz, M. E. Kiris, and G. Oturanc, “Approximate solutions of generalized pantograph equations by the differential transform method,” International Journal of Nonlinear Sciences and Numerical Simulation, vol. 8, no. 2, pp. 159–164, 2007. View at: Google Scholar
9. M. Sezer and A. Akyüz-Daşcıoğlu, “A Taylor method for numerical solution of generalized pantograph equations with linear functional argument,” Journal of Computational and Applied Mathematics, vol. 200, no. 1, pp. 217–225, 2007. View at: Publisher Site | Google Scholar | MathSciNet
10. E. Tohidi, A. H. Bhrawy, and K. Erfani, “A collocation method based on Bernoulli operational matrix for numerical solution of generalized pantograph equation,” Applied Mathematical Modelling. Simulation and Computation for Engineering and Environmental Systems, vol. 37, no. 6, pp. 4283–4294, 2013.
11. S. Yuzbası, N. Sahin, and M. Sezer, “A Bessel collocation method for numerical solution of generalized pantograph equations,” Numerical Methods for Partial Differential Equations, vol. 28, no. 4, pp. 1105–1123, 2011. View at: Publisher Site | Google Scholar | MathSciNet
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16. M. Sezer, S. Yalçınbas, and M. Gülsu, “A Taylor polynomial approach for solving generalized pantograph equations with nonhomogenous term,” International Journal of Computer Mathematics, vol. 85, no. 7, pp. 1055–1063, 2008. View at: Publisher Site | Google Scholar | MathSciNet
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Copyright © 2014 Ayşe Betül Koç et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. | 3,246 | 13,810 | {"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} | 2.59375 | 3 | CC-MAIN-2022-21 | longest | en | 0.889122 |
https://ericmjl.github.io/blog/2017/2/8/numba-my-first-attempt-at-being-serious-with-it/ | 1,632,201,222,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00158.warc.gz | 291,345,235 | 5,033 | # Numba: My first attempt at being serious with it
written by on 2017-02-08 | tags: numba open source data science optimization coding snippets
This evening, I saw a Tweet about using numba, and I thought, it's about time I give it a proper shot. I had been solving some dynamic programming problems just for fun, and I thought this would be a good test case for numba's capabilities.
The DP problem I was trying to solve was that of collecting apples on a grid. Here's how the problem is posed:
I have a number of apples distributed randomly on a grid. I start at the top-left hand corner, and I'm only allowed to move downwards or to the right. Along the way, I pick up apples. What's the maximum number of apples I can pick up along the way?
This is a classic 2-dimensional DP problem. I simulated some random integers:
n = 200
arr = np.random.randint(low=0, high=100, size=n**2).reshape(n, n)
I then wrote out my solution, and wrapped it in two versions of the function call: one native and one numba-JIT'd.
# Let's collect apples.
from numba import jit
@jit
def collect_apples(arr):
sum_apples = np.zeros(shape=arr.shape)
for row in range(arr.shape[0]):
for col in range(arr.shape[1]):
if col != 0:
val_left = arr[row, col - 1]
else:
val_left = 0
if row != 0:
val_up = arr[row - 1, col]
else:
val_up = 0
sum_apples[row, col] = arr[row, col] + max(val_left, val_up)
return sum_apples
def collect_apples_nonjit(arr):
sum_apples = np.zeros(shape=arr.shape)
for row in range(arr.shape[0]):
for col in range(arr.shape[1]):
if col != 0:
val_left = arr[row, col - 1]
else:
val_left = 0
if row != 0:
val_up = arr[row - 1, col]
else:
val_up = 0
sum_apples[row, col] = arr[row, col] + max(val_left, val_up)
return sum_apples
Here's the performance results:
%%timeit
collect_apples(arr)
The slowest run took 4.27 times longer than the fastest. This could mean that an intermediate result is being cached. 10000 loops, best of 3: 99.7 µs per loop
%%timeit
collect_apples_nonjit(arr)
10 loops, best of 3: 50.3 ms per loop
Wow! Over 500-fold speedup! All obtained for free using the @jit decorator. | 592 | 2,110 | {"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": 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} | 3.40625 | 3 | CC-MAIN-2021-39 | longest | en | 0.884374 |
https://goprep.co/in-figure-5-31-ab-ef-find-x-and-y-i-1nkw34 | 1,618,529,405,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088264.43/warc/CC-MAIN-20210415222106-20210416012106-00165.warc.gz | 355,198,290 | 26,968 | In figure 5.31 AB
Given:
AB || EF
Construction.
Draw a line parallel to both AB and EF from the meeting point of their transverse
Theory:
If 2 lines are parallel then alternate interior angles are equal
As constructed line will divide y into 2 different parts
As the constructed line is parallel to AB
By interior alternate angles
The upper part of y = 125°
As the constructed line is parallel to EF
By interior alternate angles
The lower part of y = 141°
Combining both parts of y we get
y = 125° + 141° = 266°
As y and x make complete circle
y + x = 360°
266° + x = 360°
x = 360° - 266° = 94°
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view all courses | 324 | 1,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-17 | latest | en | 0.848471 |
https://studylib.net/doc/8117717/section-6.7-hyperbolic-functions | 1,618,856,226,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038916163.70/warc/CC-MAIN-20210419173508-20210419203508-00237.warc.gz | 656,248,437 | 13,489 | # Section 6.7 Hyperbolic Functions
```Section 6.7
Difference Equations
to
Differential Equations
Hyperbolic Functions
The final class of functions we will consider are the hyperbolic functions. In a sense these
functions are not new to us since they may all be expressed in terms of the exponential
function and its inverse, the natural logarithm function. However, we will see that they
have many interesting and useful properties.
Definition
by
For any real number x, the hyperbolic sine of x, denoted sinh(x), is defined
1 x
(e − e−x )
2
and the hyperbolic cosine of x, denoted cosh(x), is defined by
sinh(x) =
cosh(x) =
1 x
(e + e−x ).
2
(6.7.1)
(6.7.2)
Note that, for any real number t,
1 t
1
(e + e−t )2 − (et − e−t )2
4
4
1 2t
1
= (e + 2et e−t + e−2t ) − (e2t − 2et e−t + e−2t )
4
4
1
= (2 + 2)
4
= 1.
cosh2 (t) − sinh2 (t) =
Thus we have the useful identity
cosh2 (t) − sinh2 (t) = 1
(6.7.3)
for any real number t. Put another way, (cosh(t), sinh(t)) is a point on the hyperbola
x2 −y 2 = 1. Hence we see an analogy between the hyperbolic cosine and sine functions and
the cosine and sine functions: Whereas (cos(t), sin(t)) is a point on the circle x2 + y 2 = 1,
(cosh(t), sinh(t)) is a point on the hyperbola x2 − y 2 = 1. In fact, the cosine and sine
functions are sometimes referred to as the circular cosine and sine functions. We shall see
many more similarities between the hyperbolic trigonometric functions and their circular
counterparts as we proceed with our discussion.
To understand the graphs of the hyperbolic sine and cosine functions, we first note
that, for any value of x,
sinh(−x) =
1 −x
(e − ex ) = − sinh(x),
2
1
(6.7.4)
c by Dan Sloughter 2000
Hyperbolic Functions
Section 6.7
10
7.5
5
2.5
-4
-2
2
4
-2.5
-5
-7.5
-10
Figure 6.7.1 Graph of y = sinh(x)
and
cosh(−x) =
1 −x
(e + ex ) = cosh(x).
2
(6.7.5)
Now for large values of x, e−x ≈ 0, in which case
sinh(x) =
and
1 x
1
(e − e−x ) ≈= ex
2
2
1
sinh(−x) = − sinh(x) ≈ − ex .
2
Thus the graph of y = sinh(x) appears as in Figure 6.7.1. Similarly, for large values of x,
cosh(x) =
1 x
1
(e + e−x ) ≈ ex
2
2
and
cosh(−x) = cosh(x) ≈
1 x
e .
2
The graph of y = cosh(x) is shown in Figure 6.7.2.
The derivatives of the hyperbolic sine and cosine functions follow immediately from
their definitions. Namely,
d 1 x
1
d
sinh(x) =
(e − e−x ) = (ex + e−x ) = cosh(x)
dx
dx 2
2
and
d
d 1 x
1
cosh(x) =
(e + e−x ) = (ex − e−x ) = sinh(x).
dx
dx 2
2
Here again we see similarities between the circular and hyperbolic sine and cosine functions.
Section 6.7
Hyperbolic Functions
3
10
8
6
4
2
-4
-2
2
4
Figure 6.7.2 Graph of y = cosh(x)
Proposition
d
sinh(x) = cosh(x).
dx
d
cosh(x) = sinh(x).
dx
As a consequence of this proposition, we also have
Z
sinh(x)dx = cosh(x) + c
and
Using the chain rule, we have
d
d
sinh2 (3x) = 2 sinh(3x) sinh(3x) = 6 sinh(3x) cosh(3x).
dx
dx
Example
Using the chain and product rules, we have
d
sinh(2x) cosh(2x) = sinh(2x)(2 sinh(2x)) + cosh(2x)(2 cosh(2x))
dx
= 2 sinh2 (2x) + 2 cosh2 (2x).
Example
(6.7.7)
(6.7.8)
Z
cosh(x)dx = sinh(x) + c.
Example
(6.7.6)
Analogous to
Z
1
sin(3x)dx = − cos(3x) + c,
3
(6.7.9)
4
Hyperbolic Functions
Section 6.7
4
2
-10
-5
5
10
-2
-4
Figure 6.7.3 Graph of y = sinh−1 (x)
we have
Z
sinh(3x)dx =
Example
It is tempting to evaluate
Z
1
cosh(3x) + c.
3
e−x sinh(x)dx
using integration by parts in the same manner that we would evaluate
Z
e−x sin(x)dx.
However, this integral is much easier if we notice that
1
1 x
−x
−x
−x
e sinh(x) = e
(e − e ) = (1 − e−2x ).
2
2
Hence
Z
−x
e
1
sinh(x)dx =
2
Z
(1 − e−2x )dx =
x 1 −2x
+ e
+ c.
2 4
d
Since dx
sinh(x) = cosh(x) > 0 for all x, the hyperbolic sine function is increasing
on the interval (−∞, ∞). Thus it has an inverse function, called the inverse hyperbolic
sine function, with value at x denoted by sinh−1 (x). Since the domain and range of the
hyperbolic sine function are both (−∞, ∞), the domain and range of the inverse hyperbolic
sine function are also both (−∞, ∞). As usual with inverse functions,
y = sinh−1 (x) if and only if sinh(y) = x.
The graph of y = sinh−1 (x) is shown in Figure 6.7.3.
(6.7.10)
Section 6.7
Hyperbolic Functions
5
Example The hyperbolic sine function and its inverse provide an alternative method
for evaluating
Z
1
√
dx.
1 + x2
Namely, if we make the substitution
x = sinh(u), −∞ < u < ∞,
dx = cosh(u)du,
then
p
1+
x2
=
q
2
1 + sinh (u) =
q
cosh2 (u) = cosh(u),
where the second equality follows from the identity cosh2 (u) − sinh2 (u) = 1 and the last
equality from the fact that cosh(u) > 0 for all u. Hence
Z
Z
Z
1
cosh(u)
√
du = du = u + c = sinh−1 (x) + c.
dx =
2
cosh(u)
1+x
The following proposition is a consequence of the previous example.
Proposition
1
d
sinh−1 (x) = √
.
dx
1 + x2
In Section 6.6 we saw, using the substitution x = tan(u), − π2 < u <
Z
p
1
2
√
dx = log x + 1 + x + c.
2
1+x
(6.7.11)
π
2,
that
Since two antiderivatives of a function can differ at most by a constant, there must exist
a constant k such that
p
−1
2
sinh (x) = log x + 1 + x + k
for all x. Evaluating both sides of this equality at x = 0, we have
0 = sinh−1 (0) = log(1) + k = k.
Thus k = 0 and
−1
sinh
p
2
(x) = log x + 1 + x (6.7.12)
for all x. Since the hyperbolic sine function is defined in terms of the exponential function,
we should not find it surprising that the inverse hyperbolic sine function may be expressed
in terms of the natural logarithm function.
d
Similarly, since dx
cosh(x) = sinh(x) > 0 for all x > 0, the hyperbolic cosine function
is increasing on the interval [0, ∞), and so has an inverse if we restrict its domain to [0, ∞).
That is, we define the inverse hyperbolic cosine function by the relationship
y = cosh−1 (x) if and only if x = cosh(y),
(6.7.13)
6
Hyperbolic Functions
Section 6.7
5
4
3
2
1
2
4
8
6
10
Figure 6.7.4 Graph of y = cosh−1 (x)
where we require y ≥ 0. Note that since cosh(x) ≥ 1 for all x, the domain of of the inverse
hyperbolic cosine function is [1, ∞). The graph of y = cosh−1 (x) is shown in Figure 6.7.4.
In Problem 3 at the end of this section you are asked to show that
Z
√
1
x2
−1
dx = cosh−1 (x) + c
for x > 1, from which the following proposition follows.
Proposition
1
d
cosh−1 (x) = √
.
dx
x2 − 1
(6.7.14)
In the same problem you are asked to show that, for x ≥ 1,
p
cosh−1 (x) = log x + x2 − 1
Example
In Section 6.6 we evaluated the integral
Z
√
1
x2 − 9
dx,
for x > 3, using the substitution x = 3 sec(u), 0 < u <
π
2.
x = 3 cosh(u), u > 0,
dx = 3 sinh(u)du
The substitution
(6.7.15)
Section 6.7
Hyperbolic Functions
7
provides a somewhat simpler approach. Namely,
Z
Z
1
3 sinh(u)
√
q
dx =
du
2
x2 − 9
9 cosh (u) − 9
Z
3 sinh(u)
q
=
du
2
3 cosh (u) − 1
Z
sinh(u)
q
=
du
sinh2 (u)
Z
sinh(u)
=
du
sinh(u)
Z
= du
=u+c
= cosh−1
x
3
+ c,
where we have used the fact that sinh(u) > 0 when u > 0.
Having defined the hyperbolic sine and cosine functions, it is possible to define four
more hyperbolic trigonometric functions in analogy with the circular trigonometric functions. Namely, the hyperbolic tangent function is given by
tanh(x) =
sinh(x)
,
cosh(x)
(6.7.16)
where −∞ < x < ∞; the hyperbolic cotangent function by
coth(x) =
cosh(x)
,
sinh(x)
(6.7.17)
where x 6= 0; the hyperbolic secant function by
sech(x) =
1
,
cosh(x)
(6.7.18)
where −∞ < x < ∞; and the hyperbolic cosecant function by
csch(x) =
1
,
sinh(x)
(6.7.19)
where x 6= 0. In Problem 5 at the end of this section you are asked to verify the following
results.
Proposition
d
tanh(x) = sech2 (x).
dx
(6.7.20)
8
Hyperbolic Functions
Section 6.7
Proposition
d
coth(x) = −csch2 (x)
dx
(6.7.21)
d
sech(x) = −sech(x) tanh(x).
dx
(6.7.22)
d
csch(x) = −csch(x) coth(x).
dx
(6.7.23)
Proposition
Proposition
Since
sinh(x)
ex − e−x
,
tanh(x) =
= x
cosh(x)
e + e−x
we have
ex − e−x
x→∞ ex + e−x
ex (1 − e−2x )
= lim x
x→∞ e (1 + e−2x )
1 − e−2x
= lim
x→∞ 1 + e−2x
=1
lim tanh(x) = lim
x→∞
and
ex − e−x
lim tanh(x) = lim x
x→−∞
x→−∞ e + e−x
e−x (e2x − 1)
= lim −x 2x
x→−∞ e
(e + 1)
2x
e −1
= lim 2x
x→−∞ e
+1
= −1.
Hence y = 1 and y = −1 are both horizontal asymptotes for the graph of y = tanh(x).
Combining this information with tanh(0) = 0 and
d
tanh(x) = sech2 (x) > 0
dx
for all x, we can see why the graph of y = tanh(x) looks as it does in Figure 6.7.5.
Since the hyperbolic tangent function is increasing on (−∞, ∞), it has an inverse, called
the inverse hyperbolic tangent function, with value at x denoted by tanh−1 (x). That is, as
usual,
y = tanh−1 (x) if and only if tanh(y) = x.
(6.7.24)
The domain of the inverse hyperbolic tangent function is (−1, 1) the range of the hyperbolic tangent function, and its range is (−∞, ∞), the domain of the hyperbolic tangent
Section 6.7
Hyperbolic Functions
9
1.5
1
0.5
-4
-2
2
4
-0.5
-1
-1.5
Figure 6.7.5 Graph of y = tanh(x)
function. Corresponding to the horizontal asymptotes of the graph of the hyperbolic tangent function, the graph of the inverse hyperbolic tangent function has vertical asymptotes
x = −1 and x = 1, as shown in Figure 6.7.6.
Example
As an alternative to using partial fractions, we may evaluate the integral
Z
1
dx
1 − x2
for −1 < x < 1 using the substitution
x = tanh(u), −∞ < u < ∞,
dx = sech2 (u)du.
Then
Z
1
dx =
1 − x2
Z
sech2 (u)
du.
1 − tanh2 (u)
Now from the identity
cosh2 (x) − sinh2 (x) = 1
we obtain
sinh2 (x)
1
cosh2 (x)
−
=
.
2
2
cosh (x) cosh (x)
cosh2 (x)
In other words,
1 − tanh2 (x) = sech2 (x).
Hence
Z
1
dx =
1 − x2
Z
sech2 (u)
du =
sech2 (u)
Z
du = u + c = tanh−1 (x) + c.
(6.7.25)
10
Hyperbolic Functions
Section 6.7
4
2
-1.5
-1
-0.5
1
0.5
1.5
-2
-4
Figure 6.7.6 Graph of y = tanh−1 (x)
Note that (6.7.25) gives us the useful identity
tanh2 (x) + sech2 (x) = 1
(6.7.26)
for all x. Moreover, we have the following proposition as a consequence of this example.
Proposition
d
1
tanh−1 (x) =
.
dx
1 − x2
(6.7.27)
If we were to use partial fractions to evaluate the integral of the previous example, we
would obtain, for −1 < x < 1,
Z
1
1
1
1
dx = log(1 + x) − log(1 − x) + c = log
2
1−x
2
2
2
1+x
1−x
+ c.
It follows that
−1
tanh
1
(x) = log
2
1+x
1−x
+k
for some constant k. Evaluating at 0, we have
0 = 0 + k.
Thus k = 0 and we have
−1
tanh
for 1 < x < 1.
1
(x) = log
2
1+x
1−x
(6.7.28)
Section 6.7
Hyperbolic Functions
Problems
1. Differentiate each of the following functions.
(a) f (x) = sinh(3x)
(c) f (t) = 3t sinh(t) cosh(2t)
(e) y(t) = 5t2 cosh2 (4t)
(b) g(t) = 3t cosh(4t)
(d) g(x) = 4x sinh(3x2 − 1)
(f) f (t) = 3 cosh2 (2t) − 13 sinh(3t2 )
2. Evaluate each of the following integrals.
Z
(a)
sinh(3x)dx
Z
(c)
sinh(z) cosh(z)dz
Z
(e)
e−2t cosh(2t)dt
Z
(g)
5t2 cosh(2t)dt
Z
cosh(4t − 3)dt
(b)
Z
(d)
3x sinh(2x)dx
Z
cosh2 (x) sinh(x)dx
Z
sinh(t)
dt
cosh2 (t)
(f)
(h)
3. (a) Use the substitution x = cosh(u), u > 0, to show that
Z
1
√
dx = cosh−1 (x) + c
2
x −1
for x > 1.
(b) Use the substitution x = sec(u), 0 < u <
Z
√
1
x2 − 1
π
2
, to show that
p
2
dx = log x + x − 1 + c
for x > 1.
(c) Using (a) and (b), show that
−1
cosh
p
2
(x) = log x + x − 1
for x > 1.
4. Evaluate the following integrals.
Z
1
√
(a)
dx
4 + x2
Z
3
√
(c)
dt
9 + 3t2
Z
(b)
√
1
dx, x > 2
x2 − 4
Z
1
√
(d)
dx, x < −1
x2 − 1
5. Verify the following derivatives.
(a)
d
tanh(x) = sech2 (x)
dx
(b)
d
coth(x) = −csch2 (x)
dx
11
12
Hyperbolic Functions
Section 6.7
d
d
sech(x) = −sech(x) tanh(x)
(d)
csch(x) = −csch(x) coth(x)
dx
dx
6. Differentiate each of the following functions.
(c)
(a) f (x) = 3x tanh(4x)
(c) h(θ) = 4 tanh2 (θ)sech(θ)
7. Evaluate each of the following integrals.
Z
(a)
tanh(x)dx
Z
1
(c)
dx
4 − x2
(b) g(t) = sech2 (3t)
(d) f (x) = 5xsech(4x) − 21 tanh3 (4x)
Z
(b)
tanh(2x)sech(2x)dx
Z
(d)
5
dt
9 − 3t2
8. Graph each of the following functions on an appropriate interval.
(a) y = sech(x)
(c) y = csch(x)
(b) y = coth(x)
(d) y = 3 tanh(4x)
``` | 4,948 | 11,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-17 | latest | en | 0.838154 |
http://www.theweatherprediction.com/habyhints/303/ | 1,709,150,978,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474744.31/warc/CC-MAIN-20240228175828-20240228205828-00042.warc.gz | 60,323,961 | 2,699 | [--MAIN HOME--] [--ALL HABYHINTS--] [--FACEBOOK PAGE--]
SKEW-T: A LOOK AT KI
METEOROLOGIST JEFF HABY
1. What is KI?
The KI (K INDEX) is an index used to assess convective potential.
2. How is KI determined?
The KI is a combination of the Vertical Totals (VT) and lower tropospheric moisture characteristics. The VT is the temperature difference between 850 and 500 mb while the moisture parameters are the 850 mb dewpoint and 700 mb dewpoint depression.
KI = (T850 - T500) + (Td850 - Tdd700)
The equation above for the KI is stating (850 mb temperature minus 500 mb temperature) plus (850 mb dewpoint minus 700 mb dewpoint depression)
The Skew-T below gives a KI value of 30 for that sounding. Here is how that number was determined:
850 mb Temperature = 20 C
500 mb Temperature = -7 C
850 mb Dewpoint = 9 C
700 mb Dewpoint Depression = 6 units of C
KI = (20 - (-7)) + (9 - 6) = 27 + 3 = 30
3. Operational significance of KI:
K INDEX
15-25 Small convective potential 26-39 Moderate convective potential 40+ High convective potential
A high 850 mb dewpoint and a low 700 mb dewpoint depression at the same time indicates there is a deep layer of warm and moist air in the lower to middle troposphere. This is very beneficial to producing instability, especially when the VT is high.
4. Pitfalls:
a. May not pick up a capping inversion that prevents storms from developing.
b. Index should not be used to determine severity of storms.
c. VT may be very high and contributes to causing a high KI even when moisture is lacking. Index will be unrealistically too unstable in these situations.
d. Works best for flat areas in low to moderate elevations. Does not work for high elevations.
e. Index value interpretation varies with season and location.
f. 700 mb dewpoint depression may be very large and thus will lead to a stable KI. Very dry air at 700 mb will not degrade the convective potential as long as there is moisture below this layer. KI is better to use when forecasting within a deep layer of mT (maritime tropical) air as opposed to a differential advection situation in which an elevated mixed layer advects over mT air. | 545 | 2,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.857793 |
https://linearprogramminghelp.com/linear-programming-meaning/ | 1,680,422,277,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950422.77/warc/CC-MAIN-20230402074255-20230402104255-00695.warc.gz | 408,742,881 | 19,145 | Linear Programming Meaning
Linear programming function means that the output (data) will be continuously determined as long as the input data are being processed. The main advantage of using this type of function is that an infinite number of inputs and corresponding outputs can be determined. Using this function, programmers can calculate the Fibonacci ratios as well as the optimal way to deal with discontinuities in data.
Basically, the main idea of linear programming comes from the fact that most of the results of a mathematical operation are already determined once the function is started. This kind of function is usually used in order to solve a mathematical problem or to give solutions to optimization problems. In other words, the main goal of linear programming is to solve a certain problem by determining its solution in terms of the desired output. For example, an optimization problem could be solved through linear programming if the output is the sum of all the individual results for some set of weights xi and yi. Another example is the Fibonacci ratio, which is calculated through linear programming by first determining the area between the lowest x value and the greatest y value. In both of these cases, the function uses numbers as inputs and then uses the function’s output as inputs for the next number as well.
There are two main ways in which the values of a linear programming function can be set. These are the integral and the independent variables. Both of these settings can be used but they come with different strengths depending on the needs of the program that will be made to use them.
Integral linear programming function means that the output will depend on the inputs that were included in the original equation. This kind of linear function is usually implemented as a mean difference and as a mean average of the results. On the other hand, independent linear programming function does not mean that an output will be directly connected to a specific input. Instead, this type of function will determine an output of a number depending on the inputs that are already determined from the original equation.
Both of these linear programming functions can be implemented in linear algebra as well. However, they will still differ greatly depending on the situation that is being dealt with. In linear algebra, you can have as many outputs as you wish, whereas in linear programming you can only have an output as the function outputs the limit of the input that was input into it. Basically, linear programming function values are just the limits of the inputs that were determined based on the original equations. You can also have derivatives depending on these limits. This simply means that as the input changes, so will the output that is generated.
A linear programming function can also be defined as a mathematical expression. This basically means that the expression is either a closed system or open system. In terms of a closed system, the function will be continuous and will follow some form of a rule. On the other hand, an open system will not be as well formed, and it will most likely involve an infinite number of steps.
The linear programming function can also be defined as a finite mathematical expression that shows the dependence of certain variables on another, yet independent variables. For example, if one is to compute the tangent lines by using a graph, then the linear programming function will be the graph of the tangent line on the x-axis. This tangent line will then depend on the values that are plotted on the y-axis.
This is basically what linear programming functions for. It allows for an easy computation of sums and mathematical expressions and solving of linear equations. All in all, linear programming functions for are used extensively in mathematics, calculus, physics, etc. | 712 | 3,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-14 | latest | en | 0.937648 |
http://www.formulaconversion.com/formulaconversioncalculator.php?convert=tablespoons_to_nanoliters | 1,498,645,527,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323604.1/warc/CC-MAIN-20170628101910-20170628121910-00335.warc.gz | 533,211,326 | 16,309 | Take advantage of the benefits of registration! As a registered user, the conversions you add to 'Your Favorites' are saved to your account so they follow you wherever you go for fast access anywhere! GPA inputs and calculations too. You can also post to our forums, create a profile, and more! Click here for our simple sign-up page
# Tablespoons to nanoliters (tbsp to nl) Metric conversion calculator
Welcome to our tablespoons to nanoliters (tbsp to nl) conversion calculator. You can enter a value in either the tablespoons or nanoliters input fields. For an understanding of the conversion process, we include step by step and direct conversion formulas. If you'd like to perform a different conversion, just select between the listed Volume units in the 'Select between other Volume units' tab below or use the search bar above. Tip: Use the swap button to switch from converting tablespoons to nanoliters to nanoliters to tablespoons.
## nanoliters (nl)
(not bookmarks)
Swap
< == >
1 tbsp = 14786764.781 nl 1 nl = 7.0E-8 tbsp
Algebraic Steps / Dimensional Analysis Formula
tbsp * 1 fl.oz. 2 tbsp * 1 l 33.814 fl.oz. * 1000000000 nl1 l = nl
Direct Conversion Formula
tbsp * 14786764.781 nl1 tbsp = nl
cubic feet 0 cubic meters 0 cubic millimeters 0 cubic yards 0 cups 0 fluid ounces 0 gallons (liquid) 0 liters 0 milliliters 0
If you would like to switch between Volume units, select from the tables below
bushels centiliters cubic centimeters cubic decimeters cubic dekameters cubic feet cubic gigameters cubic hectometers cubic inches cubic kilometers cubic megameters cubic meters cubic micrometers cubic miles cubic millimeters cubic nanometers cubic yards cups cups (metric) deciliters dekaliters fluid ounces gallons (dry) gallons (liquid) gigaliters gills hectoliters imperial gallon kiloliters liters megaliters microliters milliliters nanoliters pecks pints (dry) pints (liquid) quarts (dry) quarts (liquid) tablespoons teaspoons < == > bushels centiliters cubic centimeters cubic decimeters cubic dekameters cubic feet cubic gigameters cubic hectometers cubic inches cubic kilometers cubic megameters cubic meters cubic micrometers cubic miles cubic millimeters cubic nanometers cubic yards cups cups (metric) deciliters dekaliters fluid ounces gallons (dry) gallons (liquid) gigaliters gills hectoliters imperial gallon kiloliters liters megaliters microliters milliliters nanoliters pecks pints (dry) pints (liquid) quarts (dry) quarts (liquid) tablespoons teaspoons
Active Users | 606 | 2,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.692083 |
http://www.enotes.com/homework-help/find-f-x-2x-arccotx-log-1-x-2-1-2-x-increasing-349613 | 1,477,497,289,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720962.53/warc/CC-MAIN-20161020183840-00103-ip-10-171-6-4.ec2.internal.warc.gz | 423,884,925 | 11,276 | # find f(x)=2x + arccotx + log{ [1+x^2]^(1/2) -x} is increasing or decreasingNeed urgent answer
mlehuzzah | Student, Graduate | (Level 1) Associate Educator
Posted on
If you have a positive number and you take the log of it, you don't necessarily have a positive number.
E.g.: ln(1/e)=-1
1/e is positive, but -1 is negative.
So if you show [1+x^2]^(1/2) -x>0 you have shown that
log{ [1+x^2]^(1/2) -x} is defined (since you can only take logs of positive numbers) but you haven't shown that number is positive.
Moreover, showing something is positive doesn't mean it is increasing.
`f(x)=2x + "arccot" (x) + "log" (sqrt(1+x^2)-x)`
To find out if this is increasing or decreasing, we take a derivative and see if it is positive or negative:
`f'(x)=2 + (-1)/(1+x^2)+ ((1)/(sqrt(1+x^2)-x))(((1)/(2))(1+x^2)^(-1/2)*2x-1)`
`=2 - (1)/(1+x^2)+(x(1+x^2)^(-1/2)-1)/(sqrt(1+x^2)-x)`
We start by doing some algebra on that last term:
`(x(1+x^2)^(-1/2)-1)/(sqrt(1+x^2)-x)`
`=((x(1+x^2)^(-1/2)-1)(sqrt(1+x^2)+x))/((sqrt(1+x^2)-x)(sqrt(1+x^2)+x))`
`=(x+x^2(1+x^2)^(-1/2)-(1+x^2)^(1/2)-x)/(1+x^2-x^2)`
Putting this back into `f'` we have:
`f'(x)=2-(1)/(1+x^2)+x+x^2(1+x^2)^(-1/2)-(1+x^2)^(1/2)-x`
`=2-(1)/(1+x^2)+(x^2)/(sqrt(1+x^2))-sqrt(1+x^2)`
`=2-(1)/(1+x^2)+(x^2sqrt(1+x^2))/(1+x^2)-(sqrt(1+x^2)(1+x^2))/(1+x^2)`
`=(2(1+x^2)-1+x^2 sqrt(1+x^2)-(1+x^2)sqrt(1+x^2))/(1+x^2)`
`=(1+2x^2-sqrt(1+x^2))/(1+x^2)`
The denominator is always positive, so it won't affect whether `f'` is positive or negative. So consider the numerator:
`1+2x^2-sqrt(1+x^2)=(1+x^2)-sqrt(1+x^2)+x^2`
Now, if `r>1` then `r> sqrt(r)`
`x^2>= 0`, and, unless `x=0` , we have `x^2>0` so `x^2+1> 1`, so `x^2+1> sqrt(x^2+1)`
(and at `x=0` we have `x^2+1=sqrt(x^2+1)` )
thus `(1+x^2)-sqrt(1+x^2)> 0` so that is positive
also, `x^2>0`, so that is also positive.
A positive plus a positive is positive
So, unless `x=0` , the numerator is positive, so `f'(x)` is positive, so `f(x)` is increasing
And at `x=0` the numerator is exactly 0, so `f'(x)=0` , so `f(x)` "flattens out" ` `
jeew-m | College Teacher | (Level 1) Educator Emeritus
Posted on
Let y=arc cotx
Then coty=x
By differntiation
d(coty)/dx = 1
-sec^2x/tan^2x *dy/dx= 1
-(1+tan^2x)/tan^2x*dy/dx =1
-(cot^2x+1)*dy/dx =1
dy/dx= -1/(1+x^2)
Let P=2x+arc cotx
dP/dx = 2-1/(1+x^2)
= (2x^2+1)/(1+x^2)
For all x ;2x^2+1>0
For all x;1+x^2 >0
So dP/dx >0 for all x.
dP/dx means the gradient of the function P. Since dP/dx>0 always P is increasing.
Let Q=log{ [1+x^2]^(1/2) -x}
We know 1+x^2>x^2
Then [1+x^2]^(1/2) >sqrt(x^2)
[1+x^2]^(1/2) >x
Then [1+x^2]^(1/2) -x>0
So log{ [1+x^2]^(1/2) -x}>0 since log is not negative.
So every value of x it is a poitive value and increasing.
So Q is increasing for all x.
But f(x) = P+Q
Since both P and Q are increasing f(x) is increasing.
Sources: | 1,251 | 2,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2016-44 | latest | en | 0.70684 |
https://www.futurelearn.com/courses/preparing-for-uni/19/steps/599653 | 1,581,949,828,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875142323.84/warc/CC-MAIN-20200217115308-20200217145308-00009.warc.gz | 748,599,741 | 24,349 | 5.23
Astronauts data
Days into mission % Reduction in body mass % Reduction in leg strength % Reduction in cardiac output
10 -2.15% -2.50% -1.50%
20 -4.30% -5.00% -3.00%
30 -6.45% -7.50% -4.50%
40 -8.60% -10.00% -6.00%
50 -10.75% -12.50% -7.50%
60 -12.90% -15.00% -9.00%
Can you think of any problems with this data? On day 60 the astronaut has lost 15 % of their leg strength.
According to this data they would then be at 30% of their original leg strength at day 120. There would be a point when they would have no leg strength at all. Calculate for body mass, leg strength and cardiac output, the point that they each reach zero.
For the astronaut with a body mass of 100 kg, what would their body mass be after 220 days in space if they did not exercise? Does this seem like a sensible calculation?
Can you suggest a better way of expressing loss of body functions over time? Do you think this linear model is sensible?
In the next step you will be answering some questions which ask you to think about how you have approached these self assessment questions and the material in this lesson in general. | 303 | 1,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.934491 |
https://www.physicsforums.com/threads/need-some-assistance-with-integration.415646/ | 1,521,694,618,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647768.45/warc/CC-MAIN-20180322034041-20180322054041-00593.warc.gz | 908,638,044 | 14,880 | # Need some assistance with integration
1. Jul 12, 2010
### Brownfractals
1. The problem statement, all variables and given/known data
∫ x^(5)-2x^(2)-1/3x^(4) dx
2. Relevant equations
I am familiar with the power rule and how to split the expression into three separate expressions and then simplifying. I just cant seem to sort out my algebra for this problem.
xn dx = xn+1
n + 1 + C if n is NOT= -1
x-1 dx = ln|x|+ C
3. The attempt at a solution
Here is where i become stuck...Please show me a step by step process to solving the remainder of this problem; that way i can analyze it more deeply. Please Point out any things that seem wrong,i would REALLY appreciate the help given. :) :
∫ (x^5/3x^4 -2x^2/3x^4 -1/3x^4)dx = ∫ 1/3x -2x^(-2) -1/3x^(-4)dx = ∫ (x^2/6 + 2/x +1/9x^3) + C <------- This is the part where i become flabbergasted. . .Its simple im sure but i just cannot see it. I was thinking about multiplying the whole expression by 18x^3 but i am not sure if that is correct.
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
Last edited: Jul 12, 2010
2. Jul 13, 2010
### Tedjn
Before we can give you help, can you clarify with parentheses what your integrand is? Is it (x5-2x2-1)/(3x4)? I ask, because it is difficult to read the problem and your work without parentheses.
3. Jul 13, 2010
### Staff: Mentor
Also, when you write stuff like this
xn dx = xn+1
n + 1 + C if n is NOT= -1
x-1 dx = ln|x|+ C
it's difficult to discern that xn means xn or that x-1 means x-1. At the very least, use ^ more consistently. | 532 | 1,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-13 | longest | en | 0.897905 |
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# Design Calculation Of Ball Mill
Lead Ore Ball Mill Design Calculation Formula. Lead angle helix angle The lead angle helix angle is dependent on and related to the diameter and the pitch of the thread This measurement can be represented by a triangle being unwound from the component The lead angle is calculated according to the formula below
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Of the bond ball mill grindability test. for all model based methods, a reliable method to calculate mill power draw for a given mill is required for the calculation of power draw. morrell 1996 proposed a mathematical model for autogenous, semi-autogenous and ball mills which is based on the motion of grinding charge inside the mill.
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• ### Ball Mill Design Calculation
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• ### Calculation Of Reduction Ratio Ball Mill
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• ### Power Draw Calculator Of Ball Mill Ball Mill
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• ### Amit 135 Lesson 7 Ball Mills Amp Circuits Mining Mill
Wet ball mill kg kwh 0.16a i-0.015 0.33 dry ball mill kg kwh 0.023a i 0.5 replacement ball size. rowland and kjos proposed the use of their equation for the determination of the initial and replacement media size. azzaroni 1981 and dunn 1989 recommended the use of the following expression for the size of the makeup media
• ### Calculation Of Ball Mill Design
Design calculations for ball mills. design calculations for ball millsall mill design power draw tool parameters values displayed with a may be changed click on a value to display an entry form or use the arrows right of a value to select from a listool behaviour this tool exhibits spreadsheetlike behaviour change a value and affected values are recalculated automaticallyor. | 2,363 | 11,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-21 | latest | en | 0.864576 |
https://marthaspdx.com/domino-a-fun-and-entertaining-game-for-the-whole-family/ | 1,719,049,748,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862310.25/warc/CC-MAIN-20240622081408-20240622111408-00598.warc.gz | 329,267,596 | 16,946 | Domino is a fun and entertaining game for the entire family. Some people like to line up the dominoes in long rows and knock them down, while others play games with the individual pieces. Dominoes can also be used as building blocks or components of Rube Goldberg machines.
The word domino is derived from the Latin dominus, meaning “lord.” It originally denoted a garment worn over a priest’s surplice or a mask in carnival season or at a masquerade. The domino piece itself is reminiscent of that garment, with the black domino being a contrast to the white of the surplice.
There are several different types of domino games and each requires its own set of rules. Most games are played with a double six or a double nine set, though larger sets are available. The individual tiles are called dominoes, bones, tickets or spinners, and they can be found in many varieties of colors.
The individual pieces in a domino set must be shuffled before a game begins. The shuffled dominoes are then called the boneyard. Once a player has an opening double, for example a three or one-pip domino, they may then pick a domino from the boneyard and begin playing that domino in their turn. In some games, players take turns picking dominoes until one player plays all of their remaining tiles and wins the hand.
Most domino games fall into two main categories, blocking and scoring games. Blocking games involve emptying your own hand while preventing the opponent from playing a tile. Scoring games, such as bergen and muggins, determine points by counting the number of dots (or pips) on the exposed ends of the dominoes in the losing players’ hands.
A domino effect can be much bigger than you might think. In a video from 2009, University of British Columbia physicist Stephen Morris showed that the first domino in a series can knock down things more than a quarter of its size. He then built up a series of 13 dominoes, each a full 1.5 times bigger than the previous one. The resulting chain spanned more than 100 feet and weighed over 100 pounds.
One of the best ways to learn how to lead is by listening to the leaders who work around you. This principle is evident in a recent episode of the hit show Undercover Boss, where Domino’s CEO Don Meij sends his top managers into several restaurants to observe how they work with employees and customers. After seeing the good and the bad of how leadership is carried out, Meij implements new changes that are designed to make Domino’s a better place for its employees. He cites that leadership is more than just management and encourages employees to take control of their own destiny by taking action when needed. In turn, this creates a happier and more productive workplace. | 586 | 2,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-26 | latest | en | 0.95725 |
https://notes.reasoning.page/html/truth-tables | 1,680,096,700,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00664.warc.gz | 481,314,832 | 13,064 | # A calculus of the absurd
#### 20.2 Truth tables
One very powerful (but perhaps not very elegant) way to reason about propositions is by writing a big table. The advantage of this is that it’s easy to ensure that we’ve considered every case (provided that we list every possible value). To do this, we must list them systematically. The convention is to combine the approach in Section 16.3 with the constraint that we want to list all variables possibilities in increasing order of size. I realise this probably makes no sense, so it’s best to proceed with an example:
• Example 20.2.1 Prove that $$A \land (A \lor B) \equiv A$$.
We want to draw a table which contains every possible value for $$A$$ and $$B$$. Note that we want to list the possibilities together, rather than separately (i.e. there are $$4$$ total values to list as there are $$2$$ possible values for $$A$$ and then for each of those possibilities there are $$2$$ possible values for $$B$$)125125 See Section 16.1.2 if you are unsure about this.. What about this weird “increasing order of size”? First, examine the table
$$A$$ $$B$$ $$A \lor B$$ $$A \land (A \lor B)$$ F F F F F T T F T F T T T T T T
If we set $$F=0$$ and $$T=1$$, and then read the column $$A$$ and that of $$B$$ and concatenate the two values (e.g. in the first row of the table, take $$A=F=0$$ and $$B=F=0$$, so their concatenation is the string $$0$$ followed by the string $$0$$ again), then we wish to list them in increasing value (this just makes everything much clearer, and also makes it harder to make mistakes). In the table above we have $$00 < 01 < 10 < 11$$. The same idea extends to larger tables.
Returning to the formula, it must be true, as the column for $$A$$ is the same as the column for $$A \land (A \lor B)$$. | 468 | 1,781 | {"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} | 4.40625 | 4 | CC-MAIN-2023-14 | latest | en | 0.886919 |
http://www.buenastareas.com/ensayos/Cuk-Converter/1456239.html | 1,529,587,185,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864148.93/warc/CC-MAIN-20180621114153-20180621134153-00101.warc.gz | 379,527,946 | 21,853 | # Cuk converter
Páginas: 3 (549 palabras) Publicado: 26 de enero de 2011
CUK CONVERTER DESIGN:
50V-200V/100V 500W 90%> EFFICIENCY
Objective
Intro Description Material Analysis TIPS
The objective is to Design a DC-DC converter,making use of the concepts, formulas, and excersices learnt during class and laboratory hours.
Overview
Intro
Description
Material
Analysis
TIPS
“Cuk Converter” was develped byprofessor Slobodan Cuk from teh California Institute of Technology
Oposite sign!!!
L1 & C2L2 work as filters for the current at the input and the output.
Basic Topology
Intro DescriptionMaterial Analysis TIPS
The basic topology of the circuit comes from the configuration of both, the buck and the boos converter.
Switch ON/OFF
Intro Description Material Analysis TIPSPosition 1
Position 2
List of Materials
Intro Intro Material Analysis TIPS
Power Schottky silicon carbide diode IRGP50B60PD1 International Rectifier Max Volt = Peak I= 600v 40A RCE(on) typ. = 61mΩVCES = IC = 600V 33A
Capacitor B32776 EPCOS Cr= 50uf Capacitor B25655 EPCOS Cr= 500uf
Voltage = Iripple= ESRyp= 450v 15A 4mohm Voltage = Imax= 450v 120A Rs= 1 m ohm
Inductors
CopperInductance= Rdc=(ohm) loss= 10mH 49m 1.82 W 15mH 129m 32.98W
Core Part # Turns= E827-35 139 E827-26 232
Vmax of C1 must be greater than Vin+Vout
Intro Description MaterialAnalysis TIPS
Intro Description Material Analysis TIPS
Schematics
Intro Description Material Analysis TIPS
Schematics
Intro Description Material Analysis
Vo D =Vd 1 − D
Vo D = Vd 1 − D
Design TIPS
Intro Description Material Analysis TIPS
1) Ripple admisible de las corrientes de entrada y salida 2) Ripple admisible de la tensión de salida Calculations
Intro Description Material Analysis Suppose we want a∆Vout
= 1.7 mvolts
(.11A)(66.66 µs ) = 539.16 µf 8(1.7 mV )
C2 =
Calculations
Intro Description Material Analysis TIPS...
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https://communities.sas.com/t5/SAS-Statistical-Procedures/Factor-Analysis/td-p/369831?nobounce | 1,510,952,298,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934803944.17/warc/CC-MAIN-20171117204606-20171117224606-00246.warc.gz | 577,614,685 | 26,367 | Solved
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# Factor Analysis
Hello
While doing factor analysis using proc factor
1) Can we have discrete and dummy variables together with continuous variables.
2) As per my understanding we can find correlation only if two variables are continuous, since using the option "corr" in proc factor returns correlation or partial correlations, which means if we have discrete or dummy variables proc factor returns partial correlation?.
Is partial correlation is used when we have discrete variables?.
please correct me if I am going wrong.
Accepted Solutions
Solution
06-23-2017 08:43 AM
Posts: 1,908
## Re: Factor Analysis
PROC FACTOR does not allow categorical (discrete) variables to be analyzed.
You can use dummy variables into PROC FACTOR, but their use and interpretation is somewhat artificial and strained, and you do so at your own risk. As always, you can use your favorite search engine and type in
dummy variables in factor analysis
All Replies
Solution
06-23-2017 08:43 AM
Posts: 1,908
## Re: Factor Analysis
PROC FACTOR does not allow categorical (discrete) variables to be analyzed.
You can use dummy variables into PROC FACTOR, but their use and interpretation is somewhat artificial and strained, and you do so at your own risk. As always, you can use your favorite search engine and type in
dummy variables in factor analysis | 309 | 1,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.829725 |
http://mathhelpforum.com/advanced-algebra/115187-isomorphism-print.html | 1,503,047,203,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104631.25/warc/CC-MAIN-20170818082911-20170818102911-00222.warc.gz | 271,100,979 | 5,611 | # Isomorphism
• Nov 17th 2009, 11:10 AM
slevvio
Isomorphism
Prove that any two cyclic groups of the same order are isomorphic.
Hint: Let G and G' be cyclic groups with generators a and a' respectively. If G and G' are both infinite show that $\phi : G \rightarrow G'$ defined by $\phi(a^m) = (a')^m$ for $m \in \mathbb{Z}$ is an isomorphism. If G and G' are both of finite order n, use $\phi$ again but note that it is necessary to prove that $\phi$ is consistently defined because it is possible to have $a^i = a^j$ when i does not equal j.
It's easy to show that phi is injective when the group is infinite since a^m = a^n means that m = n in a cyclic group of infinite order, but I am having trouble showing phi is injective with the finite case and any help would be appreciated. How do I show it is consistently defined? Thanks again everyone
• Nov 17th 2009, 11:18 AM
tonio
Quote:
Originally Posted by slevvio
Prove that any two cyclic groups of the same order are isomorphic.
Hint: Let G and G' be cyclic groups with generators a and a' respectively. If G and G' are both infinite show that $\phi : G \rightarrow G'$ defined by $\phi(a^m) = (a')^m$ for $m \in \mathbb{Z}$ is an isomorphism. If G and G' are both of finite order n, use $\phi$ again but note that it is necessary to prove that $\phi$ is consistently defined because it is possible to have $a^i = a^j$ when i does not equal j.
It's easy to show that phi is injective when the group is infinite since a^m = a^n means that m = n in a cyclic group of infinite order, but I am having trouble showing phi is injective with the finite case and any help would be appreciated. How do I show it is consistently defined? Thanks again everyone
$a'\,^m=1\Longrightarrow n\mid m \Longrightarrow m=kn\Longrightarrow a^m=(a^n)^k=1^k=1\Longrightarrow Ker\,\phi = 1$
Tonio
• Nov 17th 2009, 02:29 PM
slevvio
thanks:)
• Nov 17th 2009, 02:49 PM
Drexel28
I am not sure what consistently defined means? Most likely well-defined?
But maybe the following is along the lines of what you want?
Problem: Let $G,G'$ be cyclic groups of order $n$. Prove that $G\cong G'$
Proof: Let $\mathcal{G}=\langle g\rangle$ be any cyclic group of order $n$. Define a mapping $\Phi:\mathbb{Z}_n\mapsto \mathcal{G}$ by $\Phi(k)=g^k$. Clearly this is well defined. To see that it is an injection note that $g^{\kappa}=g^{\kappa-\kappa_1}=e$, which means that $n|\kappa-\kappa_1$ but since $-(n-1)\le \kappa-\kappa_1\le n-1$ this will only be true if $\kappa-\kappa_1=0\implies \kappa=\kappa_1$. To see that this mapping is surjective let $g'\in\mathcal{G}$, then $g'=g^{m}$ for some $0\le m \le n-1$. Therefore $g'=\Phi(m)$. Lastly, we must show that $\Phi$ is a homomorphism, but this is trivial since $\Phi(\tau+\tau_1)=g^{\tau+\tau_1}=g^{\tau}g^{\tau_ 1}=\Phi(\tau)\Phi(\tau_1)$. Therefore any cyclic grou of order $n$ is isomorphic to $\mathbb{Z}_n$, but since $\cong$ is an equivalence relation (you can check this yourself) we see that $G\cong \mathbb{Z}_n\text{ and }G'\cong\mathbb{Z}_n\implies G\cong G'$ by transivity.
• Nov 18th 2009, 10:20 AM
slevvio
yeah thanks guys, it's very strange as it says well defined in all my notes and examples then suddenly it starts talking about consistently definedeness
• Nov 18th 2009, 10:47 AM
slevvio
Two Isomorphism Questions
Hello I am struggling with these 2 questions.
1) Prove that $S_3$ and $D_6$ (symmetry group of the equilateral triangle) are isomorphic.
It's quite clear that they are but is there a way to show this without having to build an isomorphism that would take a long time as I would have to show every pair of permutations f,g in $S_3$ would give $\phi (fg) = \phi (f) \phi (g)$
I have this theorem in my notes: Every group G is isomorphic to a subgroup of $S_x$ for some choice of set X. [X = { bijections f: f: X -> X }]
In particular, if $|G| < \infty$ then G and H are isomorphic for some $H \le S_n$ for some $n \in \mathbb{N}$.
Am I supposed to use this? Or can I just draw a Cayley table of both groups and say look the two groups are algebraically indistinguishable and hence are isomorphic? I found a few proofs on the internet but I don't understand the notation or theorems they use.
2)Let G be a group and suppose that G has subgroups H and K for which the following conditions all hold:
(a) G = HK, (b) $H \cap K = {1}$, (c) $H \lhd G$, (d) $K \lhd G$.
Show also that $G \cong H x K$.
HINT: Show first that any element of G can be uniquely expressed in the form hk, where $h \in H, k \in K$. Then, by considering elements of the form $h^{-1}k^{-1}hk$ where $h \in H, k \in K$ show that every element in H commutes with every element of K. Finally by considering a suitable mapping from H X K to G obtain the required isomorphism.
OK I can make the isomorphism from H x K to G but in proving it is bijective I need to use commutativity and uniqueness of elements in G so how do i prove these 2 things ( in red ) ? I have pages and pages of h's and k's. I was trying to show that $h^{-1}k^{-1}hk$was in H and in K and therefore in $H \cap K = {1}$
And of course if $h^{-1}k^{-1}hk = 1$ then $hk = kh$
Any help and discussion would be appreciated. :)
• Nov 18th 2009, 12:08 PM
Drexel28
Quote:
Originally Posted by slevvio
Hello I am struggling with these 2 questions.
1) Prove that $S_3$ and $D_6$ (symmetry group of the equilateral triangle) are isomorphic.
It's quite clear that they are but is there a way to show this without having to build an isomorphism that would take a long time as I would have to show every pair of permutations f,g in $S_3$ would give $\phi (fg) = \phi (f) \phi (g)$
I have this theorem in my notes: Every group G is isomorphic to a subgroup of $S_x$ for some choice of set X. [X = { bijections f: f: X -> X }]
In particular, if $|G| < \infty$ then G and H are isomorphic for some $H \le S_n$ for some $n \in \mathbb{N}$.
Am I supposed to use this? Or can I just draw a Cayley table of both groups and say look the two groups are algebraically indistinguishable and hence are isomorphic? I found a few proofs on the internet but I don't understand the notation or theorems they use.
2)Let G be a group and suppose that G has subgroups H and K for which the following conditions all hold:
(a) G = HK, (b) $H \cap K = {1}$, (c) $H \lhd G$, (d) $K \lhd G$.
Show also that $G \cong H x K$.
HINT: Show first that any element of G can be uniquely expressed in the form hk, where $h \in H, k \in K$. Then, by considering elements of the form $h^{-1}k^{-1}hk$ where $h \in H, k \in K$ show that every element in H commutes with every element of K. Finally by considering a suitable mapping from H X K to G obtain the required isomorphism.
OK I can make the isomorphism from H x K to G but in proving it is bijective I need to use commutativity and uniqueness of elements in G so how do i prove these 2 things ( in red ) ? I have pages and pages of h's and k's. I was trying to show that $h^{-1}k^{-1}hk$was in H and in K and therefore in $H \cap K = {1}$
And of course if $h^{-1}k^{-1}hk = 1$ then $hk = kh$
Any help and discussion would be appreciated. :)
1. $D_{2n}\subseteq S_n$. Also, $\left|D_{2n}\right|=2n, \left|S_{n}\right|=n!$. So $\left|S_{3}\right|=3!=6$ and $\left|D_{6}\right|=6$ and since $D_{6}\subseteq S_{3}$ the conclusion follows.
2
• Nov 19th 2009, 02:13 AM
slevvio
why is $D_{2n} \subset S_n$? And why does their order being the same make them isomorphic?
• Nov 19th 2009, 06:29 AM
Drexel28
Quote:
Originally Posted by slevvio
why is $D_{2n} \subset S_n$? And why does their order being the same make them isomorphic?
First of all, I used bad notation. I should have put $D_{2n}\le S_n$. Secondly, $D_{2n}$ is a permutation group, in other words it is isomorphic to a subgroup of the symmetric group on n letters. Think of it as $S_{n}$ is "how many ways can I permute the numbers $\left\{1,\cdots,n\right\}$" and $D_{2n}$ as "how many ways can I permute $\left\{1,\cdots,n\right\}$ so that each number still shares it's original neighbors?". So we can easily define an isomorphism between the subsetof $S_n$ that's "neighbor preserving" and $D_{2n}$. But a subgroup with exactly the same elements as the group cannot be proper, which forces it to be improper. | 2,531 | 8,248 | {"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": 87, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.937531 |
https://en.wikipedia.org/wiki/Reservoir_sampling | 1,726,788,069,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00423.warc.gz | 207,669,236 | 30,262 | # Reservoir sampling
Reservoir sampling is a family of randomized algorithms for choosing a simple random sample, without replacement, of k items from a population of unknown size n in a single pass over the items. The size of the population n is not known to the algorithm and is typically too large for all n items to fit into main memory. The population is revealed to the algorithm over time, and the algorithm cannot look back at previous items. At any point, the current state of the algorithm must permit extraction of a simple random sample without replacement of size k over the part of the population seen so far.
## Motivation
Suppose we see a sequence of items, one at a time. We want to keep ten items in memory, and we want them to be selected at random from the sequence. If we know the total number of items n and can access the items arbitrarily, then the solution is easy: select 10 distinct indices i between 1 and n with equal probability, and keep the i-th elements. The problem is that we do not always know the exact n in advance.
## Simple: Algorithm R
A simple and popular but slow algorithm, Algorithm R, was created by Jeffrey Vitter.[1]
Initialize an array ${\displaystyle R}$ indexed from ${\displaystyle 1}$ to ${\displaystyle k}$, containing the first k items of the input ${\displaystyle x_{1},...,x_{k}}$. This is the reservoir.
For each new input ${\displaystyle x_{i}}$, generate a random number j uniformly in ${\displaystyle \{1,...,i\}}$. If ${\displaystyle j\in \{1,...,k\}}$, then set ${\displaystyle R[j]:=x_{i}.}$ Otherwise, discard ${\displaystyle x_{i}}$.
Return ${\displaystyle R}$ after all inputs are processed.
This algorithm works by induction on ${\displaystyle i\geq k}$.
Proof
When ${\displaystyle i=k}$, Algorithm R returns all inputs, thus providing the basis for a proof by mathematical induction.
Here, the induction hypothesis is that the probability that a particular input is included in the reservoir just before the ${\displaystyle (i+1)}$-th input is processed is ${\displaystyle {\frac {k}{i}}}$ and we must show that the probability that a particular input is included in the reservoir is ${\displaystyle {\frac {k}{i+1}}}$ just after the ${\displaystyle (i+1)}$-th input is processed.
Apply Algorithm R to the ${\displaystyle (i+1)}$-th input. Input ${\displaystyle x_{i+1}}$ is included with probability ${\displaystyle {\frac {k}{i+1}}}$ by definition of the algorithm. For any other input ${\displaystyle x_{r}\in \{x_{1},...,x_{i}\}}$, by the induction hypothesis, the probability that it is included in the reservoir just before the ${\displaystyle (i+1)}$-th input is processed is ${\displaystyle {\frac {k}{i}}}$. The probability that it is still included in the reservoir after ${\displaystyle x_{i+1}}$ is processed (in other words, that ${\displaystyle x_{r}}$ is not replaced by ${\displaystyle x_{i+1}}$) is ${\displaystyle {\frac {k}{i}}\times \left(1-{\frac {1}{i+1}}\right)={\frac {k}{i+1}}}$. The latter follows from the assumption that the integer ${\displaystyle j}$ is generated uniformly at random; once it becomes clear that a replacement will in fact occur, the probability that ${\displaystyle x_{r}}$ in particular is replaced by ${\displaystyle x_{i+1}}$ is ${\displaystyle {\frac {k}{i+1}}{\frac {1}{k}}={\frac {1}{i+1}}}$.
We have shown that the probability that a new input enters the reservoir is equal to the probability that an existing input in the reservoir is retained. Therefore, we conclude by the principle of mathematical induction that Algorithm R does indeed produce a uniform random sample of the inputs.
While conceptually simple and easy to understand, this algorithm needs to generate a random number for each item of the input, including the items that are discarded. The algorithm's asymptotic running time is thus ${\displaystyle O(n)}$. Generating this amount of randomness and the linear run time causes the algorithm to be unnecessarily slow if the input population is large.
This is Algorithm R, implemented as follows:
(* S has items to sample, R will contain the result *)
ReservoirSample(S[1..n], R[1..k])
// fill the reservoir array
for i := 1 to k
R[i] := S[i]
end
// replace elements with gradually decreasing probability
for i := k+1 to n
(* randomInteger(a, b) generates a uniform integer from the inclusive range {a, ..., b} *)
j := randomInteger(1, i)
if j <= k
R[j] := S[i]
end
end
end
## Optimal: Algorithm L
If we generate ${\displaystyle n}$ random numbers ${\displaystyle u_{1},...,u_{n}\sim U[0,1]}$ independently, then the indices of the smallest ${\displaystyle k}$ of them is a uniform sample of the k-subsets of ${\displaystyle \{1,...,n\}}$.
The process can be done without knowing ${\displaystyle n}$:
Keep the smallest ${\displaystyle k}$ of ${\displaystyle u_{1},...,u_{i}}$ that has been seen so far, as well as ${\displaystyle w_{i}}$, the index of the largest among them. For each new ${\displaystyle u_{i+1}}$, compare it with ${\displaystyle u_{w_{i}}}$. If ${\displaystyle u_{i+1}, then discard ${\displaystyle u_{w_{i}}}$, store ${\displaystyle u_{i+1}}$, and set ${\displaystyle w_{i+1}}$ to be the index of the largest among them. Otherwise, discard ${\displaystyle u_{i+1}}$, and set ${\displaystyle w_{i+1}=w_{i}}$.
Now couple this with the stream of inputs ${\displaystyle x_{1},...,x_{n}}$. Every time some ${\displaystyle u_{i}}$ is accepted, store the corresponding ${\displaystyle x_{i}}$. Every time some ${\displaystyle u_{i}}$ is discarded, discard the corresponding ${\displaystyle x_{i}}$.
This algorithm still needs ${\displaystyle O(n)}$ random numbers, thus taking ${\displaystyle O(n)}$ time. But it can be simplified.
First simplification: it is unnecessary to test new ${\displaystyle u_{i+1},u_{i+2},...}$ one by one, since the probability that the next acceptance happens at ${\displaystyle u_{i+l}}$ is ${\displaystyle (1-u_{w_{i}})^{l-1}\,u_{w_{i}}=(1-u_{w_{i}})^{l-1}-(1-u_{w_{i}})^{l}}$, that is, the interval ${\displaystyle l}$ of acceptance follows a geometric distribution.
Second simplification: it's unnecessary to remember the entire array of the smallest ${\displaystyle k}$ of ${\displaystyle u_{1},...,u_{i}}$ that has been seen so far, but merely ${\displaystyle w}$, the largest among them. This is based on three observations:
• Every time some new ${\displaystyle x_{i+1}}$ is selected to be entered into storage, a uniformly random entry in storage is discarded.
• ${\displaystyle u_{i+1}\sim U[0,w]}$
• ${\displaystyle w_{i+1}}$ has the same distribution as ${\displaystyle \max\{u_{1},...,u_{k}\}}$, where all ${\displaystyle u_{1},...,u_{k}\sim U[0,w]}$ independently. This can be sampled by first sampling ${\displaystyle u\sim U[0,1]}$, then taking ${\displaystyle w\cdot u^{\frac {1}{k}}}$.
This is Algorithm L,[2] which is implemented as follows:
(* S has items to sample, R will contain the result *)
ReservoirSample(S[1..n], R[1..k])
// fill the reservoir array
for i = 1 to k
R[i] := S[i]
end
(* random() generates a uniform (0,1) random number *)
W := exp(log(random())/k)
while i <= n
i := i + floor(log(random())/log(1-W)) + 1
if i <= n
(* replace a random item of the reservoir with item i *)
R[randomInteger(1,k)] := S[i] // random index between 1 and k, inclusive
W := W * exp(log(random())/k)
end
end
end
This algorithm computes three random numbers for each item that becomes part of the reservoir, and does not spend any time on items that do not. Its expected running time is thus ${\displaystyle O(k(1+\log(n/k)))}$,[2] which is optimal.[1] At the same time, it is simple to implement efficiently and does not depend on random deviates from exotic or hard-to-compute distributions.
## With random sort
If we associate with each item of the input a uniformly generated random number, the k items with the largest (or, equivalently, smallest) associated values form a simple random sample.[3] A simple reservoir-sampling thus maintains the k items with the currently largest associated values in a priority queue.
(*
S is a stream of items to sample
S.Current returns current item in stream
S.Next advances stream to next position
min-priority-queue supports:
Count -> number of items in priority queue
Minimum -> returns minimum key value of all items
Extract-Min() -> Remove the item with minimum key
Insert(key, Item) -> Adds item with specified key
*)
ReservoirSample(S[1..?])
H := new min-priority-queue
while S has data
r := random() // uniformly random between 0 and 1, exclusive
if H.Count < k
H.Insert(r, S.Current)
else
// keep k items with largest associated keys
if r > H.Minimum
H.Extract-Min()
H.Insert(r, S.Current)
end
S.Next
end
return items in H
end
The expected running time of this algorithm is ${\displaystyle O(n+k\log k\log(n/k))}$ and it is relevant mainly because it can easily be extended to items with weights.
## Weighted random sampling
This method, also called sequential sampling, is incorrect in the sense that it does not allow to obtain a priori fixed inclusion probabilities. Some applications require items' sampling probabilities to be according to weights associated with each item. For example, it might be required to sample queries in a search engine with weight as number of times they were performed so that the sample can be analyzed for overall impact on user experience. Let the weight of item i be ${\displaystyle w_{i}}$, and the sum of all weights be W. There are two ways to interpret weights assigned to each item in the set:[4]
1. In each round, the probability of every unselected item to be selected in that round is proportional to its weight relative to the weights of all unselected items. If X is the current sample, then the probability of an item ${\displaystyle i\notin X}$ to be selected in the current round is ${\displaystyle \textstyle w_{i}/(W-\sum _{j\in X}{w_{j}})}$.
2. The probability of each item to be included in the random sample is proportional to its relative weight, i.e., ${\displaystyle w_{i}/W}$. Note that this interpretation might not be achievable in some cases, e.g., ${\displaystyle k=n}$.
### Algorithm A-Res
The following algorithm was given by Efraimidis and Spirakis that uses interpretation 1:[5]
(*
S is a stream of items to sample
S.Current returns current item in stream
S.Weight returns weight of current item in stream
S.Next advances stream to next position
The power operator is represented by ^
min-priority-queue supports:
Count -> number of items in priority queue
Minimum() -> returns minimum key value of all items
Extract-Min() -> Remove the item with minimum key
Insert(key, Item) -> Adds item with specified key
*)
ReservoirSample(S[1..?])
H := new min-priority-queue
while S has data
r := random() ^ (1/S.Weight) // random() produces a uniformly random number in (0,1)
if H.Count < k
H.Insert(r, S.Current)
else
// keep k items with largest associated keys
if r > H.Minimum
H.Extract-Min()
H.Insert(r, S.Current)
end
end
S.Next
end
return items in H
end
This algorithm is identical to the algorithm given in Reservoir Sampling with Random Sort except for the generation of the items' keys. The algorithm is equivalent to assigning each item a key ${\displaystyle r^{1/w_{i}}}$ where r is the random number and then selecting the k items with the largest keys. Equivalently, a more numerically stable formulation of this algorithm computes the keys as ${\displaystyle -\ln(r)/w_{i}}$ and select the k items with the smallest keys.[6][failed verification]
### Algorithm A-ExpJ
The following algorithm is a more efficient version of A-Res, also given by Efraimidis and Spirakis:[5]
(*
S is a stream of items to sample
S.Current returns current item in stream
S.Weight returns weight of current item in stream
S.Next advances stream to next position
The power operator is represented by ^
min-priority-queue supports:
Count -> number of items in the priority queue
Minimum -> minimum key of any item in the priority queue
Extract-Min() -> Remove the item with minimum key
Insert(Key, Item) -> Adds item with specified key
*)
ReservoirSampleWithJumps(S[1..?])
H := new min-priority-queue
while S has data and H.Count < k
r := random() ^ (1/S.Weight) // random() produces a uniformly random number in (0,1)
H.Insert(r, S.Current)
S.Next
end
X := log(random()) / log(H.Minimum) // this is the amount of weight that needs to be jumped over
while S has data
X := X - S.Weight
if X <= 0
t := H.Minimum ^ S.Weight
r := random(t, 1) ^ (1/S.Weight) // random(x, y) produces a uniformly random number in (x, y)
H.Extract-Min()
H.Insert(r, S.Current)
X := log(random()) / log(H.Minimum)
end
S.Next
end
return items in H
end
This algorithm follows the same mathematical properties that are used in A-Res, but instead of calculating the key for each item and checking whether that item should be inserted or not, it calculates an exponential jump to the next item which will be inserted. This avoids having to create random variates for each item, which may be expensive. The number of random variates required is reduced from ${\displaystyle O(n)}$ to ${\displaystyle O(k\log(n/k))}$ in expectation, where ${\displaystyle k}$ is the reservoir size, and ${\displaystyle n}$ is the number of items in the stream.[5]
### Algorithm A-Chao
Warning: the following description is wrong, see Chao's original paper and the discussion here.
Following algorithm was given by M. T. Chao uses interpretation 2:[7] and Tillé (2006).[8]
(*
S has items to sample, R will contain the result
S[i].Weight contains weight for each item
*)
WeightedReservoir-Chao(S[1..n], R[1..k])
WSum := 0
// fill the reservoir array
for i := 1 to k
R[i] := S[i]
WSum := WSum + S[i].Weight
end
for i := k+1 to n
WSum := WSum + S[i].Weight
p := S[i].Weight / WSum // probability for this item
j := random(); // uniformly random between 0 and 1
if j <= p // select item according to probability
R[randomInteger(1,k)] := S[i] //uniform selection in reservoir for replacement
end
end
end
For each item, its relative weight is calculated and used to randomly decide if the item will be added into the reservoir. If the item is selected, then one of the existing items of the reservoir is uniformly selected and replaced with the new item. The trick here is that, if the probabilities of all items in the reservoir are already proportional to their weights, then by selecting uniformly which item to replace, the probabilities of all items remain proportional to their weight after the replacement.
Note that Chao doesn't specify how to sample the first k elements. He simple assumes we have some other way of picking them in proportion to their weight. Chao: "Assume that we have a sampling plan of fixed size with respect to S_k at time A; such that its first-order inclusion probability of X_t is π(k; i)".
#### Algorithm A-Chao with Jumps
Similar to the other algorithms, it is possible to compute a random weight j and subtract items' probability mass values, skipping them while j > 0, reducing the number of random numbers that have to be generated.[4]
(*
S has items to sample, R will contain the result
S[i].Weight contains weight for each item
*)
WeightedReservoir-Chao(S[1..n], R[1..k])
WSum := 0
// fill the reservoir array
for i := 1 to k
R[i] := S[i]
WSum := WSum + S[i].Weight
end
j := random() // uniformly random between 0 and 1
pNone := 1 // probability that no item has been selected so far (in this jump)
for i := k+1 to n
WSum := WSum + S[i].Weight
p := S[i].Weight / WSum // probability for this item
j -= p * pNone
pNone := pNone * (1 - p)
if j <= 0
R[randomInteger(1,k)] := S[i] //uniform selection in reservoir for replacement
j = random()
pNone := 1
end
end
end
## Relation to Fisher–Yates shuffle
Suppose one wanted to draw k random cards from a deck of cards. A natural approach would be to shuffle the deck and then take the top k cards. In the general case, the shuffle also needs to work even if the number of cards in the deck is not known in advance, a condition which is satisfied by the inside-out version of the Fisher–Yates shuffle:[9]
(* S has the input, R will contain the output permutation *)
Shuffle(S[1..n], R[1..n])
R[1] := S[1]
for i from 2 to n do
j := randomInteger(1, i) // inclusive range
R[i] := R[j]
R[j] := S[i]
end
end
Note that although the rest of the cards are shuffled, only the first k are important in the present context. Therefore, the array R need only track the cards in the first k positions while performing the shuffle, reducing the amount of memory needed. Truncating R to length k, the algorithm is modified accordingly:
(* S has items to sample, R will contain the result *)
ReservoirSample(S[1..n], R[1..k])
R[1] := S[1]
for i from 2 to k do
j := randomInteger(1, i) // inclusive range
R[i] := R[j]
R[j] := S[i]
end
for i from k + 1 to n do
j := randomInteger(1, i) // inclusive range
if (j <= k)
R[j] := S[i]
end
end
end
Since the order of the first k cards is immaterial, the first loop can be removed and R can be initialized to be the first k items of the input. This yields Algorithm R.
## Limitations
Reservoir sampling makes the assumption that the desired sample fits into main memory, often implying that k is a constant independent of n. In applications where we would like to select a large subset of the input list (say a third, i.e. ${\displaystyle k=n/3}$), other methods need to be adopted. Distributed implementations for this problem have been proposed.[10]
## References
1. ^ a b Vitter, Jeffrey S. (1 March 1985). "Random sampling with a reservoir" (PDF). ACM Transactions on Mathematical Software. 11 (1): 37–57. CiteSeerX 10.1.1.138.784. doi:10.1145/3147.3165. S2CID 17881708.
2. ^ a b Li, Kim-Hung (4 December 1994). "Reservoir-Sampling Algorithms of Time Complexity O(n(1+log(N/n)))". ACM Transactions on Mathematical Software. 20 (4): 481–493. doi:10.1145/198429.198435. S2CID 15721242.
3. ^ Fan, C.; Muller, M.E.; Rezucha, I. (1962). "Development of sampling plans by using sequential (item by item) selection techniques and digital computers". Journal of the American Statistical Association. 57 (298): 387–402. doi:10.1080/01621459.1962.10480667. JSTOR 2281647.
4. ^ a b Efraimidis, Pavlos S. (2015). "Weighted Random Sampling over Data Streams". Algorithms, Probability, Networks, and Games. Lecture Notes in Computer Science. Vol. 9295. pp. 183–195. arXiv:1012.0256. doi:10.1007/978-3-319-24024-4_12. ISBN 978-3-319-24023-7. S2CID 2008731.
5. ^ a b c Efraimidis, Pavlos S.; Spirakis, Paul G. (2006-03-16). "Weighted random sampling with a reservoir". Information Processing Letters. 97 (5): 181–185. doi:10.1016/j.ipl.2005.11.003.
6. ^ Arratia, Richard (2002). Bela Bollobas (ed.). "On the amount of dependence in the prime factorization of a uniform random integer". Contemporary Combinatorics. 10: 29–91. CiteSeerX 10.1.1.745.3975. ISBN 978-3-642-07660-2.
7. ^ Chao, M. T. (1982). "A general purpose unequal probability sampling plan". Biometrika. 69 (3): 653–656. doi:10.1093/biomet/69.3.653.
8. ^ Tillé, Yves (2006). Sampling Algorithms. Springer. ISBN 978-0-387-30814-2.
9. ^ National Research Council (2013). Frontiers in Massive Data Analysis. The National Academies Press. p. 121. ISBN 978-0-309-28781-4.
10. ^ Reservoir Sampling in MapReduce | 5,165 | 19,318 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 82, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-38 | latest | en | 0.897923 |
http://mysmartphonekit.mobileeducationkit.net/index.php/program-9/ | 1,623,708,731,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00616.warc.gz | 35,839,518 | 5,922 | Define a class with a generator which can iterate the numbers, which are divisible by 7, between a given range 0 and n.
Solution:
def putNumbers(n):
i = 0
while i<n:
j=i
i=i+1
if j%7==0:
yield j
for i in reverse(100):
print i | 72 | 226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-25 | longest | en | 0.837623 |
https://en.wikisource.org/wiki/Page:EB1911_-_Volume_22.djvu/717 | 1,566,241,524,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314904.26/warc/CC-MAIN-20190819180710-20190819202710-00115.warc.gz | 444,281,302 | 12,010 | # Page:EB1911 - Volume 22.djvu/717
701
PYTHAGORAS
squares (Euclid l. 47). Commonly called the “ theorem of Pythagoras, " is attributed to him by many authorities, of whom the oldest is Vitruvius.1 (12) One of the methods of finding right-angled triangles whose sides can be expressed in numbers (Pythagorean triangles)-that setting out from the odd numbers-is referred to Pythagoras by Heron of Alexandria and Proclus? (13) The discovery of irrational quantities is ascribed to Pythagoras by Eudemus (Procl. op. cil. p. 65). (14) The three proportions-arithmetical, geometrical and harmonica-were known to Pythagorasi (15) Iamblichus' says, “ Formerly, in the time of Pythagoras and the mathematicians under him, there were three means only-the arithmetical, the geometrical and the third in order, which was known by the name sub-contrary (bare:/avria), but which Archytas and Hippasus designated the harmonica, since it appeared to include the ratios concerning harmony and melody.” (16) The so-called most perfect or musical proportion, e.g. 6:8::9:12, which comprehends in it all the former ratios, according to Iamblichusf' is said to be an invention of the Babylonians and to have been first brought into Greece by Pythagoras. (17) Arithmetical progressions were treated by the Pythagoreans, and it appears from a passage in Lucian that Pythagoras himself had considered the special case of triangular numbers: Pythagoras asks some one, “How do you count?" He replies, “One, two, three, four." Pythagoras, interrupting, says, “Do you see? what you take to be four, that is ten and a perfect triangle and our oath."' (18) The odd numbers were called by the Pythagoreans “ gnomons, "' and were regarded as generating, inasmuch as by the addition of successive gnomons-consisting each of an odd number of unit squares—to the original square unit or monad the square form was preserved. (19) In like manner, if the simplest oblong (éreponrmes), consisting of two unit squares or monads in juxtaposition, be taken and four unit squares be placed about it after the manner of a gnomon, and then in like manner six, eight unit squares be placed in succession, the oblong form will be preserved. (20) Another of his doctrines was, that of all solid figures the sphere was the most beautiful, and of all plane figures the circle.” (21) According to Iamblichus the Pythagoreans are said to have found the quadrature of the circle.” him as a symbol of health. It is said to have obtained its special name from the letters u, -y 1, 0 (=¢i), a. having been written at its prominent vertices.
De arch. ix.; Praef. 5, 6, 7. Amongst other authorities are Diogenes Laertius (viii. 11), Proclus (op. cit., p. 426), and Plutarch (ut supra, 6). Plutarch, however, attributes to the Egyptians the knowledge of this theorem in the particular case where the sides are 3, 4, and 5 (De Is. et Osir. c. 56),
2 Heron Alex. Geam. et stereom. rel., ed. F. Hultsch, pp. 56, 146; Procl. op. cit. p. 428. The method of Pythagoras is as follows: he took an odd number as the lesser side; then, having squared this number and diminished the square by unity, he took half the remainder as the greater side, and by adding unity to this number he obtained the hypotenuse, e.g. 3, 4, 5; 5, 12, 13. Nicom. Ger. Inlrod. Ar. c. xxii.
In Nicomachi arithmelicam, ed. S. Tennulius, p. 141. 5 Op. cil. p. 168. As an example of this proportion Nicomachus and, after him, lamblichus give the numbers 6, 8, 9, 12, the harmonica and arithmetical means between two numbers forming a geometric proportion with the numbers themselves az-2213-2:-@:b a-H2 2
Iamblichus further relates (loc. cit.) that many Pythagoreans made use of this proportion, as Aristaeus of Crotona, Timaeus of Locri, Philolaus and Archytas of Tarentum and many others, and after them Plato in his Timaeus (see Nicom. Inst. arithm., ed. Ast, p. 153, and Animadversiones, pp. 327-329; and Iambl. op. cit. p. 172 se .).
fl
° Bfwv 1rp&'0'LS, 4, l. 317, Cd. C. I3COlI)ltZ. 7 Fvdipwv means that by which anything is known or “ criterion "; its oldest concrete signification seems to be the carpenter's square (norma) by which a right angle is known. Hence it came to denote a perpendicular, of which, indeed, it was the archaic name (Proclus, op. cit. p. 283). Gnonion is also an instrument for measuring altitudes, by means of which the meridian can be found; it denotes, further, the index or style of a sundial, the shadow of which points out the hours. In geometry it means the square or rectangle about the diagonal of a square or rectangle, together with the two complements, on account of the resemblance of the figure to a carpenter's square; and then, more generally, the similar figure with regard to any parallelogram, as defined by Euclid II. def. 2. Again, in a still more general signification, it means the figure which, being added to any figure, preserves the original form. See Heron, Dejiniliones (59). Vi/hen gnomons are added successively in this manner to a square monad, the first gnomon may be regarded as that consisting of three square monads, and is indeed the constituent of a simple Greek fret; the second of five square monads, &c.; hence we have the gnomonic numbers.
3 Diag. Laért. De vit Pyth. viii. 19.
Simplicius, In Arislotelis physicorum libro.: quattuor priores comment aria, ed. H. Diels, p. 60.
On examining the purely geometrical work of Pythagoras and his early disciples, as given in the preceding extracts, we observe that it is much concerned with the geometry of areas, and we are indeed struck with its E yptian character. This appears in the theorem (3) concerning the Elling up a plane with regular figures for floors or walls covered with tiles of various colours were common in Egypt; in the construction of the regular solids (8), for some of them are found in Egyptian architecture; in the problems concerning the application of areas (5); and lastly, in the theorem of Pythagoras (II), coupled with his rule for the construction of rightanglcd triangles in numbers (12). We learn from Plutarch that the Egyptians were acquainted with the geometrical fact that a. triangle whose sides contain three, four and five parts is right angled, and that the square of the greatest side is equal to the squares of the sides containing the right angle. It is probable too that this theorem was known to them in the simple case where the right-angled triangle is isosceles, inasmuch as it would be at once suggested by the contemplation of a floor covered with square tiles -the square on the diagonal and the sum of the squares on the sides contain each four of the right-angled triangles into which one of the squares is divided by its diagonal. It is easy now to see how the problem to construct a square which shall be equal to the sum of two squares could, in some cases, be solved numerically. From the observation of a chequered board it would be perceived that the element in the successive formation of squares is the gnomon or carpenter's square. Each gnomon consists of an odd number of squares, and the successive gnomons correspond to the successive odd numbers, and include, therefore, all odd squares. Suppose, now, two squares are given, one consisting of sixteen and the other of nine unit squares, and that it is proposed to form from them another square. It is evident that the square consisting of nine unit squares can take the form of the fourth gnomon, which, being placed round the former square, will generate a new square containing twenty-five unit squares. Similarly it may have been observed that the twelfth gnomon, consisting of twenty-five unit squares, could be transformed into a square each of whose sides contains five units, and thus it may have been seen conversely that the latter s uare, by taking the gnomonic or generating form with respect to flue square on twelve units as base, would produce the square of thirteen units, and so on. This method required only to be generalized in order to enable Pythagoras to arrive at his rule for finding right-angled triangles whose sides can be expressed in numbers, which, we are told, sets out from the odd numbers. The nth square together with the nth gnomon forms the (n+1)th square; if the nth gnomon contains m' unit squares, rn being an odd number, we have 2n-I-1 =m2, . '.n=§ (m2-1), which gives the rule of Pythagoras.
The general proof of Euclid I. 47 is attributed to Pythagoras, but we have the express statement of Proclus (op. cil. p. 426) that this theorem was not proved in the first instance as it is in the Elements. The following simple and natural way of arriving at the theorem is suggested by Bretschneider after Camerer.1° A square can be dissected into the sum of two squares and two equal rectangles, as in Euclid II. 4; these two rectangles can, by drawing their diagonals, be decomposed into four equal right-angled triangles, the sum of the sides of each being equal to the side of the square; again, these four right-angled triangles can be placed so that a vertex of each shall be in one of the corners of the square in such a way that a greater and less side are in continuation. The original square is thus dissected into the four triangles as before and the figure within, which is the square on the hypotenuse. This square, therefore, must be equal to the sum of the squares on the sides of the right-angled triangle.
It is well known that the Pythagoreans were much occupied with the construction of regular polygons and solids, which in their cosmology played an essential part as the fundamental forms of the elements of the universe. We can trace the origin of these mathematical speculations in the theorem (3) that “the plane around a point is completely filled by six equilateral triangles, four squares, or three regular hexagons." Plato also makes the Pythagorean Timaeus explain—“ Each straight-lined figure consists of triangles, but all triangles can be dissected into rectangular ones which are either isosceles or scalene. Among the latter the most beautiful is that out of the doubling of which an equilateral arises, or in which the square of the greater perpendicular is three times that of the smaller, or in which the smaller perpendicular is half the hypotenuse. But two or four right-angled isosceles triangles, properly put together, form the square; two or six of the most beautiful scalene right-angled triangles form the equilateral triangle; and out of these two figures arise the solids which correspond with the four elements of the real world, the tetrahedron, octahedron, icosahedron and the cube ” 11 (Timaeus, 53, 54, 55). The construction of the regular solids is distinctly ascribed to Pythagoras himself by Eudemus (8). Of these five 1° See Bretsch. Die Georn. 'vor Enklides, p. 82; Camerer, Euclidis elem. i. 444, and the references given there.
“ The dodecahedron was assigned to the fifth element, quinta pars, aether, or, as some think, to the universe. (See PHILOLAUS.) | 2,758 | 11,003 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.962214 |
https://slideplayer.com/slide/13448438/ | 1,638,938,506,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363437.15/warc/CC-MAIN-20211208022710-20211208052710-00180.warc.gz | 578,019,424 | 19,731 | Meters, Grams and Liters
Presentation on theme: "Meters, Grams and Liters"— Presentation transcript:
Meters, Grams and Liters
The Metric System Meters, Grams and Liters
The Metric System The metric system is a measurement system based on our decimal (base 10) number system. Other countries and all scientists and engineers use the metric system for measurement.
Metric Prefixes Prefix Symbol Factor Number Factor Word Kilo- k 1000
Metric Units The metric system has prefix modifiers that are multiples of 10. Prefix Symbol Factor Number Factor Word Kilo- k 1000 Thousand Hecto- h 100 Hundred Deca- da or dk 10 Ten Unit m, l, or g 1 One Deci- d .1 Tenth Centi- c .01 Hundredth Milli- m .001 thousandth
Meters Meters measure length or distance
One millimeter is about the thickness of a dime.
Meters One centimeter is about the width of a large paper clip
Meters A meter is about the width of a doorway
Meters A kilometer is about six city blocks or 10 football fields.
1.6 kilometers is about 1 mile
Gram Grams are used to measure mass or the weight of an object.
Grams A milligram weighs about as much as a grain of salt.
Grams 1 gram weighs about as much as a small paper clip.
1 kilogram weighs about as much as 6 apples or 2 pounds.
Liters Liters measure liquids or capacity. 2 Liter Soda
Liter 1 milliliter is about the amount of one drop
Liter 1 liter is half of a 2 liter bottle of Coke or other soda
What unit would you use to measure each of the following items?
Which unit would you use to measure the length of this bicycle?
km m cm mm
Which unit would you use to measure the mass of a penny?
km g cL mg
Which unit would you use to measure the water in an aquarium?
L m cL mg
Which unit would you use to measure the mass of a feather?
L m cL mg
Which unit would you use to measure the mass of a student desk?
kg g cL mg
Which unit would you use to measure the mass of a whole watermelon?
kg g cL mg
Which unit would you use to measure the mass of an egg?
kg g mm mg
Which unit would you use to measure a can of soup?
kL L mm mL
Which unit would you use to measure this glass of milk?
kL L cL mL
Which unit would you use to measure the distance across Kansas?
km m cm mL
Which unit would you use to measure the height of a tree?
km m cm mm
Which unit would you use to measure the length of a bracelet?
km m cm mm
Changing Metric Units To change from one unit to another in the metric system you simply multiply or divide by a power of 10.
Place Values of Metric Prefixes
Move the decimal point to the right to multiply. Move the decimal point to the left to divide.
Commonly used Metric Units of Length, Mass and Volume
Length mm cm m km (small) (big) Mass Mg g kg (small) (Big) Volume mL L (small) (Big)
To change from a larger unit to a smaller unit, you need to multiply.
km m (x 1000) 1000 m = 1 km m cm (x 100) 100 cm = 1 m cm mm (x 10) 10 mm = 1 cm Length Example- 15 km = ____m 15 x 1000 = 15,000 m
Mass g mg (x 1000) 1,000 mg = 1kg kg g (x 1000) 1,000 g = 1kg Example- 9 g = ____ mg 9 x 1000 = 9,000 mg
Volume L mL (x 1000) 1,000 mL = 1 L Example- 60 L = ___ mL 60 x 1000 = 60,000 mL
To change from smaller units to larger units you divide by a power of ten.
Length Example- 50 cm = ____ m 50 ÷ 100 = .5 m 75 mm = ___ cm 75 ÷ 10 = 7.5 cm mm cm (÷ 10) 10 mm = 1 cm cm m (÷ 100) 100 cm = 1m m km (÷ 1000) 1,000 m = 1km
Mass mg g (÷ 1000) 1,000 mg = 1kg g kg (÷ 1000) 1,000 g = 1kg Example- 600 mg = ____ g 600 ÷ 1000 = .6 g
Volume mL L (÷ 1000) 1,000 mL = 1 L Example- 500 mL = ____ L 500 ÷ 1000 = .5 L
Setting up a Proportion or Conversion Ratio
60 cm = ____ m -Write your conversion factor 100 cm = 1 m -Set up your proportion/conversion ratio 60 cm X -Cross multiply and solve for X X = 60 100 cm = 1 m 60 cm X X = .6 m
Similar presentations | 1,184 | 3,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2021-49 | latest | en | 0.925161 |
https://calendarmaniacs.com/hours-from-time/when-is-9-hours-from-8am.html | 1,722,676,832,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00101.warc.gz | 130,947,050 | 2,912 | 9 hours from 8am
Here we have calculated what time it will be 9 hours from 8am. In other words, what is 8am plus 9 hours? It does not matter if it is 8am today or any other day from the past or future.
To clarify, when we say 9 hours from 8am we mean 9 hours after 8am or 9 hours forward from 8am.
We of course took into account that there are twenty-four hours in a day, which include twelve hours in the am and twelve hours in the pm. 9 hours from 8am:
5pm
Hours From Time Calculator
9 hours from 8am is not all we have calculated. Please submit a similar question for us below.
hours from
10 hours from 8am
Now you know the time at 9 hours after 8am. Go here for the next question on our list that we have figured out for you. | 201 | 736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-33 | latest | en | 0.96278 |
https://web2.0calc.com/questions/5w-2-y-6-2w-5-y | 1,591,503,803,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523476.97/warc/CC-MAIN-20200607013327-20200607043327-00593.warc.gz | 594,013,221 | 6,296 | +0
# 5w^-2 y^6 * 2w^5 y
0
377
1
5w^-2 y^6 * 2w^5 y
simplify
Feb 1, 2018
#1
+109527
+2
5w^-2 y^6 * 2w^5 y
$$5w^{-2} y^6 \times 2w^5 y\\5w^{-2} y^6 \times 2w^5 y^1\\ =5\times 2w^{-2+5} y^{6+1}\\ =10w^{3} y^{7}\\$$
.
Feb 1, 2018
#1
+109527
+2
$$5w^{-2} y^6 \times 2w^5 y\\5w^{-2} y^6 \times 2w^5 y^1\\ =5\times 2w^{-2+5} y^{6+1}\\ =10w^{3} y^{7}\\$$ | 230 | 356 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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} | 3.625 | 4 | CC-MAIN-2020-24 | latest | en | 0.32988 |
https://www.proprofs.com/quiz-school/story.php?title=take-me-to-moon-facts-info-about-our-nearest-neighbour | 1,721,454,699,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515020.57/warc/CC-MAIN-20240720052626-20240720082626-00689.warc.gz | 810,367,885 | 101,138 | # Take Me To The Moon - Facts And Info About Our Nearest Neighbour
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Student submitted questions about the moon to test your knowledge. Use whatever resources you have available (e. G. A knowledgable person, the internet, or books) and see if you can answer these questions correctly.
• 1.
### If you can jump 3 feet high (92 centimeters) on the Earth, how high could you jump on the Moon?
• A.
Just over 18 feet (about 5½ meters).
• B.
2 feet [ 60 cm]
• C.
1 foot [30 cm]
• D.
7 foot [ 2.1 m]
A. Just over 18 feet (about 5½ meters).
Explanation
How high can you jump on the Moon?
The gravitational field of the Moon at its surface only
162.2 cm/sec/sec
----------------- = 0.16
983 cm /sec/sec
or about 1/6th of what we feel on the surface of Earth in terms of the acceleration of gravity, so this means that for the same energy expended, you should be able to jump about 6 times higher. I can jump about 20 inches, so that means on the Moon I could jump 6 x 20 = 120 inches or 10 feet straight up!
http://www.daviddarling.info/childrens_encyclopedia/Moon_QA.html
Rate this question:
• 2.
### Who was the first person on the moon?
• A.
Elmo
• B.
Neil Armstrong
• C.
Buzz Lightyear
• D.
Buzz Aldron
B. Neil Armstrong
Explanation
http://www.nasa.gov/audience/forstudents/9-12/index.html
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• 3.
### How far is the moon from the earth?a) 384,400 kmb) 100,000 kmc) 500,000 kmd) 345,123 km
• A.
384,400 km
• B.
100,000 km
• C.
500,000 km
• D.
345,123 km
A. 384,400 km
Explanation
http://www.solarviews.com/eng/moon.htm
Rate this question:
• 4.
### Why is there no air on the moon?
• A.
Gases are drawn away from the moon by the earth's gravity.
• B.
There once were gases, but they turned to stone in the cold lunar nights.
• C.
Gases are lost to space because of the moon's low gravity.
• D.
There never were any gases on the moon.
C. Gases are lost to space because of the moon's low gravity.
Explanation
Experiments performed by Apollo astronauts were able to confirm that the moon does have a very thin atmosphere.
The Moon has an atmosphere, but it is very tenuous. Gases in the lunar atmosphere are easily lost to space. Because of the Moon's low gravity, light atoms such as helium receive sufficient energy from solar heating that they escape in just a few hours. Heavier atoms take longer to escape, but are ultimately ionized by the Sun's ultraviolet radiation, after which they are carried away from the Moon by solar wind.
Because of the rate at which atoms escape from the lunar atmosphere, there must be a continuous source of particles to maintain even a tenuous atmosphere. Sources for the lunar atmosphere include the capture of particles from solar wind and the material released from the impact of comets and meteorites. For some atoms, particularly helium and argon, outgassing from the Moon's interior may also be a source.
http://spacemath.gsfc.nasa.gov/weekly/4Page26.pdf
http://lunarscience.arc.nasa.gov/kids/moon_air
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• 5.
### How often are the Earth, Moon and Sun in a straight line?
• A.
2 weeks
• B.
2 days
• C.
4 weeks
• D.
1 day
A. 2 weeks
Explanation
When the Moon, earth and Sun are in a straight line, their gravitational fields act together, causing king tides. These occur every two weeks, as the moon revolves around the Earth.
Eclipses (solar or lunar) occur when the Earth, Moon and Sun are in an almost exact straight line. They're sort of in a straight line every new and full Moon, but because the plane of the Moon's orbit around the Earth is slightly tilted relative to the plane of the Earth's orbit around the Sun, they don't usually line up exactly enough to make an eclipse.
http://www.llun.net/astronomy/MoonEarthSun.html
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• 6.
### How thick is the crust of the Moon?
• A.
71 km
• B.
65 km
• C.
70 km
• D.
68 km
D. 68 km
Explanation
The correct answer is 68 km. The crust of the Moon is approximately 68 km thick. This is based on scientific studies and data collected from lunar missions. The Moon's crust is composed of various types of rock and is thinner compared to the Earth's crust. Understanding the thickness of the Moon's crust is important for studying its geological history and formation processes.
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• 7.
### Where does the moon come from?
• A.
It was aways there
• B.
As a result of a massive earthquake part of Earth broke off on became the moon.
• C.
The moon was an asteroid that was caught inEarth's gravitational field.
• D.
A small planet hit Earth. The rebound brought with it enough material into orbit around our planet to form the moon.
D. A small planet hit Earth. The rebound brought with it enough material into orbit around our planet to form the moon.
Explanation
Many astronomers think that the Moon was formed after a Mars-sized object smashed into Earth over four and a half billion years ago. This collision caused material from both Earth and the colliding object to be thrown into orbit around Earth. This material eventually gathered together to form the Moon. At first the Moon was closer to Earth than it is now. Over time, it gradually moved farther away. The Moon is still moving away from Earth at a rate of about 1.5 inches or 3.8 centimeters a year.
http://lunarscience.arc.nasa.gov/kids/moonform
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• 8.
### How old is the moon?
• A.
9.7 trillion years
• B.
4.5 billion years
• C.
12 years
• D.
5.4 billion years
B. 4.5 billion years
Explanation
By studying data from rocks that the Apollo astronauts brought back from the Moon, scientists have found that the Moon is about four and a half billion years old, which makes it about the same age as Earth. The rocks to the right are among the oldest rocks brought back from the Moon.
http://www.woodlands-junior.kent.sch.uk/time/moon/facts.htm
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• 9.
### What was the Roman goddess of the moon's name?
• A.
Aries the moon goddess
• B.
Sisals the moon goddess
• C.
Diana the moon goddess
• D.
Amanda a moon goddess
C. Diana the moon goddess
Explanation
his evening the waxing crescent Moon will have aged by three days since its conjunction with the Sun. About one-eighth of its apparent disc will be well illuminated. The rest of it should be detectable due to earthshine. Artists and philosophers have long deemed this to be the most pleasing shape of the Moon.
The three-day-old Moon has traditionally been known as Diana's Bow (as in bow & arrow.) Diana was a beautiful virgin goddess in the ancient Roman pantheon (Artemis to the Greeks.) She was believed to rule both hunting and the Moon. Hence, her using a crescent Moon as a bow for her arrow combines aspects of her two domains. In Disney's 1940 classic animated feature, Fantasia, she is depicted in just that pose.
http://cs.astronomy.com/asycs/forums/p/13795/271002.aspx
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• 10.
### Roughly how long does it take for the moon to finish its cycle?
• A.
Half a year
• B.
• C.
Month
• D.
Year
C. Month
Explanation
The moon takes roughly a month to complete its cycle. This is known as the lunar month or the synodic month. During this time, the moon goes through its different phases, starting from the new moon, then waxing crescent, first quarter, waxing gibbous, full moon, waning gibbous, third quarter, and finally waning crescent before starting the cycle again. Each phase lasts for about a week, resulting in a total cycle of approximately 29.5 days.
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Related Topics | 2,078 | 8,460 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-30 | latest | en | 0.849308 |
affareryanp.web.app | 1,702,097,821,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00173.warc.gz | 103,760,764 | 6,856 | # Deriváty arcsinhu
I have been trying to find this derivative for nearly an hour and have had no luck. I know what the answer is supposed to be, but I keep getting 1/(x^2sqrt(x^2 - 1)). The answer is really close, but there should not be an x^2 in the final answer. PLEASE HELP!
In this case, y = f(x) = mx + b, for real numbers m and b, and the slope m is given by Tables of Formulas for Derivatives. A table of formulas for the first derivatives of common functions used in mathematics is presented. Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels. Dec 21, 2020 · The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot.
It can be easier to apply the definition of arcsine: $$x=\sin(\arcsin(x))$$ The “rule of inversion” ensures you that the derivative of the arcsine exists (with a condition that I'll deal with later) so you can differentiate both sides using the chain rule: $$1=\cos(\arcsin(x))\arcsin'(x)$$ Therefore $$\arcsin'(x)=\frac{1}{\cos(\arcsin(x))}$$ The condition I mentioned above is, of Please Subscribe here, thank you!!! https://goo.gl/JQ8NysDerivation of arcsinh(x) the Inverse Hyperbolic Sine Function Dec 28, 2020 · This is intended as a guide to assist those who must occasionally calculate derivatives in generally non-mathematical courses such as economics, and can also be used as a guide for those just starting to learn calculus. Proof for the derivative of the inverse sine of x. Jun 29, 2018 · (A new question of the week) In Monday’s post about fallacies in calculus, one of them used the definition of the derivative (or rather, misused it).
## Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history
Generally, a specific functional group of the compound participates in the derivatization reaction and transforms the educt to a derivate of deviating reactivity, solubility, boiling point, melting point, … Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. Girls Vs. Boys (Part 36) St Georges A100 Offer Holders 2021 Scottish Universities Masters in Social Work Applicants Derivative of arctan(x) Let’s use our formula for the derivative of an inverse function to find the deriva tive of the inverse of the tangent function: y = tan−1 x = arctan x. We simplify the equation by taking the tangent of both sides: Let f(x) = arcsin(x) + arccos(x) We first prove that f(x) is a constant function.
### 1. What are Derivative Instruments? A derivative is an instrument whose value is derived from the value of one or more underlying, which can be commodities, precious metals, currency, bonds, stocks, stocks indices, etc.
In case 3, there’s a tangent […] In this chapter we will cover many of the major applications of derivatives. Applications included are determining absolute and relative minimum and maximum function values (both with and without constraints), sketching the graph of a function without using a computational aid, determining the Linear Approximation of a function, L’Hospital’s Rule (allowing us to compute some limits we Specific Recommendations General Recommendations Implementation Notes This page: http://www.ecse.rpi.edu Last modified: You are running from , Number of hits since 5 In this chapter we introduce Derivatives.
I get the first derivative and I think the second one's correct too, but I tend to get lost from there on. Could anyone please help me with this ? Greetings, I'm working (playing) on a problem involving approximating the arcsin() function. I've attmpted to verify the known derivative of the arcsin function (d(arcsin))/dx = 1/sqrt(1-z^2) I know I have a mistake in my derivation. I've attached an electronic copy of my work We take the positive sign because cos y is positive for all values of y in the range of y = arcsin x, which is the 1st and 4th quadrants.(Topic 19 of Trigonometry.)Problem 2. An alternative method that doesn't require knowing the derivative of arcsin(x): y = arcsin(3x) sin(y) = 3x Differentiate both sides with respect to x.
Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. 1357 views around the world You can reuse this answer Creative Commons License Mar 07, 2007 Mar 23, 2017 Oct 16, 2016 Mar 27, 2008 Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. The derivative of any inverse trig function should not be memorized because it implies that you are memorizing a lot of other things, like the power reduction formulas, instead of deriving and understanding them. What is the derivative of [math]ar Please Subscribe here, thank you!!! https://goo.gl/JQ8NysDerivation of arcsinh(x) the Inverse Hyperbolic Sine Function Derivative of arcsin(x^3) http://www.rootmath.org | Calculus 1We use implicit differentiation to take the derivative of the inverse sine function: arcsin(x). It can be easier to apply the definition of arcsine: $$x=\sin(\arcsin(x))$$ The “rule of inversion” ensures you that the derivative of the arcsine exists (with a condition that I'll deal with later) so you can differentiate both sides using the chain rule: $$1=\cos(\arcsin(x))\arcsin'(x)$$ Therefore $$\arcsin'(x)=\frac{1}{\cos(\arcsin(x))}$$ The condition I mentioned above is, of Jun 02, 2017 This is a good result but we need some additional ideas to make the answer complete. What about the allowed values for x?Clearly, the derivative … 13.
The derivative of y = arccsc x. I T IS NOT NECESSARY to memorize the derivatives of this Lesson. Rather, the student should know now to derive them. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. Tables of Formulas for Derivatives. A table of formulas for the first derivatives of common functions used in mathematics is presented.
Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. 1357 views around the world You can reuse this answer Creative Commons License Mar 07, 2007 Mar 23, 2017 Oct 16, 2016 Mar 27, 2008 Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. The derivative of any inverse trig function should not be memorized because it implies that you are memorizing a lot of other things, like the power reduction formulas, instead of deriving and understanding them.
Apr 07, 2012 · d/dx [arcsin (3x)/x] = [3x/√ (1 - 9x²) - arcsin (3x)]/x² by the quotient rule. It can be easier to apply the definition of arcsine: $$x=\sin(\arcsin(x))$$ The “rule of inversion” ensures you that the derivative of the arcsine exists (with a condition that I'll deal with later) so you can differentiate both sides using the chain rule: $$1=\cos(\arcsin(x))\arcsin'(x)$$ Therefore $$\arcsin'(x)=\frac{1}{\cos(\arcsin(x))}$$ The condition I mentioned above is, of Please Subscribe here, thank you!!! https://goo.gl/JQ8NysDerivation of arcsinh(x) the Inverse Hyperbolic Sine Function Dec 28, 2020 · This is intended as a guide to assist those who must occasionally calculate derivatives in generally non-mathematical courses such as economics, and can also be used as a guide for those just starting to learn calculus.
změňte své peníze ve španělštině
trojúhelníková klouzavá průměrná pásma | 1,968 | 8,239 | {"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} | 4 | 4 | CC-MAIN-2023-50 | latest | en | 0.907854 |
https://pedagogue.app/activities-to-teach-students-about-repeating-decimals/ | 1,718,976,961,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00835.warc.gz | 395,063,763 | 25,975 | # Activities to Teach Students About Repeating Decimals
Repeating decimals can be a challenging concept for students to grasp, but there are many ways that teachers can make the topic more engaging and interactive. In this article, we will explore several activities to teach students about repeating decimals.
1. Fraction to Decimal Conversion Game
In this game, students will practice converting fractions to decimals. Begin by writing several fractions on the board or a piece of paper. Then, ask the students to convert these fractions to decimals. To make the game more engaging, use a timer to see how quickly the students can complete the task. Finally, have the students identify which of the decimals they produced are repeating decimals.
2. Repeating Decimal Bingo
In this activity, students will play a game of Bingo to identify repeating decimals. Begin by creating Bingo cards that feature repeating decimals. When calling out the numbers, use a mix of repeating decimals and non-repeating decimals. To make the game more challenging, you can also include fractions that convert to repeating decimals. For example, 2/3 is equal to 0.6666….
3. Decimal Place Value Manipulatives
One way to help students understand the concept of repeating decimals is to use manipulatives to represent decimal numbers. Use base 10 blocks or decimal cubes to show students how repeating decimals represent a pattern of numbers that continues infinitely. Encourage students to experiment with different repeating decimals, observing how the pattern changes when the decimals are multiplied or divided by different numbers.
4. Decimal Scavenger Hunt
In this activity, students will search for decimals around the room or school and identify which decimals are repeating decimals. Provide students with a list of decimals to find, or have them come up with their own decimals to search for. Once they have identified a repeating decimal, have them explain how they know it is repeating.
5. Decimal War Card Game
In this card game, students play a version of War with decimal cards. Each player draws a card and announces the decimal they drew. The player with the larger decimal wins the cards and adds them to their pile. To make the game more challenging, include decimals that are repeating and have students identify which ones they are.
Overall, repeating decimals can be a confusing concept for students, but with the right activities, teachers can make the topic more engaging and understandable. Encourage your students to experiment with different decimals and patterns, and don’t be afraid to let them take the lead in their learning. With a little creativity, you can help your students master the concept of repeating decimals and develop a love of math. | 525 | 2,769 | {"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} | 4.46875 | 4 | CC-MAIN-2024-26 | latest | en | 0.913584 |
https://www.thermofisher.com/jp/en/home/life-science/pcr/real-time-pcr/qpcr-education/absolute-vs-relative-quantification-for-qpcr.html | 1,524,203,519,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937114.2/warc/CC-MAIN-20180420042340-20180420062340-00605.warc.gz | 918,470,910 | 26,022 | ## Absolute vs relative quantification at a glance
When calculating the results of your real-time PCR (qPCR) experiment, you can use either absolute or relative quantification.
Absolute quantification
(digital PCR method)
Absolute quantification(standard curve method) Relative quantification
Overview In absolute quantification using digital PCR, no known standards are needed. The target of interest can be directly quantified with precision determined by number of digital PCR replicates. In absolute quantification using the standard curve method, you quantitate unknowns based on a known quantity. First you create a standard curve; then you compare unknowns to the standard curve and extrapolate a value. In relative quantification, you analyze changes in gene expression in a given sample relative to another reference sample (such as an untreated control sample).
Example
Quantify copies of rare allele present in heterogeneous mixtures.
Count the number of cell equivalents in sample by targeting genomic DNA.
Determine absolute number of viral copies present in a given sample without reference to a standard.
Correlating viral copy number with a disease state.
Measuring gene expression in response to a drug.
In this example, you would compare the level of gene expression of a particular gene of interest in a chemically-treated sample relative to the level of gene expression in an untreated sample.
## Absolute quantification using the digital PCR method
Digital PCR works by partitioning a sample into many individual real-time PCR reactions; some portion of these reactions contain the target molecule (positive), while others do not (negative). Following PCR analysis, the fraction of negative answers is used to generate an absolute answer for the exact number of target molecules in the sample, without reference to standards or endogenous controls.
Figure 1: Digital PCR uses the ratio of positive (white) to negative (black) PCR reactions to count the number of target molecules.
## Absolute quantification using the standard curve method
The standard curve method for absolute quantification is similar to the standard curve method for relative quantification, except the absolute quantities of the standards must first be known by some independent means.
Figure 2: Amplification plot and standard curve for absolute quantification
Critical guidelines
The guidelines below are critical for proper use of the standard curve method for absolute quantification:
• It is important that the DNA or RNA be a single, pure species. For example, plasmid DNA prepared from E. coli often is contaminated with RNA, which increases the A260 measurement and inflates the copy number determined for the plasmid.
• Accurate pipetting is required because the standards must be diluted over several orders of magnitude. Plasmid DNA or in vitro transcribed RNA must be concentrated in order to measure an accurate A260 value. This concentrated DNA or RNA must then be diluted 106–1012 -fold to be at a concentration similar to the target in biological samples.
• The stability of the diluted standards must be considered, especially for RNA. Divide diluted standards into small aliquots, store at –80 °C, and thaw only once before use.
It is generally not possible to use DNA as a standard for absolute quantification of RNA because there is no control for the efficiency of the reverse transcription step.
Standards
The absolute quantities of the standards must first be known by some independent means. Plasmid DNA and in vitro transcribed RNA are commonly used to prepare absolute standards. Concentration is measured by A260 and converted to the number of copies using the molecular weight of the DNA or RNA.
## Relative quantification
Calculation methods for relative quantification
Relative quantification can be performed with data from all real-time PCR instruments. The calculation methods used for relative quantification are:
• Standard curve method
• Comparative CT method
Figure 3: Relative quantification
## Which method should I use?
Digital PCR method Standard curve method Comparative CT method Overview Nucleic acid quantification utilizes theory of limiting sample dilution that is spread across many technical replicate PCR reactions. Absolute quantification is determined by ratio of number of negative versus total reactions. This type of analysis is different from Ct and delta Ct comparisons, and instead allows each assayed target to be quantified independently without the need for reference standards. It is easy to prepare standard curves for relative quantification because quantity is expressed relative to some basis sample (called a calibrator), such as an untreated control. For all experimental samples, you determine target quantity from the standard curve and divide by the target quantity of the calibrator. Thus, the calibrator becomes the 1× sample, and all other quantities are expressed as an n-fold difference relative to the calibrator. This method compares the Ct value of one target gene to another (using the formula: 2ΔΔCT)—for example, an internal control or reference gene (e.g., housekeeping gene)—in a single sample. Advantages No need to rely on references or standards. Desired precision can be achieved by increasing total number of PCR replicates. Highly tolerant to inhibitors. Capable of analyzing complex mixtures. Unlike traditional qPCR, digital PCR provides a linear response to the number of copies present to allow for small-fold change differences to be detected Running the target and endogenous control amplifications in separate tubes and using the standard curve method of analysis requires the least amount of optimization and validation. You don't need a standard curve and can increase throughput because wells no longer need to be used for the standard curve samples. This also eliminates dilution errors made in creating the standard curve samples.You can amplify the target and endogenous control in the same tube, increasing throughput and reducing pipetting errors. When RNA is the template, performing amplification in the same tube provides some normalization against variables such as RNA integrity and reverse transcription efficiencies. Experimental validation Validated digital PCR results with side-by-side comparison to a well-characterized sample with known copy number. See Advantages above. You have to run a validation experiment to show that the efficiencies of the target and endogenous control amplifications are approximately equal. To amplify the target and endogenous control in the same tube, limiting primer concentrations must be identified and shown not to affect CT values. Critical guidelines It is important to use low-binding plastics as much as possible throughout experimental set-up. Since digital PCR emphasizes assaying limiting dilution, any sample that sticks to intermediate set-up equipment will be lost and skew results. We recommend using low binding 2.0 mL tubes for dilutions and low-retention pipette tips. It is important to know the digital area of the desired sample to be tested. If unknown, consult user guide for help in determining copy number of known species (gDNA) or perform a preliminary screening experiment with multiple dilutions of sample/assay combination to determine optimal digital concentration to ensure meaningful data attained. Sample should not be kept at low concentration for extended periods of time, nor exposed to excessive freeze-thawing. Carriers have not been shown to be as important to reproducibility as much as using non-stick plastics for experimental set-up. Careful planning of dilutions is desired to minimize variability due to dilution scheme. It is important that stock RNA or DNA be accurately diluted, but the units used to express this dilution are irrelevant. If two-fold dilutions of a total RNA preparation from a control cell line are used to construct a standard curve, the units could be the dilution values 1, 0.5, 0.25, 0.125, and so on. By using the same stock RNA or DNA to prepare standard curves for multiple plates, the relative quantities determined can be compared across the plates. You can use a DNA standard curve for relative quantification of RNA. Doing this assumes that the reverse transcription efficiency of the target is the same in all samples, but the exact value of this efficiency need not be known. For quantification normalized to an endogenous control, standard curves are prepared for both the target and the endogenous reference. For each experimental sample, the amount of target and endogenous reference is determined from the appropriate standard curve. Then, the target amount is divided by the endogenous reference amount to obtain a normalized target value. Again, one of the experimental samples is the calibrator, or 1× sample. Each of the normalized target values is divided by the calibrator normalized target value to generate the relative expression levels. For the comparative CT method to be valid, the efficiency of the target amplification (your gene of interest) and the efficiency of the reference amplification (your endogenous control) must be approximately equal. Endogenous control Digital PCR does not rely on the presence of endogenous controls for reference measurements. Amplification of an endogenous control may be performed to standardize the amount of sample RNA or DNA added to a reaction. For the quantification of gene expression, researchers have used ß-actin, glyceraldehyde-3-phosphate dehydrogenase (GAPDH), ribosomal RNA (rRNA), or other RNAs as an endogenous control. Standards Because digital PCR uses the fraction of negative to total replicates to determine an absolute count of molecules, no standards are required. Because the sample quantity is divided by the calibrator quantity, the unit from the standard curve drops out. Thus, all that is required of the standards is that their relative dilutions be known. For relative quantification, this means any stock RNA or DNA containing the appropriate target can be used to prepare standards. | 1,933 | 10,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-17 | latest | en | 0.847921 |
http://community.fantasyflightgames.com/index.php?showtopic=63881 | 1,412,181,062,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663467.44/warc/CC-MAIN-20140930004103-00119-ip-10-234-18-248.ec2.internal.warc.gz | 71,673,263 | 16,242 | # Warp jump from/to inside a solar system
5 replies to this topic
### #1 wolph42
wolph42
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Posted 01 May 2012 - 01:00 AM
In follow up to this post:
It's unlikely that your players ever will take the chance, but ORKS just might do this. So here a suggestion of the percentile mishap (= BOOM)
Assumptions:
- Warp exit/entry at edge: change of mishap 0% (concerning the gravity well)
- Warp exit/entry close to the star: chance of mishap 50% (again: concerning the gravity well). Initially I was thinking 100% but as habital planets are typically 1 AU away in a typically 40 AU solar system would mean that jumping close to the habital planet is about 100% chance on mishap. That with the fact that mishap = warpdrive explodes, in other words: end of story for the ship!! AND the reference earlier mentioned of the ork fleet jumping in on a star fleet in orbit, would have meant the instant decimation of the entire fleet. Hence: 50% chance on blow-up.
The formula to calculate this is very simple:
Chance on mishap = (7*DFE/SR)^2 %
here is:
SR = Solar system Radius (e.g. 30 AU, solar system of Terra)
DFE = distance from edge of the solar system where you jump in. E.g. with an with an SR=30, 0 is the edge and 30 is inside the sun. Entry at halfway = 15 AU:
Mishap = (7*15/30)^2 = 12%.
If you like to keep things simple and assume EVERY solar system of size 50AU then the formula becomes really simple:
Chance on Mishap = DFE^2 / 50 %
To resolve simply throw a 1d100 if you roll UNDER the set score: BOOM
Some chances (assuming SR = 50)
Jumping in at:
Distance from edge - %
0 (jump at the edge) - 0%
6 - 1%
10 - 2%
20 - 8%
30 - 18%
40 - 32%
49 (jumping close to a habital planet which are typically at 1 to 3 AU away from its sun) - 48%
50 - 50%
### #2 Fresnel
Fresnel
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Posted 19 May 2012 - 04:01 AM
If a navigator could exit at a Lagrangian point it might be safer. However this might be a hellish difficulty.
### #3 wolph42
wolph42
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Posted 29 May 2012 - 10:41 PM
Fresnel said:
If a navigator could exit at a Lagrangian point it might be safer. However this might be a hellish difficulty.
hmm interesting notion… I guess that in that case L1 to L3 are right out as they are unstable and too small to ever make that jump (in addition 1 and 2 are REALLY close to the planet), which leaves L4 and 5, which are stable and quite a bit larger then 1,2 and 3. If you would jump there then you would be at 1AU distance from the planet. L4 and 5 are actually pretty big, about 1AU length and half that in width so its not THAT hard. What makes it hard in general to find the right spot at all is the fact of the time distortion in the warp. In order to jump right next to a planet or in a lagrange point you need to know the time (as that decides the position of the planet and LP in regard of its star).
So basically you want to jump at a distance of 1AU from the star without much error, lets say max 10% so within 0.9 to 1.1 AU AND you need to know the exit time. Both these rolls are part of the 5 navigation steps.
First you need to know the time, now the LP are pretty big so there's always a 30% chance that you jump into one. And I would rule that if you jump in from farther then from the edge of the solar system as beyond that its really completely random at what exact time you jump in. Should you jump first to the edge and then from the edge to the LP then I would rule a very hard Navigation Warp test (-30) for the correct estimate, however since there is a 30% chance that you get it right anyway I would simply rule a Navigation Warp Test (+0). Every degree of fail increases the difficulty of the 'leaving the warp' with one step, more then 3 DoF means your NOT jumping in the LP.
'Charting course' is negligible (again assuming jumping from edge) so ordinary warp
'steering vessel' will again influence the time and space exit which are vital so again very hard test (-30) navigation warp and again 3 DOF means NOT jumping in the LP and every DoF increases the difficulty with one step on leaving the warp.
When that's done you need to roll 'leaving the warp test' which is usually at -20, this time modified with the DoF of the other rolls.
Should you actually succeed this test then this would half the chance of BOOM to 24% lowering it further with 1% for every degree of success. Fail would mean you arrive at roughly 1AU from the sun: 48%. Should you fail this warp with 6+ DoF then you jump straight into the sun.
### #4 Sna
Sna
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Posted 21 June 2012 - 12:40 AM
Lagrange points are cosmically not extremely large. One of the systems my RT ran into had a stabilization gate that allowed Lagrange point jumping, but other then that only one Ork vessel and two Chaos vessels ever attempted it (one failing its check miserably and completely torn apart as a result). The above rules are a nice system, I'll keep it in mind if my players ever get desperate enough.
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Posted 21 June 2012 - 01:44 PM
even if they are hard to jump IN to. think of the time saved by jumping OUT from them.
### #6 wolph42
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Posted 24 June 2012 - 09:48 PM
even if they are hard to jump IN to. think of the time saved by jumping OUT from them.
hmm… very very good point. Typically something my players would bring along when I would have proposed this house rule.
However given the fact that jumping out of lagrange spots is still not described anywhere in the rules (although that would require some astronomy/physics knowledge) and thus we sort of can expect that its not common practice. If its not common practice then this can have two origins:
1. unknown (that is the empire is unaware of LP's and thus don't use them.) If thats the case then the players won't know about them either and thus they're effectively useless.
2. (too?) dangerous. Although the BOOM % is much lower then jumping in… its still there, though Im inclined to make it a lot slower. Basically I would say
a. its very hard to plot a course form there -30 nav tests to plot to the edge of the solar, -60 if you try to plot immediately to a further destination.
b. there is still a BOOM %… 10%? maybe relate it to the ship size… still have a test with 1% modification per DoS / DoF ?
© 2013 Fantasy Flight Publishing, Inc. Fantasy Flight Games and the FFG logo are ® of Fantasy Flight Publishing, Inc. All rights reserved. | 1,678 | 6,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-41 | longest | en | 0.917081 |
https://community.targit.com/hc/en-us/community/posts/11953393428252-Creating-Price-Forecast-and-Calculation-in-Calculation-Fields-Based-on-Field-Content-in-Frontend?sort_by=votes | 1,720,957,572,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00617.warc.gz | 152,752,862 | 10,225 | # Creating Price Forecast and Calculation in Calculation Fields Based on Field Content in Frontend
I have a desire to be able to create a price and calculation via the frontend in calculation fields. Below are data for 'Organisation Level 4' and 'Visited time.' I would like to be able to create a forecast of expected billing if I could specify an average price per 'Organisation Level 4.'?
Org Niv 4 Price
HJP Nord 385
HJP Syd 390
HJP Vest 370
HJP Øst 375
• Hi Lars, you can try this:
In my example, Casual Clothing Retail (d1) was calculated *1,1 and Calsual Cloting Wholesale (d2) was calculated with 1,2.
My formula is:
if allcount(d-1, d1:0, m1) = 1 then sum(d-1, d1, m1) * 1,1
else if allcount(d-1, d1:0, m1) = 2 then sum(d-1, d2, m1) * 1,2
else 999
• Hi Marlene.
Thank you for the creative suggestion, which, of course, means that I need to figure out in which positions the individual organizations are displayed.
I've tried updating my small sheet with sample data and can see that it's also multiplying on the total. It has also done that in your example, where the total is not the sum of Retail and Wholesale, but instead Total x 1,1. Is there any way to fix that?
• Thank you for the great suggestions :-)
I will work on them further.
Happy New Year
• If you can't be sure of the position of a certain dimension member, you may also use a modified version of Marlene's suggestion:
`if allcount(d-1, d1:0, m1) = allcount(d-1, d1:@"[HJP Nord]", m1) then sum(d-1, @"[HJP Nord]", m1) * 1,1else if allcount(d-1, d1:0, m1) = allcount(d-1, d1:@"[HJP Øst]", m1) then sum(d-1, @"[HJP Øst]", m1) * 1,2else 999BR / Ole`
• Thanks a lot, Ole - I will try this :-)
Best regards / Lars | 518 | 1,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-30 | latest | en | 0.874258 |
https://www.scribd.com/doc/187431290/Experiment-101-RESOLUTION-OF-FORCES | 1,563,888,967,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529406.97/warc/CC-MAIN-20190723130306-20190723152306-00013.warc.gz | 827,333,209 | 45,911 | You are on page 1of 2
# Analysis In the first trial, the result from the force table are F4 is 110 g and the angle
is 326 thus the actual being 146. In the experiment, if the angle is moved even slightly or the mass is decreased or increased even in the smallest value, the ring will move away from the center. The percentage error of the graphical method when compared to the result from the force table is 0% and 0.6849%, for the resultant and the angle respectively. When the values from the force table are compared to the results from the component method, the percentage error is 1.1854% and 0.8119%, for the resultant and the angle respectively. In the second trial, the resultant is 115 g and the actual is 91. The percentage error of the graphical method when compared to the result from the force table is 0% and 2.1978%, for the resultant and the angle respectively. While the percentage error for the component method is 0.9565% and 0.4286%, for the resultant and the angle respectively. Conclusion Based on the experiment, the resultant force can be determined using Graphical Method (polygon method) and Analytical Method (component method) and also by using the force table. The accuracy of the three methods is quite high. The graphical method is the most accurate in determining the resultant while the analytical method is the best of the three in finding the direction because of its low percentage error. Human error is one of the probable sources of errors in the experiment. The measurements were within 2% difference and sometimes 0%, so the force table was reasonably accurate.
Related Application One important application of this principle is in the recreational sport of sail boating. Sailboats encounter a force of wind resistance due to the impact of the moving air molecules against the sail. This force of wind resistance is directed perpendicular to the face of the sail, and as such is often directed at an angle to the direction of the sailboat's motion. The actual direction of this force is dependent upon the orientation of the sail. To determine the influence of the wind resistance force in the direction of motion, that force will have to be resolved into two components - one in the direction that the sailboat is moving and the other in a direction perpendicular to the sailboat's motion.
Source: http://www.physicsclassroom.com/class/vectors/U3L3b.cfm | 508 | 2,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.936675 |
https://algebraworksheets.co/quadratic-functions-worksheet-answers-algebra-2/ | 1,657,191,595,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104690785.95/warc/CC-MAIN-20220707093848-20220707123848-00274.warc.gz | 133,631,484 | 14,538 | Quadratic Functions Worksheet Answers Algebra 2 – The aim of Algebra Worksheets is to help students learn the basics of mathematics in an easy method. This particular study area focuses on studying mathematical symbols and the rules that govern their manipulation. It is the central thread that unites mathematics, geometry, and physics. It’s an essential component of your educational. This section will teach you how to use algebra, focusing on the most important subjects. These worksheets can help you increase your proficiency as an undergraduate.
Besides learning how to work out equations, these worksheets teach students to solve word problems involving algebra. They can use fractions, integers, as well as decimals. They can also rewrite equations and evaluate formulas. These worksheets are offered in customary and metric units. These worksheets include many different linear equations as well as the ability to identify the x-intercept and solve word problems that involve parallellines.
You can learn quadratic equations by using these top algebra worksheets. Many of them can be printed out, so you don’t need to download additional software. These worksheets are designed to aid students master the basics of algebra without the need to employ a calculator, or spend time searching for the right answer. The worksheets are designed for reinforcement of concepts and to enhance your ability. Below are some great tools to help you develop your mathematical capabilities.
These worksheets for algebra provide students with a complete understanding of the fundamentals of math. They show students how to make use of fractions in order to represent numbers. They also show students how to solve equations containing several unknown variables. This is a crucial aspect of learning math. Therefore, these printable tools are a huge help in helping you to advance your study. These worksheets can assist you to comprehend equations more clearly. If you require assistance with increasing your math knowledge be sure to have a good set of worksheets.
Free worksheets for algebra are a great option to build your math skills. By using these free worksheets, you can practice the fundamental principles of math. They can help you learn the fundamentals of algebra as well as apply the principles of the subject to your everyday life. They are a fantastic source of information in the event that you are a student. They will help you become an even better student and productive. These printables will be popular with your students!
There are also free worksheets that will aid in the development of your math abilities. For instance, if you’re unfamiliar with algebra, you could be confused about how to utilize graphs. You can find worksheets available for download to help you learn the subject. It’s simple to download and browse these worksheets and begin working in your math classes today. There are numerous options available on the internet. If you have a high schooler and you want to choose a math lesson to teach your child. | 550 | 3,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-27 | latest | en | 0.945307 |
https://rdrr.io/cran/jackalope/man/indels.html | 1,582,780,475,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146647.82/warc/CC-MAIN-20200227033058-20200227063058-00060.warc.gz | 501,587,810 | 14,365 | # indels: Insertions and deletions (indels) specification In jackalope: A Swift, Versatile Phylogenomic and High-Throughput Sequencing Simulator
## Description
Construct necessary information for insertions and deletions (indels) that will be used in `create_variants`.
## Usage
`1` ```indels(rate, max_length = 10, a = NULL, rel_rates = NULL) ```
## Arguments
`rate` Single number specifying the overall indel rate among all lengths. `max_length` Maximum length of indels. Defaults to `10`. `a` Extra parameter necessary for generating rates from a Lavalette distribution. See Details for more info. Defaults to `NULL`. `rel_rates` A numeric vector of relative rates for each indel length from 1 to the maximum length. If provided, all arguments other than `rate` are ignored. Defaults to `NULL`.
## Details
All indels require the `rate` parameter, which specifies the overall indels rate among all lengths. The `rate` parameter is ultimately combined with a vector of relative rates among the different lengths of indels from 1 to the maximum possible length. There are three different ways to specify/generate relative-rate values.
1. Assume that rates are proportional to `exp(-L)` for indel length `L` from 1 to the maximum length (Albers et al. 2011). This method will be used if the following arguments are provided:
• `rate`
• `max_length`
2. Generate relative rates from a Lavalette distribution (Fletcher and Yang 2009), where the rate for length `L` is proportional to `{L * max_length / (max_length - L + 1)}^(-a)`. This method will be used if the following arguments are provided:
• `rate`
• `max_length`
• `a`
3. Directly specify values by providing a numeric vector of relative rates for each insertion/deletion length from 1 to the maximum length. This method will be used if the following arguments are provided:
• `rate`
• `rel_rates`
## Value
An `indel_info` object, which is an R6 class that wraps the info needed for the `create_variants` function. It does not allow the user to directly manipulate the info inside, as that should be done using this function. You can use the `rates()` method to view the indel rates by size.
## References
Albers, C. A., G. Lunter, D. G. MacArthur, G. McVean, W. H. Ouwehand, and R. Durbin. 2011. Dindel: accurate indel calls from short-read data. Genome Research 21:961–973.
Fletcher, W., and Z. Yang. 2009. INDELible: a flexible simulator of biological sequence evolution. Molecular Biology and Evolution 26:1879–1888.
## Examples
``` 1 2 3 4 5 6 7 8 9 10``` ```# relative rates are proportional to `exp(-L)` for indel # length `L` from 1 to 5: indel_rates1 <- indels(0.1, max_length = 5) # relative rates are proportional to Lavalette distribution # for length from 1 to 10: indel_rates2 <- indels(0.2, max_length = 10, a = 1.1) # relative rates are all the same for lengths from 1 to 100: indel_rates3 <- indels(0.2, rel_rates = rep(1, 100)) ```
jackalope documentation built on Dec. 1, 2019, 1:17 a.m. | 796 | 2,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.723747 |
https://oeis.org/A009594 | 1,718,317,553,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861517.98/warc/CC-MAIN-20240613221354-20240614011354-00219.warc.gz | 389,314,914 | 3,889 | The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A009594 Expansion of e.g.f. sinh(sin(x)*exp(x)). 1
0, 1, 2, 3, 12, 77, 382, 1519, 6328, 37753, 266522, 1627131, 7776516, 32335877, 186528214, 1552283815, 8927485936, -15949155215, -942524148046, -10331486791053, -66361022595972, -416998255691971, -6076708664221010 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 LINKS G. C. Greubel, Table of n, a(n) for n = 0..450 FORMULA E.g.f.: sinh(sin(x)*exp(x)). MATHEMATICA With[{nmax = 50}, CoefficientList[Series[Sinh[Sin[x]*Exp[x]], {x, 0, nmax}], x]*Range[0, nmax]!] (* G. C. Greubel, Jan 30 2018 *) PROG (PARI) x='x+O('x^30); concat([0], Vec(serlaplace(sinh(sin(x)*exp(x))))) \\ G. C. Greubel, Jan 30 2018 CROSSREFS Sequence in context: A009269 A012396 A013012 * A074179 A012586 A088223 Adjacent sequences: A009591 A009592 A009593 * A009595 A009596 A009597 KEYWORD sign,easy AUTHOR R. H. Hardin EXTENSIONS Extended with signs by Olivier Gérard, Mar 15 1997 STATUS approved
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Last modified June 13 17:32 EDT 2024. Contains 373391 sequences. (Running on oeis4.) | 533 | 1,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.635413 |
https://betterlesson.com/lesson/reflection/14426/template-and-class-structure | 1,511,081,949,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805466.25/warc/CC-MAIN-20171119080836-20171119100836-00160.warc.gz | 585,564,036 | 23,822 | ## Reflection: Lesson Planning Super Practice with Angles and Algebra - Section 1: Setting the Pace
super practice reflection.mp4
Super Practice with Angles and Algebra
# Super Practice with Angles and Algebra
Unit 5: Lines, Angles, and Algebraic Reasoning
Lesson 14 of 16
## Big Idea: Students can only master a topic if they are given the opportunities needed to practice the topic.
Print Lesson
10 teachers like this lesson
Standards:
Subject(s):
Math, Geometry, supplementary angles, vertical angles, adjacent angles, angles and lines
85 minutes
### Shaun Errichiello
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Windermere, FL
Environment: Urban | 314 | 1,367 | {"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} | 3.03125 | 3 | CC-MAIN-2017-47 | latest | en | 0.76899 |
https://math.stackexchange.com/questions/2218184/doubt-in-application-of-weierstrass-theorem-showing-that-d-is-compact-and-f | 1,563,732,255,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527089.77/warc/CC-MAIN-20190721164644-20190721190644-00496.warc.gz | 466,180,194 | 35,806 | # Doubt in application of Weierstrass Theorem - Showing that $D$ is compact and $f$ is continuous
I have the following problem:
Let the constraint set be $D=\{(x,y) \in \Re^2 \;| x^2 + y^2=1\}$, show that the $f(x,y)=x^2-y^2$ is a continuous function on $D$ and also that $D$ is compact.
Showing that $\lim_{(x,y)\rightarrow(a,b)}f(x,y)$ is equal to $f(a,b)$ suffices to prove the continuity?
To compactness is obvious that $D$ is bounded, by $B(0,2)$ for instance. But how to show that is closed?
• For your Q on proving the continuity: YES. – DanielWainfleet Apr 4 '17 at 23:14
On D being closed: Two methods:
$(\;(x_n,y_n)\;)_n$ converges to $(x,y) \implies (\;(x-x_n)^2+(y-y_n)^2\;)_n$ converges to $0\implies (\lim_n x_n=x$ and $\lim_ny_n=y)\implies (\lim_n x_n^2=x^2$ and $\lim_ny_n^2=y^2)$.
If $(x_n,y_n)\in D$ and $(\;(x_n,y_n)\;)_n$ converges to $(x,y)$ then $$|x^2+y^2-1|=|(x_n^2+y_n^2-1)+(x^2-x_n^2)+(y^2-y_n)^2 |=$$ $$=|(x^2-x_n^2)+(y-y_n^2)|\leq |(x^2-x_n^2)|+|(y^2-y_n^2)|$$ which is $<r$ for any $r>0$ if $n$ is big enough. So $|x^2+y^2-1|$ is less than any $r>0$, so $x^2+y^2-1=0$. So $(x,y)\in D.$
Another method: Show that the complement of $D$ is open. With $\|(x,y)\|=\sqrt {x^2+y^2}$ we have the Triangle Inequality: For $u,v\in \mathbb R^2$ we have $\|u+v\|\leq \|u\|+\|v\|.$
If $u\in \mathbb R^2$ with $\|u\|=1+s\ne 1$ then the open ball $B(u,|s|/2)$ is disjoint from $D$ because
(i). If $s<0$ then $$v\in B(u,|s|/2)\implies \|v\|=\|(v-u)+u\|\leq \|v-u\|+\|u\|<$$ $$<|s|/2+(1+s)=|s|/2+(1-|s|)=1-|s|/2< 1.$$
(ii). If $s>0$ then $$v\in B(u,|s|/2)\implies \|v\|=\|u-(u-v)\|\geq \|u\|-\|u-v\|>$$ $$>(1+s)-|s|/2=(1+|s|)-|s|/2=1+|s|/2>1.$$ | 739 | 1,668 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2019-30 | latest | en | 0.723553 |
https://www.physicsforums.com/threads/finding-moment-of-inertia-using-integration.704590/ | 1,582,893,617,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147154.70/warc/CC-MAIN-20200228104413-20200228134413-00059.warc.gz | 840,891,480 | 21,213 | # Finding Moment of Inertia Using Integration
## Homework Statement
A cylinder with radius R and mass M has density that increases linearly with distance r from the cylinder axis, ρ = αr, where α is a positive constant. Calculate the moment of inertia of the cylinder about a longitudinal axis through its center in terms of M and R.
This is from my textbook, but my textbook has only explained how to find inertias from uniform masses leading up to this question. What's even worse is that this question is given the easiest rating.
## Homework Equations
I = Ʃ(mi*r^2i)
Normally I would find the integral of r^2*dm, but my book says that it is only for a uniform distribution of mass.
## The Attempt at a Solution
I = Ʃ(mi*r^2i) = m(1)*r^2(1) + m(2)*r^2(2) + ...
I just cannot figure out how to incorporate an increasing density into this.
Last edited:
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SteamKing
Staff Emeritus
Homework Helper
You can't use your proposed formula for non-constant density. You must go back to the integral definition of mass moment of inertia.
If the axis of rotation is the z-axis, then the MMOI about the z-axis is:
???
I couldn't find non-uniform increasing density in the link you provided. I feel fairly comfortable with simple objects that have a uniform density. But I just have no idea what to with a changing density. Here's my work.
I = ∫r^2*dm
ρ = αr (according to the question)
ρ = αr = dm/dV then dm = α*r*dV
I = ∫r^2*α*r*dV
dV = 2*∏*L*r*dr
I = ∫r^2*α*r*2*∏*L*r*dr
= ∫r^4*α*2*∏*L*dr
= α*2*∏*L*∫r^4*dr
= (ρ/r)*2*∏*L*∫r^4*dr
= (ρ/r)*2*∏*L*(r^5)/5
= ρ*2*∏*L*(r^4)/5
= (M/V)*2*∏*L*(r^4)/5
= (M*2*∏*L*r^4)/(5*∏*L*R^2)
= (2*M*R^2)/5
My textbook has I = (3*M*R^2)/5. I do not feel confident in my answer.
TSny
Homework Helper
Gold Member
Here's my work.
I = ∫r^2*dm
ρ = αr (according to the question)
ρ = αr = dm/dV then dm = α*r*dV
I = ∫r^2*α*r*dV
dV = 2*∏*L*r*dr
I = ∫r^2*α*r*2*∏*L*r*dr
= ∫r^4*α*2*∏*L*dr
= α*2*∏*L*∫r^4*dr
I think everything's fine up to here. Of course, you should have definite limits of integration in mind for the integral.
But your next step shown below is odd.
= (ρ/r)*2*∏*L*∫r^4*dr
You can't replace the constant α by ρ/r because r is your variable of integration. Leave the integral in terms of α. But then you will need to do a separate integration to relate α to M and R.
I think everything's fine up to here. Of course, you should have definite limits of integration in mind for the integral.
But your next step shown below is odd.
You can't replace the constant α by ρ/r because r is your variable of integration.
The reason why I did this is because the question said that α is a constant. So then I just made sure that α = p/r stayed as a constant. I feel quite lost.
Leave the integral in terms of α. But then you will need to do a separate integration to relate α to M and R.
I don't why I would do that, and I don't know how I would do that. I feel like this question is beyond anything that we learnt up until now in the textbook. But it is a question among many relatively easy ones that the chapter has a direct explanation for.
TSny
Homework Helper
Gold Member
The reason why I did this is because the question said that α is a constant. So then I just made sure that α = p/r stayed as a constant. I feel quite lost.
OK, let's just take it step by step. You had a very good start where you got to the integral
I = ∫(r2)(αr)(L2∏rdr) = α2∏L∫r4dr.
(Note that we have formatting buttons for superscripts and subscripts that you can use to make the expressions easier to read.)
Now, α is just a constant factor that you can leave alone for now. Go ahead and evaluate the integral with the correct lower and upper limits for r.
ok, let's just take it step by step. You had a very good start where you got to the integral
i = ∫(r2)(αr)(l2∏rdr) = α2∏l∫r4dr.
(note that we have formatting buttons for superscripts and subscripts that you can use to make the expressions easier to read.)
now, α is just a constant factor that you can leave alone for now. Go ahead and evaluate the integral with the correct lower and upper limits for r.
I = α*2*∏*l*∫r4dr = (α*2*∏*L*r5)/5
I can't figure out how to show the proper scripts for the lower and upper limits for the integral. The lower is 0 and the upper is R.
Last edited:
TSny
Homework Helper
Gold Member
I = α*2*∏*l*∫r4dr = (α*2*∏*L*r5)/5
[EDIT] Shouldn't your result for the integral be expressed in terms of the limits of integration?
I am just trying to figure out how to show that.
TSny
Homework Helper
Gold Member
Don't worry about showing the limits of integration on the integral. Just worry about using those limits when evaluating the integral.
$\int$$^{R}_{0}$ How did you get your subscripts on top and on the bottom like that.
TSny
Homework Helper
Gold Member
Let's not worry about how to type in the limits on the integral. What is the answer to $\int_0^R r^4dr$?
Ok, it is r = R, and r = 0, right?
R5/5
TSny
Homework Helper
Gold Member
Ok, it is r = R, and r = 0, right?
Those are the upper and lower limits of the integral. Do you know how to evaluate an integral at its limits? So far, you have gotten to the point of evaluating
$I = a2\pi L \int_0^Rr^4dr$.
Can you carry out the integration and evaluate at the limits?
TSny
Homework Helper
Gold Member
R5/5
Good. So, what is your expression for the moment of inertia at this point?
Good. So, what is your expression for the moment of inertia at this point?
I = a*2*∏*L*R5/5
TSny
Homework Helper
Gold Member
I = a*2*∏*L*R5/5
OK. Now you just need to figure out how to express the constant $a$ in terms of $M$ and $R$. Can you use the density function to set up another integral that would represent the total mass of the cylinder?
OK. Now you just need to figure out how to express the constant $a$ in terms of $M$ and $R$. Can you use the density function to set up another integral that would represent the total mass of the cylinder?
I = a*2*∏*L*R5/5 and α = ρ/r = m/(∏*L*r3)
= (m*2*∏*L*R5)/(∏*L*r3*5)
= (m*2*R2)/5
Hmmm, I'm stuck.
TSny
Homework Helper
Gold Member
I = a*2*∏*L*R5/5 and α = ρ/r = m/(∏*L*r3)
You don't want to substitute α = ρ/r. That would put the variable r into the expression. But I should be a fixed constant. Note that r and R have distinct meanings, the first is a variable while the second is a constant.
The total mass M must be the sum (integral) of all the elements of mass dm: M = ∫dm. You know how to substitute for dm in terms of the density function and dV. Try it here and see if you can evaluate the integral.
You don't want to substitute α = ρ/r. That would put the variable r into the expression. But I should be a fixed constant. Note that r and R have distinct meanings, the first is a variable while the second is a constant.
The total mass M must be the sum (integral) of all the elements of mass dm: M = ∫dm. You know how to substitute for dm in terms of the density function and dV. Try it here and see if you can evaluate the integral.
ρ = dm/dV then dm = ρ*dV
So M = ρ∫dV = ρ∫2*L*r*dr, oh boy, I have a feeling that I am not on the right track.
TSny
Homework Helper
Gold Member
ρ = dm/dV then dm = ρ*dV
So M = ρ∫dV = ρ∫2*L*r*dr
Oops. You can't pull ρ outside of the integral M = ∫ρdV, it's a function: ρ = αr. Also, you left out a factor of ∏ in the volume element: dV = 2∏*L*r*dr.
Oops. You can't pull ρ outside of the integral M = ∫ρdV, it's a function: ρ = αr. Also, you left out a factor of ∏ in the volume element: dV = 2∏*L*r*dr.
Ahhhh I think I got it.
M = ∫ρ*L*∏*r*dr
M = ∫α*r2*2*L*∏*dr
M = (α*2*L*∏*r3)/3
α = (3*M)/(2*L*∏*r3)
I = α*2*∏*L*R5/5 Now I will substitute in for α to bring M into the equation.
I = (3*M)/(2*L*∏*r3)*(2*∏*L*R5/5)
I = 3*M*R2/5
Is this correct?
TSny
Homework Helper
Gold Member
Ahhhh I think I got it.
Yes, I think that's essentially it. When you do the integral M = ∫α*r2*2*L*∏*dr you should evaluate the result at the limits r = 0 and r = R. So, instead of writing the result as M = (α*2*L*∏*r3)/3, it should be (α*2*L*∏*R3)/3. When you solve for α, it should be written in terms of R instead of r.
Otherwise, looks good!
Yes, I think that's essentially it. When you do the integral M = ∫α*r2*2*L*∏*dr you should evaluate the result at the limits r = 0 and r = R. So, instead of writing the result as M = (α*2*L*∏*r3)/3, it should be (α*2*L*∏*R3)/3. When you solve for α, it should be written in terms of R instead of r.
Otherwise, looks good!
Thank-you so much! | 2,659 | 8,514 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-10 | latest | en | 0.916978 |
https://www.math10.com/forum/viewtopic.php?f=5&t=5857 | 1,542,528,186,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744320.70/warc/CC-MAIN-20181118073231-20181118095231-00141.warc.gz | 907,635,251 | 5,985 | # What type of Math do I use to to describe this situation
### What type of Math do I use to to describe this situation
Everyone has heard the question, "When will I use _________ in real life?", when referring to math topics. I have the opposite question, what kind of math can I use to describe this real life phenomenon. I feel like it has something to do with series, convergence, and divergence, probably with calculus and perhaps a natural log thrown in somewhere. I don't know what to search for in google, but I haven't had luck so far with the searches I've tried.
Here is a sample situation:
A lighting unit connected to a power supply by a long cable draws a constant amount of electrical Power ( [P]ower = [I]Current * [V]olts ) after it stabilizes. The power supply provides a constant [V]oltage, and the cable has (for our purposes) a constant [R]esistance. The voltage from the supply to the light will drop along the cable based on how much current is drawn (Ohm's law V = I * R). With the light off it draws no current, so there is no drop across the cable, and the voltage at the light will be the same as at the supply.
Once the light starts drawing current, there will be a voltage drop along the cable, which will cause the voltage at the light to drop, which in turn causes the light to draw more current to maintain constant power. But now the light is drawing more current, which causes a larger voltage drop, which causes the light to draw more current, which causes a larger voltage drop, which causes . . . .
I can easily work out what current draw the light will stabilize by setting this up in the lab and testing it, and I can do it in software using loops, but I'd like to know what tools I need to work this out mathematically.
Another good example would be maintaining constant brightness on an LED. LEDs get dimmer as they warm up, they warm up as you supply more current, and you need to supply more current to maintain constant brightness because they are dimming.
Thanks!
Guest | 447 | 2,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-47 | longest | en | 0.961512 |
https://www.physicsforums.com/threads/help-with-matrix-proof.638263/ | 1,660,099,453,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571097.39/warc/CC-MAIN-20220810010059-20220810040059-00048.warc.gz | 813,999,863 | 14,087 | # Help with matrix proof
pyroknife
the matrices A and B are
invertible symmetric matrices and AB = BA.
Show that A*B^-1 is symmetric
(A*B^-1)^T
=A^T * (B^-1)^T
=A^T * (B^T)^-1
Since A and B are symmetric
=A*B^-1
Is this right? Is (B^-1)^T = (B^T)^-1?
Homework Helper
the matrices A and B are
invertible symmetric matrices and AB = BA.
Show that A*B^-1 is symmetric
(A*B^-1)^T
=A^T * (B^-1)^T
=A^T * (B^T)^-1
Since A and B are symmetric
=A*B^-1
Is this right? Is (B^-1)^T = (B^T)^-1?
Yes, that is true, but "(A*B^-1)^T= A^T*(B^-1)^T" isn't.
Rather, (A*B^{-1})^T= (B^{-1})^T*B^T.
pyroknife
Yes, that is true, but "(A*B^-1)^T= A^T*(B^-1)^T" isn't.
Rather, (A*B^{-1})^T= (B^{-1})^T*B^T.
Oh I memorized the identity wrong. (AB^T)=B^t*A^T
pyroknife
The solution in the book first proves
IF AB=BA, then B^-1 * A *B=A, so B^-1*A=AB^-1
For the last type, the "B^-1*A=AB^-1", part how did they go from B^-1 * A *B=A to
B^-1*A=AB^-1.
Did they divide both sides by B? | 399 | 968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2022-33 | latest | en | 0.852726 |
http://math.stackexchange.com/questions/696380/proving-that-certain-enriched-representable-functors-are-isomorphic | 1,469,823,142,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831771.10/warc/CC-MAIN-20160723071031-00034-ip-10-185-27-174.ec2.internal.warc.gz | 162,530,033 | 19,050 | # Proving that certain enriched representable functors are isomorphic
Let $\mathcal{V}$ be a closed symmetric monoidal category. One is supposed to show that if $\underline{C}$ is an enriched category, $x, y \in \underline{C}$ are objects, then if they are isomorphic in the underlying category $C$ of $\underline{C}$, the representable $\mathcal{V}$-functors $$\underline{C}(x,-), \underline{C}(y,-): \underline{C} \rightarrow \underline{\mathcal{V}}$$ are $\mathcal{V}$-isomorphic. I have tried to construct a natural isomorphism of enriched functors, but i am having real problems with showing that the construction really is an enriched natural transformation. Any hints or solutions?
-
You need to start with the enriched composition of $\underline{\mathcal{C}}$ and restrict along the morphism $I \to \underline{\mathcal{C}}(x, y)$ that picks out the particular isomorphism $x \to y$ you are given. – Zhen Lin Mar 2 '14 at 17:10
@ZhenLin Right, that is what I started with - but how to show this is enriched natural? – user101036 Mar 2 '14 at 17:20
Use enriched associativity, of course. – Zhen Lin Mar 2 '14 at 18:25
Would you mind elaborating? I have filled 12 pages of commutative diagrams so it would be greatly appreciated. – user101036 Mar 2 '14 at 18:28
@ZhenLin Or just a sketch would be appreciated when and if you have the time! – user101036 Mar 2 '14 at 20:12
Commutative diagrams are hard to draw here, so let me outline the steps in words instead. Suppose $f : x \to y$ and $g : y \to x$ are mutually inverse morphisms in the underlying category of $\underline{\mathcal{C}}$. For ease of notation I will pretend $\mathcal{V}$ is a strict monoidal category.
1. Recall that the morphisms $a \to b$ in the underlying category of $\underline{\mathcal{C}}$ are defined to be morphisms $I \to \underline{\mathcal{C}} (a, b)$ in $\mathcal{V}$, with the evident induced composition. Writing $\mu : \underline{\mathcal{C}} (b, c) \otimes \underline{\mathcal{C}} (a, b) \to \underline{\mathcal{C}} (a, c)$ for the composition in $\underline{\mathcal{C}}$ and $\eta_a : I \to \underline{\mathcal{C}} (a, a)$ for the identity morphisms, that means $\mu \circ (f \otimes g) = \eta_y$ and $\mu \circ (g \otimes f) = \eta_x$.
2. Consider $(g^*)_z = \mu \circ (g \otimes \mathrm{id}) : \underline{\mathcal{C}} (x, z) \to \underline{\mathcal{C}} (y, z)$ and $(f^*)_z = \mu \circ (f \otimes \mathrm{id}) : \underline{\mathcal{C}} (y, z) \to \underline{\mathcal{C}} (x, z)$. By enriched associativity, $(g^*)_z \circ (f^*)_z = (\eta_y^*)_z$ and $(f^*)_z \circ (g^*)_z = (\eta_x^*)_z$, and enriched unitality means $(\eta_x^*)_z = \mathrm{id}$ and $(\eta_y^*)_z = \mathrm{id}$, so $(f^*)_z$ and $(g^*)_z$ are indeed isomorphisms in $\mathcal{V}$.
3. Finally, we must show enriched naturality of $(f^*)_z$ and $(g^*)_z$ in $z$; by symmetry it suffices to consider just $(f^*)_z$. The claim is simply that, for all $w$, we have $\mu \circ (\mathrm{id} \otimes (f^*)_z) = (f^*)_w \circ \mu$ as morphisms $\underline{\mathcal{C}} (z, w) \otimes \underline{\mathcal{C}} (y, z) \to \underline{\mathcal{C}} (y, w)$. This again is a consequence of enriched associativity.
-
I am fine with 1 and 2, but I don't see why 3 coincides with the definition for the for an enriched natural transformation. On pg. 35 , def . 3.5.8 of math.harvard.edu/~eriehl/cathtpy.pdf one defines it as a certain commutative square. When we're in V, I see that we can use adjointness to simplify it somewhat, so I agree that in that diagram, first going right, then down (i.e $(\alpha_y)_\ast \circ F_{x,y})$ is the same by adjointness to $f^\ast_w \circ \mu$. But why is the other direction what you wrote? I am sure it's elementary, but I am a bit confused. – user101036 Mar 3 '14 at 12:58
And I just want to add that I am very grateful for the time you are taking and sharing your knowledge. From what I understand, $G_{x,y}$ in our case is the transpose of the composition morphism. It probably follows by adjointness that it transforms as you said but I can't see why. – user101036 Mar 3 '14 at 12:59
Yes, use the tensor–hom adjunction to reduce it to my equation. – Zhen Lin Mar 3 '14 at 13:54
Sorry, but why does it reduce? I realize this should follow easily but I can't see why as in the diagram $(f)^\ast_z \circ G_{x,y}$ where $G_{x,y}$ is adjoint to composition reduces under the adjunction to $\mu \otimes (id \otimes (f^\ast)_z)$. – user101036 Mar 3 '14 at 17:01
This is a fact about two-variable adjunctions (in ordinary categories). Try working out in the case $\mathcal{V} = \mathbf{Set}$ first. – Zhen Lin Mar 3 '14 at 17:31 | 1,434 | 4,614 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2016-30 | latest | en | 0.80213 |
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