text:
ElMaestro
★★★

Denmark,
2019-06-18 17:22
(1073 d 23:35 ago)

Posting: # 20337
Views: 5,471

## Recalled Paracetamol, a question for CMC gurus [Off Topic]

Hi all,

I noticed the news about recalled paracetamol .

If I get it right, the issue has been detected as a visible discolouration of the tablets .

Fungal species usually will not and do not overgrow anything anywhere unless there is a damp environment. So, I am guessing there is a lot of humidity in the affected bottles. This is not a little drop of water. Having fungus overgrowing tablets may imply a deluge of Biblical proportions.

It could come from the containers themselves or from the tablets. It could have arisen before, during, after the bottle filling. What is your best guess, how and/or when did the water enter this game in the first place?

A post-tabletting drum left in open air and suffering a rain shower? LOD testing done on some of the fortunate tablets prior to release? Or tested before the rain shower. And so forth, let me hear some wild speculation.

Pass or fail!
ElMaestro
Helmut
★★★

Vienna, Austria,
2019-06-18 18:40
(1073 d 22:17 ago)

@ ElMaestro
Posting: # 20338
Views: 4,757

## Recalled Paracetamol, a question for CMC gurus

Hi ElMaestro,

» A post-tabletting drum left in open air and suffering a rain shower? LOD testing done on some of the fortunate tablets prior to release? Or tested before the rain shower. And so forth, let me hear some wild speculation.

Manufactured in the UK. Climate change? Monsoon-like conditions?

Dif-tor heh smusma 🖖
Helmut Schütz

The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
nobody
nothing

2019-06-18 20:44
(1073 d 20:13 ago)

@ Helmut
Posting: # 20339
Views: 4,697

## Recalled Paracetamol, a question for CMC gurus

Anyone with a photo of this "1x1000" tablet pots? Would love to see...

Kindest regards, nobody
ElMaestro
★★★

Denmark,
2019-06-19 14:56
(1073 d 02:00 ago)

@ nobody
Posting: # 20342
Views: 4,647

## Recalled Paracetamol, a question for CMC gurus

Pass or fail!
ElMaestro
Helmut
★★★

Vienna, Austria,
2019-06-19 15:24
(1073 d 01:32 ago)

@ ElMaestro
Posting: # 20343
Views: 4,611

## Recalled Paracetamol, a question for CMC gurus

Hi ElMaestro?

Green mold? Isn’t that some kind of algae?

Dif-tor heh smusma 🖖
Helmut Schütz

The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
ElMaestro
★★★

Denmark,
2019-06-19 16:03
(1073 d 00:54 ago)

@ Helmut
Posting: # 20344
Views: 4,622

## Recalled Paracetamol, a question for CMC gurus

Haha,

» Green mold? Isn’t that some kind of algae?

Algae are good for you This is apparently a combination product then.

Pass or fail!
ElMaestro
nobody
nothing

2019-06-19 16:04
(1073 d 00:53 ago)

@ ElMaestro
Posting: # 20345
Views: 4,592

## Recalled Paracetamol, a question for CMC gurus

Fake News, this is a Danish backyard, not in the UK... VERY SAD!

Kindest regards, nobody
ElMaestro
★★★

Denmark,
2019-06-19 16:31
(1073 d 00:26 ago)

@ nobody
Posting: # 20346
Views: 4,620

## Recalled Paracetamol, a question for CMC gurus

» Fake News, this is a Danish backyard, not in the UK... VERY SAD!

Remember that companies are allowed to write "manufactured in the UK" on the PIL even if it is produced in India or Denmark or Tierra del Fuego, as long as the release is from someone in the UK.
So don't you come here and call my gonococci and slime molds and beautiful algae sad

Pass or fail!
ElMaestro
nobody
nothing

2019-06-19 16:41
(1073 d 00:16 ago)

@ ElMaestro
Posting: # 20347
Views: 4,650

## Recalled Paracetamol, a question for CMC gurus

...you mean all these high-quality paracetamol products are produced in India, China, Indonesia or where ever and then simply LABELED as UK-products?

Wouldn't that be shocking? Some kind of... ...hhhmmmm.... fraught? 8-O

Kindest regards, nobody
Helmut
★★★

Vienna, Austria,
2019-06-19 16:48
(1073 d 00:09 ago)

@ nobody
Posting: # 20348
Views: 4,593

## Recalled Paracetamol, a question for CMC gurus

Hi nobody,

» ...you mean all these high-quality paracetamol products are produced in India, China, Indonesia or where ever and then simply LABELED as UK-products?
»
» Wouldn't that be shocking? Some kind of... ...hhhmmmm.... fraught? 8-O

Acc. to EEA legislation: Manufactured in country A → released by a “qualified person” in European county B → officially manufactured in B. That’s it. Correct me if I’m wrong.

Dif-tor heh smusma 🖖
Helmut Schütz

The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes
ElMaestro
★★★

Denmark,
2019-06-19 17:10
(1072 d 23:47 ago)

@ Helmut
Posting: # 20349
Views: 4,593

## Recalled Paracetamol, a question for CMC gurus

Hi nobody and Hötzi,

» » Wouldn't that be shocking? Some kind of... ...hhhmmmm.... fraught? 8-O

Yes I find it highly misleading, too.

» Acc. to EEA legislation: Manufactured in country A → released by a “qualified person” in European county B → officially manufactured in B. That’s it. Correct me if I’m wrong.

That is exactly what it means.
I will be interviewed about such matters next week by a journalist who is sick and tired of some of the garbage we have to suffer in EU because the rules and regulations may be perceived as totally obscure and very much intended to protect regulators rather than patients.

What's more, you can't ask your doctor or at the pharmacy. They don't have such info either. For example,
"I don't want a Valsartan product where the active ingredient is from China because I heard it may be contaminated with carcinogens." - noone can help.
"I don't want a product manufactured in India, regardless of where it is released" - noone can help.
"I don't want a product approved on basis of a BE study done at GVK Bio or Semler" - noone can help.

Pass or fail!
ElMaestro
nobody
nothing

2019-06-19 17:40
(1072 d 23:17 ago)

(edited by nobody on 2019-06-19 18:12)
@ ElMaestro
Posting: # 20350
Views: 4,577

## Recalled Paracetamol, a question for CMC gurus

...if you don't want any products "produced" in China or India then start growing your own Herba Chamomill. in your backyard, otherwise you're lost...

This legislation is imho 100% triggered by the manufacturers (generic+originators) to "improve economic conditions world-wide and trigger important investments in developing countries" (or what it would read like in the investors brochure).

___

PS:

 MedWatch - The FDA Safety Information and Adverse Event Reporting Program A MedWatch Safety Alert was just added to the FDA Recalls web page. TOPIC: All Unexpired Sterile Drug Product Lots by Premier Pharmacy Labs: Recall - Due to Lack of Sterility Assurance AUDIENCE: Patient, Health Professional, Pharmacy ISSUE: The unexpired sterile drug product lots are being recalled due to concerns presented during the latest FDA inspection including insufficient environmental controls, potential cross contamination and lack of product specific process validations. ....

Just the latest of these emails dropping in here frequently. It doesn'T matter where your medicine is fu** up eeehm produced. It'S the same sh** everywhere.

From their homepage:

"
STERILE COMPOUNDING PHARMACY

Premier Pharmacy Labs is committed to sterile compounding excellence. We were one of the first pharmacies in the United States to register as a 503B outsourcing facility (see FDA 503B). Premier has increased its resources and revenues towards expanding the sterile operations of our business. Adhering to GMP manufacturing standards is something new to all compounding pharmacies across the US and Premier is pushing to stay ahead of the mark with continued investment and innovation.
"

Kindest regards, nobody
nobody
nothing

2019-06-28 11:51
(1064 d 05:06 ago)

@ nobody
Posting: # 20365
Views: 4,381

## Recalled Paracetamol, a question for CMC gurus

PPS:

The FDA found termites, dead pests, impurities and other serious GMP violations at Bristol-Myers Squibb’s drug facility in Manati, Puerto Rico, during a Jan. 15-Feb. 6 inspection.

The agency investigators discovered that a media fill conducted at the facility resulted in a vial that was murky and contaminated with bacteria. The firm identified the contaminants as mites. As a result, lots of lyophilized drug products were placed on hold.

Kindest regards, nobody
jag009
★★★

NJ,
2019-07-01 02:41
(1061 d 14:16 ago)

@ ElMaestro
Posting: # 20367
Views: 4,278

## Recalled Paracetamol, a question for CMC gurus

Hi El,

» ... "I don't want a product approved on basis of a BE study done at GVK Bio or Semler" - noone can help.

Yup. You are stuck. Welcome to the... ^#@*!

John


text:
# Yoneda lemma

(Redirected from Yoneda embedding)

In mathematics, specifically in category theory, the Yoneda lemma is an abstract result on functors of the type morphisms into a fixed object. It is a vast generalisation of Cayley's theorem from group theory (viewing a group as a particular kind of category with just one object and only isomorphisms). It allows the embedding of any category into a category of functors (contravariant set-valued functors) defined on that category. It also clarifies how the embedded category, of representable functors and their natural transformations, relates to the other objects in the larger functor category. It is an important tool that underlies several modern developments in algebraic geometry and representation theory. It is named after Nobuo Yoneda.

## Generalities

The Yoneda lemma suggests that instead of studying the (locally small) category ${\displaystyle {\mathcal {C}}}$ , one should study the category of all functors of ${\displaystyle {\mathcal {C}}}$  into ${\displaystyle \mathbf {Set} }$  (the category of sets with functions as morphisms). ${\displaystyle \mathbf {Set} }$  is a category we think we understand well, and a functor of ${\displaystyle {\mathcal {C}}}$  into ${\displaystyle \mathbf {Set} }$  can be seen as a "representation" of ${\displaystyle {\mathcal {C}}}$  in terms of known structures. The original category ${\displaystyle {\mathcal {C}}}$  is contained in this functor category, but new objects appear in the functor category, which were absent and "hidden" in ${\displaystyle {\mathcal {C}}}$ . Treating these new objects just like the old ones often unifies and simplifies the theory.

This approach is akin to (and in fact generalizes) the common method of studying a ring by investigating the modules over that ring. The ring takes the place of the category ${\displaystyle {\mathcal {C}}}$ , and the category of modules over the ring is a category of functors defined on ${\displaystyle {\mathcal {C}}}$ .

## Formal statement

Yoneda's lemma concerns functors from a fixed category ${\displaystyle {\mathcal {C}}}$  to the category of sets, ${\displaystyle \mathbf {Set} }$ . If ${\displaystyle {\mathcal {C}}}$  is a locally small category (i.e. the hom-sets are actual sets and not proper classes), then each object ${\displaystyle A}$  of ${\displaystyle {\mathcal {C}}}$  gives rise to a natural functor to ${\displaystyle \mathbf {Set} }$  called a hom-functor. This functor is denoted:

${\displaystyle h^{A}=\mathrm {Hom} (A,-)}$ .

The (covariant) hom-functor ${\displaystyle h^{A}}$  sends ${\displaystyle X}$  to the set of morphisms ${\displaystyle \mathrm {Hom} (A,X)}$  and sends a morphism ${\displaystyle f\colon X\to Y}$  to the morphism ${\displaystyle f\circ -}$  (composition with ${\displaystyle f}$  on the left) that sends a morphism ${\displaystyle g}$  in ${\displaystyle \mathrm {Hom} (A,X)}$  to the morphism ${\displaystyle f\circ g}$  in ${\displaystyle \mathrm {Hom} (A,Y)}$ . That is,

${\displaystyle h^{A}(f)=\mathrm {Hom} (A,f){\mbox{, or}}}$
${\displaystyle h^{A}(f)(g)=f\circ g}$ .

Let ${\displaystyle F}$  be an arbitrary functor from ${\displaystyle {\mathcal {C}}}$  to ${\displaystyle \mathbf {Set} }$ . Then Yoneda's lemma says that:

For each object ${\displaystyle A}$  of ${\displaystyle {\mathcal {C}}}$ , the natural transformations from ${\displaystyle h^{A}}$  to ${\displaystyle F}$  are in one-to-one correspondence with the elements of ${\displaystyle F(A)}$ . That is,

${\displaystyle \mathrm {Nat} (h^{A},F)\cong F(A)}$ .

Moreover this isomorphism is natural in ${\displaystyle A}$  and ${\displaystyle F}$  when both sides are regarded as functors from ${\displaystyle \mathbf {Set} ^{\mathcal {C}}\times {\mathcal {C}}}$  to ${\displaystyle \mathbf {Set} }$ . (Here the notation ${\displaystyle \mathbf {Set} ^{\mathcal {C}}}$  denotes the category of functors from ${\displaystyle {\mathcal {C}}}$  to ${\displaystyle \mathbf {Set} }$ .)

Given a natural transformation ${\displaystyle \Phi }$  from ${\displaystyle h^{A}}$  to ${\displaystyle F}$ , the corresponding element of ${\displaystyle F(A)}$  is ${\displaystyle u=\Phi _{A}(\mathrm {id} _{A})}$ ;[a] and given an element ${\displaystyle u}$  of ${\displaystyle F(A)}$ , the corresponding natural transformation is given by ${\displaystyle \Phi (f)=F(f)(u)}$ .

There is a contravariant version of Yoneda's lemma, which concerns contravariant functors from ${\displaystyle {\mathcal {C}}}$  to ${\displaystyle \mathbf {Set} }$ . This version involves the contravariant hom-functor

${\displaystyle h_{A}=\mathrm {Hom} (-,A),}$

which sends ${\displaystyle X}$  to the hom-set ${\displaystyle \mathrm {Hom} (X,A)}$ . Given an arbitrary contravariant functor ${\displaystyle G}$  from ${\displaystyle {\mathcal {C}}}$  to ${\displaystyle \mathbf {Set} }$ , Yoneda's lemma asserts that

${\displaystyle \mathrm {Nat} (h_{A},G)\cong G(A).}$

### Naming conventions

The use of ${\displaystyle h^{A}}$  for the covariant hom-functor and ${\displaystyle h_{A}}$  for the contravariant hom-functor is not completely standard. Many texts and articles either use the opposite convention or completely unrelated symbols for these two functors. However, most modern algebraic geometry texts starting with Alexander Grothendieck's foundational EGA use the convention in this article.[b]

The mnemonic "falling into something" can be helpful in remembering that ${\displaystyle h_{A}}$  is the contravariant hom-functor. When the letter ${\displaystyle A}$  is falling (i.e. a subscript), ${\displaystyle h_{A}}$  assigns to an object ${\displaystyle X}$  the morphisms from ${\displaystyle X}$  into ${\displaystyle A}$ .

### Proof

The proof of Yoneda's lemma is indicated by the following commutative diagram:

This diagram shows that the natural transformation ${\displaystyle \Phi }$  is completely determined by ${\displaystyle \Phi _{A}(\mathrm {id} _{A})=u}$  since for each morphism ${\displaystyle f\colon A\to X}$  one has

${\displaystyle \Phi _{X}(f)=(Ff)u}$ .

Moreover, any element ${\displaystyle u\in F(A)}$  defines a natural transformation in this way. The proof in the contravariant case is completely analogous.

### The Yoneda embedding

An important special case of Yoneda's lemma is when the functor ${\displaystyle F}$  from ${\displaystyle {\mathcal {C}}}$  to ${\displaystyle \mathbf {Set} }$  is another hom-functor ${\displaystyle h^{B}}$ . In this case, the covariant version of Yoneda's lemma states that

${\displaystyle \mathrm {Nat} (h^{A},h^{B})\cong \mathrm {Hom} (B,A).}$

That is, natural transformations between hom-functors are in one-to-one correspondence with morphisms (in the reverse direction) between the associated objects. Given a morphism ${\displaystyle f\colon B\to A}$  the associated natural transformation is denoted ${\displaystyle \mathrm {Hom} (f,-)}$ .

Mapping each object ${\displaystyle A}$  in ${\displaystyle {\mathcal {C}}}$  to its associated hom-functor ${\displaystyle h^{A}=\mathrm {Hom} (A,-)}$  and each morphism ${\displaystyle f\colon B\to A}$  to the corresponding natural transformation ${\displaystyle \mathrm {Hom} (f,-)}$  determines a contravariant functor ${\displaystyle h^{-}}$  from ${\displaystyle {\mathcal {C}}}$  to ${\displaystyle \mathbf {Set} ^{\mathcal {C}}}$ , the functor category of all (covariant) functors from ${\displaystyle {\mathcal {C}}}$  to ${\displaystyle \mathbf {Set} }$ . One can interpret ${\displaystyle h^{-}}$  as a covariant functor:

${\displaystyle h^{-}\colon {\mathcal {C}}^{\text{op}}\to \mathbf {Set} ^{\mathcal {C}}.}$

The meaning of Yoneda's lemma in this setting is that the functor ${\displaystyle h^{-}}$  is fully faithful, and therefore gives an embedding of ${\displaystyle {\mathcal {C}}^{\mathrm {op} }}$  in the category of functors to ${\displaystyle \mathbf {Set} }$ . The collection of all functors ${\displaystyle \{h^{A}|A\in C\}}$  is a subcategory of ${\displaystyle \mathbf {Set} ^{\mathcal {C}}}$ . Therefore, Yoneda embedding implies that the category ${\displaystyle {\mathcal {C}}^{\mathrm {op} }}$  is isomorphic to the category ${\displaystyle \{h^{A}|A\in C\}}$ .

The contravariant version of Yoneda's lemma states that

${\displaystyle \mathrm {Nat} (h_{A},h_{B})\cong \mathrm {Hom} (A,B).}$

Therefore, ${\displaystyle h_{-}}$  gives rise to a covariant functor from ${\displaystyle {\mathcal {C}}}$  to the category of contravariant functors to ${\displaystyle \mathbf {Set} }$ :

${\displaystyle h_{-}\colon {\mathcal {C}}\to \mathbf {Set} ^{{\mathcal {C}}^{\mathrm {op} }}.}$

Yoneda's lemma then states that any locally small category ${\displaystyle {\mathcal {C}}}$  can be embedded in the category of contravariant functors from ${\displaystyle {\mathcal {C}}}$  to ${\displaystyle \mathbf {Set} }$  via ${\displaystyle h_{-}}$ . This is called the Yoneda embedding.

### Representable functor

The Yoneda embedding essentially states that for every (locally small) category, objects in that category can be represented by presheaves, in a full and faithful manner. That is,

${\displaystyle \mathrm {Nat} (h_{A},P)\cong P(A).}$

for a presheaf P. Many common categories are, in fact, pre-sheaves, and on closer inspection, prove to be sheaves, and, as such examples are commonly topological in nature, they can be seen to be topoi in general. The Yoneda lemma provides a point of leverage by which the topological structure of a category can be studied and understood.

## Preadditive categories, rings and modules

A preadditive category is a category where the morphism sets form abelian groups and the composition of morphisms is bilinear; examples are categories of abelian groups or modules. In a preadditive category, there is both a "multiplication" and an "addition" of morphisms, which is why preadditive categories are viewed as generalizations of rings. Rings are preadditive categories with one object.

The Yoneda lemma remains true for preadditive categories if we choose as our extension the category of additive contravariant functors from the original category into the category of abelian groups; these are functors which are compatible with the addition of morphisms and should be thought of as forming a module category over the original category. The Yoneda lemma then yields the natural procedure to enlarge a preadditive category so that the enlarged version remains preadditive — in fact, the enlarged version is an abelian category, a much more powerful condition. In the case of a ring ${\displaystyle R}$ , the extended category is the category of all right modules over ${\displaystyle R}$ , and the statement of the Yoneda lemma reduces to the well-known isomorphism

${\displaystyle M\cong \mathrm {Hom} _{R}(R,M)}$    for all right modules ${\displaystyle M}$  over ${\displaystyle R}$ .

## Relationship to Cayley's theorem

As stated above, the Yoneda lemma may be considered as a vast generalization of Cayley's theorem from group theory. To see this, let ${\displaystyle {\mathcal {C}}}$  be a category with a single object ${\displaystyle *}$  such that every morphism is an isomorphism (i.e. a groupoid with one object). Then ${\displaystyle G=\mathrm {Hom} _{\mathcal {C}}(*,*)}$  forms a group under the operation of composition, and any group can be realized as a category in this way.

In this context, a covariant functor ${\displaystyle {\mathcal {C}}\to \mathbf {Set} }$  consists of a set ${\displaystyle X}$  and a group homomorphism ${\displaystyle G\to \mathrm {Perm} (X)}$ , where ${\displaystyle \mathrm {Perm} (X)}$  is the group of permutations of ${\displaystyle X}$ ; in other words, ${\displaystyle X}$  is a G-set. A natural transformation between such functors is the same thing as an equivariant map between ${\displaystyle G}$ -sets: a set function ${\displaystyle \alpha \colon X\to Y}$  with the property that ${\displaystyle \alpha (g\cdot x)=g\cdot \alpha (x)}$  for all ${\displaystyle g}$  in ${\displaystyle G}$  and ${\displaystyle x}$  in ${\displaystyle X}$ . (On the left side of this equation, the ${\displaystyle \cdot }$  denotes the action of ${\displaystyle G}$  on ${\displaystyle X}$ , and on the right side the action on ${\displaystyle Y}$ .)

Now the covariant hom-functor ${\displaystyle \mathrm {Hom} _{\mathcal {C}}(*,-)}$  corresponds to the action of ${\displaystyle G}$  on itself by left-multiplication (the contravariant version corresponds to right-multiplication). The Yoneda lemma with ${\displaystyle F=\mathrm {Hom} _{\mathcal {C}}(*,-)}$  states that

${\displaystyle \mathrm {Nat} (\mathrm {Hom} _{\mathcal {C}}(*,-),\mathrm {Hom} _{\mathcal {C}}(*,-))\cong \mathrm {Hom} _{\mathcal {C}}(*,*)}$ ,

that is, the equivariant maps from this ${\displaystyle G}$ -set to itself are in bijection with ${\displaystyle G}$ . But it is easy to see that (1) these maps form a group under composition, which is a subgroup of ${\displaystyle \mathrm {Perm} (G)}$ , and (2) the function which gives the bijection is a group homomorphism. (Going in the reverse direction, it associates to every ${\displaystyle g}$  in ${\displaystyle G}$  the equivariant map of right-multiplication by ${\displaystyle g}$ .) Thus ${\displaystyle G}$  is isomorphic to a subgroup of ${\displaystyle \mathrm {Perm} (G)}$ , which is the statement of Cayley's theorem.

## History

Yoshiki Kinoshita stated in 1996 that the term "Yoneda lemma" was coined by Saunders Mac Lane following an interview he had with Yoneda.[1]

1. ^ Recall that ${\displaystyle \Phi _{A}:\mathrm {Hom} (A,A)\to F(A)}$  so the last expression is well-defined and sends a morphism from ${\displaystyle A}$  to ${\displaystyle A}$ , to an element in ${\displaystyle F(A)}$ .
2. ^ A notable exception to modern algebraic geometry texts following the conventions of this article is Commutative algebra with a view toward algebraic geometry / David Eisenbud (1995), which uses ${\displaystyle h_{A}}$  to mean the covariant hom-functor. However, the later book The geometry of schemes / David Eisenbud, Joe Harris (1998) reverses this and uses ${\displaystyle h_{A}}$  to mean the contravariant hom-functor.

format: plain


text:
# 题目

## Problem Description

Zero has an old printer that doesn’t work well sometimes. As it is antique, he still like to use it to print articles. But it is too old to work for a long time and it will certainly wear and tear, so Zero use a cost to evaluate this degree.
One day Zero want to print an article which has $$N$$ words, and each word $$i$$ has a cost $$C_i$$ to be printed. Also, Zero know that print $$k$$ words in one line will cost
$$(\sum\limits_{k=1}^kC_i)^2+M$$
$$M$$ is a const number.
Now Zero want to know the minimum cost in order to arrange the article perfectly.

## Input

There are many test cases. For each test case, There are two numbers N and M in the first line (0 ≤ n ≤ 500000, 0 ≤ M ≤ 1000). Then, there are N numbers in the next 2 to N + 1 lines. Input are terminated by EOF.

## Output

A single number, meaning the mininum cost to print the article.

5 5
5
9
5
7
5

230

# 题解

$$dp[i]=dp[j]+M+(sum[i]-sum[j])^2$$

$$dp[j]-dp[k]+2\times sum[i]\times (sum[k]-sum[j])+sum[j]^2-sum[k]^2<0$$

\frac{(dp[j]+sum[j])^2-(dp[k]+sum[k])^2}{2\times (sum[j]-sum[k])}
\lt sum[i]

$$\frac{y_j-y_k}{x_j-x_k}\lt sum[i]$$

1，用一个单调队列来维护解集。
2，假设队列中从头到尾已经有元素a b c。那么当d要入队的时候，我们维护队列的上凸性质，即如果g[d,c]<g[c,b]，那么就将c点删除。直到找到g[d,x]>=g[x,y]为止，并将d点加入在该位置中。


text:
# Tips and Tricks for Simplification Section in Bank Exam

By | April 15, 2017

Simplification plays a vital role in the quantitative aptitude section of Bank Exams. It is one the most scoring section. Candidates who are applying for IBPS Exam should prepare tips and tricks for simplification section in bank examthis section properly to qualify through the overall as well as the sectional cut-off. Simplification questions are easy and take less time to solve.

To solve this simplification section the candidates must be well acquainted with the BODMAS rule:
B – Brackets []
O – Of
D – Division
M – Multiplication
S – Subtraction
Topics in simplification section are:
Number System
Square & cube
HCF & LCM
Surds & Indices
Fractions & Decimals

tips and tricks for simplification section in bank exam

Number System: this is the basis of quantitative ability which is to be attempted first. All counting numbers are termed as natural numbers. To solve these type of question the basic definitions of the number system and the division methods along with the divisible factors need to known properly. These are very simple if the roots of any numerical are known to the candidate. The division remainder rules, sum rules and the divisibility are to be solved with attention because candidate generally loses marks due to their silly mistakes. For example:

890 – 696 ÷ 3=?
32.33
658
-547.6
68.508
None of these
Solution: By BODMAS rule.

Square & Cube: This includes square root, cube root and factorization as well. Practice is required to master this section. The basic concept of perfect square numbers which are 1,4,9,6,5 and non-perfect square numbers like 2,3,7,8 are to be known. The square root and cube of the numbers including the factorization method should be at the tip. Practice will ensure faster calculation along with accuracy so that the time and accuracy in the time of the upcoming bank exam are perfect. For example: square root of 9 is 3. Cube root of 27 is 3.

tips and tricks for simplification section in bank exam

HCF & LCM: there are basically two methods to solve problems on HCF & LCM which are factorization method and division method. In factorization method, each number has to be the product of the prime factors. Then the product of least powers gives the HCF.

Forexample: HCF of 108,288,360 108 gives $2^{2} \times 3^{3}$ , 288 gives $2^{5} \times$ 32 and 360 gives $23 \times 5 \times 32$

HCF is 22 x 32= 36

Solution: 36 by factorization method
To find theLCM of 6,12,18,24 = 2 x 3 x 3 x 2 x 2 x 3 x 1 which is 72.

Solution: 72 by division method

Surds and Indices

this section is all about remembering the different laws of surds and indices to solve the problems. The candidate needs to be well versed with the formulas. This section is almost like arithmetic formulas without those the problems cannot be solved. It will be better if the candidate makes a note of these rules and revise it daily through some problem which will help them to increase their efficiency in solving this kind of sums.

For example: Rule of indices like $a^{m} \times a^{n} = a^{m+n}$

Rule of surds like $\sqrt[n]{a} = a^{\frac{1}{n}}$

Fractions and Decimals: this section consists of different types of decimal and fraction which is to be known properly to solve the sums from this section. The candidates are advised to practice through previous years question paper so that they will be well acquainted with what type of questions are likely to come from this section. Mock test papers will also help to practice with accuracy.
For example: Multiplication of a decimal- 0.06 x 0.3 0.40 = 0.00720
Conversion of decimal into fraction – 0.16 = (16-1)/90 = 1/6

The simplification techniques are easy to practice so to score good marks this section must not be ignored. This section needs to be cleared first keeping all the tips in mind.


text:
# standard or recommended font for vocabulary or gloss terms

I'm writing notes for my students and in the present section, the bulk of the content is defining terms. To help make the notes more practical, I put the technical terms being defined in bold so that it 'pops' out on the page, when skimming.

I'd like to enhance this effect by also changing the font of these terms (so new font + bold), and wondered if there was a standard font that was paired with the default Tex font (I'm using extarticle). I imagine it would be slightly larger and sans-seriff. N.B. I'm not actually using a glossary, these notes are effectively expanded glossaries.

• Why not just use \sffamily or \textsf{} in addition to \bfseries or \textbf{}. You could also make it larger but I wouldn't recommend this if you are using the terms inline because it will throw off line spacing etc. I guess one question is: which font are you using for the rest of the document? If you are using the default Computer Modern Roman, I'd stick with the sans variant as they are designed to complement each other. Similarly for Latin Modern Roman and its sans counterpart. A Minimum Working Example would help give a sense of how exactly you are using it and what might work. – cfr Mar 3 '14 at 21:54

Rather than pepper your text with explicit calls to \textbf and \textsf, you might find it easier to define a new command to mark your keywords. For example

\newcommand{\key}[1]{{\bfseries\sffamily #1}}

then you can mark your key words like this: here is a new \key{term} to remember.

This way, if you later regret your choice of font, you only have to change it in one place. Moreover, if you later decide you want to add more function (like adding the key words to an index), you only have to modify the \key command.

NB: Put the \newcommand in the preamble, before the \begin{document} line. Here's a complete example.

\documentclass{article}
\newcommand{\key}[1]{{\bfseries\sffamily #1}}
\begin{document}
In particular, any associated \key{supporting element} necessitates that urgent
consideration be applied to possible bidirectional logical relationship
approaches.  Conversely, any associated supporting element recognizes other
systems' importance and the necessity for possible bidirectional logical
relationship approaches.  However, a \key{service-oriented paradigm} is further
compounded when taking into account the evolution of specifications over a
given time period.
\end{document}

• Thanks - that's an awesome solution! Thanks for taking the time to put it together, much appreciated! – Rax Adaam Mar 4 '14 at 22:10


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An air bubble (radius 0.4 mm) ...

# An air bubble (radius 0.4 mm) rises up in water. If the coefficient of viscosity of water be 1xx10^(-3)kg//(m-s), then determine the terminal speed of the bubble density of air is negligible

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Updated On: 27-06-2022

Text Solution

 0.84 m//s   3.5 m//s  0.48 m//s  0.35 m//s

Solution : At terminal speed,  F = B , 6 pi eta r v = 4//3 pi r ^ 3 rho _w g ,  solving  v = 0. 348 m//s

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Transcript

11 so we are given the question which there is an air bubble of radius 0.4 m m Rises up in water if the coefficient of viscosity of water be 1 into 10 raise to minus 3 kg per metre second terminal velocity of the bubble density a terminal velocity of the bubble density of air is negligible alright so we have been given a way that case of terminal velocity so let's write a formula for terminal velocity for a body which is first let's consider what we have studied in the part that is let's suppose that in a fluid from body is win moving downward it is having a density of row and here it is liquid is having a density of Sigma previous studies that this body will Experience At track course a word that is FD and a buoyant force or we can say after just a light

from here we have calculated the formula for terminal velocity v t as to buy 9 into square into a row - of Sigma / Pita this is the formula which we have studied for a body which is moving downwards in a employed but here is a case where the body is moving upward or we can say the bubble is rising upward in air so here we will see the terminal velocity will be negative Elite so in our case the terminal velocity will be negative because the directions is twin rivers bubble is moving upward so what we are the expression we are getting will be v t is equals to 2 by 9 square and it comes out to be Sigma - of pro Elite and / it are the negative sign if you take It Inside show

this will become positive and this will become negative so we had to keep this in mind that is the density of Ab confusion and writing density of bubble is that it is the density of the material August theory cancer density of bubble in our case here and this is the density of air in our case wallet so you write a density of air show that should be no confusion we have to keep this in mind when the body is moving downward and it as a terminal velocity is bro this expression has to be used for the downward motion and writing and for permission terminal velocity will become negative now I just to have to substitute the value and we can get the desired result alright so let's put the value of SO2 by 9 in 2G can be taken as

10 an hour is being given 0.4 10 raise to minus 3 square as it is substituting the values over then we will solve the above has been given to be as negligible and density of water this is density of water sorry show nearest density of water and here it is density of air bubble this is a below write a density of water we can take it has done this 23 kg per metre cube alright and -40 you are not writing over there and the whole thing / ETA ETA has been given to be as 1 into 10 raise to minus 3 movie all the value now we have to solve it and get the answers so we can write a ab 9 into 10 into this can be done as 16 into 10 raise to minus of 8 into 10 raise to 3 into 10 raise to 3

solving it further we will get 67 - 108 can be done to 1926 3 into 10 raise to minus one this value comes out to be 3232 19 into 10 raise to minus one or we can say that 3.2 19 and that values comes out to be zero point approximately 3-4 metre per second this is the value of terminal velocity we are getting in the correct option matching with this is option d i hope you understood the concept and thank you for watching