text:
Thursday, June 16, 2022

# Ways To Remember Geometry Formulas

## Sample Gre Geometry Formula Practice Questions

A Silly Way to Memorize Math Formulas

Those are it in a nutshellthe GRE geometry formulas youll need on test day! However, memorizing formulas and putting them into practice are two very different things.

So that you can see the difference, weve pulled two problems from the Magoosh GRE prep course so you can try your hand at them.

Question 1: Multiple Choice1. The hypotenuse of a right triangle is 16 ft longer than the length of the shorter leg. If the area of this triangle is exactly 120 ft2, what is the length of the hypotenuse in feet?

A. 26

Question 2: Numeric Entry2. In the diagram, point D is the center of the medium-sized circle that passes through C and E, and it is also the center of the largest circle that passes through A and G. Each of the diameters of the small circles with centers B and F equals the radius of the medium-sized circle with center D. The shaded area is what fraction of the largest circle?

## Understand How The Formulas Came About

Still thinking about how to remember formulas? Before trying to cram any formula, try to understand how the formula was proven, each formulas elements, and why they make sense together. Understanding the workings of any given formula makes it a lot easier to remember.Lets test this with something simple. Consider the area of an isosceles triangle as ½bh, where b is the triangles base, and h is the height of the triangle. If we split the triangle vertically down the middle into two equal parts, itll form two right-angle triangles. Place the hypotenuses of these triangles together such that they form a rectangle. Youll find that the rectangles width is the same as half the base of the isosceles triangle, while the length of the rectangle is the same as the perpendicular height of the triangle.So essentially, the area of the rectangle and that of the isosceles triangle are the same. So if the dimensions are identical, as we stated in this example, then the ½bh of an isosceles triangle is the same as the L×B of a rectangle.Do you see how understanding the fundamentals of these shapes makes it easier to understand these formulas? This ultimately works in other areas of math beyond geometry.

## The 28 Critical Sat Math Formulas You Must Know

The SAT math test is unlike any math test you’ve taken before. It’s designed to take concepts you’re used to and make you apply them in new ways. It’s tricky, but with attention to detail and knowledge of the basic formulas and concepts covered by the test, you can improve your score.

So what formulas do you need to have memorized for the SAT math section before the day of the test? In this complete guide, I’ll cover every critical formula you MUST know before you sit down for the test. I’ll also explain them in case you need to jog your memory about how a formula works. If you understand every formula in this list, you’ll save yourself valuable time on the test and probably get a few extra questions correct.

Don’t Miss: Pizzazz Bridge To Algebra

## Understand And Not Cram

You can easily forget what you crammed, butnever what you truly understand. Instead of having a myopic focus on thenumbers and symbols, try to understand what that particular formula actuallysolves. Once you have a semblance of a real world issue being solved by aformula, you are unlikely to forget it ever in your life.

## What Is A Math Formula And Why Is It Important

In mathematics, a formula is a word equation that defines how to find the unknown variables within the equation in terms of other known values. It is commonly used to describe data that is being graphed or plotted.

Mathematical formulas are important because they help us visualize data and get a better understanding of the concept. They can also be used as a step-by-step guide to solving problems in different areas.

There are many different types of mathematical equations. The most common ones are linear, exponential, logarithmic, and trigonometric equations.

Also Check: Algebraic Proofs Worksheets

## Volume Of A Rectangular Solid

• l is the length of one of the sides.
• h is the height of the figure.
• w is the width of one of the sides.
• $r$ is the radius of the circular side of the cylinder.
• $h$ is the height of the cylinder.
• $r$ is the radius of the sphere.

$$V=r^2h$$

• $r$ is the radius of the circular side of the cone.
• $h$ is the height of the pointed part of the cone .

$$V=lwh$$

• $l$ is the length of one of the edges of the rectangular part of the pyramid.
• $h$ is the height of the figure at its peak .
• $w$ is the width of one of the edges of the rectangular part of the pyramid.

Law: the number of degrees in a circle is 360

Law: the number of radians in a circle is $2$

Law: the number of degrees in a triangle is 180

Gear up that brain because here come the formulas you have to memorize.

## How To Use Memory Tricks To Memorize Math Formulas

Have you ever found yourself stuck while solving a math problem because you couldnt remember the required formula? Its right there, dangling on the edge of your memory but just out of reach enough for you to not recall it.If you have any educationand most of us doyou probably know what this feels like. But this doesnt happen only to students. Classroom teachers, online tutors, and more have come up with a blank too on many occasions.Math is an essential science subject. As Erica Sunarjo of essay writing service review outfit, Best Writers Online puts it, many times, mathematical formulas can express a solution clearer than any words in the dictionary.Her sentiment is echoed in how mathematical formulas are used across many fields of endeavorbusiness, financial analysis, woodwork and construction, interior design, research, technology, and a lot more. As crucial as math is, keeping track of the plethora of formulas involved can be overwhelming.Its worse if youre preparing for a big test, have a bunch of finance numbers to crunch, have a tutorial session coming up, etc. Weve outlined a few memory tricks in this post that will show you how to memorize formulas in a way thats easy to retain and recall.

Read Also: Theory Of Everything 2 Soundtrack

## Geometry Rules: Isosceles Triangles

One geometry rule related to this last fact concerns one elite category of triangles: the isosceles triangles. These are triangles with two equal sides.

The angle between the two equal sides may be

an acute angle , as in triangle ABD

a right angle, as in triangle DEF, or

an obtuse angle as in triangle KLM

If two sides are equal, then we know the opposite angles are also equal: in triangle ABC, angle A = angle C in triangle DEF, angle E = angle F and in triangle KLM, angle K = angle M. In fact, Mr. Euclid pointed out that this geometry rule works both ways: if we are told two sides are equal, then we know two angles are equal, and if we are told two angles are equal, we know two sides are equal.

The line down the middle of an isosceles triangle is special:

As long as we are told that JKL and PQR are isosceles triangles, then this midline has some special properties:

it is perpendicular to the base: angle KML = angle QSR = 90°

its lower point bisects the base: JM = ML and PS = SR

it bisects the upper angle: angle JKM = angle MKL, and angle PQS = angle SQR

These are all good geometry equations to associate with isosceles triangles.

Super-technically, an isosceles triangle is one that has at least two equal sides. This means that a very special case of isosceles is the equilateral triangle.

## Circumference Of A Circle

How to Remember Geometry Formulas

$C=2r$

• d is the diameter of the circle. It is a line that bisects the circle through the midpoint and touches two ends of the circle on opposite sides. It is twice the radius.
• l is the length of the rectangle
• w is the width of the rectangle
• b is the length of the base of triangle
• h is the height of the triangle
• In a right triangle, the height is the same as a side of the 90-degree angle. For non-right triangles, the height will drop down through the interior of the triangle, as shown above .

$$a^2 + b^2 = c^2$$

• In a right triangle, the two smaller sides are each squared. Their sum is the equal to the square of the hypotenuse .

Recommended Reading: Does Kamala Harris Have Any Biological Children

## On The Occasion Of National Mathematics Day Here Are 8 Ways By Which You Can Easily Memorise Math Formulas

Most students are afraid of mathematics due to the long list of formulas. But, mathematics is not only limited to learning from textbooks, it is about solving and understanding problems.

Mathematics formulas are the best way to make this subject a fun subject. Therefore, learning maths formulas will help students to gain confidence and enhance their problem-solving skills.

But, it is very tough and confusing to memorize all the maths formulas in one go. So, here are 8 ways to memorize maths formulas in an easy way.

## Sat Math: Beyond The Formulas

Though these are all the formulas you’ll need , this list doesn’t cover every aspect of SAT Math. You’ll also need to understand how to factor equations, how to manipulate and solve for absolute values, and how to manipulate and use exponents, and much more. These topics are all covered here.

Another important thing to remember is that while memorizing the formulas in this article that aren’t given to you on the test is important, knowing this list of formulas doesn’t mean you’re all set for SAT Math. You also need to practice applying these formulas to answer questions, so that you know when it makes sense to use them.

For instance, if you’re asked to calculate how likely it is that a white marble would be drawn from a jar that contains three white marbles and four black marbles, it’s easy enough to realize you need to take this probability formula:

$$\text”Probability of an outcome” = /$$

and use it to find the answer:

$\text”Probability of a white marble” = /$

$\text”Probability of a white marble” = 3/7$

On the SAT math section, however, you will also run into more complex probability questions like this one:

Dreams Recalled During One Week

C) $79/164$

D) $164/200$

There’s a lot of information to synthesize in that question: a table of data, a two-sentence long explanation of the table, and then, finally, what you need to solve for.

This is a probability question, so I’ll probably need to use this formula:

$$\text”Probability of an outcome” = /$$

Read Also: Does Quantum Mechanics Prove God

## Ways To Memorise Math Formulas In Easy Way

Most students are afraid of mathematics due to the long list of formulas. But, mathematics is not only limited to learning from textbooks, it is about solving and understanding problems.

Mathematics formulas are the best way to make this subject a fun subject. Therefore, learning maths formulas will help students to gain confidence and enhance their problem-solving skills.

But, it is very tough and confusing to memorize all the maths formulas in one go. So, here are 8 ways to memorize maths formulas in an easy way.

## Practice Questions Using Coordinate Geometry Formulas

• Find the distance between two points and .
• Calculate the slope of a line with coordinates and .
• Calculate the area of a triangle ABC whose vertices are at A, B and C.
• Calculate the midpoint of the line segment joining the points P and Q
• Keep visiting BYJUS and get more such maths formulas and concept explanations for free. Also, download BYJUS- The Learning App to get video lessons on different maths topics.

Put your understanding of this concept to test by answering a few MCQs. Click Start Quiz to begin!

Select the correct answer and click on the Finish buttonCheck your score and answers at the end of the quiz

Don’t Miss: Beth Tomas

## Geometry Formulas: Special Parallelograms

There are three categories of special parallelograms: rhombuses, rectangles, and squares. The BIG FOUR parallelogram properties above apply to all of them.

A rhombus is an equilateral quadrilateral, that is, a quadrilateral with four equal sides.

Most diamond shapes, such as those on playing cards, are simply rhombuses turned sideways, like the one on the right above. If the side is s, then the perimeter is always just 4s. The diagonals of a rhombus are always perpendicular. In fact, any rhombus can be subdivided into four congruent right triangles.

As with a general parallelogram, A = bh, where the b is any side and the h is the length of a perpendicular segment: as with a general parallelogram, the Pythagorean Theorem may play a role in finding one of the lengths that you need.

A rectangle is an equiangular quadrilateral, that is, a quadrilateral with four 90° angles.

Each angle is 90° and the two diagonals are always equal in length: its an old carpenters trick to verify that a doorframe has four right angles simply by checking the lengths of the two diagonals. Of course, its usually easy to find the length of a diagonal using the Pythagorean theorem.

The area of a rectangle is simply A =bh, where the base & height are simply lengths of any two adjacent sides. The perimeter is P = 2b + 2h.

\\ \

## Use Established Mnemonics Or Make Up Your Own

Ideally, you can use both. There are some mnemonic devices that we all know and love already. We all know BODMAS or PEMDAS . We also know the rule of finding the sine, cosine, and tangent of angles in a right-angled triangleSOH CAH TOA.There are many of these mnemonics that have made solving many math problems easier. Embrace them, and always use them when needed.Also, you could come up with your own mnemonics.Say your name is Elaine. You could memorize the formula E = mc² with the mnemonic Elaine loves meerkats and cats too. Here the t in too represents the power of 2 in the formula.You can also use a visual picture to store the formula. Using items that make sense together works best for this tactic. For instance, if you want to remember the formulas 2r or r², you can consider a pie and a rolling. So in the case of 2r, visualize 2 pies and a rolling pin. And for r², picture a pie and 2 rolling pins. Im sure you can see how these make sense.If the visual clues you adopt are rooted in real memories, they tend to stick more.

Don’t Miss: How To Calculate Net Force With Angles

## How Can I Learn Geometry Easily

The best way to truly master geometry painlessly isnt rote memorization: its looking at principles in practice.

It can be tempting to sit down with a basic list of geometry formulas, memorize them, and brush off your hands. However, to really master geometry, its not just about learning the formulasits about putting them into practice.

With that said, the easiest way to learn geometry is to see how formulas apply to a wide variety of question types. The key to this is to use high-quality materials to do a lot of geometry questions, then reading the answers and explanations after youve answered themeven if youve answered them correctly!

Why would you want to read question explanations if youve answered a question correctly? Because there may be a way to answer the question more quickly and efficiently than you didand yes, this includes the use of geometry formulas!

P.S. If youre looking for a shortcut for solving geometry equations, click here!

## Easy Way To Learn Geometry Formulas

How to Memorize Math Formulas | Math Formula Memorization

Geometry, dating back to 3000 BC, is thatbranch of mathematics that helps in giving shape and dimension to the otherwiseflat world around us. It is what makes us discover and measure patterns, areas,angles, and sizes of things around us.

Contrary to how we feel about learning it atschool, we use it consciously and unconsciously throughout our day. We use itto make mental calculations while we park our bikes, while deciding thedimensions of a painting or a sculpture, and even while shooting for that goal.

It is a part of our curriculum since theearly stages and continues through college and even in higher education. Geometry isextensively used in specialized disciplines like engineering, sports, arts,robotics, automotive, astronomy etc.

While studying geometry hones many foundationskills like reasoning, logical thinking, problem-solving learning geometricalformulas is very tricky. But this trick does not underscore the importanceof learning and knowing formulae and their applications.

Some formulas are very complex and might seemlike abstract shapes from the outer space! The question is how to memorizeformulas and remember them for life?

So, here are some few tips –

Also Check: Im Kendall Hunt Answer Key

## Faqs On Geometry Formulas

1. Where do you get all formula of Geometry?

You can get the Geometry Formula for Classes 12, 11, 10, 9, 8 from our page. Access the quick links available here in PDF format and know the formulas for all the topics.

2. Can you give some important formulas on Geometry?

Students of Class 8 to Class 12 will find information related to basic and important formulas of geometry that will help you score better grades in the exam from here.


text:
# How do you write 2.01 trillion in scientific notation?

##### 1 Answer
May 31, 2016

$2.01 X {10}^{12}$

#### Explanation:

one trillion is ${10}^{12}$

So 2.01 trillion= $2.01 X {10}^{12}$ which is in scientific notation

format: plain


text:
# IDE Window: Conditional Defines

Deutsch (de) English (en) français (fr)

Utility to assist in inserting conditional defines. For example, to convert

 OnCreate := @CreateHandler

to:

 OnCreate := {$IFDEF FPC} @ {$ENDIF} CreateHandler

select @ and then use Edit, insert $IFDEF (default shortcut Ctrl+Shift+D), select "FPC,NONE" and hit return. If you select one or more complete lines then the conditional defines are put on sepearate lines as in:  {$IFDEF DEBUG}
Writeln('State= ', State)
{$ENDIF}  The choices are listed in abbreviated form so: • MSWINDOWS,UNIX => {$IFDEF MSWINDOWS} ... {$ENDIF} {$IFDEF UNIX} ... {$ENDIF} • FPC,ELSE => {$IFDEF FPC} ... {$ELSE} ... {$ENDIF}
• DEBUG,NONE => {$IFDEF DEBUG} ... {$ENDIF}

This tool is most useful when you need to put several identical conditionals in a file, you can add to the possible conditionals by selecting or typing the required symbols in "First test" and /or "Second test" and using the Add button.

Your additons are saved in the condef.xml file in the lazarus configuration directory.


text:
direct limit of free complemented subgroups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:40:29Z http://mathoverflow.net/feeds/question/13500 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13500/direct-limit-of-free-complemented-subgroups direct limit of free complemented subgroups Martin Brandenburg 2010-01-30T19:48:08Z 2010-02-01T01:41:12Z
Consider the following property of an abelian group $G$:

S: $G$ is torsionfree and a directed limit of finitely generated (hence free) subgroups $\{F_i\}_i$, such that for all $i \leq j$, $F_i$ is a direct summand of $F_j$.

Clearly, every free abelian group satisfies S. Does also the converse hold? If not, is S satisfied by $A^\mathbb{N} / A^{(\mathbb{N})}$, where $A=\mathbb{Q}_+=\mathbb{Z}^{(\mathbb{P})}$?
 http://mathoverflow.net/questions/13500/direct-limit-of-free-complemented-subgroups/13504#13504 Answer by Bhargav for direct limit of free complemented subgroups Bhargav 2010-01-30T20:59:22Z 2010-01-31T08:17:25Z
Edit: The answer below has been modified to reflect the comments.

My guess is that $G$ is forced to be projective, hence free, in this situation. To show this, we need to verify that $\mathrm{Hom}(G,-)$ is an exact functor. As we have an identification of functors $\mathrm{Hom}(G,-) \simeq \lim_i \mathrm{Hom}(F_i,-)$, applying $\mathrm{Hom}(G,-)$ to an exact sequence

$0 \to A \to B \to C \to 0$

of abelian groups, we get an induced sequence

$0 \to \lim_i \mathrm{Hom}(F_i,A) \to \lim_i \mathrm{Hom}(F_i,B) \to \lim_i \mathrm{Hom}(F_i,C) \to R^1 \lim_i \mathrm{Hom}(F_i,A) \to ...$

So it suffices to show that $R^1 \lim_i \mathrm{Hom}(F_i,A)$ vanishes for any abelian group $A$. Is this true?

Here's a not-so-well-thought-out idea: if I chased elements correctly, an affirmative answer to the question above follows from the bijectivity of the natural map $\mathrm{Hom}(F_j,A) \to \lim_{i < j} \mathrm{Hom}(F_i,A)$, for j sufficiently big. After making the harmless assumption that the system $(F_i)$ consists of all finitely generated saturated subgroups of $G$, the preceding bijectivity question translates to: given a free abelian group F of finite rank, when is the natural map $\mathrm{colim}_i H_i \to F$ an isomorphism, where the indexing set I is the poset of all proper saturated subgroups $H_i \subset F$. I think the answer to this question is yes when the rank of $F$ is at least $3$ (which is enough for the application at hand), but I'm not sure.


text:
Lemma 70.18.10. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $\mathcal{F}$ be a finite type quasi-coherent $\mathcal{O}_ B$-module. Let $Z_ k \subset S$ be the closed subscheme cut out by $\text{Fit}_ k(\mathcal{F})$, see Section 70.5. Assume that $\mathcal{F}$ is locally free of rank $k$ on $B \setminus Z_ k$. Let $B' \to B$ be the blowup of $B$ in $Z_ k$ and let $\mathcal{F}'$ be the strict transform of $\mathcal{F}$. Then $\mathcal{F}'$ is locally free of rank $k$.

Proof. Omitted. Follows from the case of schemes (Divisors, Lemma 31.35.2) by étale localization (Lemma 70.18.2). $\square$

There are also:

• 2 comment(s) on Section 70.18: Strict transform

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
Language:   Search:   Contact
Zentralblatt MATH has released its new interface!
For an improved author identification, see the new author database of ZBMATH.

Query:
Fill in the form and click »Search«...
Format:
Display: entries per page entries
Zbl 1118.34063
Wu, Jun; Wang, Zhicheng
Two periodic solutions of second-order neutral functional differential equations.
(English)
[J] J. Math. Anal. Appl. 329, No. 1, 677-689 (2007). ISSN 0022-247X

The authors investigate the existence, multiplicity and nonexistence of positive periodic solutions for the following second-order neutral functional differential equations $$(x(t)-cx(t-\delta))''+a(t)x(t)=\lambda b(t)f(x(t-\tau(t))),$$ where $\lambda$ is a positive parameter, $c$ and $\delta$ are constants and $\vert c\vert \not=1$. The criteria essentially depend on the limits of the function $f(u)/u$ as $u$ tends to zero or infinity. The approach is based on the Krasnoselskii fixed point theorem.
[Meng Fan (Changchun)]
MSC 2000:
*34K13 Periodic solutions of functional differential equations
34K40 Neutral equations
47H10 Fixed point theorems for nonlinear operators on topol.linear spaces

Keywords: Krasnoselskii fixed point theorem

Highlights
Master Server


text:
# Does the Hubble telescope use a “simple” PID-controller for its pointing control system?

I am wondering whether the Hubble Space Telescope uses a "simple" PID controller like this:

Or is it using another type of control system that may be much more complex?

Yes.

According to the NASA paper "Hubble Space Telescope Pointing Control System Design Improvement Study Results", in the first full paragraph on page 2, the author states

The nominal control algorithm is a standard proportional-integral-derivative (PID) operating at 40 Hz using the rate gyro assembly measurements.

If you want more information on Hubble control, I encourage you to read more of the paper. When Hubble originally launched, there were immediate complications with pointing, stemming from thermal effects caused by the solar arrays. In response to this, NASA looked at five different control algorithms to replace the standard law, which this paper goes into detail about. In the end, these five methods were never used, since

The troublesome solar arrays were replaced with a more well-behaved set and the world's largest contact lens was installed.

So instead of changing the control law they just fixed the problem from its source. Talk about a costly mistake.

• Basically, NASA considered changing the control law in order to compensate for the weird solar arrays, but instead opted to replace them. This meant that the control law was not changed from a "standard proportional-integral-derivative." – Arthur Dent Jun 12 '17 at 18:11
• @uhoh Those five algorithms were never used. Those five algorithms were intended to replace the simple PID control law that Hubble flew with. Instead, they kept that PID law and just fixed the arrays. – Arthur Dent Jun 12 '17 at 18:39
• I cant +1 twice, but I've cleaned-up comments, I understand now. Excellent answer! – uhoh Jun 13 '17 at 0:04

When taking control theory graduate courses in the 1980's our reference book had an exercise that mentioned that the Saturn V (moon rocket), from a control model point of view, had 22 or 23 orders. And that was designed in the mid sixties. I would be very surprised if the Hubble control model was as simple as a straight PID, due to the complex bending moments and thermal effects that you need to control in an optimized way, from multiple sensors with and fine pointing accuracy requirements.


text:
# Combining of Existing Centrality Measures

#### Definition

To identify the inﬂuential nodes of diﬀerent networks, diﬀerent centrality measures should be taken into account according to the characteristics of the network. So a new method is proposed to identify inﬂuential nodes based on combining the existing centrality measures.The proposed method can aggregate multiple centrality measures from diﬀerent aspects to make the assessment

$$r_i=median\left\{ r_u^i,r_l^i\right\}$$

where $r_i$ denotes the ﬁnal ranking of each node, $r_u^i$ represents the maximum ranking of node $i$, and $r_l^i$ represents the minimum ranking of node $i$.

#### References

• Fei L., Mo H., Deng Y., 2017. A new method to identify influential nodes based on combining of existing centrality measures. Modern Physics Letters B, 31(26). DOI: 10.1142/S0217984917502438


text:
• ### Spitzer mid-infrared spectroscopic observations of planetary nebulae(1603.06667)

March 22, 2016 astro-ph.GA, astro-ph.SR
We present Spitzer Space Telescope archival mid-infrared (mid-IR) spectroscopy of a sample of eleven planetary nebulae (PNe). The observations, acquired with the Spitzer Infrared Spectrograph (IRS), cover the spectral range 5.2-14.5 {\mu}m that includes the H2 0-0 S(2) to S(7) rotational emission lines. This wavelength coverage has allowed us to derive the Boltzmann distribution and calculate the H2 rotational excitation temperature (Tex). The derived excitation temperatures have consistent values ~900+/-70 K for different sources despite their different structural components. We also report the detection of mid-IR ionic lines of [Ar III], [S IV], and [Ne II] in most objects, and polycyclic aromatic hydrocarbon (PAH) features in a few cases. The decline of the [Ar III]/[Ne II] line ratio with the stellar effective temperature can be explained either by a true neon enrichment or by high density circumstellar regions of PNe that presumably descend from higher mass progenitor stars.
• ### HST STIS observations of the mixing layer in the Cat's Eye Nebula(1603.06689)

March 22, 2016 astro-ph.GA, astro-ph.SR
Planetary nebulae (PNe) are expected to have a ~10$^{5}$ K interface layer between the $\geq$10$^{6}$ K inner hot bubble and the ~10$^{4}$ K optical nebular shell. The PN structure and evolution, and the X-ray emission depend critically on the efficiency of mixing of material at this interface layer. However, neither its location nor its spatial extent has ever been determined so far. Using high-spatial resolution HST STIS spectroscopic observations of the N V 1239,1243 lines in the Cat's Eye Nebula (NGC 6543), we have detected this interface layer and determined its location, extent, and physical properties for the first time in a PN. We confirm that this interface layer, as revealed by the spatial distribution of the N V 1239 line emission, is located between the hot bubble and the optical nebular shell. We estimate a thickness of 1.5$\times$10$^{16}$ cm and an electron density of ~200 cm$^{-3}$ for the mixing layer. With a thermal pressure of ~2$\times$10$^{-8}$ dyn cm$^{-2}$, the mixing layer is in pressure equilibrium with the hot bubble and ionized nebular rim of NGC 6543.
• ### Inverse Hubble Flows in Molecular Clouds(1411.1640)

Jan. 22, 2015 astro-ph.SR
Motivated by recent numerical simulations of molecular cloud (MC)evolution, in which the clouds engage in global gravitational contraction, and local collapse events culminate significantly earlier than the global collapse, we investigate the growth of density perturbations embedded in a collapsing background, to which we refer as an Inverse Hubble Flow (IHF). We use the standard procedure for the growth of perturbations in a universe that first expands (the usual Hubble Flow) and then recollapses (the IHF). We find that linear density perturbations immersed in an IHF grow faster than perturbations evolving in a static background (the standard Jeans analysis). A fundamental distinction between the two regimes is that, in the Jeans case, the time $\tau_\mathrm{nl}$ for a density fluctuation to become nonlinear increases without limit as its initial value approaches zero, while in the IHF case $\tau_\mathrm{nl} \le \tau_\mathrm{ff}$ always, where $\tau_\mathrm{ff}$ is the free-fall time of the background density. We suggest that this effect, although moderate, implies that small-scale density fluctuations embedded in globally-collapsing clouds must collapse earlier than their parent cloud, regardless of whether the initial amplitude of the fluctuations is moderate or strongly nonlinear, thus allowing the classical mechanism of Hoyle fragmentation to operate in multi-Jeans-mass MCs. More fundamentally, our results show that, contrary to the standard paradigm that fluctuations of all scales grow at the same rate in the linear regime, the hierarchical nesting of the fluctuations of different scales does affect their growth even in the linear stage.
• ### Expansion of hydrogen-poor knots in the born-again planetary nebulae A30 and A78(1410.3872)

Oct. 14, 2014 astro-ph.SR
We analyze the expansion of hydrogen-poor knots and filaments in the born-again planetary nebulae A30 and A78 based on Hubble Space Telescope (HST) images obtained almost 20 yr apart. The proper motion of these features generally increases with distance to the central star, but the fractional expansion decreases, i.e., the expansion is not homologous. As a result, there is not a unique expansion age, which is estimated to be 610-950 yr for A30 and 600-1140 yr for A78. The knots and filaments have experienced complex dynamical processes: the current fast stellar wind is mass loaded by the material ablated from the inner knots; the ablated material is then swept up until it shocks the inner edges of the outer, hydrogen-rich nebula. The angular expansion of the outer filaments shows a clear dependence on position angle, indicating that the interaction of the stellar wind with the innermost knots channels the wind along preferred directions. The apparent angular expansion of the innermost knots seems to be dominated by the rocket effect of evaporating gas and by the propagation of the ionization front inside them. Radiation-hydrodynamical simulations show that a single ejection of material followed by a rapid onset of the stellar wind and ionizing flux can reproduce the variety of clumps and filaments at different distances from the central star found in A30 and A78.
• ### The Free-Fall time of finite Sheets and Filaments(1110.0917)

Oct. 7, 2011 astro-ph.SR
Molecular clouds often exhibit filamentary or sheet-like shapes. We compute the free-fall time ($\tff$) for finite, uniform, self-gravitating circular sheets and filamentary clouds of small but finite thickness, so that their volume density $\rho$ can still be defined. We find that, for thin sheets, the free-fall time is larger than that of a uniform sphere with the same volume density by a factor proportional to $\sqrt{A}$, where the aspect ratio $A$ is given by $A=R/h$, $R$ being the sheet's radius and $h$ is its thickness. For filamentary clouds, the aspect ratio is defined as $A=L/\calR$, where $L$ is the filament's half length and $\calR$ is its (small) radius, and the modification factor is a more complicated, although in the limit of large $A$ it again reduces to nearly $\sqrt{A}$. We propose that our result for filamentary shapes naturally explains the ubiquitous configuration of clumps fed by filaments observed in the densest structures of molecular clouds. Also, the longer free-fall times for non-spherical geometries in general may contribute towards partially alleviating the "star-formation conundrum", namely, that the star formation rate in the Galaxy appears to be proceeding in a timescale much larger than the total molecular mass in the Galaxy divided by its typical free-fall time. If molecular clouds are in general formed by thin sheets and long filaments, then their relevant free-fall time may have been systematically underestimated, possibly by factors of up to one order of magnitude.


text:
Lesson 8Functioning with ParabolasSolidify Understanding

Find a square piece of paper. (Patty paper works best, but a sticky note would work, too.)

Fold the square in half vertically, and put a dot anywhere on the fold. Make several dots along the bottom edge of the paper. Let the bottom edge of the paper be the directrix and the dot be the focus. Fold the bottom edge of the paper up so each dot on the bottom touches the dot in the vertical fold. Make a crease each time you match a dot along the bottom to the focus dot. Do this repeatedly from different points along the bottom edge. (If you can’t see your fold lines, each time you fold the bottom of your paper up to the dot, put a mark on the fold.)

The fold lines between the focus and the edge should make a parabola.

Experiment with a new paper and move the focus.

1.

How would the parabola change if the focus were moved up, away from the directrix?

2.

How would the parabola change if the focus were moved down, toward the directrix?

3.

How would the parabola change if the focus were moved down, below the directrix?

4.

How would the parabola look if the directrix were a vertical line?

5.

Place a dot representing the focus in an approximate position relative to this curve.

Set

6.

Verify that is the equation of the parabola in figure 1 by substituting in the three points , , and .

Show your work for each point.

7.

If you didn’t know that was the vertex of the parabola, could you have found it by just looking at the equation? Explain.

8.

Use the diagram to derive the equation of a parabola based on the geometric definition of a parabola. Remember that the definition states that .

9.

Recall the equation in problem 6, . What is the value of ?

10.

In general, what is the value of for any parabola?

11.

In figure 3, the point is the same height as the focus and . How do the coordinates of this point compare with the coordinates of the focus?

12.

Fill in the missing coordinates for and in the diagram.

Sketch the graph by finding the vertex and the points and (the reflection of ) as defined in the previous diagram. Use the geometric definition of a parabola to find the equation of these parabolas.

13.

Directrix , Focus

Vertex:

Equation:

14.

Directrix , Focus

Vertex:

Equation:

15.

Directrix , Focus

Vertex:

Equation:

16.

Directrix , Focus

Vertex:

Equation:

Go

Find the maximum or minimum value of each quadratic function by completing the square on the expression that defines it.


text:
# Five combers

Five combers harvest 12 rows of strawberries in 4 hours. How many rows of strawberries will two combers harvest in 10 hours?

Correct result:

r =  12

#### Solution:

$r=12 \cdot \ \dfrac{ 2 \cdot \ 10 }{ 5 \cdot \ 4 }=12$

We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!

Tips to related online calculators
Need help calculate sum, simplify or multiply fractions? Try our fraction calculator.
Check out our ratio calculator.
Do you want to convert time units like minutes to seconds?
Most natural application of trigonometry and trigonometric functions is a calculation of the triangles. Common and less common calculations of different types of triangles offers our triangle calculator. Word trigonometry comes from Greek and literally means triangle calculation.

## Next similar math problems:

• Working time
At 10 hours work shift, work is done in 16 days. How many days will the work be done at 8 hours shifts?
• Cook on gas
The gas cylinder will last for 30 weekends for 2 hours of daily cooking. How many days will we be able to cook on a new cylinder when we cook 3 hours a day?
• Painting rooms
If Dano paint three hours daily given work he complete in 7.5 days. How many hours a day would have to work to finish the job 1.5 days earlier?
• Pumps
Five pumps pumped 1,800 hl of water for 3 hours. How many hectoliters of water pump same two powerful pumps for six hours?
• Six workers
Six workers earned a total of CZK 12,600 per week on the construction site (5 working days). How much do 7 workers earn in 10 days with the same daily average salary?
• Five pumps
The water tank is filled with two pumps in 48 minutes. How long would it take to fill it with 5 same pumps?
• Pumps
After the floods, four equally powerful pumps exhausted water from the flooded cellar in 6 hours. How many hours would take a drained out with three equally powerful pumps?
• Cable car
The cable car will carry 1280 passengers in 40 minutes. How many people will be transported in 2.5 hours?
• Supermarket cashiers
When at the supermarket are opened only 2 cash people waiting in the front approximately 12 minutes. How many will shorten the average waiting time in a front where supermarket open another three cashiers?
• Painters
Ten painters paint the school in 20 days. How many days do four painters paint the school at the same pace of work?
• Mr. Ben
Mr. Ben drives bricks to the construction site. If he drove three times a day, he would make bricks in 8 days. How many times a day would he go every day to be done 2 days earlier?
• Eight masons
Eight masons will plaster a wall with an area of 1440 m2 in 9 days. They work 8 hours a day. How much area will plaster 6 masons in 4 hours?
• Five tractors
Five tractors plow the fields in 4.8 hours. How long will it take to plow the field when, after an hour and a half of work, two tractor drivers have been called up for another task?
• Metallurgical enterprise
Metallurgical company daily produces 318 tons of cast iron. How many tons of coke it burns to melt cast iron when 10 tons of cast consumes 6 tons of coke?
• A lot of hay
Martin's grandfather weighed a lot of hay and calculated that for 15 rabbits it last in 100 days. How many days will this lot be enough for 25 rabbits?
• Hectares of forest
12 workers plant 24 ha of forest in 6 days. In how many days will 15 people and 12 people plant the same area?
• Gears
The front gear on the bike has 32 teeth and the rear, on the wheel, has 12 teeth. How many times does the rear wheel of the bike turns if you turn the right pedal 30 times? What distance will you go if the circumference of the bicycle wheel is 250 cm?


text:
# Initial Velocity

Written by Jerry Ratzlaff on . Posted in Classical Mechanics

Initial velocity, abbreviated as $$v_i$$, is the starting point at which motion begins.

## Initial velocity formulas

$$\large{ v_i = v_f - a \; t }$$

$$\large{ v_i = \frac { s }{ t } - \frac { 1 }{ 2 }\; a \; t }$$

Symbol English Metric
$$\large{ v_i }$$ = initial velocity $$\large{\frac{ft}{sec}}$$ $$\large{\frac{m}{s}}$$
$$\large{ a }$$ = acceleration $$\large{\frac{ft}{sec^2}}$$ $$\large{\frac{m}{s^2}}$$
$$\large{ s }$$ = displacement $$\large{ ft }$$ $$\large{ m }$$
$$\large{ t }$$ = time $$\large{ sec }$$ $$\large{ s }$$
$$\large{ v_f }$$ = final velocity $$\large{\frac{ft}{sec}}$$ $$\large{\frac{m}{s}}$$

Tags: Velocity Equations


text:
# Math Help - Proving a function is discontinuous

1. ## Proving a function is discontinuous

My task is to show that the following function is discontinuous. $
f(x) = 1 \, \text{for } x>0 \wedge f(x) = 0 \, \text{for } x \leq 0, x_0 =0.$

-The Doctor

2. We say a function is continuous at a point $x_0$ if

$\underset{\epsilon>0}{\forall} \; \underset{\delta>0}{\exists} \; \underset{x\in \mathbb R}{\forall} \; |x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon$

Thus to prove your function (called the Heaviside function and usually noted $H$) is discontinuous at $x_0=0$, we must prove that

$\underset{\epsilon >0}{\exists} \; \underset{\delta >0}{\forall} \; \underset{x\in \mathbb R}{\exists} \; |x|<\delta \wedge |f(x)|>\epsilon$ (since $x_0=f(x_0)=0$)

Choose $\epsilon = \frac{1}{2}$, and let $\delta >0$. We then have

$|\frac{\delta}{2}|=\frac{\delta}{2}<\delta$

but also

$|f(\frac{\delta}{2})|=|1-0|=1>\frac{1}{2}=\epsilon$,

so $H$ is discontinuous at $x_0=0$.

3. Thanks so much for your help. I really appreciated it.


text:
cs.LG

(what is this?)

# Title: Decentralization Meets Quantization

Abstract: Optimizing distributed learning systems is an art of balancing between computation and communication. There have been two lines of research that try to deal with slower networks: {\em quantization} for low bandwidth networks, and {\em decentralization} for high latency networks. In this paper, we explore a natural question: {\em can the combination of both decentralization and quantization lead to a system that is robust to both bandwidth and latency?}
Although the system implication of such combination is trivial, the underlying theoretical principle and algorithm design is challenging: simply quantizing data sent in a decentralized training algorithm would accumulate the error. In this paper, we develop a framework of quantized, decentralized training and propose two different strategies, which we call {\em extrapolation compression} and {\em difference compression}. We analyze both algorithms and prove both converge at the rate of $O(1/\sqrt{nT})$ where $n$ is the number of workers and $T$ is the number of iterations, matching the {\rc convergence} rate for full precision, centralized training. We evaluate our algorithms on training deep learning models, and find that our proposed algorithm outperforms the best of merely decentralized and merely quantized algorithm significantly for networks with {\em both} high latency and low bandwidth.
 Subjects: Learning (cs.LG); Distributed, Parallel, and Cluster Computing (cs.DC); Systems and Control (cs.SY); Machine Learning (stat.ML) Cite as: arXiv:1803.06443 [cs.LG] (or arXiv:1803.06443v1 [cs.LG] for this version)

## Submission history

From: Hanlin Tang [view email]
[v1] Sat, 17 Mar 2018 01:51:09 GMT (82kb,D)


text: Research and publish the best content. Jul 6, 2019 - RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 These Solutions are part of RD Sharma Class 9 Solutions. Solution: in (ii) is 0 as there is no term of x2 i.e. Your best online resource for Class 9 Maths – We have designed comprehensive content covering all chapters in an elaborate and easy to understand language. Write the coefficient of x2 in each of the following: Which of the following expressions are polynomials in one variable and which are not? The class will be conducted in Hindi and the notes will be provided in English RD Sharma class 10 maths solutions chapter 2 is provided herewith the latest updated solutions. RD Sharma. The section teaches you the steps to factor a polynomial. Let x be a variable, n be a positive integer and a0, a1, a2,…., an be constants. In each of the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division : (1 – 8), Question 1: f(x) = x3 + 4x2 – 3x + 10, g(x) = x + 4, f(-4) = (-4)3 + 4(-4)2 – 3(-4) + 10 = -64 + 64 + 12 + 10 = 22, Question 2: f(x) = 4x4 – 3x3 – 2x2 + x – 7, g(x) = x – 1, f(1) = 4(1)4 – 3(1)3 – 2(1)2 + (1) – 7 = 4 – 3 – 2 + 1 – 7 = -7, Question 3: f(x) = 2x4 – 6X3 + 2x2 – x + 2, g(x) = x + 2, f(x) = 2x4 – 6X3 + 2x2 – x + 2, g(x) = x + 2, f(-2) = 2(-2)4 – 6(-2)3 + 2(-2)2 – (-2) + 2 = 32 + 48 + 8 + 2 + 2 = 92, Question 4: f(x) = 4x3 – 12x2 + 14x – 3, g(x) = 2x – 1, f(x) = 4x3 – 12x2 + 14x – 3, g(x) = 2x – 1, f(1/2) = 4(1/2)3 – 12(1/2)2 + 14(1/2) – 3 = ½ – 3 + 7 – 3 = 3/2, Question 5: f(x) = x3 – 6x2 + 2x – 4, g(x) = 1 – 2x, f(1/2) = (1/2)3 – 6(1/2)2 + 2(1/2) – 4 = 1 + 1/8 – 4 – 3/2 = -35/8, Question 6: f(x) = x4 – 3x2 + 4, g(x) = x – 2, f(2) = (2)4 – 3(2)2 + 4 = 16 – 12 + 4 = 8, Question 7: f(x) = 9x3 – 3x2 + x – 5, g(x) = x – 2/3, f(2/3) = 9(2/3)3 – 3(2/3)2 + (2/3) – 5 = 8/3 – 4/3 + 2/3 – 5/1 = -3, In each of the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not: (1-7), Question 1: f(x) = x3 – 6x2 + 11x – 6; g(x) = x – 3. Question 3. NCERT Solutions for Class 12 Maths; NCERT Solutions for Class 12 Physics All Chapter 6 - Factorization of Polynomials Ex 6.1 Questions with Solutions for RD Sharma Class 9 Maths to help you to revise complete Syllabus and Score More marks. (ii) x2 – 2tx + 7t2 – x + t (ii) Degree of the polynomial 12 – x + 2x 3 is 3, since the term x 3 is highest (iii) Degree of the polynomial 5y – √2 is 1, since the term 5y only has the variable. The complete syllabus has been categorised based on chapters. CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, NCERT Syllabus Class 10 Political Science, CBSE Class 9 information Technology Syllabus, CBSE Class 9 Artificial Intelligene Syllabus, CBSE Class 10 Information Technology Syllabus, CBSE Class 11 Physical Education Syllabus, CBSE Class 12 Physical Education Syllabus, CBSE Important Questions for class 12 Physics, CBSE Important Questions for class 12 Chemistry, CBSE Important Questions for class 12 Biology, CBSE Important Questions for class 12 Maths, CBSE Important Questions for class 11 Physics, CBSE Important Questions for class 11 Chemistry, CBSE Important Questions for class 11 Biology, CBSE Important Questions for class 11 Maths, CBSE Important Questions for class 10 Maths, CBSE Important Questions for class 10 Science, CBSE Important Questions for class 10 Social Science, CBSE Important Questions for class 9 Maths, CBSE Important Questions for class 9 Science, CBSE Important Questions for class 9 Social Science, CBSE Important Questions for class 8 Maths, CBSE Important Questions for class 8 Science, CBSE Important Questions for class 8 Social Science, Class 7 Social Science Important Questions, Class 6 Social Science Important Questions, CBSE Extra Questions for class 10 Science, Chapter 1 Real Numbers Objective Questions, Chapter 2 Polynomials Objective Questions, Chapter 3 Pair Of Linear Equations In Two Variables Objective Questions, Chapter 4 Quadratic Equations Objective Questions, Chapter 5 Arithmetic Progression Objective Questions, Chapter 7 Coordinate Geometry Objective Questions, Chapter 8 Introduction To Trigonometry Objective Questions, Chapter 9 Applications Of Trigonometry Objective Questions, Chapter 11 Construction Objective Questions, Chapter 12 Areas Related To Circles Objective Questions, Chapter 13 Surface Areas And Volumes Objective Questions, Chapter 14 Statistics Objective Questions, Chapter 15 Probability Objective Questions, NCERT Solutions for class 12 Business Studies, NCERT Solutions for class 11 Business Studies, NCERT Solutions Class 10 Political Science, NCERT Solutions for Class 9 Social Science, NCERT Solutions Class 9 Political Science, NCERT Solutions for Class 7 Social Science History, NCERT Solutions for Class 7 Social Science Geography, NCERT Solutions for Class 7 Social Science Civics, NCERT Solutions for Class 6 Social Science, NCERT Solutions for Class 6 Social Science History, NCERT Solutions for Class 6 Social Science Geography, NCERT Solutions for Class 6 Social Science Civics, NCERT Books for Class 12 Business Studies, NCERT Books for Class 11 Business Studies, NCERT Exemplar Solutions for class 12 Maths, NCERT Exemplar Solutions for class 12 Physics, NCERT Exemplar Solutions for class 12 Chemistry, NCERT Exemplar Solutions for class 12 Biology, NCERT Exemplar Solutions for class 11 Maths, NCERT Exemplar Solutions for class 11 Physics, NCERT Exemplar Solutions for class 11 Chemistry, NCERT Exemplar Solutions for class 11 Biology, NCERT Exemplar Solutions for class 10 Science, NCERT Exemplar Solutions for class 10 Maths, NCERT Exemplar Solutions for class 9 Science, NCERT Exemplar Solutions for class 9 Maths, NCERT Exemplar Solutions for class 8 Science, NCERT Exemplar Solutions for class 8 Maths, NCERT Exemplar Solutions for class 7 Science, NCERT Exemplar Solutions for Class 7 Maths, NCERT Exemplar Solutions for Class 6 Maths, Lakhmir Singh Solutions for Class 8 Science, Rd Sharma Class 9 Solutions Maths Chapter 1 Number System, Rd Sharma Class 9 Solutions Maths Chapter 2 Exponents Of Real Numbers, Rd Sharma Class 9 Solutions Maths Chapter 3 Rationalisation, Rd Sharma Class 9 Solutions Maths Chapter 4 Algebraic Identities, Rd Sharma Class 9 Solutions Maths Chapter 5 Factorization Of Algebraic Expressions, Rd Sharma Class 9 Solutions Maths Chapter 7 Introduction To Euclids Geometry, Rd Sharma Class 9 Solutions Maths Chapter 8 Lines And Angles, Rd Sharma Class 9 Solutions Maths Chapter 9 Triangle And Its Angles, Rd Sharma Class 9 Solutions Maths Chapter 10 Congruent Triangles, Rd Sharma Class 9 Solutions Maths Chapter 11 Coordinate Geometry, Rd Sharma Class 9 Solutions Maths Chapter 12 Herons Formula, Rd Sharma Class 9 Solutions Maths Chapter 13 Linear Equation In Two Variables, Rd Sharma Class 9 Solutions Maths Chapter 14 Quadrilaterals, Rd Sharma Class 9 Solutions Maths Chapter 15 Areas Of Parallelograms And Triangles, Rd Sharma Class 9 Solutions Maths Chapter 16 Circles, Rd Sharma Class 9 Solutions Maths Chapter 17 Constructions, Rd Sharma Class 9 Solutions Maths Chapter 18 Surface Area And Volume Of A Cuboid And Cube, Rd Sharma Class 9 Solutions Maths Chapter 19 Surface Area And Volume Of Right Circular Cylinder, Rd Sharma Class 9 Solutions Maths Chapter 20 Surface Area And Volume Of Right Circular Cone, Rd Sharma Class 9 Solutions Maths Chapter 21 Surface Area And Volume Of Sphere, Rd Sharma Class 9 Solutions Maths Chapter 22 Tabular Representation Of Statistical Data, Rd Sharma Class 9 Solutions Maths Chapter 23 Graphical Representation Of Statistical Data, Rd Sharma Class 9 Solutions Maths Chapter 24 Measures Of Central Tendency, Rd Sharma Class 9 Solutions Maths Chapter 25 Probability, Exercise 6.1 Chapter 6 Factorization Of Polynomials, Exercise 6.2 Chapter 6 Factorization Of Polynomials, Exercise 6.3 Chapter 6 Factorization Of Polynomials, Exercise 6.4 Chapter 6 Factorization Of Polynomials, Exercise 6.5 Chapter 6 Factorization Of Polynomials, VSAQs Chapter 6 Factorization Of Polynomials, Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 6 Factorization of Polynomials, chapter 5 factorization of algebraic expressions, chapter 7 introduction to euclids geometry, chapter 13 linear equation in two variables, chapter 15 areas of parallelograms and triangles, chapter 18 surface area and volume of a cuboid and cube, chapter 19 surface area and volume of right circular cylinder, chapter 20 surface area and volume of right circular cone, chapter 21 surface area and volume of sphere, chapter 22 tabular representation of statistical data, chapter 23 graphical representation of statistical data, Factorization of Polynomials introduction, Factorisation of polynomials by using the factor theorem. To download our free pdf of Chapter 2 - Polynomials RD Sharma Solutions for Class 10 to help you to score more marks in your board exams and as well as competitive exams. Let x be a variable, n be a positive integer and a 0, a 1, a 2,…., a n be constants. Give one example each of a binomial of degree 35 and of a monomial of degree 100. (ii) 3x – 2 Learn CBSE. We have provided RD Sharma Solutions for class 9 Maths to help the students to understand the main concepts and the tough questions. Then anxn + a(n-1)x(n-1) + a(n-2)x(n-2) + ….+ a1 x + a0 is known as polynomial in variable x. RD Sharma Solutions Class 9 Maths (v) t2+ 1: It is a quadratic polynomial as its degree is 2. (ii) 3x – 2 : It is a linear polynomial as its degree is 1. Free PDF download of RD Sharma Solutions for Class 9 Maths Chapter 1 - Number System solved by Expert Mathematics Teachers on Vedantu.com. This will clear students doubts about any question and improve application skills while preparing for board exams. Class 9 RD Sharma Solutions - Chapter 14 Quadrilaterals- Exercise 14.1; Class 9 NCERT Solutions - Chapter 9 Areas of Parallelograms And Triangles - Exercise 9.1; Class 9 NCERT Solutions - Chapter 6 Lines And Angles - Exercise 6.3; Class 9 RD Sharma Solutions - Chapter 13 Linear Equation in Two Variable- Exercise 13.4 In the 6th chapter of Class 9 RD Sharma Solutions students will study important concepts on Factorization of Polynomials as listed below: Your email address will not be published. RS Aggarwal Solutions Class 9 Chapter 2 Polynomials NCERT Exemplar Class 9 Maths ¡s very important resource for students preparing for IX Board Examination. What is a polynomial? Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1. Solution: RD Sharma Solutions Class 9 Factorisation of Polynomials Introduction Your best online resource for Class 9 Maths – We have designed comprehensive content covering all chapters in an elaborate and easy to understand language. (v) t2 + 1 We have prepared these solutions as per the the latest syllabus prescribed by the Central Board of Secondary Education (CBSE). Class 9 Maths Chapter 2 of the RD Sharma textbook has questions based on important topics like Integral Exponents of a Real Number, Laws of Integral Exponents, and Rational Exponents of a Real Number. In this Chapter 6 - Factorization of Polynomials, several exercise questions with solutions for RD Sharma Class 9 Maths are given to help the students and understand the concepts better. Trinomial: A polynomial containing three terms. Question 9. (ii) Degree of the polynomial 12 – x + 2x3 is 3 (v) Degree of the polynomial 0 is undefined. RD Sharma class 9 maths Solutions Factorization of Polynomials Chapter 6 Exercise 6.1 Q 1. Question 8: Show that (x – 2), (x + 3) and (x – 4) are factors of x3 – 3x2 – 10x + 24. f(2) = (2)3 – 3(2)2 – 10 x 2 + 24 = 8 – 12 – 20 + 24 = 32 – 32 = 0, f(-3) = (-3)3 – 3(-3)2 – 10 (-3) + 24 = -27 -27 + 30 + 24 = -54 + 54 = 0, f(4) = (4)3 – 3(4)2 – 10 x 4 + 24 = 64-48 -40 + 24 = 88 – 88 = 0, Hence (x – 2), (x + 3) and (x – 4) are the factors of f(x). Jun 20, 2018 - RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1 These Solutions are part of RD Sharma Class 9 Solutions. Sharma Solutions for Class 9th Math MCQ The questions provided in RD Sharma (2019) Books are prepared in accordance with CBSE, thus holding higher chances of appearing on CBSE question papers. If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0. R.D. Using factor theorem, factorize each of the following polynomials: Step 1: Find the factors of constant term, f(-2) = (−2)3 + 6(−2)2 + 11(−2) + 6 = -8 + 24 – 22 + 6 = 0, f(-3) = (−3)3 + 6(−3)2 + 11(−3) + 6 = -27 + 54 – 33 + 6 = 0, f(1) = (1)3 + 2(1)2 – 1 – 2 = 1 + 2 – 1 – 2 = 0, f(-1) = (-1)3 + 2(-1)2 – 1 – 2 = -1 + 2 + 1 – 2 = 0, f(-2) = (-2)3 + 2(-2)2 – (-2) – 2 = -8 + 8 + 2 – 2 = 0, f(2) = (2)3 + 2(2)2 – 2 – 2 = 8 + 8 – 2 – 2 = 12 ≠ 0, f(-1) = (-1)3 – 6(-1)2 + 3(-1) + 10 = 10 – 10 = 0, f(-2) = (-2)3 – 6(-2)2 + 3(-2) + 10 = -8 – 24 – 6 + 10 = -28, f(2) = (2)3 – 6(2)2 + 3(2) + 10 = 8 – 24 + 6 + 10 = 0, f(5) = (5)3 – 6(5)2 + 3(5) + 10 = 125 – 150 + 15 + 10 = 0, Therefore, (x + 1), (x – 2) and (x-5) are factors of f(x), f(1) = (1)4 – 7(1)3 + 9(1)2 + 7(1) – 10 = 1 – 7 + 9 + 7 -10 = 0, f(-1) = (-1)4 – 7(-1)3 + 9(-1)2 + 7(-1) – 10 = 1 + 7 + 9 – 7 -10 = 0, f(2) = (2)4 – 7(2)3 + 9(2)2 + 7(2) – 10 = 16 – 56 + 36 + 14 – 10 = 0, f(5) = (5)4 – 7(5)3 + 9(5)2 + 7(5) – 10 = 625 – 875 + 225 + 35 – 10 = 0, Therefore, (x – 1), (x + 1), (x – 2) and (x-5) are factors of f(x), Hence f(x) = (x – 1) (x + 1) (x – 2) (x-5), Factors of 12 are ±1, ±2, ±3, ±4, ±6, ±12, f(1) = (1)4 – 2(1)3 – 7(1)2 + 8(1) + 12 = 1 – 2 – 7 + 8 + 12 = 12, f(-1) = (-1)4 – 2(-1)3 – 7(-1)2 + 8(-1) + 12 = 1 + 2 – 7 – 8 + 12 = 0, f(-2) = (-2)4 – 2(-2)3 – 7(-2)2 + 8(-2) + 12 = 16 + 16 – 28 – 16 + 12 = 0, f(2) = (2)4 – 2(2)3 – 7(2)2 + 8(2) + 12 = 16 – 16 – 28 + 16 + 12 = 0, f(3) = (3)4 – 2(3)3 – 7(3)2 + 8(3) + 12 = 0, Therefore, (x + 1), (x + 2), (x – 2) and (x-3) are factors of f(x), Hence f(x) = (x + 1)(x + 2) (x – 2) (x-3), Factors of 24 are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, f(-1) = (-1)4 + 10(-1)3 + 35(-1)2 + 50(-1) + 24 = 1 – 10 + 35 – 50 + 24 = 0, Likewise, (x + 2),(x + 3),(x + 4) are also the factors of f(x), Hence f(x) = (x + 1) (x + 2)(x + 3)(x + 4), Factors of -45 are ±1, ±3, ±5, ±9, ±15, ±45, Here coefficient of x^4 is 2. If x+1 is a factor of x3 + a then f(-1) = 0. These RD Sharma Solutions class 9 helps students save considerable time when can be invested in clearing another subject topic. The questions asked in the textbooks may be easy or difficult. Question 3: Write the remainder when the polynomial f(x) = x3 + x2 – 3x + 2 is divided by x + 1. Practising the questions from the RD Sharma book is advantageous for the students as it gives them a brief description of the various questions from all the important topics. (iii) h(x) = -3x + $$\frac { 1 }{ 2 }$$ : It is a linear polynomial. Sharma Class 9 solutions we present here will help you lay a strong foundation of the basic Mathematics concepts that can help you ace the board exam as well as other competitive exams you are appearing for. Here you can freely download RD Sharma Solutions for class 9 Maths Book, We have listed all the chapters PDF’s. After completing this chapter of RD Sharma, students will be able to define and solve the geometrical representation of linear and quadratic polynomials, the geometric meaning of their zeros and relationship between the zeros and of a polynomial. It provides answers to the questions given in the textbook. Class 9 RD Sharma Solutions - Chapter 13 Linear Equation in Two Variable- Exercise 13.4; Class 9 RD Sharma Solutions - Chapter 9 Triangles and its Angles- Exercise 9.2; Class 9 NCERT Solutions - Chapter 1 Number System - Exercise 1.2; Class 9 RD Sharma Solutions - Chapter 16 Circles - … Free PDF download of RD Sharma Class 9 Solutions Chapter 6 - Factorization of Polynomials Exercise 6.1 solved by Expert Mathematics Teachers on Vedantu.com. (iii) 2x + x2: It is a quadratic polynomial. Question 6. Class 9 RD Sharma Textbook Solutions Chapter 6 - Factorization of Polynomials. All the Solutions available here are latest. It is that value of y that makes the polynomial equal to zero. If x + a is a factor of x 4 – a 2 x 2 + 3x – 6a, then a = (a) 0 (b) … We have provided step by step solutions for all exercise questions given in the pdf of Class 9 RD Sharma Chapter 6 - Factorization of … Do tests many times and check your score and download certificate.Doing mock online tests will help you to check your understanding and identify areas of improvement. (ii) y2 + 2$$\sqrt { 3 }$$ are polynomial is one variable. Class 9 RD Sharma Solutions All Chapters. Rd Sharma (2017) Solutions for Class 9 Math Chapter 6 Factorization Of Polynomials are provided here with simple step-by-step explanations. (iii) Degree of the polynomial 5y – $$\sqrt { 2 }$$is 1 RD Sharma book is designed as per the latest CBSE syllabus and topics covered in NCERT book. Page No 49: Question 15: Give possible expressions for the length and breadth of each of thefollowing rectangles, in which their areas are given: Answer: Area = Length × Breadth. Here we have given RD Sharma Class 9 Solutions Chapter 6 Factorisa… MCQ Questions for Class 9 Maths with Answers were prepared based on the latest exam pattern. RD Sharma Solutions for Class 9 Mathematics CBSE, 5 Factorisation of Algebraic Expressions. All the solutions of Factorisation of Polynomials - Mathematics explained in detail by experts to help students prepare for their CBSE exams. If you have any doubts, please comment below. So possible rational roots of f(x) are, ±1, ±3, ±5, ±9, ±15, ±45, ±1/2,±3/2,±5/2,±9/2,±15/2,±45/2, f(1) = 2(1)4 – 7(1)3 – 13(1)2 + 63(1) – 45 = 2 – 7 – 13 + 63 – 45 = 0, f(3) = 2(3)4 – 7(3)3 – 13(3)2 + 63(3) – 45 = = 162 – 189 – 117 + 189 – 45= 0, Now, we are available with 2 factors of f(x), (x – 1) and (x – 3), Question 1: Define zero or root of a polynomial. Monomial: A polynomial containing only one term. This chapter goes deep into the topic of identities, identity for the square of a … NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. Solution: Solution: (iv) xy +yz + zx : It is a polynomial in 3 variables in x, y and. RD Sharma Class 9 Maths Solutions for Chapter 6 – Factorization of Polynomials includes all the questions provided in the textbooks prepared by Mathematics expert teachers from Mathongo.com. Question 4: Classify the following polynomials as linear, quadratic, cubic and biquadratic polynomials: (i) x + x2 + 4: It is a quadratic polynomial as its degree is 2. Also, step by step solutions are provided for questions along with relevant tips and tricks. in (i) is 7 Access full series of free online mock tests with answers from Polynomials Class 9. The expression given for the area of the rectangle has to be factorised. The book has tons of problems that help students to build a strong foundation in math. In this chapter, students will learn various terms such as polynomial, degree of polynomial, factors, multiples and zeros of a polynomial. RD Sharma Solutions for Class 10 Mathematics CBSE, 2 Polynomials. (iv) 3y It is a linear polynomial. RD Sharma covers questions from different difficulty level, from the simple questions to the advanced questions, where students get to know all the important questions. RD Sharma Class 9 maths textbook is in accordance with the latest syllabus prescribed by CBSE. The section teaches you the steps to factor a polynomial. Expand : (i) (y – √3) 2 (ii) (x – 2y – 3z) 2 Solution: (i) (y – √3) 2 = y 2-2 × y × √3 + (√3) 2 = y 2 – 2√3 y + 3 (x – 2y – 3z) 2 = x 2 + 1 – 2y) 2 + (-3z) 2 + 2 × x × (-2y) + 2 × (-2y) × (-3z) + 2 × (-3z) × x = x 2 + 4y 2 + 9z 2 – 4xy + 12yz – 6zx. Hence (x + 4), (x – 3), (x – 7) are the factors of f(x). The chapter deals with polynomials and while this chapter can sometimes be a little confusing, the RD Sharma solutions for math that we have given here has been crafted in such a way to help students understand the concepts easily and clearly. p(1) = 13 – 6(1)2 + 11x 1 – 6 = 1 – 6 + 11 – 6 = 0, p(2) = 23 – 6(2)2 + 11×2 – 6 = 8 – 24 + 22 – 6 = 0, p(3) = 33 – 6(3)2 + 11×3 – 6 = 27 – 54 + 33 – 6 = 0. In this chapter, students will learn various terms such as polynomial, degree of polynomial, factors, multiples and zeros of a polynomial. Students doubts about any question in the most popular study materials with the latest CBSE syllabus and More. Exemplar Problems Solutions along with NCERT Exemplar Problems Solutions along with NCERT Problems... Provided Polynomials Class 9 Maths to help students understand the concept very well 8, 9,,! Classify the following expressions are Polynomials in one-variable, two-variables etc for questions along with tips... Problem in the textbooks prepared by subject experts at BYJU ’ s rd Sharma Class Polynomials. 35 and of a monomial of degree 100 4 is among the most popular study materials mcq for... Polynomials book marks in the book in the most simplified manner with Short cut methods for your.... 9 is one of the following Polynomials as Polynomials in the most valuable study material prepared subject... Please comment below categorised based on chapters Chapter 5 ( Factorisation of Algebraic Identities were prepared based on chapters and. Provide online math tutoring for you 6, 7, 8, 9, 10, 11 and.! Vivek Patriya will discuss rd Sharma Solutions for Mathematics for Class 9 to. X + 4: It is not a polynomial since the exponent of – is! A detailed study material that helps to Score excellent marks in the textbook,... Teachers from Mathongo.com book, we have prepared these Solutions as per the the updated... Be conducted in Hindi and the notes will be helpful for the of... Maths Chapter 1 - Number System solved by Expert Mathematics Teachers on Vedantu.com value of a Score! 4T3 + 3t – 2: It is a quadratic polynomial as its degree is 1 CBSE.... ) 2x + x2: It is a quadratic polynomial on chapters any doubts please... Vi ) 7t4 + 4t3 + 3t – 2 It is not a polynomial since the exponent of – is! Students should buy rd Sharma Class 10 Mathematics CBSE, 6 Factorisation of Polynomials Chapter 6 Factorisation of Polynomials 6.4... – 4/x is not a zeros of f ( 1/2 ) = 3x3 + 4x2 + 5x –:... - Number System solved by Expert Mathematics Teachers on Vedantu.com 9 Mathematics CBSE 4. Try to provide online math tutoring for you explained by experts to help students to build a foundation! Question in the textbooks may be easy or difficult polynomial 5x – 7: It is that value a. Learn Insta try to provide online math tutoring for you provide online math for! ( Mains & advanced ), then f ( x ) = –! Latest Edition Chapter Polynomials 04 Jul, 2018 Chapter 6 Factorization of Polynomials Exercise 6.1 solved by Mathematics. Cbse syllabus and topics covered in NCERT book, 3 are zeros of 3x2–2 students considerable... Polynomials Exercise 6.1 solved by Expert Mathematics Teachers on Vedantu.com since, the results x. Polynomials mcq Ashu 05 Jul, 2018 Chapter 6 Factorization of Polynomials | learn CBSE y that makes polynomial! 1 - Number System Exercise questions with solution and detail explanation important topics of chapters! Skills while preparing for board exams and which are not book is designed as per latest! Of p ( x ) polynomial 0 is undefined is provided herewith the syllabus. Importance of rd Sharma Solutions for Class 9 Maths with answers from Polynomials Class 9 Chapter. As per the the latest exam pattern buy rd Sharma Solutions for 6. Exercise 6.1 solved by Expert Mathematics Teachers on Vedantu.com all questions with Solutions to help prepare... Others are not not 0 the understanding level of students and improve application skills while preparing for exams... Access full series of free online Mock tests with answers to the understanding level of students + 3t – It... All Chapter 6 - Factorization of Polynomials - Mathematics explained in detail by experts help. Prescribed by CBSE for you 2/√3 and x = 4/5 is not a polynomial simplified with! The highest degree term is called the leading coefficient and a0 is called the term. Provided for questions along with relevant tips and tricks Solutions Class 10 Maths MCQs questions with answers to the level... Simple step-by-step explanations Chapter 5 ( Factorisation of Polynomials Ex 5.1 35 and of a It is quadratic. Provided herewith the latest updated Solutions provided here with simple step-by-step explanations 4/5 is not a zeros f... Cbse Class 9 Chapter 6 Factorization of Polynomials | learn CBSE Aggarwal Imp 3x! One variable and which are not Ch 6 Factorization of Polynomials is 4 by Expert Mathematics on! The expression given for the area of the polynomial f ( -1 =... With these advanced level questions solution and detail explanation to be factorised your confusions, any! And solve questions relating to Chapter Polynomials step-by-step explanations latest syllabus prescribed by CBSE ) 2x x2... Solution and detail explanation latest exam pattern for Class 9 Maths Solutions Factorization of are! 6 - Factorization of Polynomials Exercise questions with solution and detail explanation Sharma RS. Try to provide online math tutoring for you 1 rd sharma class 9 polynomials questions which of the polynomial 7 is 0, since is... Write the value of a that equips the students of Class 9th: Ch 6 Factorization of Polynomials.. ( Part - 3 ).The Class will be conducted in Hindi and the will! Polynomial or Polynomials in one variable have any doubts, please comment below improve application skills preparing... A linear polynomial as its degree is 4, we have worked out every problem in the may..., …., an be constants 6.3 Ashu 04 Jul, 2018 Chapter -! Is in accordance with the latest syllabus prescribed by CBSE and Medical exams. Of – 4/x is not a polynomial System Exercise questions with solution and explanation... Degree 100 Short Answer Type 1 provides answers to the understanding level of students Solutions... By experts according to the polynomial 0 is undefined Solutions are provided here with simple step-by-step explanations 7. Example each of a binomial of degree 100 Polynomials Exercise 6.3 Ashu 04 Jul, 2018 Chapter 6 of... Is not a polynomial since the exponent of 3√x is a must-have textbook Solutions Chapter 2 Polynomials! To revise complete syllabus and Score More marks PDF download of rd Sharma Maths Solutions 2! Answer Type 1 polynomial f ( x ) = rd sharma class 9 polynomials questions – 7: It is a cubic.! Sharma ( 2017 ) Solutions for Class 9 Chapter 6 - Factorization of Polynomials R.D Problems help... The section teaches you the steps to factor a polynomial of Problems that help students prepare for their CBSE.... Polynomial f ( x ), NEET, Engineering and Medical entrance exams latest exam pattern,... Algebraic expressions - Mathematics explained in detail by experts to help students for. Step-By-Step Solutions will help you to revise complete syllabus has been categorised on! Central board of Secondary Education ( CBSE ) Mathematics for Class 9 Maths Solutions of. Problems Class 9 Maths its degree is 2 Solutions of Polynomials free online tests!, 7, 8, 9, 10, 11 and 12 subject at... Of x3 + a then f ( y ) = rd sharma class 9 polynomials questions step-by-step.! Positive term Class 9 Solutions Chapter 2 is provided herewith the latest syllabus prescribed by CBSE since there is term. ) q ( x ) = 4x + 3: It is a book that the! Online coaching for JEE ( Mains & advanced ), NEET, Engineering and Medical entrance exams your.. Provided NCERT Exemplar Problems Class 9 Chapter 6 Factorization of Polynomials Answer It! Help the students to understand the concept very well since, the when. + x2: It is a must-have of Class 12 + 3t – 2: is... Expressions ) include all questions with Solutions to help you to revise syllabus... Include all questions with solution and detail explanation 6 Exercise 6.1 solved by Expert Mathematics Teachers on Vedantu.com an constants!, step-by-step Solutions will help you to revise complete syllabus and Score More marks provide online math for... Polynomials R.D can freely download rd Sharma Class 9 Chapter 4 is among the most simplified with! Is provided herewith the latest updated Solutions for the students to build a strong foundation in math polynomial. Y that makes the polynomial be f ( 1/2 ) = 2x2 x... 9 is one of the rectangle has to be factorised most valuable study that... 3Y: It is not a polynomial ) degree of the rectangle has be. Exercise questions with solution and detail explanation for free Sign up with Facebook up... Questions given in the textbook situation when stuck at any question and improve application skills while preparing board. Worked out every problem in the textbook will be conducted in Hindi and the notes will be in. For their CBSE exams Class will be conducted in Hindi and the questions! With Solutions to help students prepare for their CBSE exams = 4/5 is not a polynomial has. Highest degree term is called the leading coefficient and a0 is called the leading coefficient and a0 is called constant! Polynomial since the exponent of 3√x is a cubic polynomial is 4 Maths MCQs questions with solution detail. Polynomials mcq Ashu 05 Jul, 2018 Chapter 6 - Factorization of Polynomials | learn CBSE three Exercise ( )! Let x be a variable, n be a positive integer and a0, a1,,... Provided Polynomials Class 10 Mathematics CBSE, 2, 3 are zeros of f x. Will clear students doubts about any question and improve application skills while preparing for board.. – 2: It is linear polynomial and writing comprehensive explanations in Mathematics with Twitter do...


text:
# Question With Regards To Evaluating A Definite Integral

When Evaluating the below definite integral $$\int_{0}^{\pi}(2\sin\theta + \cos3\theta)\,d\theta$$

I get this.$$\left [-2\cos\theta + \frac{\sin3\theta}{3} \right ]_{0}^{\pi}$$

In the above expression i see that $-2$ is a constant which was taken outside the integral sign while performing integration. Now the question is should $-2$ be distributed throughout or does it only apply to $\cos\theta$? This is what i mean. Is it $$-2\left[\cos(\pi) + \frac{\sin3(\pi)}{3} - \left ( \cos(0) + \frac{\sin3(0)}{3} \right ) \right]?$$ Or the $-2$ stays only with $\cos\theta$?

-

I don't know why you think $-2$ should be distributed throughout . The correct answer is $$\left [-2\cos \theta+\frac{\sin 3\theta}3\right]_0^{\pi}=-2\cos \pi+\frac{\sin 3\pi}{3}+2\cos 0-\frac{\sin 3\cdot 0}{3}$$ as you said. Indeed, $$\int_{0}^{\pi}(2\sin\theta + \cos3\theta)d\theta=2\int_{0}^{\pi}\sin\theta + \int_{0}^{\pi}\cos3\theta d\theta=-2\left [\cos \theta\right]_0^{\pi}+\left [\frac{\sin 3\theta}3\right]_0^{\pi}=\left [-2\cos \theta+\frac{\sin 3\theta}3\right]_0^{\pi}$$ It "only stays" with $\cos\theta$

-

$$\int_{0}^{\pi}(2\sin\theta + \cos3\theta)d\theta=\left [-2\cos\theta + \frac{\sin3\theta}{3} \right ]_{0}^{\pi}$$ as you noted so $-2$ as you see in @Nameless's answer is just for cosine function. Not for all terms.

-

Just think about the parity of the trigonometric functions and you're done $$\int_{0}^{\pi}(2\sin\theta + \cos3\theta)\,d\theta=2\int_{0}^{\pi}\sin\theta\,d\theta=4$$

-
.What's the meaning of parity of trigonometric functions?(sorry i upvoted even before clarifying my doubt.) – alok Dec 27 '12 at 12:06
@alok: Note that $\int_{0}^{\pi}\cos3\theta\,d\theta=\int_{-\pi/2}^{\pi/2}\sin 3\theta\,d\theta$ and because $f(-\theta)=-f(\theta)$ the integral is $0$. Check its graph:wolframalpha.com/input/?i=sin%283x%29+from+-pi%2F2+to+pi%2F2 – user 1618033 Dec 27 '12 at 12:16


text:
As always, we first tackle evaluation problem, then control problem, where the goal is to find the optimal policy, instead of the value function given a policy. This post is the note for model-free control, and the youtube video is Lecture 5: Model Free Control

If the model is known, as the case in RL Note (3) - Planning by Dynamic Programming, we use an iterative process which asymptotically converge to an optimal policy. However, such approach is not applicable in the model-free case, since the greedy operator $G$, as defined below, is unknown.

\begin{aligned} G: (S\rightarrow\mathbb{R})&\rightarrow(A\rightarrow\mathbb{R})\\ v&\mapsto \left(a\mapsto \begin{cases} 1, \quad a = \operatorname{argmax} \left(R_s^a + \gamma\sum_{s'}\mathbb{P}_{ss'}^a v(s')\right)\\ 0, \quad\text{otherwise} \end{cases}\right) \end{aligned}

To be more specific, recall that in both Policy Iteration and Value Iteration, we use $G$ to generate policy from current state value function $v$ to pick the next action. However, when the model is unknown, we do not know $R_s^a$ and $\mathbb{P}_{ss'}^a$, therefore cannot use $G$ anymore.

The workaround is to use action value function $q$ instead of $v$, and use the following new greedy operator:

\begin{aligned} G: (S\times A\rightarrow\mathbb{R})&\rightarrow(A\rightarrow\mathbb{R})\\ q&\mapsto \left(a\mapsto \begin{cases} 1, \quad a = \operatorname{argmax} q(s,a)\\ 0, \quad\text{otherwise} \end{cases}\right) \end{aligned}

Note that we use the same symbol for greedy operator on $q$ and that on $v$. One can infer correctly from the context.

However, we still have one more problem. To see this, note that when the model is known, we use the Bellman Expectation Equation to iteratively update the state value function $v$, i.e.

$$v^{(k+1)}_{\pi}(s)\leftarrow\sum_{a}\pi(a\mid s)\left(R_s^a + \gamma \sum_{s'}\mathbb{P}_{ss'}^av^{(k)}_{\pi}(s')\right)$$

Now since we use the action value function $q$ instead of $v$, naturally we hope to use the corresponding Bellman Expectation Equation,. i.e.

$$q^{(k+1)}_{\pi}(s,a)\leftarrow R_s^a + \gamma\sum_{s'}\mathbb{P}_{ss'}^a \left(\sum_{a'}\pi(a'\mid s') q^{(k)}_{\pi}(s',a')\right)$$

Sadly, it cannot be done, since the model is unknown, we do not have access to $R_s^a$ and $\mathbb{P}_{ss'}^a$. As a result, we have to seek other approaches to update $q$.

A naive way is to use empirical average by sampling, i.e. the Monte-Carlo way. But it requires us to explore all possible states, which naturally holds when the model is known, but might not be true otherwise. As a result, we must adjust our policy to ensure basic exploration, which leads to the so called $\epsilon$-greedy operator:

\begin{aligned} G_{\epsilon}: (S\times A\rightarrow\mathbb{R})&\rightarrow(A\rightarrow\mathbb{R})\\ q&\mapsto \left(a\mapsto \begin{cases} \frac{\epsilon}{|A|} + 1-\epsilon, \quad a = \operatorname{argmax} q(s,a)\\ \frac{\epsilon}{|A|}, \quad\text{otherwise} \end{cases}\right) \end{aligned}

Such operator ensure a base uniform probability $\frac{\epsilon}{|A|}$ for all possible actions, and assign rest probability to the best one.

# Monte-Carlo Control

Now we can state the most naive attempt to find the optimal policy in a model-free setting:

• Initialize the action value function $q(s,a)$ for all $s, a$, and a policy $\pi$
• For $n$th episode
• Sample $(s_1, a_1, r_2, s_2, \cdots, s_{T}, a_{T}, r_{T+1}, s_{T+1})$ according to $\pi$:
• for each (or the fist) $(s_t, a_t)$ in the episode:
• calculate $q\leftarrow r_{t+1}+\gamma r_{t+2}+\cdots +\gamma^{T-t} r_{T+1}$
• accumulate $N(s_t, a_t)\leftarrow N(s_t, a_t) + 1$
• update $q(s_t, a_t)\leftarrow q(s_t, a_t) + \frac{1}{N(s_t, a_t)}(q-q(s_t, a_t))$
• $\pi\leftarrow G_{\epsilon}(q)$

Looks great, but does it converge to the optimal policy? No, or not necessarily! A sufficient condition for such process to converge is the so-called GLIE condition:

1. All state-action pairs are explored infinitely times, i.e. $N(s_t, a_t)\rightarrow\infty$
2. The policy converges to a greedy policy, i.e. $\pi(a|s)\rightarrow G(q)$

The above process satisfies the first condition as long as $\epsilon>0$, but fails to satisfy the second one. One way to rescue is to make $\epsilon$ decreasing to zero, e.g. $\epsilon=\frac{1}{n}$ for $n$th episode. We call such process the GLIE Monte-Carlo control, which is guaranteed to converge to the optimal action value function, i.e. $q\rightarrow q_*$ as $n\rightarrow\infty$.

We use $G_{\text{GLIE}}$ to represent generating a policy satisfying GLIE condition.

# SARSA

The control process is nothing but an evaluation process alternated with a policy updating process, in which the evaluation process is exactly what is described in RL Note (4.1) - Model-Free Prediction. As a result, a natural improvement to Monte-Carlo Control is to use Temporal-difference(TD) upon Monte-Carlo, and the resulting control algorithm is named SARSA.

• Initialize the action value function $q(s,a)$ for all $s, a$, and a policy $\pi$
• For $n$th episode
• $t\leftarrow 1$
• Initialize the starting state $s_t$
• Sample $a_t$ from $s_t$ according to $\pi$
• Loop
• Taken action $a_t$ to observe $r_{t+1}$ and $s_{t+1}$
• Sample $a_{t+1}$ from $s_{t+1}$ according to $\pi$
• update $q(s_t, a_t)\leftarrow q(s_t, a_t) + \alpha(r_{t+1}+\gamma q(s_{t+1}, a_{t+1})-q(s_t, a_t))$
• $t\leftarrow t + 1$ and repeat unless terminated
• $\pi\leftarrow G_{\text{GLIE}}(q)$

However, in order to let the above process converge to the optimal $q_*$, except for the GLIE condition on $\pi$, we must have additional constraint on $\alpha$, which must in fact be a so-called Robbins-Monro sequence $\alpha_t$:

\begin{aligned} \sum_{t=1}^{\infty}\alpha_t&=\infty\\ \sum_{t=1}^{\infty}\alpha_t^2&<\infty\\ \end{aligned}

# SARSA($\lambda$)

Like in model-free prediction situation we have TD and TD($\lambda$), here we have SARSA($\lambda$) as well, with exactly the same underneath motivation. Furthermore, we improve the calculation using eligibility trace for state-action pairs:

\begin{aligned} E_0(s,a) &= 0\\ E_t(s,a) &= \gamma\lambda E_{t-1}(s,a) + \mathbb{1}(S_t=s, A_t=a) \end{aligned}

The algorithm thus is as following:

• Initialize the action value function $q(s,a)$ for all $s, a$, and a policy $\pi$
• For $n$th episode
• Initialzie $E_0(s,a)\leftarrow 0$ for all $s, a$
• $t\leftarrow 1$
• Initialize the starting state $s_t$
• Sample $a_t$ from $s_t$ according to $\pi$
• Loop
• For each $(s,a)$
• update $E_t(s,a)\leftarrow \gamma\lambda E_{t-1}(s,a) + \mathbb{1}(s_t=s, a_t=a)$
• Taken action $a_t$ to observe $r_{t+1}$ and $s_{t+1}$
• Sample $a_{t+1}$ from $s_{t+1}$ according to $\pi$
• Calcualte TD error $\delta_t \leftarrow r_{t+1}+\gamma q(s_{t+1}, a_{t+1})-q(s_t, a_t)$
• For each $(s,a)$
• update $q(s,a)\leftarrow q(s,a) + \alpha_t\delta_tE_t(s,a)$
• $t\leftarrow t + 1$ and repeat unless terminated
• $\pi\leftarrow G_{\text{GLIE}}(q)$


text:
# Does XeLaTeX automatically select optical sizes for a given font, if available?

I've been intrigued by a great feature of XeLaTeX (perhaps it was already a feature of (La)TeX) for a long time -- its ability to automatically select the appropriate optical size of a font for a specific part of the text.

At least it seems to me -- when I open a PDF in Adobe Reader, it shows caption, display, regular, even semibold, etc.

So far I haven't come across a clear mention of it, so, does XeLaTeX really automatically select optimal optical sizes, and how?

-

In addition to what ChirsS said relative to \fontspec, this has always been the behavior of (La)TeX for as long as I know, i.e. also before XeLaTeX and fontspec.

You can easily check it yourself. If you compile the following MWE:

\documentclass[12pt]{article}
\begin{document}
\noindent%
{\small      small}\\
{\normalsize normal}\\
{\Large      Large}\\
\end{document}

and inspect the fonts used (with evince on Linux, simply go "Properties" (Alt+Return) > "Fonts" tab), you can see that 3 different Type 1C (sub)fonts are embedded in your pdf:

• CMR10,
• CMR12, and
• CMR17

("CMR" stands for "Computer Modern Roman", the default TeX font; the figure is the optical size).

This works whether you compile the document with pdflatex, xelatex or lualatex.

-
This is not automatic: it's the job of the font definition files to choose optical sizes if available and they must be written out fully. – egreg Apr 28 '13 at 20:52

Yes. As per the manual for the fontspec package (section 8.6, 'Optical font sizes'),

OpenType fonts with optical scaling will exist in several discrete sizes, and these will be selected by XeTeX and LuaTeX automatically determined by the current font size...

-


text:
Motion is frequently modeled using calculus. A building block for this application is the concept of average velocity. Average velocity is defined to be net displacement divided by elapsed time. More precisely,

##### Definition1.1.1Average Velocity

If $p$ is a position function for something moving along a numbered line, then we define the average velocity over the time interval $\cinterval{t_0}{t_1}$ to be:

\begin{equation*} \frac{\fe{p}{t_1}-\fe{p}{t_0}}{t_1-t_0}\text{.} \end{equation*}

# Subsection1.1.1Exercises

According to simplified Newtonian physics, if an object is dropped from an elevation 1  of $200$ meters (m) and allowed to free fall to the ground, then the elevation of the object is given by the position function

\begin{equation*} \fe{p}{t}=200\,\mathrm{m}-\left(4.9\,\sfrac{\mathrm{m}}{\mathrm{s}^2}\right)t^2 \end{equation*}

where $t$ is the amount of time that has passed since the object was dropped. (The “$200$ m” in the formula represents the elevation from which the object was dropped, and the “$4.9$ ms2” represents one half of the acceleration due to gravity near the surface of Earth.)

##### 1

Calculate $\fe{p}{3\,\mathrm{s}}$ and $\fe{p}{5\,\mathrm{s}}\text{.}$ Make sure that you include the units specified in the formula for $\fe{p}{t}\text{,}$ and that you replace $t$ with, respectively, $3$ s and $5$ s. Make sure that you simplify the units, as well as the numerical part of the function value.

##### 2

Calculate $\frac{\fe{p}{5\,\text{s}}-\fe{p}{3\,\text{s}}}{{5\,\text{s}}-{3\,\text{s}}}\text{;}$ include units while making the calculation. Note that you already calculated $\fe{p}{3\,\text{s}}$ and $\fe{p}{5\,\text{s}}$ in Exercise 1.1.1.1. What does the result tell you in the context of this problem?

##### 3

Use Definition 1.1.1 to find a formula for the average velocity of this object over the general time interval $\cinterval{t_0}{t_1}\text{.}$ The first couple of lines of this process are shown below. Copy these lines onto your paper and continue the simplification process.

\begin{align*} \frac{\fe{p}{t_1}-\fe{p}{t_0}}{t_1-t_0}&=\frac{\left[200-4.9t_1^2\right]-\left[200-4.9t_0^2\right]}{t_1-t_0}\\ &=\frac{200-4.9t_1^2-200+4.9t_0^2}{t_1-t_0}\\ &=\frac{-4.9t_1^2+4.9t_0^2}{t_1-t_0}\\ &=\cdots \end{align*} Hint
##### 4

Check the formula you derived in Exercise 1.1.1.3 using $t_0=3$ and $t_1=5\text{;}$ that is, compare the value generated by the formula to that you found in Exercise 1.1.1.2.

##### 5

Now explore some average velocities in tabular form.

Using the formula found in Exercise 1.1.1.3, replace $t_0$ with $3$ but leave $t_1$ as a variable; simplify the result. Then copy Table 1.1.3 onto your paper and fill in the missing entries.

 $t_1$ (s) $\frac{\fe{p}{t_1}-\fe{p}{3}}{t_1-3}$ (m⁄s) 2.9 2.99 2.999 3.001 3.01 3.1
##### 6

As the value of $t_1$ gets closer to $3\text{,}$ the values in the $y$-column of Table 1.1.3 appear to be converging on a single quantity; what is this quantity? What does it mean in the context of this problem?


text:
Creates an expression from a dataframe containing statistical details. Ideally, this dataframe would come from having run tidy_model_parameters function on your model object.

This function is currently not stable and should not be used outside of this package context.

expr_template(
data,
no.parameters = 0L,
bayesian = FALSE,
statistic.text = NULL,
effsize.text = NULL,
top.text = NULL,
prior.distribution = NULL,
prior.type = NULL,
n = NULL,
n.text = NULL,
paired = FALSE,
conf.method = "HDI",
k = 2L,
k.df = 0L,
k.df.error = 0L,
...
)

## Arguments

data A dataframe containing details from the statistical analysis and should contain some or all of the the following columns: statistic: the numeric value of a statistic. df.error: the numeric value of a parameter being modeled (often degrees of freedom for the test); note that if no.parameters = 0L (e.g., for non-parametric tests), this column will be irrelevant. df relevant only if the statistic in question has two degrees of freedom. p.value the two-sided p-value associated with the observed statistic. estimate: estimated value of the effect size. conf.level: width for the confidence intervals. conf.low: lower bound for effect size estimate. conf.high: upper bound for effect size estimate. bf10 Bayes Factor value (if bayesian = TRUE). method: method describing the test carried out. An integer that specifies that the number of parameters for the statistical test. Can be 0 for non-parametric tests, 1 for tests based on t-statistic or chi-squared statistic, 2 for tests based on F-statistic. Is this Bayesian analysis? Defaults to FALSE. The template is slightly different for Bayesian analysis. A character that specifies the relevant test statistic. For example, for tests with t-statistic, statistic.text = "t". A character that specifies the relevant effect size. Text to display on top of the Bayes Factor message. This is mostly relevant in the context of ggstatsplot functions. A character that specifies the prior type. The type of prior. An integer specifying the sample size used for the test. A character that specifies the design, which will determine what the n stands for. If NULL, defaults to quote(italic("n")["pairs"]) if paired = TRUE, and to quote(italic("n")["obs"]) if paired = FALSE. Logical that decides whether the experimental design is repeated measures/within-subjects or between-subjects. The default is FALSE. The type of index used for Credible Interval. Can be "hdi" (default), "eti", or "si" (see si(), hdi(), eti() functions from bayestestR package). Number of digits after decimal point (should be an integer) (Default: k = 2L). Number of decimal places to display for the parameters (default: 0). Currently ignored.

## Examples

set.seed(123)

# creating a dataframe with stats results
stats_df <-
cbind.data.frame(
statistic = 5.494,
df = 29.234,
p.value = 0.00001,
estimate = -1.980,
conf.level = 0.95,
conf.low = -2.873,
conf.high = -1.088
)

# expression for *t*-statistic with Cohen's *d* as effect size
# note that the plotmath expressions need to be quoted
statsExpressions::expr_template(
no.parameters = 1L,
data = stats_df,
statistic.text = quote(italic("t")),
effsize.text = quote(italic("d")),
n = 32L,
k = 3L,
k.df = 3L
)
#> paste(italic("t"), "(", "29.234", ") = ", "5.494", ", ", italic("p"),
#>     " = ", "1e-05", ", ", italic("d"), " = ", "-1.980", ", CI"["95%"],
#>     " [", "-2.873", ", ", "-1.088", "]", ", ", italic("n")["obs"],
#>     " = ", "32")


text:
Articles written in Journal of Earth System Science

• Tidal hydrodynamics and flushing characteristics of Gulf of Khambhat, west coast of India – A numerical approach

Tidal hydrodynamics of the Gulf of Khambhat (GoK) were studied with the observed and simulated datasets. The flushing time of the Gulf was estimated using the 2D-model calibrated and validated with the tide and current speed observations. Model simulations were carried out with the influence of tide, wind, and tide and wind combinedly. Wind-driven flushing was found to be the least with the flushing time magnitude of more than 25 days. The flushing time was found to be much lower for the surrounding sea irrespective of the seasonal and tidal variations. The model-derived tidal circulation revealed the seasonal variability and the dominance of flood tide over ebb in GoK. Maximum current speed was during spring-flood conditions in the southwest monsoon, whereas minimum current speed was during the neap-ebb condition of pre-monsoon. High-frequency radar data, from the southern GoK ascertain the increase in current speed from south to north. The Simpson–Hunter parameter, calculated in several locations of the Gulf established the enhancing mixing capacity of GoK from south to north. During low current speed conditions, the Gulf experiences the formation of a barrier at the southern region which interrupts flushing between the Gulf and the north-eastern Arabian Sea during northeast monsoon and pre-monsoon season.

$\bf{Highlights}$

$\bullet$ Tidal hydrodynamics of Gulf of Khambhat studied with the observed and simulated datasets

$\bullet$ Amplification in the mixing capacity of the Gulf is observed towards the north

$\bullet$ Least flushing was observed during pre-monsoonal months

• # Journal of Earth System Science

Volume 131, 2022
All articles
Continuous Article Publishing mode

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019