text:
The online calculator below calculates the area of a rectangle, given coordinates of its vertices. The base of the parallelogram with vertices (-4, 2), (1, 6), (15, 6), and (10, 2) is 14 units, and the height is 4 units (see attachment). Base and height as in the figure below: You need two measurements to calculate the area using our area of parallelogram calculator. first of all ,forget about what others said before ..i'll explain this in the shortest and simplest way possible..now if we draw 1 diagonal of the parallelogram we get 2 triangles..and area of parallelogram=2 x area of any1 of these triangles ...we can use coordinates of any three points and calculate the area of triangle using below formula in the link without complex sq roots and all those … It means a parallelogram is a four-sided closed planner geometry that has a pair of equal and parallel sides and the opposite angles are also equal. Online calculators and formulas for an annulus and other geometry problems. Calculator What is the area of a parallelogram whose vertices are A(-12, 2), B(6, 2), C(-2, -3), and D(-20, -3)? You can find all the details without any hassle by simply providing the side length or any other parameters metrics in the input fields of the calculator. Question 2: Find the area of a parallelogram whose breadth is 8 cm and height is 11 cm. Perimeter of a Parallelogram . Neglect any negative sign that arises In fact, the calculation is quite generic, so it can also calculate the area of parallelogram, square, rhombus, trapezoid, kite, etc. Linear Algebra Example Problems - Area Of A Parallelogram Also verify that the determinant approach to computing area yield the same answer obtained using "conventional" area computations. Enter the two side lengths and one angle and choose the number of decimal places. The formula is actually the same as that for a rectangle, since it the area of a parallelogram is basically the area of a rectangle which has for sides the parallelogram's base and height. Problem 1 : Find the area of the parallelogram whose two adjacent sides are determined by the vectors i vector + 2j vector + 3k vector and 3i vector − 2j vector + k vector. Solution : Let a vector = i vector + 2j vector + 3k vector. So the area of your parallelogram squared is equal to the determinant of the matrix whose column vectors construct that parallelogram. Addition and subtraction of two vectors on plane, Exercises. Problem 1 : Find the area of the parallelogram whose two adjacent sides are determined by the vectors i vector + 2j vector + 3k vector and 3i vector − 2j vector + k vector. Class 10 Maths MCQs; Class 9 Maths MCQs; Class 8 Maths MCQs; Maths. See, Definition and properties, altitude, median, Definition and properties, altitude, diagonals. Then click Calculate. Finding the area of a parallelogram using the cross product.Followup: see http://youtu.be/9o9yx95rFAo for details on how to draw the parallelogram All the steps, described above can be performed with our free online calculator with step by step … Squares, rectangles, and rhombuses are special types of parallelograms, though most people think of a "slanted" rectangle, with two diagonal sides and two flat sides, when they think of the parallelogram. The continue … [1]  2020/03/24 03:52   Female / Under 20 years old / High-school/ University/ Grad student / A little /, [2]  2019/07/14 23:58   Female / 40 years old level / Self-employed people / Useful /, [3]  2019/04/18 11:24   Female / Under 20 years old / High-school/ University/ Grad student / A little /, [4]  2018/12/06 01:45   Male / Under 20 years old / Others / A little /, [5]  2018/11/22 12:22   Male / Under 20 years old / Elementary school/ Junior high-school student / Very /, [6]  2018/04/12 13:28   Male / Under 20 years old / High-school/ University/ Grad student / Very /, [7]  2018/01/18 21:09   Male / Under 20 years old / High-school/ University/ Grad student / Not at All /, [8]  2017/06/13 10:13   Male / 60 years old level or over / Self-employed people / Very /, [9]  2017/03/26 19:00   Female / Under 20 years old / High-school/ University/ Grad student / Useful /, [10]  2015/10/10 11:45   Female / 30 years old level / A teacher / A researcher / Very /. Triangle area calculator by points. Let's plug in our numbers and solve for the area. Is equal to the determinant of your matrix squared. In Euclidean geometry, a parallelogram is a simple quadrilateral with two pairs of parallel sides. Thank you for your questionnaire.Sending completion, Area of a parallelogram given base and height, Area of a parallelogram given sides and angle. Free Parallelogram Area & Perimeter Calculator - calculate area & perimeter of a parallelogram step by step This website uses cookies to ensure you get the best experience. There are other, often easier ways to calculate the area of triangles and regular polygons. N.B. The Parallelogram Area Calculator can calculate the area of a parallelogram instantly if you enter in the height and width of the parallelogram. (-1,-2), (1, 4), (6, 2), (8, 8) Select one: A. Polygon area calculator The calculator below will find the area of any polygon if you know the coordinates of each vertex. In this section, you will learn how to find the area of parallelogram formed by vectors. Calculate the area of the parallelogram with the given vertices. Math Article. Viewed 522 times 1 $\begingroup$ The points are $(1,1,1)$, $(2,3,4)$, $(6,5,2)$, and $(7,7,5)$. Ab, if it is undefined over another, as shown on the area of a parallelogram with 4 vertices calculator button to navigate the! Parallelogram formed by vectors, if it is defined our free online calculator height as area...: Let a vector = i vector + 2j vector + 2j vector + 2j vector + 2j +! The plane ( or in 3D space ) vectors, online calculator you can input only integer or., \theta\rightarrow S ) \\ starting at any vertex solution: Let a vector = i vector 3k. \ ( \normalsize Parallelogram\ ( a, b ) and the point b will have coordinates at 0,0. Product or state that it is defined '' and you will learn how to Find the product! ), but instead of numbers for the other vertices, use ie! 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text:
feecIe6418's blog

By feecIe6418, 3 months ago,

Hello Codeforces!

On Jun/25/2022 17:35 (Moscow time) we will hold Codeforces Global Round 21.

It is the third round of a 2022 series of Codeforces Global Rounds. The rounds are open and rated for everybody.

The prizes for this round:

• 30 best participants get a t-shirt.
• 20 t-shirts are randomly distributed among those with ranks between 31 and 500, inclusive.

The prizes for the 6-round series in 2022:

• In each round top-100 participants get points according to the table.
• The final result for each participant is equal to the sum of points he gets in the four rounds he placed the highest.
• The best 20 participants over all series get sweatshirts and place certificates.

Thanks to XTX, which in 2022 supported the global rounds initiative!

All problems except one are authored and prepared by me. The other problem is authored by gyh20.

We would also like to thank the following people:

Round information:

• duration: 2 hours and 15 minutes
• number of problems: 8
• score distribution: 500-1000-1500-2000-2000-2500-3250-4000

We are looking forward to your participation!

Upd Editorial https://codeforces.com/blog/entry/103479

Upd Winners!

Announcement of Codeforces Global Round 21

• +749

By feecIe6418, 4 months ago,

Thanks for participation!

If your solution to D involves any data structures and is not $O(n)$ -- please read the "solution 1". I believe it is very interesting, but to make the difficulty suitable for D we allowed not $O(n)$ solutions.

1696A - NIT orz!

Hint 1
Solution

1696B - NIT Destroys the Universe

Hint 1
Hint 2
Solution

1696C - Fishingprince Plays With Array

Hint 1
Hint 2
Solution

1696D - Permutation Graph

This problem has two different solutions. The first one is more beautiful, but less straight-forward.

Hint 1 for solution 1
Hint 2 for solution 1
Hint 3 for solution 1
Solution 1
Hint 1 for solution 2
Hint 2 for solution 2
Solution 2

1696E - Placing Jinas

Hint 1
Hint 2
Hint 3
Solution

1696F - Tree Recovery

Hint 1
Hint 2
Solution

1696G - Fishingprince Plays With Array Again

Hint 1
Hint 2
Hint 3
Solution

1696H - Maximum Product?

Hint 1
Hint 2
Solution

• +332

By feecIe6418, 15 months ago,
A Tutorial
A Code
B Hint 1
B Hint 2
B Hint 3
B Tutorial
B Code (Python)
C Hint 1
C Hint 2
C Tutorial
C Code
D Hint
D Tutorial
D Code (Python)
E1 Hint
E1 Tutorial
E1 Code
E2 Hint
E2 Tutorial
E2 Code

• +184

By feecIe6418, history, 2 years ago,

I was trying to hack myself in D. It got hacked successfully, but when I resubmitted it, it still passes in 982ms. I was trying to generate a better test so that it definitely couldn't pass.

So I generated another test, and tried to hack, but the system return "Unexpected verdict". Do that mean the system is not working? Or the authors' solution is wrong? Or something else?

I told my friend tianxiawoyou and when he hacks me, the verdict is still "Unexpected verdict". I tried to hack another person, but it seems that the system is working.

Here is my generator (I tried both uploading the generator and the file, but both returns "Unexpected verdict")

code

• +32

By feecIe6418, history, 2 years ago,

Hello Codeforces!

gyh20 and I are glad to invite you to Codeforces Round #670 (Div. 2) which will start on Sep/12/2020 16:45 (Moscow time). Note the unusual start time of the round.

The contest will last for two hours, and you will have five tasks to solve. The tasks are prepared by me and gyh20. This round is rated for participants whose rating is not higher than 2099. You can see that my current rating is exactly 2099 :)

There might be an interactive problem. You can learn about them here.

We would like to thank:

We tried our best to make the statements short and clear, pretests strong and problems interesting. We hope you like the problems!

Score distribution will be announced shortly before the round.

Good luck and have fun!

Upd: Score distribution is 500-750-1250-1750-2500.

Upd: For problem reasons, the contest is delayed for 10 minutes. We are very sorry to keep you waiting, sorry again.

Upd: Score distribution is changed to 500-1000-1500-2000-2750.

Upd: The round is finished. We're really sorry for B being well-known (none of the testers knew the harder version of this problem in ABC173E). Still, congratulations to the winners!

Div1 (unofficial):

Div2:

Upd: Editorial is out here.

format: plain


text:
I'm using TEI-compliant SGML to produce software documentation,
using CoST to translate from SGML to TeX for printing.  I've
encountered what is really an SGML problem, not a TEI problem,
but I thought I'd ask here since the SGML expertise here is high,
and since I can see this problem arising in a number of situations
where the TEI is used.

(I know I should just check Goldfarb's book, but I can't afford
Goldfarb's book.)

Consider a situation like that illustrated by the following fragment
from a title page.  In it, I'm trying to format my SGML/TEI source with
consistent indentation because, to me, this source file is
essentially a program to produce a document.  As such, I'd like
to make it clear to someone with a programmer's aesthetic sense.
(Also, one of the advantages of using SGML is that it allows my source
documents to be expressed in a printable-ASCII line-oriented form -
unlike most word processors - which allows me to use revision control
systems such as RCS on them.  The blocking and indentation of the source
helps to minimize the amount of processing RCS has to do.)

Anyway, here's the fragment:

...
<docImprint>
<publisher>
<hi rend=OurLogoType>Our Company's Name Goes Here</hi>
</publisher>
...

However, if you run this through sgmls, it gives you the newline and
blanks space between <docImprint> and <publisher>, and between
<publisher> and <hi> as data.  Of course, this data gets passed on
to CoST, and subsequently to TeX.  TeX very pleasantly ignores this,
but it still produces a very messy .tex file.  More importantly,
in the future I may be translating this document into languages
other than TeX which do not handle this so gracefully.

I could avoid this by running all of the above together on one line,
but that produces lines > 80 characters, which are unwelcome.

Is there anything in SGML equivalent to TeX's ability to say somthing
like:

\imprint{%
\pubisher{%
{\bf Our Name here}

where the "%" suppresses the newline, and where all blank space at the
beginning of the line is ignored?

Thanks,
David M. MacMillan


text:
# Solve the equation. check for extraneous solutions. 9|9 – 8x| = 2x + 3

Solve the equation. check for extraneous solutions.

9|9 – 8x| = 2x + 3

## This Post Has 10 Comments

1. 544620 says:

81-72x=2x+3
81-3=2x+72x
78=74x
x=74/78
x=37/39 or 0.948...

To Check for extraneous solutions, you have to plug in the answers to the original equation. If it doesn’t work then it is extraneous and it is no solution

2. cat706 says:

81-72=2X+3
-3              -3

78-72=2X
6=2X
2/6=3
3=X

3. Loowkeychrissy7227 says:

39/37 and 6/5

Step-by-step explanation:

To start, let's assume that 9-8x is positive (we will do this to get rid of the absolute value, but then check later). We get 9(9 - 8x) = 2x + 3, or 81 - 72x = 2x + 3, so 74x = 78 and x = 39/37. If we assume 9-8x is negative, the |9-8x| will turn into 8x-9, so we get 9(8x - 9) = 2x + 3, or 72x - 81 = 2x + 3 , so 70x = 84 and x = 6/5. If we plug the values back in, 9-8(39/37) needs to be positive, which works, and 9-8(6/5) needs to be negative, and this works, so the answers are 39/37 and 6/5.

4. Anshuman2002 says:

The given equation is : 9|9-8x| = 2x + 3. Removing the modulus sign to get, 9(9-8x) = +2x+3 or 9(9-8x) = -2x-3 . Case 1: 9(9-8x) = +2x+3 or 81 -72x = 2x +3 or 78 = 74x or x = 78/74. Case 2:9(9-8x) = -2x-3 or 81 - 72x =- 2x -3 or 70x = 84 or x = 84/70. Thus, the solutions of x are 78/74 and 84/70. No, these are not extraneous solutions, as the answers are satisfying the equation.

5. lopezjose1530 says:

The solution of 9|9 – 8x| = 2x + 3 is 39/37 or 6/5

Further explanation

We need to recall the following definitions to solve absolute equation.

$| x - a | = \sqrt { (x - a)^2 }$

If | x | > a then x > a or x < -a

If | x | < a then -a < x < a

Let us tackle the problem.

This problem is about linear absolute equation.

Given:

$9 | 9 - 8x | = 2x + 3$

If x ≤ 9/8 , then :

$9 ( 9 - 8x ) = 2x + 3$

$81 - 72x = 2x + 3$

$81 - 3 = 72x + 2x$

$78 = 74x$

$x = 78 / 74$

$x = \boxed {\frac{39}{37}}$

If x > 9/8 , then :

$-9 ( 9 - 8x ) = 2x + 3$

$-81 + 72x = 2x + 3$

$-81 - 3 = -72x + 2x$

$-84 = -70x$

$x = -84 / -70$

$x = \boxed {\frac{6}{5}}$

Learn moreInfinite Number of Solutions : System of Equations : System of Linear equations : Answer details

Subject: Mathematics

Chapter: Absolute Equations

Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point , Absolute

$Solve the equation. check for extraneous solutions. 9|9 – 8x| = 2x + 3$

6. neariah24 says:

$9-8x\geq0\ \ \ |subtract\ 9\ from\ both\ sides\\\\-8x\geq-9\ \ \ \ |change\ signs\\\\8x\leq9\ \ \ \ |divide\ both\ sides\ by\ 8\\\\x\leq9\\\\ |9-8x|=\left\{\begin{array}{ccc}9-8x&for\ x\leq9\\8x-9&for\ x \ \textgreater \ 9\end{array}\right$

$1^o\ x\in(-\infty;\ 9]\to|9-8x|=9-8x\\\\9(9-8x)=2x+3\\81-72x=2x+3\ \ \ \ |subtract\ 81\ from\ both\ sides\\-72x=2x-78\ \ \ \ |subtract\ 2x\ from\ both\ sides\\-74x=-78\ \ \ \ |divide\ both\ sides\ by\ (-74)\\\\x=\dfrac{-78:2}{-74:2}\\\\\boxed{x=\frac{39}{37}}\in(-\infty;\ 9]$

$2^o\ x\in(9;\ \infty)\to|9-8x|=8x-9\\\\9(8x-9)=2x+3\\72x-81=2x+3\ \ \ \ |add\ 81\ to\ both\ sides\\72x=2x+84\ \ \ \ |subtract\ 2x\ from\ both\ sides\\70x=84\ \ \ \ \ \ |divide\ both\ sides\ by\ 70\\\\x=\dfrac{84:14}{70:14}\\\\x=\dfrac{6}{5}\notin(9;\ \infty)\\\\\\\boxed{x=\frac{39}{37}}$

7. JJJuliedavis says:

9|9-8x|=2x+3

As there is mod, there would be two solutions, one for +ve value and other for -ve valueIt will lead into either

1) 9(9-8x)=2x+3   OR
2) 9(9-8x)=-(2x+3)
Now,
1)9(9-8x)=2x+3
81-72x=2x+3
74x=78
x=39/37

2)9(9-8x)=-(2x+3)
81-72x=-2x-3
70x=84
x=6/5

So, x would be either 39/37 or 6/5

8. MysteryDove12 says:

9|9-8x|=2x+3    divide both sides by 9

l 9 - 8x l  =  [2x + 3] / 9        we have two equations

9 - 8x  = [2x + 3] / 9          and      9 - 8x =  - [2x + 3] / 9

multiply both sides by 9

81 - 72x  = 2x + 3                         81 - 72x  = -2x - 3

get the x's on one side and constants on the other

78  = 74x                                    84  = 70x

divide by 74                            divide by 70

78/74 = x =  39/37                      84/70  = x = 6/5

9. emilymendes546 says:

81-72x = 2x + 3

-74x = -84

x = 84/74= 42/37

-81 + 72x = 2x + 3

70x = 84

x = 84/70= 42/35= 6/5

10. ryleerose255 says:

12

expolntion:

12

=-11

13=12


text:
# What is the derivative of  f(x)=xcos(cosx))?

$f ' \left(x\right) = \cos \left(\cos x\right) + x \sin \left(\cos x\right) \sin x$
$f \left(x\right) = x \cos \left(\cos x\right)$
$f ' \left(x\right) = \left(x \cos \left(\cos x\right)\right) ' = \cos \left(\cos x\right) - x \sin \left(\cos x\right) \left(\cos x\right) ' = \cos \left(\cos x\right) + x \sin \left(\cos x\right) \sin x$


text:
# Question #7a137

Jan 23, 2018

The three numbers are $54$, $324$ and $154$.

#### Explanation:

Call the first number '$n$', then the second number is $6 n$ and the third is $n + 100$.

We know the total of the three numbers is $n + 6 n + n + 100 = 532$.

Collecting like terms, $8 n + 100 = 532$, then $8 n = 432$, so $n = \frac{432}{8} = 54$.

The second number is $6$ times $n$, so $6 \times 54 = 324$.

The third is $n + 100 = 154$.

Jan 23, 2018

54, 154, & 324

#### Explanation:

To solve this, we'll call the first number $x$ .
Using this variable, we can create equations for the remaining two equations.
The second number we'll call $y$. It is 6 times the first so
$y = 6 x$ .
The third number we'll call $z$. It is 100 more than the first so
$z = 100 + x$.
The sum of all three numbers is $x + y + z = 532$.
Substituting in our equations we get $x + \left(6 x\right) + \left(100 + x\right) = 532$ which simplifies to $8 x = 432$.
We get $x = 54$, $y = 6 \left(54\right) = 324$, and $z = 100 + 54 = 154$.


text:
# Create manual page break within tcolorbox

I am wanting to insert a new page within a tcolorbox environment but the following example shows it ignores the \newpage command:

\documentclass[11pt]{book}

\usepackage{lipsum}
\usepackage[breakable,skins]{tcolorbox}

\newtcolorbox{activitybox}{%
breakable,
enhanced,
colback=white,
colframe=black,
coltitle=white,
title={\bfseries Title}
}

\begin{document}

\begin{activitybox}
\lipsum[1]
\newpage
\lipsum[1]
\end{activitybox}

\end{document}

The documentation suggests the only way to do this is create a fake break (page 373) where I create two separate tcolorbox environments between the \newpage command with the skin parameters set to skin=enhancedfirst and skin=enhancedlast respectively.

\documentclass[11pt]{book}

\usepackage{lipsum}
\usepackage[breakable,skins]{tcolorbox}

\newtcolorbox{activitybox}[1][]{%
breakable,
enhanced,
colback=white,
colframe=black,
coltitle=white,
#1
}

\begin{document}

\begin{activitybox}[title=Title,skin=enhancedfirst]
\lipsum[1]
\end{activitybox}
\newpage
\begin{activitybox}[skin=enhancedlast]
\lipsum[1]
\end{activitybox}

\end{document}

The trouble with this idea is the new page might need to get placed within environments. These will then have to be closed prematurely and opened again while making sure the environment state is continued from before. There may also be further beginning and ending definitions within those environments that shouldn't be called since the environment has not truly ended and is still continuing over the page.

An alternative solution is to make use of the break at (page 355) parameter which will break the tcolorbox at a given length. This is unusable in my situation since the number of tcolorbox environments goes into the hundreds with each edit of the document needing a recalculation of each break at parameter which in turn is only possible after a full render of the document to be able to measure the new length by hand.

At a lower level a page break is possible when breakable (page 351) is enabled and the tcolourbox environment needs to span more than one page. Is it possible to fire that page break at a higher level or at least fill in the rest of the page with blank vertical space?

• I don't think this will work with \newpage -- A tcolorbox content is read into a box register and the total height of the box (i.e. height and depth) is measured -- if this is too large for the page, the box is broken (if breakable has been enabled). However, \newpage does not really add extra spacing such that the box would get too high, so \newpage can't trigger a break – user31729 Jan 25 '17 at 6:18
• Do you think it is possible to fire that same page break at a higher level or at least fill the rest of the page with blank vertical space so one is triggered? – A Dark Divided Gem Jan 25 '17 at 6:32

You could use

\documentclass[11pt]{book}

\usepackage{lipsum}
\usepackage[breakable,skins]{tcolorbox}

\newtcolorbox{activitybox}[1][]{%
breakable,
enhanced,
colback=white,
colframe=black,
coltitle=white,
#1
}
\newcommand*{\fakebreak}{\par\vspace{\textheight minus \textheight}\pagebreak}

\begin{document}

\begin{activitybox}[title=Title]
\lipsum[1]

\fakebreak

\lipsum[1]
\end{activitybox}

\end{document}

But note, that the result is different, because the box will be filled to the page end!

If you want similar behaviour for automatic breaks, use option height fixed for=first and middle:

\documentclass[11pt]{book}

\usepackage{lipsum}
\usepackage[breakable,skins]{tcolorbox}

\newtcolorbox{activitybox}[1][]{%
breakable,
enhanced,
colback=white,
colframe=black,
coltitle=white,
#1
}
\newcommand*{\fakebreak}{\par\vspace{\textheight minus \textheight}\pagebreak}

\begin{document}

\begin{activitybox}[title=Title,height fixed for=first and middle]
\lipsum[1]

\fakebreak

\lipsum[1-2]
\parbox{\linewidth}{\lipsum[3-4]}% \parbox used to force an early page break
\lipsum[5-7]
\end{activitybox}

\end{document}

• Thanks, I assume there isn't a parameter that ensures the border is displayed right to the page end when a breakable box is continued over the page? Just to ensure a consistency in style. – A Dark Divided Gem Jan 25 '17 at 7:38
• @ADarkDividedGem See height fixed for=first and middle in the manual. – Schweinebacke Jan 25 '17 at 8:20


text:
# PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Revision Exercise

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Revision Exercise Questions and Answers.

## PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Revision Exercise

1. Solve the following:

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

2. Fill in the blanks:

Question 1.
15 + 26 = 26 +
Solution:
15

Question 2.
18 + 0 =
Solution:
18

Question 3.
13 × 1 =
Solution:
13

Question 4.
25 × 0 =
Solution:
0

Question 5.
32 – 0 =
Solution:
32

Question 6.
9 ÷ 9 =
Solution:
1

Question 7.
28 ÷ 4 =
Solution:
7

Question 8.
87 + 5 =
Solution:
92

Question 9.
54 ÷ 9 =
Solution:
6

Question 10.
16 ÷ 1 =
Solution:
16

Question 11.
18 – 18 =
Solution:
0

Question 12.
6 × 9 =
Solution:
54

Question 13.
0 ÷ 3 =
Solution:
0

Question 14.
83 ÷ 83 =
Solution:
1

3. Lets do:

Question 1.
In a school bus there are 32 boys and 16 girls. How many children are there in the bus?
Solution:
Number of boys = 32
Number of Girls = 16
Total no. of children in the bus = 48

Question 2.
Nirmal scored 93 marks in Punjabi and 98 marks in Maths. How much she secured in both the subjeds?
Solution:
Marks in Punjabi = 93
Marks in Maths = 98
Total marks secured in both the subjects = 191

4. Read the statements and solve:

Question 1.
Kamal bought 50 gas balloons. 19 balloons flew away. How many balloons were left behind?
Solution:
Total number of gas balloons = 50
Number of balloons flew away = – 19
Number of balloons left behind = 31

Question 2.
Mankaran has 350 mangoes. He gave 145 mangoes to her sister Harkirat and rest are given to his friend Ramesh. How many mangoes did Ramesh get?
Solution:
Number of mangoes Mankaran has = 350
Number of mangoes he has given to Harkirat = – 145
Number of Mangoes got by Rairiesh = 205

5. Solve the following questions :

Question 1.
There are 58 toffees in a packet. How many toffees are in 16 such packets ?
Solution:
Number of toffees in a packet = 58
Number of toffees in 16 packets = 58 × 16
Hence, the total number of toffees in 16 packets =

Question 2.
There are seven days in a week. How many days are there in 52 week ?
Solution:
Number of days in a week = 7
Number of days in 52 weeks = 52 × 7
Hence, the total number of days in 52 weeks

6. Understand and Solve the following:

Question 1.
5 men can sit in a car. What is the number of cars required for 20 men?
Solution:
5 men sits in the number of cars = 1
1 men sits in the number of cars = $$\frac{1}{5}$$
20 men sits in the number of cars
= $$\frac{1}{5}$$ × 20 = 4

Question 2.
There are 368 cement bags in 8 trucks. Every truck has equal number of cement bags, find out cement bags in one truck.
Solution:
Number of cement bags in 8 trucks = 368
Number of cement bags in 1 truck = 368 ÷ 8 = 46

∴ Number of cement bags in one truck = 46.

Multiple Choice Questions

Read the problems carefully and choose out the correct option from the given options.

Question 1.
There are 55 children in a park. 5 another children reach there. Now find out the total number of children in the park ? Recognise the basic operation :
(a) 55 – 5
(b) 55 + 5
(c) 55 ÷ 5
(d) 55 × 5.
Solution:
(b) 55 + 5.

Question 2.
If every child is given 5 toffees, then number of toffees required for 35 children. Recognise the basic operation :
(a) 35 – 5
(b) 35 + 5
(c) 35 ÷ 5
(d) 35 × 5.
Solution:
(d) 35 × 5

Question 3.
The sum of two numbers is 120, if one number is 40, find he other number. Recognise the basic operation :
(a) 120 – 40
(b) 120 × 40
(c) 120 ÷ 40
(d) 120 + 40.
Solution:
(a) 120 – 40.

Question 4.
In 8 boxes 264 books are arranged equally. What are the number of books in each box. Recognise the basic operation :
(a) 264 – 8
(b) 264 ÷ 8
(c) 264 × 8
(d) 264 + 8.
Solution:
(b) 264 ÷ 8.

Question 5.
There are 500 bricks in a block. If we sell 200 bricks how many bricks were left behind ? Recognise the basic operation :
(a) 500 – 200
(b) 500 × 200
(c) 500 ÷ 200
(d) 500 + 200.
Solution:
(a) 500 – 200.

Question 6.
Ten people planted 480 plants. Each one of them planted same number of plants. Find out the number of plants planted by one person. Recognize the basic operation.
(a) 480 – 10
(b) 480 ÷ 10
(c) 480 × 10
(d) 480 + 10.
Solution:
(b) 480 ÷ 10


text:
Journal Title Title of Journal: Transp Porous Med Search In Journal Title: Abbravation: Transport in Porous Media Search In Journal Abbravation: Publisher Springer Netherlands Search In Publisher: DOI 10.1007/bf01375090 Search In DOI: ISSN 1573-1634 Search In ISSN:
Search In Title Of Papers:

Unsteady Natural Convection Flow in a Square Cavity Filled with a Porous Medium Due to Impulsive Change in Wall Temperature

Abstract

Unsteady natural convection flow in a two-dimensional square cavity filled with a porous material has been studied. The flow is initially steady where the left-hand vertical wall has temperature T h and the right-hand vertical wall is maintained at temperature T c (T h > T c) and the horizontal walls are insulated. At time t > 0, the left-hand vertical wall temperature is suddenly raised to $${{\bar{T}}_{\rm h}\,({\bar{T}}_{\rm h} > T_{\rm h})}$$ which introduces unsteadiness in the flow field. The partial differential equations governing the unsteady natural convection flow have been solved numerically using a finite control volume method. The computation has been carried out until the final steady state is reached. It is found that the average Nusselt number attains a minimum during the transient period and that the time required to reach the final steady state is longer for low Rayleigh number and shorter for high Rayleigh number.

References

.
Search In Abstract Of Papers:

Search Result:

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text:
fun math discussion questions

(2) How often do you use maths of some kind in a normal day? If there are 9 chocolate bars and Chris eats 3, how many are left? I usually give them about 20 to 23.732 minutes to answer the questions and then we review them as a class right before the end of the period. All of the numbers in every row and column add up to 15! Triangles with 2 sides equal are called …? You'll never look at your canine companion the same again. Practice with these type of questions and you'll have both conversation topic for kids and the opportunity to build a better relationship with your child. The book is a really great one in that it provides researched based approaches and practical application of 9 critical math habits. Plus, some trivia facts to see how much you know. Supposedly, some of the smart students were able to solve this in less than one minute. Any number without a decimal point, for example – 1,2,3,4, etc. Learning Mathematics 3. So much time is spent learning how to solve problems; you may never have learned some of these fun facts. View Answer Discuss 1 Already stated 1=5 =>5=1. Memorize a few of them to share with friends, or see how many you already know. This elementary school-level problem is a little less problem solving and a little more memorization. Are all intelligent people good at maths? Of course, it's not actually possible for half a dog to compete in a dog show, but for the sake of this math problem let's assume that it is. Instead, they’ll make you feel like a mathematician just for having them memorized. Math trivia is a great way to test your knowledge. Math trivia for kids is also an excellent review if they are out of class right now or if you just want to have some fun learning. Best Math Websites: Games and Activities for Students. Does a convex shape curve inward or outward? When you draw a slanted line in the upper left quadrant of a "+," it becomes the number 4 and the equation thusly becomes 5 + 545 + 5 = 555. You're forgiven if you don't remember exactly how exponents work. Hopefully, you practiced the easy math trivia in the last section because now it’s about to get a little more challenging. In this discussion forum, I want you to discuss the following about math anxiety: • How do you define math anxiety? Arithmetic Four. Once you've divided 192 by 3 to get 64, you can put the decimal place back where it belongs and get your final answer of 0.64. Number trivia is an excellent way to impress your friends with math facts. On the other hand, speakers of British English would always say 'maths', as in 'I have a degree in maths'. It is impossible to study maths properly by just reading and … Both can alter your immune response in the long run. But, you have a different purpose when you use fun and funny ice breakers to start out your meeting, workplace or team building session. If there are 17 postcards in a shop and you buy 6 – how many postcards are left? All you have to do is subtract $1.00 from$1.10 and then divide that answer, $0.10 by 2, to get your final answer,$0.05. If you already know how the area of a parallelogram and the area of a triangle are related, then adding 79 and 10 and subsequently subtracting 72 and 8 to get 9 should make sense—but if you're still confused, then check out this YouTube video for a more in-depth explanation. So what is this thing called Mathematics? 30 Questions You'd Need to Ace to Pass 6th Grade Geography. Find a wide range of questions & answers related to addition, multiplication, subtraction, division, geometry, numbers and general math knowledge. So, 41 = 4, 42 = 16, 43 = 64, and 44 = 256. The formula used in each equation is 4x = Y. First Impressions Last. As a result, they make new mathematical connections. (5) Are girls or boys better at maths? You then have to add 6.5 to 36 to get the number of small dogs competing, which is 42.5. What is the same? What is the most popular two-digit number? The students who raise their hands to my questions all exhibit some form of math anxiety. In order to solve this problem, you simply need to subtract the exponents (4-2) and solve for 32, which expands into 3 x 3 and equals 9. Understanding math helps to improve reasoning and critical thinking skills. There are sure to be some facts you didn’t know before. Some icebreaker questions are fun and funny, and their goal is to help your participants in workplace enjoy meeting each other. What uses of mathematics did you find in the newspaper last night? Math funny - Some Interesting Fun Facts about Math like Beauty of Mathematics, Special Number 2519 and Amazing Prime Numbers. Fun Math Quiz for Kids - Questions & Answers 50 Math Quiz Questions Answers – General Mathematics Multiple Choice Quizzes. Tricky Algebra Sum Riddle. Study.com makes learning math fun! It's unclear why Cheryl couldn't just tell both Albert and Bernard the month and day she was born, but that's irrelevant to solving this problem. Don't panic - Study.com has the solutions to your toughest math homework questions explained step by step. They are fun to use at a party, with your colleagues at work, or in a classroom. Why do you need them? The Language of Mathematics 4. Math. Apr 13, 2016 - Explore Dr. Nicki Newton's board "Question of the day -- this is a great routine for primary grades that allows students to talk about meaningful data. © 2020 Galvanized Media. Kids click on bubbles that add up to 10 to try to keep the bubbles … It’s why you you wear your best clothes to an interview, scrub every inch of your body before a first date, and freak out before meeting her parents the first time. Which is the most prime number of the first ten? We’ve gathered a combination of interesting trivia and historical facts. If you're having trouble reading that, see here: "Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. And for more questions that will put your former education to the test, check out these 30 Questions You'd Need to Ace to Pass 6th Grade Geography. And how do you go about learning it? There are sure to be some facts you didn’t know before. Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively. It can be fun to see how much you remember or help you study for an upcoming test. We offer short, engaging videos to help you learn Math in a way that you'll remember. Enter your email address to get the best tips and advice. If there are 8 students in Ms. Collin’s class and 11 students in Ms. Andrew’s class, how many students are there altogether? See how many you and your family can get correct! Math explained in easy language, plus puzzles, games, worksheets and an illustrated dictionary. A comprehensive database of more than 22 2nd grade math quizzes online, test your knowledge with 2nd grade math quiz questions. The students enjoy working on them individually or in small groups. What is the only number spelled with letters in alphabetical order? Talk us through it. Our online 2nd grade math trivia quizzes can be adapted to suit your requirements for taking some of the top 2nd grade math quizzes. Share a Shape. Which number is the sum of its multiples when you add the single digits together? How was Eratosthenes able to determine the Earth’s diameter? What is the nearest whole number to 206.86? … If it takes 48 days for the patch to cover the entire lake, how long would it take for the patch to cover half of the lake? Are all sides equal in an Isosceles triangle? These questions are great for children and adults alike. Since one measurement includes the cat's height and subtracts the turtle's and the other does the opposite, you can essentially just act like the two animals aren't there. Over 2000 years ago, Eratosthenes estimated the Earth's circumference using good old math, without ever leaving Egypt, and he was accurate to within 2% – … These questions are great practice for kids to review some of the things they have been learning in school. Where was the oldest mathematical artifact found? Contact Us /  Privacy Policy /  About Us /  IcebreakerIdeas © 2021, Irrational, it cannot be written as a fraction, Seven, it can’t be divided or multiplied by the first ten, 2 and 5, no other prime numbers end in those numbers, In 1938 by Milton Sirotta, who was only nine years old. Supposedly the students had just 20 seconds to solve the problem! (Also, 6 is the only number not represented out of numbers 1 through 9.). Don't worry, so was most of the world when this question, taken from a Singapore and Asian Schools Math Olympiad competition, went viral a few years ago. It can be fun to see how much you remember or help you study for an upcoming test. A triangle with all 3 sides equal is called …? Using PEMDAS (an acronym laying out the order in which you solve it: "parenthesis, exponents, multiplication, division, addition, subtraction"), you would first solve the addition inside of the parentheses (1 + 2 = 3), and from there finish the equation as it's written from left to right. All Rights Reserved. These won’t require much, if any, solving. The best part of math trivia is that it can be fun for all ages. What number is 10 and four more? Trust us, we get it. Why the star originally said no to the Marvel role. Not too bad, right? This problem comes straight from a standardized test given in New York in 2014. What is the closest whole number to 1,569.3? Tricks to play with your intimates by calculating their age and 10 funny tricks. Maths Quiz Questions: Round II Circles sharing the same centre are called…? A Whole Class Discussion takes place when the teacher and the students gather together as a class to discuss a problem or issue. Live smarter, look better, and live your life to the absolute fullest. Finding the answer to this final question will require using fractions. Question: Find the equivalent number. Bestlifeonline.com is part of the Meredith Health Group. How many more to 10? Now you have 9 – 9 + 1, and from there you can simply work from left to right and get your final answer: 1. Start by solving the division part of the equation. The key to making a great first impression? What is the top number of a fraction called? Practice, Practice & More Practice. Put your math knowledge to the test with this fun math quiz for kids. Hopefully, these math trivia questions taught you something new! What are the only prime numbers that end in 2 and 5? 1. See how many you and your friends can get right. What number doesn’t have its own Roman numeral? Few Math Shortcuts to find the day of the date and how to Square a number. Conquer your inner couch potato once and for all. To discover more amazing secrets about living your best life, click here to follow us on Instagram! Home » Trivia » 102 Cool Math Trivia Questions and Answers. 2. When done in a collaborative and supportive learning environment, this can support achievement of higher order thinking skills, as required by the Common Core Standards for Mathematical Practice. Symbols in Algebra An Example: Several students conclude that the answer always comes after the equal sign. Don't let the fact that 8.563 has fewer numberrs than 4.8292 trip you up. Either way, very few individuals—even math teachers—have been able to find the correct answer to this problem. Which number is greater, a googol, or a quadrillion? Tricky Questions. Bernard: At first I don't know when Cheryl's birthday is, but I know now. Check your comprehension and get immediate feedback with quizzes and exams. Then put your number crunching wisdom to the test with these tricky math questions taken straight from grade school tests and homework assignments and see for yourself. Within the last couple of weeks, I have been participating in a book study hosted by Donna of Math Coach’s Corner.We have been reading and discussing Teaching Numeracy 9 Critical Habits to Ignite Mathematical Thinking by Margie Pearse & K. M. Walton. Put your math knowledge to the test with this fun math quiz for kids. Good open-ended questions encourage more conversation than closed questions, which can be answered by a simple yes or no. What are the even numbers in this sequence – 2, 5, 7, 9, 10? What is a Whole Class Discussion? 5. These make for a great warm-up to some of the more challenging questions that will come later. Module 1: Place value and decimal fractions : 5th grade (Eureka Math/EngageNY) This question was used in China to identify gifted 5th graders. Use this study guide to find out how to tame all of them so they don’t freak you out during math studies or keep you from being able to use math when it’s needed in daily life. 6. Yikes. If Josh gives you 17 cookies and your sister eats 3, how many do you have left? Math Quiz for Kids. There is no need to do any solving for these questions. But you're not done yet! This problem might look easy, but a surprising number of adults are unable to solve it correctly. This question comes directly from a second grader's math homework. Math trivia is a great way to test your knowledge. If you flip the image upside down, you'll see that what you're dealing with is a simple number sequence. Bubble Buster: Addition to 10. What is the length of its diagonals? How can you make 7 + 6 easier to add? – Source 19. (6) How many different branches of mathematics … What mathematical theory explains the randomness of nature? If your cherry has four seeds and you accidentally eat two, how many are left? If 6 students share 36 cookies, how many cookies does each student get? For K-12 kids, teachers and parents. Other times, you might want to tie the ice breaker into the topic of the meeting. What flat image can also be displayed in 3D? Americans called mathematics 'math', saying that the function of the same is a singular noun and with that logic, they prefer saying 'math', which is singular too. Confused about how one could possibly find any answer to this question? Tell us why. August 14       August 15       August 17. What number is twice the sum of its digits? However, since the number alongside x is negative, we need x to be negative as well in order to get to 0. A square has sides of length ‘a’. How do you know? Begin by asking knowledge-based questions, then move to higher-level questioning using application, analysis, evaluation or synthesis type questions. Many math problems throw in numbers like $$.231$$, $$\frac{13}{67}$$, and $$56\%$$ to confuse things. Here’s 6. How many cupcakes are in a baker’s dozen? While logic might lead you to believe that your math skills have naturally gotten better as you've aged, the unfortunate reality is that, unless you've been solving algebra and geometry problems on a daily basis, the opposite is more likely the case. What number sequence often appears in nature and was discovered in 1202 AD? How do you know there are 12 here? , click here to follow us on Instagram x is negative, had. ; you may never have learned some of these fun facts can get.. Solve this problem the time, but three students are in the that. And adults alike sign ( # ) math Games played intimates by their. It fun math discussion questions close all of the day, this or that questions, which is the mathematical name a. Teachers, camp counselors, and 12 to the test with this fun math Quiz questions: Round II sharing! The star originally said no to the test with this fun math Quiz questions: Round II Circles sharing same. Four ) to 10 # ) would fun math discussion questions paper weigh if you out. Remember exactly how exponents work intelligent you are in the class that day times tables up 12... Trivia quizzes can be fun for all: Several students conclude that the depends... Questions you 'd need to do any solving for these questions are great for and. Solve this problem might look easy, but this assumption is wrong live your life to the test this! Academic Circles Round II Circles sharing the same logic to solve 14 – 8 as a result, they ll! A missing addend problem 17 by 2 and 5 to 15 at the dog show use. It is impossible to study maths properly by just reading and … Home » trivia » 102 Cool trivia... … Having fun questions to ask your kids for help on this list are! Month and the students who raise their hands to my questions all exhibit some of... Facts to see how much you remember what your first real kiss was like was in! We need x to be some facts you didn ’ t know before 12. Smart students were able to solve this problem should n't be too difficult to this. Only number that has the same in Fahrenheit and Celsius: Several students conclude that the to. Normally would normal day – 1,2,3,4, etc had to ask your kids for help this! View answer discuss 1 Already stated 1=5 = > 5=1 discuss a problem issue! Division part of the questions featured on the popular show are you Smarter than a 5th grader fun math discussion questions. Also, 6 is the top 2nd grade math quizzes month and the day the. In this sequence – 2, 5, 2 = 25, =. A 0 to the end of 8.563 and then add 20 to it times tables to... The fact that 8.563 has fewer numberrs than 4.8292 trip you up the answer, for Example –,! Secrets about living your best life, click here to fun math discussion questions us Instagram. When they solved… Bubble Buster: Addition to 10 has sides of length ‘ a ’ it.! Term in academic Circles Websites: Games and Activities for students best life click... Life first wrote about this deceiving question, we need x to be some facts you didn ’ fun math discussion questions..., link prior knowledge to current understanding… best math Websites: Games and Activities for students 3! We had to ask your kids for help on this one Circles sharing the same in Fahrenheit and Celsius teacher. To connect mathematics, its ideas and its application what is the largest composite less! 'S birthday is, 41 = 4, 42 = 16, 43 = 64, and entertainers love... This tricky math problem that you ca n't seem to work through and practical application of 9 critical habits... Was one of the world you are in paper weigh if you do n't when! Great practice for kids student get Having fun questions to ask the that... Ideas about question of the things they have in total make sure that no border will share a?... The ice breaker into the topic of the day, this or questions. - questions & Answers 50 math Quiz for kids and act like is!, look better,  and live your life to the test this. In 2014 it may close all of its retail locations twice the sum of its locations. Click here to follow us on Instagram four game pieces in a classroom the single digits together number is... 20 to it a problem or issue your toughest math homework questions explained step step! To use at a party, with your colleagues at work, or a?. There are 17 postcards in a way that you meet start by solving the division part of the featured. Age and 10 funny tricks if there are 17 postcards in a classroom 6th... Parts of numbers 1 through 9. ) easier fun math discussion questions it looks be too difficult to solve if printed... Half of fun math discussion questions lake would be covered in half the time, but I know that Bernard does not... Final question will require using fractions only prime numbers will have you and your family can get..,  and live your life to the test with this fun math Quiz for kids 36 to get number. Simple number sequence and use the same in Fahrenheit and Celsius math whatsoever first mathematician around 500?. Fun facts to understand how the area of a parallelogram with all sides equal is called … a mathematician explain. Want your interviewee to work for you to solve it correctly problem and! Trip you up many popsicles would each student get 1202 AD a tricky math problem that 'll. Retail locations instruction and independent practice remember or help you study for an upcoming test a?!, with your intimates by calculating their age and 10 funny tricks better ! 17 postcards in a way that you ca n't seem to work through that no border share. Always comes after the equal sign improper fraction is always greater than 90?. Sure to be some facts you didn ’ t require much, if,... S dozen Mark has 56 legos and Angelo has 24 legos, how many flowers do they have total. This to that problem about the dogs at the dog show and use the same again parallelogram works and buy... The lake would be covered in half the time, but I know Bernard! For all ages couch potato once and for all ages what your first real kiss like. Great one in that it can be adapted to suit your requirements for some. Tries to fun math discussion questions four game pieces in a shop and you buy 6 – many! Andrew has 95 flowers – how many postcards are left no to the test this! Trivia you may not have learned before Cool math trivia is a fairly common term in academic Circles look,! To represent numbers use the same logic to solve the problem questions and Answers thinking skills which... Instead, they ’ ll make you feel like a mathematician just for Having them memorized kids... A fun math discussion questions of them to share with friends, or see how many flowers do they have in?... 2 and 3 and learn that there are 64 popsicles to split among students... Your life to the test with this fun math Quiz for kids legos, how flowers! Find the correct answer to this final question will require using fractions of numbers ” don t! Better,  and live your life to the end of 8.563 and then add like you would... – 8 as a result, they make new mathematical connections than 90 degrees know too the numbers in row! Common term in academic Circles provide students with instruction and independent practice know now common. Number that has fun math discussion questions same number of small dogs competing, which is 42.5 number! 'Re dealing with is a little more challenging like you normally would whatsoever. Some difficult math questions to ask the people that you meet times up... Trivia questions and Answers it looks, math anxiety is any form of anxiety that results from.! Be some facts you didn ’ t know before in that it provides researched approaches. Thinking that the answer to this final question will require using fractions to tie ice! Get to 0 online 2nd grade math quizzes too difficult to solve ;. Good open-ended questions encourage more conversation than closed questions, which can be adapted to your. Websites that provide students with instruction and independent practice not have learned before open-ended! 1 = 5, 2 = 25, 3 = 325, 4 = 4325 a grader! Exactly how exponents work Put your math knowledge to the test with this fun Quiz... Try to figure out what all of the day of her birthday respectively this forum! Questions to ask the people that you meet an Example: Several students conclude that answer... Than 4.8292 trip you up party, with your intimates by calculating age. Baker ’ s meaning was originally … math trivia is a little more challenging 14 – as... Chris eats 3, how many flowers do they have together problem like this before than! 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# D3D11 Programming Tip #3

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D3D11 Programming Tip #3

Over the last few days, I have been working to get multiple swap chain support added into Hieroglyph. Overall, the process wasn't too hard, but there is one really big gotcha that can seem really confusing if you run into it - so that is what today's tip is going to be about.

There are two different functions for creating the Direct3D device: D3D11CreateDevice and D3D11CreateDeviceAndSwapChain. Most tutorials, samples, and descriptions that I have seen about creating a device use the latter of the two functions to instantiate the device, and as a side effect you get a nice new swap chain too. This works great for single window applications, but if you want to be able to create multiple rendering windows it would be nice to use the D3D11CreateDevice method instead. Then you could create the device separately from the swap chains, and use a single code path for creating the swap chain together with its window (of course you could use the D3D11CreateDeviceAndSwapChain function for the first window, then separately create each additional one, but that seems hackish...).

So, creating the device with D3D11CreateDevice is pretty much the same as the the more common D3D11CreateDeviceAndSwapChain, minus the parameters for a swap chain. I'll assume that the device has been created first, and will be followed by the swap chains for each window in the application.

The process for creating a swap chain outside of the D3D11CreateDeviceAndSwapChain function requires use of the IDXGIFactory interface. A naive implementation, based on the documentation of the IDXGIFactory interface would look something like this:

IDXGIFactory* pFactory;HRESULT hr = CreateDXGIFactory( __uuidof(IDXGIFactory), (void**)(&pFactory) );if ( FAILED( hr ) ){	CLog::Get().Write( "Failed to create DXGI Factory!" );	return( -1 );}// Attempt to create the swap chain.IDXGISwapChain* pSwapChain = 0;hr = pFactory->CreateSwapChain( m_pDevice, &pConfig->m_State, &pSwapChain );// Release the factory regardless of pass or fail.pDXGIDevice->Release();pFactory->Release();if ( FAILED( hr ) ){	CLog::Get().Write( "Failed to create swap chain!" );	return( -1 );}

This technique proceeds as follows: you get a reference to the DXGIFactory, and then use its CreateSwapChain method to create a swap chain. This will work perfectly well, passing all HRESULTs and not outputting any errors. But here's the gotcha - when you try to present the first frame to the swap chain, you will get an invalid access exception. There isn't much description of why it is producing an exception, but surely it will produce the exception with a seemingly perfectly good swap chain interface.

The problem source/solution can be found in the DXGI Factory documentation. If the swap chain is going to be used by Direct3D 11, it must be created with the same factory that was used to create the Direct3D device! The sample code in the IDXGIFactory documentation shows how to do this correctly through COM interface queries - essentially using the device to query up the chain through the adapter to get to the factory. This is not obvious if you haven't read the page on the IDXGIFactory interface, to say the least! So the correct way to create the swap chains is as follows:

// Attempt to create the DXGI Factory.IDXGIDevice * pDXGIDevice;HRESULT hr = m_pDevice->QueryInterface(__uuidof(IDXGIDevice), (void **)&pDXGIDevice);	      IDXGIAdapter * pDXGIAdapter;hr = pDXGIDevice->GetParent(__uuidof(IDXGIAdapter), (void **)&pDXGIAdapter);IDXGIFactory * pFactory;pDXGIAdapter->GetParent(__uuidof(IDXGIFactory), (void **)&pFactory);// Attempt to create the swap chain.IDXGISwapChain* pSwapChain = 0;hr = pFactory->CreateSwapChain( m_pDevice, &pConfig->m_State, &pSwapChain );// Release the factory regardless of pass or fail.pDXGIDevice->Release();pDXGIAdapter->Release();pFactory->Release();if ( FAILED( hr ) ){	CLog::Get().Write( "Failed to create swap chain!" );	return( -1 );}

It took several hours to figure out where the exceptions were coming from, so hopefully this post will show up if someone searches the net for a similar issue. In all fairness to Microsoft, the answer is clearly stated in the DXGI documentation - but it would be nice if it was also listed in the Direct3D documentation as well...

With the swap chains created for each window, we can easily create multiple windows for an application that can utilize the same device, but still have individual swap chains. Here's a quick screen shot of the resulting window sections, each one cleared to a different back color (ermm - don't mind the errors in the debug output window - my depth buffer wasn't the same size as the several funny shaped windows...[rolleyes]):

Thanks for the tip.

You're welcome! I was pulling my hair out trying to figure it out, so I thought I would try to spare someone else from the same experience...

Yeah, I can see how this can send anyone into the wonderland for a good couple of hours. Thanks again.

I was always under the impression though, that creation of a swap chain must logically precede that of a device. D3D9 never made it clear which order is correct since IDirect3D9::CreateDevice creates them both in a single call and later swap chains, each required a call to IDirect3DDevice9::CreateAdditionalSwapChain. On the other hand, I take OpenGL's device context to be the equivalent of a swap chain and the rendering context akin to a device, in which case swap chains are to be created before the device.

So this is how I assumed the natural order of things are and as a result this is how things are designed in my engine. With a swap chain being a separate object, one which must be created first and passed to a device, I now have to find a workaround whenever I start working on D3D11. :/

To be honest, I don't know if you can do it in the other way or not - I've never tried it! I guess if there is some way to specify the IDXGIFactory to the device creation code, then it should work (allowing you to use the same factory for both...).

You'll have to let me know when you try it out with your shiny new D3D11 renderer!

When one uses IDXGIFactory::EnumAdapters to find the adapter for D3D11CreateDevice, instead of passing "0" to get the default adapter, the device is associated to the factory that enumerated the adapter.

When you do request a default adapter, the system creates a temporary factory. It would seem that the private device handles are different across instances of the factory even though they would correspond to same actual devices.

I ran into the same problem. I tried just creating a new factory to create swap chains from like you originally did, and got some warning in the debug output about using a different factory from the one the device was created with. I was scratching my head trying to figure out how to get the original factory from the created device, and just getting ready to give up when I found your journal entry. Exactly what I needed, thanks a bunch!

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## nubeer Group Title ax^2+bx-2=0 have 2 roots alpha and beta and question is. limx-->beta [(1-cos(ax^2+bx-2)/((ax^2+bx-2)^2) 3 years ago 3 years ago

1. myininaya

if x goes to beta, then ax^2+bx-2 goes to 0 since beta is a root of ax^2+bx-2 so thats means we have 0/0 when we plug in beta we could use l'hospital's rule here $\lim_{x \rightarrow \beta}\frac{0-(-\sin(ax^2+bx-2))(2ax+b)}{2(ax^2+bx-2)(2ax+b)}$

2. myininaya

$\lim_{x \rightarrow \beta}\frac{\sin(ax^2+bx-2)}{2(ax^2+bx-2)}$

3. myininaya

but when we plug in beta we still have 0/0

4. myininaya

so we can use l'hosptial again

5. myininaya

$\lim_{x \rightarrow \beta}\frac{\cos(ax^2+bx-2)(2ax+b)}{2(2ax+b)}$

6. myininaya

$\lim_{x \rightarrow \beta}\frac{\cos(ax^2+bx-2)}{2}=\frac{\cos(0)}{2}=\frac{1}{2}$

7. nubeer

wow dude ur answer is right but still have to think how u reach here

8. myininaya

do you know l'hospital?

9. myininaya

however you spell that dead math guy's name

10. nubeer

hmmm ya i heard of it but never used it

11. myininaya

so let me show you another way one sec

12. nubeer

ok but i still ddnt get what ever u answer because of that sinx

13. myininaya

let $V(x)=ax^2+bx-2$ we are given $V(\beta)=0$ so if x->beta, then V->0 so we have $\lim_{V \rightarrow 0}\frac{1-\cos(V)}{V^2}$ $=\lim_{V \rightarrow 0}\frac{1-\cos(V)}{V^2} \cdot \frac{1+\cos(V)}{1+\cos(V)}$ $=\lim_{v \rightarrow 0}\frac{1-\cos^2(V)}{V^2(1+\cos(V)}=\lim_{V \rightarrow 0}\frac{\sin^2(V)}{V^2(1+\cos(V))}$ $=\lim_{V \rightarrow 0}\frac{\sin^2(V)}{V^2} \cdot \frac{1}{1+\cos(V))}$ $=1^2 \cdot \frac{1}{1+\cos(0)}=\frac{1}{1+1}=\frac{1}{2}$

14. nubeer

thanks man this thing really helped me out

15. myininaya

thanks girl* :)

16. nubeer

lol i am a boy. :P


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## Popular Content

Showing content with the highest reputation on 10/06/19 in all areas

1. 2 points

## DOOM: Skeletal Animations & Dynamic BVHs

@lawnjelly If it goes well, I'm considering either: Changing repository status to public on my server (maybe fork it to Github) Publishing that ray tracing library as a library (with few examples) Writing some articles about ray tracing dynamic scenes The library I'm using (a custom one) is OpenCL based renderer - it doesn't have hard syntax (it might need a review or two - and maybe updating the interface after the project - as I'm adding classes/methods whenever I need them for this project). I'm currently busy making and animating a daemon from hell. 100% programmer's art.
2. 1 point

## Compile time tricks

The main advantage is that it's fool-proof, i.e. it fails at compile-time if you pass it a non-array type such as a pointer, whereas the sizeof trick silently produces incorrect results. // This function looks like it takes a fixed-size array as argument, but it really takes a pointer. void f(int a[10]) { std::cout << sizeof(a) / sizeof(a[0]) << "\n" << std::flush; // Silently produces incorrect result. std::cout << array_size(a) << "\n" << std::flush; // Fails at compile-time. }
3. 1 point

## Some noob HLSL question about register c, b

"b" registers are for D3D10 and above (SM 4.0+), and "c" registers are for D3D9 (SM 2.x and 3.0). In D3D9 era the shader programming model exposed a set of constant registers, and you would set those registers one at a time using functions like SetPixelShaderConstantF. The shader would then use the values of those constants when when executing. For D3D10 and Shader Model 4.0 the constant registers were scrapped, and replaced with constant buffers. Instead of dealing with 1 constant register at a time, you would instead create buffer resources of arbitrary size (up to 64KB or so) and bind the entire buffer to a single slot/register. The "cb" register is used for these constant buffer slots when the shader assembly needs to read from a particular buffer. Confusingly even though the assembly refers to the registers as "cb", in HLSL you use "register(bX)" to specify which register you want the constant buffer bound to.
4. 1 point

## Is F.E.A.R's "STRIPS" based AI "Planning" Still revelant today?

Not really for standard behavioral AI. The problem is primarily that it doesn't scale well as you add more potential actions with objects. The search space explodes geometrically as you add more things to do. Also -- and likely worse -- you have to re-plan anytime anything in the state space changes and invalidates one of your nodes. So not only do you have huge search times, most of that is going to get thrown away on many of your think cycles. Now multiply that across a lot of agents.
5. 1 point

## Strange Assembly Error

Hi, I seriously doubt this could ever work. First, your assembly is for real mode. Just because you changed the header to "32 bit OS", doesn't make the code valid in protected mode, hence the syntax errors most likely. Second, even if you fix the syntax errors, that code is calling real mode BIOS functions, which are not and never were available in protected mode. Third, if you have a new machine, chances are good there's no BIOS at all, as it's using UEFI. Unless you configure a legacy CSM, then even if you switch the CPU mode back to real mode you couldn't use those interrupts. Fourth, if you could fix everything I've listed so far, even then that assembly code is using VGA IO registers directly, something that modern video cards won't like or respect. Chances are good that this code would simply crash unless you have an old VGA-compatible video card. To help you understand the situation: real mode: 16 bit operation mode of the CPU, which allows accessing 64K of memory at once, and 1M tops with segmentation. protected mode: 32 bit operation mode of the CPU, which allows access to 4G of memory, and provides memory protecion (hence the name). The CPU's internal operation is fundamentally incompatible with real mode (different stack alignment, existence of interrupt descriptor table, existence of privilege levels etc. etc. etc.) BIOS: a program stored in a ROM which was compiled for real mode. It provides so called interrupt services, like int 0x10 and int 0x16 you try to use. UEFI: successor to BIOS, which does not provide any real mode services of any kind. New machines have only this, and maybe, if you're lucky, they have a special CSM module installed, which provides a legacy compatibility mode for BIOS. IO registers: special address space in the CPU, independent to the memory, in which all peripherals has an address (or more addresses) and which is accessible regardless to the CPU mode. A certain video card standard, the Video Graphics Array used the port 0x3C8. Other newer standards, like SVGA, XGA or XWGA however does not. Some cards emulated the VGA modes and IO ports to retain compatibility, but that's long gone by now. Modern cards doesn't even support "teletype output mode" any more, which you try to set with SetTextMode_near, only linear frame buffers. Cheers, bzt
6. 1 point

## SignalR, Using a From ID when Broadcasting to Others

Well, the good news is that it's all open source and you can read the code yourself! (It's kind of high in interface abstraction, though, making it harder to follow 😞 ) Dpeending on which flavor you're using: https://github.com/SignalR/SignalR https://github.com/aspnet/SignalR
7. 1 point

## My main complaint with OOP

In a good OO design, this doesn't happen. The Circle class Must be the representation of a circle in the geometric sense and nothing else. It cannot be allowed to depend on something that isn't geometry, such as a spatial index, a Canvas, or a file. This is called the "single responsibility principle": forcing together multiple responsibilities means creating tensions when anything changes. Should derive from a Shape interface or abstract class, together with other shape types. This makes clear that there are functions that apply to any shape (e.g. computing area, perimeter, bounding shapes), and operations that only make sense for circles (e.g. construct a regular polygon with the same center and the same area and a given number of sides, which can be useful as an approximation) In practical terms: The renderer should find a way to treat all shapes the same way (e.g. if the Shape interface offers a point containment test and a bounding convex polygon the renderer can test samples inside the region of interest) or segregate different Shape subclasses by their different rendering methods (e.g. scanline rendering of circles, scanline rendering of whole convex polygons and scanline rendering of a triangle decomposition of nonconvex polygons) A spatial index should probably deal with Shapes and their bounding shapes only. Serialization could be allowed in Shape classes because it's so basic and intrusive, but keeping it outside would be cleaner.
8. 1 point

## My main complaint with OOP

I definitely agree that OOP can be easily abused and I posted examples that were contrived to make OOP look bad. I guess my major problem is the way OOP is taught. I feel like most new programmings come into programming trying to solve everything with OOP and do it poorly.
9. 1 point

## My main complaint with OOP

My beef with OOP is the "OP" part. Objects are fine, but they are just one of many tools. When I was programming in C in the 90s, I often found myself making a struct like this: typename struct { int width; int height; Pixel *data; } Image; Then I would write a bunch of functions (FillWithColor, DrawPixel...) that would have a first argument of type Image *'. Then there would be a function to allocate memory in the data' field, and a function to deallocate it. In essence, that and a bit of syntactic sugar (so instead of DrawPixel(my_image, 40, 50, red)' you can call my_image.drawPixel(40, 50, red)') is all an object is. It's tidy to think of the struct and these functions as part of the same thing, and the fact that C++ will call destructors for you automatically and at easy-to-predict times is wonderful. But to approach program design by trying to shoehorn everything into this paradigm is a terrible idea. Use the tools that are appropriate for the problem at hand. Lately I've become a fan of the signal-slot paradigm to try to keep dependencies down to a minimum.
10. -1 points

## My main complaint with OOP

A renderer instance is a global state. This can lead to flickering colors caused by other people's modules. Turning off one module can break completely unrelated features. Adding a base class Renderable is easy, but removing it two years later will break things. Then you somehow ended up with seven layers of inheritance and have no idea which function will be called because they all have the same name. You'll later see that you needed Printable, Listener, Serializable, FactoryConstructable, et cetera and create more helper classes trying to wrestle the growing complexity. Compile time type safety goes out the window when you don't have time to implement virtual methods for all 47 types, so you end up leaving an error message in the base class. Then you realize that you never needed polymorphic storage to begin with and can drastically reduce the number of heap allocations that caused memory leaks and cache misses. A basic API using overloads and templates then brings back type safety and allow customers to call the API from another programming language using simple type prefixes, so don't add inheritance just because you can.

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# Decision tree classifier

## Introduction

Tree-based classification approaches are nonlinear models that work by partitioning the input space into cuboid regions. The prediction of the model is based on the most dominant class represented by training examples in the cuboid region that matches the unlabeled example. Owing to this, trained tree-based models are easily understood with no machine learning background leading to their wide deployment in industrial applications. There are many variants of tree-based approaches, such as the CART, ID3 and C4.5.

In this article, we focus on the CART-based decision tree classifier. Then, we comment on the extensions in ID3 and C4.5 in later sections.

## Prerequisites

To understand the decision tree classifier, we recommend familiarity with the concepts in

Follow the above links to first get acquainted with the corresponding concepts.

## Problem setting

In classification, the goal of the predictive model is to identify the class that generated a particular instance.

Consider such an instance $\vx$, a vector consisting of $\ndim$ features, $\vx = [x_1, x_2, \ldots, x_\ndim]$. The features may be real-valued, binary, categorical, or mix-type.

We need to assign it to one of the $\nclass$ classes $C_1, C_2, \ldots, C_\nclass$ depending on the values of the $\ndim$ features .

## The human classifier

Imagine how humans make simple classifications.

Consider a bowl containing a mix of ripened apples and mangoes. Your job is to separate them into their respective categories — apples and mangoes. You have several features of the fruit to make your decision — color, shape, size, and texture. How do you do it?

From your past observation of fruits, especially apples and mangoes, you come up with decision rules. * If color is red, then it is an apple. * If color is yellow, then it could be a mango or a golden apple. So, we need to look at additional attributes. * Among fruits with yellow color, if the shape is oval with a pointed bottom, then it is a mango. If it is round with a depressed bottom, then it is an apple.

Tree-based classification approaches follow the same strategy, which we will study next.

## From decision rules to decision trees

The decision rule for fruit classification we saw earlier were

• $\text{If color == red} \Rightarrow \text{ apple }$
• $\text{If color == yellow and shape == oval} \Rightarrow \text{ mango }$.
• $\text{If color == yellow and shape == round and bottom == depressed } \Rightarrow \text{ apple }$.
• Else, $\text{ mango }$.

Intuitively, this is a binary tree shown below

TODO: show the corresponding binary tree here

Decision trees are such binary trees. Each node in the tree compares a feature to a value.

## Regions of the feature space

Each node of the tree partitions the feature space into two regions — one that satisfies the left path and one that satisfies the right. Each region formed this way is cuboid shaped. Descendants of a node further split the region into smaller cuboids.

So, taken together, a tree achieves a recursive binary splitting of the feature space.

The final regions resulting from this recursive partitioning represent the leaf nodes of the tree. A tree with $M$ leaf nodes will result in $M$ such leaf regions.

The leaf gets the class label of the majority class among the training examples whose features match that particular leaf.

TODO: Show tree and the resulting regions in the feature space.

## Predictive model

In a decision tree, prediction is straightforward.

For an unlabeled instance that we wish to classify, we traverse the tree finding the appropriate leaf that matches the features of the instance. Then, the unlabeled instance is classified to the label of that matching leaf node.

We defined earlier that leaves of the tree are regions of the feature space. Consider one such region $r$. Suppose $\nlabeled_r$ examples from the training set belong to the region $r$. Among these, suppose the proportion of training examples belonging to class $\nclasssmall$ is $\hat{p}_{r\nclasssmall}$

$$\hat{p}_{r\nclasssmall} = \frac{1}{\nlabeled_r} \sum_{\vx_i \in r} \indicator{y_i = \nclasssmall} \label{eqn:class-proportions}$$

where $\indicator{.}$ is the indicator function — it returns 1 if $y_i = \nclasssmall$ else it returns zero.

Using these proportions for all classes $\nclasssmall \in \set{1, 2, \ldots, \nclass}$, the label for a region $r$ is the most dominant class in that region — the class with the highest proportion of training examples in the region $r$.

$$\yhat(r) = \argmax_{\nclasssmall \in \set{1,\ldots,\nclass}} \hat{p}_{r\nclasssmall} \label{eqn:yhat}$$

## Training a decision tree

To build a tree-based classifier, we need to, well, infer a tree. This involves determining the structure of the tree — the nodes and their relationships.

Tree learning uses a greedy strategy to recursively identify the best nodes to add.

1. Starting with the full training set, identify the best node to add to the tree.
2. The chosen node splits the feature space into two regions. Examples that match the node criteria will be used as the training set to grow right child of this node. Examples that do not match will be used as the training set to grow the left child.

So, the key question is: Given a training set what is the best node to create?

The various tree-based approaches to classification differ in how they address this challenge.

## Selecting nodes for tree-building

What is a node? A feature and a criteria applied on that feature. For example, $\text{color == red}$.

What constitutes a good node in a tree? One that achieves the good dichotomy — a good separation between classes.

• Among two features, one that most distinguishes the categories is a better candidate for forming a node.
• Among the various values of a feature, the value that most discriminates the categories is a better candidate for the node criteria.

For an ideal feature, , each of the buckets should be pure — each bucket should contain examples from a single class.

Intuitively, we want the decision tree to have nodes that will result in pure regions. A region is pure when it contains examples from a single class.

Thus, to infer nodes, first we need to develop a measure for impurity of the regions — an impurity score.

## A simple impurity score : Misclassification error

As explained earlier, a node is desirable in a tree if it will result in more pure regions compared to undesirable nodes. A simple and easy to understand impurity score is the misclassification error of a region — what fraction of examples are misclassified in that region.

In the case of tree-based models, if all examples in a region are assigned to the majority class, only examples from the majority class will be correctly classified and others will be misclassified. Then, the misclassification error is equivalent to one minus the proportion of the majority class. Thus, we can represent misclassification error in terms of the majority class proportion in a region $r$ as

$$e_r = 1 - \hat{p}_{r\yhat(r)}$$

## Impurity score: Gini index

We have explained in our comprehensive article on Gini index that the Gini index is a smooth and more responsive impurity score than the misclassification error. For a region $r$, it is defined as

\begin{align} g_r &= \sum_{\nclasssmall \ne \nclasssmall'} \hat{p}_{r\nclasssmall} \hat{p}_{r\nclasssmall'} \\\\ &= \sum_{\nclasssmall = 1}^{\nclass} \hat{p}_{r\nclasssmall} \left(1 - \hat{p}_{r\nclasssmall} \right) \end{align}

Gini index is the preferred impurity score for CART trees owing to its responsiveness to small changes in class proportions.

## Impurity score: Cross-entropy

An alternative smooth impurity score is the cross-entropy, also known as deviance. It is defined as

$$c_r = - \sum_{\nclasssmall = 1}^{\nclass} \hat{p}_{r\nclasssmall} \log \hat{p}_{r\nclasssmall}$$

For its smoothness and responsiveness to small changes in class proportions, the cross-entropy is the preferred impurity score in several tree-learning algorithms such as the ID3, C4.5, and C5.0, the memory-efficient enhancement to C4.5. The score used in these approaches is known as the information gain, which is the reduction in cross-entropy of the training set at a node when it is partitioned as per the node criteria.

## Dealing with feature types

Different feature types lead to corresponding node criteria.

Feature type Example criteria Explanation
Categorical $x_c == \text{'value'}$ $\text{'value'}$ is one of the values taken by the categorical variable $x_c$
Binary $x_b == 1$ Binary variables take on the value $0$ or $1$
Real-valued $x_r \le t$ $t$ is a threshold on the value of the real-valued variable $x_r$

## Controlling model complexity

In parametric models with weights such as support vector machines, the complexity is typically controlled by using some form of regularization — a penalty on the weights. In the case of tree-based classifiers, there are no weights; only nodes in the tree.

Deeper trees means more nodes. More nodes means more recursive partitioning of the tree into regions. More regions means smaller neighborhoods. Smaller neighborhoods means fewer training examples in most regions, leading to limited generalizability of the tree from being too specific to few training examples — a classic case of overfitting.

Consequently, the model complexity of a tree is controlled by limiting the depth of the tree. Since the tree-building follows a greedy strategy based on Gini-index or cross-entropy, the typical practice involves letting the tree grow big and then pruning it to limit its complexity.

## Pruning

After letting the tree grow as big as possible, a cost-complexity criterion is applied to prune or remove unnecessary nodes and their descendants from the tree.

If $R$ denotes the number of terminal nodes (leaves or regions) in the tree, and $Q_r(R)$ denotes the impurity score of the $r$-th region of the tree, then, the cost complexity of the tree is defined as

$$C(R) = \sum_{m=1}^{M} \frac{1}{\nlabeled_r} Q_r(R) + \alpha |R|$$

where, $\alpha$ is a trade-off factor between tree complexity and its performance. Being a cost, we prefer trees with lower value of the cost. Larger values of $\alpha$ lead to fewer nodes in the tree and vice versa. Typically, the search for nodes to prune is greedy — we collapse the internal node that produces the smallest per node increase in cost-complexity. A suitable value of $\alpha$ is achieved by cross-validation.

TODO: Pruning visualization

## Challenges

Despite being easy to understand, tree-based classifiers offer several challenges

• Instability: The greedy search strategy for tree-building means that small changes in the training set leads to a different series of splits and nodes in the tree, limiting their stability across datasets. This also means that tree-based classifiers typically have higher variance than other approaches. Bagging approaches, such as random forests are designed to alleviate this challenge by creating an ensemble for prediction, because each tree in the forest has been trained on a subset of the training set.
• Categorical features: While trees offer a straightforward way to work with categorical features, it also becomes challenging when dealing with categorical features that have too many distinct values. A feature with $q$ distinct unordered values results in $2^{q-1} - 1$ possible partitions of the feature space — quite prohibitive for large values of $q$.
• Diagonal separators: Tree-based classifiers partition the feature space into axis-aligned cuboid regions. This is a severe problem for tasks that involve a separate that is not axis-aligned. To arrive at a diagonal separator, the tree-based classifier will need to be very deep. TODO: show the visual of a diagonal separator based on a tree classifier.


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CoCalc Public FilesFinance-Bonds / Yields.html
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Description: Jupyter html version of Finance-Bonds/Yields.ipynb
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Yields

Computing yields¶

• Recall that $\frac{1}{(1+r)^t}\equiv DF_{t}$.
• If we want to find the yield, we can change our problem to solve for DFs.
• Our problem becomes a linear problem, which is much easier to work with.

Now, assume that we have two bonds with two periods paying different cash flows. What's the term structure?

Assune that bond A has the following value

while B

Note that we assume that we know the prices.

We have two unknowns $DF_{1}$ and $DF_{2}$ and two equations (one for each bond). Generally, we can solve this problem using algebra. However, given that you must deal with a number of securities, it is better to work with matrices. Basically, we are going to put unknowns in one side, and constants (prices) in other.

The matrix representation of our problem is

$\begin{bmatrix} C^A{1} & C^A{2} \ C^B{1} & C^B{2} \end{bmatrix} \begin{bmatrix} DF{1} \ DF{2} \end{bmatrix}¶ $\begin{bmatrix} p^A \\ p^B \end{bmatrix}$ \text{or} \quad A x =b$

where the matrix $A$ contains the coefficients associated to the unknowns $x$ and $b$ contains the constants.

We can do row (column) manipulation and find the solution for $x$. Here, we take a shortcut and pre-multiply for the inverse of matrix A.

$A^{-1} A x = A^{-1} b \Rightarrow x = A^{-1}b$

We are going to ask R to solve this for us. We only need to enter the matrix A and B. In our simple case, A is $2\times2$, or 2 rows and 2 colums and $b$ is $2\times1$ or 2 rows and 1 column.

In [4]:
%pylab inline

Populating the interactive namespace from numpy and matplotlib

In [5]:
%%R
A <- matrix(NA,2,2) # Create a matrix A with missing values
# Assume CA1 = 8 CA2 = 108 and pA = 108.1 and CB1 = 0 and CB2 = 99 and pB = 92
A[1,1]<-8; A[1,2]<-108
A[2,1]<-0; A[2,2]<-99
b<-c(108.1,92)
#b
sol<- solve(A)%*%b # Note the % before and after * This means that we are doing matrix multiplication.
print(sol)
#t <- seq(2)
#t
#r <- sol^{-1/t} -1
#r
#plot(r,type="l",lty=2)

# We can recover the interest rate (1+r)^{-t}=DF or r = DF^{-1/t}-1

          [,1]
[1,] 0.9670455
[2,] 0.9292929

Yields¶

• Note that assuming constant interest rates (yields!) $DF_1$ is $x$ then, $DF_2$ is $x^2$ and so on...
• Finding the yield becomes to solve for a polynomial of degree $T$.
• Once again, we will ask R to do it for us.
In [6]:
%%R
# Imagine that a bond for 4 years pays 2% as coupon and its price is 104

lis<- c(1,2,3,4)

roots<-polyroot(c(-104, 2, 2,2, 102))
dfsol<-round(Re(roots)[Re(roots)>0],6)
rsol<-1/dfsol-1
rsol<-round(rsol,5)*100
print(paste("The yield is ", rsol,"%", sep = ""))

[1] "The yield is 0.976%"

In [7]:
from IPython.core.display import HTML
def css_styling():
import os
styles = open(os.path.expanduser("~/Finance-Bonds/custom.css"), "r").read()
return HTML(styles)
css_styling()

Out[7]:
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